Universe in Problems - User contributions [en]

Friedman-Lemaitre-Robertson-Walker (FLRW) metric

2015-02-03T16:46:53Z

Igor: /* Problem 20: Christoffel symbols for FLRW metric */

[[Category:Dynamics of the Expanding Universe|3]]
__TOC__
<div id="equ16"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1: expanding baloon ===
Consider two points $A$ and $B$ on a two-dimensional sphere with radius $a(t)$ depending on time. Find the distance between the points $r_{AB}$, as measured along the surface of the sphere, and their relative velocity $v_{AB}={dr_{AB}}/{dt}$.
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[[File:2_16.jpg|center|thumb|400px|]]
<p style="text-align: left;">When the radius of the sphere grows with time as $a(t)$, the angle $\theta_{AB}$ between two arbitrary points $A$ and $B$ is constant. Therefore the distance between the points changes as
$r_{AB}(t) = a(t)\theta _{AB}$ and relative velocity is $v_{AB} = \dot r_{AB} = \dot a(t)\theta _{AB} = \frac{\dot a}{a}r_{AB}.$
On denoting $\frac{\dot a}{a} \equiv H(t),$ one recovers the Hubble's law.</p>
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<div id="equ17"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2: scale factor and Hubble's parameter ===
The comoving reference frame is defined so that matter is at rest in it, and the distance $\chi_{AB}$ between any two points $A$ and $B$ is constant. Show that in a homogeneous and isotropic Universe the proper (physical) distance $r_{AB}$ between two points is
related to the comoving one as
\[r_{AB}=a(t)\cdot \chi_{AB},\]
where quantity $a$ is called the scale factor and it can depend on time only. Integrate the Hubble's law and find $a(t)$.
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<p style="text-align: left;">Suppose $a=a(r,t)$. Then we differentiate $r_{AB}$ with respect to time to obtain
\[V_{AB}=\chi_{AB}\; \dot{a}
=r_{AB}\frac{\dot{a}}{a},\]
so it is easy to see that
\[H=\frac{\dot{a}}{a},\]
where the scale factor $a(r,t)$ can be expressed in terms of the Hubble's parameter as
\[a(t) = a_{0}\exp \left( \int H(t)dt \right),\]
and thus it can depend solely on time.

The scale factor represents an analogue of radius of the two-dimensional sphere from the previous problem. Its normalization is arbitrary and it determines the unit of length in the comoving reference frame. If the normalization is fixed then the scale factor determines distance between objects or observers at a given moment of time. The comoving distance between them $\chi_{AB}$ is analogous to the angle $\theta_{AB}$ from the previous problem and it can be treated as a Lagrangian (comoving) coordinate of the point $B$ in the reference frame centered in point $A$.</p>
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<div id="equ21"></div><div style="border: 1px solid #AAA; padding:5px;">

=== Problem 3: FLRW metric ===
Consider a spacetime with homogeneous and isotropic spatial section of constant time $dt=0$. Show that in the comoving coordinates its metric necessarily has the form of the Friedman-Lemaître-Robertson-Walker (FLRW)$^*$ metric:
\begin{equation}\label{FLRW1}
ds^2=dt^2-a^2(t)
\left\{ d\chi^2+\Sigma^2(\chi)
(d\theta^2+\sin^2\theta d\varphi^2)\right\},
\end{equation}
where
\[\Sigma^2(\chi)=
\left\{\begin{array}{lcl}
\sin^2\chi \\%\qquad \; \; k=+1\\
\chi^2 \\%\qquad \qquad k=0\\
\sinh^2 \chi. \\%\qquad k=-1,\\
\end{array}\right.\]
The time coordinate $t$, which is the proper time for the comoving matter, is referred to as cosmic (or cosmological) time.

$^*$Depending on geographical or historical preferences, named after a subset of the four scientists: Alexander Friedman (also spelled Friedmann), Georges Lemaître, Howard Percy Robertson and Arthur Geoffrey Walker. Thus abbreviations FRW, RW or FL are also used.
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<p style="text-align: left;">
Spatial isotropy implies spherical symmetry. In this case the spatial line element in comoving coordinates $(\chi,\vartheta,\phi)$ takes the form:
\[dl ^2
= d\chi^2 + f^2 (\chi)
\left( d\vartheta ^2
+ \sin^2\vartheta d\phi ^2 \right),\]
where $f(\chi)$ is a real-valued function which must satisfy the condition $f(\chi)\approx \chi $ at $\chi \rightarrow 0$ in the case of non-singular metrics.

Let us fix the angle $\theta$ and consider the triangle $DGE$ in the plane $\theta=\pi/2$, shown on the figure: </p>

[[File:FLRW1.png|center|thumb|400px|Geometry in an isotropic and homogeneous space, the section $\theta=\pi/2$.]]

<p style="text-align: left;">Here $DH=HE=\chi$, $HA=\delta$, the distance $DE$ and angle $\gamma$ are assumed to be small. Angles $GDH$ and $GEH$ are both equal to $\gamma$ due to homogeneity and isotropy. Note also that \begin{equation}\label{21-1}
EF \simeq EF' = f( 2\chi )\gamma = f( \chi)\beta
\end{equation}
and
\begin{equation}\label{21-2}
AC = \gamma f(\chi+\delta) = AB + BC
=\gamma f(\chi-\delta) + \beta f(\delta).
\end{equation}

Using (\ref{21-1}), we exclude the ratio $\beta /\gamma $ from (\ref{21-2}) and in the limit $\delta \to 0$ obtain the equation
\begin{equation}\label{21-3}
\frac{df}{d\chi}
=\frac{f( 2\chi)}{2f(\chi)}\cdot
\frac{df}{d\chi}\Big|_{\chi=0}.
\end{equation}
We are interested in its solutions which satisfy $df(\chi)/d\chi\big|_{\chi=0}=1$. It is easy to verify that functions $\chi$, $\sin\chi$ and $\sinh\chi$ are all such solutions. We note also, that if $f(\chi)$ is a solution of the equation (\ref{21-3}), then $f(\chi/\alpha)$, where $\alpha$ is an arbitrary constant, is also a solution. If we assume that $f$ is analytic and expand it into series over $\chi$, it is easy to show that up to rescaling $\chi\to \chi/\alpha$ all the possible solutions of (\ref{21-3}) are given by the following three:
\[f(\chi)=\Sigma(\chi)=\left\{\begin{array}{l}
\sin \chi\\ \chi\\ \sinh \chi\;.
\end{array}\right.\]
Without violation of homogeneity and isotropy, the line element of the three-dimensional space can be multiplied by an arbitrary function of time, and thus the general form of the spacetime metric is
\[ds^2=dt^2-a^2(t)
\left\{ d\chi^2+f^2(\chi)
(d\theta^2+\sin^2\theta d\varphi^2)\right\}.\]</p>
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<div id="equ20"></div><div style="border: 1px solid #AAA; padding:5px;">

=== Problem 4: another representation ===
Show that the FLRW metric (\ref{FLRW1}) can be presented in the form
\begin{equation}\label{FLRW2}
ds^2=dt^2-a^2(t) \left\{
\frac{dr^2}{1-kr^2}+r^2
(d\theta^2+\sin^2\theta d\varphi^2) \right\},
\end{equation}
where $k=0,\pm1$ is the sign of [[#equ28n|spatial curvature]].
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<p style="text-align: left;"> In each of the three cases we introduce a new variable
\[r\equiv \Sigma(\chi)=\left\{\begin{array}{lcl}
\sin\chi \\%\qquad \; \; k=+1\\
\chi \\%\qquad \qquad k=0\\
\sinh\chi \\%\qquad k=-1\\
\end{array}\right.,\qquad
dr=\left\{\begin{array}{lcl}
\cos\chi d\chi\\
d\chi\qquad\\
\cosh\chi d\chi\\
\end{array},\right.\]
so that
\[ d\chi^2 =\frac{dr^2}{1-kr^2},
\quad\text{where}\quad
k=\left\{\begin{array}{rcl}
+1&\quad\text{if}\quad& r=\sin\chi \\
0&\quad\text{if}\quad& r=\chi\\
-1&\quad\text{if}\quad& r=\sinh\chi
\end{array}\right.\]
and finally obtain
\[ ds^2=dt^2-a^2(t) \left\{
\frac{dr^2}{1-kr^2}+r^2
(d\theta^2+\sin^2\theta d\varphi^2) \right\}.\]</p>
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<div id="equ57"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5: sign of spatial curvature ===
Show that only the sign of spatial curvature has physical meaning, as renormalization of the scale factor rescales the curvature.
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<p style="text-align: left;">Suppose we have metric
\[ ds^2=dt^2-\alpha^2(t) \Big\{
\frac{dr^2}{1-k \beta^{2}r^2}+r^2
(d\theta^2+\sin^2\theta d\varphi^2) \Big\},\]
where $\beta=const$. Then we can always introduce $\chi=\beta r$ and $a(t)=\alpha(t)/\beta$ and bring the metric to the canonical form (\ref{FLRW2}). Note, that this has sense only if $k\neq 0$.</p>
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<div id="equ56"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6: spatially flat Universe ===
Why is the normalization of the scale factor not fixed for a spatially flat Universe, for which $k=0$?
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<p style="text-align: left;">It follows from the fact that in the case of a flat Universe there is no spatial scale to normalize the scale factor by.</p>
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<div id="equ22"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7: geometry of the closed Universe ===
Consider a closed Universe (with $k=+1$) and find the length of equator and full volume of its spatial section $dt=0$.
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<p style="text-align: left;">In the case of closed Universe the spatial line element takes the form
\[dl^2=a^2(t)\left\{d\chi^2
+\sin^{2}\chi(d\theta^2+\sin^2\theta d\phi^2)\right\}.\]
Then the area of a sphere with radius $\chi$, circumscribed around the center, equals to
\[s(\chi)= 4\pi a^2(t) \sin^2(\chi),\]
and the equator's length is
\[l=2\pi a(t) \sin(\chi).\]
It can be seen from the relation for $s(\chi)$ that $s$ reaches maximum $s_{max} = 4\pi a^2(t)$ at $\chi=\pi/2$, then decreases and turns again to zero at $\chi = \pi$. Thus the value $\chi = \pi$ gives the maximum value the coordinate $\chi$ can take.

The volume inside a sphere of radius $\chi$ equals to
\[V =\int\limits_0^{\chi} s(\chi') a(t) d\chi'
= 4\pi a^3(t) \int\limits_0^{\chi} \sin^2(\chi') d\chi'
= 4\pi a^3(t)\left(\frac 1 2 \chi
- \frac 1 4 \sin(2\chi) \right)\]
and monotonically grows with $\chi$. At $\chi = \pi$ one obtains the total volume of the closed world:
\[V_{universe} = 2 \pi^2 a^3.\]</p>
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<div id="equ23"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8: electric charge of the Universe ===
Present arguments in favor of the affirmation that the electric charge of a closed Universe should be exactly zero.
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<p style="text-align: left;">A charge is a source of electric field. In Euclidean space the electric field lines begin on a charge and then either end on a charge of the opposite sign or go to infinity. However there is no infinity in the closed world.

Let us fix some arbitrary electric field distribution $\vec E$ in the closed world and find the corresponding charge density $\varepsilon _e$ using the equation $ \mbox{div}\vec E = 4\pi \varepsilon _e$. It will always turn out that the total charge equals to zero, i.e. $Z = \int \varepsilon_e dV = 0$, as in the absence of infinity the lines of force always start from one charge and necessarily end on another charge of the opposite sign, which thus neutralizes the former charge.</p>
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<div id="equ24"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9: Hubble's law ===
Using the FLRW metric, derive the Hubble's law.
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<p style="text-align: left;">In an expanding Universe the physical distance $D$ between the observer in the origin of reference frame and a receding galaxy is measured along the surface of constant time $dt=0$. If one considers the radial distance then $d\theta=d\phi=0$, so
\[D(t) = a(t)\chi.\]
Evaluating time derivative and taking into account that $\dot \chi = 0,$ one obtains
\[ V \equiv \frac{dD(t)}{dt} = \dot a(t)\chi
= \frac{\dot a(t)}{a(t)}a(t)\chi = H(t)D.\]</p>
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<div id="equ64"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10: conformal time ===
Conformal time $\eta$ is defined as
\[dt=a(\eta)d\eta.\]
It can be interpreted as the time measured by a clock that decelerates along with the expansion of the Universe. Rewrite the FLRW metric in conformal time. Show that the logarithmic derivative of the scale factor with respect to conformal time determines its evolution in the physical time.
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<p style="text-align: left;">On substitution $dt=a d\eta$ we get
\begin{equation}\label{FLRWconformal}
d{s^2} = a^2(\eta )
\Big\{ d\eta ^2 - \big[ d\chi ^2
+ \Sigma ^2(\chi )\left( d\theta ^2 +
\sin^2\theta\; d\varphi ^2\right)\big]\Big\}.
\end{equation}
Then
\[\dot{a}=\frac{da}{dt}=\frac{da}{ad\eta}
=\frac{d\,\ln a}{d\eta}.\]</p>
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<div id="equ71"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11: comoving-conformal coordinates ===
Express the FLRW metric in comoving coordinates and conformal time. Show that in the case $k=0$ it is conformally flat, i.e. it can be made flat (pseudo-Euclidean) by means of global stretching.
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<p style="text-align: left;">In comoving coordinates the FLRW metric has the form
\[d{s^2} = d{t^2} - a^2(t)\delta _{ij}dx^{i}dx^{j},\]
then the change of variables $dt=a(\eta)d\eta$ leads to metric
\[ds^2 = a^2(\eta )
\left[ d\eta ^2 - \delta _{ij}dx^idx^j \right],\]
and it follows that $g_{\mu \nu } = a^2(\eta )\eta _{\mu \nu }$, where $\eta _{\mu \nu }$ is the Minkowski metric.</p>
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<div id=""></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12: proper coordinates ===
Express the FLRW metric in proper coordinates.
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<p style="text-align: left;">
\begin{align}
ds^2& = dt^2 - a^2 (t)\left( {dr^2 + r^2 d\Omega ^2 } \right), \\
d\Omega ^2 &= d\theta ^2 + \sin ^2 \theta d\phi ^2 ; \\
R &= a(t)r, \\
dR &= \dot ardt + adr, \\
dr &= \frac{1}{a}\left( {dR - HRdt} \right); \\
\end{align}
\[ds^2 = \left[ {1 - R^2 H^2 (t)} \right]dt^2 + 2RH(t)dRdt - dR^2 - R^2 d\Omega ^2 .\]
</p>
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<div id="equ66n"></div><div style="border: 1px solid #AAA; padding:5px;">

=== Problem 13: conformal time algebra ===
Consider an arbitrary function of time $f(t)$ and express $\dot{f}$ and $\ddot{f}$ in terms of derivatives with respect to conformal time.
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<p style="text-align: left;">Let us denote differentiation with respect to time $t$ by a dot, and with respect to conformal time $\eta$ by a prime. Then
\begin{align*}
&\dot f = \frac{f'}{a};\\
&\ddot f=\frac{f''}{a^2} - \frac{f'a'}{a^3}
= \frac{f''-\mathcal{H} f'}{a^2},
\quad\text{where}\quad
\mathcal{H}\equiv \frac{a'}{a}
\end{align*}
is the Hubble' constant in conformal time.</p>
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<div id="equ65"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 14: photon's geodesics in flat case===
Obtain the equation of a photon's worldline in terms of conformal time for the case of the isotropic and spatially flat Universe.
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<p style="text-align: left;">A worldline of a photon is defined by equation $d{s^2} =0$. It is sufficient to consider only radial trajectories with the observer in the origin of coordinate frame. Using the metric (\ref{FLRWconformal}), one obtains
\[\chi =\pm\eta +const.\]</p>
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<div id="equGeo1"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15: photon's geodesics in general case (!) ===
Derive the equations of geodesics in terms of conformal time and comoving coordinates for the case of radial motion in the FLRW metric.
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<p style="text-align: left;">Consider the FLRW metrics in coordinates $(\eta,\chi,\theta,\varphi)$ (\ref{FLRWconformal}). In the case of radial motion $u^{\theta}=u^{\varphi}=0$, so one needs only the components of connection with the indices equal to $\eta$ and $\chi$. They are evaluated using the [[Equations_of_General_Relativity#GammaChristoffel|explicit formula in terms of the metric]]. The non-zero components are the following:
\[{\Gamma^{\eta}}_{\eta\eta}
={\Gamma^{\eta}}_{\chi\chi}
={\Gamma^{\chi}}_{\eta\chi}=\frac{a'}{a},\]
where the prime denotes the derivative with respect to $\eta$.

Then the equation for geodesic $u^{\mu}(\lambda)$ reduces to
\[\frac{du^{\eta}}{d\lambda}
+{\Gamma^{\eta}}_{\eta\eta}(u^{\eta})^{2}
+{\Gamma^{\eta}}_{\chi\chi}(u^{\chi})^{2}=0;
\qquad
\frac{du^{\chi}}{d\lambda}
+2{\Gamma^{\chi}}_{\eta\chi}u^{\eta}u^{\chi}=0,\]
and after substitution of explicit expressions for the Christoffel symbols one gets
\begin{equation}\label{equGeo1-eq1}
\frac{du^{\eta}}{d\lambda}
+\frac{a'}{a}\Big[(u^{\eta})^{2}+(u^{\chi})^{2}\Big]=0
;\qquad
\frac{du^{\chi}}{d\lambda}
+2\frac{a'}{a}u^{\eta}u^{\chi}=0.
\end{equation}
These equations are complemented by the normalization condition for the $4$-velocity:
\begin{equation}\label{equGeo1-eq2}
\epsilon^{2}=u^{\mu}u_{\mu}
=g_{\mu\nu}u^{\mu}u^{\nu}
=g_{\eta\eta}(u^{\eta})^{2}
+g_{\chi\chi}(u^{\chi})^{2}
=a^{2}\big[(u^{\eta})^{2}-(u^{\chi})^{2}\big],
\end{equation}
where $\epsilon^2 =0$ for massless particles (photons) and $\epsilon^2 =1$ for massive ones.

Using equation (\ref{equGeo1-eq2}), one can rewrite the geodesic equations for massive and massless particles (\ref{equGeo1-eq1}) in the form
\[\frac{du^{\eta}}{d\lambda}
+\frac{a'}{a}(u^{\eta})^{2}=\epsilon^{2}\frac{a'}{a^3};
\qquad
\frac{du^{\chi}}{d\lambda}
+2\frac{a'}{a}u^{\eta}u^{\chi}=0.\]
Taking into account that
\[a' u^{\eta}=\frac{da}{d\eta}\frac{d\eta}{d\lambda}
=\frac{da}{d\lambda},\]
one can present the right hand side in the form of total derivative and reduce them to
\begin{align}\label{equGeo1-eq3}
\frac{d}{d\lambda}\big(a^{2} u^{\eta}\big)
=&+\frac{d}{d\lambda}\big(u_{\eta}\big)
=\epsilon^{2}\mathcal{H}
;\\ \label{equGeo1-eq4}
\frac{d}{d\lambda}\big(a^{2} u^{\chi}\big)
=&-\frac{d}{d\lambda}\big(u_{\chi}\big)=0.
\end{align}
Here we also took into account that $u_{\eta}=a^{2}u^{\eta}$, $u_{\chi}=-a^{2}u^{\chi}$.

It is easy to see that for photons with $\epsilon^{2}=0$ and $u^{\chi}=\pm u^{\eta}$ the two equations coinside. The particles's momenta, both for massive and massless ones, are always conserved in coordinates $(\eta,\chi)$ (for massive particles $p_{\chi}=mc u_{\chi}$, for photons $k_{\chi}\sim u_{\chi}$). It is as it ought to be, as the FLRW metric is spatially homogeneous. It means that the photon's energy is conserved as well, but in the case of massive particles the Hubble's constant serves as a source of energy. It should be stressed that, though the obtained result is obviously physically meaningful, $u_{\eta}$ and $u_{\chi}$ are not the energy and momentum as measured by a comoving observer. Regarding this see the next problem.</p>
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<div id="equGeo2"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 16: cosmological redshift (!) ===
A comoving observer is the one that is at rest in the comoving coordinates. He sees the Universe as isotropic, and can also be called an isotropic observer. Show that the frequency of a photon and velocity of a free particle, as measured by a comoving observer$^*$ at time $t$, are proportional to $1/a(t)$.

$^*$We will refer to these quantities as to the "physical" energy and momentum of a particle, to stress that they are the ones directly measured in the most natural way.
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<p style="text-align: left;">The equations for radial geodesics (\ref{equGeo1-eq2}--\ref{equGeo1-eq4}) reduce to the form
\begin{equation}\label{equGeo2-eq1}
u_{\chi}\equiv\pi_{0}=const;\qquad
(u^{\eta})^{2}=(u^{\chi})^{2}+\epsilon^{2}/a^{2}.
\end{equation}

The physical energy and momentum are those measured by a comoving observer. In coordinates $(\eta,\chi)$ its $4$-velocity is
\[\tilde{u}^{\mu}=\tfrac{1}{a}\delta^{\mu}_{0}
=\tfrac{1}{a}\,(1,0,0,0),\]
where the factor $1/a$ is necessary to satisfy the normalization conditions
\[1=\tilde{u}^{\mu}\tilde{u}_{\mu}
=g_{\eta\eta}(\tilde{u}^{\eta})^{2}
=a^{2}(\tilde{u}^{\eta})^{2}.\]

The photon's frequency measured by this observer [[Equations_of_General_Relativity#equ_oto1a|equals to]]
\[\omega_{ph}=k_{\mu}\tilde{u}^{\mu}
=k_{\eta}\tilde{u}^{\eta}
\sim \frac{k_{\chi}}{a}\sim \frac{u_{\chi}}{a}
\sim \frac{1}{a}.\]

From the normalization condition for massive particles it follows that
\[u^{\eta}=\sqrt{(u^{\chi})^2+1/a^2},\]
so the measured energy for a particle of unit mass (here $c=1$) is
\[E_{m}^{(1)}=u^{\mu}\tilde{u}_{\mu}
=g_{\eta\eta}\tilde{u}^{\eta}u^{\eta}
=a^{2}\frac{1}{a}u^{\eta}
=\sqrt{a^{2}(u^\chi)^2+1}
=\sqrt{\frac{\pi_{0}^{2}}{a^2}+1}.\]
Then its physical momentum equals to
\[p_{ph}^{(1)}=\sqrt{E^2 -1}=\frac{\pi_{0}}{a},\]
and for the physical velocity one obtains
\[v_{ph}=\frac{p}{E}
=\frac{\pi_0}{\sqrt{\pi_0^2+a^{2}}}.\]</p>
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<div id="equ70"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 17: redshift and emission time ===
Express the detected redshift of a photon as a function of the cosmic time $t$ at the moment of its emission and vice versa: express the time $t$ and conformal time $\eta$ at the moment of its emission in terms of its registered redshift.
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<p style="text-align: left;">By definition, the redshift is
\begin{equation}\label{RedshiftDefinition}
z =\frac{\omega_{emit}-\omega_{obs}}
{\omega_{obs}},
\end{equation}
and if $\omega\sim 1/a$ then
\begin{equation}\label{Redshift}
1+z=\frac{\omega_{emit}}{\omega_{obs}}
=\frac{a_{obs}}{a_{emit}}.
\end{equation}
Let at the moment of observation $a_{obs}=1$ and we consider the dependence of redshift $z(t)$ and scale factor $a(t)$ on the time of emission $t$:
\begin{equation}\label{1+z}
1+z=\frac{1}{a(t)}.
\end{equation}
Then
\[dz=-\frac{da}{a^{2}}=-\frac{H}{a}dt
=-(1+z)Hdt=-H\,dz,\]
and after integration one finds
\[ t = \int\limits_z^\infty \frac{dz}{(1 + z)H(z)};
\quad
\eta=\int\limits_z^\infty\frac{dz}{H(z)}.\]
The integration constant is chosen so that $z\to\infty$ for $t\to0$. Thus the history of the Universe in the cosmic and conformal times is expressed in terms of redshift.</p>
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<div id="equ69"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 18: scale factor and conformal time ===
Obtain the relation between the scale factor and conformal time using the properties of conformal time interval.
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<p style="text-align: left;">As the comoving distance between the source and observer does not change, the conformal time interval between two light signals at the point of emission [[#equ65|coincides with that at the point of detection]]. Using the definition of conformal time $dt=a(\eta)d\eta$, one obtains
\[\left. \frac{\Delta t}{a} \right|_{emit}
= \left. \frac{\Delta t}{a}\right|_{obs}.\]
It follows now that
\[\omega _{obs}a_0 = \omega _{emit}a(t).\]

After substitution into the definition (\ref{RedshiftDefinition}), one has again $a(z) = \frac{1}{1 + z}$.</p>
</div>
</div></div>

<div id="equ72"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 19: closed and open universes ===
Is it possible for an open Universe to evolve into a closed one or vice versa?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">No.</p>
</div>
</div></div>

<div id="equ27"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 20: Christoffel symbols for FLRW metric ===
Calculate all connection coefficients (Christoffel symbols) for the FLRW metric.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">We use [[Equations_of_General_Relativity#GammaChristoffel|explicit expressions]] for ${\Gamma^{i}}_{kl}$
\begin{equation}\label{GammaGen}
\Gamma _{\alpha \beta }^{\mu }= \frac{1}{2}g^{\mu \nu }\left[
\frac{\partial g_{\alpha \nu }}{\partial x^{\beta }}+\frac{\partial g_{\beta \nu }}{\partial x^{\alpha }}-\frac{\partial g_{\alpha \beta }}{\partial x^{\nu }} \right].
\end{equation}
Here we use Greek alphabet for spacetime indices $\alpha,\beta,\nu=0,1,2,3$ and Latin for spatial indices $i,j=1,2,3$.

First, we consider the flat case. Then
\[\Gamma _{\alpha \beta }^{0}
=-\frac{1}{2}
\frac{\partial {{g}_{\alpha \beta }}}{\partial t}.\]

The derivatives are non-zero only if $\alpha ,\beta $ are spatial indices $i,j=1,2,3$:
\[\Gamma _{00}^{0}=0,\quad \Gamma _{0i}^{0}
=\Gamma _{i0}^{0}=0,\quad \Gamma _{ij}^{0}
={{\delta }_{ij}}\dot{a}a.\]

The symbols $\Gamma _{\alpha \beta }^{i}$ are non-zero only if one of the indices is spatial and the other is temporal
\[\Gamma _{0j}^{i}=\Gamma _{j0}^{i}
={{\delta }_{ij}}\frac{{\dot{a}}}{a}.\]

In the general case we present the FLRW metric in the form
\[d{{s}^{2}}=d{{t}^{2}}-{{a}^{2}}(t)
\gamma_{ij}(x)d{{x}^{i}}d{{x}^{j}},\]
where the non-zero components of the three-dimensional metric $\gamma_{ij}$ are
\[\gamma_{11}= \gamma_{rr} = \frac{1}{1-k r^2},\quad \gamma_{22}=\gamma_{\theta \theta}=r^2;\quad \gamma_{33} = \gamma_{\varphi \varphi } = r^2\sin^2\theta.\]

The non-zero components of the metric tensor ${{g}_{\alpha \beta }}$ and its inverse ${{g}^{\alpha \beta }}$ are the following:
\[ g_{00}= 1,~g_{ij}=- a^2(t)\gamma_{ij};
\quad g^{00}=1,
\quad g^{ij}=-\frac{1}{a^2(t)}{\gamma^{ij}}.\]
Using again the expression (\ref{GammaGen}), for the non-zero Christoffel symbols we obtain
\[ \Gamma _{0j}^{i}
=\frac{1}{2}g^{ik}\frac{\partial g_{jk}}{\partial t}
= \frac{\dot{a}}{a}\delta_{ij};
\quad\Gamma _{ij}^0=a\dot{a}{\gamma _{ij}}.\]
The symbols $\Gamma _{jk}^{i}$ are calculated also with the help of (\ref{GammaGen}), where instead of the metric tensor $g^{\mu\nu}$ one uses the three-dimensional metric $\gamma_{ij}$. Therefore all the non-zero Christoffel symbols generated by the FLRW-metric are
\begin{align*}
&\Gamma _{11}^{0}=\frac{a\dot{a}}{1-k{{r}^{2}}};\quad
\Gamma _{22}^{0}= a\dot{a}{r^2};\quad
\Gamma _{33}^{0}=a\dot{a}{r^2}\sin^2\theta\\
&\Gamma_{10}^1=\Gamma _{01}^1 =\frac{\dot a}{a};
\quad
\Gamma _{11}^1 =\frac{kr}{1-kr^2};\quad
\Gamma _{22}^1=-r(1-kr^2);\quad
\Gamma _{33}^1=-r(1-kr^2)\sin^2\theta;\\
&\Gamma^{2}_{20}=\Gamma^{2}_{02}
= \frac{\dot{a}}{a};\quad
\Gamma^{2}_{21}=\Gamma^{2}_{12}=\frac 1 r ;\quad
\Gamma^{2}_{33}=-\sin\theta\cos\theta;\\
&\Gamma^{3}_{30}=\Gamma^{3}_{03}
=\frac{\dot{a}}{a};\quad
\Gamma^{3}_{31}=\Gamma^{3}_{13}=\frac 1 r ;\quad
\Gamma^{3}_{32}=\Gamma^{3}_{23}
=\cot\theta.
\end{align*}</p>
</div>
</div></div>

<div id="equ28"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 21: Ricci tensor and scalar ===
Derive the components of Ricci tensor, scalar curvature and the trace of energy-momentum tensor for the FLRW metric.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">From (see first equation in solution of [[Equations_of_General_Relativity#equ_oto6|problem]]) the components of Ricci tensor are given by
\[R_{\nu \sigma }
= \Gamma _{\nu \sigma ,\mu }^\mu
- \Gamma _{\nu \mu ,\sigma }^\mu
+\Gamma_{\nu \sigma }^{\alpha}
\Gamma _{\alpha \mu }^\mu
- \Gamma _{\nu \mu }^\alpha
\Gamma _{\alpha \sigma }^{\mu}.\]
Using the Christoffel symbols obtained in the previous problem, one gets
\begin{align*}
&R_{00} = - 3\frac{\ddot a}{a}; \\
&R_{ij} = \left( \ddot aa + 2\dot a^2 + 2k\right)\gamma _{ij}; \\
&R_{11} = \frac{\left( {\ddot aa + 2\dot a^2 + 2k} \right)}{1 - kr^2 };\\
&R_{22} = \left( \ddot aa + 2\dot a^2 + 2k \right)r^2 ; \\
&R_{33} = \left( \ddot aa + 2\dot a^2 + 2k \right)r^2 \sin ^2 \theta.
\end{align*}
All other components of the tensor are equal to zero. It is useful also to calculate components of $R_\mu ^\nu$
\begin{align*}
&R_\mu ^\nu = g^{\nu \alpha } R_{\alpha \mu } ; \\
&R_0^0 = g^{00} R_{00} = R_{00}= - 3\frac{\ddot{ a}}{a}; \\
&R_1^1 = R_2^2 = R_3^3= - \frac{\ddot aa + 2\dot a^2 + 2k}{a^2}.
\end{align*}</p>
</div>
</div></div>

<div id="equ28n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 22: spatial curvature ===
Obtain the components of the Ricci tensor and scalar curvature ${}^{(3)}R$ of the spatial section $t=const$ of the FLRW metric. Show that $k=sign^{(3)}R$ if ${}^{(3)}R\neq 0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Using the results of the previous problem one obtains
\begin{align*}
& R = g^{\mu \nu } R_{\mu \nu }
= g^{00} R_{00} + g^{ij} R_{ij}
= R_{00} - \frac{1}{a^2}\gamma ^{ij} R_{ij};\\
&R_{ij} =
\left(\ddot{a}a+2\dot{a}^2+2k\right)\gamma _{ij};\\
&R = - 6\left( \frac{\ddot a}{a}
+ \frac{\dot a^2}{a^2} + \frac{k}{a^2} \right).
\end{align*}

The spatial curvature can be obtained from $R$ by formal substitution $g_{\alpha\beta}\to-g_{\alpha\beta}$ and $a=const$ (thus $R^{0}_{0}$ also turns to zero). The scalar spacetime curvature then equals to $R=-6k/a^2$. If one changes the sign of the metric, then $\Gamma_{ikl}$ changes its sign too, but ${\Gamma^{i}}_{kl}=g^{ij}\Gamma_{j,kl}$ does not, as well as the curvature tensor ${R^{i}}_{klm}$ and Ricci tensor $R_{km}={R^{i}}_{kim}$, but the scalar curvature $R=g^{km}R_{km}$ changes its sign again. Then the spatial curvature equals to ${}^{(3)}R=6k/a^{2}$, and thus $k$ coincides with its sign.</p>
</div>
</div></div>

<div id="equ29n"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 23: cosmological energy-momentum tensor ===
Derive the components and trace of the energy-momentum tensor which satisfies the cosmological principle.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">In order to obtain the explicit expression for the energy-momentum tensor in a curved spacetime, we consider first the case of flat spacetime. The energy-momentum tensor for isotropic distribution of matter at rest as a whole in the absence of internal rotations is known to have the following form
\begin{equation}
T^{\mu\nu}=
\left
(\begin{array}{cccc}
\rho & 0& 0& 0\\
0 & p& 0& 0\\
0 & 0& p& 0\\
0 & 0& 0& p\\
\end{array}
\right) \label{1_equ29n}
\end{equation}
Let us generalize this expression for the case of matter in motion. In the latter case, in addition to dependence on energy density and pressure, the energy-momentum tensor must also depend on the $4$-velocity vector $u^\mu$. In order to obtain this dependence, note that in the comoving frame the $4$-velocity vector equals to
\[u^\mu = (1,0,0,0).\]
Therefore if we define the tensor quantity
\begin{equation}
(\rho+p)u^\mu u^\nu - p\eta^{\mu\nu},
\label{2_equ29n}
\end{equation}
in the comoving frame it will coincide with the energy-momentum tensor (\ref{2_equ29n}). As both quantities transform as tensors, they coincide in any other frame.
<br/>
The simplest way to generalize the expression (\ref{1_equ29n}) for the case of curved space is to replace the Minkowski space metric $\eta^{\mu\nu}$ by an arbitrary one $g^{\mu\nu} $. Indeed, for any given point of the spacetime there [[Equations_of_General_Relativity#equ_oto3|exists a locally Lorentzian reference frame]], in which the metric tensor locally coincides with the Minkowski tensor, and the energy-momentum tensor for matter takes the form (\ref{2_equ29n}). After transition to arbitrary reference frame one arrives to:
\begin{equation}
T^{\mu\nu}= (\rho+p)u^\mu u^\nu - pg^{\mu\nu} . \label{3_equ29n}
\end{equation}
It is worth noting that in general the expression (\ref{3_equ29n}) is valid only in the case of weak gravity; otherwise the expression for the energy-momentum tensor may contain additional terms depending on the curvature tensor.<br/>

Note also that the simplest way to present explicitly the components of energy-momentum tensor (\ref{3_equ29n}) is the following
\begin{equation}
T_\mu ^\nu = g^{\nu \alpha } T_{\alpha \mu } = (\rho+p)u^\nu u_\mu - pg_\mu^\nu = (\rho+p) u^\nu u_\mu - p\delta_\mu^\nu ,\label{4_equ29n}
\end{equation}
and therefore
\begin{equation}
T_\mu ^\nu =
\left
(\begin{array}{cccc}
\rho & 0& 0& 0\\
0 & -p& 0& 0\\
0 & 0& -p& 0\\
0 & 0& 0& -p\\
\end{array}
\right),
\end{equation}
and so the trace of $T^{\mu}_{\nu}$ is
\[T\equiv T_\mu ^\mu = \rho - 3p.\]</p>
</div>
</div></div>

The Milne Universe

2014-02-27T00:00:17Z

Igor: /* Problem 2: alternative derivation */

[[Category:Dynamics of the Universe in the Big Bang Model|5]]
__NOTOC__
<div id="Milne1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1: the solution ===
Find the solutions of the Friedman equations for the Milne Universe: the expanding Universe with $\rho\to 0$ and $k=-1$. Why is it necessarily spatially open? What is the scalar curvature of this spacetime?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">The Friedman equations with $\rho\to 0$ reduce to
\begin{align*}
(\dot{a})^{2}=-k,\\
\ddot{a}=0,
\end{align*}
and thus we have that $k=-1$, and the solution for the scale factor is $a=t$ (or $a(t)=ct$ if we preserve $c$).

The metric of the Milne Universe therefore has the form
\begin{equation}\label{Milne-metric}
ds^{2}=dt^{2}
-t^{2}\big[dr^{2}+\sinh^{2}r d\Omega^{2}\big].
\end{equation}
Using the [[Friedman-Lemaitre-Robertson-Walker (FLRW) metric#equ28n|expression for the scalar curvature]] in the FLRW metric, we get
\[R = - 6\left( \frac{\ddot a}{a}
+ \frac{\dot a^2}{a^2} + \frac{k}{a^2} \right)=0,\]
as it should be in a spacetime with no matter whatsoever.</p>
</div>
</div></div>

<div id="Milne2"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 2: alternative derivation ===
Let us consider the Minkowski spacetime in spherical spatial coordinates $(T,R,\theta,\phi)$. Let at some moment of initial explosion a cloud of particles emerge from the origin with all possible velocities $v<c$ in all directions, which stay constant. Their mass is considered negligible, so that they do not interact and do not affect the underlying spacetime. The larger is the velocity of a particle, the further away from the origin it is at a given moment of time, so the velocity of a particle $v$, or alternatively its "rapidity"
\[r=\text{artanh}\, v
\equiv\tfrac{1}{2}\ln\frac{1+v}{1-v}\]
serve as radial coordinates in the region $R<T$. Let $\tau$ be the proper time of the particle. Show that the region $R<T$ in coordinates $(\tau,r,\theta,\phi)$ ''is'' the Milne Universe$^*$.

$^*$In fact, this is the way Milne in his papers of 1935 and 1948 introduced this spacetime, trying to show that Big Bang can be described by pure kinematics and in the frame of Special Theory of Relativity only. This is in general not possible, but his renowned example is very instructive.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">The velocity and Lorentz-factor of a particle expressed through its rapidity are
\[v=\tanh r,\qquad \gamma=\cosh r.\]
Then taking into account the time dilation
\[\tau=\gamma^{-1}T=T/\cosh r,\]
and the obvious $R=vT$, we obtain
\begin{align}
T&=\tau \cosh r,\nonumber\\
\label{Milne-Mink}
R&=\tau \sinh r.
\end{align}
Now it is not hard to see that
\[dT^{2}-dR^{2}=d\tau^{2}-\tau^{2}dr^{2},\]
and the metric of the original Minkowski spacetime turns out to describe the Milne Universe:
\[ds^{2}=dT^{2}-dR^{2}-R^{2}d\Omega^{2}
=d\tau^{2}
-\tau^{2}\big[dr^{2}+\sinh^2 r d\Omega^{2}\big].\]

As the Milne Universe is just a re-parametrized Minkowski, not only its scalar curvature is zero, but the curvature tensor as well (as should be in the absence of any matter). Its spatial slices $T=const$ are, obviously, flat, while the slices $\tau=const$, which are ''hyperboloids'' $T^{2}-R^{2}=\tau^{2}$, have constant ''negative'' (scalar) curvature $k=-1$.

It should be noted however, that though the (part of) spacetime is the same, the Milne and Minkowski metrics describe quite different frames of reference, and the observables in them (redshifts for example) will be different too.</p>
</div>
</div></div>

<div id="Milne3"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 3: deeper relation ===
Let the density of matter in the Milne Universe (in the comoving frame) be small but finite. Find the dependence of (number) density on the distance to the horizon $R=T$ in the Minkowski spacetime (the laboratory frame with regard to the experiment of the Big Bang), if the distribution in the Milne Universe is homogeneous. What is the total number of particles (galaxies) in each of the frames of reference?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">a) In the Milne frame the spatial metric of the slice $\tau=const$ is
\[dl^{2}_{\tau}
=\tau^2 dr^2 +\tau^2 \sinh^{2}r d\Omega^2,\]
so the volume element is (integrating over $d\Omega$)
\[dV_{\tau}=4\pi \tau^{3}\sinh^{2}r\, dr.\]
If $n_{\tau}=const$ is the particles' number density, the total number of particles within a sphere of radius (radial coordinate, to be precise) $r_{\star}$ is
\[N_{\tau}(r_{\star})=4\pi\;
\underbrace{n_{\tau}(\tau)\tau^{3}}_{const}
\int\limits_{0}^{r_{\star}}dr\,\sinh^{2}r
\xrightarrow[r_{\star}\to\infty]{}\infty,\]
which is natural, as we assume the number density is constant at a given time on an unbounded hyperboloid of infinite volume.

b) In the Minkowski frame the spatial metric of the slice $T=const$ is
\[dl^{2}_{T}=dR^{2}+R^{2}d\Omega,\]
so the volume element is $dV_{T}=4\pi R^2 dR$. We can rewrite it in terms of $r$, while keeping in mind that we are talking of the slice $T=const$, and thus from (\ref{Milne-Mink}) $R=T\tanh r$ and $dR=Tdr/\cosh^{2}r$ we get
\[dV_{T}=4\pi R^{2}dR
=4\pi T^{3}\frac{\sinh^{2}r\,dr}{\cosh^{4}r}
=4\pi \tau^{3}\frac{\sinh^{2}r\,dr}{\cosh r}.\]

As $r$ is the ''comoving'' coordinate, and the number of particles is conserved, it stays the same in the layer between $r$ and $r+dr$ regardless of the chosen spatial slice, so
\[dN[r,dr]=n_{\tau}dV_{\tau}=n_{T}dV_{T}\]
and
\[n_{T}=n_{\tau}\cosh r =n_{\tau}\gamma
=\frac{n_{\tau}}{\sqrt{1-R^2 /T^2}}.\]
We could have obtained this result by simple observation that the number (or mass) density in the Minkowski frame changes due to the contraction of the radial distance between the particles by the Lorentz-factor $\gamma$.

However, when considering the slice $T=const$, we now also have to take into account that $\tau$ is a function of $R$:
\[\tau^{2}=T^{2}-R^{2},\]
so due to $n_{\tau}\tau^{3}=const$, we get
\begin{equation}\label{Milne-density}
n_{T}=n_{0}\frac{(T_{0}/T)^{3}}
{\big[1-R^2 /T^2\big]^{2}},
\end{equation}
which leads to the correct divergence of the integral for the total number of particles on the horizon $R=T$:
\begin{align*}
N_{T}&=\int n_{T}dV_{T}
=4\pi\;n_{0}T_{0}^{3} \int\limits_{0}^{T}
\frac{R^2\,dR /T^3}{(1-R^2 /T^2)^{2}}=\\
&= \Big\| \xi=R/T\;\;\Big\|
=4\pi\;n_{\tau}\tau^{3} \int\limits_{0}^{1}
\frac{\xi^{2}\,d\xi}{(1-\xi^2)^{2}}=\\
&=\Big\| \xi=\tanh r\Big\|
=4\pi\;n_{\tau}\tau^{3} \int\limits_{0}^{\infty}
\sinh^{2}r\,dr\to\infty.
\end{align*}

In terms of mass density the divergence of $\rho_{T}$ at the horizon means that even for arbitrarily small density $\rho_{0}$ in the comoving frame the density in the laboratory frame becomes arbitrarily large close enough to the horizon. Thus in the neighborhood of the horizon we cannot actually consistently neglect the mass distribution and the curvature of spacetime it induces, and even for small $\rho_0$ we cannot define the world time $T$ in all of the originally Minkowski spacetime.</p>
</div>
</div>
</div>

The Milne Universe

2014-02-26T23:59:34Z

Igor: /* Problem 1: the solution */

[[Category:Dynamics of the Universe in the Big Bang Model|5]]
__NOTOC__
<div id="Milne1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1: the solution ===
Find the solutions of the Friedman equations for the Milne Universe: the expanding Universe with $\rho\to 0$ and $k=-1$. Why is it necessarily spatially open? What is the scalar curvature of this spacetime?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">The Friedman equations with $\rho\to 0$ reduce to
\begin{align*}
(\dot{a})^{2}=-k,\\
\ddot{a}=0,
\end{align*}
and thus we have that $k=-1$, and the solution for the scale factor is $a=t$ (or $a(t)=ct$ if we preserve $c$).

The metric of the Milne Universe therefore has the form
\begin{equation}\label{Milne-metric}
ds^{2}=dt^{2}
-t^{2}\big[dr^{2}+\sinh^{2}r d\Omega^{2}\big].
\end{equation}
Using the [[Friedman-Lemaitre-Robertson-Walker (FLRW) metric#equ28n|expression for the scalar curvature]] in the FLRW metric, we get
\[R = - 6\left( \frac{\ddot a}{a}
+ \frac{\dot a^2}{a^2} + \frac{k}{a^2} \right)=0,\]
as it should be in a spacetime with no matter whatsoever.</p>
</div>
</div></div>

<div id="Milne2"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 2: alternative derivation ===
Let us consider the Minkowski spacetime in spherical spatial coordinates $(T,R,\theta,\phi)$. Let at some moment of initial explosion a cloud of particles emerge from the origin with all possible velocities $v<c$ in all directions, which stay constant. Their mass is considered negligible, so that they do not interact and do not affect the underlying spacetime. The larger is the velocity of a particle, the further away from the origin it is at a given moment of time, so the velocity of a particle $v$, or alternatively its "rapidity"
\[r=\text{artanh}\, v
\equiv\tfrac{1}{2}\ln\frac{1+v}{1-v}\]
serve as radial coordinates in the region $R<T$. Let $\tau$ be the proper time of the particle. Show that the region $R<T$ in coordinates $(\tau,r,\theta,\phi)$ ''is'' the Milne Universe$^*$.

$^*$In fact, this is the way Milne in his papers of 1935 and 1948 introduced this spacetime, trying to show that Big Bang can be described by pure kinematics and in the frame of Special Theory of Relativity only. This is in general not possible, but his renowned example is very instructive.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">The velocity and Lorentz-factor of a particle expressed through its rapidity are
\[v=\tanh r,\qquad \gamma=\cosh r.\]
Then taking into account the time dilation
\[\tau=\gamma^{-1}T=T/\cosh r,\]
and the obvious $R=vT$, we obtain
\begin{align}
T&=\tau \cosh r,\nonumber\\
\label{Milne-Mink}
R&=\tau \sinh r.
\end{align}
Now it is not hard to see that
\[dT^{2}-dR^{2}=d\tau^{2}-\tau^{2}dr^{2},\]
and the metric of the original Minkowski spacetime turns out to describe the Milne Universe:
\[ds^{2}=dT^{2}-dR^{2}-R^{2}d\Omega^{2}
=d\tau^{2}
-\tau^{2}\big[dr^{2}+r^{2}d\Omega^{2}\big].\]

As the Milne Universe is just a re-parametrized Minkowski, not only its scalar curvature is zero, but the curvature tensor as well (as should be in the absence of any matter). Its spatial slices $T=const$ are, obviously, flat, while the slices $\tau=const$, which are ''hyperboloids'' $T^{2}-R^{2}=\tau^{2}$, have constant ''negative'' (scalar) curvature $k=-1$.

It should be noted however, that though the (part of) spacetime is the same, the Milne and Minkowski metrics describe quite different frames of reference, and the observables in them (redshifts for example) will be different too.</p>
</div>
</div></div>

<div id="Milne3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3: deeper relation ===
Let the density of matter in the Milne Universe (in the comoving frame) be small but finite. Find the dependence of (number) density on the distance to the horizon $R=T$ in the Minkowski spacetime (the laboratory frame with regard to the experiment of the Big Bang), if the distribution in the Milne Universe is homogeneous. What is the total number of particles (galaxies) in each of the frames of reference?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">a) In the Milne frame the spatial metric of the slice $\tau=const$ is
\[dl^{2}_{\tau}
=\tau^2 dr^2 +\tau^2 \sinh^{2}r d\Omega^2,\]
so the volume element is (integrating over $d\Omega$)
\[dV_{\tau}=4\pi \tau^{3}\sinh^{2}r\, dr.\]
If $n_{\tau}=const$ is the particles' number density, the total number of particles within a sphere of radius (radial coordinate, to be precise) $r_{\star}$ is
\[N_{\tau}(r_{\star})=4\pi\;
\underbrace{n_{\tau}(\tau)\tau^{3}}_{const}
\int\limits_{0}^{r_{\star}}dr\,\sinh^{2}r
\xrightarrow[r_{\star}\to\infty]{}\infty,\]
which is natural, as we assume the number density is constant at a given time on an unbounded hyperboloid of infinite volume.

b) In the Minkowski frame the spatial metric of the slice $T=const$ is
\[dl^{2}_{T}=dR^{2}+R^{2}d\Omega,\]
so the volume element is $dV_{T}=4\pi R^2 dR$. We can rewrite it in terms of $r$, while keeping in mind that we are talking of the slice $T=const$, and thus from (\ref{Milne-Mink}) $R=T\tanh r$ and $dR=Tdr/\cosh^{2}r$ we get
\[dV_{T}=4\pi R^{2}dR
=4\pi T^{3}\frac{\sinh^{2}r\,dr}{\cosh^{4}r}
=4\pi \tau^{3}\frac{\sinh^{2}r\,dr}{\cosh r}.\]

As $r$ is the ''comoving'' coordinate, and the number of particles is conserved, it stays the same in the layer between $r$ and $r+dr$ regardless of the chosen spatial slice, so
\[dN[r,dr]=n_{\tau}dV_{\tau}=n_{T}dV_{T}\]
and
\[n_{T}=n_{\tau}\cosh r =n_{\tau}\gamma
=\frac{n_{\tau}}{\sqrt{1-R^2 /T^2}}.\]
We could have obtained this result by simple observation that the number (or mass) density in the Minkowski frame changes due to the contraction of the radial distance between the particles by the Lorentz-factor $\gamma$.

However, when considering the slice $T=const$, we now also have to take into account that $\tau$ is a function of $R$:
\[\tau^{2}=T^{2}-R^{2},\]
so due to $n_{\tau}\tau^{3}=const$, we get
\begin{equation}\label{Milne-density}
n_{T}=n_{0}\frac{(T_{0}/T)^{3}}
{\big[1-R^2 /T^2\big]^{2}},
\end{equation}
which leads to the correct divergence of the integral for the total number of particles on the horizon $R=T$:
\begin{align*}
N_{T}&=\int n_{T}dV_{T}
=4\pi\;n_{0}T_{0}^{3} \int\limits_{0}^{T}
\frac{R^2\,dR /T^3}{(1-R^2 /T^2)^{2}}=\\
&= \Big\| \xi=R/T\;\;\Big\|
=4\pi\;n_{\tau}\tau^{3} \int\limits_{0}^{1}
\frac{\xi^{2}\,d\xi}{(1-\xi^2)^{2}}=\\
&=\Big\| \xi=\tanh r\Big\|
=4\pi\;n_{\tau}\tau^{3} \int\limits_{0}^{\infty}
\sinh^{2}r\,dr\to\infty.
\end{align*}

In terms of mass density the divergence of $\rho_{T}$ at the horizon means that even for arbitrarily small density $\rho_{0}$ in the comoving frame the density in the laboratory frame becomes arbitrarily large close enough to the horizon. Thus in the neighborhood of the horizon we cannot actually consistently neglect the mass distribution and the curvature of spacetime it induces, and even for small $\rho_0$ we cannot define the world time $T$ in all of the originally Minkowski spacetime.</p>
</div>
</div>
</div>

Friedman-Lemaitre-Robertson-Walker (FLRW) metric

2014-02-26T09:14:34Z

Igor: /* Problem 20: Christoffel symbols for FLRW metric */

[[Category:Dynamics of the Expanding Universe|3]]
__TOC__
<div id="equ16"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1: expanding baloon ===
Consider two points $A$ and $B$ on a two-dimensional sphere with radius $a(t)$ depending on time. Find the distance between the points $r_{AB}$, as measured along the surface of the sphere, and their relative velocity $v_{AB}={dr_{AB}}/{dt}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
[[File:2_16.jpg|center|thumb|400px|]]
<p style="text-align: left;">When the radius of the sphere grows with time as $a(t)$, the angle $\theta_{AB}$ between two arbitrary points $A$ and $B$ is constant. Therefore the distance between the points changes as
$r_{AB}(t) = a(t)\theta _{AB}$ and relative velocity is $v_{AB} = \dot r_{AB} = \dot a(t)\theta _{AB} = \frac{\dot a}{a}r_{AB}.$
On denoting $\frac{\dot a}{a} \equiv H(t),$ one recovers the Hubble's law.</p>
</div>
</div></div>

<div id="equ17"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2: scale factor and Hubble's parameter ===
The comoving reference frame is defined so that matter is at rest in it, and the distance $\chi_{AB}$ between any two points $A$ and $B$ is constant. Show that in a homogeneous and isotropic Universe the proper (physical) distance $r_{AB}$ between two points is
related to the comoving one as
\[r_{AB}=a(t)\cdot \chi_{AB},\]
where quantity $a$ is called the scale factor and it can depend on time only. Integrate the Hubble's law and find $a(t)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Suppose $a=a(r,t)$. Then we differentiate $r_{AB}$ with respect to time to obtain
\[V_{AB}=\chi_{AB}\; \dot{a}
=r_{AB}\frac{\dot{a}}{a},\]
so it is easy to see that
\[H=\frac{\dot{a}}{a},\]
where the scale factor $a(r,t)$ can be expressed in terms of the Hubble's parameter as
\[a(t) = a_{0}\exp \left( \int H(t)dt \right),\]
and thus it can depend solely on time.

The scale factor represents an analogue of radius of the two-dimensional sphere from the previous problem. Its normalization is arbitrary and it determines the unit of length in the comoving reference frame. If the normalization is fixed then the scale factor determines distance between objects or observers at a given moment of time. The comoving distance between them $\chi_{AB}$ is analogous to the angle $\theta_{AB}$ from the previous problem and it can be treated as a Lagrangian (comoving) coordinate of the point $B$ in the reference frame centered in point $A$.</p>
</div>
</div></div>

<div id="equ21"></div><div style="border: 1px solid #AAA; padding:5px;">

=== Problem 3: FLRW metric ===
Consider a spacetime with homogeneous and isotropic spatial section of constant time $dt=0$. Show that in the comoving coordinates its metric necessarily has the form of the Friedman-Lemaître-Robertson-Walker (FLRW)$^*$ metric:
\begin{equation}\label{FLRW1}
ds^2=dt^2-a^2(t)
\left\{ d\chi^2+\Sigma^2(\chi)
(d\theta^2+\sin^2\theta d\varphi^2)\right\},
\end{equation}
where
\[\Sigma^2(\chi)=
\left\{\begin{array}{lcl}
\sin^2\chi \\%\qquad \; \; k=+1\\
\chi^2 \\%\qquad \qquad k=0\\
\sinh^2 \chi. \\%\qquad k=-1,\\
\end{array}\right.\]
The time coordinate $t$, which is the proper time for the comoving matter, is referred to as cosmic (or cosmological) time.

$^*$Depending on geographical or historical preferences, named after a subset of the four scientists: Alexander Friedman (also spelled Friedmann), Georges Lemaître, Howard Percy Robertson and Arthur Geoffrey Walker. Thus abbreviations FRW, RW or FL are also used.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Spatial isotropy implies spherical symmetry. In this case the spatial line element in comoving coordinates $(\chi,\vartheta,\phi)$ takes the form:
\[dl ^2
= d\chi^2 + f^2 (\chi)
\left( d\vartheta ^2
+ \sin^2\vartheta d\phi ^2 \right),\]
where $f(\chi)$ is a real-valued function which must satisfy the condition $f(\chi)\approx \chi $ at $\chi \rightarrow 0$ in the case of non-singular metrics.

Let us fix the angle $\theta$ and consider the triangle $DGE$ in the plane $\theta=\pi/2$, shown on the figure: </p>

[[File:FLRW1.png|center|thumb|400px|Geometry in an isotropic and homogeneous space, the section $\theta=\pi/2$.]]

<p style="text-align: left;">Here $DH=HE=\chi$, $HA=\delta$, the distance $DE$ and angle $\gamma$ are assumed to be small. Angles $GDH$ and $GEH$ are both equal to $\gamma$ due to homogeneity and isotropy. Note also that \begin{equation}\label{21-1}
EF \simeq EF' = f( 2\chi )\gamma = f( \chi)\beta
\end{equation}
and
\begin{equation}\label{21-2}
AC = \gamma f(\chi+\delta) = AB + BC
=\gamma f(\chi-\delta) + \beta f(\delta).
\end{equation}

Using (\ref{21-1}), we exclude the ratio $\beta /\gamma $ from (\ref{21-2}) and in the limit $\delta \to 0$ obtain the equation
\begin{equation}\label{21-3}
\frac{df}{d\chi}
=\frac{f( 2\chi)}{2f(\chi)}\cdot
\frac{df}{d\chi}\Big|_{\chi=0}.
\end{equation}
We are interested in its solutions which satisfy $df(\chi)/d\chi\big|_{\chi=0}=1$. It is easy to verify that functions $\chi$, $\sin\chi$ and $\sinh\chi$ are all such solutions. We note also, that if $f(\chi)$ is a solution of the equation (\ref{21-3}), then $f(\chi/\alpha)$, where $\alpha$ is an arbitrary constant, is also a solution. If we assume that $f$ is analytic and expand it into series over $\chi$, it is easy to show that up to rescaling $\chi\to \chi/\alpha$ all the possible solutions of (\ref{21-3}) are given by the following three:
\[f(\chi)=\Sigma(\chi)=\left\{\begin{array}{l}
\sin \chi\\ \chi\\ \sinh \chi\;.
\end{array}\right.\]
Without violation of homogeneity and isotropy, the line element of the three-dimensional space can be multiplied by an arbitrary function of time, and thus the general form of the spacetime metric is
\[ds^2=dt^2-a^2(t)
\left\{ d\chi^2+f^2(\chi)
(d\theta^2+\sin^2\theta d\varphi^2)\right\}.\]</p>
</div>
</div></div>

<div id="equ20"></div><div style="border: 1px solid #AAA; padding:5px;">

=== Problem 4: another representation ===
Show that the FLRW metric (\ref{FLRW1}) can be presented in the form
\begin{equation}\label{FLRW2}
ds^2=dt^2-a^2(t) \left\{
\frac{dr^2}{1-kr^2}+r^2
(d\theta^2+\sin^2\theta d\varphi^2) \right\},
\end{equation}
where $k=0,\pm1$ is the sign of [[#equ28n|spatial curvature]].
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> In each of the three cases we introduce a new variable
\[r\equiv \Sigma(\chi)=\left\{\begin{array}{lcl}
\sin\chi \\%\qquad \; \; k=+1\\
\chi \\%\qquad \qquad k=0\\
\sinh\chi \\%\qquad k=-1\\
\end{array}\right.,\qquad
dr=\left\{\begin{array}{lcl}
\cos\chi d\chi\\
d\chi\qquad\\
\cosh\chi d\chi\\
\end{array},\right.\]
so that
\[ d\chi^2 =\frac{dr^2}{1-kr^2},
\quad\text{where}\quad
k=\left\{\begin{array}{rcl}
+1&\quad\text{if}\quad& r=\sin\chi \\
0&\quad\text{if}\quad& r=\chi\\
-1&\quad\text{if}\quad& r=\sinh\chi
\end{array}\right.\]
and finally obtain
\[ ds^2=dt^2-a^2(t) \left\{
\frac{dr^2}{1-kr^2}+r^2
(d\theta^2+\sin^2\theta d\varphi^2) \right\}.\]</p>
</div>
</div></div>

<div id="equ57"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5: sign of spatial curvature ===
Show that only the sign of spatial curvature has physical meaning, as renormalization of the scale factor rescales the curvature.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Suppose we have metric
\[ ds^2=dt^2-\alpha^2(t) \Big\{
\frac{dr^2}{1-k \beta^{2}r^2}+r^2
(d\theta^2+\sin^2\theta d\varphi^2) \Big\},\]
where $\beta=const$. Then we can always introduce $\chi=\beta r$ and $a(t)=\alpha(t)/\beta$ and bring the metric to the canonical form (\ref{FLRW2}). Note, that this has sense only if $k\neq 0$.</p>
</div>
</div></div>

<div id="equ56"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6: spatially flat Universe ===
Why is the normalization of the scale factor not fixed for a spatially flat Universe, for which $k=0$?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">It follows from the fact that in the case of a flat Universe there is no spatial scale to normalize the scale factor by.</p>
</div>
</div></div>

<div id="equ22"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7: geometry of the closed Universe ===
Consider a closed Universe (with $k=+1$) and find the length of equator and full volume of its spatial section $dt=0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">In the case of closed Universe the spatial line element takes the form
\[dl^2=a^2(t)\left\{d\chi^2
+\sin^{2}\chi(d\theta^2+\sin^2\theta d\phi^2)\right\}.\]
Then the area of a sphere with radius $\chi$, circumscribed around the center, equals to
\[s(\chi)= 4\pi a^2(t) \sin^2(\chi),\]
and the equator's length is
\[l=2\pi a(t) \sin(\chi).\]
It can be seen from the relation for $s(\chi)$ that $s$ reaches maximum $s_{max} = 4\pi a^2(t)$ at $\chi=\pi/2$, then decreases and turns again to zero at $\chi = \pi$. Thus the value $\chi = \pi$ gives the maximum value the coordinate $\chi$ can take.

The volume inside a sphere of radius $\chi$ equals to
\[V =\int\limits_0^{\chi} s(\chi') a(t) d\chi'
= 4\pi a^3(t) \int\limits_0^{\chi} \sin^2(\chi') d\chi'
= 4\pi a^3(t)\left(\frac 1 2 \chi
- \frac 1 4 \sin(2\chi) \right)\]
and monotonically grows with $\chi$. At $\chi = \pi$ one obtains the total volume of the closed world:
\[V_{universe} = 2 \pi^2 a^3.\]</p>
</div>
</div></div>

<div id="equ23"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8: electric charge of the Universe ===
Present arguments in favor of the affirmation that the electric charge of a closed Universe should be exactly zero.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">A charge is a source of electric field. In Euclidean space the electric field lines begin on a charge and then either end on a charge of the opposite sign or go to infinity. However there is no infinity in the closed world.

Let us fix some arbitrary electric field distribution $\vec E$ in the closed world and find the corresponding charge density $\varepsilon _e$ using the equation $ \mbox{div}\vec E = 4\pi \varepsilon _e$. It will always turn out that the total charge equals to zero, i.e. $Z = \int \varepsilon_e dV = 0$, as in the absence of infinity the lines of force always start from one charge and necessarily end on another charge of the opposite sign, which thus neutralizes the former charge.</p>
</div>
</div></div>

<div id="equ24"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9: Hubble's law ===
Using the FLRW metric, derive the Hubble's law.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">In an expanding Universe the physical distance $D$ between the observer in the origin of reference frame and a receding galaxy is measured along the surface of constant time $dt=0$. If one considers the radial distance then $d\theta=d\phi=0$, so
\[D(t) = a(t)\chi.\]
Evaluating time derivative and taking into account that $\dot \chi = 0,$ one obtains
\[ V \equiv \frac{dD(t)}{dt} = \dot a(t)\chi
= \frac{\dot a(t)}{a(t)}a(t)\chi = H(t)D.\]</p>
</div>
</div></div>

<div id="equ64"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10: conformal time ===
Conformal time $\eta$ is defined as
\[dt=a(\eta)d\eta.\]
It can be interpreted as the time measured by a clock that decelerates along with the expansion of the Universe. Rewrite the FLRW metric in conformal time. Show that the logarithmic derivative of the scale factor with respect to conformal time determines its evolution in the physical time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">On substitution $dt=a d\eta$ we get
\begin{equation}\label{FLRWconformal}
d{s^2} = a^2(\eta )
\Big\{ d\eta ^2 - \big[ d\chi ^2
+ \Sigma ^2(\chi )\left( d\theta ^2 +
\sin^2\theta\; d\varphi ^2\right)\big]\Big\}.
\end{equation}
Then
\[\dot{a}=\frac{da}{dt}=\frac{da}{ad\eta}
=\frac{d\,\ln a}{d\eta}.\]</p>
</div>
</div></div>

<div id="equ71"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11: comoving-conformal coordinates ===
Express the FLRW metric in comoving coordinates and conformal time. Show that in the case $k=0$ it is conformally flat, i.e. it can be made flat (pseudo-Euclidean) by means of global stretching.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">In comoving coordinates the FLRW metric has the form
\[d{s^2} = d{t^2} - a^2(t)\delta _{ij}dx^{i}dx^{j},\]
then the change of variables $dt=a(\eta)d\eta$ leads to metric
\[ds^2 = a^2(\eta )
\left[ d\eta ^2 - \delta _{ij}dx^idx^j \right],\]
and it follows that $g_{\mu \nu } = a^2(\eta )\eta _{\mu \nu }$, where $\eta _{\mu \nu }$ is the Minkowski metric.</p>
</div>
</div></div>

<div id=""></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12: proper coordinates ===
Express the FLRW metric in proper coordinates.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\begin{align}
ds^2& = dt^2 - a^2 (t)\left( {dr^2 + r^2 d\Omega ^2 } \right), \\
d\Omega ^2 &= d\theta ^2 + \sin ^2 \theta d\phi ^2 ; \\
R &= a(t)r, \\
dR &= \dot ardt + adr, \\
dr &= \frac{1}{a}\left( {dR - HRdt} \right); \\
\end{align}
\[ds^2 = \left[ {1 - R^2 H^2 (t)} \right]dt^2 + 2RH(t)dRdt - dR^2 - R^2 d\Omega ^2 .\]
</p>
</div>
</div></div>

<div id="equ66n"></div><div style="border: 1px solid #AAA; padding:5px;">

=== Problem 13: conformal time algebra ===
Consider an arbitrary function of time $f(t)$ and express $\dot{f}$ and $\ddot{f}$ in terms of derivatives with respect to conformal time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let us denote differentiation with respect to time $t$ by a dot, and with respect to conformal time $\eta$ by a prime. Then
\begin{align*}
&\dot f = \frac{f'}{a};\\
&\ddot f=\frac{f''}{a^2} - \frac{f'a'}{a^3}
= \frac{f''-\mathcal{H} f'}{a^2},
\quad\text{where}\quad
\mathcal{H}\equiv \frac{a'}{a}
\end{align*}
is the Hubble' constant in conformal time.</p>
</div>
</div></div>

<div id="equ65"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 14: photon's geodesics in flat case===
Obtain the equation of a photon's worldline in terms of conformal time for the case of the isotropic and spatially flat Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">A worldline of a photon is defined by equation $d{s^2} =0$. It is sufficient to consider only radial trajectories with the observer in the origin of coordinate frame. Using the metric (\ref{FLRWconformal}), one obtains
\[\chi =\pm\eta +const.\]</p>
</div>
</div></div>

<div id="equGeo1"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15: photon's geodesics in general case (!) ===
Derive the equations of geodesics in terms of conformal time and comoving coordinates for the case of radial motion in the FLRW metric.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Consider the FLRW metrics in coordinates $(\eta,\chi,\theta,\varphi)$ (\ref{FLRWconformal}). In the case of radial motion $u^{\theta}=u^{\varphi}=0$, so one needs only the components of connection with the indices equal to $\eta$ and $\chi$. They are evaluated using the [[Equations_of_General_Relativity#GammaChristoffel|explicit formula in terms of the metric]]. The non-zero components are the following:
\[{\Gamma^{\eta}}_{\eta\eta}
={\Gamma^{\eta}}_{\chi\chi}
={\Gamma^{\chi}}_{\eta\chi}=\frac{a'}{a},\]
where the prime denotes the derivative with respect to $\eta$.

Then the equation for geodesic $u^{\mu}(\lambda)$ reduces to
\[\frac{du^{\eta}}{d\lambda}
+{\Gamma^{\eta}}_{\eta\eta}(u^{\eta})^{2}
+{\Gamma^{\eta}}_{\chi\chi}(u^{\chi})^{2}=0;
\qquad
\frac{du^{\chi}}{d\lambda}
+2{\Gamma^{\chi}}_{\eta\chi}u^{\eta}u^{\chi}=0,\]
and after substitution of explicit expressions for the Christoffel symbols one gets
\begin{equation}\label{equGeo1-eq1}
\frac{du^{\eta}}{d\lambda}
+\frac{a'}{a}\Big[(u^{\eta})^{2}+(u^{\chi})^{2}\Big]=0
;\qquad
\frac{du^{\chi}}{d\lambda}
+2\frac{a'}{a}u^{\eta}u^{\chi}=0.
\end{equation}
These equations are complemented by the normalization condition for the $4$-velocity:
\begin{equation}\label{equGeo1-eq2}
\epsilon^{2}=u^{\mu}u_{\mu}
=g_{\mu\nu}u^{\mu}u^{\nu}
=g_{\eta\eta}(u^{\eta})^{2}
+g_{\chi\chi}(u^{\chi})^{2}
=a^{2}\big[(u^{\eta})^{2}-(u^{\chi})^{2}\big],
\end{equation}
where $\epsilon^2 =0$ for massless particles (photons) and $\epsilon^2 =1$ for massive ones.

Using equation (\ref{equGeo1-eq2}), one can rewrite the geodesic equations for massive and massless particles (\ref{equGeo1-eq1}) in the form
\[\frac{du^{\eta}}{d\lambda}
+\frac{a'}{a}(u^{\eta})^{2}=\epsilon^{2}\frac{a'}{a^3};
\qquad
\frac{du^{\chi}}{d\lambda}
+2\frac{a'}{a}u^{\eta}u^{\chi}=0.\]
Taking into account that
\[a' u^{\eta}=\frac{da}{d\eta}\frac{d\eta}{d\lambda}
=\frac{da}{d\lambda},\]
one can present the right hand side in the form of total derivative and reduce them to
\begin{align}\label{equGeo1-eq3}
\frac{d}{d\lambda}\big(a^{2} u^{\eta}\big)
=&+\frac{d}{d\lambda}\big(u_{\eta}\big)
=\epsilon^{2}\mathcal{H}
;\\ \label{equGeo1-eq4}
\frac{d}{d\lambda}\big(a^{2} u^{\chi}\big)
=&-\frac{d}{d\lambda}\big(u_{\chi}\big)=0.
\end{align}
Here we also took into account that $u_{\eta}=a^{2}u^{\eta}$, $u_{\chi}=-a^{2}u^{\chi}$.

It is easy to see that for photons with $\epsilon^{2}=0$ and $u^{\chi}=\pm u^{\eta}$ the two equations coinside. The particles's momenta, both for massive and massless ones, are always conserved in coordinates $(\eta,\chi)$ (for massive particles $p_{\chi}=mc u_{\chi}$, for photons $k_{\chi}\sim u_{\chi}$). It is as it ought to be, as the FLRW metric is spatially homogeneous. It means that the photon's energy is conserved as well, but in the case of massive particles the Hubble's constant serves as a source of energy. It should be stressed that, though the obtained result is obviously physically meaningful, $u_{\eta}$ and $u_{\chi}$ are not the energy and momentum as measured by a comoving observer. Regarding this see the next problem.</p>
</div>
</div></div>

<div id="equGeo2"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 16: cosmological redshift (!) ===
A comoving observer is the one that is at rest in the comoving coordinates. He sees the Universe as isotropic, and can also be called an isotropic observer. Show that the frequency of a photon and velocity of a free particle, as measured by a comoving observer$^*$ at time $t$, are proportional to $1/a(t)$.

$^*$We will refer to these quantities as to the "physical" energy and momentum of a particle, to stress that they are the ones directly measured in the most natural way.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">The equations for radial geodesics (\ref{equGeo1-eq2}--\ref{equGeo1-eq4}) reduce to the form
\begin{equation}\label{equGeo2-eq1}
u_{\chi}\equiv\pi_{0}=const;\qquad
(u^{\eta})^{2}=(u^{\chi})^{2}+\epsilon^{2}/a^{2}.
\end{equation}

The physical energy and momentum are those measured by a comoving observer. In coordinates $(\eta,\chi)$ its $4$-velocity is
\[\tilde{u}^{\mu}=\tfrac{1}{a}\delta^{\mu}_{0}
=\tfrac{1}{a}\,(1,0,0,0),\]
where the factor $1/a$ is necessary to satisfy the normalization conditions
\[1=\tilde{u}^{\mu}\tilde{u}_{\mu}
=g_{\eta\eta}(\tilde{u}^{\eta})^{2}
=a^{2}(\tilde{u}^{\eta})^{2}.\]

The photon's frequency measured by this observer [[Equations_of_General_Relativity#equ_oto1a|equals to]]
\[\omega_{ph}=k_{\mu}\tilde{u}^{\mu}
=k_{\eta}\tilde{u}^{\eta}
\sim \frac{k_{\chi}}{a}\sim \frac{u_{\chi}}{a}
\sim \frac{1}{a}.\]

From the normalization condition for massive particles it follows that
\[u^{\eta}=\sqrt{(u^{\chi})^2+1/a^2},\]
so the measured energy for a particle of unit mass (here $c=1$) is
\[E_{m}^{(1)}=u^{\mu}\tilde{u}_{\mu}
=g_{\eta\eta}\tilde{u}^{\eta}u^{\eta}
=a^{2}\frac{1}{a}u^{\eta}
=\sqrt{a^{2}(u^\chi)^2+1}
=\sqrt{\frac{\pi_{0}^{2}}{a^2}+1}.\]
Then its physical momentum equals to
\[p_{ph}^{(1)}=\sqrt{E^2 -1}=\frac{\pi_{0}}{a},\]
and for the physical velocity one obtains
\[v_{ph}=\frac{p}{E}
=\frac{\pi_0}{\sqrt{\pi_0^2+a^{2}}}.\]</p>
</div>
</div></div>

<div id="equ70"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 17: redshift and emission time ===
Express the detected redshift of a photon as a function of the cosmic time $t$ at the moment of its emission and vice versa: express the time $t$ and conformal time $\eta$ at the moment of its emission in terms of its registered redshift.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">By definition, the redshift is
\begin{equation}\label{RedshiftDefinition}
z =\frac{\omega_{emit}-\omega_{obs}}
{\omega_{obs}},
\end{equation}
and if $\omega\sim 1/a$ then
\begin{equation}\label{Redshift}
1+z=\frac{\omega_{emit}}{\omega_{obs}}
=\frac{a_{obs}}{a_{emit}}.
\end{equation}
Let at the moment of observation $a_{obs}=1$ and we consider the dependence of redshift $z(t)$ and scale factor $a(t)$ on the time of emission $t$:
\begin{equation}\label{1+z}
1+z=\frac{1}{a(t)}.
\end{equation}
Then
\[dz=-\frac{da}{a^{2}}=-\frac{H}{a}dt
=-(1+z)Hdt=-H\,dz,\]
and after integration one finds
\[ t = \int\limits_z^\infty \frac{dz}{(1 + z)H(z)};
\quad
\eta=\int\limits_z^\infty\frac{dz}{H(z)}.\]
The integration constant is chosen so that $z\to\infty$ for $t\to0$. Thus the history of the Universe in the cosmic and conformal times is expressed in terms of redshift.</p>
</div>
</div></div>

<div id="equ69"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 18: scale factor and conformal time ===
Obtain the relation between the scale factor and conformal time using the properties of conformal time interval.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">As the comoving distance between the source and observer does not change, the conformal time interval between two light signals at the point of emission [[#equ65|coincides with that at the point of detection]]. Using the definition of conformal time $dt=a(\eta)d\eta$, one obtains
\[\left. \frac{\Delta t}{a} \right|_{emit}
= \left. \frac{\Delta t}{a}\right|_{obs}.\]
It follows now that
\[\omega _{obs}a_0 = \omega _{emit}a(t).\]

After substitution into the definition (\ref{RedshiftDefinition}), one has again $a(z) = \frac{1}{1 + z}$.</p>
</div>
</div></div>

<div id="equ72"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 19: closed and open universes ===
Is it possible for an open Universe to evolve into a closed one or vice versa?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">No.</p>
</div>
</div></div>

<div id="equ27"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 20: Christoffel symbols for FLRW metric ===
Calculate all connection coefficients (Christoffel symbols) for the FLRW metric.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">We use [[Equations_of_General_Relativity#GammaChristoffel|explicit expressions]] for ${\Gamma^{i}}_{kl}$
\begin{equation}\label{GammaGen}
\Gamma _{\alpha \beta }^{\mu }= \frac{1}{2}g^{\mu \nu }\left[
\frac{\partial g_{\alpha \nu }}{\partial x^{\beta }}+\frac{\partial g_{\beta \nu }}{\partial x^{\alpha }}-\frac{\partial g_{\alpha \beta }}{\partial x^{\nu }} \right].
\end{equation}
Here we use Greek alphabet for spacetime indices $\alpha,\beta,\nu=0,1,2,3$ and Latin for spatial indices $i,j=1,2,3$.

First, we consider the flat case. Then
\[\Gamma _{\alpha \beta }^{0}
=-\frac{1}{2}
\frac{\partial {{g}_{\alpha \beta }}}{\partial t}.\]

The derivatives are non-zero only if $\alpha ,\beta $ are spatial indices $i,j=1,2,3$:
\[\Gamma _{00}^{0}=0,\quad \Gamma _{0i}^{0}
=\Gamma _{i0}^{0}=0,\quad \Gamma _{ij}^{0}
={{\delta }_{ij}}\dot{a}a.\]

The symbols $\Gamma _{\alpha \beta }^{i}$ are non-zero only if one of the indices is spatial and the other is temporal
\[\Gamma _{0j}^{i}=\Gamma _{j0}^{i}
={{\delta }_{ij}}\frac{{\dot{a}}}{a}.\]

In the general case we present the FLRW metric in the form
\[d{{s}^{2}}=d{{t}^{2}}-{{a}^{2}}(t)
\gamma_{ij}(x)d{{x}^{i}}d{{x}^{j}},\]
where the non-zero components of the three-dimensional metric $\gamma_{ij}$ are
\[\gamma_{11}= \gamma_{rr} = \frac{1}{1-k r^2},\quad \gamma_{22}=\gamma_{\theta \theta}=r^2;\quad \gamma_{33} = \gamma_{\varphi \varphi } = r^2\sin^2\theta.\]

The non-zero components of the metric tensor ${{g}_{\alpha \beta }}$ and its inverse ${{g}^{\alpha \beta }}$ are the following:
\[ g_{00}= 1,~g_{ij}=- a^2(t)\gamma_{ij};
\quad g^{00}=1,
\quad g^{ij}=-\frac{1}{a^2(t)}{\gamma^{ij}}.\]
Using again the expression (\ref{GammaGen}), for the non-zero Christoffel symbols we obtain
\[ \Gamma _{0j}^{i}
=\frac{1}{2}g^{ik}\frac{\partial g_{jk}}{\partial t}
= \frac{\dot{a}}{a}\delta_{ij};
\quad\Gamma _{ij}^0=a\dot{a}{\gamma _{ij}}.\]
The symbols $\Gamma _{jk}^{i}$ are calculated also with the help of (\ref{GammaGen}), where instead of the metric tensor $g^{\mu\nu}$ one uses the three-dimensional metric $\gamma_{ij}$. Therefore all the non-zero Christoffel symbols generated by the FLRW-metric are
\begin{align*}
&\Gamma _{11}^{0}=\frac{a\dot{a}}{1-k{{r}^{2}}};\quad
\Gamma _{22}^{0}= a\dot{a}{r^2};\quad
\Gamma _{33}^{0}=a\dot{a}{r^2}\sin^2\theta\\
&\Gamma_{10}^1=\Gamma _{01}^1 =\frac{\dot a}{a};
\quad
\Gamma _{11}^1 =\frac{kr}{1-kr^2};\quad
\Gamma _{22}^1=-r(1-kr^2);\quad
\Gamma _{33}^1=-r(1-kr^2)\sin^2\theta;\\
&\Gamma^{2}_{20}=\Gamma^{2}_{02}
= \frac{\dot{a}}{a};\quad
\Gamma^{2}_{21}=\Gamma^{2}_{12}=\frac 1 r ;\quad
\Gamma^{2}_{33}=-\sin\theta\cos\theta;\\
&\Gamma^{3}_{30}=\Gamma^{3}_{03}
=\frac{\dot{a}}{a};\quad
\Gamma^{3}_{31}=\Gamma^{3}_{13}=\frac 1 r ;\quad
\Gamma^{3}_{32}=\Gamma^{3}_{23}
=\coth\theta.
\end{align*}</p>
</div>
</div></div>

<div id="equ28"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 21: Ricci tensor and scalar ===
Derive the components of Ricci tensor, scalar curvature and the trace of energy-momentum tensor for the FLRW metric.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">From (see first equation in solution of [[Equations_of_General_Relativity#equ_oto6|problem]]) the components of Ricci tensor are given by
\[R_{\nu \sigma }
= \Gamma _{\nu \sigma ,\mu }^\mu
- \Gamma _{\nu \mu ,\sigma }^\mu
+\Gamma_{\nu \sigma }^{\alpha}
\Gamma _{\alpha \mu }^\mu
- \Gamma _{\nu \mu }^\alpha
\Gamma _{\alpha \sigma }^{\mu}.\]
Using the Christoffel symbols obtained in the previous problem, one gets
\begin{align*}
&R_{00} = - 3\frac{\ddot a}{a}; \\
&R_{ij} = \left( \ddot aa + 2\dot a^2 + 2k\right)\gamma _{ij}; \\
&R_{11} = \frac{\left( {\ddot aa + 2\dot a^2 + 2k} \right)}{1 - kr^2 };\\
&R_{22} = \left( \ddot aa + 2\dot a^2 + 2k \right)r^2 ; \\
&R_{33} = \left( \ddot aa + 2\dot a^2 + 2k \right)r^2 \sin ^2 \theta.
\end{align*}
All other components of the tensor are equal to zero. It is useful also to calculate components of $R_\mu ^\nu$
\begin{align*}
&R_\mu ^\nu = g^{\nu \alpha } R_{\alpha \mu } ; \\
&R_0^0 = g^{00} R_{00} = R_{00}= - 3\frac{\ddot{ a}}{a}; \\
&R_1^1 = R_2^2 = R_3^3= - \frac{\ddot aa + 2\dot a^2 + 2k}{a^2}.
\end{align*}</p>
</div>
</div></div>

<div id="equ28n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 22: spatial curvature ===
Obtain the components of the Ricci tensor and scalar curvature ${}^{(3)}R$ of the spatial section $t=const$ of the FLRW metric. Show that $k=sign^{(3)}R$ if ${}^{(3)}R\neq 0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Using the results of the previous problem one obtains
\begin{align*}
& R = g^{\mu \nu } R_{\mu \nu }
= g^{00} R_{00} + g^{ij} R_{ij}
= R_{00} - \frac{1}{a^2}\gamma ^{ij} R_{ij};\\
&R_{ij} =
\left(\ddot{a}a+2\dot{a}^2+2k\right)\gamma _{ij};\\
&R = - 6\left( \frac{\ddot a}{a}
+ \frac{\dot a^2}{a^2} + \frac{k}{a^2} \right).
\end{align*}

The spatial curvature can be obtained from $R$ by formal substitution $g_{\alpha\beta}\to-g_{\alpha\beta}$ and $a=const$ (thus $R^{0}_{0}$ also turns to zero). The scalar spacetime curvature then equals to $R=-6k/a^2$. If one changes the sign of the metric, then $\Gamma_{ikl}$ changes its sign too, but ${\Gamma^{i}}_{kl}=g^{ij}\Gamma_{j,kl}$ does not, as well as the curvature tensor ${R^{i}}_{klm}$ and Ricci tensor $R_{km}={R^{i}}_{kim}$, but the scalar curvature $R=g^{km}R_{km}$ changes its sign again. Then the spatial curvature equals to ${}^{(3)}R=6k/a^{2}$, and thus $k$ coincides with its sign.</p>
</div>
</div></div>

<div id="equ29n"></div><div style="border: 1px solid #AAA; padding:5px;">
=== Problem 23: cosmological energy-momentum tensor ===
Derive the components and trace of the energy-momentum tensor which satisfies the cosmological principle.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">In order to obtain the explicit expression for the energy-momentum tensor in a curved spacetime, we consider first the case of flat spacetime. The energy-momentum tensor for isotropic distribution of matter at rest as a whole in the absence of internal rotations is known to have the following form
\begin{equation}
T^{\mu\nu}=
\left
(\begin{array}{cccc}
\rho & 0& 0& 0\\
0 & p& 0& 0\\
0 & 0& p& 0\\
0 & 0& 0& p\\
\end{array}
\right) \label{1_equ29n}
\end{equation}
Let us generalize this expression for the case of matter in motion. In the latter case, in addition to dependence on energy density and pressure, the energy-momentum tensor must also depend on the $4$-velocity vector $u^\mu$. In order to obtain this dependence, note that in the comoving frame the $4$-velocity vector equals to
\[u^\mu = (1,0,0,0).\]
Therefore if we define the tensor quantity
\begin{equation}
(\rho+p)u^\mu u^\nu - p\eta^{\mu\nu},
\label{2_equ29n}
\end{equation}
in the comoving frame it will coincide with the energy-momentum tensor (\ref{2_equ29n}). As both quantities transform as tensors, they coincide in any other frame.
<br/>
The simplest way to generalize the expression (\ref{1_equ29n}) for the case of curved space is to replace the Minkowski space metric $\eta^{\mu\nu}$ by an arbitrary one $g^{\mu\nu} $. Indeed, for any given point of the spacetime there [[Equations_of_General_Relativity#equ_oto3|exists a locally Lorentzian reference frame]], in which the metric tensor locally coincides with the Minkowski tensor, and the energy-momentum tensor for matter takes the form (\ref{2_equ29n}). After transition to arbitrary reference frame one arrives to:
\begin{equation}
T^{\mu\nu}= (\rho+p)u^\mu u^\nu - pg^{\mu\nu} . \label{3_equ29n}
\end{equation}
It is worth noting that in general the expression (\ref{3_equ29n}) is valid only in the case of weak gravity; otherwise the expression for the energy-momentum tensor may contain additional terms depending on the curvature tensor.<br/>

Note also that the simplest way to present explicitly the components of energy-momentum tensor (\ref{3_equ29n}) is the following
\begin{equation}
T_\mu ^\nu = g^{\nu \alpha } T_{\alpha \mu } = (\rho+p)u^\nu u_\mu - pg_\mu^\nu = (\rho+p) u^\nu u_\mu - p\delta_\mu^\nu ,\label{4_equ29n}
\end{equation}
and therefore
\begin{equation}
T_\mu ^\nu =
\left
(\begin{array}{cccc}
\rho & 0& 0& 0\\
0 & -p& 0& 0\\
0 & 0& -p& 0\\
0 & 0& 0& -p\\
\end{array}
\right),
\end{equation}
and so the trace of $T^{\mu}_{\nu}$ is
\[T\equiv T_\mu ^\mu = \rho - 3p.\]</p>
</div>
</div></div>

Energy conditions and the Raychaudhuri equation

2013-12-09T18:28:02Z

Igor: /* Energy conditions */

[[Category:Dynamics of the Universe in the Big Bang Model|7]]
__TOC__

==Energy conditions==
<p style="text-align: left;">S. Carroll writes$^{*}$: '' "Sometimes it is useful to think about Einstein's equation without specifying the theory of matter from which $T^{\mu\nu}$ is derived. This leaves us with a great deal of arbitrariness; consider for example the question, What metrics obey Einstein's equation? In the absence of some constraints on $T^{\mu\nu}$, the answer is any metric at all; simply take the metric of your choice, compute the Einstein tensor $G^{\mu\nu}$ for this metric, and then demand that $T^{\mu\nu}$ be equal to $G^{\mu\nu}$. It will automatically be conserved, by the Bianchi identity. Our real concern is with the existence of solutions to Einstein's equation in the presence of "realistic" sources of energy and momentum, whatever that means. One strategy is to consider specific kinds of sources, such as scalar fields, dust, or electromagnetic fields. However, we occasionally wish to understand properties of Einstein's equations that hold for a variety of different sources. In this circumstance it is convenient to impose energy conditions that limit the arbitrariness of $T^{\mu\nu}$." ''</p>

<p style="text-align: left;">The energy conditions are formulated in coordinate-independent way, but in the context of cosmology they are most useful in application to the energy-momentum tensor of a perfect fluid$^*$:
\[\begin{array}{lcc}
\text{Name}&\text{Statement}&\text{For perfect fluid}\\
\text{Weak}\phantom{\Big|}& T_{\mu\nu}v^{\mu}v^{\nu}\geq0&
\rho\geq 0,\quad \rho+p\geq 0;\\[0.2cm]
\text{Null}& T_{\mu\nu}k^{\mu}k^{\nu}\geq 0&
\rho+p\geq 0;\\[0.2cm]
\text{Strong}&
(T_{\mu\nu}-\tfrac{1}{2}Tg_{\mu\nu})
v^{\nu}v^{\nu}\geq 0\quad&
\quad\rho+p\geq 0,\quad \rho+3p\geq 0;\\[0.2cm]
\text{Dominant}& \quad T^{\mu}_{\nu}v^{\nu}\;
\mbox{is non-spacelike and future-directed}\quad &
\rho\geq |p\,|.
\end{array}\]
The conditions are assumed to hold for arbitrary timelike vectors $v^{\mu}$ and arbitrary null vectors $k^{\mu}$.</p>

<p style="text-align: left;"> $^*$For more detailed discussion see textbooks:

Carroll S. ''Spacetime and geometry: an introduction to General Relativity''. AW, 2003; ISBN 0805387323, 525p.,

Poisson E. ''A relativist's toolkit''. CUP, 2004; ISBN 0521830915, 248p. (ch 2)</p>

<div id="EnCond1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1: energy conditions for perfect fluid ===
Derive the energy conditions for the perfect fluid, shown in the last column, from the coordinate-independent formulations.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let us use the locally Euclidean frame (such that $g_{\mu\nu}=\eta_{\mu\nu}$) comoving with the fluid, so that $u^{\mu}$ in
\begin{equation}\label{EnCond_PFl}
T^{\mu\nu}=\rho u^{\mu}u^{\nu}-pg^{\mu\nu}
\end{equation}
has the coordinates $u^{\mu}=(1,0,0,0)$. Next we choose the $x$-coordinate to be in the direction of the arbitrarily chosen timelike $v^\mu$ or null $k^\mu$ vector. Then those can be parametrized as
\begin{equation}\label{EnCond_VKparametrization}
v^{\mu}=\gamma(1,v,0,0),\qquad k^{\mu}=(1,1,0,0),
\end{equation}
where $v<1$ is the velocity of the particle in the comoving frame, and $\gamma=(1-v^2)^{-1/2}$ its Lorentz factor. We remember that normalization of a null vector is arbitrary.<br/>
'''1)''' ''Weak energy condition (WEC).'' <br/>Plugging in (\ref{EnCond_VKparametrization}), and taking into account that $v^{\mu}v_{\mu}=1$, we obtain
\[ T_{\mu\nu}v^{\mu}v^{\nu}
=\gamma^{2}(\rho+p)-p
=\gamma^{2}(\rho+v^{2}p)\geq0.\]
Since this must hold for any $v\in[0,1)$, it follows that
\[\rho\geq0,\quad \rho+p\geq 0.\]
<br/>
'''2)''' ''Null energy condition (NEC).'' <br/>In the same way using $k^{\mu}$ from (\ref{EnCond_VKparametrization}) with $k_{\mu}k^{\mu}=0$, we arrive to
\[T_{\mu\nu}k^{\mu}k^{\nu}
=\rho+p\geq 0,\]
which is already what we wanted to prove. We can also obtain this from the weak energy condition by removing $\gamma$ (normalization is not fixed) and formally replacing $v\to 1$.<br/>
'''3)''' ''Strong energy condition (SEC):''
\begin{align*}
&\big[T_{\mu\nu}-\tfrac{1}{2}Tg_{\mu\nu}\big]
v^{\mu}v^{\nu}=
\big[(\rho+p)u_{\mu}u_{\nu}
-\tfrac{1}{2}(\rho-p) g_{\mu\nu}\big]
v^{\mu}v^{\nu}=\\
&\quad=\gamma^{2}(\rho+p)-\tfrac{1}{2}(\rho-p)
=\tfrac{1}{2}\gamma^{2}
\big\{(\rho+3p)+v^{2}(\rho-p)\big\}\geq 0.
\end{align*}
As this must hold for $v\in[0,1)$, we get the conditions
\[\rho+3p\geq 0,\qquad \rho+p\geq 0.\]
Note that in the limit $v=1-0$ ($v$ tends to $1$ from below) the SEC is rewritten as
\[\rho(1-0)+p(1+0)\geq 0,\]
so the non-strict inequality sign remains in $\rho+p\geq 0$.

The strong energy condition is in the most simple way formulated for the Ricci tensor:
\[R_{\mu\nu}v^{\mu}v^{\nu}\geq 0,\]
This is the reason it is used in the focusing theorem discussed below.<br/>
'''4)''' ''Dominant energy condition (DEC).''<br/>The vector $w^{\mu}=T^{\mu}_{\nu}v^{\nu}$ is the momentum density as measured by the observer with 4-velocity $v^{\mu}$. In the chosen coordinate frame
\[w^{\mu}=\gamma (\rho+p)u^{\mu}-pv^{\mu}
=\gamma (\rho,-pv).\]
As this vector has to be future-directed, we have $\rho\geq 0$ and by demanding for it to be non-spacelike, we arrive to
\[w^{\mu}w_{\mu}
=\gamma^{2}\big(\rho^{2}-v^{2}p^{2})\geq 0.\]
Thus the DEC for a perfect fluid can be written as
\[|p|\leq \rho,\]
which also implies $\rho\geq 0$.</p>
</div>
</div></div>

<div id="EnCond2"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 2: weak or strong? ===
Does the weak energy condition follow from the strong one? Which of the energy conditions imply the others?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
No. Small negative $\rho$ with large positive $p$ obey the SEC and NEC but violate WEC and DEC.</p>

<p style="text-align: left;">Despite the naming, which is a bit confusing here, the ''null'' energy condition is the weakest, ''not'' the weak energy condition$^*$. The weak, strong and dominant conditions all imply NEC, but of all three only the dominant energy condition implies the weak one.</p>

<p style="text-align: left;">All other variants of matter, obeying one condition but violating another, are hypothetically possible.</p>

<p style="text-align: left;">$^*$Strictly speaking, the weak energy condition allows $\rho+p=0$, which is prohibited by the null condition, but from here on we will not usually distinguish the strict and non-strict equalities. However, this makes a difference when considering the cosmological constant (see the chapter on dark energy).</p>
</div>
</div></div>

<div id="EnCond3"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 3: energy conditions through geometry ===
Express the energy conditions in terms of scale factor and its derivatives.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">With the help of Friedman equations we can express density and pressure through the scale factor and its derivatives
\[\rho = \frac{3}{8\pi G}
\left[\Big(\frac{\dot a}{a}\Big)^{2}
+ \frac{k}{a^2}\right],\quad
p = - \frac{1}{8\pi G}\left[2\frac{\ddot a}{a}
+\Big(\frac{\dot a}{a}\Big)^{2}+\frac{k}{a^2} \right],\]
and then rewrite the energy conditions in the form
\[\begin{array}{ll}
\text{Null:}&\displaystyle
\phantom{\text{NEC}}
- \frac{\ddot a}{a}
+\Big(\frac{\dot a}{a}\Big)^{2}
+ \frac{k}{a^2} \geq 0;\\[0.4cm]
\text{Weak:}&\displaystyle \text{NEC and}\quad
\Big(\frac{\dot a}{a}\Big)^{2}
+\frac{k}{a^2} \geq 0;\\[0.4cm]
\text{Strong:}&\displaystyle\text{NEC and}\quad
\frac{\ddot a}{a} \leq 0.
\end{array}\]
The strong energy condition follows immediately from the second Friedman the expansion of the universe is decelerating equation. We should stress, that the SEC implies that the Unverse is decelerating and this conclusion holds regardless to whether the Universe is open, closed or flat.

For the dominant EC we have two variants: $\rho\geq p\geq 0$ and $-p\geq \rho\geq 0$, which lead to
\[\text{Dominant}:\quad\left\{
\begin{array}{l}\displaystyle
\frac{\ddot a}{a}+\frac{1}{2}
\left[\Big(\frac{\dot a}{a}\Big)^{2}
+\frac{k}{a^2}\right]\leq 0,\\[0.3cm] \displaystyle
\frac{\ddot a}{a}+2\left[
\Big(\frac{\dot a}{a}\Big)^{2}+\frac{k}{a^2}\right]\geq0,
\end{array}\right.
\quad\text{or}\quad\left\{
\begin{array}{l}\displaystyle
\frac{\ddot a}{a}+\tfrac{1}{2}
\left[\Big(\frac{\dot a}{a}\Big)^{2}
+\frac{k}{a^2}\right]\geq0,\\[0.3cm]
\displaystyle \frac{\ddot a}{a}
-\Big(\frac{\dot a}{a}\Big)^{2}\geq0,
\end{array}\right..\]</p>
</div>
</div></div>

<div id="EnCond4"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 4: and in terms of redshift ===
Express the null, weak and strong energy conditions in terms of the Hubble parameter and redshift.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Using the definitions $H=\dot{a}/a$ and relation to the redshift $a_{0}/a=(z+1)$, we get
\[\begin{array}{ll}
\text{Null:}&\displaystyle
\phantom{\text{NEC and}}
\frac{\partial H^2}{\partial z}
\ge -\frac{2k\left(1 + z\right)}{a_0^2};\\[0.4cm]
\text{Weak:}&\displaystyle\text{NEC and}\quad
\frac{k\left(1+z\right)^2}{a_0^2H^2}\geq -1;
\\[0.4cm]
\text{Strong:}&\displaystyle\text{NEC and}\quad
\frac{\partial \ln H}{\partial z}\geq
\frac{1}{1+z}.
\end{array}\]</p>
</div>
</div></div>

<div id="EnCond5"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 5: restrictions on deceleration parameter ===
Find the restrictions that the energy conditions impose on the deceleration parameter in a flat Universe with $\rho>0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Dividing one Friedman equation by the other, we get
\[q=-\frac{\ddot{a}/a}{H^2}
=\frac{1}{2}\big(1+3p/\rho\big)=\frac{1}{2}(1+3w).\]
Then from the energy conditions, if we assume $\rho>0$, we straightforwardly obtain
\begin{eqnarray*}
\text{NEC or WEP:} & 1+w\geq 0 &
\Rightarrow\quad q \geq - 1; \\
\text{SEC:} &1+3w\geq 0 &
\Rightarrow\quad q \geq 0;\\
\text{DEC:} & |w|<1&
\Rightarrow\; -1\leq q \leq 2\end{eqnarray*}</p>
</div>
</div></div>

==Raychaudhuri equation==

<div id="Ray1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6: projection operators ===
Consider a timelike curve $x^{\mu}(\tau)$ and find the projection operators on its tangent vector and on its orthogonal complement.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">The decomposition of unit operator into projectors is
\[\delta^{\mu}_{\nu}=u^{\mu}u_{\nu}+h^{\mu}_{\nu},\]
or in terms of metric
\[g_{\mu\nu}=u_{\mu\nu}+h_{\mu\nu},\quad
\mbox{where}\quad
h_{\mu\nu}=g_{\mu\nu}-u_{\mu}u_{\nu}.\]
One can see that $u^{\mu}u_{\nu}$ and $h^{\mu}_{\nu}$ are indeed projection operators from their action on $u^{\mu}$ ($u^{\mu}u_{\mu}=1$):
\[(u^{\mu}u_{\nu})u^{\nu}=u^{\mu},\quad
h^{\mu}_{\nu}u^{\nu}=0,\]
and on any vector $e^{\mu}$, orthogonal to $u^{\mu}$ ($u^{\mu}e_{\mu}=0$):
\[(u^{\mu}u_{\nu})e^{\nu}=0,\quad
h^{\mu}_{\nu}e^{\nu}=e^{\mu}.\]
At the same time the quantity $h_{\mu\nu}$ acts as the metric in the orthogonal complement to linear span of $u^{\mu}$.
\[h_{\mu\nu}e^{\mu}f^{\nu}
=g_{\mu\nu}e^{\mu}e^{\nu}.\]</p>
</div>
</div></div>

<div id="Ray2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7: geodesic deviation ===
A ''congruence'' is a set of curves having the property that each point in a given region belongs to one and only one curve of the set. Consider a congruence of timelike geodesics. Let us mark two infinitely close geodesics and look at their relative evolution along their length. Let $\xi^{\mu}$ be the infinitesimal $4$-vector that is directed normal to one of the curves towards the other. Show that
\[\frac{d\xi_{\nu}}{d\tau}
=B_{\nu\mu}\xi^{\mu},\]
where
\[B_{\nu\mu}=u_{\nu;\mu},\]
is a three-dimensional spacelike tensor orthogonal to $u^{\mu}$, and $\tau$ is the parameter along the geodesic.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let $\lambda$ be the parameter along a curve with tangent vector $\xi^{\mu}$. Then
\[B^{\nu\mu}\xi_{\mu}
=\xi^{\mu}\nabla_{\mu}u^{\nu}
=\frac{du^{\nu}}{d\lambda}
=\frac{dx^{\nu}}{d\tau d\lambda}
=\frac{dx^{\nu}}{d\lambda d\tau}
=\frac{d\xi^{\nu}}{d\tau}\]
and
\[\frac{d\xi_{nu}}{d\tau}=B_{\nu\mu}\xi^{\mu}.\]

From the geodesic equation and normalization condition
\[\left\{\begin{array}{l}
B_{\mu\nu}u^{\nu}=u^{\nu}\nabla_{\nu}u_{\mu}=0,\\
B_{\mu\nu}u^{\mu}=u^{\mu}\nabla_{\nu}u_{\mu}
=\tfrac{1}{2}\nabla_{\nu}(u^{\mu}u_{\mu})
\sim\nabla_{\nu} (1)=0.
\end{array}\right.\]

On the other hand, we can always choose a coordinate frame in such a way that locally $g_{\mu\nu}=\eta_{\mu\nu}$, and by additional Lorentz boost we can direct the time along $u^{\mu}$. If ${e^{\mu}}_{i}$ ($i,j=1,2,3$) are the orthonormal unit vectors of spatial directions, such that $h_{\mu\nu}{e^{\mu}}_{i}{e^{\nu}}_{j}=\eta_{ij}=-\delta_{ij}$, then due to orthogonality $B$ is expressible as
\[B^{\mu\nu}=B_{ij}{e^{\mu}}_{i}{e^{\nu}}_{j},\]
so in particular we obtain $B_{\mu\nu}B^{\mu\nu}=B_{ij}B^{ij}\geq 0$.</p>
</div>
</div></div>

<div id="Ray3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8: tensor decomposition ===
Show that any tensor field of second rank defined on an $n-$dimensional Riemannian manifold with positive definite metric $g_{\mu\nu}$ can be uniquely decomposed into
\begin{equation}\label{TensorDecomposition}
B_{\mu\nu}=\frac{1}{n}\Theta g_{\mu\nu}
+\sigma_{\mu\nu}+\omega_{\mu\nu},
\end{equation}
where $\Theta={B^{\mu}}_{\mu}$, $\sigma_{\mu\nu}$ is the symmetric traceless part of $B_{\mu\nu}$, and $\omega_{\mu\nu}$ is the antisymmetric part of $B_{\mu\nu}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let us decompose $B_{\mu\nu}$ into the symmetric and anti-symmetric parts (symmetrization is denoted by parenthesis $()$ and anti-symmetrization by brackets $[]$):
\[B_{\mu\nu}
=\frac{B_{\mu\nu}+B_{\nu\mu}}{2}
+\frac{B_{\mu\nu}-B_{\nu\mu}}{2}
=B_{(\mu\nu)}+B_{[\mu\nu]}.\]
Then search for the representation of the symmetrical part in the form
\[B_{(\mu\nu)}=\alpha g_{\mu\nu}+\sigma_{\mu\nu},
\quad\mbox{where}\quad
\sigma^{\mu}_{\mu}=0.\]
The contraction is $\Theta=B^{\mu}_{\mu}=\alpha \delta^{\mu}_{\mu}=\alpha n$, so we obtain $\alpha=\Theta/n$ and thus derive the representation of $B_{\mu\nu}$ we had sought for.</p>
</div>
</div></div>

<div id="Ray4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9: geometrical meaning in 3D ===
Let there be a congruence in a three-dimensional Riemannian manifold. What is the geometric meaning of $\Theta$, $\sigma$ and $\omega$ for $B_{\mu\nu}=u_{\nu;\mu}$?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let the vector $\xi^{\mu}$ be a two-dimensional vector that lies in the space $S$ orthogonal to the tangent vector, and let us consider the pipe of curves from the congruence in the neighborhood of the given curve and the evolution of its form with the parameter $\tau$. Each of the generating lines of the pipe is parametrized by the corresponding vector $\xi^{\mu}$.</p>

<p style="text-align: left;">a) Let $B_{\mu\nu}=\frac{\Theta}{2}g_{\mu\nu}$. Then
\[B^{\mu}_{\nu}=\frac{\Theta}{2}
\begin{pmatrix} 1&0\\0&1\end{pmatrix}
\quad\mbox{and}\quad
\frac{d\xi^{\mu}}{d\tau}
=\frac{\Theta}{2}\xi^{\mu},\]
so the pipe's section is merely stretched $(1+\Theta/2)$ times. The area $S$ of the pipe's section changes as $dS/S=d(|\xi|^2)/|\xi|^2 = \Theta d\tau$, so
\[\Theta=\frac{1}{S}\frac{dS}{d\tau}.\]

b) The matrix of the anti-symmetric part is
\[{\omega^{\mu}}_{\nu}=\begin{pmatrix}
0&\omega\\-\omega&0
\end{pmatrix},\]
and it describes the torsion of the pipe, without changing its section area.

c) The matrix of the symmetric traceless part can be presented in the form
\[\sigma^{\mu}_{\nu}=
\begin{pmatrix}\sigma_{+}&\sigma_{\times}\\
\sigma_{\times}&-\sigma_{+}\end{pmatrix}
=\sigma_{+}
\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}
+\sigma_{\times}
\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}
\begin{pmatrix}0&-1\\ -1&0\end{pmatrix}.\]
The first term describes expansion in one direction in $(1+\sigma_{+})$ times, and contraction by the same factor in the perpendicular direction. The second term is the superposition of the same deformation without change of volume and a rotation. So the general deformation is a shear in arbitrary direction, without change of section area.</p>
</div>
</div></div>

<div id="Ray5"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 10: Landau-Raychaudhuri equation ===
Derive the Raychaudhuri equation for a congruence of timelike geodesics in spacetime
\begin{equation}\label{Raychaudhuri}
\frac{d\Theta}{d\tau}=
-\frac{1}{3}\Theta^{2}
-\sigma_{\mu\nu}\sigma^{\mu\nu}
+\omega_{\mu\nu}\omega^{\mu\nu}
-R_{\mu\nu}u^{\mu}u^{\mu}.
\end{equation}
Here $\Theta$, $\sigma_{\mu\nu}$ and $\omega_{\mu\nu}$ are the components (\ref{TensorDecomposition}) of decomposition of $B_{\mu\nu}=u_{\mu;\nu}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let us first consider the evolution of tensor $B_{\mu\nu}$ along the geodesic: changing the order of the covariant derivatives and using one of the definitions of the curvature tensor, we get
\begin{align*}
\frac{dB_{\mu\nu}}{d\tau}&
=u^{\lambda}\nabla_{\lambda}\nabla_{\nu}u_{\mu}
=\\&
=u^{\lambda}\nabla_{\nu}\nabla_{\lambda}u_{\mu}
+u^{\lambda}{R^{\sigma}}_{\mu\nu\lambda}u_{\sigma}
=\\&
=\nabla_{\nu}(u^{\lambda}\nabla_{\lambda}u_{\mu})
-(\nabla_{\nu}u^{\lambda})(\nabla_{\lambda}u_{\mu})
-R_{\rho\mu\sigma\lambda}u^{\rho}u^{\sigma}
=\\&
=-{B_{\nu}}^{\lambda}B_{\lambda\mu}
-R_{\mu\rho\lambda\sigma}u^{\rho}u^{\sigma}.
\end{align*}

On contraction we obtain
\[\frac{d\Theta}{d\tau}=-B^{\mu\nu}B_{\nu\mu}
-R_{\mu\nu}u^{\mu}u^{\nu}.\]
Now rewrite the contraction $B_{\mu\nu}B_{\nu\mu}$ in terms of components of $B$'s decomposition and take into account that pairwise contractions are zero due to the symmetries and due to $h^{\mu\nu}\sigma_{\mu\nu}=g^{\mu\nu}\sigma_{\mu\nu}=0$:
\[B^{\mu\nu}B_{\nu\mu}
=\frac{\Theta^2}{9}h^{\mu\nu}h_{\mu\nu}
+\sigma^{\mu\nu}\sigma_{\mu\nu}
-\omega^{\mu\nu}\omega_{\mu\nu}\]
(the last term goes with the opposit sign due to anti-symmetry of $\omega^{\mu\nu}\omega_{\nu\mu}=-\omega^{\mu\nu}\omega_{\mu\nu}$), so taking into account that $h^{\mu\nu}h_{\mu\nu}=3$, we derive the Raychaudhuri equation (\ref{Raychaudhuri}).</p>
</div>
</div>
</div>

<div id="Ray6"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 11: congruence orthogonal to a hypersurface ===
Show that for a congruence of geodesics orthogonal to a family of hypersurfaces $\omega_{\mu\nu}=0$. Prove further, that in case the strong energy condition $R_{\mu\nu}u^{\mu}u^{\nu}\geq0$ holds (see [[#EnCond1|energy conditions]] and the next problem), then the following (the focusing theorem) also is true: if $\Theta=\Theta_{0}<0$ at some initial moment, then in a finite period of proper time $\Theta$ diverges and tends to $-\infty$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Orthogonality of a congruence to a family of hypersurfaces means that there is a scalar field $\phi(x)$, the level lines of which $\phi(x)=const$ define this family and $u_{\mu}=\alpha \phi_{,\nu}$. Then
\[\omega_{\mu\nu}=B_{[\mu\nu]}=u_{[\mu;\nu]}
=\alpha\phi_{[;\mu;\nu]}+\alpha_{[,\nu}\phi_{,\mu]}
=\alpha_{[,\nu}\phi_{,\mu]}
\sim \alpha_{[,\nu}u_{\mu]}
=\alpha_{,\nu}u_{\mu}-\alpha_{,\mu}u_{\nu}.\]
But on the other hand
\[0=\omega_{\mu\nu}u^{\nu}
=\alpha_{,\nu}u^{\nu}u_{\mu}-\alpha_{,\mu},\]
so $\alpha_{,\mu}\sim u_{\mu}$. But then $\omega_{\mu\nu}\sim u_{[\mu}u_{\nu]}=0$, and thus torsion turns to zero.

Then in case the strong energy condition $R_{\mu\nu}u^{\mu}u^{\nu}$ holds, there are only non-positive terms left in the right hand part of the Raychaudhuri equation and
\[\frac{d\Theta}{d\tau}\leq
-\frac{1}{3}\Theta^{2}.\]
On integration
\[\frac{1}{\Theta}-\frac{1}{\Theta_0}
\geq \frac{\Delta\tau}{3}\quad\Leftrightarrow\quad
\Theta\leq \frac{\Theta_{0}}
{1+\Delta\tau \Theta_{0}/3},\]
so if at some moment $\Theta_{0}<0$, then for $\Delta\tau\to |\Theta_{0}|/3$ we obtain from the inequality that $\Theta\to -\infty$, i.e. the geodesics of the congruence are focused into a point.

The focusing theorem implies that the geodesics form a caustics, which in general does not necessarily mean a singularity. Some further elaboration is needed, making use of the focusing theorem and usually some energy conditions, in order to prove that a singularity actually takes place$^{*}$</p>

<p style="text-align: left;">$^{*}$ For more on this better see textbooks Hawking S.W., Ellis G.F.R. ''The large scale structure of space-time'', CUP, 1973 (ISBN 0521099064) and Wald R.M, ''General relativity.'' U. Chicago, 1984, 505p (ISBN 0226870332).</p>
</div>
</div></div>

<div id="Ray7"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 12: Raychaudhuri equation for FLRW ===
Write out the Raychaudhuri equation for the geodesics of comoving matter in the FLRW Universe and show that it is reduced to the second Friedman equation.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let the metric have the form
\[ds^{2}=dt^{2}
-a^{2}(t)h_{ij}dx^{i}dx^{j}.\]
Consider the congruence of geodesics orthogonal to the hypersurfaces of constant cosmic time $t$, so that torsion $\omega_{\mu\nu}$ is zero and $u_{\mu}=\partial_{\mu}t=\delta_{\mu}^{t}$. Then
\[B_{\mu\nu}=u_{\nu;\mu}=u_{\nu,\mu}
+{\Gamma^{\lambda}}_{\mu\nu}u_{\lambda}
={\Gamma^{0}}_{\mu\nu}=\Gamma_{0,\mu\nu}
=-\tfrac{1}{2}\partial_{t}g_{\mu\nu}\]
and for the spatial components, which are the only ones different from zero,
\[B_{ij}
=-\tfrac{1}{2}\frac{\partial_{t}a^2}{a^2}(-h_{ij})
=\frac{\dot{a}}{a}h_{ij}.\]
We see from the explicit notation (as $t$ is also the natural parameter along the geodesics) that
\[\Theta=3\frac{\dot{a}}{a},\quad
\sigma_{\mu\nu}=0,\quad
\omega_{\mu\nu}=0.\]
and the Raychaudhuri equation (\ref{Raychaudhuri}) is rewritten as
\[\frac{d}{dt}\Big(3\frac{\dot{a}}{a}\Big)
=-3\Big(\frac{\dot{a}}{a}\Big)^{2}-R_{00},\]
thus
\[R_{00}=-\dot{\Theta}-\Theta^{2}/3
=-3\Big[\frac{d}{dt}\Big(\frac{\dot{a}}{a}\Big)
+\Big(\frac{\dot{a}}{a}\Big)^{2}\Big]
=-3\frac{\ddot{a}}{a}.\]
On substitution of this into the $\binom{0}{0}$ component of the Einstein's equation
\[R_{00}
=\frac{8\pi G}{c^4}\big(T_{00}-\tfrac{1}{2}Tg_{00})
=\frac{4\pi G}{c^4}\big(\varepsilon +3p\big),\]
we obtain the second Friedman equation.</p>

<p style="text-align: left;">We see now that the focusing theorem in the cosmological context implies that the geodesics of comoving matter must converge at some time in the past, as is already known to be a major feature of the solutions of the Friedman equations. Though in general a caustic of geodesics does not necessarily mean a singularity, in this case the considered geodesics are actually the geodesics of all the comoving matter in the Universe, so their focusing actually implies the Big Band singularity. The theorem becomes inapplicable, however, at small times, when particles' interaction has to be taken into account. This is where the more general singularity theorems work$^{*}$</p>

<p style="text-align: left;">$^{*}$ For more on this better see textbooks Hawking S.W., Ellis G.F.R. ''The large scale structure of space-time'', CUP, 1973 (ISBN 0521099064) and Wald R.M, ''General relativity.'' U. Chicago, 1984, 505p (ISBN 0226870332).</p>
</div>
</div></div>

==Sudden Future Singularities==
The following problems are composed in the spirit of [http://arxiv.org/abs/gr-qc/0403084 John D. Barrow, Sudden Future Singularities, arXiv:0403084v3].

<div id="sing1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 13: a sudden singularity ===
Let us consider the possibility of ''sudden future singularities''. The "suddenness" implies that they occur at some time in the future, while both the scale factor and the Hubble constant remain bounded and separated from zero:
\[a\to a_{s}\neq 0,\infty,
\qquad H\to H_{s}\neq 0,\infty.\]
What scalars can in principle become unbounded in this scenario?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">The first Friedman equation implies that under the imposed conditions of "unexpected-ness" density must be finite. However, $p$ can diverge along with $\ddot{a}$ and $\dot{\rho}$:
\[\frac{\ddot{a}}{a}\sim 4\pi k\,p,
\qquad \dot{\rho}\sim -3Hp.\]</p>
</div>
</div></div>

<div id="sing2"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 14: asymptotics ===
Consider a solution of Friedman equations of the form
\[a(t)=A+Bt^{q}+C(t_{s}-t)^{n},\]
where $A,B,q,n>0$ and $C$ are some free constants. What values of $q$ and $n$ are compatible with the sudden singularity of the previous problem?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Fixing the zero time by condition $a(0)=0$ and using the freedom of rescaling the scale factor, we can rewrite the solution as
\[a(t)=1+(a_{s}-1)\Big(\frac{t}{t_s}\Big)^{q}
-\Big(1-\frac{t}{t_s}\Big)^{n},\]
where $a_{s}=a(t_{s})$.
Then as $t\to t_{s}-0$ we have
\begin{equation}\label{sing_eq}
\ddot{a}=\frac{q(q-1)}{t_s^q}(a_{s}-1)t^{q-2}
-\frac{n(n-1)}{t_s^2}\Big(1-\frac{t}{t_s}\Big)^{n-2}
\to -\infty.
\end{equation}
The solution with singularity exists on the interval $0<t<t_s$ for
\[1<n<2 \quad\text{and}\quad 0<q\leq 1.\]
Note that $n>1$ is needed for $\dot{a}$ to stay finite and $n<2$ for $\ddot{a}$ to diverge. For $2<n<3$ the values of $\ddot{a}$ and $p$ would remain bounded, while $\dddot{a}$ and $\dot{p}$ diverge.</p>
</div>
</div></div>

<div id="sing3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15: energy conditions ===
Is any energy condition violated by the solutions with the sudden future singularity? What physical constraint on matter can be introduced that would prevent it?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Due to the second Friedman equation, we see that as (\ref{sing_eq}), both $\rho$ and $(\rho+3p)$ remain positive, so all common energy conditions are satisfied. The considered singularity is possible, however, only if pressure is allowed to be unbounded at finite values of density. If we demand that $p<C\rho$ for some $C>0$, for example, along with the common conditions $\rho>0$ and $\rho+3p>0$, then the sudden future singularity is eliminated.</p>
</div>
</div></div>

Energy conditions and the Raychaudhuri equation

2013-12-09T18:27:27Z

Igor: /* Problem 1: energy conditions for perfect fluid */

[[Category:Dynamics of the Universe in the Big Bang Model|7]]
__TOC__

==Energy conditions==
<p style="text-align: left;">S. Carroll writes$^{*}$: '' "Sometimes it is useful to think about Einstein's equation without specifying the theory of matter from which $T^{\mu\nu}$ is derived. This leaves us with a great deal of arbitrariness; consider for example the question, What metrics obey Einstein's equation? In the absence of some constraints on $T^{\mu\nu}$, the answer is any metric at all; simply take the metric of your choice, compute the Einstein tensor $G^{\mu\nu}$ for this metric, and then demand that $T^{\mu\nu}$ be equal to $G^{\mu\nu}$. It will automatically be conserved, by the Bianchi identity. Our real concern is with the existence of solutions to Einstein's equation in the presence of "realistic" sources of energy and momentum, whatever that means. One strategy is to consider specific kinds of sources, such as scalar fields, dust, or electromagnetic fields. However, we occasionally wish to understand properties of Einstein's equations that hold for a variety of different sources. In this circumstance it is convenient to impose energy conditions that limit the arbitrariness of $T^{\mu\nu}$." ''</p>

<p style="text-align: left;">The energy conditions are formulated in coordinate-independent way, but in the context of cosmology they are most useful in application to the energy-momentum tensor of a perfect fluid$^*$:
\[\begin{array}{lcc}
\text{Name}&\text{Statement}&\text{For perfect fluid}\\
\text{Weak}\phantom{\Big|}& T_{\mu\nu}v^{\mu}v^{\nu}\geq0&
\rho\geq 0,\quad \rho+p>0;\\[0.2cm]
\text{Null}& T_{\mu\nu}k^{\mu}k^{\nu}\geq 0&
\rho+p\geq 0;\\[0.2cm]
\text{Strong}&
(T_{\mu\nu}-\tfrac{1}{2}Tg_{\mu\nu})
v^{\nu}v^{\nu}\geq 0\quad&
\quad\rho+p\geq 0,\quad \rho+3p\geq 0;\\[0.2cm]
\text{Dominant}& \quad T^{\mu}_{\nu}v^{\nu}\;
\mbox{is non-spacelike and future-directed}\quad &
\rho\geq |p\,|.
\end{array}\]
The conditions are assumed to hold for arbitrary timelike vectors $v^{\mu}$ and arbitrary null vectors $k^{\mu}$.</p>

<p style="text-align: left;"> $^*$For more detailed discussion see textbooks:

Carroll S. ''Spacetime and geometry: an introduction to General Relativity''. AW, 2003; ISBN 0805387323, 525p.,

Poisson E. ''A relativist's toolkit''. CUP, 2004; ISBN 0521830915, 248p. (ch 2)</p>

<div id="EnCond1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1: energy conditions for perfect fluid ===
Derive the energy conditions for the perfect fluid, shown in the last column, from the coordinate-independent formulations.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let us use the locally Euclidean frame (such that $g_{\mu\nu}=\eta_{\mu\nu}$) comoving with the fluid, so that $u^{\mu}$ in
\begin{equation}\label{EnCond_PFl}
T^{\mu\nu}=\rho u^{\mu}u^{\nu}-pg^{\mu\nu}
\end{equation}
has the coordinates $u^{\mu}=(1,0,0,0)$. Next we choose the $x$-coordinate to be in the direction of the arbitrarily chosen timelike $v^\mu$ or null $k^\mu$ vector. Then those can be parametrized as
\begin{equation}\label{EnCond_VKparametrization}
v^{\mu}=\gamma(1,v,0,0),\qquad k^{\mu}=(1,1,0,0),
\end{equation}
where $v<1$ is the velocity of the particle in the comoving frame, and $\gamma=(1-v^2)^{-1/2}$ its Lorentz factor. We remember that normalization of a null vector is arbitrary.<br/>
'''1)''' ''Weak energy condition (WEC).'' <br/>Plugging in (\ref{EnCond_VKparametrization}), and taking into account that $v^{\mu}v_{\mu}=1$, we obtain
\[ T_{\mu\nu}v^{\mu}v^{\nu}
=\gamma^{2}(\rho+p)-p
=\gamma^{2}(\rho+v^{2}p)\geq0.\]
Since this must hold for any $v\in[0,1)$, it follows that
\[\rho\geq0,\quad \rho+p\geq 0.\]
<br/>
'''2)''' ''Null energy condition (NEC).'' <br/>In the same way using $k^{\mu}$ from (\ref{EnCond_VKparametrization}) with $k_{\mu}k^{\mu}=0$, we arrive to
\[T_{\mu\nu}k^{\mu}k^{\nu}
=\rho+p\geq 0,\]
which is already what we wanted to prove. We can also obtain this from the weak energy condition by removing $\gamma$ (normalization is not fixed) and formally replacing $v\to 1$.<br/>
'''3)''' ''Strong energy condition (SEC):''
\begin{align*}
&\big[T_{\mu\nu}-\tfrac{1}{2}Tg_{\mu\nu}\big]
v^{\mu}v^{\nu}=
\big[(\rho+p)u_{\mu}u_{\nu}
-\tfrac{1}{2}(\rho-p) g_{\mu\nu}\big]
v^{\mu}v^{\nu}=\\
&\quad=\gamma^{2}(\rho+p)-\tfrac{1}{2}(\rho-p)
=\tfrac{1}{2}\gamma^{2}
\big\{(\rho+3p)+v^{2}(\rho-p)\big\}\geq 0.
\end{align*}
As this must hold for $v\in[0,1)$, we get the conditions
\[\rho+3p\geq 0,\qquad \rho+p\geq 0.\]
Note that in the limit $v=1-0$ ($v$ tends to $1$ from below) the SEC is rewritten as
\[\rho(1-0)+p(1+0)\geq 0,\]
so the non-strict inequality sign remains in $\rho+p\geq 0$.

The strong energy condition is in the most simple way formulated for the Ricci tensor:
\[R_{\mu\nu}v^{\mu}v^{\nu}\geq 0,\]
This is the reason it is used in the focusing theorem discussed below.<br/>
'''4)''' ''Dominant energy condition (DEC).''<br/>The vector $w^{\mu}=T^{\mu}_{\nu}v^{\nu}$ is the momentum density as measured by the observer with 4-velocity $v^{\mu}$. In the chosen coordinate frame
\[w^{\mu}=\gamma (\rho+p)u^{\mu}-pv^{\mu}
=\gamma (\rho,-pv).\]
As this vector has to be future-directed, we have $\rho\geq 0$ and by demanding for it to be non-spacelike, we arrive to
\[w^{\mu}w_{\mu}
=\gamma^{2}\big(\rho^{2}-v^{2}p^{2})\geq 0.\]
Thus the DEC for a perfect fluid can be written as
\[|p|\leq \rho,\]
which also implies $\rho\geq 0$.</p>
</div>
</div></div>

<div id="EnCond2"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 2: weak or strong? ===
Does the weak energy condition follow from the strong one? Which of the energy conditions imply the others?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
No. Small negative $\rho$ with large positive $p$ obey the SEC and NEC but violate WEC and DEC.</p>

<p style="text-align: left;">Despite the naming, which is a bit confusing here, the ''null'' energy condition is the weakest, ''not'' the weak energy condition$^*$. The weak, strong and dominant conditions all imply NEC, but of all three only the dominant energy condition implies the weak one.</p>

<p style="text-align: left;">All other variants of matter, obeying one condition but violating another, are hypothetically possible.</p>

<p style="text-align: left;">$^*$Strictly speaking, the weak energy condition allows $\rho+p=0$, which is prohibited by the null condition, but from here on we will not usually distinguish the strict and non-strict equalities. However, this makes a difference when considering the cosmological constant (see the chapter on dark energy).</p>
</div>
</div></div>

<div id="EnCond3"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 3: energy conditions through geometry ===
Express the energy conditions in terms of scale factor and its derivatives.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">With the help of Friedman equations we can express density and pressure through the scale factor and its derivatives
\[\rho = \frac{3}{8\pi G}
\left[\Big(\frac{\dot a}{a}\Big)^{2}
+ \frac{k}{a^2}\right],\quad
p = - \frac{1}{8\pi G}\left[2\frac{\ddot a}{a}
+\Big(\frac{\dot a}{a}\Big)^{2}+\frac{k}{a^2} \right],\]
and then rewrite the energy conditions in the form
\[\begin{array}{ll}
\text{Null:}&\displaystyle
\phantom{\text{NEC}}
- \frac{\ddot a}{a}
+\Big(\frac{\dot a}{a}\Big)^{2}
+ \frac{k}{a^2} \geq 0;\\[0.4cm]
\text{Weak:}&\displaystyle \text{NEC and}\quad
\Big(\frac{\dot a}{a}\Big)^{2}
+\frac{k}{a^2} \geq 0;\\[0.4cm]
\text{Strong:}&\displaystyle\text{NEC and}\quad
\frac{\ddot a}{a} \leq 0.
\end{array}\]
The strong energy condition follows immediately from the second Friedman the expansion of the universe is decelerating equation. We should stress, that the SEC implies that the Unverse is decelerating and this conclusion holds regardless to whether the Universe is open, closed or flat.

For the dominant EC we have two variants: $\rho\geq p\geq 0$ and $-p\geq \rho\geq 0$, which lead to
\[\text{Dominant}:\quad\left\{
\begin{array}{l}\displaystyle
\frac{\ddot a}{a}+\frac{1}{2}
\left[\Big(\frac{\dot a}{a}\Big)^{2}
+\frac{k}{a^2}\right]\leq 0,\\[0.3cm] \displaystyle
\frac{\ddot a}{a}+2\left[
\Big(\frac{\dot a}{a}\Big)^{2}+\frac{k}{a^2}\right]\geq0,
\end{array}\right.
\quad\text{or}\quad\left\{
\begin{array}{l}\displaystyle
\frac{\ddot a}{a}+\tfrac{1}{2}
\left[\Big(\frac{\dot a}{a}\Big)^{2}
+\frac{k}{a^2}\right]\geq0,\\[0.3cm]
\displaystyle \frac{\ddot a}{a}
-\Big(\frac{\dot a}{a}\Big)^{2}\geq0,
\end{array}\right..\]</p>
</div>
</div></div>

<div id="EnCond4"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 4: and in terms of redshift ===
Express the null, weak and strong energy conditions in terms of the Hubble parameter and redshift.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Using the definitions $H=\dot{a}/a$ and relation to the redshift $a_{0}/a=(z+1)$, we get
\[\begin{array}{ll}
\text{Null:}&\displaystyle
\phantom{\text{NEC and}}
\frac{\partial H^2}{\partial z}
\ge -\frac{2k\left(1 + z\right)}{a_0^2};\\[0.4cm]
\text{Weak:}&\displaystyle\text{NEC and}\quad
\frac{k\left(1+z\right)^2}{a_0^2H^2}\geq -1;
\\[0.4cm]
\text{Strong:}&\displaystyle\text{NEC and}\quad
\frac{\partial \ln H}{\partial z}\geq
\frac{1}{1+z}.
\end{array}\]</p>
</div>
</div></div>

<div id="EnCond5"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 5: restrictions on deceleration parameter ===
Find the restrictions that the energy conditions impose on the deceleration parameter in a flat Universe with $\rho>0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Dividing one Friedman equation by the other, we get
\[q=-\frac{\ddot{a}/a}{H^2}
=\frac{1}{2}\big(1+3p/\rho\big)=\frac{1}{2}(1+3w).\]
Then from the energy conditions, if we assume $\rho>0$, we straightforwardly obtain
\begin{eqnarray*}
\text{NEC or WEP:} & 1+w\geq 0 &
\Rightarrow\quad q \geq - 1; \\
\text{SEC:} &1+3w\geq 0 &
\Rightarrow\quad q \geq 0;\\
\text{DEC:} & |w|<1&
\Rightarrow\; -1\leq q \leq 2\end{eqnarray*}</p>
</div>
</div></div>

==Raychaudhuri equation==

<div id="Ray1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6: projection operators ===
Consider a timelike curve $x^{\mu}(\tau)$ and find the projection operators on its tangent vector and on its orthogonal complement.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">The decomposition of unit operator into projectors is
\[\delta^{\mu}_{\nu}=u^{\mu}u_{\nu}+h^{\mu}_{\nu},\]
or in terms of metric
\[g_{\mu\nu}=u_{\mu\nu}+h_{\mu\nu},\quad
\mbox{where}\quad
h_{\mu\nu}=g_{\mu\nu}-u_{\mu}u_{\nu}.\]
One can see that $u^{\mu}u_{\nu}$ and $h^{\mu}_{\nu}$ are indeed projection operators from their action on $u^{\mu}$ ($u^{\mu}u_{\mu}=1$):
\[(u^{\mu}u_{\nu})u^{\nu}=u^{\mu},\quad
h^{\mu}_{\nu}u^{\nu}=0,\]
and on any vector $e^{\mu}$, orthogonal to $u^{\mu}$ ($u^{\mu}e_{\mu}=0$):
\[(u^{\mu}u_{\nu})e^{\nu}=0,\quad
h^{\mu}_{\nu}e^{\nu}=e^{\mu}.\]
At the same time the quantity $h_{\mu\nu}$ acts as the metric in the orthogonal complement to linear span of $u^{\mu}$.
\[h_{\mu\nu}e^{\mu}f^{\nu}
=g_{\mu\nu}e^{\mu}e^{\nu}.\]</p>
</div>
</div></div>

<div id="Ray2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7: geodesic deviation ===
A ''congruence'' is a set of curves having the property that each point in a given region belongs to one and only one curve of the set. Consider a congruence of timelike geodesics. Let us mark two infinitely close geodesics and look at their relative evolution along their length. Let $\xi^{\mu}$ be the infinitesimal $4$-vector that is directed normal to one of the curves towards the other. Show that
\[\frac{d\xi_{\nu}}{d\tau}
=B_{\nu\mu}\xi^{\mu},\]
where
\[B_{\nu\mu}=u_{\nu;\mu},\]
is a three-dimensional spacelike tensor orthogonal to $u^{\mu}$, and $\tau$ is the parameter along the geodesic.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let $\lambda$ be the parameter along a curve with tangent vector $\xi^{\mu}$. Then
\[B^{\nu\mu}\xi_{\mu}
=\xi^{\mu}\nabla_{\mu}u^{\nu}
=\frac{du^{\nu}}{d\lambda}
=\frac{dx^{\nu}}{d\tau d\lambda}
=\frac{dx^{\nu}}{d\lambda d\tau}
=\frac{d\xi^{\nu}}{d\tau}\]
and
\[\frac{d\xi_{nu}}{d\tau}=B_{\nu\mu}\xi^{\mu}.\]

From the geodesic equation and normalization condition
\[\left\{\begin{array}{l}
B_{\mu\nu}u^{\nu}=u^{\nu}\nabla_{\nu}u_{\mu}=0,\\
B_{\mu\nu}u^{\mu}=u^{\mu}\nabla_{\nu}u_{\mu}
=\tfrac{1}{2}\nabla_{\nu}(u^{\mu}u_{\mu})
\sim\nabla_{\nu} (1)=0.
\end{array}\right.\]

On the other hand, we can always choose a coordinate frame in such a way that locally $g_{\mu\nu}=\eta_{\mu\nu}$, and by additional Lorentz boost we can direct the time along $u^{\mu}$. If ${e^{\mu}}_{i}$ ($i,j=1,2,3$) are the orthonormal unit vectors of spatial directions, such that $h_{\mu\nu}{e^{\mu}}_{i}{e^{\nu}}_{j}=\eta_{ij}=-\delta_{ij}$, then due to orthogonality $B$ is expressible as
\[B^{\mu\nu}=B_{ij}{e^{\mu}}_{i}{e^{\nu}}_{j},\]
so in particular we obtain $B_{\mu\nu}B^{\mu\nu}=B_{ij}B^{ij}\geq 0$.</p>
</div>
</div></div>

<div id="Ray3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8: tensor decomposition ===
Show that any tensor field of second rank defined on an $n-$dimensional Riemannian manifold with positive definite metric $g_{\mu\nu}$ can be uniquely decomposed into
\begin{equation}\label{TensorDecomposition}
B_{\mu\nu}=\frac{1}{n}\Theta g_{\mu\nu}
+\sigma_{\mu\nu}+\omega_{\mu\nu},
\end{equation}
where $\Theta={B^{\mu}}_{\mu}$, $\sigma_{\mu\nu}$ is the symmetric traceless part of $B_{\mu\nu}$, and $\omega_{\mu\nu}$ is the antisymmetric part of $B_{\mu\nu}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let us decompose $B_{\mu\nu}$ into the symmetric and anti-symmetric parts (symmetrization is denoted by parenthesis $()$ and anti-symmetrization by brackets $[]$):
\[B_{\mu\nu}
=\frac{B_{\mu\nu}+B_{\nu\mu}}{2}
+\frac{B_{\mu\nu}-B_{\nu\mu}}{2}
=B_{(\mu\nu)}+B_{[\mu\nu]}.\]
Then search for the representation of the symmetrical part in the form
\[B_{(\mu\nu)}=\alpha g_{\mu\nu}+\sigma_{\mu\nu},
\quad\mbox{where}\quad
\sigma^{\mu}_{\mu}=0.\]
The contraction is $\Theta=B^{\mu}_{\mu}=\alpha \delta^{\mu}_{\mu}=\alpha n$, so we obtain $\alpha=\Theta/n$ and thus derive the representation of $B_{\mu\nu}$ we had sought for.</p>
</div>
</div></div>

<div id="Ray4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9: geometrical meaning in 3D ===
Let there be a congruence in a three-dimensional Riemannian manifold. What is the geometric meaning of $\Theta$, $\sigma$ and $\omega$ for $B_{\mu\nu}=u_{\nu;\mu}$?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let the vector $\xi^{\mu}$ be a two-dimensional vector that lies in the space $S$ orthogonal to the tangent vector, and let us consider the pipe of curves from the congruence in the neighborhood of the given curve and the evolution of its form with the parameter $\tau$. Each of the generating lines of the pipe is parametrized by the corresponding vector $\xi^{\mu}$.</p>

<p style="text-align: left;">a) Let $B_{\mu\nu}=\frac{\Theta}{2}g_{\mu\nu}$. Then
\[B^{\mu}_{\nu}=\frac{\Theta}{2}
\begin{pmatrix} 1&0\\0&1\end{pmatrix}
\quad\mbox{and}\quad
\frac{d\xi^{\mu}}{d\tau}
=\frac{\Theta}{2}\xi^{\mu},\]
so the pipe's section is merely stretched $(1+\Theta/2)$ times. The area $S$ of the pipe's section changes as $dS/S=d(|\xi|^2)/|\xi|^2 = \Theta d\tau$, so
\[\Theta=\frac{1}{S}\frac{dS}{d\tau}.\]

b) The matrix of the anti-symmetric part is
\[{\omega^{\mu}}_{\nu}=\begin{pmatrix}
0&\omega\\-\omega&0
\end{pmatrix},\]
and it describes the torsion of the pipe, without changing its section area.

c) The matrix of the symmetric traceless part can be presented in the form
\[\sigma^{\mu}_{\nu}=
\begin{pmatrix}\sigma_{+}&\sigma_{\times}\\
\sigma_{\times}&-\sigma_{+}\end{pmatrix}
=\sigma_{+}
\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}
+\sigma_{\times}
\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}
\begin{pmatrix}0&-1\\ -1&0\end{pmatrix}.\]
The first term describes expansion in one direction in $(1+\sigma_{+})$ times, and contraction by the same factor in the perpendicular direction. The second term is the superposition of the same deformation without change of volume and a rotation. So the general deformation is a shear in arbitrary direction, without change of section area.</p>
</div>
</div></div>

<div id="Ray5"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 10: Landau-Raychaudhuri equation ===
Derive the Raychaudhuri equation for a congruence of timelike geodesics in spacetime
\begin{equation}\label{Raychaudhuri}
\frac{d\Theta}{d\tau}=
-\frac{1}{3}\Theta^{2}
-\sigma_{\mu\nu}\sigma^{\mu\nu}
+\omega_{\mu\nu}\omega^{\mu\nu}
-R_{\mu\nu}u^{\mu}u^{\mu}.
\end{equation}
Here $\Theta$, $\sigma_{\mu\nu}$ and $\omega_{\mu\nu}$ are the components (\ref{TensorDecomposition}) of decomposition of $B_{\mu\nu}=u_{\mu;\nu}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let us first consider the evolution of tensor $B_{\mu\nu}$ along the geodesic: changing the order of the covariant derivatives and using one of the definitions of the curvature tensor, we get
\begin{align*}
\frac{dB_{\mu\nu}}{d\tau}&
=u^{\lambda}\nabla_{\lambda}\nabla_{\nu}u_{\mu}
=\\&
=u^{\lambda}\nabla_{\nu}\nabla_{\lambda}u_{\mu}
+u^{\lambda}{R^{\sigma}}_{\mu\nu\lambda}u_{\sigma}
=\\&
=\nabla_{\nu}(u^{\lambda}\nabla_{\lambda}u_{\mu})
-(\nabla_{\nu}u^{\lambda})(\nabla_{\lambda}u_{\mu})
-R_{\rho\mu\sigma\lambda}u^{\rho}u^{\sigma}
=\\&
=-{B_{\nu}}^{\lambda}B_{\lambda\mu}
-R_{\mu\rho\lambda\sigma}u^{\rho}u^{\sigma}.
\end{align*}

On contraction we obtain
\[\frac{d\Theta}{d\tau}=-B^{\mu\nu}B_{\nu\mu}
-R_{\mu\nu}u^{\mu}u^{\nu}.\]
Now rewrite the contraction $B_{\mu\nu}B_{\nu\mu}$ in terms of components of $B$'s decomposition and take into account that pairwise contractions are zero due to the symmetries and due to $h^{\mu\nu}\sigma_{\mu\nu}=g^{\mu\nu}\sigma_{\mu\nu}=0$:
\[B^{\mu\nu}B_{\nu\mu}
=\frac{\Theta^2}{9}h^{\mu\nu}h_{\mu\nu}
+\sigma^{\mu\nu}\sigma_{\mu\nu}
-\omega^{\mu\nu}\omega_{\mu\nu}\]
(the last term goes with the opposit sign due to anti-symmetry of $\omega^{\mu\nu}\omega_{\nu\mu}=-\omega^{\mu\nu}\omega_{\mu\nu}$), so taking into account that $h^{\mu\nu}h_{\mu\nu}=3$, we derive the Raychaudhuri equation (\ref{Raychaudhuri}).</p>
</div>
</div>
</div>

<div id="Ray6"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 11: congruence orthogonal to a hypersurface ===
Show that for a congruence of geodesics orthogonal to a family of hypersurfaces $\omega_{\mu\nu}=0$. Prove further, that in case the strong energy condition $R_{\mu\nu}u^{\mu}u^{\nu}\geq0$ holds (see [[#EnCond1|energy conditions]] and the next problem), then the following (the focusing theorem) also is true: if $\Theta=\Theta_{0}<0$ at some initial moment, then in a finite period of proper time $\Theta$ diverges and tends to $-\infty$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Orthogonality of a congruence to a family of hypersurfaces means that there is a scalar field $\phi(x)$, the level lines of which $\phi(x)=const$ define this family and $u_{\mu}=\alpha \phi_{,\nu}$. Then
\[\omega_{\mu\nu}=B_{[\mu\nu]}=u_{[\mu;\nu]}
=\alpha\phi_{[;\mu;\nu]}+\alpha_{[,\nu}\phi_{,\mu]}
=\alpha_{[,\nu}\phi_{,\mu]}
\sim \alpha_{[,\nu}u_{\mu]}
=\alpha_{,\nu}u_{\mu}-\alpha_{,\mu}u_{\nu}.\]
But on the other hand
\[0=\omega_{\mu\nu}u^{\nu}
=\alpha_{,\nu}u^{\nu}u_{\mu}-\alpha_{,\mu},\]
so $\alpha_{,\mu}\sim u_{\mu}$. But then $\omega_{\mu\nu}\sim u_{[\mu}u_{\nu]}=0$, and thus torsion turns to zero.

Then in case the strong energy condition $R_{\mu\nu}u^{\mu}u^{\nu}$ holds, there are only non-positive terms left in the right hand part of the Raychaudhuri equation and
\[\frac{d\Theta}{d\tau}\leq
-\frac{1}{3}\Theta^{2}.\]
On integration
\[\frac{1}{\Theta}-\frac{1}{\Theta_0}
\geq \frac{\Delta\tau}{3}\quad\Leftrightarrow\quad
\Theta\leq \frac{\Theta_{0}}
{1+\Delta\tau \Theta_{0}/3},\]
so if at some moment $\Theta_{0}<0$, then for $\Delta\tau\to |\Theta_{0}|/3$ we obtain from the inequality that $\Theta\to -\infty$, i.e. the geodesics of the congruence are focused into a point.

The focusing theorem implies that the geodesics form a caustics, which in general does not necessarily mean a singularity. Some further elaboration is needed, making use of the focusing theorem and usually some energy conditions, in order to prove that a singularity actually takes place$^{*}$</p>

<p style="text-align: left;">$^{*}$ For more on this better see textbooks Hawking S.W., Ellis G.F.R. ''The large scale structure of space-time'', CUP, 1973 (ISBN 0521099064) and Wald R.M, ''General relativity.'' U. Chicago, 1984, 505p (ISBN 0226870332).</p>
</div>
</div></div>

<div id="Ray7"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 12: Raychaudhuri equation for FLRW ===
Write out the Raychaudhuri equation for the geodesics of comoving matter in the FLRW Universe and show that it is reduced to the second Friedman equation.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Let the metric have the form
\[ds^{2}=dt^{2}
-a^{2}(t)h_{ij}dx^{i}dx^{j}.\]
Consider the congruence of geodesics orthogonal to the hypersurfaces of constant cosmic time $t$, so that torsion $\omega_{\mu\nu}$ is zero and $u_{\mu}=\partial_{\mu}t=\delta_{\mu}^{t}$. Then
\[B_{\mu\nu}=u_{\nu;\mu}=u_{\nu,\mu}
+{\Gamma^{\lambda}}_{\mu\nu}u_{\lambda}
={\Gamma^{0}}_{\mu\nu}=\Gamma_{0,\mu\nu}
=-\tfrac{1}{2}\partial_{t}g_{\mu\nu}\]
and for the spatial components, which are the only ones different from zero,
\[B_{ij}
=-\tfrac{1}{2}\frac{\partial_{t}a^2}{a^2}(-h_{ij})
=\frac{\dot{a}}{a}h_{ij}.\]
We see from the explicit notation (as $t$ is also the natural parameter along the geodesics) that
\[\Theta=3\frac{\dot{a}}{a},\quad
\sigma_{\mu\nu}=0,\quad
\omega_{\mu\nu}=0.\]
and the Raychaudhuri equation (\ref{Raychaudhuri}) is rewritten as
\[\frac{d}{dt}\Big(3\frac{\dot{a}}{a}\Big)
=-3\Big(\frac{\dot{a}}{a}\Big)^{2}-R_{00},\]
thus
\[R_{00}=-\dot{\Theta}-\Theta^{2}/3
=-3\Big[\frac{d}{dt}\Big(\frac{\dot{a}}{a}\Big)
+\Big(\frac{\dot{a}}{a}\Big)^{2}\Big]
=-3\frac{\ddot{a}}{a}.\]
On substitution of this into the $\binom{0}{0}$ component of the Einstein's equation
\[R_{00}
=\frac{8\pi G}{c^4}\big(T_{00}-\tfrac{1}{2}Tg_{00})
=\frac{4\pi G}{c^4}\big(\varepsilon +3p\big),\]
we obtain the second Friedman equation.</p>

<p style="text-align: left;">We see now that the focusing theorem in the cosmological context implies that the geodesics of comoving matter must converge at some time in the past, as is already known to be a major feature of the solutions of the Friedman equations. Though in general a caustic of geodesics does not necessarily mean a singularity, in this case the considered geodesics are actually the geodesics of all the comoving matter in the Universe, so their focusing actually implies the Big Band singularity. The theorem becomes inapplicable, however, at small times, when particles' interaction has to be taken into account. This is where the more general singularity theorems work$^{*}$</p>

<p style="text-align: left;">$^{*}$ For more on this better see textbooks Hawking S.W., Ellis G.F.R. ''The large scale structure of space-time'', CUP, 1973 (ISBN 0521099064) and Wald R.M, ''General relativity.'' U. Chicago, 1984, 505p (ISBN 0226870332).</p>
</div>
</div></div>

==Sudden Future Singularities==
The following problems are composed in the spirit of [http://arxiv.org/abs/gr-qc/0403084 John D. Barrow, Sudden Future Singularities, arXiv:0403084v3].

<div id="sing1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 13: a sudden singularity ===
Let us consider the possibility of ''sudden future singularities''. The "suddenness" implies that they occur at some time in the future, while both the scale factor and the Hubble constant remain bounded and separated from zero:
\[a\to a_{s}\neq 0,\infty,
\qquad H\to H_{s}\neq 0,\infty.\]
What scalars can in principle become unbounded in this scenario?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">The first Friedman equation implies that under the imposed conditions of "unexpected-ness" density must be finite. However, $p$ can diverge along with $\ddot{a}$ and $\dot{\rho}$:
\[\frac{\ddot{a}}{a}\sim 4\pi k\,p,
\qquad \dot{\rho}\sim -3Hp.\]</p>
</div>
</div></div>

<div id="sing2"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 14: asymptotics ===
Consider a solution of Friedman equations of the form
\[a(t)=A+Bt^{q}+C(t_{s}-t)^{n},\]
where $A,B,q,n>0$ and $C$ are some free constants. What values of $q$ and $n$ are compatible with the sudden singularity of the previous problem?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Fixing the zero time by condition $a(0)=0$ and using the freedom of rescaling the scale factor, we can rewrite the solution as
\[a(t)=1+(a_{s}-1)\Big(\frac{t}{t_s}\Big)^{q}
-\Big(1-\frac{t}{t_s}\Big)^{n},\]
where $a_{s}=a(t_{s})$.
Then as $t\to t_{s}-0$ we have
\begin{equation}\label{sing_eq}
\ddot{a}=\frac{q(q-1)}{t_s^q}(a_{s}-1)t^{q-2}
-\frac{n(n-1)}{t_s^2}\Big(1-\frac{t}{t_s}\Big)^{n-2}
\to -\infty.
\end{equation}
The solution with singularity exists on the interval $0<t<t_s$ for
\[1<n<2 \quad\text{and}\quad 0<q\leq 1.\]
Note that $n>1$ is needed for $\dot{a}$ to stay finite and $n<2$ for $\ddot{a}$ to diverge. For $2<n<3$ the values of $\ddot{a}$ and $p$ would remain bounded, while $\dddot{a}$ and $\dot{p}$ diverge.</p>
</div>
</div></div>

<div id="sing3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15: energy conditions ===
Is any energy condition violated by the solutions with the sudden future singularity? What physical constraint on matter can be introduced that would prevent it?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Due to the second Friedman equation, we see that as (\ref{sing_eq}), both $\rho$ and $(\rho+3p)$ remain positive, so all common energy conditions are satisfied. The considered singularity is possible, however, only if pressure is allowed to be unbounded at finite values of density. If we demand that $p<C\rho$ for some $C>0$, for example, along with the common conditions $\rho>0$ and $\rho+3p>0$, then the sudden future singularity is eliminated.</p>
</div>
</div></div>

Causal Structure

2013-12-08T18:00:47Z

Igor:

[[Category:Horizons|4]]

__TOC__

The causal structure is determined by propagation of light and is best understood in terms of ''conformal diagrams''. In this section we construct and analyze those for a number of important model cosmological solutions (which are assumed to be already known), following mostly the exposition of <ref name=Mukhanov/>.

In terms of comoving distance $\tilde{\chi}$ and conformal time $\tilde{\eta}$ (in this section they are denoted by tildes) the two-dimensional radial part of the FLRW metric takes form
\begin{equation}
ds_{2}^{2}=a^{2}(\tilde{\eta})\big[d\tilde{\eta}^2 -d\tilde{\chi}^2\big].
\label{ds2Dconf}
\end{equation}
In the brackets here stands the line element of two-dimensional Minkowski flat spacetime. Coordinate transformations that preserve the \emph{conformal} form of the metric
\begin{equation*}
ds_2^2 =\Omega^{2}(\eta,\chi)\big[d\eta^2 -d\chi^2 \big],
\end{equation*}
are called conformal transformations, and the corresponding coordinates $(\eta,\chi)$ -- conformal coordinates.

<div id="Horizon39"></div>
=== Problem 1 ===
Show that it is always possible to construct $\eta(\tilde{\eta},\tilde{\chi})$, $\chi(\tilde{\eta},\tilde{\chi})$, such that the conformal form of metric (\ref{ds2Dconf}) is preserved, but $\eta$ and $\chi$ are bounded and take values in some finite intervals. Is the choice of $(\eta,\chi)$ unique?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Suppose $\tilde{\eta},\tilde{\chi}$ span infinite or semi-infinite values. Then we can always make the following sequence of coordinate transformations:
<div style="text-align: left;">
* Pass to null coordinates
\begin{equation}
u=\tilde{\eta}-\tilde{\chi},\qquad v=\tilde{\eta}+\tilde{\chi};
\end{equation}
* Bring their range of values to a finite interval by some appropriate function, i.e.
\begin{equation}
U=\arctan u,\qquad V=\arctan v.
\end{equation}
* Go back to timelike and spacelike coordinates (this is not really necessary at this point and is done mostly for aesthetic reasons):
\begin{equation}
T=V+U,\qquad R=V-U.
\end{equation}
Now the range of $(T,R)$ obviously covers some bounded region on the plane, while the radial part of the line element preserves its conformal form:
\begin{equation}
ds_2^2 \sim d\tilde{\eta}^2-d\tilde{\chi}^2 \sim du\, dv \sim dU\, dV \sim dT^2 -dR^2 .
\end{equation}
As the choice of function $\arctan$ was rather arbitrary (though convenient), the choice of conformal coordinates is not unique.
</div>
</p>
</div>
</div>

In this section we will reserve notation $\eta$ and $\chi$ and name "conformal coordinates/variables" to such variables that can only take values in a bounded region on $\mathbb{R}^2$; $\tilde{\eta}$ and $\tilde{\chi}$ can span infinite or semi-infinite intervals. Spacetime diagram in terms of conformal variables $(\eta,\chi)$ is called conformal diagram. Null geodesics $\eta=\pm \chi + const$ are diagonal straight lines on conformal diagrams.

<div id="Horizon40"></div>
=== Problem 2 ===
Construct the conformal diagram for the closed Universe filled with
* radiation;
* dust;
* mixture of dust and radiation.

Show the particle and event horizons for the observer at the origin $\chi=0$ (it will be assumed hereafter that the horizons are always constructed with respect to this chosen observer).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* Solution for $a(\eta)$ in a radiation dominated closed Universe is
\begin{equation}
a=a_m \sin\eta ,\qquad \eta\in(0,\pi),\quad \chi \in [0,\pi].
\end{equation}
As the ranges spanned by $\eta$ and $\chi$ are finite, they are already conformal coordinates. The conformal diagram is a square $\eta,\chi\in [0,\pi]$. Edges $\eta=0$ and $\eta=\pi$ correspond to the Big Bang and Big Crunch singularities respectively; worldline $\chi=\pi$ is the point on the three-sphere that is situated at the opposite pole with respect to observer at $\chi=0$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Rad.png|center|thumb|365px|Conformal diagram for the radiation dominated closed Universe. Particle horizon is shown in blue, event horizon in red.]]
|}

The particle horizon is
\[\eta=\chi,\]
and event horizon is
\[\eta=\eta_{max}-\chi =\pi -\chi.\]
Both exist for all cosmological times: by the finite moment of the Big Crunch $\eta=\pi$ the event horizon is collapsed into a point (which is natural, as there is no more time left), while the particle horizon extends to the whole Universe. Thus only at the finite moment the whole of the Universe becomes observable. The farther the point, though, the younger will it look, of course, and the opposite pole will only be "observed" by our observer at the last moment of the Universe as it was at its creation.

* Solution for $a(\eta)$ in a dust dominated closed Universe is
\begin{equation}
a=a_m (1-\cos\eta) ,\qquad \eta\in(0,2\pi),\quad \chi \in [0,\pi].
\end{equation}
The difference from the previous case is that $\eta$ spans twice the range, and $\eta_{max}=2\pi =2\chi_{max}$.

Therefore the event horizon is given by
\[\eta=\eta_{max}-\chi=2\pi -\chi,\]
so it exists only in the second, contracting, phase $\eta>\pi$. The particle horizon is given by $\eta=\chi$ again, but now it exists only during the expanding phase $\eta<\pi$. It encloses the full Universe at the moment of maximal expansion $\eta=\pi$ and for later times does not exist.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Dust.png|center|thumb|365px|Conformal diagrams of the dust dominated closed Universe. Thin lines show light rays that realize the first and second images of a galaxy at different cosmic times, from opposite directions.]]
|}

* Though the full analytic solution is more complicated, it is clear that the main features remain the same as in the previous considered cases. At early and late times, close to the singularities, the dynamics is determined by the radiation component. If there is enough dust, then at large enough scale factors (which may or may not be achieved depending on the initial conditions), which would correspond to the epoch around the maximal expansion, it will be dominating. The influence of dust is that dynamics is slowed down, so that depending on the ratio of densities
\[\eta_{max}\in [\pi,2\pi].\]
Thus qualitatively the picture will be the same as in a dust dominated Universe: the conformal diagram is a rectangle, the event horizon exists only starting from some time $\eta_{e}=\eta_{max}-\pi$, while particle horizon, on the contrary, vanishes at $\eta_{p}=\pi$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-RadDust.png|center|thumb|365px|Conformal diagram for a closed Universe filled with a mix of dust and radiation.]]
|}
</div>
</p>
</div>
</div>

<div id="Horizon41"></div>

=== Problem 3 ===
'''Closed dS.''' Construct the conformal diagram for the de Sitter space in the closed sections coordinates. Provide reasoning that this space is (null) geodesically complete, i.e. every (null) geodesic extends to infinite values of affine parameters at both ends.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The scale factor in the closed dS Universe is
\begin{equation}
a(t)=H_\Lambda^{-1}\cosh (H_\Lambda t),\qquad t\in (-\infty, +\infty),
\end{equation}
so on integration, for conformal time we obtain
\begin{equation}
\eta (t)=\int\limits_{-\infty}^{t}\frac{dt}{a(t)}=\arcsin\big[\tanh (H_\Lambda t)\big]+\frac{\pi}{2} \in (0,\pi).
\end{equation}
We choose the integration constant here so that $\eta=0$ corresponds to $t=-\infty$ and $\eta=\pi$ to $t=+\infty$. The full metric then can be written as
\begin{equation}
ds^{2}_{dS}=\frac{H_\Lambda^2}{\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\sin^2 \chi d\Omega^2 \big]. \label{dSclosed}
\end{equation}

As the values of $\eta$ span a finite interval, $(\eta,\chi)$ are already conformal coordinates. The conformal diagram is again a square
\begin{equation}
\eta\in [0,\pi],\qquad \chi \in[0,\pi],
\end{equation}
with the difference from the radiation dominated Universe that the edges $\eta=0,\pi$ do not represent a singularity anymore, but instead correspond to infinite (and regular) past and future respectively. Both horizons are given again by
\begin{equation}
\eta_{e}=\pi-\chi,\qquad \eta_{p}=\chi
\end{equation}
and exist at all times.

The spacelike boundaries of the conformal diagram correspond to $t\to \pm\infty$, and therefore to infinite values of affine parameter. This can be shown if one remembers the general formula for the cosmological redshift:
\begin{equation}
\text{const}=\omega a =\frac{dt}{d\lambda}a,\quad \Rightarrow\quad
\lambda=\text{const}\cdot\int\limits^{t}dt \cosh(H_\Lambda t)
\underset{t\to\pm\infty}{\longrightarrow}\infty .
\end{equation}
The timelike boundary of the diagram corresponds to the opposite pole, there is no real boundary there, the same as on a sphere: as particles propagate across the pole, their radial coordinate begins to decrease again, while the worldline on the conformal diagram is reflected from $\chi=\pi$. Thus by definition the spacetime is (null) geodesically complete.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-closed.png|center|thumb|365px|The de Sitter Universe. The closed sections coordinates cover the whole space, which is geodesically complete. There are no singularities: the horizontal boundaries of the diagram correspond to infinite past and future.]]
|}
</p>
</div>
</div>

<div id="Horizon42"></div>
=== Problem 4 ===
'''Static dS.''' Rewrite the metric of de Sitter space (\ref{dSclosed}) in terms of "static coordinates" $T,R$:
\begin{equation}
\tanh (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad
H_\Lambda R =\frac{\sin\chi}{\sin\eta}.
\end{equation}
* What part of the conformal diagram in terms of $(\eta,\chi)$ is covered by the static coordinate chart $(T,R)$?
* Express the horizon's equations in terms of $T$ and $R$
* Draw the surfaces of constant $T$ and $R$ on the conformal diagram.
* Write out the coordinate transformation between $(\eta,\chi)$ and $(T,R)$ in the regions where $|\cos\eta|>|\cos\chi|$. Explain the meaning of $T$ and $R$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us introduce dimensionless coordinates $t=H_\Lambda T$ and $r=H_\Lambda R$. The inverse relations then are
\begin{equation}
\sin^2 \eta =\frac{1}{\cosh^2 t-r^2 \sinh^2 t},
\quad \sin^2 \chi =\frac{r^2}{\cosh^2 t-r^2 \sinh^2 t}.
\end{equation}
Using them, after some algebra from (\ref{dSclosed}) we get
\begin{equation}
ds_{dS}^2 =\big[1-H_\Lambda^2 R^2\big]dT^2 -\frac{dR^2}{1-H_\Lambda^2 R^2}-R^2 d\Omega^2 , \label{dSstatic}
\end{equation}
which resembles the Schwarzschild line element (and this is not a coincidence).
</p>

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-static.png|center|thumb|365px|The de Sitter Universe with contour lines of the "static" coordinates $(T,R)$. The solid lines are $T=const$, and the dashed ones $R=const$. The coordinates become singular on both horizons, so the whole spacetime is divided by them into four sectors, each separately covered by the regular coordinate chart $(R,T)$. Spacetime is actually static only in the left and right sectors (I and III), where $T$ is a timelike coordinate and $R$ spacelike: the static patches are bounded by the horizons.]]
|}

<p style="text-align: left;">
* As $|\tanh (H_\Lambda T)|<1$, we have $|\cos\eta| \leq |\cos\chi|$. This condition cuts out two out of four sectors from the square conformal diagram, I and III: one is $\eta\in [\chi, \pi-\chi]$ and the other is $\eta\in [\pi-\chi, \chi]$. In both $|\sin\eta|\geq |\sin\chi|$, so $R \leq H_{\Lambda}^{-1}$;
* The particle horizon $\eta=\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=-\infty$. It is one part of the boundary of the region (in two parts) covered by coordinates $(T,R)$. The event horizon $\eta=\pi-\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=+\infty$ and is the other part of the boundary of this region.
* $T=const$ is $\cos\eta =t\cos\chi $ and $R=const$ is $\sin\eta =r^{-1}\sin\chi$.
* In regions II and IV the needed relation is obtained if we simply replace $\tanh$ with $\coth$ in the first relation:
\begin{equation}
\coth (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad H_\Lambda R=\frac{\sin\chi}{\sin\eta},
\end{equation}
as can be checked explicitly by substitution into (\ref{dSstatic}), which again gives (\ref{dSclosed}). In these regions $R$ is a timelike coordinate, and $T$ is spacelike. The geodesics of comoving massive particles are $\chi=const$, one of them $\chi=\pi/2$ corresponds to $T=0$. Thus in the lower part of the diagram, where $R\in (+\infty,H_\Lambda^{-1})$, the spacetime is contracting; in the upper part, where $R\in (H_\Lambda^{-1},\infty)$, it is expanding. The coordinate frame is not static. The relations between $(T,R)$ and $(\eta,\chi)$ in static and non-static regions mirros those in Schwarzshild black hole solution between the static and the global (Kruskal-Szekeres) coordinates. Here $\eta$ and $\chi$ are the global coordinates.
</p>
</div>
</div>

<div id="Horizon43"></div>

=== Problem 4 ===
'''Flat dS.''' The scale factor in flat de Sitter is $a(t)=H_\Lambda^{-1} e^{H_\Lambda t}$.

* Find the range of values spanned by conformal time $\tilde{\eta}$ and comoving distance $\tilde{\chi}$ in the flat de Sitter space
* Verify that coordinate transformation
\begin{equation}
\tilde{\eta}=\frac{-\sin\eta}{\cos\chi-\cos\eta},\qquad
\tilde{\chi}=\frac{\sin\chi}{\cos\chi-\cos\eta}
\end{equation}
bring the metric to the form of that of de Sitter in closed slicing (it is assumed that $\tilde{\eta}=0$ is chosen to correspond to infinite future).
* Which part of the conformal diagram is covered by the coordinate chart $(\tilde{\eta},\tilde{\chi})$? Is the flat de Sitter space geodesically complete?
* Where are the particle and event horizons in these coordinates?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As $t\in(-\infty,+\infty)$,
\begin{equation}
\tilde{\eta} =\int\frac{dt}{a(t)}=H_\Lambda \int\limits_{+\infty}^{t}dt\; e^{-H_\Lambda t} =-e^{-H_\Lambda t} \in (-\infty,0).
\end{equation}
Here we choose $+\infty$ as the lower limit, because at $-\infty$ the integral diverges.
<div style="text-align: left;>
* Direct calculation yields (\ref{dSclosed}), with $\eta \in (0,\pi)$, $\chi\in (0,\pi)$;
* The upper triangle $\eta>\chi$, above the particle horizon, on which $\tilde{\eta}\to -\infty$. It is not geodesically complete, as geodesics are cut at the particle horizon.
* The particle horizon is the boundary of the patch of full dS space covered by flat slicing coordinates, and event horizon in these coordinates exists only in the latter part of evolution, for $\eta >\pi/2$.
</div>
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-flat.png|center|thumb|365px|The de Sitter Universe in flat sections' coordinates, which cover only half of it. The boundary -- the particle horizon -- consists of three different infinities.]]
|}
</p>
</div>
</div>

<div id="Horizon44"></div>
=== Problem 5 ===
'''Infinities.''' What parts of the spacetime's boundary on the conformal diagram of flat de Sitter space corresponds to

* spacelike infinity $i^0$, where $\tilde{\chi}\to +\infty$;
* past timelike infinity $i^-$, where $\tilde{\eta}\to -\infty$ and from which all timelike worldlines emanate
* past lightlike infinity $J^-$, from which all null geodesics emanate?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* The one point at the left of the diagram;
* the one point at the bottom;
* the particle horizon.
</div>
</p>
</div>
</div>

<div id="Horizon45"></div>
=== Problem 6 ===
'''Open dS.''' Consider the de Sitter space in open slicing, in which $a(t)=H_\Lambda \sinh (H_\Lambda t)$, so conformal time is
\begin{equation}
\tilde{\eta} =\int\limits_{+\infty}^{t} \frac{dt}{a(t)},
\end{equation}
where again the lower limit is chosen so that the integral is bounded.

* Find $\tilde{\eta}(t)$ and verify that coordinate transformation from $(\tilde{\eta},\tilde{\chi})$ to $\eta,\chi$, such that
\begin{equation}
\tanh\tilde{\eta}=\frac{-\sin\eta}{\cos\cos\chi},\qquad
\tanh\tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
brings the metric to the form of de Sitter in closed slicing.
* What are the ranges spanned by $(\tilde{\eta},\tilde{\chi})$ and $(\eta,\chi)$? Which part of the conformal diagram do they cover?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
After getting
\begin{equation}
\sinh \tilde{\eta}=-\frac{1}{\sinh(H_\Lambda t)},
\end{equation}
the first part is checked straightforwardly; in the open de Sitter $\tilde{\chi}\in [0,+\infty)$, and $\tilde{\eta}\in(-\infty,0)$. The region covered by coordinates $(\tilde{\eta},\tilde{\chi})$ is $\{\eta>\chi+\pi/2\}$, only one eighth part of the full diagram.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-open.png|center|thumb|365px|The de Sitter Universe in open sections' coordinates, which cover only $1/8^{\text{th}}$ of the full diagram.]]
|}
</p>
</div>
</div>

<div id="Horizon46"></div>
=== Problem 7 ===
'''Minkowski 1.''' Rewrite the Minkowski metric in terms of coordinates $(\eta,\chi)$, which are related to $(t,r)$ by the relation
\begin{equation}
\tanh \tilde{\eta}=\frac{\sin\eta}{\cos\chi},\qquad
\tanh \tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
that mirrors the one between the open and closed coordinates of de Sitter. Construct the conformal diagram and determine different types of infinities. Are there new ones compared to the flat de Sitter space?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Coordinate transformation gives
\begin{equation}
ds^2 =\frac{1}{\cos^2 \chi -\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\Psi^{2}(\eta,\chi)d\Omega^2 \big].
\end{equation}
Here $r\in [0,+\infty)$ and $t\in (-\infty,+\infty)$. Comparing with the relation between $(\tilde{\eta},\tilde{\chi})$ with conformal coordinates $(\eta,\chi)$ in the open de Sitter universe, where $\tilde{\eta}\in (-\infty,0)$, we see that the difference is that $\tilde{\eta}$ spans $(-\infty,0)$, while now $t$ spans twice the range, $(-\infty,\infty)$. Therefore the conformal diagram is composed of two triangles, one the same as for open de Sitter and one for its time-reversed copy. Accordingly there now appear future timelike infinity $i^+$ and future lightlike infinity $J^+$.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |The Minkowski spacetime in two different pairs of conformal coordinates and the full set of infinities. Thin dashed and solid lines show the images of coordinate grid $(t,r)$. The one on the right corresponds to another choice of conformal coordinates, discussed in the next problem,

|[[File:Conf-Mink1.png|center|thumb|365px]]
|[[File:Conf-Mink2.png|center|thumb|365px]]
|}
</p>
</div>
</div>

<div id="Horizon47"></div>

=== Problem 8 ===
'''Minkowski 2.''' The choice of conformal coordinates is not unique. Construct the conformal diagram for Minkowski using the universal scheme: first pass to null coordinates, then bring their span to finite intervals with $\arctan$ (one of the possible choices), then pass again to timelike and spacelike coordinates.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
We start from spherical coordinates
\begin{equation}
ds^2 =dt^2 -dr^2 -r^2 d\Omega^2 .
\end{equation}
<div style="text-align: left;">
* The first step is introducing null coordinates
\begin{equation}
u=t-r,\quad v=t+r ,
\end{equation}
so that
\begin{equation}
ds^2 =4du\,dv -\frac{(v-u)^2}{4}d\Omega^2 .
\end{equation}
* Then bring the range of values to finite intervals
\begin{equation}
U=\arctan u, \qquad V=\arctan v,
\end{equation}
so that
\begin{equation}
ds^2 =\frac{1}{4\cos^2 U \cos^2 V}\big[4dU\,dV -\sin^2 (V-U)d\Omega^2\big].
\end{equation}
The whole spacetime is simply the half of the square $U,V\in (-\pi/2,\pi/2)$, in which $r>0$, i.e. $v>u\quad \Leftrightarrow\quad V>U$: on the plane $(U,V)$ it is the triangle \begin{equation}
-\pi/2<U<V<\pi/2.
\end{equation}
* Finally, go back to spacelike and timelike coordinates
\begin{equation}
T=V+U,\qquad R=V-U
\end{equation}
so that metric becomes
\begin{equation}
ds^2 =\frac{1}{[\cos T +\cos R]^2}\big[dT^2 -dR^2 -\sin^2 R\; d\Omega^2\big].
\end{equation}
The triangle is shrunken by $\sqrt{2}$ and rotated by $3\pi/4$ clockwise, thus turning into
\begin{equation}
\{R>0,\quad |T|<\pi/2 -R\}.
\end{equation}
This is the same form as obtained by the other construction (up to scaling, which is purely decorative).
</div>
</p>
</div>
</div>

<div id="Horizon48"></div>
=== Problem 9 ===
Draw the conformal diagram for the Milne Universe and show which part of Minkowski space's diagram it covers.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Minkowski metric $ds^2 =dT^2 -dR^2$, rewritten in terms of $(\tau, r)$ such that
\begin{equation}
T=\tau \cosh r ,\quad R=\tau \sinh r ,
\end{equation}
is the metric of the Milne Universe. As Minkowski space is complete, the Milne Universe then is a \emph{part} of Minkowski in different variables. This part is where $T>R$. The boundary $T=R$ is the future light cone, so the whole Milne Universe is represented by the triangle homothetic to the whole space but 4 times smaller in area, with the common node of future infinity.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Milne.png|center|thumb|365px|The Milne Universe on Minkowski's conformal diagram. It has the same shape and shares the same future infinity, but covers one eighth of the area. The boundary is the past horizon.]]
|}
</p>
</div>
</div>

<div id="Horizon49"></div>
=== Problem 10 ===
Consider open or flat Universe filled with matter that satisfies strong energy condition $\varepsilon +3p>0$. What are the coordinate ranges spanned by the comoving coordinate $\tilde{\chi}$ and conformal time $\tilde{\eta}$? Compare with the Minkowski metric and construct the diagram. Identify the types of infinities and the initial Big Bang singularity.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For flat Universe $\tilde{\chi}=r$, for the open $\tilde{\chi}=\sinh r$, so in both cases $\tilde{\chi}\in (0,+\infty)$.

Strong energy condition implies that $w>-1/3$, so
\begin{equation}
\rho \sim a^{-3(1+w)}=a^{-n},
\end{equation}
where $n>2$. From the first Friedman equation then after simple manipulations we obtain that
\begin{equation}
\frac{da}{dt}\sim a^{-\theta},
\end{equation}
where $\theta$ is some positive number. Therefore both
\begin{equation}
t\sim \int da\; a^{\theta},
\quad\text{and}\quad
\eta =\int \frac{da}{a}a^{\theta}
\end{equation}
converge at $a\to 0$ and diverge at $a\to \infty$. Consequently, the integration constant can be chosen so that $\eta\in (0,+\infty)$.

The conformal structure is the same as that of the \emph{upper} half of Minkowski spacetime. The Big Bang singularity at $\eta=0$ is at the cut, and there are spacelike infinity, future infinity and future null infinity.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-SEC.png|center|thumb|365px|Conformal diagram for an open or flat Universe filled with matter which satisfies the strong energy condition.]]
|}

</p>
</div>
</div>

<div id="Horizon50"></div>
=== Problem 11 ===
Draw the conformal diagram for open and flat Universes with power-law scale factor $a(t)\sim t^{n}$, with $n>1$. This is the model for the power-law inflation. Check whether the strong energy condition is satisfied.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As seen in the previous problem, if strong energy condition were satisfied, we would have $\dot{a}\sim a^{-\theta}$ with some positive $\theta$; this is not the case, so the condition is violated. As $t\in (0,+\infty)$,
\begin{equation}
\eta\sim \int \frac{dt}{t^n}
\end{equation}
diverges at small $a$ (thus also small $t$) and converges at $a\to \infty$ (thus as large $t$). So integration constant can be chosen so that $\eta\in (-\infty,0)$.

The conformal structure is the same as \emph{lower half} of Minkowski spacetime. The cut is regular future infinity, and from Minkowski there are spacelike infinity, past null infinity, and past infinity. The point of past infinity corresponds to Big Bang and is singular.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Inflation.png|center|thumb|365px|Conformal diagram for an open or flat Universe filled with matter which violates the strong energy condition (SEC), such as in the case of power-law inflation.]]
|}
</p>
</div>
</div>

<references>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN~0521563984</ref>
</references>

File:Conf-Inflation.png

2013-12-08T17:57:26Z

Igor: Conformal diagram for the power-law Inflation in an open Universe.

Conformal diagram for the power-law Inflation in an open Universe.

File:Conf-SEC.png

2013-12-08T17:55:40Z

Igor: Conformal diagram for an open Universe filled with matter satisfying the strong energy condition.

Conformal diagram for an open Universe filled with matter satisfying the strong energy condition.

File:Conf-Milne.png

2013-12-08T17:54:39Z

Igor: Milne Universe as part of Minkowski spacetime: conformal diagram

Milne Universe as part of Minkowski spacetime: conformal diagram

Causal Structure

2013-12-08T17:53:08Z

Igor: /* Problem 7 */

[[Category:Horizons|4]]

__TOC__

The causal structure is determined by propagation of light and is best understood in terms of ''conformal diagrams''. In this section we construct and analyze those for a number of important model cosmological solutions (which are assumed to be already known), following mostly the exposition of <ref name=Mukhanov/>.

In terms of comoving distance $\tilde{\chi}$ and conformal time $\tilde{\eta}$ (in this section they are denoted by tildes) the two-dimensional radial part of the FLRW metric takes form
\begin{equation}
ds_{2}^{2}=a^{2}(\tilde{\eta})\big[d\tilde{\eta}^2 -d\tilde{\chi}^2\big].
\label{ds2Dconf}
\end{equation}
In the brackets here stands the line element of two-dimensional Minkowski flat spacetime. Coordinate transformations that preserve the \emph{conformal} form of the metric
\begin{equation*}
ds_2^2 =\Omega^{2}(\eta,\chi)\big[d\eta^2 -d\chi^2 \big],
\end{equation*}
are called conformal transformations, and the corresponding coordinates $(\eta,\chi)$ -- conformal coordinates.

<div id="Horizon39"></div>
=== Problem 1 ===
Show that it is always possible to construct $\eta(\tilde{\eta},\tilde{\chi})$, $\chi(\tilde{\eta},\tilde{\chi})$, such that the conformal form of metric (\ref{ds2Dconf}) is preserved, but $\eta$ and $\chi$ are bounded and take values in some finite intervals. Is the choice of $(\eta,\chi)$ unique?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Suppose $\tilde{\eta},\tilde{\chi}$ span infinite or semi-infinite values. Then we can always make the following sequence of coordinate transformations:
<div style="text-align: left;">
* Pass to null coordinates
\begin{equation}
u=\tilde{\eta}-\tilde{\chi},\qquad v=\tilde{\eta}+\tilde{\chi};
\end{equation}
* Bring their range of values to a finite interval by some appropriate function, i.e.
\begin{equation}
U=\arctan u,\qquad V=\arctan v.
\end{equation}
* Go back to timelike and spacelike coordinates (this is not really necessary at this point and is done mostly for aesthetic reasons):
\begin{equation}
T=V+U,\qquad R=V-U.
\end{equation}
Now the range of $(T,R)$ obviously covers some bounded region on the plane, while the radial part of the line element preserves its conformal form:
\begin{equation}
ds_2^2 \sim d\tilde{\eta}^2-d\tilde{\chi}^2 \sim du\, dv \sim dU\, dV \sim dT^2 -dR^2 .
\end{equation}
As the choice of function $\arctan$ was rather arbitrary (though convenient), the choice of conformal coordinates is not unique.
</div>
</p>
</div>
</div>

In this section we will reserve notation $\eta$ and $\chi$ and name "conformal coordinates/variables" to such variables that can only take values in a bounded region on $\mathbb{R}^2$; $\tilde{\eta}$ and $\tilde{\chi}$ can span infinite or semi-infinite intervals. Spacetime diagram in terms of conformal variables $(\eta,\chi)$ is called conformal diagram. Null geodesics $\eta=\pm \chi + const$ are diagonal straight lines on conformal diagrams.

<div id="Horizon40"></div>
=== Problem 2 ===
Construct the conformal diagram for the closed Universe filled with
* radiation;
* dust;
* mixture of dust and radiation.

Show the particle and event horizons for the observer at the origin $\chi=0$ (it will be assumed hereafter that the horizons are always constructed with respect to this chosen observer).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* Solution for $a(\eta)$ in a radiation dominated closed Universe is
\begin{equation}
a=a_m \sin\eta ,\qquad \eta\in(0,\pi),\quad \chi \in [0,\pi].
\end{equation}
As the ranges spanned by $\eta$ and $\chi$ are finite, they are already conformal coordinates. The conformal diagram is a square $\eta,\chi\in [0,\pi]$. Edges $\eta=0$ and $\eta=\pi$ correspond to the Big Bang and Big Crunch singularities respectively; worldline $\chi=\pi$ is the point on the three-sphere that is situated at the opposite pole with respect to observer at $\chi=0$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Rad.png|center|thumb|365px|Conformal diagram for the radiation dominated closed Universe. Particle horizon is shown in blue, event horizon in red.]]
|}

The particle horizon is
\[\eta=\chi,\]
and event horizon is
\[\eta=\eta_{max}-\chi =\pi -\chi.\]
Both exist for all cosmological times: by the finite moment of the Big Crunch $\eta=\pi$ the event horizon is collapsed into a point (which is natural, as there is no more time left), while the particle horizon extends to the whole Universe. Thus only at the finite moment the whole of the Universe becomes observable. The farther the point, though, the younger will it look, of course, and the opposite pole will only be "observed" by our observer at the last moment of the Universe as it was at its creation.

* Solution for $a(\eta)$ in a dust dominated closed Universe is
\begin{equation}
a=a_m (1-\cos\eta) ,\qquad \eta\in(0,2\pi),\quad \chi \in [0,\pi].
\end{equation}
The difference from the previous case is that $\eta$ spans twice the range, and $\eta_{max}=2\pi =2\chi_{max}$.

Therefore the event horizon is given by
\[\eta=\eta_{max}-\chi=2\pi -\chi,\]
so it exists only in the second, contracting, phase $\eta>\pi$. The particle horizon is given by $\eta=\chi$ again, but now it exists only during the expanding phase $\eta<\pi$. It encloses the full Universe at the moment of maximal expansion $\eta=\pi$ and for later times does not exist.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Dust.png|center|thumb|365px|Conformal diagrams of the dust dominated closed Universe. Thin lines show light rays that realize the first and second images of a galaxy at different cosmic times, from opposite directions.]]
|}

* Though the full analytic solution is more complicated, it is clear that the main features remain the same as in the previous considered cases. At early and late times, close to the singularities, the dynamics is determined by the radiation component. If there is enough dust, then at large enough scale factors (which may or may not be achieved depending on the initial conditions), which would correspond to the epoch around the maximal expansion, it will be dominating. The influence of dust is that dynamics is slowed down, so that depending on the ratio of densities
\[\eta_{max}\in [\pi,2\pi].\]
Thus qualitatively the picture will be the same as in a dust dominated Universe: the conformal diagram is a rectangle, the event horizon exists only starting from some time $\eta_{e}=\eta_{max}-\pi$, while particle horizon, on the contrary, vanishes at $\eta_{p}=\pi$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-RadDust.png|center|thumb|365px|Conformal diagram for a closed Universe filled with a mix of dust and radiation.]]
|}
</div>
</p>
</div>
</div>

<div id="Horizon41"></div>

=== Problem 3 ===
'''Closed dS.''' Construct the conformal diagram for the de Sitter space in the closed sections coordinates. Provide reasoning that this space is (null) geodesically complete, i.e. every (null) geodesic extends to infinite values of affine parameters at both ends.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The scale factor in the closed dS Universe is
\begin{equation}
a(t)=H_\Lambda^{-1}\cosh (H_\Lambda t),\qquad t\in (-\infty, +\infty),
\end{equation}
so on integration, for conformal time we obtain
\begin{equation}
\eta (t)=\int\limits_{-\infty}^{t}\frac{dt}{a(t)}=\arcsin\big[\tanh (H_\Lambda t)\big]+\frac{\pi}{2} \in (0,\pi).
\end{equation}
We choose the integration constant here so that $\eta=0$ corresponds to $t=-\infty$ and $\eta=\pi$ to $t=+\infty$. The full metric then can be written as
\begin{equation}
ds^{2}_{dS}=\frac{H_\Lambda^2}{\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\sin^2 \chi d\Omega^2 \big]. \label{dSclosed}
\end{equation}

As the values of $\eta$ span a finite interval, $(\eta,\chi)$ are already conformal coordinates. The conformal diagram is again a square
\begin{equation}
\eta\in [0,\pi],\qquad \chi \in[0,\pi],
\end{equation}
with the difference from the radiation dominated Universe that the edges $\eta=0,\pi$ do not represent a singularity anymore, but instead correspond to infinite (and regular) past and future respectively. Both horizons are given again by
\begin{equation}
\eta_{e}=\pi-\chi,\qquad \eta_{p}=\chi
\end{equation}
and exist at all times.

The spacelike boundaries of the conformal diagram correspond to $t\to \pm\infty$, and therefore to infinite values of affine parameter. This can be shown if one remembers the general formula for the cosmological redshift:
\begin{equation}
\text{const}=\omega a =\frac{dt}{d\lambda}a,\quad \Rightarrow\quad
\lambda=\text{const}\cdot\int\limits^{t}dt \cosh(H_\Lambda t)
\underset{t\to\pm\infty}{\longrightarrow}\infty .
\end{equation}
The timelike boundary of the diagram corresponds to the opposite pole, there is no real boundary there, the same as on a sphere: as particles propagate across the pole, their radial coordinate begins to decrease again, while the worldline on the conformal diagram is reflected from $\chi=\pi$. Thus by definition the spacetime is (null) geodesically complete.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-closed.png|center|thumb|365px|The de Sitter Universe. The closed sections coordinates cover the whole space, which is geodesically complete. There are no singularities: the horizontal boundaries of the diagram correspond to infinite past and future.]]
|}
</p>
</div>
</div>

<div id="Horizon42"></div>
=== Problem 4 ===
'''Static dS.''' Rewrite the metric of de Sitter space (\ref{dSclosed}) in terms of "static coordinates" $T,R$:
\begin{equation}
\tanh (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad
H_\Lambda R =\frac{\sin\chi}{\sin\eta}.
\end{equation}
* What part of the conformal diagram in terms of $(\eta,\chi)$ is covered by the static coordinate chart $(T,R)$?
* Express the horizon's equations in terms of $T$ and $R$
* Draw the surfaces of constant $T$ and $R$ on the conformal diagram.
* Write out the coordinate transformation between $(\eta,\chi)$ and $(T,R)$ in the regions where $|\cos\eta|>|\cos\chi|$. Explain the meaning of $T$ and $R$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us introduce dimensionless coordinates $t=H_\Lambda T$ and $r=H_\Lambda R$. The inverse relations then are
\begin{equation}
\sin^2 \eta =\frac{1}{\cosh^2 t-r^2 \sinh^2 t},
\quad \sin^2 \chi =\frac{r^2}{\cosh^2 t-r^2 \sinh^2 t}.
\end{equation}
Using them, after some algebra from (\ref{dSclosed}) we get
\begin{equation}
ds_{dS}^2 =\big[1-H_\Lambda^2 R^2\big]dT^2 -\frac{dR^2}{1-H_\Lambda^2 R^2}-R^2 d\Omega^2 , \label{dSstatic}
\end{equation}
which resembles the Schwarzschild line element (and this is not a coincidence).
</p>

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-static.png|center|thumb|365px|The de Sitter Universe with contour lines of the "static" coordinates $(T,R)$. The solid lines are $T=const$, and the dashed ones $R=const$. The coordinates become singular on both horizons, so the whole spacetime is divided by them into four sectors, each separately covered by the regular coordinate chart $(R,T)$. Spacetime is actually static only in the left and right sectors (I and III), where $T$ is a timelike coordinate and $R$ spacelike: the static patches are bounded by the horizons.]]
|}

<p style="text-align: left;">
* As $|\tanh (H_\Lambda T)|<1$, we have $|\cos\eta| \leq |\cos\chi|$. This condition cuts out two out of four sectors from the square conformal diagram, I and III: one is $\eta\in [\chi, \pi-\chi]$ and the other is $\eta\in [\pi-\chi, \chi]$. In both $|\sin\eta|\geq |\sin\chi|$, so $R \leq H_{\Lambda}^{-1}$;
* The particle horizon $\eta=\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=-\infty$. It is one part of the boundary of the region (in two parts) covered by coordinates $(T,R)$. The event horizon $\eta=\pi-\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=+\infty$ and is the other part of the boundary of this region.
* $T=const$ is $\cos\eta =t\cos\chi $ and $R=const$ is $\sin\eta =r^{-1}\sin\chi$.
* In regions II and IV the needed relation is obtained if we simply replace $\tanh$ with $\coth$ in the first relation:
\begin{equation}
\coth (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad H_\Lambda R=\frac{\sin\chi}{\sin\eta},
\end{equation}
as can be checked explicitly by substitution into (\ref{dSstatic}), which again gives (\ref{dSclosed}). In these regions $R$ is a timelike coordinate, and $T$ is spacelike. The geodesics of comoving massive particles are $\chi=const$, one of them $\chi=\pi/2$ corresponds to $T=0$. Thus in the lower part of the diagram, where $R\in (+\infty,H_\Lambda^{-1})$, the spacetime is contracting; in the upper part, where $R\in (H_\Lambda^{-1},\infty)$, it is expanding. The coordinate frame is not static. The relations between $(T,R)$ and $(\eta,\chi)$ in static and non-static regions mirros those in Schwarzshild black hole solution between the static and the global (Kruskal-Szekeres) coordinates. Here $\eta$ and $\chi$ are the global coordinates.
</p>
</div>
</div>

<div id="Horizon43"></div>

=== Problem 4 ===
'''Flat dS.''' The scale factor in flat de Sitter is $a(t)=H_\Lambda^{-1} e^{H_\Lambda t}$.

* Find the range of values spanned by conformal time $\tilde{\eta}$ and comoving distance $\tilde{\chi}$ in the flat de Sitter space
* Verify that coordinate transformation
\begin{equation}
\tilde{\eta}=\frac{-\sin\eta}{\cos\chi-\cos\eta},\qquad
\tilde{\chi}=\frac{\sin\chi}{\cos\chi-\cos\eta}
\end{equation}
bring the metric to the form of that of de Sitter in closed slicing (it is assumed that $\tilde{\eta}=0$ is chosen to correspond to infinite future).
* Which part of the conformal diagram is covered by the coordinate chart $(\tilde{\eta},\tilde{\chi})$? Is the flat de Sitter space geodesically complete?
* Where are the particle and event horizons in these coordinates?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As $t\in(-\infty,+\infty)$,
\begin{equation}
\tilde{\eta} =\int\frac{dt}{a(t)}=H_\Lambda \int\limits_{+\infty}^{t}dt\; e^{-H_\Lambda t} =-e^{-H_\Lambda t} \in (-\infty,0).
\end{equation}
Here we choose $+\infty$ as the lower limit, because at $-\infty$ the integral diverges.
<div style="text-align: left;>
* Direct calculation yields (\ref{dSclosed}), with $\eta \in (0,\pi)$, $\chi\in (0,\pi)$;
* The upper triangle $\eta>\chi$, above the particle horizon, on which $\tilde{\eta}\to -\infty$. It is not geodesically complete, as geodesics are cut at the particle horizon.
* The particle horizon is the boundary of the patch of full dS space covered by flat slicing coordinates, and event horizon in these coordinates exists only in the latter part of evolution, for $\eta >\pi/2$.
</div>
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-flat.png|center|thumb|365px|The de Sitter Universe in flat sections' coordinates, which cover only half of it. The boundary -- the particle horizon -- consists of three different infinities.]]
|}
</p>
</div>
</div>

<div id="Horizon44"></div>
=== Problem 5 ===
'''Infinities.''' What parts of the spacetime's boundary on the conformal diagram of flat de Sitter space corresponds to

* spacelike infinity $i^0$, where $\tilde{\chi}\to +\infty$;
* past timelike infinity $i^-$, where $\tilde{\eta}\to -\infty$ and from which all timelike worldlines emanate
* past lightlike infinity $J^-$, from which all null geodesics emanate?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* The one point at the left of the diagram;
* the one point at the bottom;
* the particle horizon.
</div>
</p>
</div>
</div>

<div id="Horizon45"></div>
=== Problem 6 ===
'''Open dS.''' Consider the de Sitter space in open slicing, in which $a(t)=H_\Lambda \sinh (H_\Lambda t)$, so conformal time is
\begin{equation}
\tilde{\eta} =\int\limits_{+\infty}^{t} \frac{dt}{a(t)},
\end{equation}
where again the lower limit is chosen so that the integral is bounded.

* Find $\tilde{\eta}(t)$ and verify that coordinate transformation from $(\tilde{\eta},\tilde{\chi})$ to $\eta,\chi$, such that
\begin{equation}
\tanh\tilde{\eta}=\frac{-\sin\eta}{\cos\cos\chi},\qquad
\tanh\tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
brings the metric to the form of de Sitter in closed slicing.
* What are the ranges spanned by $(\tilde{\eta},\tilde{\chi})$ and $(\eta,\chi)$? Which part of the conformal diagram do they cover?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
After getting
\begin{equation}
\sinh \tilde{\eta}=-\frac{1}{\sinh(H_\Lambda t)},
\end{equation}
the first part is checked straightforwardly; in the open de Sitter $\tilde{\chi}\in [0,+\infty)$, and $\tilde{\eta}\in(-\infty,0)$. The region covered by coordinates $(\tilde{\eta},\tilde{\chi})$ is $\{\eta>\chi+\pi/2\}$, only one eighth part of the full diagram.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-open.png|center|thumb|365px|The de Sitter Universe in open sections' coordinates, which cover only $1/8^{\text{th}}$ of the full diagram.]]
|}
</p>
</div>
</div>

<div id="Horizon46"></div>
=== Problem 7 ===
'''Minkowski 1.''' Rewrite the Minkowski metric in terms of coordinates $(\eta,\chi)$, which are related to $(t,r)$ by the relation
\begin{equation}
\tanh \tilde{\eta}=\frac{\sin\eta}{\cos\chi},\qquad
\tanh \tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
that mirrors the one between the open and closed coordinates of de Sitter. Construct the conformal diagram and determine different types of infinities. Are there new ones compared to the flat de Sitter space?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Coordinate transformation gives
\begin{equation}
ds^2 =\frac{1}{\cos^2 \chi -\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\Psi^{2}(\eta,\chi)d\Omega^2 \big].
\end{equation}
Here $r\in [0,+\infty)$ and $t\in (-\infty,+\infty)$. Comparing with the relation between $(\tilde{\eta},\tilde{\chi})$ with conformal coordinates $(\eta,\chi)$ in the open de Sitter universe, where $\tilde{\eta}\in (-\infty,0)$, we see that the difference is that $\tilde{\eta}$ spans $(-\infty,0)$, while now $t$ spans twice the range, $(-\infty,\infty)$. Therefore the conformal diagram is composed of two triangles, one the same as for open de Sitter and one for its time-reversed copy. Accordingly there now appear future timelike infinity $i^+$ and future lightlike infinity $J^+$.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |The Minkowski spacetime in two different pairs of conformal coordinates and the full set of infinities. Thin dashed and solid lines show the images of coordinate grid $(t,r)$. The one on the right corresponds to another choice of conformal coordinates, discussed in the next problem,

|[[File:Conf-Mink1.png|center|thumb|365px]]
|[[File:Conf-Mink2.png|center|thumb|365px]]
|}
</p>
</div>
</div>

<div id="Horizon47"></div>

=== Problem 8 ===
'''Minkowski 2.''' The choice of conformal coordinates is not unique. Construct the conformal diagram for Minkowski using the universal scheme: first pass to null coordinates, then bring their span to finite intervals with $\arctan$ (one of the possible choices), then pass again to timelike and spacelike coordinates.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
We start from spherical coordinates
\begin{equation}
ds^2 =dt^2 -dr^2 -r^2 d\Omega^2 .
\end{equation}
<div style="text-align: left;">
* The first step is introducing null coordinates
\begin{equation}
u=t-r,\quad v=t+r ,
\end{equation}
so that
\begin{equation}
ds^2 =4du\,dv -\frac{(v-u)^2}{4}d\Omega^2 .
\end{equation}
* Then bring the range of values to finite intervals
\begin{equation}
U=\arctan u, \qquad V=\arctan v,
\end{equation}
so that
\begin{equation}
ds^2 =\frac{1}{4\cos^2 U \cos^2 V}\big[4dU\,dV -\sin^2 (V-U)d\Omega^2\big].
\end{equation}
The whole spacetime is simply the half of the square $U,V\in (-\pi/2,\pi/2)$, in which $r>0$, i.e. $v>u\quad \Leftrightarrow\quad V>U$: on the plane $(U,V)$ it is the triangle \begin{equation}
-\pi/2<U<V<\pi/2.
\end{equation}
* Finally, go back to spacelike and timelike coordinates
\begin{equation}
T=V+U,\qquad R=V-U
\end{equation}
so that metric becomes
\begin{equation}
ds^2 =\frac{1}{[\cos T +\cos R]^2}\big[dT^2 -dR^2 -\sin^2 R\; d\Omega^2\big].
\end{equation}
The triangle is shrunken by $\sqrt{2}$ and rotated by $3\pi/4$ clockwise, thus turning into
\begin{equation}
\{R>0,\quad |T|<\pi/2 -R\}.
\end{equation}
This is the same form as obtained by the other construction (up to scaling, which is purely decorative).
</div>
</p>
</div>
</div>

<div id="Horizon48"></div>
=== Problem 9 ===
Draw the conformal diagram for the Milne Universe and show which part of Minkowski space's diagram it covers.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Minkowski metric $ds^2 =dT^2 -dR^2$, rewritten in terms of $(\tau, r)$ such that
\begin{equation}
T=\tau \cosh r ,\quad R=\tau \sinh r ,
\end{equation}
is the metric of the Milne Universe. As Minkowski space is complete, the Milne Universe then is a \emph{part} of Minkowski in different variables. This part is where $T>R$. The boundary $T=R$ is the future light cone, so the whole Milne Universe is represented by the triangle homothetic to the whole space but 4 times smaller in area, with the common node of future infinity.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Milne.png|center|thumb|365px|The Milne Universe on Minkowski's conformal diagram. It has the same shape and shares the same future infinity, but covers one eighth of the area. The boundary is the past horizon.]]
|}
</p>
</div>
</div>

<div id="Horizon49"></div>
=== Problem 10 ===
Consider open or flat Universe filled with matter that satisfies strong energy condition $\varepsilon +3p>0$. What are the coordinate ranges spanned by the comoving coordinate $\tilde{\chi}$ and conformal time $\tilde{\eta}$? Compare with the Minkowski metric and construct the diagram. Identify the types of infinities and the initial Big Bang singularity.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For flat Universe $\tilde{\chi}=r$, for the open $\tilde{\chi}=\sinh r$, so in both cases $\tilde{\chi}\in (0,+\infty)$.

Strong energy condition implies that $w>-1/3$, so
\begin{equation}
\rho \sim a^{-3(1+w)}=a^{-n},
\end{equation}
where $n>2$. From the first Friedman equation then after simple manipulations we obtain that
\begin{equation}
\frac{da}{dt}\sim a^{-\theta},
\end{equation}
where $\theta$ is some positive number. Therefore both
\begin{equation}
t\sim \int da\; a^{\theta},
\quad\text{and}\quad
\eta =\int \frac{da}{a}a^{\theta}
\end{equation}
converge at $a\to 0$ and diverge at $a\to \infty$. Consequently, the integration constant can be chosen so that $\eta\in (0,+\infty)$.

The conformal structure is the same as that of the \emph{upper} half of Minkowski spacetime. The Big Bang singularity at $\eta=0$ is at the cut, and there are spacelike infinity, future infinity and future null infinity.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |Conformal diagrams for open or flat Universes. The one on the left is for ones filled with matter which satisfies the strong energy condition (SEC), and the one on the right for ones in which SEC does not hold, such as in case of power-law inflation..

|[[File:Conf-SEC.png|center|thumb|365px]]
|[[File:Conf-Inflation.png|center|thumb|365px]]
|}

</p>
</div>
</div>

<div id="Horizon50"></div>
=== Problem 11 ===
Draw the conformal diagram for open and flat Universes with power-law scale factor $a(t)\sim t^{n}$, with $n>1$. This is the model for the power-law inflation. Check whether the strong energy condition is satisfied.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As seen in the previous problem, if strong energy condition were satisfied, we would have $\dot{a}\sim a^{-\theta}$ with some positive $\theta$; this is not the case, so the condition is violated. As $t\in (0,+\infty)$,
\begin{equation}
\eta\sim \int \frac{dt}{t^n}
\end{equation}
diverges at small $a$ (thus also small $t$) and converges at $a\to \infty$ (thus as large $t$). So integration constant can be chosen so that $\eta\in (-\infty,0)$.

The conformal structure is the same as \emph{lower half} of Minkowski spacetime. The cut is regular future infinity, and from Minkowski there are spacelike infinity, past null infinity, and past infinity. The point of past infinity corresponds to Big Bang and is singular.
</p>
</div>
</div>

<references>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN~0521563984</ref>
</references>

File:Conf-Mink2.png

2013-12-08T17:50:04Z

Igor: Minkowski spacetime: conformal diagram (different choice of conformal variables)

Minkowski spacetime: conformal diagram (different choice of conformal variables)

File:Conf-Mink1.png

2013-12-08T17:48:59Z

Igor: Minkowski spacetime: conformal diagram

Minkowski spacetime: conformal diagram

File:Conf-dS-open.png

2013-12-08T17:48:02Z

Igor: The de Sitter Universe in open slicing: conformal diagram

The de Sitter Universe in open slicing: conformal diagram

File:Conf-dS-flat.png

2013-12-08T17:47:15Z

Igor: The de Sitter Universe in flat slicing: conformal diagram

The de Sitter Universe in flat slicing: conformal diagram

Causal Structure

2013-12-08T17:45:22Z

Igor: /* Problem 4 */ alignment?

[[Category:Horizons|4]]

__TOC__

The causal structure is determined by propagation of light and is best understood in terms of ''conformal diagrams''. In this section we construct and analyze those for a number of important model cosmological solutions (which are assumed to be already known), following mostly the exposition of <ref name=Mukhanov/>.

In terms of comoving distance $\tilde{\chi}$ and conformal time $\tilde{\eta}$ (in this section they are denoted by tildes) the two-dimensional radial part of the FLRW metric takes form
\begin{equation}
ds_{2}^{2}=a^{2}(\tilde{\eta})\big[d\tilde{\eta}^2 -d\tilde{\chi}^2\big].
\label{ds2Dconf}
\end{equation}
In the brackets here stands the line element of two-dimensional Minkowski flat spacetime. Coordinate transformations that preserve the \emph{conformal} form of the metric
\begin{equation*}
ds_2^2 =\Omega^{2}(\eta,\chi)\big[d\eta^2 -d\chi^2 \big],
\end{equation*}
are called conformal transformations, and the corresponding coordinates $(\eta,\chi)$ -- conformal coordinates.

<div id="Horizon39"></div>
=== Problem 1 ===
Show that it is always possible to construct $\eta(\tilde{\eta},\tilde{\chi})$, $\chi(\tilde{\eta},\tilde{\chi})$, such that the conformal form of metric (\ref{ds2Dconf}) is preserved, but $\eta$ and $\chi$ are bounded and take values in some finite intervals. Is the choice of $(\eta,\chi)$ unique?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Suppose $\tilde{\eta},\tilde{\chi}$ span infinite or semi-infinite values. Then we can always make the following sequence of coordinate transformations:
<div style="text-align: left;">
* Pass to null coordinates
\begin{equation}
u=\tilde{\eta}-\tilde{\chi},\qquad v=\tilde{\eta}+\tilde{\chi};
\end{equation}
* Bring their range of values to a finite interval by some appropriate function, i.e.
\begin{equation}
U=\arctan u,\qquad V=\arctan v.
\end{equation}
* Go back to timelike and spacelike coordinates (this is not really necessary at this point and is done mostly for aesthetic reasons):
\begin{equation}
T=V+U,\qquad R=V-U.
\end{equation}
Now the range of $(T,R)$ obviously covers some bounded region on the plane, while the radial part of the line element preserves its conformal form:
\begin{equation}
ds_2^2 \sim d\tilde{\eta}^2-d\tilde{\chi}^2 \sim du\, dv \sim dU\, dV \sim dT^2 -dR^2 .
\end{equation}
As the choice of function $\arctan$ was rather arbitrary (though convenient), the choice of conformal coordinates is not unique.
</div>
</p>
</div>
</div>

In this section we will reserve notation $\eta$ and $\chi$ and name "conformal coordinates/variables" to such variables that can only take values in a bounded region on $\mathbb{R}^2$; $\tilde{\eta}$ and $\tilde{\chi}$ can span infinite or semi-infinite intervals. Spacetime diagram in terms of conformal variables $(\eta,\chi)$ is called conformal diagram. Null geodesics $\eta=\pm \chi + const$ are diagonal straight lines on conformal diagrams.

<div id="Horizon40"></div>
=== Problem 2 ===
Construct the conformal diagram for the closed Universe filled with
* radiation;
* dust;
* mixture of dust and radiation.

Show the particle and event horizons for the observer at the origin $\chi=0$ (it will be assumed hereafter that the horizons are always constructed with respect to this chosen observer).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* Solution for $a(\eta)$ in a radiation dominated closed Universe is
\begin{equation}
a=a_m \sin\eta ,\qquad \eta\in(0,\pi),\quad \chi \in [0,\pi].
\end{equation}
As the ranges spanned by $\eta$ and $\chi$ are finite, they are already conformal coordinates. The conformal diagram is a square $\eta,\chi\in [0,\pi]$. Edges $\eta=0$ and $\eta=\pi$ correspond to the Big Bang and Big Crunch singularities respectively; worldline $\chi=\pi$ is the point on the three-sphere that is situated at the opposite pole with respect to observer at $\chi=0$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Rad.png|center|thumb|365px|Conformal diagram for the radiation dominated closed Universe. Particle horizon is shown in blue, event horizon in red.]]
|}

The particle horizon is
\[\eta=\chi,\]
and event horizon is
\[\eta=\eta_{max}-\chi =\pi -\chi.\]
Both exist for all cosmological times: by the finite moment of the Big Crunch $\eta=\pi$ the event horizon is collapsed into a point (which is natural, as there is no more time left), while the particle horizon extends to the whole Universe. Thus only at the finite moment the whole of the Universe becomes observable. The farther the point, though, the younger will it look, of course, and the opposite pole will only be "observed" by our observer at the last moment of the Universe as it was at its creation.

* Solution for $a(\eta)$ in a dust dominated closed Universe is
\begin{equation}
a=a_m (1-\cos\eta) ,\qquad \eta\in(0,2\pi),\quad \chi \in [0,\pi].
\end{equation}
The difference from the previous case is that $\eta$ spans twice the range, and $\eta_{max}=2\pi =2\chi_{max}$.

Therefore the event horizon is given by
\[\eta=\eta_{max}-\chi=2\pi -\chi,\]
so it exists only in the second, contracting, phase $\eta>\pi$. The particle horizon is given by $\eta=\chi$ again, but now it exists only during the expanding phase $\eta<\pi$. It encloses the full Universe at the moment of maximal expansion $\eta=\pi$ and for later times does not exist.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Dust.png|center|thumb|365px|Conformal diagrams of the dust dominated closed Universe. Thin lines show light rays that realize the first and second images of a galaxy at different cosmic times, from opposite directions.]]
|}

* Though the full analytic solution is more complicated, it is clear that the main features remain the same as in the previous considered cases. At early and late times, close to the singularities, the dynamics is determined by the radiation component. If there is enough dust, then at large enough scale factors (which may or may not be achieved depending on the initial conditions), which would correspond to the epoch around the maximal expansion, it will be dominating. The influence of dust is that dynamics is slowed down, so that depending on the ratio of densities
\[\eta_{max}\in [\pi,2\pi].\]
Thus qualitatively the picture will be the same as in a dust dominated Universe: the conformal diagram is a rectangle, the event horizon exists only starting from some time $\eta_{e}=\eta_{max}-\pi$, while particle horizon, on the contrary, vanishes at $\eta_{p}=\pi$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-RadDust.png|center|thumb|365px|Conformal diagram for a closed Universe filled with a mix of dust and radiation.]]
|}
</div>
</p>
</div>
</div>

<div id="Horizon41"></div>

=== Problem 3 ===
'''Closed dS.''' Construct the conformal diagram for the de Sitter space in the closed sections coordinates. Provide reasoning that this space is (null) geodesically complete, i.e. every (null) geodesic extends to infinite values of affine parameters at both ends.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The scale factor in the closed dS Universe is
\begin{equation}
a(t)=H_\Lambda^{-1}\cosh (H_\Lambda t),\qquad t\in (-\infty, +\infty),
\end{equation}
so on integration, for conformal time we obtain
\begin{equation}
\eta (t)=\int\limits_{-\infty}^{t}\frac{dt}{a(t)}=\arcsin\big[\tanh (H_\Lambda t)\big]+\frac{\pi}{2} \in (0,\pi).
\end{equation}
We choose the integration constant here so that $\eta=0$ corresponds to $t=-\infty$ and $\eta=\pi$ to $t=+\infty$. The full metric then can be written as
\begin{equation}
ds^{2}_{dS}=\frac{H_\Lambda^2}{\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\sin^2 \chi d\Omega^2 \big]. \label{dSclosed}
\end{equation}

As the values of $\eta$ span a finite interval, $(\eta,\chi)$ are already conformal coordinates. The conformal diagram is again a square
\begin{equation}
\eta\in [0,\pi],\qquad \chi \in[0,\pi],
\end{equation}
with the difference from the radiation dominated Universe that the edges $\eta=0,\pi$ do not represent a singularity anymore, but instead correspond to infinite (and regular) past and future respectively. Both horizons are given again by
\begin{equation}
\eta_{e}=\pi-\chi,\qquad \eta_{p}=\chi
\end{equation}
and exist at all times.

The spacelike boundaries of the conformal diagram correspond to $t\to \pm\infty$, and therefore to infinite values of affine parameter. This can be shown if one remembers the general formula for the cosmological redshift:
\begin{equation}
\text{const}=\omega a =\frac{dt}{d\lambda}a,\quad \Rightarrow\quad
\lambda=\text{const}\cdot\int\limits^{t}dt \cosh(H_\Lambda t)
\underset{t\to\pm\infty}{\longrightarrow}\infty .
\end{equation}
The timelike boundary of the diagram corresponds to the opposite pole, there is no real boundary there, the same as on a sphere: as particles propagate across the pole, their radial coordinate begins to decrease again, while the worldline on the conformal diagram is reflected from $\chi=\pi$. Thus by definition the spacetime is (null) geodesically complete.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-closed.png|center|thumb|365px|The de Sitter Universe. The closed sections coordinates cover the whole space, which is geodesically complete. There are no singularities: the horizontal boundaries of the diagram correspond to infinite past and future.]]
|}
</p>
</div>
</div>

<div id="Horizon42"></div>
=== Problem 4 ===
'''Static dS.''' Rewrite the metric of de Sitter space (\ref{dSclosed}) in terms of "static coordinates" $T,R$:
\begin{equation}
\tanh (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad
H_\Lambda R =\frac{\sin\chi}{\sin\eta}.
\end{equation}
* What part of the conformal diagram in terms of $(\eta,\chi)$ is covered by the static coordinate chart $(T,R)$?
* Express the horizon's equations in terms of $T$ and $R$
* Draw the surfaces of constant $T$ and $R$ on the conformal diagram.
* Write out the coordinate transformation between $(\eta,\chi)$ and $(T,R)$ in the regions where $|\cos\eta|>|\cos\chi|$. Explain the meaning of $T$ and $R$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us introduce dimensionless coordinates $t=H_\Lambda T$ and $r=H_\Lambda R$. The inverse relations then are
\begin{equation}
\sin^2 \eta =\frac{1}{\cosh^2 t-r^2 \sinh^2 t},
\quad \sin^2 \chi =\frac{r^2}{\cosh^2 t-r^2 \sinh^2 t}.
\end{equation}
Using them, after some algebra from (\ref{dSclosed}) we get
\begin{equation}
ds_{dS}^2 =\big[1-H_\Lambda^2 R^2\big]dT^2 -\frac{dR^2}{1-H_\Lambda^2 R^2}-R^2 d\Omega^2 , \label{dSstatic}
\end{equation}
which resembles the Schwarzschild line element (and this is not a coincidence).
</p>

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-static.png|center|thumb|365px|The de Sitter Universe with contour lines of the "static" coordinates $(T,R)$. The solid lines are $T=const$, and the dashed ones $R=const$. The coordinates become singular on both horizons, so the whole spacetime is divided by them into four sectors, each separately covered by the regular coordinate chart $(R,T)$. Spacetime is actually static only in the left and right sectors (I and III), where $T$ is a timelike coordinate and $R$ spacelike: the static patches are bounded by the horizons.]]
|}

<p style="text-align: left;">
* As $|\tanh (H_\Lambda T)|<1$, we have $|\cos\eta| \leq |\cos\chi|$. This condition cuts out two out of four sectors from the square conformal diagram, I and III: one is $\eta\in [\chi, \pi-\chi]$ and the other is $\eta\in [\pi-\chi, \chi]$. In both $|\sin\eta|\geq |\sin\chi|$, so $R \leq H_{\Lambda}^{-1}$;
* The particle horizon $\eta=\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=-\infty$. It is one part of the boundary of the region (in two parts) covered by coordinates $(T,R)$. The event horizon $\eta=\pi-\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=+\infty$ and is the other part of the boundary of this region.
* $T=const$ is $\cos\eta =t\cos\chi $ and $R=const$ is $\sin\eta =r^{-1}\sin\chi$.
* In regions II and IV the needed relation is obtained if we simply replace $\tanh$ with $\coth$ in the first relation:
\begin{equation}
\coth (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad H_\Lambda R=\frac{\sin\chi}{\sin\eta},
\end{equation}
as can be checked explicitly by substitution into (\ref{dSstatic}), which again gives (\ref{dSclosed}). In these regions $R$ is a timelike coordinate, and $T$ is spacelike. The geodesics of comoving massive particles are $\chi=const$, one of them $\chi=\pi/2$ corresponds to $T=0$. Thus in the lower part of the diagram, where $R\in (+\infty,H_\Lambda^{-1})$, the spacetime is contracting; in the upper part, where $R\in (H_\Lambda^{-1},\infty)$, it is expanding. The coordinate frame is not static. The relations between $(T,R)$ and $(\eta,\chi)$ in static and non-static regions mirros those in Schwarzshild black hole solution between the static and the global (Kruskal-Szekeres) coordinates. Here $\eta$ and $\chi$ are the global coordinates.
</p>
</div>
</div>

<div id="Horizon43"></div>

=== Problem 4 ===
'''Flat dS.''' The scale factor in flat de Sitter is $a(t)=H_\Lambda^{-1} e^{H_\Lambda t}$.

* Find the range of values spanned by conformal time $\tilde{\eta}$ and comoving distance $\tilde{\chi}$ in the flat de Sitter space
* Verify that coordinate transformation
\begin{equation}
\tilde{\eta}=\frac{-\sin\eta}{\cos\chi-\cos\eta},\qquad
\tilde{\chi}=\frac{\sin\chi}{\cos\chi-\cos\eta}
\end{equation}
bring the metric to the form of that of de Sitter in closed slicing (it is assumed that $\tilde{\eta}=0$ is chosen to correspond to infinite future).
* Which part of the conformal diagram is covered by the coordinate chart $(\tilde{\eta},\tilde{\chi})$? Is the flat de Sitter space geodesically complete?
* Where are the particle and event horizons in these coordinates?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As $t\in(-\infty,+\infty)$,
\begin{equation}
\tilde{\eta} =\int\frac{dt}{a(t)}=H_\Lambda \int\limits_{+\infty}^{t}dt\; e^{-H_\Lambda t} =-e^{-H_\Lambda t} \in (-\infty,0).
\end{equation}
Here we choose $+\infty$ as the lower limit, because at $-\infty$ the integral diverges.
<div style="text-align: left;>
* Direct calculation yields (\ref{dSclosed}), with $\eta \in (0,\pi)$, $\chi\in (0,\pi)$;
* The upper triangle $\eta>\chi$, above the particle horizon, on which $\tilde{\eta}\to -\infty$. It is not geodesically complete, as geodesics are cut at the particle horizon.
* The particle horizon is the boundary of the patch of full dS space covered by flat slicing coordinates, and event horizon in these coordinates exists only in the latter part of evolution, for $\eta >\pi/2$.
</div>
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-flat.png|center|thumb|365px|The de Sitter Universe in flat sections' coordinates, which cover only half of it. The boundary -- the particle horizon -- consists of three different infinities.]]
|}
</p>
</div>
</div>

<div id="Horizon44"></div>
=== Problem 5 ===
'''Infinities.''' What parts of the spacetime's boundary on the conformal diagram of flat de Sitter space corresponds to

* spacelike infinity $i^0$, where $\tilde{\chi}\to +\infty$;
* past timelike infinity $i^-$, where $\tilde{\eta}\to -\infty$ and from which all timelike worldlines emanate
* past lightlike infinity $J^-$, from which all null geodesics emanate?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* The one point at the left of the diagram;
* the one point at the bottom;
* the particle horizon.
</div>
</p>
</div>
</div>

<div id="Horizon45"></div>
=== Problem 6 ===
'''Open dS.''' Consider the de Sitter space in open slicing, in which $a(t)=H_\Lambda \sinh (H_\Lambda t)$, so conformal time is
\begin{equation}
\tilde{\eta} =\int\limits_{+\infty}^{t} \frac{dt}{a(t)},
\end{equation}
where again the lower limit is chosen so that the integral is bounded.

* Find $\tilde{\eta}(t)$ and verify that coordinate transformation from $(\tilde{\eta},\tilde{\chi})$ to $\eta,\chi$, such that
\begin{equation}
\tanh\tilde{\eta}=\frac{-\sin\eta}{\cos\cos\chi},\qquad
\tanh\tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
brings the metric to the form of de Sitter in closed slicing.
* What are the ranges spanned by $(\tilde{\eta},\tilde{\chi})$ and $(\eta,\chi)$? Which part of the conformal diagram do they cover?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
After getting
\begin{equation}
\sinh \tilde{\eta}=-\frac{1}{\sinh(H_\Lambda t)},
\end{equation}
the first part is checked straightforwardly; in the open de Sitter $\tilde{\chi}\in [0,+\infty)$, and $\tilde{\eta}\in(-\infty,0)$. The region covered by coordinates $(\tilde{\eta},\tilde{\chi})$ is $\{\eta>\chi+\pi/2\}$, only one eighth part of the full diagram.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-open.png|center|thumb|365px|The de Sitter Universe in open sections' coordinates, which cover only $1/8^{\text{th}}$ of the full diagram.]]
|}
</p>
</div>
</div>

<div id="Horizon46"></div>
=== Problem 7 ===
'''Minkowski 1.''' Rewrite the Minkowski metric in terms of coordinates $(\eta,\chi)$, which are related to $(t,r)$ by the relation
\begin{equation}
\tanh \tilde{\eta}=\frac{\sin\eta}{\cos\chi},\qquad
\tanh \tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
that mirrors the one between the open and closed coordinates of de Sitter. Construct the conformal diagram and determine different types of infinities. Are there new ones compared to the flat de Sitter space?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Coordinate transformation gives
\begin{equation}
ds^2 =\frac{1}{\cos^2 \chi -\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\Psi^{2}(\eta,\chi)d\Omega^2 \big].
\end{equation}
Here $r\in [0,+\infty)$ and $t\in (-\infty,+\infty)$. Comparing with the relation between $(\tilde{\eta},\tilde{\chi})$ with conformal coordinates $(\eta,\chi)$ in the open de Sitter universe, where $\tilde{\eta}\in (-\infty,0)$, we see that the difference is that $\tilde{\eta}$ spans $(-\infty,0)$, while now $t$ spans twice the range, $(-\infty,\infty)$. Therefore the conformal diagram is composed of two triangles, one the same as for open de Sitter and one for its time-reversed copy. Accordingly there now appear future timelike infinity $i^+$ and future lightlike infinity $J^+$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |The Minkowski spacetime in two different pairs of conformal coordinates and the full set of infinities. Thin dashed and solid lines show the images of coordinate grid $(t,r)$.

|[[File:Conf-Mink1.png|center|thumb|365px]]
|[[File:Conf-Mink2.png|center|thumb|365px]]
|}
</p>
</div>
</div>

<div id="Horizon47"></div>
=== Problem 8 ===
'''Minkowski 2.''' The choice of conformal coordinates is not unique. Construct the conformal diagram for Minkowski using the universal scheme: first pass to null coordinates, then bring their span to finite intervals with $\arctan$ (one of the possible choices), then pass again to timelike and spacelike coordinates.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
We start from spherical coordinates
\begin{equation}
ds^2 =dt^2 -dr^2 -r^2 d\Omega^2 .
\end{equation}
<div style="text-align: left;">
* The first step is introducing null coordinates
\begin{equation}
u=t-r,\quad v=t+r ,
\end{equation}
so that
\begin{equation}
ds^2 =4du\,dv -\frac{(v-u)^2}{4}d\Omega^2 .
\end{equation}
* Then bring the range of values to finite intervals
\begin{equation}
U=\arctan u, \qquad V=\arctan v,
\end{equation}
so that
\begin{equation}
ds^2 =\frac{1}{4\cos^2 U \cos^2 V}\big[4dU\,dV -\sin^2 (V-U)d\Omega^2\big].
\end{equation}
The whole spacetime is simply the half of the square $U,V\in (-\pi/2,\pi/2)$, in which $r>0$, i.e. $v>u\quad \Leftrightarrow\quad V>U$: on the plane $(U,V)$ it is the triangle \begin{equation}
-\pi/2<U<V<\pi/2.
\end{equation}
* Finally, go back to spacelike and timelike coordinates
\begin{equation}
T=V+U,\qquad R=V-U
\end{equation}
so that metric becomes
\begin{equation}
ds^2 =\frac{1}{[\cos T +\cos R]^2}\big[dT^2 -dR^2 -\sin^2 R\; d\Omega^2\big].
\end{equation}
The triangle is shrunken by $\sqrt{2}$ and rotated by $3\pi/4$ clockwise, thus turning into
\begin{equation}
\{R>0,\quad |T|<\pi/2 -R\}.
\end{equation}
This is the same form as obtained by the other construction (up to scaling, which is purely decorative).
</div>
</p>
</div>
</div>

<div id="Horizon48"></div>
=== Problem 9 ===
Draw the conformal diagram for the Milne Universe and show which part of Minkowski space's diagram it covers.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Minkowski metric $ds^2 =dT^2 -dR^2$, rewritten in terms of $(\tau, r)$ such that
\begin{equation}
T=\tau \cosh r ,\quad R=\tau \sinh r ,
\end{equation}
is the metric of the Milne Universe. As Minkowski space is complete, the Milne Universe then is a \emph{part} of Minkowski in different variables. This part is where $T>R$. The boundary $T=R$ is the future light cone, so the whole Milne Universe is represented by the triangle homothetic to the whole space but 4 times smaller in area, with the common node of future infinity.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Milne.png|center|thumb|365px|The Milne Universe on Minkowski's conformal diagram. It has the same shape and shares the same future infinity, but covers one eighth of the area. The boundary is the past horizon.]]
|}
</p>
</div>
</div>

<div id="Horizon49"></div>
=== Problem 10 ===
Consider open or flat Universe filled with matter that satisfies strong energy condition $\varepsilon +3p>0$. What are the coordinate ranges spanned by the comoving coordinate $\tilde{\chi}$ and conformal time $\tilde{\eta}$? Compare with the Minkowski metric and construct the diagram. Identify the types of infinities and the initial Big Bang singularity.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For flat Universe $\tilde{\chi}=r$, for the open $\tilde{\chi}=\sinh r$, so in both cases $\tilde{\chi}\in (0,+\infty)$.

Strong energy condition implies that $w>-1/3$, so
\begin{equation}
\rho \sim a^{-3(1+w)}=a^{-n},
\end{equation}
where $n>2$. From the first Friedman equation then after simple manipulations we obtain that
\begin{equation}
\frac{da}{dt}\sim a^{-\theta},
\end{equation}
where $\theta$ is some positive number. Therefore both
\begin{equation}
t\sim \int da\; a^{\theta},
\quad\text{and}\quad
\eta =\int \frac{da}{a}a^{\theta}
\end{equation}
converge at $a\to 0$ and diverge at $a\to \infty$. Consequently, the integration constant can be chosen so that $\eta\in (0,+\infty)$.

The conformal structure is the same as that of the \emph{upper} half of Minkowski spacetime. The Big Bang singularity at $\eta=0$ is at the cut, and there are spacelike infinity, future infinity and future null infinity.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |Conformal diagrams for open or flat Universes. The one on the left is for ones filled with matter which satisfies the strong energy condition (SEC), and the one on the right for ones in which SEC does not hold, such as in case of power-law inflation..

|[[File:Conf-SEC.png|center|thumb|365px]]
|[[File:Conf-Inflation.png|center|thumb|365px]]
|}

</p>
</div>
</div>

<div id="Horizon50"></div>
=== Problem 11 ===
Draw the conformal diagram for open and flat Universes with power-law scale factor $a(t)\sim t^{n}$, with $n>1$. This is the model for the power-law inflation. Check whether the strong energy condition is satisfied.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As seen in the previous problem, if strong energy condition were satisfied, we would have $\dot{a}\sim a^{-\theta}$ with some positive $\theta$; this is not the case, so the condition is violated. As $t\in (0,+\infty)$,
\begin{equation}
\eta\sim \int \frac{dt}{t^n}
\end{equation}
diverges at small $a$ (thus also small $t$) and converges at $a\to \infty$ (thus as large $t$). So integration constant can be chosen so that $\eta\in (-\infty,0)$.

The conformal structure is the same as \emph{lower half} of Minkowski spacetime. The cut is regular future infinity, and from Minkowski there are spacelike infinity, past null infinity, and past infinity. The point of past infinity corresponds to Big Bang and is singular.
</p>
</div>
</div>

<references>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN~0521563984</ref>
</references>

File:Conf-dS-static.png

2013-12-08T17:40:36Z

Igor: The de Sitter Universe in static slicing: conformal diagram

The de Sitter Universe in static slicing: conformal diagram

Causal Structure

2013-12-08T17:39:53Z

Igor: /* Problem 2 */

[[Category:Horizons|4]]

__TOC__

The causal structure is determined by propagation of light and is best understood in terms of ''conformal diagrams''. In this section we construct and analyze those for a number of important model cosmological solutions (which are assumed to be already known), following mostly the exposition of <ref name=Mukhanov/>.

In terms of comoving distance $\tilde{\chi}$ and conformal time $\tilde{\eta}$ (in this section they are denoted by tildes) the two-dimensional radial part of the FLRW metric takes form
\begin{equation}
ds_{2}^{2}=a^{2}(\tilde{\eta})\big[d\tilde{\eta}^2 -d\tilde{\chi}^2\big].
\label{ds2Dconf}
\end{equation}
In the brackets here stands the line element of two-dimensional Minkowski flat spacetime. Coordinate transformations that preserve the \emph{conformal} form of the metric
\begin{equation*}
ds_2^2 =\Omega^{2}(\eta,\chi)\big[d\eta^2 -d\chi^2 \big],
\end{equation*}
are called conformal transformations, and the corresponding coordinates $(\eta,\chi)$ -- conformal coordinates.

<div id="Horizon39"></div>
=== Problem 1 ===
Show that it is always possible to construct $\eta(\tilde{\eta},\tilde{\chi})$, $\chi(\tilde{\eta},\tilde{\chi})$, such that the conformal form of metric (\ref{ds2Dconf}) is preserved, but $\eta$ and $\chi$ are bounded and take values in some finite intervals. Is the choice of $(\eta,\chi)$ unique?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Suppose $\tilde{\eta},\tilde{\chi}$ span infinite or semi-infinite values. Then we can always make the following sequence of coordinate transformations:
<div style="text-align: left;">
* Pass to null coordinates
\begin{equation}
u=\tilde{\eta}-\tilde{\chi},\qquad v=\tilde{\eta}+\tilde{\chi};
\end{equation}
* Bring their range of values to a finite interval by some appropriate function, i.e.
\begin{equation}
U=\arctan u,\qquad V=\arctan v.
\end{equation}
* Go back to timelike and spacelike coordinates (this is not really necessary at this point and is done mostly for aesthetic reasons):
\begin{equation}
T=V+U,\qquad R=V-U.
\end{equation}
Now the range of $(T,R)$ obviously covers some bounded region on the plane, while the radial part of the line element preserves its conformal form:
\begin{equation}
ds_2^2 \sim d\tilde{\eta}^2-d\tilde{\chi}^2 \sim du\, dv \sim dU\, dV \sim dT^2 -dR^2 .
\end{equation}
As the choice of function $\arctan$ was rather arbitrary (though convenient), the choice of conformal coordinates is not unique.
</div>
</p>
</div>
</div>

In this section we will reserve notation $\eta$ and $\chi$ and name "conformal coordinates/variables" to such variables that can only take values in a bounded region on $\mathbb{R}^2$; $\tilde{\eta}$ and $\tilde{\chi}$ can span infinite or semi-infinite intervals. Spacetime diagram in terms of conformal variables $(\eta,\chi)$ is called conformal diagram. Null geodesics $\eta=\pm \chi + const$ are diagonal straight lines on conformal diagrams.

<div id="Horizon40"></div>
=== Problem 2 ===
Construct the conformal diagram for the closed Universe filled with
* radiation;
* dust;
* mixture of dust and radiation.

Show the particle and event horizons for the observer at the origin $\chi=0$ (it will be assumed hereafter that the horizons are always constructed with respect to this chosen observer).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* Solution for $a(\eta)$ in a radiation dominated closed Universe is
\begin{equation}
a=a_m \sin\eta ,\qquad \eta\in(0,\pi),\quad \chi \in [0,\pi].
\end{equation}
As the ranges spanned by $\eta$ and $\chi$ are finite, they are already conformal coordinates. The conformal diagram is a square $\eta,\chi\in [0,\pi]$. Edges $\eta=0$ and $\eta=\pi$ correspond to the Big Bang and Big Crunch singularities respectively; worldline $\chi=\pi$ is the point on the three-sphere that is situated at the opposite pole with respect to observer at $\chi=0$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Rad.png|center|thumb|365px|Conformal diagram for the radiation dominated closed Universe. Particle horizon is shown in blue, event horizon in red.]]
|}

The particle horizon is
\[\eta=\chi,\]
and event horizon is
\[\eta=\eta_{max}-\chi =\pi -\chi.\]
Both exist for all cosmological times: by the finite moment of the Big Crunch $\eta=\pi$ the event horizon is collapsed into a point (which is natural, as there is no more time left), while the particle horizon extends to the whole Universe. Thus only at the finite moment the whole of the Universe becomes observable. The farther the point, though, the younger will it look, of course, and the opposite pole will only be "observed" by our observer at the last moment of the Universe as it was at its creation.

* Solution for $a(\eta)$ in a dust dominated closed Universe is
\begin{equation}
a=a_m (1-\cos\eta) ,\qquad \eta\in(0,2\pi),\quad \chi \in [0,\pi].
\end{equation}
The difference from the previous case is that $\eta$ spans twice the range, and $\eta_{max}=2\pi =2\chi_{max}$.

Therefore the event horizon is given by
\[\eta=\eta_{max}-\chi=2\pi -\chi,\]
so it exists only in the second, contracting, phase $\eta>\pi$. The particle horizon is given by $\eta=\chi$ again, but now it exists only during the expanding phase $\eta<\pi$. It encloses the full Universe at the moment of maximal expansion $\eta=\pi$ and for later times does not exist.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Dust.png|center|thumb|365px|Conformal diagrams of the dust dominated closed Universe. Thin lines show light rays that realize the first and second images of a galaxy at different cosmic times, from opposite directions.]]
|}

* Though the full analytic solution is more complicated, it is clear that the main features remain the same as in the previous considered cases. At early and late times, close to the singularities, the dynamics is determined by the radiation component. If there is enough dust, then at large enough scale factors (which may or may not be achieved depending on the initial conditions), which would correspond to the epoch around the maximal expansion, it will be dominating. The influence of dust is that dynamics is slowed down, so that depending on the ratio of densities
\[\eta_{max}\in [\pi,2\pi].\]
Thus qualitatively the picture will be the same as in a dust dominated Universe: the conformal diagram is a rectangle, the event horizon exists only starting from some time $\eta_{e}=\eta_{max}-\pi$, while particle horizon, on the contrary, vanishes at $\eta_{p}=\pi$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-RadDust.png|center|thumb|365px|Conformal diagram for a closed Universe filled with a mix of dust and radiation.]]
|}
</div>
</p>
</div>
</div>

<div id="Horizon41"></div>

=== Problem 3 ===
'''Closed dS.''' Construct the conformal diagram for the de Sitter space in the closed sections coordinates. Provide reasoning that this space is (null) geodesically complete, i.e. every (null) geodesic extends to infinite values of affine parameters at both ends.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The scale factor in the closed dS Universe is
\begin{equation}
a(t)=H_\Lambda^{-1}\cosh (H_\Lambda t),\qquad t\in (-\infty, +\infty),
\end{equation}
so on integration, for conformal time we obtain
\begin{equation}
\eta (t)=\int\limits_{-\infty}^{t}\frac{dt}{a(t)}=\arcsin\big[\tanh (H_\Lambda t)\big]+\frac{\pi}{2} \in (0,\pi).
\end{equation}
We choose the integration constant here so that $\eta=0$ corresponds to $t=-\infty$ and $\eta=\pi$ to $t=+\infty$. The full metric then can be written as
\begin{equation}
ds^{2}_{dS}=\frac{H_\Lambda^2}{\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\sin^2 \chi d\Omega^2 \big]. \label{dSclosed}
\end{equation}

As the values of $\eta$ span a finite interval, $(\eta,\chi)$ are already conformal coordinates. The conformal diagram is again a square
\begin{equation}
\eta\in [0,\pi],\qquad \chi \in[0,\pi],
\end{equation}
with the difference from the radiation dominated Universe that the edges $\eta=0,\pi$ do not represent a singularity anymore, but instead correspond to infinite (and regular) past and future respectively. Both horizons are given again by
\begin{equation}
\eta_{e}=\pi-\chi,\qquad \eta_{p}=\chi
\end{equation}
and exist at all times.

The spacelike boundaries of the conformal diagram correspond to $t\to \pm\infty$, and therefore to infinite values of affine parameter. This can be shown if one remembers the general formula for the cosmological redshift:
\begin{equation}
\text{const}=\omega a =\frac{dt}{d\lambda}a,\quad \Rightarrow\quad
\lambda=\text{const}\cdot\int\limits^{t}dt \cosh(H_\Lambda t)
\underset{t\to\pm\infty}{\longrightarrow}\infty .
\end{equation}
The timelike boundary of the diagram corresponds to the opposite pole, there is no real boundary there, the same as on a sphere: as particles propagate across the pole, their radial coordinate begins to decrease again, while the worldline on the conformal diagram is reflected from $\chi=\pi$. Thus by definition the spacetime is (null) geodesically complete.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-closed.png|center|thumb|365px|The de Sitter Universe. The closed sections coordinates cover the whole space, which is geodesically complete. There are no singularities: the horizontal boundaries of the diagram correspond to infinite past and future.]]
|}
</p>
</div>
</div>

<div id="Horizon42"></div>
=== Problem 4 ===
'''Static dS.''' Rewrite the metric of de Sitter space (\ref{dSclosed}) in terms of "static coordinates" $T,R$:
\begin{equation}
\tanh (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad
H_\Lambda R =\frac{\sin\chi}{\sin\eta}.
\end{equation}
* What part of the conformal diagram in terms of $(\eta,\chi)$ is covered by the static coordinate chart $(T,R)$?
* Express the horizon's equations in terms of $T$ and $R$
* Draw the surfaces of constant $T$ and $R$ on the conformal diagram.
* Write out the coordinate transformation between $(\eta,\chi)$ and $(T,R)$ in the regions where $|\cos\eta|>|\cos\chi|$. Explain the meaning of $T$ and $R$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us introduce dimensionless coordinates $t=H_\Lambda T$ and $r=H_\Lambda R$. The inverse relations then are
\begin{equation}
\sin^2 \eta =\frac{1}{\cosh^2 t-r^2 \sinh^2 t},
\quad \sin^2 \chi =\frac{r^2}{\cosh^2 t-r^2 \sinh^2 t}.
\end{equation}
Using them, after some algebra from (\ref{dSclosed}) we get
\begin{equation}
ds_{dS}^2 =\big[1-H_\Lambda^2 R^2\big]dT^2 -\frac{dR^2}{1-H_\Lambda^2 R^2}-R^2 d\Omega^2 , \label{dSstatic}
\end{equation}
which resembles the Schwarzschild line element (and this is not a coincidence).

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-static.png|center|thumb|365px|The de Sitter Universe with contour lines of the "static" coordinates $(T,R)$. The solid lines are $T=const$, and the dashed ones $R=const$. The coordinates become singular on both horizons, so the whole spacetime is divided by them into four sectors, each separately covered by the regular coordinate chart $(R,T)$. Spacetime is actually static only in the left and right sectors (I and III), where $T$ is a timelike coordinate and $R$ spacelike: the static patches are bounded by the horizons.]]
|}

<div style="text-align: left;>
* As $|\tanh (H_\Lambda T)|<1$, we have $|\cos\eta| \leq |\cos\chi|$. This condition cuts out two out of four sectors from the square conformal diagram, I and III: one is $\eta\in [\chi, \pi-\chi]$ and the other is $\eta\in [\pi-\chi, \chi]$. In both $|\sin\eta|\geq |\sin\chi|$, so $R \leq H_{\Lambda}^{-1}$;
* The particle horizon $\eta=\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=-\infty$. It is one part of the boundary of the region (in two parts) covered by coordinates $(T,R)$. The event horizon $\eta=\pi-\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=+\infty$ and is the other part of the boundary of this region.
* $T=const$ is $\cos\eta =t\cos\chi $ and $R=const$ is $\sin\eta =r^{-1}\sin\chi$.
* In regions II and IV the needed relation is obtained if we simply replace $\tanh$ with $\coth$ in the first relation:
\begin{equation}
\coth (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad H_\Lambda R=\frac{\sin\chi}{\sin\eta},
\end{equation}
as can be checked explicitly by substitution into (\ref{dSstatic}), which again gives (\ref{dSclosed}). In these regions $R$ is a timelike coordinate, and $T$ is spacelike. The geodesics of comoving massive particles are $\chi=const$, one of them $\chi=\pi/2$ corresponds to $T=0$. Thus in the lower part of the diagram, where $R\in (+\infty,H_\Lambda^{-1})$, the spacetime is contracting; in the upper part, where $R\in (H_\Lambda^{-1},\infty)$, it is expanding. The coordinate frame is not static. The relations between $(T,R)$ and $(\eta,\chi)$ in static and non-static regions mirros those in Schwarzshild black hole solution between the static and the global (Kruskal-Szekeres) coordinates. Here $\eta$ and $\chi$ are the global coordinates.
</div>
</p>
</div>
</div>

<div id="Horizon43"></div>
=== Problem 4 ===
'''Flat dS.''' The scale factor in flat de Sitter is $a(t)=H_\Lambda^{-1} e^{H_\Lambda t}$.

* Find the range of values spanned by conformal time $\tilde{\eta}$ and comoving distance $\tilde{\chi}$ in the flat de Sitter space
* Verify that coordinate transformation
\begin{equation}
\tilde{\eta}=\frac{-\sin\eta}{\cos\chi-\cos\eta},\qquad
\tilde{\chi}=\frac{\sin\chi}{\cos\chi-\cos\eta}
\end{equation}
bring the metric to the form of that of de Sitter in closed slicing (it is assumed that $\tilde{\eta}=0$ is chosen to correspond to infinite future).
* Which part of the conformal diagram is covered by the coordinate chart $(\tilde{\eta},\tilde{\chi})$? Is the flat de Sitter space geodesically complete?
* Where are the particle and event horizons in these coordinates?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As $t\in(-\infty,+\infty)$,
\begin{equation}
\tilde{\eta} =\int\frac{dt}{a(t)}=H_\Lambda \int\limits_{+\infty}^{t}dt\; e^{-H_\Lambda t} =-e^{-H_\Lambda t} \in (-\infty,0).
\end{equation}
Here we choose $+\infty$ as the lower limit, because at $-\infty$ the integral diverges.
<div style="text-align: left;>
* Direct calculation yields (\ref{dSclosed}), with $\eta \in (0,\pi)$, $\chi\in (0,\pi)$;
* The upper triangle $\eta>\chi$, above the particle horizon, on which $\tilde{\eta}\to -\infty$. It is not geodesically complete, as geodesics are cut at the particle horizon.
* The particle horizon is the boundary of the patch of full dS space covered by flat slicing coordinates, and event horizon in these coordinates exists only in the latter part of evolution, for $\eta >\pi/2$.
</div>
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-flat.png|center|thumb|365px|The de Sitter Universe in flat sections' coordinates, which cover only half of it. The boundary -- the particle horizon -- consists of three different infinities.]]
|}
</p>
</div>
</div>

<div id="Horizon44"></div>
=== Problem 5 ===
'''Infinities.''' What parts of the spacetime's boundary on the conformal diagram of flat de Sitter space corresponds to

* spacelike infinity $i^0$, where $\tilde{\chi}\to +\infty$;
* past timelike infinity $i^-$, where $\tilde{\eta}\to -\infty$ and from which all timelike worldlines emanate
* past lightlike infinity $J^-$, from which all null geodesics emanate?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* The one point at the left of the diagram;
* the one point at the bottom;
* the particle horizon.
</div>
</p>
</div>
</div>

<div id="Horizon45"></div>
=== Problem 6 ===
'''Open dS.''' Consider the de Sitter space in open slicing, in which $a(t)=H_\Lambda \sinh (H_\Lambda t)$, so conformal time is
\begin{equation}
\tilde{\eta} =\int\limits_{+\infty}^{t} \frac{dt}{a(t)},
\end{equation}
where again the lower limit is chosen so that the integral is bounded.

* Find $\tilde{\eta}(t)$ and verify that coordinate transformation from $(\tilde{\eta},\tilde{\chi})$ to $\eta,\chi$, such that
\begin{equation}
\tanh\tilde{\eta}=\frac{-\sin\eta}{\cos\cos\chi},\qquad
\tanh\tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
brings the metric to the form of de Sitter in closed slicing.
* What are the ranges spanned by $(\tilde{\eta},\tilde{\chi})$ and $(\eta,\chi)$? Which part of the conformal diagram do they cover?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
After getting
\begin{equation}
\sinh \tilde{\eta}=-\frac{1}{\sinh(H_\Lambda t)},
\end{equation}
the first part is checked straightforwardly; in the open de Sitter $\tilde{\chi}\in [0,+\infty)$, and $\tilde{\eta}\in(-\infty,0)$. The region covered by coordinates $(\tilde{\eta},\tilde{\chi})$ is $\{\eta>\chi+\pi/2\}$, only one eighth part of the full diagram.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-open.png|center|thumb|365px|The de Sitter Universe in open sections' coordinates, which cover only $1/8^{\text{th}}$ of the full diagram.]]
|}
</p>
</div>
</div>

<div id="Horizon46"></div>
=== Problem 7 ===
'''Minkowski 1.''' Rewrite the Minkowski metric in terms of coordinates $(\eta,\chi)$, which are related to $(t,r)$ by the relation
\begin{equation}
\tanh \tilde{\eta}=\frac{\sin\eta}{\cos\chi},\qquad
\tanh \tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
that mirrors the one between the open and closed coordinates of de Sitter. Construct the conformal diagram and determine different types of infinities. Are there new ones compared to the flat de Sitter space?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Coordinate transformation gives
\begin{equation}
ds^2 =\frac{1}{\cos^2 \chi -\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\Psi^{2}(\eta,\chi)d\Omega^2 \big].
\end{equation}
Here $r\in [0,+\infty)$ and $t\in (-\infty,+\infty)$. Comparing with the relation between $(\tilde{\eta},\tilde{\chi})$ with conformal coordinates $(\eta,\chi)$ in the open de Sitter universe, where $\tilde{\eta}\in (-\infty,0)$, we see that the difference is that $\tilde{\eta}$ spans $(-\infty,0)$, while now $t$ spans twice the range, $(-\infty,\infty)$. Therefore the conformal diagram is composed of two triangles, one the same as for open de Sitter and one for its time-reversed copy. Accordingly there now appear future timelike infinity $i^+$ and future lightlike infinity $J^+$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |The Minkowski spacetime in two different pairs of conformal coordinates and the full set of infinities. Thin dashed and solid lines show the images of coordinate grid $(t,r)$.

|[[File:Conf-Mink1.png|center|thumb|365px]]
|[[File:Conf-Mink2.png|center|thumb|365px]]
|}
</p>
</div>
</div>

<div id="Horizon47"></div>
=== Problem 8 ===
'''Minkowski 2.''' The choice of conformal coordinates is not unique. Construct the conformal diagram for Minkowski using the universal scheme: first pass to null coordinates, then bring their span to finite intervals with $\arctan$ (one of the possible choices), then pass again to timelike and spacelike coordinates.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
We start from spherical coordinates
\begin{equation}
ds^2 =dt^2 -dr^2 -r^2 d\Omega^2 .
\end{equation}
<div style="text-align: left;">
* The first step is introducing null coordinates
\begin{equation}
u=t-r,\quad v=t+r ,
\end{equation}
so that
\begin{equation}
ds^2 =4du\,dv -\frac{(v-u)^2}{4}d\Omega^2 .
\end{equation}
* Then bring the range of values to finite intervals
\begin{equation}
U=\arctan u, \qquad V=\arctan v,
\end{equation}
so that
\begin{equation}
ds^2 =\frac{1}{4\cos^2 U \cos^2 V}\big[4dU\,dV -\sin^2 (V-U)d\Omega^2\big].
\end{equation}
The whole spacetime is simply the half of the square $U,V\in (-\pi/2,\pi/2)$, in which $r>0$, i.e. $v>u\quad \Leftrightarrow\quad V>U$: on the plane $(U,V)$ it is the triangle \begin{equation}
-\pi/2<U<V<\pi/2.
\end{equation}
* Finally, go back to spacelike and timelike coordinates
\begin{equation}
T=V+U,\qquad R=V-U
\end{equation}
so that metric becomes
\begin{equation}
ds^2 =\frac{1}{[\cos T +\cos R]^2}\big[dT^2 -dR^2 -\sin^2 R\; d\Omega^2\big].
\end{equation}
The triangle is shrunken by $\sqrt{2}$ and rotated by $3\pi/4$ clockwise, thus turning into
\begin{equation}
\{R>0,\quad |T|<\pi/2 -R\}.
\end{equation}
This is the same form as obtained by the other construction (up to scaling, which is purely decorative).
</div>
</p>
</div>
</div>

<div id="Horizon48"></div>
=== Problem 9 ===
Draw the conformal diagram for the Milne Universe and show which part of Minkowski space's diagram it covers.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Minkowski metric $ds^2 =dT^2 -dR^2$, rewritten in terms of $(\tau, r)$ such that
\begin{equation}
T=\tau \cosh r ,\quad R=\tau \sinh r ,
\end{equation}
is the metric of the Milne Universe. As Minkowski space is complete, the Milne Universe then is a \emph{part} of Minkowski in different variables. This part is where $T>R$. The boundary $T=R$ is the future light cone, so the whole Milne Universe is represented by the triangle homothetic to the whole space but 4 times smaller in area, with the common node of future infinity.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Milne.png|center|thumb|365px|The Milne Universe on Minkowski's conformal diagram. It has the same shape and shares the same future infinity, but covers one eighth of the area. The boundary is the past horizon.]]
|}
</p>
</div>
</div>

<div id="Horizon49"></div>
=== Problem 10 ===
Consider open or flat Universe filled with matter that satisfies strong energy condition $\varepsilon +3p>0$. What are the coordinate ranges spanned by the comoving coordinate $\tilde{\chi}$ and conformal time $\tilde{\eta}$? Compare with the Minkowski metric and construct the diagram. Identify the types of infinities and the initial Big Bang singularity.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For flat Universe $\tilde{\chi}=r$, for the open $\tilde{\chi}=\sinh r$, so in both cases $\tilde{\chi}\in (0,+\infty)$.

Strong energy condition implies that $w>-1/3$, so
\begin{equation}
\rho \sim a^{-3(1+w)}=a^{-n},
\end{equation}
where $n>2$. From the first Friedman equation then after simple manipulations we obtain that
\begin{equation}
\frac{da}{dt}\sim a^{-\theta},
\end{equation}
where $\theta$ is some positive number. Therefore both
\begin{equation}
t\sim \int da\; a^{\theta},
\quad\text{and}\quad
\eta =\int \frac{da}{a}a^{\theta}
\end{equation}
converge at $a\to 0$ and diverge at $a\to \infty$. Consequently, the integration constant can be chosen so that $\eta\in (0,+\infty)$.

The conformal structure is the same as that of the \emph{upper} half of Minkowski spacetime. The Big Bang singularity at $\eta=0$ is at the cut, and there are spacelike infinity, future infinity and future null infinity.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |Conformal diagrams for open or flat Universes. The one on the left is for ones filled with matter which satisfies the strong energy condition (SEC), and the one on the right for ones in which SEC does not hold, such as in case of power-law inflation..

|[[File:Conf-SEC.png|center|thumb|365px]]
|[[File:Conf-Inflation.png|center|thumb|365px]]
|}

</p>
</div>
</div>

<div id="Horizon50"></div>
=== Problem 11 ===
Draw the conformal diagram for open and flat Universes with power-law scale factor $a(t)\sim t^{n}$, with $n>1$. This is the model for the power-law inflation. Check whether the strong energy condition is satisfied.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As seen in the previous problem, if strong energy condition were satisfied, we would have $\dot{a}\sim a^{-\theta}$ with some positive $\theta$; this is not the case, so the condition is violated. As $t\in (0,+\infty)$,
\begin{equation}
\eta\sim \int \frac{dt}{t^n}
\end{equation}
diverges at small $a$ (thus also small $t$) and converges at $a\to \infty$ (thus as large $t$). So integration constant can be chosen so that $\eta\in (-\infty,0)$.

The conformal structure is the same as \emph{lower half} of Minkowski spacetime. The cut is regular future infinity, and from Minkowski there are spacelike infinity, past null infinity, and past infinity. The point of past infinity corresponds to Big Bang and is singular.
</p>
</div>
</div>

<references>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN~0521563984</ref>
</references>

File:Conf-dS-closed.png

2013-12-08T17:35:23Z

Igor: The de Sitter Universe in closed slicing: conformal diagram

The de Sitter Universe in closed slicing: conformal diagram

Causal Structure

2013-12-08T17:34:26Z

Igor: /* Problem 2 */

[[Category:Horizons|4]]

__TOC__

The causal structure is determined by propagation of light and is best understood in terms of ''conformal diagrams''. In this section we construct and analyze those for a number of important model cosmological solutions (which are assumed to be already known), following mostly the exposition of <ref name=Mukhanov/>.

In terms of comoving distance $\tilde{\chi}$ and conformal time $\tilde{\eta}$ (in this section they are denoted by tildes) the two-dimensional radial part of the FLRW metric takes form
\begin{equation}
ds_{2}^{2}=a^{2}(\tilde{\eta})\big[d\tilde{\eta}^2 -d\tilde{\chi}^2\big].
\label{ds2Dconf}
\end{equation}
In the brackets here stands the line element of two-dimensional Minkowski flat spacetime. Coordinate transformations that preserve the \emph{conformal} form of the metric
\begin{equation*}
ds_2^2 =\Omega^{2}(\eta,\chi)\big[d\eta^2 -d\chi^2 \big],
\end{equation*}
are called conformal transformations, and the corresponding coordinates $(\eta,\chi)$ -- conformal coordinates.

<div id="Horizon39"></div>
=== Problem 1 ===
Show that it is always possible to construct $\eta(\tilde{\eta},\tilde{\chi})$, $\chi(\tilde{\eta},\tilde{\chi})$, such that the conformal form of metric (\ref{ds2Dconf}) is preserved, but $\eta$ and $\chi$ are bounded and take values in some finite intervals. Is the choice of $(\eta,\chi)$ unique?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Suppose $\tilde{\eta},\tilde{\chi}$ span infinite or semi-infinite values. Then we can always make the following sequence of coordinate transformations:
<div style="text-align: left;">
* Pass to null coordinates
\begin{equation}
u=\tilde{\eta}-\tilde{\chi},\qquad v=\tilde{\eta}+\tilde{\chi};
\end{equation}
* Bring their range of values to a finite interval by some appropriate function, i.e.
\begin{equation}
U=\arctan u,\qquad V=\arctan v.
\end{equation}
* Go back to timelike and spacelike coordinates (this is not really necessary at this point and is done mostly for aesthetic reasons):
\begin{equation}
T=V+U,\qquad R=V-U.
\end{equation}
Now the range of $(T,R)$ obviously covers some bounded region on the plane, while the radial part of the line element preserves its conformal form:
\begin{equation}
ds_2^2 \sim d\tilde{\eta}^2-d\tilde{\chi}^2 \sim du\, dv \sim dU\, dV \sim dT^2 -dR^2 .
\end{equation}
As the choice of function $\arctan$ was rather arbitrary (though convenient), the choice of conformal coordinates is not unique.
</div>
</p>
</div>
</div>

In this section we will reserve notation $\eta$ and $\chi$ and name "conformal coordinates/variables" to such variables that can only take values in a bounded region on $\mathbb{R}^2$; $\tilde{\eta}$ and $\tilde{\chi}$ can span infinite or semi-infinite intervals. Spacetime diagram in terms of conformal variables $(\eta,\chi)$ is called conformal diagram. Null geodesics $\eta=\pm \chi + const$ are diagonal straight lines on conformal diagrams.

<div id="Horizon40"></div>
=== Problem 2 ===
Construct the conformal diagram for the closed Universe filled with
* radiation;
* dust;
* mixture of dust and radiation.

Show the particle and event horizons for the observer at the origin $\chi=0$ (it will be assumed hereafter that the horizons are always constructed with respect to this chosen observer).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* Solution for $a(\eta)$ in a radiation dominated closed Universe is
\begin{equation}
a=a_m \sin\eta ,\qquad \eta\in(0,\pi),\quad \chi \in [0,\pi].
\end{equation}
As the ranges spanned by $\eta$ and $\chi$ are finite, they are already conformal coordinates. The conformal diagram is a square $\eta,\chi\in [0,\pi]$. Edges $\eta=0$ and $\eta=\pi$ correspond to the Big Bang and Big Crunch singularities respectively; worldline $\chi=\pi$ is the point on the three-sphere that is situated at the opposite pole with respect to observer at $\chi=0$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" | Conformal diagrams of radiation (left) and dust (right) dominated closed Universes. Particle horizon is shown in blue, event horizon in red. Thin lines in the dust dominated universe show light rays that realize the first and second images of a galaxy at different cosmic times, from opposite directions.
|[[File:Conf-Rad.png|center|thumb|365px]]
|[[File:Conf-Dust.png|center|thumb|365px]]
|}

The particle horizon is
\[\eta=\chi,\]
and event horizon is
\[\eta=\eta_{max}-\chi =\pi -\chi.\]
Both exist for all cosmological times: by the finite moment of the Big Crunch $\eta=\pi$ the event horizon is collapsed into a point (which is natural, as there is no more time left), while the particle horizon extends to the whole Universe. Thus only at the finite moment the whole of the Universe becomes observable. The farther the point, though, the younger will it look, of course, and the opposite pole will only be "observed" by our observer at the last moment of the Universe as it was at its creation.

* Solution for $a(\eta)$ in a dust dominated closed Universe is
\begin{equation}
a=a_m (1-\cos\eta) ,\qquad \eta\in(0,2\pi),\quad \chi \in [0,\pi].
\end{equation}
The difference from the previous case is that $\eta$ spans twice the range, and $\eta_{max}=2\pi =2\chi_{max}$.

Therefore the event horizon is given by
\[\eta=\eta_{max}-\chi=2\pi -\chi,\]
so it exists only in the second, contracting, phase $\eta>\pi$. The particle horizon is given by $\eta=\chi$ again, but now it exists only during the expanding phase $\eta<\pi$. It encloses the full Universe at the moment of maximal expansion $\eta=\pi$ and for later times does not exist.

* Though the full analytic solution is more complicated, it is clear that the main features remain the same as in the previous considered cases. At early and late times, close to the singularities, the dynamics is determined by the radiation component. If there is enough dust, then at large enough scale factors (which may or may not be achieved depending on the initial conditions), which would correspond to the epoch around the maximal expansion, it will be dominating. The influence of dust is that dynamics is slowed down, so that depending on the ratio of densities
\[\eta_{max}\in [\pi,2\pi].\]
Thus qualitatively the picture will be the same as in a dust dominated Universe: the conformal diagram is a rectangle, the event horizon exists only starting from some time $\eta_{e}=\eta_{max}-\pi$, while particle horizon, on the contrary, vanishes at $\eta_{p}=\pi$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" | Conformal diagram for a closed Universe filled with a mix of dust and radiation.
|[[File:Conf-RadDust.png|center|thumb|365px]]
|}
</div>
</p>
</div>
</div>

<div id="Horizon41"></div>

=== Problem 3 ===
'''Closed dS.''' Construct the conformal diagram for the de Sitter space in the closed sections coordinates. Provide reasoning that this space is (null) geodesically complete, i.e. every (null) geodesic extends to infinite values of affine parameters at both ends.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The scale factor in the closed dS Universe is
\begin{equation}
a(t)=H_\Lambda^{-1}\cosh (H_\Lambda t),\qquad t\in (-\infty, +\infty),
\end{equation}
so on integration, for conformal time we obtain
\begin{equation}
\eta (t)=\int\limits_{-\infty}^{t}\frac{dt}{a(t)}=\arcsin\big[\tanh (H_\Lambda t)\big]+\frac{\pi}{2} \in (0,\pi).
\end{equation}
We choose the integration constant here so that $\eta=0$ corresponds to $t=-\infty$ and $\eta=\pi$ to $t=+\infty$. The full metric then can be written as
\begin{equation}
ds^{2}_{dS}=\frac{H_\Lambda^2}{\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\sin^2 \chi d\Omega^2 \big]. \label{dSclosed}
\end{equation}

As the values of $\eta$ span a finite interval, $(\eta,\chi)$ are already conformal coordinates. The conformal diagram is again a square
\begin{equation}
\eta\in [0,\pi],\qquad \chi \in[0,\pi],
\end{equation}
with the difference from the radiation dominated Universe that the edges $\eta=0,\pi$ do not represent a singularity anymore, but instead correspond to infinite (and regular) past and future respectively. Both horizons are given again by
\begin{equation}
\eta_{e}=\pi-\chi,\qquad \eta_{p}=\chi
\end{equation}
and exist at all times.

The spacelike boundaries of the conformal diagram correspond to $t\to \pm\infty$, and therefore to infinite values of affine parameter. This can be shown if one remembers the general formula for the cosmological redshift:
\begin{equation}
\text{const}=\omega a =\frac{dt}{d\lambda}a,\quad \Rightarrow\quad
\lambda=\text{const}\cdot\int\limits^{t}dt \cosh(H_\Lambda t)
\underset{t\to\pm\infty}{\longrightarrow}\infty .
\end{equation}
The timelike boundary of the diagram corresponds to the opposite pole, there is no real boundary there, the same as on a sphere: as particles propagate across the pole, their radial coordinate begins to decrease again, while the worldline on the conformal diagram is reflected from $\chi=\pi$. Thus by definition the spacetime is (null) geodesically complete.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-closed.png|center|thumb|365px|The de Sitter Universe. The closed sections coordinates cover the whole space, which is geodesically complete. There are no singularities: the horizontal boundaries of the diagram correspond to infinite past and future.]]
|}
</p>
</div>
</div>

<div id="Horizon42"></div>
=== Problem 4 ===
'''Static dS.''' Rewrite the metric of de Sitter space (\ref{dSclosed}) in terms of "static coordinates" $T,R$:
\begin{equation}
\tanh (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad
H_\Lambda R =\frac{\sin\chi}{\sin\eta}.
\end{equation}
* What part of the conformal diagram in terms of $(\eta,\chi)$ is covered by the static coordinate chart $(T,R)$?
* Express the horizon's equations in terms of $T$ and $R$
* Draw the surfaces of constant $T$ and $R$ on the conformal diagram.
* Write out the coordinate transformation between $(\eta,\chi)$ and $(T,R)$ in the regions where $|\cos\eta|>|\cos\chi|$. Explain the meaning of $T$ and $R$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us introduce dimensionless coordinates $t=H_\Lambda T$ and $r=H_\Lambda R$. The inverse relations then are
\begin{equation}
\sin^2 \eta =\frac{1}{\cosh^2 t-r^2 \sinh^2 t},
\quad \sin^2 \chi =\frac{r^2}{\cosh^2 t-r^2 \sinh^2 t}.
\end{equation}
Using them, after some algebra from (\ref{dSclosed}) we get
\begin{equation}
ds_{dS}^2 =\big[1-H_\Lambda^2 R^2\big]dT^2 -\frac{dR^2}{1-H_\Lambda^2 R^2}-R^2 d\Omega^2 , \label{dSstatic}
\end{equation}
which resembles the Schwarzschild line element (and this is not a coincidence).

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-static.png|center|thumb|365px|The de Sitter Universe with contour lines of the "static" coordinates $(T,R)$. The solid lines are $T=const$, and the dashed ones $R=const$. The coordinates become singular on both horizons, so the whole spacetime is divided by them into four sectors, each separately covered by the regular coordinate chart $(R,T)$. Spacetime is actually static only in the left and right sectors (I and III), where $T$ is a timelike coordinate and $R$ spacelike: the static patches are bounded by the horizons.]]
|}

<div style="text-align: left;>
* As $|\tanh (H_\Lambda T)|<1$, we have $|\cos\eta| \leq |\cos\chi|$. This condition cuts out two out of four sectors from the square conformal diagram, I and III: one is $\eta\in [\chi, \pi-\chi]$ and the other is $\eta\in [\pi-\chi, \chi]$. In both $|\sin\eta|\geq |\sin\chi|$, so $R \leq H_{\Lambda}^{-1}$;
* The particle horizon $\eta=\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=-\infty$. It is one part of the boundary of the region (in two parts) covered by coordinates $(T,R)$. The event horizon $\eta=\pi-\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=+\infty$ and is the other part of the boundary of this region.
* $T=const$ is $\cos\eta =t\cos\chi $ and $R=const$ is $\sin\eta =r^{-1}\sin\chi$.
* In regions II and IV the needed relation is obtained if we simply replace $\tanh$ with $\coth$ in the first relation:
\begin{equation}
\coth (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad H_\Lambda R=\frac{\sin\chi}{\sin\eta},
\end{equation}
as can be checked explicitly by substitution into (\ref{dSstatic}), which again gives (\ref{dSclosed}). In these regions $R$ is a timelike coordinate, and $T$ is spacelike. The geodesics of comoving massive particles are $\chi=const$, one of them $\chi=\pi/2$ corresponds to $T=0$. Thus in the lower part of the diagram, where $R\in (+\infty,H_\Lambda^{-1})$, the spacetime is contracting; in the upper part, where $R\in (H_\Lambda^{-1},\infty)$, it is expanding. The coordinate frame is not static. The relations between $(T,R)$ and $(\eta,\chi)$ in static and non-static regions mirros those in Schwarzshild black hole solution between the static and the global (Kruskal-Szekeres) coordinates. Here $\eta$ and $\chi$ are the global coordinates.
</div>
</p>
</div>
</div>

<div id="Horizon43"></div>
=== Problem 4 ===
'''Flat dS.''' The scale factor in flat de Sitter is $a(t)=H_\Lambda^{-1} e^{H_\Lambda t}$.

* Find the range of values spanned by conformal time $\tilde{\eta}$ and comoving distance $\tilde{\chi}$ in the flat de Sitter space
* Verify that coordinate transformation
\begin{equation}
\tilde{\eta}=\frac{-\sin\eta}{\cos\chi-\cos\eta},\qquad
\tilde{\chi}=\frac{\sin\chi}{\cos\chi-\cos\eta}
\end{equation}
bring the metric to the form of that of de Sitter in closed slicing (it is assumed that $\tilde{\eta}=0$ is chosen to correspond to infinite future).
* Which part of the conformal diagram is covered by the coordinate chart $(\tilde{\eta},\tilde{\chi})$? Is the flat de Sitter space geodesically complete?
* Where are the particle and event horizons in these coordinates?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As $t\in(-\infty,+\infty)$,
\begin{equation}
\tilde{\eta} =\int\frac{dt}{a(t)}=H_\Lambda \int\limits_{+\infty}^{t}dt\; e^{-H_\Lambda t} =-e^{-H_\Lambda t} \in (-\infty,0).
\end{equation}
Here we choose $+\infty$ as the lower limit, because at $-\infty$ the integral diverges.
<div style="text-align: left;>
* Direct calculation yields (\ref{dSclosed}), with $\eta \in (0,\pi)$, $\chi\in (0,\pi)$;
* The upper triangle $\eta>\chi$, above the particle horizon, on which $\tilde{\eta}\to -\infty$. It is not geodesically complete, as geodesics are cut at the particle horizon.
* The particle horizon is the boundary of the patch of full dS space covered by flat slicing coordinates, and event horizon in these coordinates exists only in the latter part of evolution, for $\eta >\pi/2$.
</div>
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-flat.png|center|thumb|365px|The de Sitter Universe in flat sections' coordinates, which cover only half of it. The boundary -- the particle horizon -- consists of three different infinities.]]
|}
</p>
</div>
</div>

<div id="Horizon44"></div>
=== Problem 5 ===
'''Infinities.''' What parts of the spacetime's boundary on the conformal diagram of flat de Sitter space corresponds to

* spacelike infinity $i^0$, where $\tilde{\chi}\to +\infty$;
* past timelike infinity $i^-$, where $\tilde{\eta}\to -\infty$ and from which all timelike worldlines emanate
* past lightlike infinity $J^-$, from which all null geodesics emanate?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* The one point at the left of the diagram;
* the one point at the bottom;
* the particle horizon.
</div>
</p>
</div>
</div>

<div id="Horizon45"></div>
=== Problem 6 ===
'''Open dS.''' Consider the de Sitter space in open slicing, in which $a(t)=H_\Lambda \sinh (H_\Lambda t)$, so conformal time is
\begin{equation}
\tilde{\eta} =\int\limits_{+\infty}^{t} \frac{dt}{a(t)},
\end{equation}
where again the lower limit is chosen so that the integral is bounded.

* Find $\tilde{\eta}(t)$ and verify that coordinate transformation from $(\tilde{\eta},\tilde{\chi})$ to $\eta,\chi$, such that
\begin{equation}
\tanh\tilde{\eta}=\frac{-\sin\eta}{\cos\cos\chi},\qquad
\tanh\tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
brings the metric to the form of de Sitter in closed slicing.
* What are the ranges spanned by $(\tilde{\eta},\tilde{\chi})$ and $(\eta,\chi)$? Which part of the conformal diagram do they cover?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
After getting
\begin{equation}
\sinh \tilde{\eta}=-\frac{1}{\sinh(H_\Lambda t)},
\end{equation}
the first part is checked straightforwardly; in the open de Sitter $\tilde{\chi}\in [0,+\infty)$, and $\tilde{\eta}\in(-\infty,0)$. The region covered by coordinates $(\tilde{\eta},\tilde{\chi})$ is $\{\eta>\chi+\pi/2\}$, only one eighth part of the full diagram.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-open.png|center|thumb|365px|The de Sitter Universe in open sections' coordinates, which cover only $1/8^{\text{th}}$ of the full diagram.]]
|}
</p>
</div>
</div>

<div id="Horizon46"></div>
=== Problem 7 ===
'''Minkowski 1.''' Rewrite the Minkowski metric in terms of coordinates $(\eta,\chi)$, which are related to $(t,r)$ by the relation
\begin{equation}
\tanh \tilde{\eta}=\frac{\sin\eta}{\cos\chi},\qquad
\tanh \tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
that mirrors the one between the open and closed coordinates of de Sitter. Construct the conformal diagram and determine different types of infinities. Are there new ones compared to the flat de Sitter space?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Coordinate transformation gives
\begin{equation}
ds^2 =\frac{1}{\cos^2 \chi -\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\Psi^{2}(\eta,\chi)d\Omega^2 \big].
\end{equation}
Here $r\in [0,+\infty)$ and $t\in (-\infty,+\infty)$. Comparing with the relation between $(\tilde{\eta},\tilde{\chi})$ with conformal coordinates $(\eta,\chi)$ in the open de Sitter universe, where $\tilde{\eta}\in (-\infty,0)$, we see that the difference is that $\tilde{\eta}$ spans $(-\infty,0)$, while now $t$ spans twice the range, $(-\infty,\infty)$. Therefore the conformal diagram is composed of two triangles, one the same as for open de Sitter and one for its time-reversed copy. Accordingly there now appear future timelike infinity $i^+$ and future lightlike infinity $J^+$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |The Minkowski spacetime in two different pairs of conformal coordinates and the full set of infinities. Thin dashed and solid lines show the images of coordinate grid $(t,r)$.

|[[File:Conf-Mink1.png|center|thumb|365px]]
|[[File:Conf-Mink2.png|center|thumb|365px]]
|}
</p>
</div>
</div>

<div id="Horizon47"></div>
=== Problem 8 ===
'''Minkowski 2.''' The choice of conformal coordinates is not unique. Construct the conformal diagram for Minkowski using the universal scheme: first pass to null coordinates, then bring their span to finite intervals with $\arctan$ (one of the possible choices), then pass again to timelike and spacelike coordinates.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
We start from spherical coordinates
\begin{equation}
ds^2 =dt^2 -dr^2 -r^2 d\Omega^2 .
\end{equation}
<div style="text-align: left;">
* The first step is introducing null coordinates
\begin{equation}
u=t-r,\quad v=t+r ,
\end{equation}
so that
\begin{equation}
ds^2 =4du\,dv -\frac{(v-u)^2}{4}d\Omega^2 .
\end{equation}
* Then bring the range of values to finite intervals
\begin{equation}
U=\arctan u, \qquad V=\arctan v,
\end{equation}
so that
\begin{equation}
ds^2 =\frac{1}{4\cos^2 U \cos^2 V}\big[4dU\,dV -\sin^2 (V-U)d\Omega^2\big].
\end{equation}
The whole spacetime is simply the half of the square $U,V\in (-\pi/2,\pi/2)$, in which $r>0$, i.e. $v>u\quad \Leftrightarrow\quad V>U$: on the plane $(U,V)$ it is the triangle \begin{equation}
-\pi/2<U<V<\pi/2.
\end{equation}
* Finally, go back to spacelike and timelike coordinates
\begin{equation}
T=V+U,\qquad R=V-U
\end{equation}
so that metric becomes
\begin{equation}
ds^2 =\frac{1}{[\cos T +\cos R]^2}\big[dT^2 -dR^2 -\sin^2 R\; d\Omega^2\big].
\end{equation}
The triangle is shrunken by $\sqrt{2}$ and rotated by $3\pi/4$ clockwise, thus turning into
\begin{equation}
\{R>0,\quad |T|<\pi/2 -R\}.
\end{equation}
This is the same form as obtained by the other construction (up to scaling, which is purely decorative).
</div>
</p>
</div>
</div>

<div id="Horizon48"></div>
=== Problem 9 ===
Draw the conformal diagram for the Milne Universe and show which part of Minkowski space's diagram it covers.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Minkowski metric $ds^2 =dT^2 -dR^2$, rewritten in terms of $(\tau, r)$ such that
\begin{equation}
T=\tau \cosh r ,\quad R=\tau \sinh r ,
\end{equation}
is the metric of the Milne Universe. As Minkowski space is complete, the Milne Universe then is a \emph{part} of Minkowski in different variables. This part is where $T>R$. The boundary $T=R$ is the future light cone, so the whole Milne Universe is represented by the triangle homothetic to the whole space but 4 times smaller in area, with the common node of future infinity.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Milne.png|center|thumb|365px|The Milne Universe on Minkowski's conformal diagram. It has the same shape and shares the same future infinity, but covers one eighth of the area. The boundary is the past horizon.]]
|}
</p>
</div>
</div>

<div id="Horizon49"></div>
=== Problem 10 ===
Consider open or flat Universe filled with matter that satisfies strong energy condition $\varepsilon +3p>0$. What are the coordinate ranges spanned by the comoving coordinate $\tilde{\chi}$ and conformal time $\tilde{\eta}$? Compare with the Minkowski metric and construct the diagram. Identify the types of infinities and the initial Big Bang singularity.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For flat Universe $\tilde{\chi}=r$, for the open $\tilde{\chi}=\sinh r$, so in both cases $\tilde{\chi}\in (0,+\infty)$.

Strong energy condition implies that $w>-1/3$, so
\begin{equation}
\rho \sim a^{-3(1+w)}=a^{-n},
\end{equation}
where $n>2$. From the first Friedman equation then after simple manipulations we obtain that
\begin{equation}
\frac{da}{dt}\sim a^{-\theta},
\end{equation}
where $\theta$ is some positive number. Therefore both
\begin{equation}
t\sim \int da\; a^{\theta},
\quad\text{and}\quad
\eta =\int \frac{da}{a}a^{\theta}
\end{equation}
converge at $a\to 0$ and diverge at $a\to \infty$. Consequently, the integration constant can be chosen so that $\eta\in (0,+\infty)$.

The conformal structure is the same as that of the \emph{upper} half of Minkowski spacetime. The Big Bang singularity at $\eta=0$ is at the cut, and there are spacelike infinity, future infinity and future null infinity.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |Conformal diagrams for open or flat Universes. The one on the left is for ones filled with matter which satisfies the strong energy condition (SEC), and the one on the right for ones in which SEC does not hold, such as in case of power-law inflation..

|[[File:Conf-SEC.png|center|thumb|365px]]
|[[File:Conf-Inflation.png|center|thumb|365px]]
|}

</p>
</div>
</div>

<div id="Horizon50"></div>
=== Problem 11 ===
Draw the conformal diagram for open and flat Universes with power-law scale factor $a(t)\sim t^{n}$, with $n>1$. This is the model for the power-law inflation. Check whether the strong energy condition is satisfied.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As seen in the previous problem, if strong energy condition were satisfied, we would have $\dot{a}\sim a^{-\theta}$ with some positive $\theta$; this is not the case, so the condition is violated. As $t\in (0,+\infty)$,
\begin{equation}
\eta\sim \int \frac{dt}{t^n}
\end{equation}
diverges at small $a$ (thus also small $t$) and converges at $a\to \infty$ (thus as large $t$). So integration constant can be chosen so that $\eta\in (-\infty,0)$.

The conformal structure is the same as \emph{lower half} of Minkowski spacetime. The cut is regular future infinity, and from Minkowski there are spacelike infinity, past null infinity, and past infinity. The point of past infinity corresponds to Big Bang and is singular.
</p>
</div>
</div>

<references>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN~0521563984</ref>
</references>

File:Conf-RadDust.png

2013-12-08T17:31:30Z

Igor: Conformal diagram for a closed Universe filled with a mixture of radiation and dust

Conformal diagram for a closed Universe filled with a mixture of radiation and dust

File:Conf-Dust.png

2013-12-08T17:30:23Z

Igor: Conformal diagram for a closed Universe with dominating non-relativistic matter (dust)

Conformal diagram for a closed Universe with dominating non-relativistic matter (dust)

File:Conf-Rad.png

2013-12-08T17:29:29Z

Igor: Conformal diagram for the closed Universe with dominating relativistic matter (radiation)

Conformal diagram for the closed Universe with dominating relativistic matter (radiation)

Kerr black hole

2013-12-08T14:38:07Z

Igor: /* Problem 4: some more simple algebra */

[[Category:Black Holes|3]]

Kerr solution$^{*}$ is the solution of Einstein's equations in vacuum that describes a rotating black hole (or the metric outside of a rotating axially symmetric body) . In the Boyer-Lindquist coordinates$^{**}$ it takes the form
\begin{align}\label{Kerr}
&&ds^2=\bigg(1-\frac{2\mu r}{\rho^2}\bigg)dt^2
+\frac{4\mu a \,r\;\sin^{2}\theta}{\rho^2}
\;dt\,d\varphi
-\frac{\rho^2}{\Delta}\;dr^2-\rho^2\, d\theta^2
+\qquad\nonumber\\
&&-\bigg(
r^2+a^2+\frac{2\mu r\,a^2 \,\sin^{2}\theta}{\rho^2}
\bigg) \sin^2 \theta\;d\varphi^2;\\
\label{Kerr-RhoDelta}
&&\text{where}\quad
\rho^2=r^2+a^2 \cos^2 \theta,\qquad
\Delta=r^2-2\mu r+a^2.
\end{align}
Here $\mu$ is the black hole's mass, $J$ its angular momentum, $a=J/\mu$; $t$ and $\varphi$ are time and usual azimuth angle, while $r$ and $\theta$ are some coordinates that become the other two coordinates of the spherical coordinate system at $r\to\infty$.

$^{*}$ R.P. Kerr, Gravitational field of a spinning mass as an example of algebraically special metrics. ''Phys. Rev. Lett.'' '''11''' (5), 237 (1963).

$^{**}$ R.H. Boyer, R.W. Lindquist. Maximal Analytic Extension of the Kerr Metric. ''J. Math. Phys'' '''8''', 265–281 (1967).

__TOC__

==General axially symmetric metric==
A number of properties of the Kerr solution can be understood qualitatively without use of its specific form. In this problem we consider the axially symmetric metric of quite general kind
\begin{equation}\label{AxiSimmMetric}
ds^2=A dt^2-B(d\varphi-\omega dt)^{2}-
C\,dr^2-D\,d\theta^{2},\end{equation}
where functions $A,B,C,D,\omega$ depend only on $r$ and $\theta$.

<div id="BlackHole53"></div>
=== Problem 1: preliminary algebra ===
Find the components of metric tensor $g_{\mu\nu}$ and its inverse $g^{\mu\nu}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The metric is:
\begin{equation}\label{AxiSimmMetricmatrix}
g_{\mu\nu}=\begin{pmatrix}
A-\omega^2 B&0&0&\omega B\\
0&-C&0&0\\
0&0&-D&0\\
\omega B&0&0&-B
\end{pmatrix}.
\end{equation}
Taking into account the structure of $g_{\mu\nu}$, for the inverse matrix we get
\begin{align*}
&g^{rr}=\frac{1}{g_{rr}};\quad
g^{\theta\theta}=\frac{1}{g_{\theta\theta}};\\
&g^{tt}=
\frac{g_{\varphi\varphi}}{|G|};\quad
g^{\varphi\varphi}=\frac{g^{tt}}{|G|};
\quad g^{t\varphi}
=-\frac{g_{t\varphi}}{|G|};\\
&\text{where}\quad
G=g_{tt}g_{\varphi\varphi}-g_{t\varphi}^{2}.
\end{align*}
Using the explicit expression for $g_{\mu\nu}$, we see that $G=-AB$ and thus finally
\begin{equation}\label{AxiSimmMetricInvmatrix}
g^{\mu\nu}=\begin{pmatrix}
1/A&0&0&\omega/A\\
0&-1/C&0&0\\
0&0&-1/D&0\\
\omega/A&0&0&\frac{\omega^2 B-A}{AB}
\end{pmatrix}
\end{equation} </p>
</div>
</div>

<div id="BlackHole54"></div>
=== Problem 2: integrals of motion ===
Write down the integrals of motion corresponding to Killing vectors $\boldsymbol{\xi}_{t}=\partial_t$ and $\boldsymbol{\xi}_{\varphi}=\partial_\varphi$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
A particle's integrals of motion are
\begin{equation}\label{AxiSimm-Integrals}
\boldsymbol{u}\cdot\boldsymbol{\xi}_t=u_{t};
\quad
\boldsymbol{u}\cdot\boldsymbol{\xi}_{\varphi}=u_{\varphi}.\end{equation}
Energy and angular momentum are defined the same way as in the Schwarzshild case
\[E=mc^{2}u_{t};\qquad L=-m u_\varphi.\] </p>
</div>
</div>

<div id="BlackHole55"></div>

=== Problem 3: Zero Angular Momentum Observer/particle ===
Find the coordinate angular velocity $\Omega=\tfrac{d\varphi}{dt}$ of a particle with zero angular momentum $u_{\mu}(\partial_{\varphi})^{\mu}=0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a particle moving in the axially symmetric field
\begin{align*}
u^{t}=&g^{t\mu}u_{\mu}=
g^{tt}u_{t}+g^{t\varphi}u_{\varphi};\\
u^{\varphi}=&g^{\varphi\mu}u_{\mu}=
g^{\varphi t}u_{t}+g^{\varphi\varphi}u_{\varphi}.
\end{align*}

Then for a particle with zero angular momentum (ZAMO) $u_{\varphi}=0$ we get
\[u^{t}=g^{tt}u_{t};
\quad u^{\varphi}=g^{t\varphi}u_{t},\]
and therefore its angular velocity is
\[\frac{d\varphi}{dt}
=\frac{d\varphi/ds}{dt/ds}=
\frac{u^{\varphi}}{u^t}=
\frac{g^{t\varphi}u_t}{g_{tt}u_t}=
\frac{\omega/A}{1/A}=\omega(r,\theta).\]
Now we see the physical meaning of the quantity $\omega(r,\theta)$. </p>
</div>
</div>

<div id="BlackHole55plus"></div>

=== Problem 4: some more simple algebra ===
Calculate $A,B,C,D,\omega$ for the Kerr metric.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us introduce notation
\begin{align*}
\Sigma^{2}=&(r^2+a^2)^{2}-a^{2}\Delta \sin^{2}\theta,\\
&\Delta=r^2 -2\mu r +a^2 ,
\end{align*}
so that for the Kerr metric $g_{\varphi\varphi}=-\tfrac{\Sigma^2}{\rho^2}\sin^{2}\theta$. After some straightforward calculations then we obtain
\begin{equation}\label{Kerr-ABCD}
A=\frac{\Delta \rho^2}{\Sigma^2},\quad
B=\frac{\Sigma^2}{\rho^2}\sin^2 \theta,\quad
C=\frac{\rho^2}{\Delta},\quad
D=\rho^2,\quad
\omega=\frac{2\mu ra}{\Sigma^2}.
\end{equation} </p>
</div>
</div>

==Limiting cases==
<div id="BlackHole56"></div>
=== Problem 5: Schwarzshild limit ===
Show that in the limit $a\to 0$ the Kerr metric turns into Schwarzschild with $r_{g}=2\mu$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For $a=0$ we'll have $\rho^2=r^2$ and $\Delta=r^{2}h(r)$. On substituting this into the Kerr metric, we see it is reduced to the Schwarzshild one. </p>
</div>
</div>

<div id="BlackHole57"></div>
=== Problem 6: Minkowski limit ===
Show that in the limit $\mu\to 0$ the Kerr metric describes Minkowski space with the spatial part in coordinates that are related to Cartesian as
\begin{align*}
&x=\sqrt{r^2+a^2}\;\sin\theta\cos\varphi,
\nonumber\\
&y=\sqrt{r^2+a^2}\;\sin\theta\sin\varphi,\\
&z=r\;\cos\theta\nonumber,\\
&\text{where}\quad
r\in[0,\infty),\quad
\theta\in[0,\pi],\quad \varphi\in[0,2\pi).\nonumber
\end{align*}
Find equations of surfaces $r=const$ and $\theta=const$ in coordinates $(x,y,z)$. What is the surface $r=0$?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the limit $\mu\to0$ we get $\Delta=a^2+r^2$ and
\begin{align}\label{KerrM=0}
ds^2&=dt^2-\frac{\rho^2}{\Delta}\;dr^2-
\rho^2\, d\theta^2-
\big(r^2+a^2\big) \sin^2 \theta\;d\varphi^2=
\nonumber\\
&=dt^2-\frac{\rho^2}{r^2+a^2}\;dr^2-
\rho^2\, d\theta^2-
\big(r^2+a^2\big) \sin^2 \theta\;d\varphi^2,\\
&\text{where}\quad \rho^2=r^2+a^2 \cos^2 \theta.
\nonumber
\end{align}
On the other hand, the Euclidean line element $ds^{2}_{eu}=dx^{2}+dy^{2}+dz^{2}$ takes the same form in spherical coordinates $(x,y,z)\to(r,\theta,\varphi)$.

Surfaces $r=const$ are ellipsoids of revolution around the z axis
\[\frac{x^2+y^2}{r^2+a^2}+\frac{z^2}{r^2}=1,\]
and the special case $r=0$ corresponds to a disk of radius $a$ in the equatorial plane. Surfaces $\theta=const$ are hyperboloids of revolution around the same axis
\[\frac{x^2+y^2}{\sin^{2}\theta}-
\frac{z^2}{\cos^2\theta}=a^2.\]
In the limit $a/r\to 0$ this coordinate system turns into the spherical one. </p>
</div>
</div>

<div id="BlackHole58"></div>
=== Problem 7: weak field rotation effect ===
Write the Kerr metric in the limit $a/r \to 0$ up to linear terms.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the zeroth order it is the Schwarzshild metric, and terms linear by $a$ are only present in $g_{t\varphi}$, thus
\begin{align}\label{KerrAsymp}
ds^2=\Big(1-\frac{2\mu}{r}\Big)dt^2-
\Big(1-\frac{2\mu}{r}\Big)^{-1}dr^2-r^2d\Omega^2
+\frac{4a\mu}{r}\sin^{2}\theta\,dt\,d\varphi
+O(a^2)=\nonumber\\
=ds^{2}_{Schw}+2r\,dt\,\omega_{\infty}\! d\varphi,
\quad\text{where}\quad
\omega_{\infty}=\frac{2a\mu}{r^3}\sin^{2}\theta.
\end{align} </p>
</div>
</div>

==Horizons and singularity==
Event horizon is a closed null surface. A null surface is a surface with null normal vector $n^\mu$:
\[n^{\mu}n_{\mu}=0.\]
This same notation means that $n^\mu$ belongs to the considered surface (which is not to be wondered at, as a null vector is always orthogonal to self). It can be shown further, that a null surface can be divided into a set of null geodesics. Thus the light cone touches it in each point: the future light cone turns out to be on one side of the surface and the past cone on the other side. This means that world lines of particles, directed in the future, can only cross the null surface in one direction, and the latter works as a one-way valve, -- "event horizon"
<div id="BlackHole59"></div>
=== Problem 8: on null surfaces ===
Show that if a surface is defined by equation $f(r)=0$, and on it $g^{rr}=0$, it is a null surface.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The normal vector $n_\mu$ to a surface $f(x)=0$ is directed along $\partial_{\mu}f$. It is null if
\[g^{\mu\nu}(\partial_{\mu}f)(\partial_{\nu}f)=0.\]

The normal to the surface $f(r)=0$ is directed along $\partial_{r}f$, i.e. $\partial_{\mu}f\sim \delta_{\mu}^{r}$ and the null condition takes the form
\[0=g^{\mu\nu}\delta^{r}_{\mu}\delta^{r}_{\nu}
=g^{rr}.\] </p>
</div>
</div>

<div id="BlackHole60"></div>
=== Problem 9: null surfaces in Kerr metric ===
Find the surfaces $g^{rr}=0$ for the Kerr metric. Are they spheres?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Equations of surfaces, on which $g^{rr}$ terms to zero and $g_{rr}$ to infinity, are
\begin{align}
\Delta=0\quad\Leftrightarrow\quad
r^2-2\mu r+a^2=0\quad\Leftrightarrow\quad
r=r_{\pm},\quad\text{where}\nonumber\\
\label{Kerr-Rhor+-}
r_{\pm}\equiv\mu\pm\sqrt{\mu^2-a^2}.
\end{align}

Although those are surfaces of constant $r$, their intrinsic metric is not spherical. Plugging $r=r_{\pm}$ into the spatial section $dt=0$ of the Kerr metric (\ref{Kerr}) ans using the relation $r_{\pm}^{2}+a^2=2\mu r_{\pm}$, which holds on the surfaces $r=r_\pm$, we obtain
\[dl^{2}_{r=r_\pm}=
\rho^{2}_{\pm}d\theta^{2}+
\Big(\frac{2\mu r_\pm}{\rho_\pm}\Big)^2
\sin^{2}\theta \,d\varphi^2
=\rho_{\pm}^{2}
\big(d\theta^2 +\sin^{2}\theta d\varphi^{2} \big)
+2a^{2}(r^2 +a^2 \cos^{2}\tfrac{\theta}{2})
\sin^{4}\theta\,d\varphi^{2}.\]
The first terms is the metric of a sphere, while the second gives additional positive contribution to the distance measured along $\varphi$. Thus if we embedded such a surface into a three-dimensional Euclidean space, we'd get something similar to an oblate ellipsoid of rotation. </p>
</div>
</div>

<div id="BlackHole61"></div>
=== Problem 10: horizon area ===
Calculate surface areas of the outer and inner horizons.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The horizon's area is
\begin{equation}\label{Kerr-HorizonSurface}
S_{\pm}=\int\rho_\pm d\theta\cdot
\frac{2\mu r_\pm}{\rho_\pm}\sin\theta d\varphi=
2\mu r_{\pm}
\int\limits_{0}^{\pi}\sin\theta d\theta
\int\limits_{0}^{2\pi}d\varphi=
8\pi \mu r_{\pm}=4\pi (r_{\pm}^{2}+a^2).
\end{equation} </p>
</div>
</div>

<div id="BlackHole62"></div>
=== Problem 11: black holes and naked singularities ===
What values of $a$ lead to existence of horizons?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Solutions of $\Delta=0$ exist when
\[a<m.\] </p>
</div>
</div>
On calculating curvature invariants, one can see they are regular on the horizons and diverge only at $\rho^2 \to 0$. Thus only the latter surface is a genuine singularity.

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=== Problem 12: $r=0$ is not a point. ===
Derive the internal metric of the surface $r=0$ in Kerr solution.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us consider the set of points $r=0$. Assuming $r=0$ and $dr=0$ in (\ref{Kerr}), from (\ref{Kerr-RhoDelta}) we obtain
\[ \rho^2=a^2 \cos^2 \theta;\quad
\Delta=a^2;\quad\Rightarrow\quad
g_{\theta\theta}=-a^2,\quad
g_{\varphi\varphi}=-a^{2}\sin^{2}\theta, \]
so metric takes the form
\begin{equation}\label{KerrR=0}
ds^{2}_{r=0}=dt^2-a^{2}\cos^{2}\theta d\theta^2
-a^{2}\sin^{2}\theta d\varphi^2=
\Big\| a\sin\theta=\eta\Big\|=
dt^{2}-\Big(d\eta^2+\eta^2 d\varphi^2 \Big).
\end{equation}
This is a flat disk of radius $a$, center $\eta=\theta=0$, with distance to the center measured by $\eta=a\sin\theta$. </p>
</div>
</div>

<div id="BlackHole64"></div>

=== Problem 13: circular singularity ===
Show that the set of points $\rho=0$ is a circle. How it it situated relative to the inner horizon?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The boundary of the disk is a circle $\eta=a$, or in original variables
\[\{r=0,\;\theta=\pi/2,\;\varphi\in[0,2\pi)\},\]
which lies beyond the inner horizon. If $a=\mu$ (extremal Kerr black hole), then $r_{-}=0$ and it lies on the horizon. </p>
</div>
</div>

==Stationary limit==
Stationary limit is a surface that delimits areas in which particles can be stationary and those in which they cannot. An infinite redshift surface is a surface such that a phonon emitted on it escapes to infinity with frequency tending to zero (and thus its redshift tends to infinity). The event horizon of the Schwarzschild solution is both a stationary limit and an infinite redshift surface (see [[Schwarzschild_black_hole#Blackness_of_black_holes| the problems on blackness of Schwarzshild black hole]]). In the general case the two do not necessarily have to coincide.

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=== Problem 14: geometry of the stationary limit surfaces in Kerr ===
Find the equations of surfaces $g_{tt}=0$ for the Kerr metric. How are they situated relative to the horizons? Are they spheres?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Equations of surfaces $g_{tt}=0$ are
\begin{align}
2\mu r=\rho^2,\quad\Leftrightarrow\quad
r^2-2\mu r+a^{2}\cos^{2}\theta=0,
\quad\Leftrightarrow\quad
r=r_{S\pm},\quad\text{where}
\nonumber\\
\label{Kerr-RS+-}
r_{S\pm}\equiv
\mu\pm\sqrt{\mu^2-a^{2}\cos^{2}\theta}.
\end{align}
Those are two axially symmetric surfaces, which in the limit $a\to 0$ meld into the horizons and tend to $r=0$ and $r=2\mu$. Their intrinsic metric is obtained by plugging $r=r_{S\pm}$ into (\ref{Kerr}) and using that $r_{S\pm}^{2}+a^{2}=2\mu r_{S\pm}+a^{2}\sin^{2}\theta$:
\[ dl^{2}_{r=r_{S\pm}}
=2\mu r_{S\pm}
( d\theta^{2}+\sin^{2}\theta\,d\varphi^2 )+
2a^{2}\sin^{4}\theta\,d\varphi^2.\]

They are not spheres either, but surfaces similar to oblate ellipsoids, if embedded into a three-dimensional Euclidean space.

Comparing (\ref{Kerr-Rhor+-}) and (\ref{Kerr-RS+-}), we see, that the inner ergosurface $r=r_{S-}$ lies entirely inside the inner horizon $r=r_-$, touching it at the poles $\theta=0,\pi$. Its intersection with the equatorial plane is the circular singularity.

The outer ergosurface $r=r_{S+}$, on the contrary, encloses the outer horizon $r=r_+$ while touching it at the poles. </p>

<gallery widths=200px caption="This is a schematic view of the Kerr solution in section $t,\theta=const$. Solid lines denote ergosurfaces, the dashed ones are horizons. The thin circle shows the value of $a$, equal to its radius.">
File:BHfig-Kerr1-05small.png|a=0.5 (slowly rotating)
File:BHfig-Kerr2-09generic.png|a=0.97 (generic astrophysical)
File:BHfig-Kerr3-10extremal.png|a=1.0 (extremal)
File:BHfig-Kerr4-11naked.png|a=1.1 (naked singularity)
</gallery>

</div>
</div>

<div id="BlackHole66"></div>
=== Problem 15: natural angular velocities ===
Calculate the coordinate angular velocity of a massless particle moving along $\varphi$ in the general axially symmetric metric (\ref{AxiSimmMetric}). There should be two solutions, corresponding to light traveling in two opposite directions. Show that both solutions have the same sign on the surface $g_{tt}=0$. What does it mean? Show that on the horizon $g^{rr}=0$ the two solutions merge into one. Which one?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us consider motion of a light ray along the angular coordinate $\varphi$, such that only $u_{t}$ and $u_{\varphi}$ are different from zero. Equation of its worldline is
\[0=ds^2=g_{tt}dt^2+2g_{t\varphi}\,dt\,d\varphi
+g_{\varphi\varphi}d\varphi^2,\]
and plugging in the metric $g_{\mu\nu}$ in terms of parameters $A,B,C,D,\omega$ from (\ref{AxiSimmMetricmatrix}), we get
\begin{equation}\label{Kerr-OmegaPM}
\Omega_{\pm}\equiv
\frac{d\varphi_{\pm}}{dt}=
-\frac{g_{t\varphi}}{g_{\varphi\varphi}}\pm
\sqrt{\Big(
\frac{g_{t\varphi}}{g_{\varphi\varphi}}
\Big)^{2}-
\frac{g_{tt}}{g_{\varphi\varphi}}}=
\omega\pm
\sqrt{\omega^{2}-
\frac{g_{tt}}{g_{\varphi\varphi}}}=
\omega\pm\sqrt{A/B}.\end{equation}
The two signs correspond to light rays emitted in two opposite directions, the prograde and the retrograde one. If
\[g_{tt}=0\]
the retrograde angular velocity turns to zero
\[\frac{d\varphi_+}{dt}=2\omega,\quad
\frac{d\varphi_-}{dt}=0,\]
and at $g_{tt}<0$ both solutions are of the same sign. This means the dragging is so strong that light cannon propagate in the direction opposite to that of black hole's rotation. In other words, the locally inertial frame on the surface $g_{tt}=0$ has linear coordinate velocity along $\varphi$ equal to the speed of light.

The root is double in case $A=0$ or $B\to\infty$. From (\ref{Kerr-ABCD}) we see that the first condition is realized on the horizon, therefore on the outer horizon $\Omega\pm=\omega$. </p>
</div>
</div>

<div id="BlackHole67"></div>
=== Problem 16: angular velocities for massive particles and rigidity of horizon's rotation ===
What values of angular velocity can be realized for a massive particle? In what region angular velocity cannot be zero? What can it be equal to near the horizon?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As worldlines of massive particles lie in the light cone, the interval of possible values of angular velocities for them is
\[ \Omega\in(\Omega_{-},\Omega_{+}).\]
Thus, angular velocity cannot be equal to zero if the value $\Omega=0$ does not belong to this interval, which holds beyond the stationary limit, in the region where$g_{tt}<0$. This is the reason for the term "stationary limit".

In the vicinity of the horizon ''all'' particles rotate with one angular velocity $\Omega_{H}=\omega\big|_{r=r_+}$, which is often called the angular velocity of the horizon:
\begin{equation}\label{Kerr-OmegaHorizon}
\Omega_{H}\equiv \omega\Big|_{r=r_+}
=\frac{2\mu ra}{\Sigma^{2}}\Big|_{r=r_+}
=\frac{a}{2\mu r_{+}}.\end{equation} </p>
</div>
</div>

<div id="BlackHole68"></div>
=== Problem 17: redshift ===
A stationary source radiates light of frequency $\omega_{em}$. What frequency will a stationary detector register? What happens if the source is close to the surface $g_{tt}=0$? What happens if the detector is close to this surface?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us consider stationary observers, with $4$-velocities directed along $\boldsymbol{\xi}_{t}$. Due to normalizing condition $\boldsymbol{u}\cdot \boldsymbol{u}=1$ we have $\boldsymbol{\xi}_{t}=\alpha \boldsymbol{u}$, where $\alpha^2=\boldsymbol{\xi}_{t}\cdot\boldsymbol{\xi}_{t}=g_{tt}$.

Then the frequency of a photon with $4$-wavevector $k^\mu$ as measured by the stationary observer is
\begin{equation}\label{StaticOmega}
\omega_{stat}=\boldsymbol{k}\cdot \boldsymbol{u}=
\frac{1}{\alpha}\boldsymbol{k}\cdot\boldsymbol{\xi}_{t}=
\frac{1}{\alpha}\omega_{0}=
\frac{\omega_{0}}{\sqrt{g_{tt}}},\end{equation}
where $\omega_{0}=\boldsymbol{k}\cdot \boldsymbol{\xi}_t$ is its frequency in the world time, which is the integral of motion. In case both emitter and detector are stationary,
\[\omega_{em}\sqrt{g_{tt}^{(em)}}=
\omega_{det}\sqrt{g_{tt}^{(det)}}.\]

Thus, the frequency of a photon emitted in the outer region (where $g_{tt}>0$), measured by a stationary observer near to the stationary limit, on which $g_{tt}\to 0$, tends to infinity. Conversely, if a stationary emitter close to the stationary limit emits a photon, its frequency measured by a stationary observer at finite (or infinite) distance, will tend to zero, and its redshift to infinity. This is why those surfaces are also called the surfaces of infinite redshift. </p>
</div>
</div>

==Ergosphere and the Penrose process==
Ergosphere is the area between the outer stationary limit and the outer horizon. As it lies before the horizon, a particle can enter it and escape back to infinity, but $g_{tt}<0$ there. This leads to the possibility of a particle's energy in ergosphere to be also negative, which leads in turn to interesting effects.

''All we need to know of the Kerr solution in this problem is that it \emph{has an ergosphere}, i.e. the outer horizon lies beyond the outer static limit, and that on the external side of the horizon all the parameters $A,B,C,D,\omega$ are positive (you can check!). Otherwise, it is enough to consider the axially symmetric metric of general form.''

<div id="BlackHole69"></div>
=== Problem 18: bounds on particle's energy ===
Let a massive particle move along the azimuth angle $\varphi$, with fixed $r$ and $\theta$. Express the first integral of motion $u_t$ through the second one $u_{\varphi}$ (tip: use the normalizing condition $u^\mu u_{\mu}=1$).

${}^{*}$ Note: relations [[Technical_warm-up#u0|((7))]] and [[Technical_warm-up#EnergyStat|((8))]] do not hold, as they were derived in assumption that $g_{00}>0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
From normalization condition
\[1=u^{\mu}u_{\mu}=g^{\mu\nu}u_{\mu}u_{\nu}\]
we obtain
\[g^{tt}(u_{t})^{2}+2g^{t\varphi}u_{t}u_{\varphi}+
\big(g^{\varphi\varphi}(u_{\varphi})^{2}-1\big)=0,\]
therefore plugging in $g_{\mu\nu}$, we have
\begin{align*}
u_{t}=&-\frac{g^{t\varphi}}{g^{tt}}u_{\varphi}\pm
\sqrt{\Big(
\frac{g^{t\varphi}}{g^{tt}}u_{\varphi}\Big)^{2}-
\frac{1}{g^{tt}}\big[
g^{\varphi\varphi}(u_{\varphi})^{2}-1\big]}=\\
&=-\omega u_{\varphi}\pm
\sqrt{\omega^{2} (u_{\varphi})^{2}-
A\Big[\frac{\omega^{2}B-A}{AB}(u_\varphi)^{2}
-1\Big]}=\\
&=-\omega u_{\varphi}+
\sqrt{A+\frac{A}{B}(u_{\varphi})^{2}}
\end{align*}
We chose the sign "$+$" here because far from the black hole, where $C,D,B,A-\omega^2 B \to 1$ and $\omega\to0$, and thus $A\to 1$, $u^t$ should tend to $+1$ when velocity turns to zero. </p>
</div>
</div>

<div id="BlackHole70"></div>
=== Problem 19: negative energy ===
Under what condition a particle can have $u_{t}<0$? In what area can it be fulfilled? Can such a particle escape to infinity?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For $u_{t}$ to be negative we need the condition $u_{\varphi}>0$ to hold (with $a>0$, so that $\omega>0$) and for the square root to be less than $\omega u_{\varphi}$. As all the coefficients $A,B,C,D$ are positive, the condition of negativity of $u_t$ can be written as
\begin{equation}\label{PenroseUlimit}
\omega u_{\varphi}>
\sqrt{A+\frac{A}{B}(u_{\varphi})^{2}}\quad
\Rightarrow\quad
(u_{\varphi})^{2}
(\omega^{2}-A/B)>A
\quad\Leftrightarrow\quad
(u_{\varphi})^{2}(-g_{tt})>AB.\end{equation}
This can only be true if $g_{tt}<0$, i.e. in the ergosphere. </p>
</div>
</div>

<div id="BlackHole71"></div>
=== Problem 20: unambiguity of negativeness ===
What is the meaning of negative energy? Why in this case (and in GR in general) energy is ''not'' defined up to an additive constant?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the Schwarzschild black hole the energy of a particle at rest near the horizon tends to zero. This means the work needed to pull away to infinity, where energy is $mc^2$, equals to its rest mass. If a particle' energy in the ergosphere is negative, it means the work needed to pull it away to infinity is ''greater'' than its rest mass. Note that in GTR the starting point for energy is not arbitrarily chosen, as any mass serves as a source of gravitational field. </p>
</div>
</div>

<div id="BlackHole72"></div>
=== Problem 21: profit! ===
Let a particle $A$ fall into the ergosphere, decay into two particles $B$ and $C$ there, and particle $C$ escape to infinity. Suppose $C$'s energy turns out to be greater than $A$'s. Find the bounds on energy and angular momentum carried by the particle $B$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The answer is for the most part already obtained above. This can happen if particle $B$'s energy is negative. The restriction on its angular momentum $l_{m}=-mu_{\varphi}$ is obtained from condition (\ref{PenroseUlimit}):
\begin{equation}\label{PenroseMomLimit}
u_{\varphi}>0,\quad
(u_{\varphi})^{2}>\frac{A}{\Omega_{+}\Omega_{-}}
\quad\Leftrightarrow\quad
l_{m}<-m
\sqrt{\frac{A}{\Omega_{+}\Omega_{-}}}.\end{equation}
Thus it should be directed in the \emph{opposite} direction to the momentum of the black hole, and negative energy is achieved on retrograde orbits. </p>
</div>
</div>

==Integrals of motion==
<div id="BlackHole73"></div>
=== Problem 22: massless particles on circular orbits ===
Find the integrals of motion for a massless particle moving along the azimuth angle $\varphi$ (i.e. $dr=d\theta=0$). What signs of energy $E$ and angular momentum $L$ are possible for particles in the exterior region and in ergosphere?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The 4-velocity of a particle moving along the azimuth angle is
\[u=u^{t}(1,0,0,\Omega).\]
For massless ones
\[0=u^{\mu}u_{\mu}=g_{tt}(u^{t})^{2}
+2g_{t\varphi}u^{t}u^{\varphi}
+g_{\varphi\varphi}(u^\varphi)^{2}=
(u^{t})^{2}\big(g_{\varphi\varphi}\Omega^{2}
+2g_{t\varphi}\Omega+g_{tt}\big).\]
In parenthesis we have the equation for $\Omega_{\pm}$ (\ref{Kerr-OmegaPM})
\[\Omega_{\pm}=\omega\pm\sqrt{A/B};\quad
\Omega_{+}+\Omega_{-}=2\omega,\quad
\Omega_{+}\Omega_{-}
=\frac{g_{tt}}{g_{\varphi\varphi}}=
\omega^{2}-A/B,\]
so
\begin{equation}\label{MasslessCircleU}
\boldsymbol{u}=u^{t}(\partial_{t}+
\Omega_{\pm}\partial_{\varphi}).\end{equation}

Then the integrals of motion are
\begin{align}
u_{t}=g_{tt}u^{t}+g_{t\varphi}u^{\varphi}&
=u^{t}(g_{tt}+g_{t\varphi}\Omega_{\pm})
=u^{t}\big[A-\omega^{2}B+
\omega B(\omega\pm\sqrt{A/B})\big]=\nonumber\\
&=u^{t}(A\pm\omega\sqrt{AB})
=\pm \Omega_{\pm}\sqrt{AB}\cdot u^{t};\nonumber\\
\label{MasslessCircleIntegrals}
u_{\varphi}=g_{t\varphi}u^{t}
+g_{\varphi\varphi}u^{\varphi}&
=u^{t}(g_{t\varphi}
+g_{\varphi\varphi}\Omega_{\pm})
=u^{t}\big[\omega B-
B(\omega\pm\sqrt{A/B})\big]
=\mp\sqrt{AB}\cdot u^{t}\\
\Rightarrow\quad&\quad
\frac{E}{L}=-\frac{u_t}{u_\varphi}=\Omega_{\pm}.
\end{align}

Note that $u^{t}$ is always positive. On one hand, it cannot turn to zero in any point of the worldline, as this would violate the normalizing condition. So in the outer region, where $g_{tt}>0$, it is obvious that $u^{t}>0$, and any particle in the ergosphere can be pulled there along some trajectory from the outside (this is obvious for massive particles, but can also be imagined for the massless ones). As $u^t$ does not turn into zero, it will remain positive.

Then in the outer region, where $\Omega_{+}\Omega_{-}<0$, a photon's energy is always positive, while its angular momentum can be of any sign. In the ergosphere, where $\Omega_{+}\Omega_{-}>0$, energy is positive and angular momentum (remember it differs in sign from $u_\varphi$) is positive on prograde orbits $\Omega=\Omega_+$; on retrograde orbits $\Omega=\Omega_-$ both energy and are negative. </p>
</div>
</div>

<div id="BlackHole74"></div>
=== Problem 23: massive particles on circular orbits ===
Calculate the same integrals for massive particles. Derive the condition for negativity of energy in terms of its angular velocity $d\varphi/dt$. In what region can it be fulfilled? Show that it is equivalent to the condition on angular momentum found in the [[Kerr_black_hole#BlackHole70|problem on negative energy]].
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For massive particles we have the same quadric equation in the normalizing condition and plugging in $g_{\varphi\varphi}=-B$, we obtain
\begin{align*}
1=&g_{\mu\nu}u^{\mu}u^{\nu}=(u^{t})^{2}
\big[g_{\varphi\varphi}\Omega^{2}
+2g_{t\varphi}\Omega+g_{tt}\big]=\\
&= (u^{t})^{2}\cdot (-B)
(\Omega-\Omega_{+})(\Omega-\Omega_{-})=
(u^{t})^{2}
\big(A-B(\Omega-\omega)^{2}\big).
\end{align*}
Then
\begin{equation}\label{MassiveCircleU}
\boldsymbol{u}=
\frac{\partial_{t}+\Omega\partial_{\varphi}}
{\sqrt{-B(\Omega-\Omega_+)(\Omega-\Omega_-)}}=
\frac{\partial_{t}+\Omega\partial_{\varphi}}
{\sqrt{A-B(\Omega-\omega)^{2}}};\quad
\Omega\in(\Omega_-,\Omega_+).\end{equation}
and the integrals of motion are
\begin{align*}
u_{t}=&u^{t}(g_{tt}+g_{t\varphi}\Omega)=
u^{t}(A+B\omega (\Omega-\omega))=
\frac{A+B\omega (\Omega-\omega)}
{\sqrt{A-B(\Omega-\omega)^{2}}};\\
u_{\varphi}=&u^{t}
(g_{t\varphi}+g_{\varphi\varphi}\Omega)=
u^{t}(\omega B-\Omega B)=
\frac{B(\omega-\Omega)}
{\sqrt{A-B(\Omega-\omega)^{2}}}.
\end{align*}

We see that angular momentum of a particle with $\Omega=\omega$ is zero:
\[L=-u_{\varphi}=0;\quad E=u_{0}=\sqrt{A}.\]

Energy is negative if
\[u_{t}<0\quad\Leftrightarrow\quad
\omega-\Omega>\frac{A/B}{\omega}
\quad\Leftrightarrow\quad
\Omega<\frac{\omega^{2}-A/B}{\omega}=
\frac{\Omega_{+}\Omega_{-}}{\omega}
=-\frac{g_{tt}}{g_{t\varphi}},\]
thus angular velocity should be \emph{less} than the critical value
\[\label{PenroseAngVelocity}
\Omega_{P}\equiv
-\frac{g_{tt}}{g_{t\varphi}}\equiv
\frac{\Omega_{+}\Omega_{-}}{\omega}
\in(\Omega_{-},\omega);\]

Clearly $\Omega_{P}>0$ only in the ergosphere, where $\Omega_{\pm}$ are of the same sign. Taking into account that $\Omega_{\pm}=\omega\pm\sqrt{A/B}$, we can put down the following chain of inequalities for the characteristic angular velocities in the ergosphere
\begin{equation}\label{ErgospereOmegas}
0<\Omega_{P}=
\frac{\Omega_{+}\Omega_{-}}{\omega}
<\omega<\Omega_{+}<2\omega.\end{equation}
It follows that the window for the Penrose process always exists.</p>
[[File:BHfig-KerrOmegas_thin.png|center|thumb|400px|$\Omega_{\pm}(\xi)$ and $\omega(\xi)$ in the equatorial plane on the Kerr solution for $\alpha=a/\mu=0.9375$. A particle's energy in the ergosphere is negative if $\Omega\in(\Omega_{-},\Omega_{P})$; $\Omega_{P}(\xi)$ in the equatorial plane is a straight line.]]

<p style="text-align: left;">
In terms of angular momentum $l_{m}=-mu_{\varphi}$ energy is negative when
\[u_{\varphi}>
\frac{B\cdot \frac{A/B}{\omega}}
{\sqrt{A-B\cdot \frac{A^2 /B^{2}}{\omega^{2}}}}=
\frac{A/\omega}
{\sqrt{A-\frac{A^2 }{B\omega^{2}}}}=
\sqrt{\frac{A}{\omega^{2}-A/B}}=
\sqrt{\frac{A}{\Omega_{+}\Omega_{-}}}=
\sqrt{A\frac{\omega}{\Omega_{P}}}\]
and finally we derive the same condition as the one obtained above:
\[l_{m}<-m\sqrt{\frac{A}{\Omega_{+}\Omega_{-}}}=
-m\sqrt{A\frac{\omega}{\Omega_P}}.\] </p>
</div>
</div>

<div id="BlackHole75"></div>

=== Problem 24: general case ===
Derive the integrals of motion for particles with arbitrary $4$-velocity $u^{\mu}$. What is the allowed interval of angular velocities $\Omega=d\varphi/dt$? Show that for any particle $(E-\tilde{\Omega} L )>0$ for any $\tilde{\Omega}\in(\Omega_{-},\Omega_{+})$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For massive particles moving in arbitrary direction the normalizing condition takes the form
\[1=(u_{t})^{2}\big(
A-B(\Omega-\omega)^{2}-C(\tfrac{dr}{dt})^{2}
-D(\tfrac{d\theta}{dt})^{2}\big).\]
Then the interval of possible angular velocities is (\ref{MassiveCircleU}):
\[\Omega\in(\Omega_{-},\Omega_{+}),
\quad\text{where}\quad
\Omega_{\pm}=\omega\pm\sqrt{A/B},\]
which should be expected. The limiting values are realized for motion along $\varphi$ in the ultrarelativistic limit $E\to\infty$.

The denominators in (\ref{MassiveCircleU}), which ensure the normalizing condition for $u^\mu$, are different now, but in the same way otherwise for the integrals of motion we obtai
\begin{align*}
&u_{t}=u^{t}\cdot
\big(A+B\omega(\Omega-\omega)\big),\\
&u_{\varphi}=-u^{t}\cdot B(\Omega-\omega),
\end{align*}
so
\[\frac{E}{L}=-\frac{u_{t}}{u_{\varphi}}
=\omega+\frac{A/B}{\Omega-\omega}.\]
Taking into account that $|\Omega-\omega|\leq \sqrt{A/B}$, we can extract from this the restrictions on $E/L$, but it is easier to achieve in a different way for all possible signs of $E$ and $L$ simultaneously. Let there be a geodesic particle with $u_{t}$ and $u_\varphi$ (those are conserving quantities, other components are arbitrary), and let there be an observer moving on a circular orbit with angular velocity $\Omega\in(\Omega_{-},\Omega_{+})$, so that his $4$-velocity is $\tilde{u}=\tilde{u}^{t}(\partial_{t}+\tilde{\Omega}\partial_{\varphi})$. The energy of the first particle as measured by this observer is the invariant
\[\tilde{E}=\tilde{u}^{\mu}u_{\mu}
=\tilde{u}^{t}
\big(u_{t}+\tilde{\Omega}u_{\varphi}\big) >0.\]
Then
\[E-\tilde{\Omega} L>0\quad\Rightarrow\quad
L<\frac{E}{\tilde{\Omega}}
\quad\text{for}\quad\forall
\tilde{\Omega}\in(\Omega_{-},\Omega_{+}),\]
which is what we wanted to prove. Observers near the event horizon can only have $\tilde{\Omega}=\Omega_{H}$, so for a particle crossing the horizon
\[L<\frac{E}{\Omega_{H}}.\] </p>
</div>
</div>

==The laws of mechanics of black holes==
If a Killing vector is null on some null hypersurface $\Sigma$, $\Sigma$ is called a Killing horizon.
<div id="BlackHole76"></div>
=== Problem 25: Killing horizons ===
Show that vector $K=\partial_{t}+\Omega_{H}\partial_{\varphi}$ is a Killing vector for the Kerr solution, and it is null on the outer horizon $r=r_{+}$. Here $\Omega_{H}=\omega\big|_{r=r_+}$ is the angular velocity of the horizon.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
First note, that due to linearity of the Killing equation $\xi_{\mu;\nu}+\xi_{\nu;\mu}=0$, a linear combination of two Killing vector fields with constant coefficients is also a Killing vector field. As $\Omega_H$ is a constant, this holds for $K^\mu$. Next,
\begin{align*}
g_{\mu\nu}K^{\mu}K^{\nu}=&g_{\mu\nu}
(\delta_{t}^{\mu}+\Omega_{H}\delta_{\varphi}^{\mu})
(\delta_{t}^{\nu}+\Omega_{H}\delta_{\varphi}^{\nu})
=g_{tt}+2g_{t\varphi}\Omega_{H}
+g_{\varphi\varphi}\Omega_{H}^2=\\
&=A-\omega^{2}B+2\omega\Omega_{H}B
-\Omega_{H}^{2}B=A-B(\omega-\Omega_H)^2.
\end{align*}
In the vicinity of the horizon $\omega\to\Omega_{H}$, and also $A\sim\Delta\to 0$ (\ref{Kerr-ABCD}), so $K^{\mu}K_{\mu}\to 0$. </p>
</div>
</div>

<div id="BlackHole77"></div>
=== Problem 26: surface gravity ===
Let us define the surface gravity for the Kerr black hole as the limit
\[\kappa=\lim\limits_{r\to r_{+}}
\frac{\sqrt{a^{\mu}a_{\mu}}}{u^0}\]
for a particle near the horizon with $4$-velocity $\boldsymbol{u}=u^{t}(\partial_{t}+\omega\partial_{\varphi})$. In the particular case of Schwarzschild metric this definition reduces to the one given in the [[Schwarzschild_black_hole#BlackHole43|corresponding problem]]. Calculate $\kappa$ for particles with zero angular momentum in the Kerr metric. What is it for the critical black hole, with $a=\mu$?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Recall the normalizing condition
\[1=u^{\mu}u_{\mu}=(u^{t})^{2}
(g_{tt}+2\Omega g_{t\varphi}
+\Omega^{2}g_{\varphi\varphi}).\]
As all the metric components depend only on $r$ and $\theta$, acceleration can be reduced to
\begin{align*}
a^{\mu}=&{\Gamma^{\mu}}_{\nu\lambda}
u^{\nu}u^{\lambda}
=(u^t)^{2}\big({\Gamma^{\mu}}_{tt}
+2\Omega{\Gamma^{\mu}}_{t\varphi}
+\Omega^{2}{\Gamma^{\mu}}_{\varphi\varphi}\big)=\\
&=(u^t)^{2}\frac{g^{\mu\nu}}{2}
\big(\!-\partial_{\nu}g_{tt}
-2\Omega\partial_{\nu}g_{t\varphi}
-\Omega^2\partial_{\nu}g_{\varphi\varphi}\big)=
-(u^t)^{2}\frac{g^{\mu\nu}}{2}\partial_{\nu}
\big(g_{tt}+2\Omega g_{t\varphi}
+\Omega^2 g_{\varphi\varphi}\big)=\\
&=-g^{\mu\nu}\tfrac{1}{2}(u^t)^{2}\partial_{\nu}
\frac{1}{(u^t)^2}=
-g^{\mu\nu}\partial_{\nu}\ln u^t.
\end{align*}
Again taking into account the symmetries, we get
\[a^2\equiv a^{\mu}a_{\mu}
=|g^{rr}|(\partial_{r}\ln u^t)^2
+|g^{\theta\theta}|(\partial_{\theta}\ln u^t)^2.\]

Let us now consider a particle with zero angular momentum $\Omega=\omega$, for which (see the previous problem and the one on [[Kerr_black_hole#BlackHole74|integrals of motion of massless particles]])
\[(u^{t})^2=\frac{1}{A}
=\Big(\frac{\rho^2 \Delta}{\Sigma^2}\Big)^{-1},\quad
\text{thus}\quad
\partial_{\mu}\ln u^{t}=-\tfrac{1}{2}
\frac{\partial_{\mu} A}{A}.\]
we are interested only in the part of $a^2$ that is divergent on the horizon, so differentiate only $\Delta$:
\[\partial_{\mu}\ln u^t \sim -\frac{1}{2}
\frac{\partial_{\mu}\Delta}{\Delta}
=-\frac{1}{2}\frac{\partial_{\mu}(r^2-2\mu r+a^2)}
{\Delta} =-\frac{r-\mu}{\Delta}\delta_{\mu}^{1}.\]
Plugging $g^{rr}=\Delta/\rho^2$, we get
\[a^2\approx\frac{(r-\mu)^2}{\rho^2 \Delta^2},\]
and on substitution of $(u^t)^2$, we obtain the surface gravity:
\begin{align}
\kappa^2 =&\lim\limits_{r\to r_{+}}
\frac{a^2}{(u^t)^2}
=\frac{(r-\mu)^2}{\Sigma^2}\Big|_{r=r_+}
\quad\Rightarrow\nonumber\\
\label{SurfaceGravity}
\kappa=&\frac{r_{+}-\mu}{r_{+}^{2}+a^2}
=\frac{r_{+}-\mu}{2\mu r_{+}}
=\frac{1}{2\mu}
\frac{\sqrt{1-\alpha^2}}{1+\sqrt{1-\alpha^2}}
=\Omega_{H}\frac{\sqrt{1-\alpha^2}}{\alpha},
\quad \alpha=\frac{a}{\mu}.
\end{align}
For the extremal Kerr solution, in the limit $\alpha\to 1$, it tends to zero. </p>
</div>
</div>

<div id="BlackHole78"></div>
=== Problem 27: horizon's area evolution ===
Find the change of (outer) horizon area of a black hole when a particle with energy $E$ and angular momentum $L$ falls into it. Show that it is always positive.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In case a particle of mass $m$ crosses the event horizon, the black hole's mass increases by $\delta \mu=E$, and its angular momentum by $\delta J=L$. Then, using the result of [[Kerr_black_hole#BlackHole75|problem on particle's integrals of motion]], for any continuous process of accretion on a black hole the following holds
\begin{equation}\label{Kerr-JM}
\delta J<\frac{\delta \mu}{\Omega_H}.\end{equation}
Note that this relation works both for positive and negative $E$ and $L$.

As $\alpha=\tfrac{a}{\mu}=\tfrac{J}{\mu^2}$, the area of the horizon is expressed through $\mu$ and $J$ this way
\[A_{+}=4\pi(r_{+}^{2}+a^{2})=8\pi\mu r_{+}=
8\pi\mu^{2}(1+\sqrt{1-\alpha^2})=
8\pi(\mu^2+\sqrt{\mu^4-J^2}),\]
and $\Omega_{H}$ can be rewritten in terms of the same quantities as
\[\Omega_{H}\equiv\omega(r_{+})
=\frac{a}{r_{+}^{2}+a^2}
=\frac{a}{2\mu r_{+}}
=\frac{J/\mu}{\mu^2+\sqrt{\mu^4-J^2}}.\]
On differentiating $A_{+}$, we obtain then
\[\delta A_{+}=\frac{2\mu A_{+}}{\sqrt{\mu^4-J^2}}
\Big\{\delta\mu-\Omega_{H}\delta J\Big\}.\]
Expressing the factor by the braces though $\alpha$, we obtain surface gravity (\ref{SurfaceGravity}):
\begin{equation}\label{BHThermodynamics}
\delta A_{+}=\frac{8\pi}{\kappa}
\Big\{\delta\mu-\Omega_{H}\delta J\Big\}.
\end{equation}
Due to condition (\ref{Kerr-JM}) the surface area of the horizon always increases:
\begin{equation}\label{AreaTheorem}
\delta A_{+}>0\end{equation} </p>
</div>
</div>

<div id="BlackHole79"></div>
=== Problem 28: irreducible mass ===
Let us define the irreducible mass $M_{irr}$ of Kerr black hole as the mass of Schwarzschild black hole with the same horizon area. Find $M_{irr}(\mu,J)$ and $\mu(M_{irr},J)$. Which part of the total mass of a black hole can be extracted from it with the help of Penrose process?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As defined,
\[A_{+}=4\pi (2M_{irr})^{2}=16\pi M_{irr}^2.\]
Then
\[M_{irr}^{2}=\tfrac{1}{2}
\Big(\mu^2 + \sqrt{\mu^4-J^2}\Big)
\quad\Leftrightarrow\quad
\mu^2=M_{irr}^{2}+\Big(\frac{J}{2M_{irr}}\Big)^{2}.\]
This relation provides interesting interpretation: the full mass of a black hole $\mu$ consists of the irreducible mass $M_{irr}$ and the rotational energy $J/2M_{irr}$, which add up squared. The second term can in principle be extracted through the Penrose process. </p>
</div>
</div>

<div id="BlackHole80"></div>
=== Problem 29: extremal limit ===
Show that an underextremal Kerr black hole (with $a<\mu$) cannot be turned into the extremal one in any continuous accretion process.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\delta\alpha=\delta\Big(\frac{J}{\mu^2}\Big)
=\frac{1}{\mu^3}
\Big[\mu \delta J-2J \delta\mu\Big],\]
and using the condition (\ref{Kerr-JM}), we get
\[\delta\alpha<\frac{2\delta\mu}{\mu}
\frac{1+\sqrt{1-\alpha^2}}{\alpha}
\cdot\sqrt{1-\alpha^2}.\]
When $\alpha\to1$ the last factor tends to zero, so $\alpha$ cannot become equal or greater than unity in any continuous accretion process. </p>
</div>
</div>

This problem's results can be presented in the form that provides far-reaching analogy with the laws of thermodynamics.

'''0:''' Surface gravity $\kappa$ is constant on the horizon of a stationary black hole. The zeroth law of thermodynamics: a system in thermodynamic equilibrium has constant temperature $T$.

'''1:''' The relation
\[\delta\mu=\frac{\kappa}{8\pi}\delta A_{+}
+\Omega_{H}\delta J\]
gives an analogy of the first law of thermodynamics, energy conservation.

'''2:''' Horizon area $A_+$ is nondecreasing. This analogy with the second law of thermodynamics hints at a correspondence between the horizon area and entropy.

'''3:''' There exists no such continuous process, which can lead as a result to zero surface gravity. This is an analogy to the third law of thermodynamics: absolute zero is unreachable.

==Particles' motion in the equatorial plane==
The following questions refer to a particle's motion in the equatorial plane $\theta=\pi/2$ of the Kerr metric.

<div id="BlackHole81"></div>
=== Problem 30: preparatory algebra ===
Put down explicit expressions for the metric components and parameters $A,B,C,D,\omega$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[g_{\mu\nu}=\begin{pmatrix}
1-\frac{2\mu}{r}&0&0&\frac{2\mu a}{r}\\
0&-\frac{r^2}{\Delta}&0&0\\
9&0&-r^2&0\\
\frac{2\mu a}{r}&0&0&
-\frac{\Sigma^2}{r^2}
\end{pmatrix},\quad
g^{\mu\nu}=\begin{pmatrix}
\frac{\Sigma^2}{r^2\Delta}
&0&0&\frac{2\mu a}{\Delta r}\\
0&-\frac{\Delta}{r^2}&0&0\\
9&0&-\frac{1}{r^2}&0\\
\frac{2\mu a}{\Delta r}&0&0&
-\frac{1}{\Delta}\Big(1-\frac{2\mu}{r}\Big)
\end{pmatrix}\]
where
\[\Delta=r^{2}+a^{2}-2\mu r;\quad
\Sigma^{2}=r^{2}(r^2+a^2)+2\mu r a^2;\]
the other expressions, for $A,B,\ldots,\omega,\Omega_\pm$ etc. are not simplified essentially. </p>
</div>
</div>

<div id="BlackHole82"></div>
=== Problem 31: zero energy particles ===
What is the angular velocity of a particle with zero energy?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For $\theta=\pi/2$ and $\rho^2=r^2$ we get
\[\label{PenroseAngVelocity2}
\Omega_{P}=\frac{\frac{2\mu r}{\rho^2}-1}
{\frac{2\mu r}{\rho^2}a\sin^2 \theta}=
\frac{1-\frac{\rho^2}{2\mu r}}{a\sin^2 \theta}=
\Big\| \rho^2=r^2,\; \sin\theta=1\Big\|
=\frac{1-\frac{r}{2\mu}}{a}.\]
It s clear also that $\Omega_P$ turns to zero at the ergosurface, where $2\mu r=\rho^2$. On the other hand, due to the system of inequalities (\ref{ErgospereOmegas}) on the horizon it should be equal to the horizon's angular velocity $\Omega_H$ (\ref{Kerr-OmegaHorizon}). Indeed, the latter can be written using $r_{+}r_{-}=a^2$ as
\[\Omega_{H}\equiv\omega\Big|_{r=r_+}
=\frac{2\mu ra}{\Sigma^2}\big|_{r=r_+}
=\frac{a}{2\mu r_+}=\frac{r_-}{2\mu a}.\]
and then we affirm that
\[\Omega_{P}|_{r_+}
=\frac{1}{a}\Big(1-\frac{r_+}{2\mu}\Big)
=\frac{1}{a}\frac{2\mu-r_+}{2\mu}
=\frac{r_-}{2\mu a}=\Omega_H.\] </p>
</div>
</div>

<div id="BlackHole83"></div>
=== Problem 32: geodesics and effective potential ===
Use the normalizing conditions for the $4$-velocity $u^{\mu}u_{\mu}=\epsilon^2$ and two conservation laws to derive geodesic equations for particles, determine the effective potential for radial motion.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integrals of motion are
\[\left\{\begin{array}{l}
E\equiv u_{t}=g_{tt}u^{t}+g_{t\varphi}u^\varphi,\\
-L\equiv u_{\varphi}=g_{t\varphi}u^{t}+
g_{\varphi\varphi}u^{\varphi}
\end{array}\right.\;\Rightarrow\;
\left\{\begin{array}{l}
u^{t}=\frac{1}{G}(g_{\varphi\varphi}E+
g_{t\varphi}L)\\
u^{\varphi}=-\frac{1}{G}(g_{tt}L-g_{t\varphi}E)
\end{array}\right.,\;\text{where}\quad
G\equiv
\begin{vmatrix} g_{tt}&g_{t\varphi}\\
g_{t\varphi}&g_{\varphi\varphi}\end{vmatrix}.\]
Plugging in the metric, we get $G=-\Delta$ and
\[ u^{t}=\frac{1}{\Delta}\Big[
\big(r^2+a^2+\frac{2\mu a^2}{r}\big)E-
\frac{2\mu a}{r} L\Big];\quad
u^\varphi =\frac{1}{\Delta}\Big[
\frac{2\mu a}{r}E +\big(1-\frac{2\mu}{r}\big)L
\Big].\]

Let us write the normalizing condition as
\[u^{\mu}u_{\mu}=\epsilon^2,\]
so that $\epsilon^2=1$ for a massive particle, and $\epsilon^2=0$ to a massless one. Then
\[\epsilon^2=g^{tt}u_{t}^{2}
+2g^{t\varphi}u_{t}u_{\varphi}
+g^{\varphi\varphi}u_{\varphi}^{2}+g^{rr}u_{r}^{2},\]
and taking into account that $g_{rr}=1/g_{rr}$,
\[\Big(\frac{dr}{ds}\Big)^{2}\equiv (u^{r})^{2}
=g^{rr}(g_{rr}u_{r})^{2}=
g^{rr}\Big(\epsilon^2-g^{tt}E^2+2g^{t\varphi}EL
-g^{\varphi\varphi}L^2\Big).\]

This can be transformed to an equation for $r(s)$ in the form
\begin{equation}\label{KerrEqPotential}
\frac{1}{2}\Big(\frac{dr}{ds}\Big)^{2}+U_{eff}=0,
\quad\text{where}\quad
U_{eff}=\frac{\epsilon^2-E^2}{2}
-\frac{\epsilon^2 \mu}{r}
+\frac{L^2-a^{2}(E^2-\epsilon^2)}{2r^2}
-\frac{\mu(L-aE)^{2}}{r^3}\end{equation}
is effective gravitational energy, with both $E$ and $L$ acting as parameters. As both of them are present in the left hand side, we can as well leave just zero on the right.

In terms of dimensionless variables
\[\xi=\frac{r}{\mu},\quad \alpha=\frac{a}{\mu},
\quad \lambda=\frac{L}{\mu}\]
the full system of equations is
\begin{align}
&\frac{1}{2}(u^r)^{2}+U_{eff}=0;\quad
U_{eff}=
\frac{\epsilon^{2}-E^2}{2}
-\frac{\epsilon^{2}}{\xi}
+\frac{\lambda^2 +\alpha^{2}(\epsilon^2-E^2)}
{2\xi^2}
-\frac{(\lambda-\alpha E)^{2}}{\xi^3}.\\
&u^{t}=\frac{\mu^2 E}{\Delta}
\Big[\xi^2+\alpha^2+\frac{2\alpha^2}{\xi}
-\frac{2\alpha}{\xi}\frac{\lambda}{E}\Big];\\
&u^\varphi =\frac{\mu E}{\Delta}
\Big[\frac{2\alpha}{\xi}+
\big(1-\frac{2}{\xi}\big)\frac{\lambda}{E}\Big].
\end{align} </p>
</div>
</div>

<div id="BlackHole84"></div>
=== Problem 33: principal null geodesics ===
Integrate the equations of motion for null geodesics with $L=aE$, investigate the asymptotes close to the horizons, limits $a\to 0$ and $a\to \mu$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
When $L=Ea$, which corresponds to the impact parameter at infinity equal to $a$, the equations for null geodesics are essentially simplified:
\begin{align*}
&(u^r)^{2}=E^2;\\
&u^{t}\equiv\frac{dt}{ds}
=\frac{E}{\Delta}\big[r^2+a^2\big];\\
&u^\varphi\equiv\frac{d\varphi}{ds}
=\frac{aE}{\Delta}.
\end{align*}
Note that this relation, $L=Ea$, singles out quite peculiar (critical) particles, for which the asymptote of the effective potential at small $\xi$ is $\sim \xi^{-2}$, as opposed to any other particle, for whichh $U_{eff}\sim \xi^{-3}$.

Then $ds=\pm dr/E$, where the plus sign corresponds to a photon falling to the center and minus to the one moving from the center, and
\begin{align*}
&\mp\varphi(r)=\mp\int d\varphi=\int\frac{a dr}{\Delta}
=\int\frac{a\, dr}{(r-r_{-})(r-r_{+})}
=\ldots=\frac{\alpha}{2\sqrt{1-\alpha^2}}
\ln \Big|\frac{r-r_-}{r-r_+}\Big|;\\
&\mp t(r)=\mp\int dt=\int \frac{dr(r^2+a^2)}{\Delta}
=\ldots
=r\mu \ln\Big|\frac{r-r_-}{r-r_+}\Big|+
\frac{\mu}{\sqrt{1-\alpha^2}}\ln|\Delta|.
\end{align*}

In the limit $r\to r_{+}$ both $\varphi(r)$ and $t(r)$ diverge as logarithms, but it is not hard to show that
\[\frac{\varphi}{t}\approx
\frac{\alpha/2\mu}{1+\sqrt{1-\alpha^2}}
=\frac{a}{2\mu r_{+}}\equiv \Omega_{H}.\]
This should be expected, as we know that at the horizon all particles should rotate with the angular velocity of the horizon.

Assuming $\alpha\to0$ we get the usual Schwarzshild solutions $\varphi=0$, $\mp t=r+\mu.$ In the opposite limit $\alpha\to 1$ the quantity $\Delta$ has a double root and recalculating the integrals, we see that $t$ and $\varphi$ now diverge not logarithmically but as $(r-r_{+})^{-1}$. </p>
</div>
</div>

<div id="BlackHole85"></div>
=== Problem 34: innermost stable circular orbits, massless case ===
Find the minimal radii of circular geodesics for massless particles, the corresponding values of integrals of motion and angular velocities. Show that of the three solutions one lies beyond the horizon, one describes motion in positive direction and one in negative direction. Explore the limiting cases of Schwarzschild $a\to0$ and extreme Kerr $a\to\mu$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a massless particle $\epsilon=0$ the equation for $r(t)$ is
\[\frac{1}{2}\Big(\frac{dr}{d\tilde{s}}\Big)^2
+U_{eff}=0,\quad\text{where}\quad
U_{eff}=-\frac{1}{2}
+\frac{\lambda^2-\alpha^2 E^2}{2\xi^2}
-\frac{(\lambda-\alpha E)^2}{\xi^3}.\]
It is convenient, as usual for massless particles, to introduce the parameter $p=L/E$, which at $r\to\infty$ has the meaning of the impact parameter.

For a circular orbit
\[U_{eff}=0,\quad \frac{dU_{eff}}{d\xi}=0.\]
The second condition gives
\[\xi_{0}=3\frac{\lambda-\alpha E}{\lambda+\alpha E}=
3\frac{\sigma-1}{\sigma+1},\quad\text{where}\quad
\sigma\equiv\frac{\lambda}{\alpha E}=
\frac{p}{a}=\frac{p}{\alpha\mu}.\]

The first one leads to a cubic equation for $\sigma$:
\[(\sigma+1)^{3}=\frac{27}{\alpha^2}(\sigma-1).\]

When $\alpha\ll 1$ (the near-Schwarzschild case) the three roots are $\sigma\approx 1+\frac{8}{27}\alpha^2,\pm \frac{3\sqrt{3}}{\alpha}$. The first one gives $\xi_0\approx\alpha^2$, i.e. this orbit lies beyond the horizons (we do not consider this region), and the other two
\[\alpha\ll1:\qquad \sigma\approx3\sqrt{3}/\alpha;
\quad\Rightarrow\quad
p\approx 3\sqrt{3}\mu,\quad\xi_{0}\approx 3.\]
Those are the familiar parameters of a Schwarzschild black hole's photon sphere of , at $r=3\mu=\frac{3}{2}r_{g}$. It is natural that in this limit the radius does not depend on the sign of $L$.

In the limit $\alpha\to1$ (extremal Kerr black hole) the equation is reduced to $(\sigma+1)^{3}=27(\sigma-1)$,
the roots of which are, as can be checked, $\sigma=2,2,-7$. Substituting this in $\xi_0$ and $p$, we get
\[p_{+}^{(extr)}\approx2\mu,\quad
\xi_{+}^{(extr)}\approx 1;\qquad
p_{-}^{(extr)}\approx-7\mu,\quad
\xi_{-}^{(extr)}\approx 4.\]
The first root corresponds to prograde photons, the second one to the retrograde ones. It can be shown that in the limit $a\to 0$ one of the two close roots, which merge at $a=1$, lies before the horizon, and the other one is beyond it.

In the general case it is convenient to rewrite the equation for $\sigma$ as
\[\nu^3=\nu-2\beta,\quad\text{where}\quad
\beta=\frac{\alpha}{3^{3/2}},\quad
\nu=\beta(\sigma+1);\qquad
\xi_{0}=3\nu^2.\]
This is a reduced equation, solved by the change of variables $\nu=a+b$ with additional constraint $3ab=1$. The resulting system for $a^{3},b^{3}$ is
\[a^{6}-2\beta a^{3}+\frac{1}{27}=0.\]
Its solution is
\[a^{3}=-\frac{\alpha\pm\sqrt{\alpha^2-1}}{3^{3/2}}=
\pm 3^{-3/2}\exp\{\pm i\omega\},\quad\text{where}
\quad \omega=\arccos \alpha
\in\Big(0,\frac{\pi}{2}\Big).\]
Extracting the root and selecting the pairs $(a,b)$ which obey the imposed condition $3ab=1$, we obtain the three roots
\[\nu_1=-\frac{2}{3}\cos\frac{\omega}{3};\quad
\nu_2=\frac{2}{3}\cos\frac{\pi-\omega}{3};\quad
\nu_3=\frac{2}{3}\cos\frac{\pi+\omega}{3}.\]

The "radii" of corresponding orbits are
\[\xi_{1}=4\cos^{2}\frac{\omega}{3};\quad
\xi_{2}=4\cos^{2}\frac{\pi-\omega}{3}\quad
\xi_{3}=4\cos^{2}\frac{\pi+\omega}{3}.\]

As $\omega\in(0,\pi/2)$, it is not hard to show that a sequence of inequalities hold
\[0<\xi_3 <1<\xi_2 <3<\xi_1 <4;\]
it also turns out that $\xi_2<\xi_{+}<\xi_3$ (in terms of $\omega$ the "radius" of the outer horizon is $\xi_{+}=1+\sin\omega$, so the problem is reduced to trigonometric inequalities). Thus, the second root always lies on the outside of the horizon, while the third one is beyond it and is unphysical.

The two remaining solutions correspond to positive and negative $p$. Taking into account that $\pi-\omega=\arccos(-\alpha)$, they both can be expressed in the form
\begin{equation}\label{KerrPhotonOrbits}
\xi_{1,2}=4\cos^{2}\Big[\frac{1}{3}
\arccos(\pm\alpha)\Big];\end{equation}
The corresponding angular momenta are
\[p_{i}=\mu\alpha\sigma_{i}
=\mu\alpha(\nu_{i}/\beta-1)
=\mu(3\sqrt{3}\nu_{i}-\alpha)\quad\Rightarrow\quad
p_{1,2}=\mp 6\mu\cos\Big[\frac{1}{3}
\arccos(\pm\alpha)\Big]-a.\]

The angular velocities for photons on circular orbits are
\[\Omega_{1,2}=\Omega_{\mp}(\xi=\xi_{1,2}),\]
where the upper sign corresponds to retrograde orbits ($\xi_1$) and the lower sign to prograde ones ($\xi_2$). </p>
</div>
</div>

<div id="BlackHole86"></div>
=== Problem 35: circular orbits for massive particles ===
Find $L^2$ and $E^2$ as functions of radii for circular geodesics of the massive particles.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us write the effective potential for massive particles in terms of $u=1/\xi$:
\[U_{eff}=\frac{1-E^2}{2}-u+\tfrac{\beta}{2}u^2
-\nu^2 u^3,\quad\text{where}\quad
\nu=\lambda-\alpha E,\quad
\beta=\lambda^{2}+\alpha^2 (1-E^2).\]
For a circular orbit (the full effective energy with the chosen potential is zero)
\[U_{eff}(u)=0,\quad {U_{eff}}'(u)=0,\]
that is
\begin{align*}
&\frac{1-E^2}{2}-u
+\frac{\beta}{2}u^2-\nu^2 u^{3}=0;\\
&-1+\beta u-3\nu^2 u^2=0.
\end{align*}
The orbit $u=u_{i}(E,\lambda)$ is stable if for the given $E$ and $\lambda$ in the neighborhood of $u_{i}$ holds $U_{eff}<0$, so that $(u^r)^{2}<0$ and there are no other solutions. Thus a stable orbit corresponds to a point in which $U_{eff}$ touches zero from above, and unstable orbits to a point in which is touches zero from below.

The condition of stability is $d^{2}U_{eff}/dr^{2}>0$. Taking into account $0=U_{eff}=U_{eff}'$, it is equivalent to $d^{2}U_{eff}/du^{2}>0$, and thus
\[6\nu^{2}u<\beta.\]

An equality would mean that the two touching points merge into an inflection point, which gives us the minimal radius of the stable orbit and the maximal radius of the unstable one.

Subtracting the second equation multiplied by $u/2$ from the first one, we get
\[\nu^2 u^3 -u=E^2-1.\]

Using $U_{eff}'=0$, we can exclude $\beta$ and $E^2$
\[\beta=3\nu^{2}u+u^{-1};\quad
E=\frac{u\nu^2 (3u-1)+1-\alpha^2 u}{2\alpha u\nu},\]
and obtain a quadratic equation for $\nu^2$:
\begin{equation}\label{KerrEq-M-Nu2Eq}
u^{2}\big[(3u-1)^{2}-4\alpha^{2}u^{3}\big]\nu^4
-2u\big[\alpha^{2}u(u+1)-(3u-1)\big]\nu^2
-(1-\alpha^2 u)^{2}=0.
\end{equation}

After straightforward calculation the discriminant can be brought to the form
\[4\alpha^2 u^3 (\alpha^2 u^2 -2u+1)^{2}.\]
When $r>r_{+}$, the expression in braces can be shown to always be positive.

The two solutions for $\nu^2$ and $E^2=(\nu^2 u^3 +1-u)$ then take the form
\begin{equation}\label{KerrEq-M-Nu2Sol}
\nu^{2}=\frac{(u^{-1/2}\pm\alpha)^2}
{1-3u\mp 2\alpha u^{3/2}},;\qquad
E^{2}=\frac{(1-2u\mp\alpha u^{3/2})^{2}}
{1-3u\mp 2\alpha u^{3/2}}.
\end{equation}
Restoring relative signs of $E$ and $\nu$ from the condition $U_{eff}'=0$, we finally obtain
\begin{equation}\label{KerrEq-M-La2Sol}
\lambda^{2}
=\frac{(u^{-1/2}\pm 2\alpha u+\alpha^2 u^{3/2})^{2}}
{1-3u\mp 2\alpha u^{3/2}}.
\end{equation} </p>
</div>
</div>

<div id="BlackHole87"></div>
=== Problem 36: innermost stable circular orbits, massive case ===
Derive equation for the minimal radius of a stable circular orbit; find the energy and angular momentum of a particle on it, the minimal radius in the limiting cases $a/\mu\to 0,1$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
An orbit's stability condition is $3\nu^2 u^2 <1$. Substituting $\beta$, it is brought to
\[3\nu^2 u^2 <1,\]
and substituting $\nu^2$, to
\[3\alpha^2 u^2 \pm 8\alpha u^{3/2}+6u-1<0.\]
In terms of $\xi$
\[\xi^2-6\xi\pm 8\alpha \sqrt{\xi}-3\alpha^2>0.\]
In the limit $\alpha\to 0$ we get $\xi>6$, i.e. $r>3\mu=\frac{3}{2}r_{g}$, which is the familiar result for Schwarzschild.</p>

[[File:BHfig-Kerr-XiMin.png|center|thumb|400px| " ''Radius'' of innermost stable circular orbit as function of $\alpha=a/\mu$: the upper curve for $L<0$, the lower one for $L>0$. The horizon is shown by the dashed line."]]

<p style="text-align: left;">In the limit $a\to 1$ different signs lead to different inequalities, which are easier to solve numerically. From the curves we see, that for the "$-$" sign (retrograde or counter-rotating orbits) the stability condition holds when $\xi>9$, and for the "$+$" sign (prograde or co-rotating orbits) when $\xi>1$. In the general case then
\begin{align*}
&''-'':\qquad \alpha\in(0,1)\;\Rightarrow\;
\xi_{min}\in(6,9);\\
&''+'':\qquad \alpha\in(0,1)\;\Rightarrow\;
\xi_{min}\in(6,1).
\end{align*}
The binding energy on the minimal stable (prograde) orbit is
\[E^2=\nu^2 u^3+1-u=u/3+1-u=1-\frac{2}{3\xi},\]
and for $\alpha\approx 1$ for the most strongly bound particle with $\xi\approx 1$
\[E_{min} \approx \sqrt{1-\frac{2}{3}}
=\frac{1}{\sqrt{3}}, \]
so the binding energy in the units of rest mass is
\[E_{acc}\approx 1-\frac{1}{\sqrt{3}}\approx 0.42.\]
In the model of $\alpha$-disk accretion on a compact object this number gives the upper limit to the accretion effectiveness, i.e. the part of a particle's rest mass that can be radiated into outer space due to dissipation in the disk caused by slow slipping of particles into the gravitational well along almost circular orbits. </p>
</div>
</div>

Category:Horizons

2013-10-31T08:39:12Z

Igor: /* References */

In this chapter we assemble the problems that concern one of the most distinctive features of General Relativity and Cosmology --- the horizons. The first part gives an elementary introduction into the concept in the cosmological context, following and borrowing heavily from the most comprehensible text by E. Harrison <ref name=Harrison />. Then we move to more formal exposition of the subject, making use of the seminal works of W. Rindler<ref name=Rindler /> and G.F.R. Ellis, T. Rothman<ref name=Ellis />, and consider first simple, and then composite models, such as $\Lambda$CDM. The fourth section elevates the rigor one more step and explores the causal structure of different simple cosmological models in terms of conformal diagrams, following mostly the efficient approach of V. Mukhanov<ref name=Mukhanov />, and inflation. The section on black holes relates the general scheme of constructing conformal diagrams for stationary black hole spacetimes, following mostly the excellent textbook of K. Bronnikov and S. Rubin<ref name=BrRubin />. The consequent sections focus on more specific topics, such as the various problems regarding the Hubble sphere, inflation and holography.

==References==
<references>
<ref name=Harrison>E. Harrison. Cosmology: the science of the Universe. CUP (1981) ISBN 978-0-521-66148-5</ref>
<ref name="Rindler">W. Rindler. Visual horizons in world-models. [http://adsabs.harvard.edu/abs/1956MNRAS.116..662R MNRAS 116 (6), 662--677 (1956) (1981)]</ref>
<ref name="Ellis">G.F.R. Ellis and T. Rothman. Lost horizons. Am. J. Phys. 61, 883 (1993)</ref>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN 0521563984</ref>
<ref name="BrRubin">K.A. Bronnikov and S.G. Rubin. Black Holes, Cosmology and Extra Dimensions. (WSPC, 2012), ISBN 978-9814374200</ref>

</references>

Category:Horizons

2013-10-31T08:36:29Z

Igor: /* References */

In this chapter we assemble the problems that concern one of the most distinctive features of General Relativity and Cosmology --- the horizons. The first part gives an elementary introduction into the concept in the cosmological context, following and borrowing heavily from the most comprehensible text by E. Harrison <ref name=Harrison />. Then we move to more formal exposition of the subject, making use of the seminal works of W. Rindler<ref name=Rindler /> and G.F.R. Ellis, T. Rothman<ref name=Ellis />, and consider first simple, and then composite models, such as $\Lambda$CDM. The fourth section elevates the rigor one more step and explores the causal structure of different simple cosmological models in terms of conformal diagrams, following mostly the efficient approach of V. Mukhanov<ref name=Mukhanov />, and inflation. The section on black holes relates the general scheme of constructing conformal diagrams for stationary black hole spacetimes, following mostly the excellent textbook of K. Bronnikov and S. Rubin<ref name=BrRubin />. The consequent sections focus on more specific topics, such as the various problems regarding the Hubble sphere, inflation and holography.

==References==
<references>
<ref name=Harrison>E. Harrison. Cosmology: the science of the Universe. CUP (1981)</ref>
<ref name="Rindler">W. Rindler. Visual horizons in world-models. [http://adsabs.harvard.edu/abs/1956MNRAS.116..662R MNRAS 116 (6), 662--677 (1956) (1981)]</ref>
<ref name="Ellis">G.F.R. Ellis and T. Rothman. Lost horizons. Am. J. Phys. 61, 883 (1993)</ref>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN 0521563984</ref>
<ref name="BrRubin">K.A. Bronnikov and S.G. Rubin. Black Holes, Cosmology and Extra Dimensions. (WSPC, 2012), ISBN 978-9814374200</ref>

</references>

Causal Structure

2013-10-09T13:47:28Z

Igor: minor edits

[[Category:Horizons|4]]

__TOC__

The causal structure is determined by propagation of light and is best understood in terms of ''conformal diagrams''. In this section we construct and analyze those for a number of important model cosmological solutions (which are assumed to be already known), following mostly the exposition of <ref name=Mukhanov/>.

In terms of comoving distance $\tilde{\chi}$ and conformal time $\tilde{\eta}$ (in this section they are denoted by tildes) the two-dimensional radial part of the FLRW metric takes form
\begin{equation}
ds_{2}^{2}=a^{2}(\tilde{\eta})\big[d\tilde{\eta}^2 -d\tilde{\chi}^2\big].
\label{ds2Dconf}
\end{equation}
In the brackets here stands the line element of two-dimensional Minkowski flat spacetime. Coordinate transformations that preserve the \emph{conformal} form of the metric
\begin{equation*}
ds_2^2 =\Omega^{2}(\eta,\chi)\big[d\eta^2 -d\chi^2 \big],
\end{equation*}
are called conformal transformations, and the corresponding coordinates $(\eta,\chi)$ -- conformal coordinates.

<div id="Horizon39"></div>
=== Problem 1 ===
Show that it is always possible to construct $\eta(\tilde{\eta},\tilde{\chi})$, $\chi(\tilde{\eta},\tilde{\chi})$, such that the conformal form of metric (\ref{ds2Dconf}) is preserved, but $\eta$ and $\chi$ are bounded and take values in some finite intervals. Is the choice of $(\eta,\chi)$ unique?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Suppose $\tilde{\eta},\tilde{\chi}$ span infinite or semi-infinite values. Then we can always make the following sequence of coordinate transformations:
<div style="text-align: left;">
* Pass to null coordinates
\begin{equation}
u=\tilde{\eta}-\tilde{\chi},\qquad v=\tilde{\eta}+\tilde{\chi};
\end{equation}
* Bring their range of values to a finite interval by some appropriate function, i.e.
\begin{equation}
U=\arctan u,\qquad V=\arctan v.
\end{equation}
* Go back to timelike and spacelike coordinates (this is not really necessary at this point and is done mostly for aesthetic reasons):
\begin{equation}
T=V+U,\qquad R=V-U.
\end{equation}
Now the range of $(T,R)$ obviously covers some bounded region on the plane, while the radial part of the line element preserves its conformal form:
\begin{equation}
ds_2^2 \sim d\tilde{\eta}^2-d\tilde{\chi}^2 \sim du\, dv \sim dU\, dV \sim dT^2 -dR^2 .
\end{equation}
As the choice of function $\arctan$ was rather arbitrary (though convenient), the choice of conformal coordinates is not unique.
</div>
</p>
</div>
</div>

In this section we will reserve notation $\eta$ and $\chi$ and name "conformal coordinates/variables" to such variables that can only take values in a bounded region on $\mathbb{R}^2$; $\tilde{\eta}$ and $\tilde{\chi}$ can span infinite or semi-infinite intervals. Spacetime diagram in terms of conformal variables $(\eta,\chi)$ is called conformal diagram. Null geodesics $\eta=\pm \chi + const$ are diagonal straight lines on conformal diagrams.

<div id="Horizon40"></div>
=== Problem 2 ===
Construct the conformal diagram for the closed Universe filled with
* radiation;
* dust;
* mixture of dust and radiation.

Show the particle and event horizons for the observer at the origin $\chi=0$ (it will be assumed hereafter that the horizons are always constructed with respect to this chosen observer).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* Solution for $a(\eta)$ in a radiation dominated closed Universe is
\begin{equation}
a=a_m \sin\eta ,\qquad \eta\in(0,\pi),\quad \chi \in [0,\pi].
\end{equation}
As the ranges spanned by $\eta$ and $\chi$ are finite, they are already conformal coordinates. The conformal diagram is a square $\eta,\chi\in [0,\pi]$. Edges $\eta=0$ and $\eta=\pi$ correspond to the Big Bang and Big Crunch singularities respectively; worldline $\chi=\pi$ is the point on the three-sphere that is situated at the opposite pole with respect to observer at $\chi=0$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" | Conformal diagrams of radiation (left) and dust (right) dominated closed Universes. Particle horizon is shown in blue, event horizon in red. Thin lines in the dust dominated universe show light rays that realize the first and second images of a galaxy at different cosmic times, from opposite directions.
|[[File:Conf-Rad.png|center|thumb|365px]]
|[[File:Conf-Dust.png|center|thumb|365px]]
|}

The particle horizon is
\[\eta=\chi,\]
and event horizon is
\[\eta=\eta_{max}-\chi =\pi -\chi.\]
Both exist for all cosmological times: by the finite moment of the Big Crunch $\eta=\pi$ the event horizon is collapsed into a point (which is natural, as there is no more time left), while the particle horizon extends to the whole Universe. Thus only at the finite moment the whole of the Universe becomes observable. The farther the point, though, the younger will it look, of course, and the opposite pole will only be "observed" by our observer at the last moment of the Universe as it was at its creation.

* Solution for $a(\eta)$ in a dust dominated closed Universe is
\begin{equation}
a=a_m (1-\cos\eta) ,\qquad \eta\in(0,2\pi),\quad \chi \in [0,\pi].
\end{equation}
The difference from the previous case is that $\eta$ spans twice the range, and $\eta_{max}=2\pi =2\chi_{max}$.

Therefore the event horizon is given by
\[\eta=\eta_{max}-\chi=2\pi -\chi,\]
so it exists only in the second, contracting, phase $\eta>\pi$. The particle horizon is given by $\eta=\chi$ again, but now it exists only during the expanding phase $\eta<\pi$. It encloses the full Universe at the moment of maximal expansion $\eta=\pi$ and for later times does not exist.

* Though the full analytic solution is more complicated, it is clear that the main features remain the same as in the previous considered cases. At early and late times, close to the singularities, the dynamics is determined by the radiation component. If there is enough dust, then at large enough scale factors (which may or may not be achieved depending on the initial conditions), which would correspond to the epoch around the maximal expansion, it will be dominating. The influence of dust is that dynamics is slowed down, so that depending on the ratio of densities
\[\eta_{max}\in [\pi,2\pi].\]
Thus qualitatively the picture will be the same as in a dust dominated Universe: the conformal diagram is a rectangle, the event horizon exists only starting from some time $\eta_{e}=\eta_{max}-\pi$, while particle horizon, on the contrary, vanishes at $\eta_{p}=\pi$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" | Conformal diagrams of radiation (left) and dust (right) dominated closed Universes. Particle horizon is shown in blue, event horizon in red. Thin lines in the dust dominated universe show light rays that realize the first and second images of a galaxy at different cosmic times, from opposite directions.
|[[File:Conf-RadDust.png|center|thumb|365px|A closed Universe filled with a mix of dust and radiation.]]
|}
</div>
</p>
</div>
</div>

<div id="Horizon41"></div>
=== Problem 3 ===
'''Closed dS.''' Construct the conformal diagram for the de Sitter space in the closed sections coordinates. Provide reasoning that this space is (null) geodesically complete, i.e. every (null) geodesic extends to infinite values of affine parameters at both ends.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The scale factor in the closed dS Universe is
\begin{equation}
a(t)=H_\Lambda^{-1}\cosh (H_\Lambda t),\qquad t\in (-\infty, +\infty),
\end{equation}
so on integration, for conformal time we obtain
\begin{equation}
\eta (t)=\int\limits_{-\infty}^{t}\frac{dt}{a(t)}=\arcsin\big[\tanh (H_\Lambda t)\big]+\frac{\pi}{2} \in (0,\pi).
\end{equation}
We choose the integration constant here so that $\eta=0$ corresponds to $t=-\infty$ and $\eta=\pi$ to $t=+\infty$. The full metric then can be written as
\begin{equation}
ds^{2}_{dS}=\frac{H_\Lambda^2}{\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\sin^2 \chi d\Omega^2 \big]. \label{dSclosed}
\end{equation}

As the values of $\eta$ span a finite interval, $(\eta,\chi)$ are already conformal coordinates. The conformal diagram is again a square
\begin{equation}
\eta\in [0,\pi],\qquad \chi \in[0,\pi],
\end{equation}
with the difference from the radiation dominated Universe that the edges $\eta=0,\pi$ do not represent a singularity anymore, but instead correspond to infinite (and regular) past and future respectively. Both horizons are given again by
\begin{equation}
\eta_{e}=\pi-\chi,\qquad \eta_{p}=\chi
\end{equation}
and exist at all times.

The spacelike boundaries of the conformal diagram correspond to $t\to \pm\infty$, and therefore to infinite values of affine parameter. This can be shown if one remembers the general formula for the cosmological redshift:
\begin{equation}
\text{const}=\omega a =\frac{dt}{d\lambda}a,\quad \Rightarrow\quad
\lambda=\text{const}\cdot\int\limits^{t}dt \cosh(H_\Lambda t)
\underset{t\to\pm\infty}{\longrightarrow}\infty .
\end{equation}
The timelike boundary of the diagram corresponds to the opposite pole, there is no real boundary there, the same as on a sphere: as particles propagate across the pole, their radial coordinate begins to decrease again, while the worldline on the conformal diagram is reflected from $\chi=\pi$. Thus by definition the spacetime is (null) geodesically complete.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-closed.png|center|thumb|365px|The de Sitter Universe. The closed sections coordinates cover the whole space, which is geodesically complete. There are no singularities: the horizontal boundaries of the diagram correspond to infinite past and future.]]
|}
</p>
</div>
</div>

<div id="Horizon42"></div>
=== Problem 4 ===
'''Static dS.''' Rewrite the metric of de Sitter space (\ref{dSclosed}) in terms of "static coordinates" $T,R$:
\begin{equation}
\tanh (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad
H_\Lambda R =\frac{\sin\chi}{\sin\eta}.
\end{equation}
* What part of the conformal diagram in terms of $(\eta,\chi)$ is covered by the static coordinate chart $(T,R)$?
* Express the horizon's equations in terms of $T$ and $R$
* Draw the surfaces of constant $T$ and $R$ on the conformal diagram.
* Write out the coordinate transformation between $(\eta,\chi)$ and $(T,R)$ in the regions where $|\cos\eta|>|\cos\chi|$. Explain the meaning of $T$ and $R$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us introduce dimensionless coordinates $t=H_\Lambda T$ and $r=H_\Lambda R$. The inverse relations then are
\begin{equation}
\sin^2 \eta =\frac{1}{\cosh^2 t-r^2 \sinh^2 t},
\quad \sin^2 \chi =\frac{r^2}{\cosh^2 t-r^2 \sinh^2 t}.
\end{equation}
Using them, after some algebra from (\ref{dSclosed}) we get
\begin{equation}
ds_{dS}^2 =\big[1-H_\Lambda^2 R^2\big]dT^2 -\frac{dR^2}{1-H_\Lambda^2 R^2}-R^2 d\Omega^2 , \label{dSstatic}
\end{equation}
which resembles the Schwarzschild line element (and this is not a coincidence).

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-static.png|center|thumb|365px|The de Sitter Universe with contour lines of the "static" coordinates $(T,R)$. The solid lines are $T=const$, and the dashed ones $R=const$. The coordinates become singular on both horizons, so the whole spacetime is divided by them into four sectors, each separately covered by the regular coordinate chart $(R,T)$. Spacetime is actually static only in the left and right sectors (I and III), where $T$ is a timelike coordinate and $R$ spacelike: the static patches are bounded by the horizons.]]
|}

<div style="text-align: left;>
* As $|\tanh (H_\Lambda T)|<1$, we have $|\cos\eta| \leq |\cos\chi|$. This condition cuts out two out of four sectors from the square conformal diagram, I and III: one is $\eta\in [\chi, \pi-\chi]$ and the other is $\eta\in [\pi-\chi, \chi]$. In both $|\sin\eta|\geq |\sin\chi|$, so $R \leq H_{\Lambda}^{-1}$;
* The particle horizon $\eta=\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=-\infty$. It is one part of the boundary of the region (in two parts) covered by coordinates $(T,R)$. The event horizon $\eta=\pi-\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=+\infty$ and is the other part of the boundary of this region.
* $T=const$ is $\cos\eta =t\cos\chi $ and $R=const$ is $\sin\eta =r^{-1}\sin\chi$.
* In regions II and IV the needed relation is obtained if we simply replace $\tanh$ with $\coth$ in the first relation:
\begin{equation}
\coth (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad H_\Lambda R=\frac{\sin\chi}{\sin\eta},
\end{equation}
as can be checked explicitly by substitution into (\ref{dSstatic}), which again gives (\ref{dSclosed}). In these regions $R$ is a timelike coordinate, and $T$ is spacelike. The geodesics of comoving massive particles are $\chi=const$, one of them $\chi=\pi/2$ corresponds to $T=0$. Thus in the lower part of the diagram, where $R\in (+\infty,H_\Lambda^{-1})$, the spacetime is contracting; in the upper part, where $R\in (H_\Lambda^{-1},\infty)$, it is expanding. The coordinate frame is not static. The relations between $(T,R)$ and $(\eta,\chi)$ in static and non-static regions mirros those in Schwarzshild black hole solution between the static and the global (Kruskal-Szekeres) coordinates. Here $\eta$ and $\chi$ are the global coordinates.
</div>
</p>
</div>
</div>

<div id="Horizon43"></div>
=== Problem 4 ===
'''Flat dS.''' The scale factor in flat de Sitter is $a(t)=H_\Lambda^{-1} e^{H_\Lambda t}$.

* Find the range of values spanned by conformal time $\tilde{\eta}$ and comoving distance $\tilde{\chi}$ in the flat de Sitter space
* Verify that coordinate transformation
\begin{equation}
\tilde{\eta}=\frac{-\sin\eta}{\cos\chi-\cos\eta},\qquad
\tilde{\chi}=\frac{\sin\chi}{\cos\chi-\cos\eta}
\end{equation}
bring the metric to the form of that of de Sitter in closed slicing (it is assumed that $\tilde{\eta}=0$ is chosen to correspond to infinite future).
* Which part of the conformal diagram is covered by the coordinate chart $(\tilde{\eta},\tilde{\chi})$? Is the flat de Sitter space geodesically complete?
* Where are the particle and event horizons in these coordinates?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As $t\in(-\infty,+\infty)$,
\begin{equation}
\tilde{\eta} =\int\frac{dt}{a(t)}=H_\Lambda \int\limits_{+\infty}^{t}dt\; e^{-H_\Lambda t} =-e^{-H_\Lambda t} \in (-\infty,0).
\end{equation}
Here we choose $+\infty$ as the lower limit, because at $-\infty$ the integral diverges.
<div style="text-align: left;>
* Direct calculation yields (\ref{dSclosed}), with $\eta \in (0,\pi)$, $\chi\in (0,\pi)$;
* The upper triangle $\eta>\chi$, above the particle horizon, on which $\tilde{\eta}\to -\infty$. It is not geodesically complete, as geodesics are cut at the particle horizon.
* The particle horizon is the boundary of the patch of full dS space covered by flat slicing coordinates, and event horizon in these coordinates exists only in the latter part of evolution, for $\eta >\pi/2$.
</div>
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-flat.png|center|thumb|365px|The de Sitter Universe in flat sections' coordinates, which cover only half of it. The boundary -- the particle horizon -- consists of three different infinities.]]
|}
</p>
</div>
</div>

<div id="Horizon44"></div>
=== Problem 5 ===
'''Infinities.''' What parts of the spacetime's boundary on the conformal diagram of flat de Sitter space corresponds to

* spacelike infinity $i^0$, where $\tilde{\chi}\to +\infty$;
* past timelike infinity $i^-$, where $\tilde{\eta}\to -\infty$ and from which all timelike worldlines emanate
* past lightlike infinity $J^-$, from which all null geodesics emanate?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div style="text-align: left;">
* The one point at the left of the diagram;
* the one point at the bottom;
* the particle horizon.
</div>
</p>
</div>
</div>

<div id="Horizon45"></div>
=== Problem 6 ===
'''Open dS.''' Consider the de Sitter space in open slicing, in which $a(t)=H_\Lambda \sinh (H_\Lambda t)$, so conformal time is
\begin{equation}
\tilde{\eta} =\int\limits_{+\infty}^{t} \frac{dt}{a(t)},
\end{equation}
where again the lower limit is chosen so that the integral is bounded.

* Find $\tilde{\eta}(t)$ and verify that coordinate transformation from $(\tilde{\eta},\tilde{\chi})$ to $\eta,\chi$, such that
\begin{equation}
\tanh\tilde{\eta}=\frac{-\sin\eta}{\cos\cos\chi},\qquad
\tanh\tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
brings the metric to the form of de Sitter in closed slicing.
* What are the ranges spanned by $(\tilde{\eta},\tilde{\chi})$ and $(\eta,\chi)$? Which part of the conformal diagram do they cover?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
After getting
\begin{equation}
\sinh \tilde{\eta}=-\frac{1}{\sinh(H_\Lambda t)},
\end{equation}
the first part is checked straightforwardly; in the open de Sitter $\tilde{\chi}\in [0,+\infty)$, and $\tilde{\eta}\in(-\infty,0)$. The region covered by coordinates $(\tilde{\eta},\tilde{\chi})$ is $\{\eta>\chi+\pi/2\}$, only one eighth part of the full diagram.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-dS-open.png|center|thumb|365px|The de Sitter Universe in open sections' coordinates, which cover only $1/8^{\text{th}}$ of the full diagram.]]
|}
</p>
</div>
</div>

<div id="Horizon46"></div>
=== Problem 7 ===
'''Minkowski 1.''' Rewrite the Minkowski metric in terms of coordinates $(\eta,\chi)$, which are related to $(t,r)$ by the relation
\begin{equation}
\tanh \tilde{\eta}=\frac{\sin\eta}{\cos\chi},\qquad
\tanh \tilde{\chi}=\frac{\sin\chi}{\cos\eta}
\end{equation}
that mirrors the one between the open and closed coordinates of de Sitter. Construct the conformal diagram and determine different types of infinities. Are there new ones compared to the flat de Sitter space?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Coordinate transformation gives
\begin{equation}
ds^2 =\frac{1}{\cos^2 \chi -\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\Psi^{2}(\eta,\chi)d\Omega^2 \big].
\end{equation}
Here $r\in [0,+\infty)$ and $t\in (-\infty,+\infty)$. Comparing with the relation between $(\tilde{\eta},\tilde{\chi})$ with conformal coordinates $(\eta,\chi)$ in the open de Sitter universe, where $\tilde{\eta}\in (-\infty,0)$, we see that the difference is that $\tilde{\eta}$ spans $(-\infty,0)$, while now $t$ spans twice the range, $(-\infty,\infty)$. Therefore the conformal diagram is composed of two triangles, one the same as for open de Sitter and one for its time-reversed copy. Accordingly there now appear future timelike infinity $i^+$ and future lightlike infinity $J^+$.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |The Minkowski spacetime in two different pairs of conformal coordinates and the full set of infinities. Thin dashed and solid lines show the images of coordinate grid $(t,r)$.

|[[File:Conf-Mink1.png|center|thumb|365px]]
|[[File:Conf-Mink2.png|center|thumb|365px]]
|}
</p>
</div>
</div>

<div id="Horizon47"></div>
=== Problem 8 ===
'''Minkowski 2.''' The choice of conformal coordinates is not unique. Construct the conformal diagram for Minkowski using the universal scheme: first pass to null coordinates, then bring their span to finite intervals with $\arctan$ (one of the possible choices), then pass again to timelike and spacelike coordinates.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
We start from spherical coordinates
\begin{equation}
ds^2 =dt^2 -dr^2 -r^2 d\Omega^2 .
\end{equation}
<div style="text-align: left;">
* The first step is introducing null coordinates
\begin{equation}
u=t-r,\quad v=t+r ,
\end{equation}
so that
\begin{equation}
ds^2 =4du\,dv -\frac{(v-u)^2}{4}d\Omega^2 .
\end{equation}
* Then bring the range of values to finite intervals
\begin{equation}
U=\arctan u, \qquad V=\arctan v,
\end{equation}
so that
\begin{equation}
ds^2 =\frac{1}{4\cos^2 U \cos^2 V}\big[4dU\,dV -\sin^2 (V-U)d\Omega^2\big].
\end{equation}
The whole spacetime is simply the half of the square $U,V\in (-\pi/2,\pi/2)$, in which $r>0$, i.e. $v>u\quad \Leftrightarrow\quad V>U$: on the plane $(U,V)$ it is the triangle \begin{equation}
-\pi/2<U<V<\pi/2.
\end{equation}
* Finally, go back to spacelike and timelike coordinates
\begin{equation}
T=V+U,\qquad R=V-U
\end{equation}
so that metric becomes
\begin{equation}
ds^2 =\frac{1}{[\cos T +\cos R]^2}\big[dT^2 -dR^2 -\sin^2 R\; d\Omega^2\big].
\end{equation}
The triangle is shrunken by $\sqrt{2}$ and rotated by $3\pi/4$ clockwise, thus turning into
\begin{equation}
\{R>0,\quad |T|<\pi/2 -R\}.
\end{equation}
This is the same form as obtained by the other construction (up to scaling, which is purely decorative).
</div>
</p>
</div>
</div>

<div id="Horizon48"></div>
=== Problem 9 ===
Draw the conformal diagram for the Milne Universe and show which part of Minkowski space's diagram it covers.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Minkowski metric $ds^2 =dT^2 -dR^2$, rewritten in terms of $(\tau, r)$ such that
\begin{equation}
T=\tau \cosh r ,\quad R=\tau \sinh r ,
\end{equation}
is the metric of the Milne Universe. As Minkowski space is complete, the Milne Universe then is a \emph{part} of Minkowski in different variables. This part is where $T>R$. The boundary $T=R$ is the future light cone, so the whole Milne Universe is represented by the triangle homothetic to the whole space but 4 times smaller in area, with the common node of future infinity.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Milne.png|center|thumb|365px|The Milne Universe on Minkowski's conformal diagram. It has the same shape and shares the same future infinity, but covers one eighth of the area. The boundary is the past horizon.]]
|}
</p>
</div>
</div>

<div id="Horizon49"></div>
=== Problem 10 ===
Consider open or flat Universe filled with matter that satisfies strong energy condition $\varepsilon +3p>0$. What are the coordinate ranges spanned by the comoving coordinate $\tilde{\chi}$ and conformal time $\tilde{\eta}$? Compare with the Minkowski metric and construct the diagram. Identify the types of infinities and the initial Big Bang singularity.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For flat Universe $\tilde{\chi}=r$, for the open $\tilde{\chi}=\sinh r$, so in both cases $\tilde{\chi}\in (0,+\infty)$.

Strong energy condition implies that $w>-1/3$, so
\begin{equation}
\rho \sim a^{-3(1+w)}=a^{-n},
\end{equation}
where $n>2$. From the first Friedman equation then after simple manipulations we obtain that
\begin{equation}
\frac{da}{dt}\sim a^{-\theta},
\end{equation}
where $\theta$ is some positive number. Therefore both
\begin{equation}
t\sim \int da\; a^{\theta},
\quad\text{and}\quad
\eta =\int \frac{da}{a}a^{\theta}
\end{equation}
converge at $a\to 0$ and diverge at $a\to \infty$. Consequently, the integration constant can be chosen so that $\eta\in (0,+\infty)$.

The conformal structure is the same as that of the \emph{upper} half of Minkowski spacetime. The Big Bang singularity at $\eta=0$ is at the cut, and there are spacelike infinity, future infinity and future null infinity.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" |Conformal diagrams for open or flat Universes. The one on the left is for ones filled with matter which satisfies the strong energy condition (SEC), and the one on the right for ones in which SEC does not hold, such as in case of power-law inflation..

|[[File:Conf-SEC.png|center|thumb|365px]]
|[[File:Conf-Inflation.png|center|thumb|365px]]
|}

</p>
</div>
</div>

<div id="Horizon50"></div>
=== Problem 11 ===
Draw the conformal diagram for open and flat Universes with power-law scale factor $a(t)\sim t^{n}$, with $n>1$. This is the model for the power-law inflation. Check whether the strong energy condition is satisfied.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As seen in the previous problem, if strong energy condition were satisfied, we would have $\dot{a}\sim a^{-\theta}$ with some positive $\theta$; this is not the case, so the condition is violated. As $t\in (0,+\infty)$,
\begin{equation}
\eta\sim \int \frac{dt}{t^n}
\end{equation}
diverges at small $a$ (thus also small $t$) and converges at $a\to \infty$ (thus as large $t$). So integration constant can be chosen so that $\eta\in (-\infty,0)$.

The conformal structure is the same as \emph{lower half} of Minkowski spacetime. The cut is regular future infinity, and from Minkowski there are spacelike infinity, past null infinity, and past infinity. The point of past infinity corresponds to Big Bang and is singular.
</p>
</div>
</div>

<references>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN~0521563984</ref>
</references>

Composite Models

2013-10-09T13:40:08Z

Igor: pr1: link to previous section

[[Category:Horizons|3]]
__TOC__
<div id="Horizon34"></div>
=== Problem 1 ===
Consider a flat universe with several components $\rho=\sum_i \rho_i$, each with density $\rho_i$ and partial pressure $p_i$ being related by the linear state equation $p_i =w_i \rho_i$. Find the particle horizon at present time $t_0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The particle horizon $L_p$ at the current moment $t_0$ is (see [[Simple_Math#LpDp|this relation]])
\[L_p =\lim\limits_{t_e \to 0} D_p (t_e ,t_0)\]
The proper distance can be written as
\[D_p = a_0\int_t^{t_0} \frac{dt'}{a(t')}
=- \int_{z(t)}^{z(t_0)} \frac{dz'}{H(z')}
=\int_0^z \frac{dz'}{H(z')}. \]
The $i$-th energy density in terms of redshift $z$ is
\[\rho_i =\rho_{0i}(1+z)^{3(1+w_i)},\]
and
\[H(z)=H_0 \sqrt{\sum \Omega_{0i}(1+z)^{3(1+w_i)}},\]
so
\begin{equation}
L_p (t_0)=H_0^{-1}\int \limits_{0}^{\infty}
\frac{dz}{\sqrt{\sum \Omega_{0i}(1+z)^{3(1+w_i)}}}
\end{equation}
</p>
</div>
</div>

<div id="Horizon35"></div>
=== Problem 2 ===
Suppose we know the current material composition of the Universe $\Omega_{i0}$, $w_i$ and its expansion rate as function of redshift $H(z)$. Find the particle horizon $L_p (z)$ and the event horizon $L_p (z)$ (i.e the distances to the respective surfaces along the surface $t=const$) at the time that corresponds to current observations with redshift $z$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the particle horizon we rewrite the previous result in terms of redshifts
\begin{equation}
L_p (z)=H^{-1} (z)\int \limits_{0}^{\infty}
\frac{dz'}{\sqrt{\sum \Omega_{i}(z)(1+z')^{3(1+w_i)}}},
\end{equation}
and take into account how the partial densities $\Omega_i$ depend on time: they are defined to satisfy
\begin{equation}
H^2 (z)=H_0^2 \sum \Omega_{i0} (1+z)^{3(1+w_i)}.
\end{equation}
at any $z$ (or, equivalently, $t$), thus $\Omega_i (z)$ by definition is the ratio of the $i$th term of the sum to the whole sum at any moment of time:
\begin{equation}
\Omega_i (z) = \Omega _{i0} \frac{H_0^2 }{H^2 (z)}(1 + z)^{3(1+w_i)}.
\end{equation}
Then for the particle horizon (and for event horizon in the same way) we obtain
\begin{align}
L_p (z) &= \frac{1}{H(z)}\int_0^\infty
\frac{dz'}{\sqrt {\sum\limits_i \Omega_i (z)(1 + z')^{3(1 +w_i )}}}; \label{LpZ}\\
L_e (z) &= \frac{1}{H(z)}\int_{-1}^{0}
\frac{dz'}{\sqrt {\sum\limits_i \Omega_i (z)(1 + z')^{3(1 +w_i )}}} \label{LeZ}.
\end{align}
</p>
</div>
</div>

<div id="Horizon36"></div>
=== Problem 3 ===
When are $L_p$ and $L_e$ equal? It is interesting to know whether both horizons might have or not the same values, and if so, how often this could happen.

<div id="Horizon37"></div>

=== Problem 4 ===
Find the particle and event horizons for any redshift $z$ in the standard cosmological model -- $\Lambda$CDM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Using the general formulae (\ref{LpZ}--\ref{LeZ}), one obtains
\begin{align}
L_p(z) &= \frac{1}{H(z)}\int_0^\infty
\frac{dz'}{\sqrt {\Omega_m (z) (1 + z')^3 + \Omega_\Lambda (z)}},\\
L_e(z) &= \frac{1}{H(z)}\int_{-1}^{0}
\frac{dz'}{\sqrt {\Omega_m (z) (1 + z')^3 + \Omega_\Lambda (z)}},\\
&\Omega_m (z)=\Omega _{m0}\frac{H_0^2}{H(z)^2} (1 + z)^3 ,\\
&\Omega_\Lambda (z) = \Omega_{\Lambda 0} \frac{H_0^2}{H(z)^2}.
\end{align}
</p>
</div>
</div>

<div id="Horizon38"></div>
=== Problem 5 ===
Express the particle $L_p (z)$ and event $L_e (z)$ horizons in $\Lambda$CDM through the hyper-geometric function.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\begin{align}
L_p (z) &= \frac{2\sqrt {A(z)}}{H_0 \sqrt {\Omega _{\Lambda 0}}} F\left(\frac{1}{2},\frac{1}{6},\frac{7}{6}; - A(z) \right),\\
L_e (z) &= \frac{1}{H_0 \sqrt {\Omega _{\Lambda 0}}} F\left(\frac{1}{2},\frac{1}{3},\frac{4}{3}; - \frac{1}{A(z)} \right),\\
&A(z) = \frac{\Omega_{\Lambda 0}}{\Omega _{m0}} (1 + z)^{-3}.
\end{align}

Composite Models

2013-10-09T13:32:37Z

Igor: /* Problem 3 */ no solution

[[Category:Horizons|3]]
__TOC__
<div id="Horizon34"></div>
=== Problem 1 ===
Consider a flat universe with several components $\rho=\sum_i \rho_i$, each with density $\rho_i$ and partial pressure $p_i$ being related by the linear state equation $p_i =w_i \rho_i$. Find the particle horizon at present time $t_0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The particle horizon $L_p$ at the current moment $t_0$ is (see \ref{LpDp})
\[L_p =\lim\limits_{t_e \to 0} D_p (t_e ,t_0)\]
The proper distance can be written as
\[D_p = a_0\int_t^{t_0} \frac{dt'}{a(t')}
=- \int_{z(t)}^{z(t_0)} \frac{dz'}{H(z')}
=\int_0^z \frac{dz'}{H(z')}. \]
The $i$-th energy density in terms of redshift $z$ is
\[\rho_i =\rho_{0i}(1+z)^{3(1+w_i)},\]
and
\[H(z)=H_0 \sqrt{\sum \Omega_{0i}(1+z)^{3(1+w_i)}},\]
so
\begin{equation}
L_p (t_0)=H_0^{-1}\int \limits_{0}^{\infty}
\frac{dz}{\sqrt{\sum \Omega_{0i}(1+z)^{3(1+w_i)}}}
\end{equation}
</p>
</div>
</div>

<div id="Horizon35"></div>
=== Problem 2 ===
Suppose we know the current material composition of the Universe $\Omega_{i0}$, $w_i$ and its expansion rate as function of redshift $H(z)$. Find the particle horizon $L_p (z)$ and the event horizon $L_p (z)$ (i.e the distances to the respective surfaces along the surface $t=const$) at the time that corresponds to current observations with redshift $z$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the particle horizon we rewrite the previous result in terms of redshifts
\begin{equation}
L_p (z)=H^{-1} (z)\int \limits_{0}^{\infty}
\frac{dz'}{\sqrt{\sum \Omega_{i}(z)(1+z')^{3(1+w_i)}}},
\end{equation}
and take into account how the partial densities $\Omega_i$ depend on time: they are defined to satisfy
\begin{equation}
H^2 (z)=H_0^2 \sum \Omega_{i0} (1+z)^{3(1+w_i)}.
\end{equation}
at any $z$ (or, equivalently, $t$), thus $\Omega_i (z)$ by definition is the ratio of the $i$th term of the sum to the whole sum at any moment of time:
\begin{equation}
\Omega_i (z) = \Omega _{i0} \frac{H_0^2 }{H^2 (z)}(1 + z)^{3(1+w_i)}.
\end{equation}
Then for the particle horizon (and for event horizon in the same way) we obtain
\begin{align}
L_p (z) &= \frac{1}{H(z)}\int_0^\infty
\frac{dz'}{\sqrt {\sum\limits_i \Omega_i (z)(1 + z')^{3(1 +w_i )}}}; \label{LpZ}\\
L_e (z) &= \frac{1}{H(z)}\int_{-1}^{0}
\frac{dz'}{\sqrt {\sum\limits_i \Omega_i (z)(1 + z')^{3(1 +w_i )}}} \label{LeZ}.
\end{align}
</p>
</div>
</div>

<div id="Horizon36"></div>
=== Problem 3 ===
When are $L_p$ and $L_e$ equal? It is interesting to know whether both horizons might have or not the same values, and if so, how often this could happen.

<div id="Horizon37"></div>

=== Problem 4 ===
Find the particle and event horizons for any redshift $z$ in the standard cosmological model -- $\Lambda$CDM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Using the general formulae (\ref{LpZ}--\ref{LeZ}), one obtains
\begin{align}
L_p(z) &= \frac{1}{H(z)}\int_0^\infty
\frac{dz'}{\sqrt {\Omega_m (z) (1 + z')^3 + \Omega_\Lambda (z)}},\\
L_e(z) &= \frac{1}{H(z)}\int_{-1}^{0}
\frac{dz'}{\sqrt {\Omega_m (z) (1 + z')^3 + \Omega_\Lambda (z)}},\\
&\Omega_m (z)=\Omega _{m0}\frac{H_0^2}{H(z)^2} (1 + z)^3 ,\\
&\Omega_\Lambda (z) = \Omega_{\Lambda 0} \frac{H_0^2}{H(z)^2}.
\end{align}
</p>
</div>
</div>

<div id="Horizon38"></div>
=== Problem 5 ===
Express the particle $L_p (z)$ and event $L_e (z)$ horizons in $\Lambda$CDM through the hyper-geometric function.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\begin{align}
L_p (z) &= \frac{2\sqrt {A(z)}}{H_0 \sqrt {\Omega _{\Lambda 0}}} F\left(\frac{1}{2},\frac{1}{6},\frac{7}{6}; - A(z) \right),\\
L_e (z) &= \frac{1}{H_0 \sqrt {\Omega _{\Lambda 0}}} F\left(\frac{1}{2},\frac{1}{3},\frac{4}{3}; - \frac{1}{A(z)} \right),\\
&A(z) = \frac{\Omega_{\Lambda 0}}{\Omega _{m0}} (1 + z)^{-3}.
\end{align}

Simple Math

2013-10-09T13:31:41Z

Igor: pr12 solution formatting

[[Category:Horizons|2]]

The problems of this section need basic understanding of Friedman equations, definitions of proper, comoving and conformal coordinates, the cosmological redshift formula and simple cosmological models (see Chapters 2 and 3).

Let us make our definitions a little more strict.

A '''particle horizon''', for a given observer $A$ and cosmic instant $t_0$ is a surface in the instantaneous three-dimensional section $t=t_0$ of space-time, which divides all comoving particles <ref>Rindler uses the term "fundamental observers".</ref> into two classes: those that have already been observable by $A$ up to time $t_0$ and those that have not.

An '''event horizon''', for a given observer $A$, is a hyper-surface in space-time, which divides all events into two non-empty classes: those that have been, are, or will be observable by $A$, and those that are forever outside of $A$'s possible powers of observation. It follows from definition that event horizon, and its existence, depend crucially on the observer's (and the whole Universe's) future as well as the past: thus it is said to be an essentially ''global concept''. It is formed by null geodesics.

The following notation is used hereafter: $L_p (t_0)$ is the proper distance from observer $A$ to its particle horizon, measured along the slice $t=t_0$. For brevity, we will also call this distance simply "the particle horizon in proper coordinates", or just "particle horizon". The corresponding comoving distance $l_p$ is the particle horizon in comoving coordinates.

Likewise, $L_e$ is the proper distance from an observer to its event horizon (or, rather, its section with the hypersurface $t=t_0$), measured also along the slice $t=t_0$. It is called "space event horizon at time $t_0$", or just the event horizon, for brevity. The respective comoving distance is denoted $l_e$.

__TOC__

<div id="Horizon12"></div>
=== Problem 1 ===
The proper distance $D_p (t_0)$ between two comoving observers is the distance measured between them at some given moment of cosmological time $t=t_0$. It is the quantity that would be obtained if all the comoving observers between the given two measure the distances between each other at one moment $t=t_0$ and then sum all of them up. Suppose one observer detects at time $t_{0}$ the light signal that was emitted by the other observer at time $t_e$. Find the proper distance between the two observers at $t_0$ in terms of $a(t)$.

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<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> By definition, measuring distance along the slice $t=const$, we have for some given time $t_0$
\begin{equation}
D_p (r, t_0)=\int |ds| = a_0 \int_0^r \frac{dr}{\sqrt {1 - k r^2} },
\quad {a_0} \equiv a (t_0)
\end{equation}

On the other hand, along the worldline of the light ray connecting the two observers $ds=0$, so we have
\[\frac{dt}{a} = \frac{dr}{\sqrt {1 - k r^2 }}.\]
Integrating along the worldline from the time of emission $t_e$ to the time of detection $t_0$, we get
\[\int_{t_e}^{t_0} \frac{dt'}{a(t')}
= \int_0^r \frac{dr}{\sqrt {1 - k r^2}} = \frac{D_p}{a_0},\]
Therefore the proper distance between two observers in an expanding Universe is
\begin{equation}
D_p (t_e, t_0)= a_0\int_{t_e}^{t_0} \frac{dt'}{a(t')}.
\end{equation}
</p>
</div>
</div>

<div id="Horizon13"></div>
=== Problem 2 ===
Show, that the proper distance $L_p$ to the particle horizon at time $t_0$ is
\begin{equation}
L_p (t_0)=\lim\limits_{t_e \to 0}D_p (t_e ,t_0). \label{LpDp}
\end{equation}
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> As define above, particle horizon is the proper distance between the observer that receives the light signal at present and the comoving particle that emitted this light at the very beginning of the Universe (which may correspond to $t\to -\infty$). Thus (\ref{LpDp}).
</p>
</div>
</div>

<div id="Horizon14"></div>
=== Problem 3 ===
The past light cone of an observer at some time $t_0$ consists of events, such that light emitted in each of them reaches the selected observer at $t_0$. Find the past light cone's equation in terms of proper distance vs. emission time $D_{plc} (t_e)$. What is its relation to the particle horizon?
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<p style="text-align: left;">
\begin{equation}
D_{plc}(t_e ,t_0) =a(t_e) \int\limits_{t_e}^{t_0} \frac{c\,dt}{a(t)}.
\label{Dplc}
\end{equation}
The corresponding comoving distance $d(t_e ,t_0)=D(t_e, t_0)/a(t_e)$ at $t_e \to t_{in}$ ($t_{in}$ is the time of creation singularity or infinite past, whichever is realized) gives us the observer that only now (at $t_0$) just becomes observable, thus determining the particle horizon at $t_0$:
\[L_p (t_0)=a(t_0) d_{plc}(t_{in}, t_0).\]
</p>
</div>
</div>

<div id="Horizon15"></div>
=== Problem 4 ===
The simplest cosmological model is the one of \emph{Einstein-de Sitter}, in which the Universe is spatially flat and filled with only dust, with $a(t)\sim t^{2/3}$. Find the past light cone distance $D_{plc}$ (\ref{Dplc}) for Einstein-de Sitter.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\begin{equation}
D_{plc}(t_e, t_0)=3c (t_e^{2/3}t_0^{1/3}-t_e ).
\end{equation}
</p>
</div>
</div>

<div id="Horizon16"></div>
=== Problem 5 ===
Demonstrate that in general $D_{plc}$ can be non-monotonic. For the case of Einstein-de Sitter show that its maximum -- the maximum emission distance -- is equal to $8/27 L_H$, while the corresponding redshift is $z=1.25$.
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<div class="NavHead">solution</div>
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<p style="text-align: left;"> From $dD_{plc}/dt_e =0$ we see that maximum exists and lies at
\begin{equation}
\frac{t_e}{t_0}=\frac{8}{27},\qquad D_{plc}^{max}=\frac{4}{9}ct_0 .
\end{equation}
The redshift of light signal emitted at maximum emission distance is then
\begin{equation}
z=\frac{a(t_0)}{a(t_e)}-1=\Big(\frac{t_0}{t_e}\Big)^{2/3}-1 =\frac{9}{4}-1=1.25 .
\end{equation}
</p>
</div>
</div>

<div id="Horizon17"></div>
=== Problem 6 ===
In a matter dominated Universe we see now, at time $t_0$, some galaxy, which is now on the Hubble sphere. At what time in the past was the photon we are registering emitted?
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> The emission event is at the intersection of the particle's worldline with the past light cone, so
\begin{equation}
\tfrac{3}{2}ct_0 \;\Big(\frac{t_e}{t_0}\Big)^{2/3}
=3c (t_e^{2/3}t_0^{1/3}-t_e ),
\end{equation}
from which
\[t_e =\frac{t_0}{8}.\]
</p>
</div>
</div>

<div id="Horizon18"></div>
=== Problem 7 ===
Show that the particle horizon in the Einstein-de Sitter model recedes at three times the speed of light.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> In Einstein-de Sitter $a\sim t^{2/3}$, thus $L_p =3ct$ and $H=2/3t$, so
\begin{equation}
\dot{L}_p =c+ \frac{2}{3t}\cdot 3ct =3c.
\end{equation}
</p>
</div>
</div>

<div id="Horizon19"></div>
=== Problem 8 ===
Does the number of observed galaxies in an open Universe filled with dust increase or decrease with time?

<div id="Horizon20"></div>
=== Problem 9 ===
Draw the past light cones $D_{plc}(t_e)$ for Einstein-de Sitter and a Universe with dominating radiation on one figure; explain their relative position.

<div id="Horizon21"></div>
=== Problem 10 ===
Find the maximum emission distance and the corresponding redshift for power law expansion $a(t)\sim (t/t_0 )^n$.
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<p style="text-align: left;"> The comoving distance along a null geodesic is $\int d\eta =\int dt/a(t)$. Then at the time of emission $t_e$ the proper distance between the comoving emitter and detector is
\begin{equation}
L (t_e ,t_0)=a(t_e )\int\limits_{t_e}^{t_0}\frac{dt}{a(t)}=\frac{t_0}{1-n}\big[x^n -1\big],
\end{equation}
where $t_0$ is the time of detection (present), and
\begin{equation}
x=\frac{t_e}{t_0}.
\end{equation}
This is the past light cone of an event at $t_0$ given in terms of proper distance and cosmic time.

The maximum of $L(t_e ,t_0)$ is at
\begin{equation}
x_m=n^{1/(1-n)}, \qquad L_{\max}=L(x_m t_0, t_0)
=\frac{t_0}{n}x_m =\frac{t_0}{n}n^{1/(1-n)}.
\end{equation}
The corresponding redshift is given by the general relation
\begin{equation}
z_{\max}+1 =\frac{a(t_0)}{a(xt_0)}=n^{n/(n-1)}.
\end{equation}

For radiation domination ($n=1/2$) we have
\begin{equation}
L_{\max}=\tfrac{1}{2}ct_0=\tfrac{1}{2}R_H,\qquad z_{\max}=1,
\end{equation}
and for matter domination ($n=2/3$)
\begin{equation}
L_{\max}=\tfrac{4}{9}ct_0 =\tfrac{8}{27}R_H ,\qquad z=1.25.
\end{equation}
</p>
</div>
</div>

<div id="Horizon22"></div>
=== Problem 11 ===
Show that the most distant point on the past light cone was exactly at the Hubble sphere at the moment of emission of the light signal that is presently registered.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> The recession velocity of the emitter at the time of emission $t_e =x_m t_0$ is
\begin{equation}
v(t_e ,L_{\max})=H(x_m t_0)\cdot L_{\max}=\frac{n}{x_m t_0}\cdot \frac{t_0}{n}x_m =1 ,
\end{equation}
so by definition the emitter (i.e. galaxy) was at that moment exactly on the Hubble sphere.
</p>
</div>
</div>

<div id="Horizon23"></div>
=== Problem 12 ===
Show that the comoving particle horizon is the age of the Universe in conformal time
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Starting from the definition
\begin{equation}
\frac{L_p (t)}{a(t)} = \int_0^t \frac{dt'}{a(t')} = \int_0^\eta d\eta ' = \eta .
\end{equation}
</p>
</div>
</div>

<div id="Horizon24"></div>
=== Problem 13 ===
Show that
\begin{align}
\frac{dL_p}{dt}&=L_p (z)H(z)+1;\\
\frac{dL_e}{dt}&=L_e (z)H(z)-1.
\end{align}
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> From the definitions (here $c=1$)
\begin{align}
\dot{L_p}&=\frac{d}{dt}\Big(a(t)\int\limits^{t}\frac{dt}{a(t)}\Big)= +1+HL_p , \label{dotLp}\\
\dot{L_e}&=\frac{d}{dt}\Big(a(t)\int\limits_{t}\frac{dt}{a(t)}\Big)= -1+HL_e .\label{dotLe}
\end{align}
</p>
</div>
</div>

<div id="Horizon25"></div>
=== Problem 14 ===
Find $\ddot{L_p}$ and $\ddot{L_e}$.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> Differentiating $L_p$ and $L_e$ from (\ref{dotLp}-\ref{dotLe}) once again, we get
\begin{align}
\frac{d^2 L_p}{dt^2} &= H(+1-q H L_p),\\
\frac{d^2 L_e}{dt^2} &= H(-1-qH L_e),
\end{align}
where $q=-\frac{\ddot{a}/a}{H^2}$ is the deceleration parameter.
</p>
</div>
</div>

<div id="Horizon26"></div>
=== Problem 15 ===
Show that observable part of the Universe expands faster than the Universe itself. In other words, the observed fraction of the Universe always increases.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> The particle horizon at distance $L_p$ recedes with velocity $\dot{L}_p$ found in the previous problem, while the galaxies at the particle horizon recede at velocity $HL_p$, hence the horizon overtakes the galaxies with the speed of light $c$.

</p>
</div>
</div>

<div id="Horizon27"></div>
=== Problem 16 ===
Show that the Milne Universe has no particle horizon.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> In the Milne Universe $a\sim t$, $H=1/t$, $q=0$, the lightcone reaches the beginning of time $t=0$ at an infinite comoving distance and there is no particle horizon. The observable universe fills the entire actual Universe and all galaxies are in principle visible. In other words, all galaxies are visible at some stage in their evolution.
</p>
</div>
</div>

<div id="Horizon28"></div>
=== Problem 17 ===
Consider a universe which started with the Big Bang, filled with one matter component. How fast must $\rho(a)$ decrease with $a$ for the particle horizon to exist in this universe?
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> The comoving particle horizon is $\int_{0}dt a^{-1}(t)$, where $t=0$ is assumed to correspond to the Big Bang singularity. Its existence depends on whether this integral converges at small times or not. Then using the Friedman equation,
\begin{equation}
dt=\frac{da}{\dot{a}}=\frac{da}{\sqrt{\rho a^2}},
\end{equation}
so the particle horizon is
\begin{equation}
l_p = \int\limits_0 \frac{da /a}{\sqrt{\rho a^2}}.
\end{equation}
The integral converges as long as $\rho a^2$ diverges at small $a$. That is, if equation of state is such that $\rho$ changes faster than $\sim 1/a^2$, then light can only propagate a finite distance between Big Bang and now. In particular, the particle horizon exists in models, which are dominated at early times either by radiation $\rho\sim 1/a^4$ or matter $\rho\sim 1/a^3$.
</p>
</div>
</div>

<div id="Horizon29"></div>
=== Problem 18 ===
Calculate the particle horizon for a universe with dominating
* radiation;
* matter.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> In radiation-dominated universe $a\sim t^{1/2}$, in matter-dominated $a\sim t^{2/3}$. Then integration of $L =a(t)\int dt/a(t)$ gives
\begin{equation}
L_{p}^{(rad)}=2t,\qquad L_p^{(mat)}=3t.
\end{equation}

</p>
</div>
</div>

<div id="Horizon30"></div>
=== Problem 19===
Consider a flat universe with one component with state equation $p =w \rho$. Find the particle horizon at present time $t_0$.
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<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> The particle horizon $L_p$ at the current moment $t_0$ is (see \ref{LpDp})
\[L_p =\lim\limits_{t_e \to 0} D_p (t_e ,t_0)\]
The proper distance can be written as
\begin{equation}
D_p = a_0\int_t^{t_0} \frac{dt'}{a(t')}
=- \int_{z(t)}^{z(t_0)} \frac{dz'}{H(z')}
=\int_0^z \frac{dz'}{H(z')}.
\end{equation}
The density in terms of redshift $z$ is
\[\rho =\rho_{0}(1+z)^{3(1+w)},\]
and
\[H(z)=H_0 \sqrt{(1+z)^{3(1+w)}},\]
so
\begin{equation}
L_p (t_0)=H_0^{-1}\int \limits_{0}^{\infty}
\frac{dz}{\sqrt{(1+z)^{3(1+w_i)}}}
\end{equation}
</p>
</div>
</div>

<div id="Horizon31"></div>
=== Problem 20 ===
Show that in a flat universe in case of domination of one matter component with equation of state $p=w\rho$, $w>-1/3$
\begin{equation}
L_{p} (z)=\frac{2}{H(z) (1+3w)},\qquad \dot{L_p}(z)=\frac{3(1+w)}{(1+3w)}.
\end{equation}
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> In the considered case
\begin{equation}
L_p (z) = \frac{1}{H(z)}\int_0^\infty \frac{dz'}{\sqrt{(1 + z')^{3(1 + w)}}}
= \frac{2}{H(z)(1 + 3w)}.
\end{equation}
Note that $L_p >0$, i.e. the horizon exists, only if $w>-1/3$.

On differentiating by time and using
\begin{equation}
H^2 = \frac{1}{3}\rho ,\qquad \dot H =- \frac{1}{2}\rho (1 + w),
\end{equation}
we get
\begin{equation}
\frac{dL_p}{dt} = \frac{3(1 + w)}{1 + 3w}.
\end{equation}

</p>
</div>
</div>

<div id="Horizon32"></div>
=== Problem 21 ===
Show that in a flat universe in case of domination of one matter component with equation of state $p=w\rho$, $w<-1/3$
\begin{equation}
L_{e} (z)=-\frac{2}{H(z) (1+3w)},\qquad \dot{L_p}(z)=-\frac{3(1+w)}{(1+3w)}.
\end{equation}
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> In the considered case
\begin{equation}
L_e (z) = \frac{1}{H(z)}\int_{-1}^{0} \frac{dz'}{\sqrt {(1 + z')^{3(1 + w)}}}
= -\frac{2}{H(z)(1 + 3w)},
\end{equation}
and
\begin{equation}
\frac{dL_e}{dt} = -\frac{3(1 + w)}{1 + 3w}.
\end{equation}

Note that $L_e >0$, i.e. the event horizon exists, only if $w<-1/3$. In particular, when we have the cosmological constant, $w=-1$, there is only the event horizon
\[L_e =H .\]
</p>
</div>
</div>

<div id="Horizon33"></div>
=== Problem 22 ===
Estimate the particle horizon size at matter-radiation equality.

<references/>

Simple English

2013-10-07T20:42:59Z

Igor:

[[Category:Horizons|1]]

A vague definition of a horizon can be accepted as the following: it is a frontier between things observable and things unobservable.


'''''Particle horizon.'''''. If the Universe has a finite age, then light travels only a finite distance in that time and the volume of space from which we can receive information at a given moment of time is limited. The boundary of this volume is called the particle horizon.

'''''Event horizon.''''' The event horizon is the complement of the particle horizon. The event horizon encloses the set of points from which signals sent at a given moment of time will never be received by an observer in the future.

Space-time diagram is a representation of space-time on a two-dimensional plane, with one timelike and one spacelike coordinate. It is typically used for spherically symmetric spacetimes (such as all homogeneous cosmological models), in which angular coordinates are suppressed.

References:<ref name=Harrison />
__TOC__

<div id="Horizon1"></div>
=== Problem 1 ===
Draw a space-time diagram that shows behaviour of worldlines of comoving observers in a
# stationary universe with beginning
# expanding universe in comoving coordinates
# expanding universe in proper coordinates

<div id="Horizon2"></div>
=== Problem 2 ===
Suppose there is a static universe with homogeneously distributed galaxies, which came into being at some finite moment of time. Draw graphically the particle horizon for some static observer.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The worldline $O$ represents our Galaxy from which we observe the universe. At the present moment we look out in the space and back in time and see other galaxies on our backward lightcone. Worldline $X$ determines the particle horizon. Objects (galaxies) beyond $X$ have not yet been observed by the present moment.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Harr1.png|center|thumb|400px| Stationary Universe with a beginning.]]
|}</p>
</div>
</div>
<div id="Horizon3"></div>
=== Problem 3 ===
How does the horizon for the given observer change with time?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

At present the observer $O$ sees no farther than $X$ (see Figure). At some later moment he sees beyond $X$ up to some worldline $Y$. The particle horizon thus recedes in the static universe and as time passes the part of the Universe we observe grows ever larger.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Harr2.png|center|thumb|365px| Evolution of horizon with time.]]
|}
</p>
</div>
</div>

<div id="Horizon4"></div>
=== Problem 4 ===
Is there an event horizon in the static Universe? What if the Universe ends at some finite time?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the Universe is eternal and galaxies shine forever, no event horizons exist. However, it does exist in a Universe that lives for some finite time (has an "end"). For an observer in such a Universe the event horizon is the lightcone built on its worldline at the last possible moment.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Harr4.png|center|thumb|400px| Stationary Universe with an ending.]]
|}</p>

<p style="text-align: left;"> Inside this ultimate lightcone are the events that have been seen by the end of the Universe, and outside are the events that can never been seen. The reason is that the lightcone cannot advance farther into time and all events outside of it remain unseen.
</p>
</div>
</div>

<div id="Horizon5"></div>
=== Problem 5 ===
''The horizon riddle.'' Consider two widely separated observers, A and B (see Figure). Suppose they have overlapping horizons, but each can apparently see things that the other cannot. We ask: Can B communicate to A information that extends A's knowledge of things beyond his horizon? If so, then a third observer C may communicate to B information that extends her horizon, which can then be communicated to A. Hence, an unlimited sequence of observers B, C, D, E,... may extend A's knowledge of the Universe to indefinite limits. According to this argument A has no true horizon. This is the horizon riddle. Try to resolve it for the static Universe.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Harr-Hor3.png|center|thumb|365px| The horizon riddle: can two observers with overlapping horizons pass information to each other regarding things outside of the other's horizon?]]
|}

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Suppose, for example, that luminous galaxies originated 10 billion years ago and the particle horizon is therefore at distance 10 billion light years. Observers A and B see each other and have overlapping horizons. Suppose A and B are separated by a distance of 6 billion light years (see Figure ).

B sends out information that travels at the speed of light and takes 6 billion years to reach A. Hence A receives from B information that was sent 6 billion years ago, when the Universe was 4 billion years old. But B's particle horizon in the past at the time when the information was sent was only 4 billion light years distant. Thus B's horizon at that time did not extend beyond A's present horizon. In other words, B communicates information to A by sending it at the speed of light on A's backward lightcone. But when B sends the information, her horizon extends no farther than A's horizon, and he cannot see farther than A.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Harr6.png|center|thumb|365px| The horizon riddle.]]
|}

</p>
</div>
</div>

<div id="Horizon6"></div>
=== Problem 6 ===
Suppose observer O in a stationary universe with beginning sees A in some direction at distance $L$ and B in the opposite direction, also at distance $L$. How large must $L$ be in order for A and B to be unaware of each other's existence at the time when they are seen by O?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Consider two visible bodies at equal distances in opposite directions from us, as shown by world lines A and B on the figure . We see these bodies, but can they see each other? Let $t$ be the time that it takes for light to travel to us from A and B. The time it takes the light to travel from A to B, or from B to A, is obviously $2t$. Hence when the Universe is older than $3t$, we not only see A and B, but they also see each other. If the Universe is younger than $3t$, and older than $t$, we see A and B, but they cannot yet see each other. There is thus a maximum distance beyond which the observed bodies A and B do not know of each other's existence. By examining the spacetime diagram, we see that this maximum distance is one third of the distance to the particle horizon. The answer to our question is that bodies at opposite directions and equal distances from us, which are larger than one third of the distance to the particle horizon, cannot at present see each other.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Harr8.png|center|thumb|365px| Two observers unaware of each other's existence by the time seen by some third observer.]]
|}

</p>
</div>
</div>

<div id="Horizon7"></div>
=== Problem 7 ===
Draw spacetime diagrams in terms of comoving coordinate and conformal time and determine whether event or particle horizons exist for:

* the universe which has a beginning and an end in conformal time. The closed Friedman universe that begins with Big Bang and ends with Big Crunch belongs to this class.

* the universe which has a beginning but no end in conformal time. The Einstein--de Sitter universe and the Friedman universe of negative curvature, which begin with a Big Bang and expand forever, belong to this class.

* the universe which has an end but no beginning in conformal time. The de Sitter and steady-state universes belong to this class.

* the universe which has no beginning and no ending in conformal time. The Einstein static and the Milne universes are members of this class.

Conformal time is the altered time coordinate $\eta=\eta (t)$, defined in such a way that lightcones on the spacetime diagram in terms of $\eta$ and comoving spatial coordinate are always straight diagonal lines, even when the universe is not stationary.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
{|class="wikitable" style="margin: 1em auto 1em auto;"
|+ align="bottom" | Spacetime diagrams for universes with and without beginnings and endings.
|[[File:Conf-Harr15.png|center|thumb|365px]]
|[[File:Conf-Harr16.png|center|thumb|365px]]
|-
|[[File:Conf-Harr17.png|center|thumb|365px]]
|[[File:Conf-Harr18.png|center|thumb|365px]]
|}
<div style="text-align:left;">
* The world line X is at the particle horizon. Notice the existence of an event horizon.
* The particle horizon is at world line X, and no event horizon exists.
* Only the event horizon exists in this case.
* There are no particle or event horizons.
</div>
</p>
</div>
</div>

<div id="Horizon8"></div>
=== Problem 8 ===
Draw the spacetime diagram in terms of comoving space and ordinary time or the universe with an end but no beginning in conformal time.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
There are universes that expand forever and yet have endings in conformal time. The de Sitter is of this kind and therefore has event horizons. The figure below  shows such a universe in a spacetime diagram of comoving space and cosmic time.

The event horizon is as shown. As the moment now advances into the infinite future, the observer’s backward lightcone approaches the event horizon more and more closely. For example, the observer O at moment $O$ sees event $a$. As the moment of observation advances into the unlimited future, the lightcone moves upward and approaches more and more slowly but never reaches the event horizon. The event horizon is the observer’s lightcone in the infinite future. Events outside this horizon can never be observed.

{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Harr22.png|center|thumb|365px|A universe with an end but no beginning in conformal time, drawn in terms of cosmic time.]]
|}
</p>
</div>
</div>

<div id="Horizon9"></div>
=== Problem 9 ===
Formulate the necessary conditions in terms of conformal time for a universe to provide a comoving observer with
* a particle horizon
* an event horizon

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

<div style="text-align:left;">
* The necessary condition for the existence of a particle horizon is that conformal time has a beginning (see items I and II of problem 7). The observer's lightcone stretches back and terminates at the lower boundary where the universe begins. When conformal time has no beginning, there is no lower boundary (see items III and IV of problem 7). In this case the lightcone stretches back without limit and intersects all world lines in the universe. In these universes there are no particle horizons. Note that beginning in conformal time does not necessarily mean beginning in ordinary time.
* The necessary condition for the existence of an event horizon is that conformal time has an ending. The event horizon in cosmology is thus nothing more than the observer's ultimate lightcone at the end of conformal time. All the events inside the event horizon (the ultimate lightcone) are at some time observed, and all events outside are never observed. Note that an end in conformal time does not necessarily mean an end in ordinary time.
</div>
</p>
</div>
</div>

<div id="Horizon10"></div>
=== Problem 10 ===
Consider two galaxies, observable at present time, $A$ and $B$. Suppose at the moment of detection of light signals from them (now) the distances to them are such that $L_{det}^{A}<L^{B}_{det}$. In other words, if those galaxies had equal absolute luminosities, the galaxy $B$ would seem to be dimmer. Is it possible for galaxy $B$ (the dimmer one) to be closer to us at the moment of its signal's emission than galaxy $A$ (the brighter one) at the moment of $A$'s signal's emission?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Yes, it is possible. The corresponding spacetime diagram is shown on the figure below . Worldlines branch out radially in all directions from the ``big bang''. Spatial slices of constant cosmic time are represented by spherical surfaces perpendicular to the world lines, while time is measured along the radial worldlines. An arbitrary worldline is chosen as the observer and labeled $O$. At some instant in time -- let it be ``now'' -- the observer's lightcone stretches out and back and intersects other worldlines such as $X$ and $Y$. Because of the expansion of space, the lightcone does not stretch out straight as in a static universe, but contracts back into the big bang. All worldlines and all backward lightcones converge into the big bang.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Harr10.png|center|thumb|365px|The reception and emission distances of galaxies $X$ and $Y$. Although galaxy $Y$ has a greater reception distance, its emission distance is smaller than that of $X$. Thus $Y$, which is now farther away than $X$, was closer to us than $X$ at the time of emission (which is different for $X$ and $Y$) of the light we now see.]]
|}
</p>
</div>
</div>

<div id="Horizon11"></div>
=== Problem 11 ===
Show on a spacetime diagram the difference in geometry of light cones in universes with and without particle horizons.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
A spacetime diagram of comoving space (in which all worldlines are parallel) and cosmic time is shown below . Some universes have particle horizons and in their case the lightcone stretches out and back to the beginning at a finite comoving distance indicated by the world line $X$ . In universes without particle horizons, the lightcone stretches out to an unlimited distance and intersects all world lines.
{|class="wikitable" style="margin: 1em auto 1em auto;"
|[[File:Conf-Harr13.png|center|thumb|365px|Lightcones in universes with and without particle horizons.]]
|}
</p>
</div>
</div>

==References==
<references>
<ref name=Harrison>E. Harrison. Cosmology: the science of the Universe. CUP (1981)</ref>
</references>

Category:Horizons

2013-10-07T20:26:24Z

Igor:

In this chapter we assemble the problems that concern one of the most distinctive features of General Relativity and Cosmology --- the horizons. The first part gives an elementary introduction into the concept in the cosmological context, following and borrowing heavily from the most comprehensible text by E. Harrison <ref name=Harrison />. Then we move to more formal exposition of the subject, making use of the seminal works of W. Rindler<ref name=Rindler /> and G.F.R. Ellis, T. Rothman<ref name=Ellis />, and consider first simple, and then composite models, such as $\Lambda$CDM. The fourth section elevates the rigor one more step and explores the causal structure of different simple cosmological models in terms of conformal diagrams, following mostly the efficient approach of V. Mukhanov<ref name=Mukhanov />, and inflation. The section on black holes relates the general scheme of constructing conformal diagrams for stationary black hole spacetimes, following mostly the excellent textbook of K. Bronnikov and S. Rubin<ref name=BrRubin />. The consequent sections focus on more specific topics, such as the various problems regarding the Hubble sphere, inflation and holography.

==References==
<references>
<ref name=Harrison>E. Harrison. Cosmology: the science of the Universe. CUP (1981)</ref>
<ref name="Rindler">W. Rindler. Visual horizons in world-models. [http://adsabs.harvard.edu/abs/1956MNRAS.116..662R MNRAS 116 (6), 662--677 (1956) (1981)]</ref>
<ref name="Ellis">G.F.R. Ellis and T. Rothman. Lost horizons. Am. J. Phys. 61, 883 (1993)</ref>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN~0521563984</ref>
<ref name="BrRubin">K.A. Bronnikov and S.G. Rubin. Black Holes, Cosmology and Extra Dimensions. (WSPC, 2012), ISBN~978-9814374200</ref>

</references>

Category:Horizons

2013-10-07T20:25:50Z

Igor:

In this chapter we assemble the problems that concern one of the most distinctive features of General Relativity and Cosmology --- the horizons. The first part gives an elementary introduction into the concept in the cosmological context, following and borrowing heavily from the most comprehensible text by E. Harrison <ref name=Harrison />. Then we move to more formal exposition of the subject, making use of the seminal works of W. Rindler<ref name=Rindler /> and G.F.R. Ellis, T. Rothman<ref name=Ellis />, and consider first simple, and then composite models, such as $\Lambda$CDM. The fourth section elevates the rigor one more step and explores the causal structure of different simple cosmological models in terms of conformal diagrams, following mostly the efficient approach of V. Mukhanov<ref name=Mukhanov />. and inflation. The section on black holes relates the general scheme of constructing conformal diagrams for stationary black hole spacetimes, following mostly the excellent textbook of K. Bronnikov and S. Rubin<ref name=BrRubin />. The consequent sections focus on more specific topics, such as the various problems regarding the Hubble sphere, inflation and holography.

==References==
<references>
<ref name=Harrison>E. Harrison. Cosmology: the science of the Universe. CUP (1981)</ref>
<ref name="Rindler">W. Rindler. Visual horizons in world-models. [http://adsabs.harvard.edu/abs/1956MNRAS.116..662R MNRAS 116 (6), 662--677 (1956) (1981)]</ref>
<ref name="Ellis">G.F.R. Ellis and T. Rothman. Lost horizons. Am. J. Phys. 61, 883 (1993)</ref>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN~0521563984</ref>
<ref name="BrRubin">K.A. Bronnikov and S.G. Rubin. Black Holes, Cosmology and Extra Dimensions. (WSPC, 2012), ISBN~978-9814374200</ref>

</references>

Category:Horizons

2013-10-07T20:25:30Z

Igor:

In this chapter we assemble the problems that concern one of the most distinctive features of General Relativity and Cosmology --- the horizons. The first part gives an elementary introduction into the concept in the cosmological context, following and borrowing heavily from the most comprehensible text by E. Harrison <ref name=Harrison />. Then we move to more formal exposition of the subject, making use of the seminal works of W. Rindler<ref name=Rindler /> and G.F.R. Ellis, T. Rothman<ref name=Ellis />, and consider first simple, and then composite models, such as $\Lambda$CDM. The fourth section elevates the rigor one more step and explores the causal structure of different simple cosmological models in terms of conformal diagrams, following mostly the efficient approach of V. Mukhanov<ref name=Muchanov />. and inflation. The section on black holes relates the general scheme of constructing conformal diagrams for stationary black hole spacetimes, following mostly the excellent textbook of K. Bronnikov and S. Rubin<ref name=BrRubin />. The consequent sections focus on more specific topics, such as the various problems regarding the Hubble sphere, inflation and holography.

==References==
<references>
<ref name=Harrison>E. Harrison. Cosmology: the science of the Universe. CUP (1981)</ref>
<ref name="Rindler">W. Rindler. Visual horizons in world-models. [http://adsabs.harvard.edu/abs/1956MNRAS.116..662R MNRAS 116 (6), 662--677 (1956) (1981)]</ref>
<ref name="Ellis">G.F.R. Ellis and T. Rothman. Lost horizons. Am. J. Phys. 61, 883 (1993)</ref>
<ref name="Mukhanov">V.F. Mukhanov. Physical foundations of cosmology (CUP, 2005) ISBN~0521563984</ref>
<ref name="BrRubin">K.A. Bronnikov and S.G. Rubin. Black Holes, Cosmology and Extra Dimensions. (WSPC, 2012), ISBN~978-9814374200</ref>

</references>

Category:Horizons

2013-09-26T11:04:21Z

Igor:

In this chapter we assemble the problems that concern one of the most distinctive features of General Relativity and Cosmology --- the horizons. The first part gives an elementary introduction into the concept in the cosmological context, following and borrowing heavily from the most comprehensible text by E. Harrison <ref name=Harrison />. Then we move to more formal exposition of the subject, making use of the seminal works of W. Rindler<ref name=Rindler /> and G.F.R. Ellis, T. Rothman<ref name=Ellis />. The third section elevates the rigor one more step and explores the causal structure of different simple cosmological models in terms of conformal diagrams, following mostly the efficient approach of V. Mukhanov<ref name=Muchanov />. The following sections focus on more specific topics, such as the various problems regarding the Hubble sphere<ref name=Melia />, more realistic composite models (including $\Lambda$CDM), and inflation. The section on black holes relates the general scheme of constructing conformal diagrams for stationary black hole spacetimes, following mostly the excellent textbook of K. Bronnikov and S. Rubin<ref name=BrRubin />.

==References==
<references>
<ref name=Harrison>E. Harrison. Cosmology: the science of the Universe. CUP (1981)</ref>
<ref name="Rindler">W. Rindler. Visual horizons in world-models. [http://adsabs.harvard.edu/abs/1956MNRAS.116..662R MNRAS 116 (6), 662--677 (1956) (1981)]</ref>
<ref name="Ellis">G.F.R. Ellis and T. Rothman. Lost horizons. Am. J. Phys. 61, 883 (1993)</ref>
<ref name="Muchanov">V.F. Muchanov. Physical foundations of cosmology (CUP, 2005) ISBN~0521563984</ref>
<ref name="BrRubin">K.A. Bronnikov and S.G. Rubin. Black Holes, Cosmology and Extra Dimensions. (WSPC, 2012), ISBN~978-9814374200</ref>
<ref name="Melia">F. Melia. The cosmic horizon. MNRAS 382 (4), 1917--1921 (2007) [http://arxiv.org/abs/0711.4181 arXiv:0711.4181];<br/>F. Melia and M. Abdelqader, The Cosmological Spacetime, Int. J. Mod. Phys. D 18, 1889 (2009) [http://arxiv.org/abs/0907.5394 arXiv:0907.5394]</ref>
</references>

Category:Holographic Universe

2013-09-23T20:05:18Z

Igor: Why is this thing with labels happening when I just fix some misprints??? damn

The traditional point of view assumed that dominating part of degrees of freedom in our World are attributed to physical fields. However it became clear soon that such concept complicates the construction of Quantum Gravity: it is necessary to introduce small distance cutoffs for all integrals in the theory in order to make it sensible. As a consequence, our world should be described on a three-dimensional discrete lattice with the period of the order of Planck length. Lately some physicists share an even more radical point of view: instead of the three-dimensional lattice, complete description of Nature requires only a two-dimensional one, situated on the space boundary of our World. This approach is based on the so-called "holographic principle". The name is related to the optical hologram, which is essentially a two-dimensional record of a
three-dimensional object. The holographic principle consists of two main statements:

1) All information contained in some region of space can be "recorded" (represented) on the boundary of that region.

2) The theory, formulated on the boundaries of the considered region of space, must have no more than one degree of freedom per Planck area:
\begin{equation}
\label{Hol_f:1}
N\le \frac{A}{A_{pl}},\quad A_{pl}=\frac{G\hbar}{c^3}.
\end{equation}
\end{enumerate}
Thus, the key piece in the holographic principle is the assumption that all the information about the Universe can be encoded on some two-dimensional surface - the holographic screen. Such approach leads to a new interpretation of cosmological acceleration and to an absolutely unusual understanding of Gravity. The gravity is understood as an entropy force, caused by variation of information connected to positions of material bodies. More precisely, the quantity of information related to matter and its position is measured in terms of entropy. Relation between the entropy and the information states that the information change is exactly the negative entropy change $\Delta I=-\Delta S$. Entropy change due to matter displacement leads to the so-called entropy force, which, as will be proven below, has the form of gravity. Its origin therefore lies in the universal tendency of any macroscopic theory to maximize the entropy. The dynamics can be constructed in terms of entropy variation and it does not depend on the details of microscopic theory. In particular, there is no fundamental field associated with the entropy force. The entropy forces are typical for macroscopic systems like colloids and biophysical systems. Big colloid molecules, placed in thermal environment of smaller particles, feel the entropy forces. Osmose is another phenomenon governed by the entropy forces.

Probably the best known example of the entropy force is the elasticity of a polymer molecule. A single polymer molecule can be modeled as a composition of many monomers of fixed length. Each monomer can freely rotate around the fixation point and choose any spacial direction. Each of such configurations has the same energy. When the polymer molecule is placed into a thermal bath, it prefers to form a ring as the entropically most preferable configuration: there are many more such configurations when the polymer molecule is short, than those when it is stretched. The statistical tendency to transit
into the maximum entropy state transforms into the macroscopic force, in the considered case - into the elastic force.

Let us consider a small piece of holographic screen and a particle of mass $m$ approaching it. According to the holographic principle, the particle affects the amount of the information (and therefore of the entropy) stored on the screen. It is natural to assume that entropy variation near the screen is linear on the displacement $\Delta x$:
\begin{equation}
\Delta S = 2\pi k_B \frac{mc}{\hbar} \Delta x. \label{delta_s}
\end{equation}
The factor $2\pi$ is introduced for convenience, which the reader will appreciate solving the problems of this Chapter. In order to understand why this quantity should be proportional to mass, let us imagine that the particle has split into two or more particles of smaller mass. Each of those particles produces its own entropy change when displaced by $\Delta x$. As entropy and mass are both additive, then it is natural that the former is proportional to the latter. According to the first law of thermodynamics, the entropy force related to information variation satisfies the equation
\begin{equation}
F\Delta x = T\Delta S. \label{delta_x}
\end{equation}
If we know the entropy gradient, which can be found from (\ref{delta_s}),
and the screen temperature, we can calculate the entropy
force.

An observer moving with acceleration $a$, feels the temperature (the Unruh temperature)
\begin{equation}
\label{Hol_f_Unruh:4}
k_B T_U=\frac{1}{2\pi}\frac\hbar c a.
\end{equation}
Let us assume that the total energy of the system equals $E$. Let us make a simple assumption that the energy is uniformly distributed over all $N$ bits of information on the holographic screen. The temperature is then defined as the average energy per bit:
\begin{equation}
E =\frac12 N k_B T. \label{average_e}
\end{equation}
Equations (\ref{delta_s})-(\ref{average_e}) allow one to describe the holographic dynamics, and as a particular case - the dynamics of the Universe, and all that without the notion of Gravity.

Category:Holographic Universe

2013-09-23T20:02:28Z

Igor:

The traditional point of view assumed that dominating part of degrees of freedom in our World are attributed to physical fields. However it became clear soon that such concept complicates the construction of Quantum Gravity: it is necessary to introduce small distance cutoffs for all integrals in the theory in order to make it sensible. As a consequence, our world should be described on a three-dimensional discrete lattice with the period of the order of Planck length. Lately some physicists share an even more radical point of view: instead of the three-dimensional lattice, complete description of Nature requires only a two-dimensional one, situated on the space boundary of our World. This approach is based on the so-called "holographic principle". The name is related to the optical hologram, which is essentially a two-dimensional record of a
three-dimensional object. The holographic principle consists of two main statements:

1) All information contained in some region of space can be "recorded" (represented) on the boundary of that region.

2) The theory, formulated on the boundaries of the considered region of space, must have no more than one degree of freedom per Planck area:
\begin{equation}
\label{Hol_f:1}
N\le \frac{A}{A_{pl}},\quad A_{pl}=\frac{G\hbar}{c^3}.
\end{equation}
Thus, the key piece in the holographic principle is the assumption that all the information about the Universe can be encoded on some two-dimensional surface - the holographic screen. Such approach leads to a new interpretation of cosmological acceleration and to an absolutely unusual understanding of Gravity. The gravity is understood as an entropy force, caused by variation of information connected to positions of material bodies. More precisely, the quantity of information related to matter and its position is measured in terms of entropy. Relation between the entropy and the information states that the information change is exactly the negative entropy change $\Delta I=-\Delta S$. Entropy change due to matter displacement leads to the so-called entropy force, which, as will be proven below, has the form of gravity. Its origin therefore lies in the universal tendency of any macroscopic theory to maximize the entropy. The dynamics can be constructed in terms of entropy variation and it does not depend on the details of microscopic theory. In particular, there is no fundamental field associated with the entropy force. The entropy forces are typical for macroscopic systems like colloids and biophysical systems. Big colloid molecules, placed in thermal environment of smaller particles, feel the entropy forces. Osmosis is another phenomenon governed by the entropy forces.

Probably the best known example of the entropy force is the elasticity of a polymer molecule. A single polymer molecule can be modeled as a composition of many monomers of fixed length. Each monomer can freely rotate around the fixation point and choose any spacial direction. Each of such configurations has the same energy. When the polymer molecule is placed into a thermal bath, it prefers to form a ring as the entropically most preferable configuration: there are many more such configurations when the polymer molecule is short, than those when it is stretched. The statistical tendency to transit
into the maximum entropy state transforms into the macroscopic force, in the considered case - into the elastic force.

Let us consider a small piece of holographic screen and a particle of mass $m$ approaching it. According to the holographic principle, the particle affects the amount of the information (and therefore of the entropy) stored on the screen. It is natural to assume that entropy variation near the screen is linear on the displacement $\Delta x$:
\begin{equation}
\Delta S = 2\pi k_B \frac{mc}{\hbar} \Delta x. \label{delta_s}
\end{equation}
The factor $2\pi$ is introduced for convenience, which the reader will appreciate solving the problems of this Chapter. In order to understand why this quantity should be proportional to mass, let us imagine that the particle has split into two or more particles of smaller mass. Each of those particles produces its own entropy change when displaced by $\Delta x$. As entropy and mass are both additive, then it is natural that the former is proportional to the latter. According to the first law of thermodynamics, the entropy force related to information variation satisfies the equation
\begin{equation}
F\Delta x = T\Delta S. \label{delta_x}
\end{equation}
If we know the entropy gradient, which can be found from (\ref{delta_s}),
and the screen temperature, we can calculate the entropy
force.

An observer moving with acceleration $a$, feels the temperature (the Unruh temperature)
\begin{equation}
\label{Hol_f_Unruh:4}
k_B T_U=\frac{1}{2\pi}\frac\hbar c a.
\end{equation}
Let us assume that the total energy of the system equals $E$. Let us make a simple assumption that the energy is uniformly distributed over all $N$ bits of information on the holographic screen. The temperature is then defined as the average energy per bit:
\begin{equation}
E =\frac12 N k_B T. \label{average_e}
\end{equation}
Equations (\ref{delta_s})-(\ref{average_e}) allow one to describe the holographic dynamics, and as a particular case - the dynamics of the Universe, and all that without the notion of Gravity.

Category:Holographic Universe

2013-09-23T19:58:48Z

Igor: Undo revision 1657 by Igor (talk)

The traditional point of view assumed that dominating part of degrees of freedom in our World are attributed to physical fields. However it became clear soon that such concept complicates the construction of Quantum Gravity: it is necessary to introduce small distance cutoffs for all integrals in the theory in order to make it sensible. As a consequence, our world should be described on a three-dimensional discrete lattice with the period of the order of Planck length. Lately some physicists share an even more radical point of view: instead of the three-dimensional lattice, complete description of Nature requires only a two-dimensional one, situated on the space boundary of our World. This approach is based on the so-called "holographic principle". The name is related to the optical hologram, which is essentially a two-dimensional record of a
three-dimensional object. The holographic principle consists of two main statements:

1) All information contained in some region of space can be "recorded" (represented) on the boundary of that region.

2) The theory, formulated on the boundaries of the considered region of space, must have no more than one degree of freedom per Planck area:
\begin{equation}
\label{Hol_f:1}
N\le \frac{A}{A_{pl}},\quad A_{pl}=\frac{G\hbar}{c^3}.
\end{equation}

Thus, the key piece in the holographic principle is the assumption that all the information about the Universe can be encoded on some two-dimensional surface - the holographic screen. Such approach leads to a new interpretation of cosmological acceleration and to an absolutely unusual understanding of Gravity. The gravity is understood as an entropy force, caused by variation of information connected to positions of material bodies. More precisely, the quantity of information related to matter and its position is measured in terms of entropy. Relation between the entropy and the information states that the information change is exactly the negative entropy change $\Delta I=-\Delta S$. Entropy change due to matter displacement leads to the so-called entropy force, which, as will be proven below, has the form of gravity. Its origin therefore lies in the universal tendency of any macroscopic theory to maximize the entropy. The dynamics can be constructed in terms of entropy variation and it does not depend on the details of microscopic theory. In particular, there is no fundamental field associated with the entropy force. The entropy forces are typical for macroscopic systems like colloids and biophysical systems. Big colloid molecules, placed in thermal environment of smaller particles, feel the entropy forces. Osmosis is another phenomenon governed by the entropy forces.

Probably the best known example of the entropy force is the elasticity of a polymer molecule. A single polymer molecule can be modeled as a composition of many monomers of fixed length. Each monomer can freely rotate around the fixation point and choose any spacial direction. Each of such configurations has the same energy. When the polymer molecule is placed into a thermal bath, it prefers to form a ring as the entropically most preferable configuration: there are many more such configurations when the polymer molecule is short, than those when it is stretched. The statistical tendency to transit
into the maximum entropy state transforms into the macroscopic force, in the considered case - into the elastic force.

Let us consider a small piece of holographic screen and a particle of mass $m$ approaching it. According to the holographic principle, the particle affects the amount of the information (and therefore of the entropy) stored on the screen. It is natural to assume that entropy variation near the screen is linear on the displacement $\Delta x$:
\begin{equation}
\Delta S = 2\pi k_B \frac{mc}{\hbar} \Delta x. \label{delta_s}
\end{equation}
The factor $2\pi$ is introduced for convenience, which the reader will appreciate solving the problems of this Chapter. In order to understand why this quantity should be proportional to mass, let us imagine that the particle has split into two or more particles of smaller mass. Each of those particles produces its own entropy change when displaced by $\Delta x$. As entropy and mass are both additive, then it is natural that the former is proportional to the latter. According to the first law of thermodynamics, the entropy force related to information variation satisfies the equation
\begin{equation}
F\Delta x = T\Delta S. \label{delta_x}
\end{equation}
If we know the entropy gradient, which can be found from (\ref{delta_s}),
and the screen temperature, we can calculate the entropy
force.

An observer moving with acceleration $a$, feels the temperature (the Unruh temperature)
\begin{equation}
\label{Hol_f_Unruh:4}
k_B T_U=\frac{1}{2\pi}\frac\hbar c a.
\end{equation}
Let us assume that the total energy of the system equals $E$. Let us make a simple assumption that the energy is uniformly distributed over all $N$ bits of information on the holographic screen. The temperature is then defined as the average energy per bit:
\begin{equation}
E =\frac12 N k_B T. \label{average_e}
\end{equation}
Equations (\ref{delta_s})-(\ref{average_e}) allow one to describe the holographic dynamics, and as a particular case - the dynamics of the Universe, and all that without the notion of Gravity.

Category:Holographic Universe

2013-09-23T19:57:59Z

Igor:

The traditional point of view assumed that dominating part of degrees of freedom in our World are attributed to physical fields. However it became clear soon that such concept complicates the construction of Quantum Gravity: it is necessary to introduce small distance cutoffs for all integrals in the theory in order to make it sensible. As a consequence, our world should be described on a three-dimensional discrete lattice with the period of the order of Planck length. Lately some physicists share an even more radical point of view: instead of the three-dimensional lattice, complete description of Nature requires only a two-dimensional one, situated on the space boundary of our World. This approach is based on the so-called "holographic principle". The name is related to the optical hologram, which is essentially a two-dimensional record of a
three-dimensional object. The holographic principle consists of two main statements:

1) All information contained in some region of space can be "recorded" (represented) on the boundary of that region.

2) The theory, formulated on the boundaries of the considered region of space, must have no more than one degree of freedom per Planck area:
\begin{equation}
\label{Hol_f:1n}
N\le \frac{A}{A_{pl}},\quad A_{pl}=\frac{G\hbar}{c^3}.
\end{equation}

Thus, the key piece in the holographic principle is the assumption that all the information about the Universe can be encoded on some two-dimensional surface - the holographic screen. Such approach leads to a new interpretation of cosmological acceleration and to an absolutely unusual understanding of Gravity. The gravity is understood as an entropy force, caused by variation of information connected to positions of material bodies. More precisely, the quantity of information related to matter and its position is measured in terms of entropy. Relation between the entropy and the information states that the information change is exactly the negative entropy change $\Delta I=-\Delta S$. Entropy change due to matter displacement leads to the so-called entropy force, which, as will be proven below, has the form of gravity. Its origin therefore lies in the universal tendency of any macroscopic theory to maximize the entropy. The dynamics can be constructed in terms of entropy variation and it does not depend on the details of microscopic theory. In particular, there is no fundamental field associated with the entropy force. The entropy forces are typical for macroscopic systems like colloids and biophysical systems. Big colloid molecules, placed in thermal environment of smaller particles, feel the entropy forces. Osmosis is another phenomenon governed by the entropy forces.

Probably the best known example of the entropy force is the elasticity of a polymer molecule. A single polymer molecule can be modeled as a composition of many monomers of fixed length. Each monomer can freely rotate around the fixation point and choose any spacial direction. Each of such configurations has the same energy. When the polymer molecule is placed into a thermal bath, it prefers to form a ring as the entropically most preferable configuration: there are many more such configurations when the polymer molecule is short, than those when it is stretched. The statistical tendency to transit
into the maximum entropy state transforms into the macroscopic force, in the considered case - into the elastic force.

Let us consider a small piece of holographic screen and a particle of mass $m$ approaching it. According to the holographic principle, the particle affects the amount of the information (and therefore of the entropy) stored on the screen. It is natural to assume that entropy variation near the screen is linear on the displacement $\Delta x$:
\begin{equation}
\Delta S = 2\pi k_B \frac{mc}{\hbar} \Delta x. \label{delta_s}
\end{equation}
The factor $2\pi$ is introduced for convenience, which the reader will appreciate solving the problems of this Chapter. In order to understand why this quantity should be proportional to mass, let us imagine that the particle has split into two or more particles of smaller mass. Each of those particles produces its own entropy change when displaced by $\Delta x$. As entropy and mass are both additive, then it is natural that the former is proportional to the latter. According to the first law of thermodynamics, the entropy force related to information variation satisfies the equation
\begin{equation}
F\Delta x = T\Delta S. \label{delta_x}
\end{equation}
If we know the entropy gradient, which can be found from (\ref{delta_s}),
and the screen temperature, we can calculate the entropy
force.

An observer moving with acceleration $a$, feels the temperature (the Unruh temperature)
\begin{equation}
\label{Hol_f_Unruh:4}
k_B T_U=\frac{1}{2\pi}\frac\hbar c a.
\end{equation}
Let us assume that the total energy of the system equals $E$. Let us make a simple assumption that the energy is uniformly distributed over all $N$ bits of information on the holographic screen. The temperature is then defined as the average energy per bit:
\begin{equation}
E =\frac12 N k_B T. \label{average_e}
\end{equation}
Equations (\ref{delta_s})-(\ref{average_e}) allow one to describe the holographic dynamics, and as a particular case - the dynamics of the Universe, and all that without the notion of Gravity.

Category:Holographic Universe

2013-09-23T19:57:10Z

Igor:

The traditional point of view assumed that dominating part of degrees of freedom in our World are attributed to physical fields. However it became clear soon that such concept complicates the construction of Quantum Gravity: it is necessary to introduce small distance cutoffs for all integrals in the theory in order to make it sensible. As a consequence, our world should be described on a three-dimensional discrete lattice with the period of the order of Planck length. Lately some physicists share an even more radical point of view: instead of the three-dimensional lattice, complete description of Nature requires only a two-dimensional one, situated on the space boundary of our World. This approach is based on the so-called "holographic principle". The name is related to the optical hologram, which is essentially a two-dimensional record of a
three-dimensional object. The holographic principle consists of two main statements:

1) All information contained in some region of space can be "recorded" (represented) on the boundary of that region.

2) The theory, formulated on the boundaries of the considered region of space, must have no more than one degree of freedom per Planck area:
\begin{equation}
\label{Hol_f:1}
N\le \frac{A}{A_{pl}},\quad A_{pl}=\frac{G\hbar}{c^3}.
\end{equation}

Thus, the key piece in the holographic principle is the assumption that all the information about the Universe can be encoded on some two-dimensional surface - the holographic screen. Such approach leads to a new interpretation of cosmological acceleration and to an absolutely unusual understanding of Gravity. The gravity is understood as an entropy force, caused by variation of information connected to positions of material bodies. More precisely, the quantity of information related to matter and its position is measured in terms of entropy. Relation between the entropy and the information states that the information change is exactly the negative entropy change $\Delta I=-\Delta S$. Entropy change due to matter displacement leads to the so-called entropy force, which, as will be proven below, has the form of gravity. Its origin therefore lies in the universal tendency of any macroscopic theory to maximize the entropy. The dynamics can be constructed in terms of entropy variation and it does not depend on the details of microscopic theory. In particular, there is no fundamental field associated with the entropy force. The entropy forces are typical for macroscopic systems like colloids and biophysical systems. Big colloid molecules, placed in thermal environment of smaller particles, feel the entropy forces. Osmosis is another phenomenon governed by the entropy forces.

Probably the best known example of the entropy force is the elasticity of a polymer molecule. A single polymer molecule can be modeled as a composition of many monomers of fixed length. Each monomer can freely rotate around the fixation point and choose any spacial direction. Each of such configurations has the same energy. When the polymer molecule is placed into a thermal bath, it prefers to form a ring as the entropically most preferable configuration: there are many more such configurations when the polymer molecule is short, than those when it is stretched. The statistical tendency to transit
into the maximum entropy state transforms into the macroscopic force, in the considered case - into the elastic force.

Let us consider a small piece of holographic screen and a particle of mass $m$ approaching it. According to the holographic principle, the particle affects the amount of the information (and therefore of the entropy) stored on the screen. It is natural to assume that entropy variation near the screen is linear on the displacement $\Delta x$:
\begin{equation}
\Delta S = 2\pi k_B \frac{mc}{\hbar} \Delta x. \label{delta_s}
\end{equation}
The factor $2\pi$ is introduced for convenience, which the reader will appreciate solving the problems of this Chapter. In order to understand why this quantity should be proportional to mass, let us imagine that the particle has split into two or more particles of smaller mass. Each of those particles produces its own entropy change when displaced by $\Delta x$. As entropy and mass are both additive, then it is natural that the former is proportional to the latter. According to the first law of thermodynamics, the entropy force related to information variation satisfies the equation
\begin{equation}
F\Delta x = T\Delta S. \label{delta_x}
\end{equation}
If we know the entropy gradient, which can be found from (\ref{delta_s}),
and the screen temperature, we can calculate the entropy
force.

An observer moving with acceleration $a$, feels the temperature (the Unruh temperature)
\begin{equation}
\label{Hol_f_Unruh:4}
k_B T_U=\frac{1}{2\pi}\frac\hbar c a.
\end{equation}
Let us assume that the total energy of the system equals $E$. Let us make a simple assumption that the energy is uniformly distributed over all $N$ bits of information on the holographic screen. The temperature is then defined as the average energy per bit:
\begin{equation}
E =\frac12 N k_B T. \label{average_e}
\end{equation}
Equations (\ref{delta_s})-(\ref{average_e}) allow one to describe the holographic dynamics, and as a particular case - the dynamics of the Universe, and all that without the notion of Gravity.

Schwarzschild black hole

2013-05-02T10:02:59Z

Igor:

[[Category:Black Holes|2]]

The spherically symmetric solution of Einstein's equations in vacuum for the spacetime metric has the form$^{*}$
\begin{align}\label{Schw}
ds^{2}=h(r)\,dt^2-h^{-1}(r)\,dr^2-r^2 d\Omega^{2},
&\qquad\mbox{where}\quad
h(r)=1-\frac{r_g}{r};\quad r_{g}=\frac{2GM}{c^{2}};\\
d\Omega^{2}=d\theta^{2}+\sin^{2}\theta\, d\varphi^{2}&\;\text{is the metric of unit sphere.}\nonumber
\end{align}
The Birkhoff's theorem$^{**}$ (1923) states, that this solution is unique up to coordinate transformations. The quantity $r_g$ is called the Schwarzschild radius, or gravitational radius, $M$ is the mass of the central body or black hole.

$^{*}$ K. Schwarzschild, On the gravitational field of a mass point according to Einstein's theory, ''Sitzungsber. Preuss. Akad. Wiss. Phys. Math. Kl.'', p.189 (1916) (there's a translation of the original paper at [http://arxiv.org/abs/physics/9905030v1 arXiv:physics/9905030v1]; please disregard the abstract/foreword, which is incorrect).

$^{**}$ G.D. Birkhoff, Relativity and Modern Physics, p.253, Harvard University Press, Cambridge (1923);
J.T. Jebsen, "Ark. Mat. Ast. Fys." (Stockholm) 15, nr.18 (1921), see also [http://arxiv.org/abs/physics/0508163 arXiv:physics/0508163v2].

__TOC__

==Simple problems==
<div id="BlackHole15"></div>
=== Problem 1: local time ===
Find the interval of local time (proper time of stationary observer) at a point $(r,\theta,\varphi)$ in terms of coordinate time $t$, and show that $t$ is the proper time of an observer at infinity. What happens when $r\to r_{g}$?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The proper time of the stationary observer is $d\tau=ds|_{dr=d\theta=d\varphi=0}$:
\[d\tau=\sqrt{g_{00}}\;dt=\sqrt{1-\frac{r_{g}}{r}}\;dt.\]
At $r\to \infty$ it coincides with $dt$, so the coordinate time $t$ can be interpreted as the proper time of a "remote" observer. At $r\to r_g$ the local time flows slower and asymptotically stops. If one of two twins were to live some time at $r\approx r_g$, he will return to his remote twin having aged less (thought he might have acquired some grey hair due to constant fear of tumbling over the horizon). </p>
</div>
</div>

<div id="BlackHole16"></div>

=== Problem 2: measuring distances ===
What is the physical distance between two points with coordinates $(r_{1},\theta,\varphi)$ and $(r_{2},\theta,\varphi)$? Between $(r,\theta,\varphi_{1})$ and $(r,\theta,\varphi_{2})$? How do these distances behave in the limit $r_{1},r\to r_{g}$?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[l_{r1-r2}=\int\limits_{r_1}^{r_2}
\frac{dr}{\sqrt{1-\frac{r_g}{r}}};\quad
l_{\varphi1-\varphi_2}=2\pi r|\varphi_1-\varphi_2|;
\quad
l_{\theta1-\theta_2}=2\pi r|\theta_1-\theta_2|.\] </p>
</div>
</div>

<div id="BlackHole17"></div>

=== Problem 3: the inner region ===
What would be the answers to the previous two questions for $r<r_g$ and why*? Why the Schwarzschild metric cannot be imagined as a system of "welded" rigid rods in $r<r_g$, as it can be in the external region?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
''This question is given not to be answered but to make one think of the answer. Correct questions and correct answers can be given in terms of a proper coordinate frame, which is regular both in $r>r_g$ and in $r<r_g$. Still, one can say something meaningful as is.''</p>

<p style="text-align: left;">
At $r<r_g$ we have $g_{00}<0$ and $g_{11}>0$, thus the $t$ coordinate is ''spatial'' and $r$ coordinate is ''temporal'' (!):
\[ds^{2}=|h|^{-1}(r)dr^{2}-|h|(r)dt^{2}
-r^{2}d\Omega^{2}.\]
The metric therefore is ''nonstationary'' in this region, depending on the temporal coordinate $r$, but homogeneous, as there is no dependence on spatial coordinates.

Then for an observer "at rest" with respect to this coordinate system we would have $dt=d\theta=d\varphi=0$, and thus
\[d\tau^2=ds^2=-\frac{dr^{2}}{1-r_{g}/r}=
\frac{r dr^{2}}{r_{g}-r}>0,
\quad\Rightarrow\quad
d\tau=\frac{\sqrt{r}dr}{\sqrt{r_{g}-r}}.\]

An observer at rest with respect to the old coordinate system $dr=d\theta=d\varphi=0$, though, does not exist, as it would be $ds^{2}<0$ for him, which corresponds to spacelike geodesics (i.e. particles traveling faster than light).

The last of the two questions cannot be answered without additional assumptions, because ''time'' $t$, which is the spatial coordinate now, in the two points is not given.

The physical distance at $d\theta=d\varphi=dr=0$ is defined as
\[dl^{2}=|h(r)|dt^2.\]
It evidently depends on time $r$.

This very fact that Schwarzschild metric is nonstationary at $r<r_g$, and that a stationary one does not exist in this region, leads to the absence of stationary observers and thus to the impossibility to imagine it "welded" of a system of stiff rods. </p>

<p style="text-align: left;">*This was actually not a very simple problem</p>
</div>
</div>

<div id="BlackHole18"></div>

=== Problem 4: acceleration ===
Calculate the acceleration of a test particle with zero velocity.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If a particle is at rest, its 4-velocity is $u^{\mu}=g_{00}^{-1/2}\delta^{\mu}_{0}$, where the factor is determined from the normalizing condition
\[1=g_{\mu\nu}u^{\mu}u^{\mu}=g_{00}(u^{0})^{2}.\]
Then the $4$-acceleration can be found from the geodesic equation:
\begin{align*}
a^{0}\equiv&\frac{du^0}{ds}=-{\Gamma^{0}}_{00}u^{0}u^{0}\sim
{\Gamma^{0}}_{00}\sim \Gamma_{0,\,00}\sim
(\partial_{0}g_{00}+\partial_{0}g_{00}
-\partial_{0}g_{00})=0;\\
a^{1}\equiv&\frac{du^1}{ds}
=-{\Gamma^{1}}_{00}u^{0}u^{0}
=-g^{11}\Gamma_{1,\,00}\;g_{00}^{-1}
=-(g_{00}g_{11})^{-1}\Gamma_{1,\,00}=\Gamma_{1,00}=
\\ &=\tfrac{1}{2}
(\partial_{0}g_{10}+\partial_{0}g_{10}
-\partial_{1}g_{00})
=-\frac{1}{2}\frac{dg_{00}}{dr}
=-\frac{h'}{2}
=\frac{r_{g}}{2r^2}.
\end{align*}
The scalar acceleration $a$ is then equal to
\[a^{2}=-g_{11}(a^{1})^{2}=\frac{(h')^{2}}{4h}
=-\frac{r_{g}^{2}}{4 r^4}
\Big(1-\frac{r_g}{r}\Big)^{-1}\]
and tends to infinity when we approach the horizon. </p>
</div>
</div>

<div id="BlackHoleExtra1"></div>

=== Problem 5: Schwarzschild is a vacuum solution ===
Show that Schwarzschild metric is a solution of Einstein's equation in vacuum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Straightforward calculation of Christoffel symbols and Ricci tensor yields the vacuum Einstein equation $R_{\mu\nu}=0$. </p>
</div>
</div>

==Symmetries and integrals of motion==
For background on Killing vectors see problems [[Equations of General Relativity#equ_oto-kill1|K1]], [[Equations of General Relativity#equ_oto-kill2|K2]], [[Equations of General Relativity#equ_oto-kill3|K3]] of chapter 2.
<div id="BlackHole19"></div>
=== Problem 6: timelike Killing vector ===
What integral of motion arises due to existence of a timelike Killing vector? Express it through the physical velocity of the particle.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If a Killing vector field is timelike we can use the coordinate frame in which $K^\mu$ is the unitary vector of the time coordinate (see problem [[Equations of General Relativity#equ_oto-kill2|K2]] of chapter 2.) $K^{\mu}=(1,0,0,0)$, while the spacelike basis vectors are orthogonal to it. Then the integral of motion is energy in the corresponding (stationary) frame $p_{\mu}K^{\mu}=p_{0}\equiv\varepsilon/c$.

Using the result [[Technical_warm-up#u0|u0]], which holds for arbitrary gravitational field, we obtain the expression for energy, which is the integral of motion in a stationary metric
\begin{equation}\label{EnergyStat}
\varepsilon=mc^{2}u_{0}=
mc^{2}\sqrt{g_{00}}\cdot\gamma=
\frac{mc^{2}\sqrt{g_{00}}}
{\sqrt{1-\frac{v^2}{c^2}}}.\end{equation} </p>
</div>
</div>

<div id="BlackHole20"></div>

=== Problem 7: Killing vectors of a sphere ===
Derive the Killing vectors for a sphere in Cartesian coordinate system; in spherical coordinates.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Killing vectors correspond to infinitesimal transformations that leave the metric invariant. Considering a two-dimensional sphere embedded in the tree-dimensional space, we can see that its symmetries are rotations. Three of them are independent: the rotations in each of the three coordinate planes.

Let us consider an infinitesimal rotation in the plane $XY$: $dx=yd\lambda,\; dy=-xd\lambda$. Thus the Killing vector is*
\[K_{1}=x\partial_y-y\partial_x.\]

Using the spherical coordinates
\[x=\sin\theta\cos\varphi;\quad y=\sin\theta\sin\varphi;\quad z=\cos\theta,\]
we obtain $\partial_{x}=-\frac{\sin\varphi}{\sin\theta}\partial_\varphi$, $\partial_y=\frac{\cos\varphi}{\sin\theta}\partial_\varphi$, and therefore $K_{1}=\partial_{\varphi}$.

Considering rotations in plains $XZ$ and $YZ$ in the same way, in the end we obtain
\[\left\{\begin{array}{l}
K_{1}=x\partial_{y}-y\partial_{x}=\partial_{\varphi}\\
K_{2}=z\partial_{x}-x\partial_{z}=
\cos\varphi\;\partial_{\theta}-\cot\theta\sin\varphi\;\partial_\varphi\\
K_{3}=z\partial_{y}-y\partial_{x}=
\sin\varphi\;\partial_{\theta}+\cot\theta\cos\varphi\;\partial_\varphi\;.
\end{array}\right.\] </p>

<p style="text-align: left;">*Hereafter we use the following notation: a 4-vector $A$ is $A=A^{\mu}\partial_{\mu}$, where $\partial_{\mu}\equiv\partial/\partial x_{\mu}$ is the "coordinate" basis, $A^{\mu}$ the coordinates of $A$ in this basis; it is not hard to verify that transformation laws for $A^\mu$ and $\partial_{\mu}$ are adjusted so that $A$ is a quantity that does not depend on a coordinate frame. This also enables us to conveniently recalculate $A^\mu$ when we change the basis. For more detail see e.g. the textbook by Carroll:</p>

<p style="text-align: left;">Carroll S., Spacetime and geometry: an introduction to General Relativity. AW, 2003, ISBN 0805387323</p>
</div>
</div>

<div id="BlackHole21"></div>

=== Problem 8: spherical symmetry of Schwarzshild ===
Verify that in coordinates $(t,r,\theta,\varphi)$ vectors
\[ \begin{array}{l}
\Omega^{\mu}=(1,0,0,0),\\
R^{\mu}=(0,0,0,1),\\
S^{\mu}=(0,0,\cos\varphi,-\cot\theta\sin\varphi),\\
T^{\mu}=(0,0,-\sin\varphi,-\cot\theta\cos\varphi)
\end{array}\]
are the Killing vectors of the Schwarzschild metric.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Vectors $\Omega^\mu$ and $R^\mu$ are Killing vectors because the metric does not depend explicitly either on $t$ or on $\varphi$. The last two correspond to the spherical symmetry (see previous problem). We can also check directly that they obey the Killing equation by evaluating the Christoffel symbols for the Schwarzschild metric. </p>
</div>
</div>

<div id="BlackHole22"></div>
=== Problem 9: planar motion ===
Show that existence of Killing vectors $S^\mu$ and $T^\mu$ leads to motion of particles in a plane.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integrals of motion corresponding to $S^\mu$ and $T^\mu$ are
\begin{align*}
S=u_{\mu}S^{\mu}=&
\;u_{2}\cos\varphi-u_{3}\cot\theta\sin\varphi;\\
T=u_{\mu}T^{\mu}=&
-u_{2}\sin\varphi-u_{3}\cot\theta\cos\varphi.
\end{align*}
Let us align the coordinate frame in such a way that the initial conditions are
\[\theta|_{t=0}=\pi/2;\quad
u_{2}|_{t=0}\equiv u_{\theta}|_{t=0}=0.\]
Then $S$ and $T$ are zero:
\[u_{2}\cos\varphi=u_{3}\cot\theta\sin\varphi;\qquad
u_{2}\sin\varphi=-u_{3}\cot\theta\cos\varphi.\]
Taking the square and adding the two equations, and also multiplying them, we obtain
\[\left\{\begin{array}{l}
u_{2}^{2}=u_{3}^{2}\cot^{2}\theta,\\
u_{2}^{2}\sin\varphi\cos\varphi=
-u_{3}^{2}\cot^{2}\theta\sin\varphi\cos\varphi
\end{array}\right.\quad\Rightarrow\quad
u_{3}^{2}\sin\varphi\cos\varphi
\cdot\cot^{2}\theta=0.\]
Then either $\varphi=const$, which means that $u_{3}=u_{2}=0$ and the motion is radial, or $\cot^{2}\theta=0$ and the motion takes place in the plane $\theta=\pi/2$. </p>
</div>
</div>

<div id="BlackHole23"></div>

=== Problem 10: stability of planar motion ===
Show that the particles' motion in the plane is stable.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Conservation of $u_{\mu}R^{\mu}$ means that $R\equiv u_{3}\equiv\varphi'=const$. It is not hard to derive from the expressions for $S$ and $T$ that
\begin{align*}
&S\sin\varphi+T\cos\varphi=R\cot\theta,\\
&S\cos\varphi-T\sin\varphi=u_2.
\end{align*}
Taking the square and adding up, we are led to
\[u_2^2 =S^2+T^2+R^2-\frac{R^2}{\sin^2 \theta},\]
then on differentiating we obtain
\[u'_{2}=R^2 \frac{\cos\theta}{\sin^{2}\theta}.\]
Let the trajectory deviate slightly from the plane $\theta=\pi/2$. Then $\theta=\pi/2+\delta\theta$ and
\[(\delta\theta)''=u'_{2}
=R^2 \frac{\cos\theta}{\sin^{2}\theta}
\approx -R^2 \delta\theta,\]
therefore $\theta$ oscillates around the stable point $\pi/2$. </p>
</div>
</div>

<div id="BlackHole24"></div>

=== Problem 11: remaining integrals of motion ===
Write down explicitly the conserved quantities $p_{\mu}\Omega^{\mu}$ and $p_{\mu}R^{\mu}$ for movement in the plane $\theta=\pi/2$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The first integral of motion is, up to a multiplier, the energy (see Eq. [[Technical_warm-up#EnergyStat|EnergyStat]]):
\begin{equation}\label{SchwInt-E1}
E=\Omega_{\mu}u^{\mu}=
g_{00}\frac{dx^{0}}{d\lambda}=h\frac{dt}{d\lambda}=
\left(1-\frac{r_g}{r}\right)\frac{dt}{d\lambda},\end{equation}
where $\lambda$ is the geodesic parameter: for a massive particle we can always choose the natural parametrization $d\lambda=ds=\gamma^{-1}\sqrt{g_{00}}\,dt$ (see Eq. [[Technical_warm-up#IntervalStaticCase|IntervalStaticCase]]) so that $E$ is energy per unit mass; recovering the multipliers, for true energy we obtain
\begin{equation}\label{SchwInt-E2}
\varepsilon_{m}\equiv mcE=
mc^{2}\sqrt{g_{00}}\,\gamma=
mc^{2}\sqrt{h}\,\gamma=
mc^{2}\sqrt{\frac{1-r_{g}/r}{1-v^{2}/c^{2}}}.\end{equation}

The second integral of motion is the angular momentum (per unit mass also, and the sign is chosen so that in the Newtonian limit we obtain the usual angular momentum)
\begin{equation}\label{SchwInt-L1}
L=-R_{\mu}u^{\mu}=-g_{33}\frac{dx^{3}}{d\lambda}=
r^{2}\sin^{2}\theta\frac{d\varphi}{d\lambda}=
r^{2}\frac{d\varphi}{d\lambda}.\end{equation}
For a massive particle we choose $\lambda=s$, then
$d\lambda=ds=\gamma^{-1}\sqrt{g_{00}}\,dt$ and
\begin{equation}\label{SchwInt-L2}
l_{m}\equiv mL=
mr^{2}\dot{\varphi} \sqrt{g_{00}}\gamma=
m r^{2}\dot{\varphi}
\sqrt{\frac{1-r_{g}/r}{1-v^{2}/c^{2}}}.\end{equation}
Note that conservation of $l_m$ is a generalization to the relativistic case of the second Kepler's law on the sweeping of equal areas per unit time $r^{2}\dot{\varphi}=const$. </p>
</div>
</div>

<div id="BlackHole25"></div>

=== Problem 12: work and mass ===
What is the work needed to pull a particle from the horizon to infinity? Will a black hole's mass change if we drop a particle with zero initial velocity from immediate proximity of the horizon?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
At $r\gg r_g$ we'll have $\varepsilon_{m}=mc^{2}$, while near the horizon $\varepsilon_{m}\to 0$. The difference is the work needed to pull the particle away from the horizon to infinity, and it equals the rest mass (energy) of the particle. No, it will not change, because the energy of the falling particle is zero. </p>
</div>
</div>

==Radial motion==
Consider a particle's radial motion: $\dot{\varphi}=\dot{\theta}=0$. In this problem one is especially interested in asymptotes of all functions as $r\to r_{g}$.

<p style="text-align: left;">Let us set $c=1$ here and henceforth measure time in the units of length, so that $x^{0}=t$, $\beta=v$, etc., and introduce the notation
\[h(r)\equiv g_{00}(r)=-\frac{1}{g_{11}(r)}=
1-\frac{r_{g}}{r}
\underset{r\to r_g+0}{\longrightarrow}+0.\]</p>

<div id="BlackHole26"></div>
=== Problem 13: null geodesics ===
Derive the equation for null geodesics (worldlines of massless particles).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Equation of null curve is $ds^2=0$. Substituting the metric with $d\theta=d\varphi=0$, we get
\[dt=\pm dr\,\sqrt{-\frac{g_{11}}{g_{00}}}
=\pm\frac{dr}{h}.\]

On integration, we obtain
\begin{equation}\label{Schw-NullGeodesic}
\pm t=r+r_{g}\ln|r-r_{g}|+const.\end{equation}
Due to symmetry it is also a geodesic. This can be verified by evaluating the Christoffel symbols for the Schwarzschild metric and writing down the geodesic equatoin in explicit form. </p>
</div>
</div>

<div id="BlackHole27"></div>
=== Problem 14: geodesic motion of massive particle ===
Use energy conservation to derive $v(r)$, $\dot{r}(r)=dr/dt$, $r(t)$ for a massive particle. Initial conditions: $g_{00}|_{\dot{r}=0}=h_{0}$ (the limiting case $h_{0}\to 1$ is especially interesting and simple).
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<p style="text-align: left;">
In the radial case the physical velocity $v$ is expressed through the $4$-velocity $u^{\mu}=(u^{0},u^{1},0,0)$ as
\begin{equation}\label{Schw-v(u)}
\beta^{2}=\frac{dx^{\alpha}dx_{\alpha}}{d\tau^{2}}=
\frac{-g_{11}(dx^{1})^{2}}{g_{00}(dx^{0})^{2}}=
h^{-2}\Big(\frac{u^1}{u^0}\Big)^2=
\Big(\frac{\dot{r}}{h}\Big)^2,\end{equation}
where dot denotes differentiation by coordinate time $t$, and thus the energy conservation law (see (\ref{SchwInt-E2})) takes the form
\[const=\Big(\frac{\varepsilon}{mc^2}\Big)^{2}=
h\gamma^{2}=\frac{h}{1-\dot{r}^{2}/h^2}.\]

Initial condition fixes the constant*:
\begin{equation}\label{Schw-h0}
\dot{r}\sim v|_{r_{0}}=0\quad
\Rightarrow\quad
\Big(\frac{\varepsilon}{mc^2}\Big)^{2}
=g_{00}(r_{0})=h(r_0)=1-\frac{r_{g}}{r_{0}}
\equiv h_{0},\end{equation}
so for the physical velocity we get
\begin{equation}\label{SchwRad-V}
\gamma^{2}=\frac{h_{0}}{h(r)}
\underset{r\to r_{g}}{\longrightarrow}\infty;
\quad \frac{v}{c}\equiv\beta=
\sqrt{1-\gamma^{-2}}=
\sqrt{1-\frac{h}{h_{0}}}
\underset{r\to r_{g}}{\longrightarrow}1.\end{equation}
It tends to infinity close to the horizon! At the same time, as
$v^{2}=\dot{r}^{2}/h^{2}$ from (\ref{Schw-v(u)}), the coordinate velocity
\begin{equation}\label{Schw-dotR}
\Big|\frac{dr}{dt}\Big|=vh=h \sqrt{1-\frac{h}{h_{0}}}=
\Big(1-\frac{r_{g}}{r}\Big)
\sqrt{1-\frac{1-\frac{r_{g}}{r}}{1-\frac{r_{g}}{r_{0}}}}
\underset{r\to r_{g}}{\thicksim}h\to 0\end{equation}
tends to zero.

After integration we obtain $t(r)$:
\begin{equation}\label{Schw-t(r)}
const\pm t=\int\limits_{r}^{r_{0}}
\frac{dr}{vh}=
\int\limits_{r}^{r_{0}}\frac{rdr}{r-r_{g}}
\bigg[1-\frac{1-r_{g}/r}{1-r_{g}/r_{0}}\bigg]^{-1/2}
\underset{\substack{r\approx r_{g}\\r_{0}\gg r_{g}}}
{\approx}\int\limits_{r}^{r_{0}}
\frac{r\,dr}{r-r_{g}}=r+r_{g}\ln|r-r_{g}|.\end{equation}
As should be expected, the asymptote is the same as for the null geodesics (for $r_{0}\gg r_{g}$).</p>

<p style="text-align: left;">*Note that $r_0$ has the meaning of a turning point only for finite motion; this, however, does not prevent us from using the same notation in the non-finite case. Then $r_0$ is determined from $\varepsilon$ by the same formula and takes negative values, while $h_{0}>1$.</p>
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=== Problem 15: radial motion in terms of proper time ===
Show that the equation of radial motion in terms of proper time of the particle is the same as in the non-relativistic Newtonian theory. Calculate the proper time of the fall from $r=r_0$ to the center. Derive the first correction in $r_{g}/r$ to the Newtonian result. Initial velocity is zero.
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<p style="text-align: left;">
The particle's proper time interval $\tau$ is $d\tau=\gamma^{-1} d\hat{t}$, where $d\hat{t}=\sqrt{h}dt$ is local time. Then
\begin{equation}\label{ShwRad-tau}
\tau=\int dt \gamma^{-1}\sqrt{h}=
\int\limits_{r}^{r_{0}}
\frac{dr}{\gamma v\sqrt{h}}=
\int\limits_{r}^{r_{0}}
\frac{dr}{v \sqrt{h_{0}}}=
\int\limits_{r}^{r_{0}}
\frac{dr}{\sqrt{h_{0}-h}}=
\int\limits_{r}^{r_{0}}
\frac{dr}{\sqrt{\frac{r_{g}}{r_{0}}-
\frac{r_{g}}{r}}}.\end{equation}
On the other hand, in the Newtonian approach from energy conservation we obtain in the same way that
\[(\dot{r})^{2}=2(E-U)=2\;\Big(E+\frac{GM}{r}\Big)
=\frac{r_{g}}{r}-\frac{r_{g}}{r_{0}},\]
which gives us the same equation of motion.

We see that the integral is regular at $r=r_{g}$, is finite and converges even at $r\to 0$. It is not hard to evaluate:
\begin{equation}\label{Schw-TauParametric}
\tau=\frac{r_0}{2}\sqrt{\frac{r_0}{r_g}}
(\eta+\sin\eta),\qquad
r=\frac{r_0}{2}(1+\cos\eta),\qquad \eta\in(0,\pi).
\end{equation}
The proper time of reaching the singularity $r=0$ is therefore
\[\tau_{f}=\frac{\pi}{2}r_{0}
\sqrt{\frac{r_{0}}{r_{g}}}.\]

The integral (\ref{Schw-t(r)}) for $t(r)$ can be rewritten as
\[t(r)=\int \frac{dr}{vh}=\sqrt{h_{0}}
\int\frac{dr}{\sqrt{h_{0}-h}}\cdot \frac{1}{h}=
\underbrace{\sqrt{1-\frac{r_{g}}{r_{0}}}}
_{\text{correction 1}}\,
\underbrace{\int\!\frac{dr}{\sqrt{\frac{r_g}{r}-\frac{r_g}{r_0}}}}_{\tau(r)}
\cdot \underbrace{\frac{1}{1-\frac{r_g}{r}}}
_{\text{correction 2}}.\]
Now we see that the first factor gives the correction taking into account time dilation in the initial point $r_0$, while the second correction accounts for local time dilation and proper time dilation due to acceleration.

If the second factor is close to unity, we can expand it in powers of $r_{g}/r$ and obtain
\[t=\sqrt{1-\frac{r_g}{r_0}}\;\;
\big[\tau+\Delta\tau\big],
\quad\text{where}\quad
\Delta\tau=\int \frac{dr\,r_{g}/r}
{\sqrt{\frac{r_g}{r}-\frac{r_g}{r_0}}}.\]
On integrating, we can express $\Delta\tau$ through $\eta$ (\ref{Schw-TauParametric}), and thus derive the equation of motion in the form analogous to (\ref{Schw-TauParametric}), but with the correction term
\begin{equation}\label{Schw-TauParametric2}
\tau=\frac{r_0}{2}\sqrt{\frac{r_0}{r_g}}\;
\Big([1+\tfrac{r_g}{r_0}]\eta+\sin\eta\Big),\qquad
r=\frac{r_0}{2}(1+\cos\eta).
\end{equation}
Evidently, this approximation works only for $r\gg r_{g}$. </p>
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=== Problem 16: ultra-relativistic limit ===
Derive the equations of radial motion in the ultra-relativistic limit.
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<p style="text-align: left;">
In the relativistic limit we should assume $\gamma\gg1$ (near $r_{g}$ this always holds), thus $v\to 1$, and from (\ref{Schw-dotR}) we get
\[\frac{dr}{dt}\approx h\quad
\Rightarrow\quad t=\int \frac{dr}{h}.\]
This is, as should be expected, the null geodesic equation.</p>
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<div id="BlackHole30"></div>

=== Problem 17: communication from near the black hole ===
A particle (observer) falling into a black hole is emitting photons, which are detected on the same radial line far away from the horizon (i.e. the photons travel from emitter to detector radially). Find $r$, $v$ and $\dot{r}$ as functions of the signal's detection time in the limit $r\to r_g$.
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<p style="text-align: left;">
The geodesics describing light signals propagating away from the horizon are given by the equation (\ref{Schw-NullGeodesic}) with the plus sign:
\begin{equation}\label{Schw-tLambda}
t=r+r_{g}\ln|r-r_{g}|+t_{\Lambda},\end{equation}
where parameter $t_\Lambda$ parametrizes this family of geodesics and gives the difference in coordinate times of signals' registration at a given $r$. At $r\gg r_g$ this is also the proper time of the remote detector. Then
\begin{equation}\label{Schw-dtL(dr)}
dt_{\Lambda}=dt-\frac{r_{g}dr}{r-r_{g}}=
dt-\frac{r_{g}dr}{rh}=
-\frac{dr}{vh}-\frac{r_{g}dr}{rh}=
-\frac{dr}{h}\Big(\frac{1}{v}+\frac{r_{g}}{r}\Big).\end{equation}
As both $v$ and $r_{g}/r$ tend to unity near the horizon, the integral diverges logarithmically, twice faster than the one for $t(r)$. The meaning of this number is the following: as $r\to r_{g}$, the geodesics of massive and massless particles asymptotically coincide, and it takes the light as much time to "escape" the potential well as it takes the particle to fall in.

Asymptotically we obtain
\begin{equation}\label{Schw-r(tL)}
-t_{\Lambda}\approx 2\int\limits^{r}
\frac{rdr}{r-r_{g}}\approx 2r_{g}\ln|r-r_{g}|
\approx 2t(r)\quad\Rightarrow\quad
r(t_{\Lambda})-r_{g}\sim
r_{g} e^{-t_{\lambda}/2r_{g}};\end{equation}
also $\gamma=h_{0}/h\approx r_{g}/(r-r_{g})$, $v(r)=\sqrt{1-h/h_{0}}\approx 1-(r-r_{g})/2r_{g}$, and $\dot{r}\approx h$, thus
\begin{equation}\label{Schw-v(tL)}
\gamma\sim e^{t_{\Lambda}/2r_{g}};\qquad
(1-v)\sim {\textstyle\frac{1}{2}}
e^{-t_{\Lambda}/2r_{g}};
\qquad \dot{r}\sim e^{-t_{\Lambda}/2r_{g}}.\end{equation}</p>
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==Blackness of black holes==
A source radiates photons of frequency $\omega_i$, its radial coordinate at the time of emission is $r=r_{em}$. Find the frequency of photons registered by a detector situated at $r=r_{det}$ on the same radial line in different situations described below. By stationary observers here, we mean stationary in the static Schwarzschild metric; "radius" is the radial coordinate $r$.

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=== Problem 18: stationary source and detector ===
The source and detector are stationary.
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<p style="text-align: left;">
The integral of motion that conserves along the light ray due to the timelike Killing vector $\Omega^{\mu}=\partial_{t}$ is energy (up to a factor):
\begin{equation}\label{Schw-OmegaConst}
const\equiv\omega_{0}=
g_{\mu\nu}\Omega^{\mu}k^{\nu}=
g_{00}k^{0}=hk^{0}.\end{equation}
For a static observer $u^{\mu}=(u^{0},0,0,0)$, while
\begin{equation}\label{Schw-u0}
1=u^{\mu}u_{\mu}=g_{\mu\nu}u^{\mu}u^{\nu}=
g_{00}(u^{0})^{2}\quad\Rightarrow\quad
u^{0}=h^{-1/2}.\end{equation}
Therefore the frequency registered by this static observer is
\begin{equation}\label{Schw-omegastat}
\omega_{stat}=g_{\mu\nu}u^{\mu}_{stat}k^{\nu}=
g_{00}u^{0}_{stat}k^{0}=\sqrt{g_{00}}k^{0}=
\sqrt{h}k^{0}=\omega_{0}h^{-1/2}.\end{equation}
Then
\begin{equation}\label{Schw-RedShift0}
\omega_{0}=\omega_{stat}(r)\sqrt{h(r)}=const,\end{equation}
so we obtain
\begin{equation}\label{Schw-RedShift}
\omega_{det}=\omega_{em}
\sqrt{\frac{g_{00}(r_{em})}{g_{00}(r_{det})}}.\end{equation}
This is gravitational redshift.</p>
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<div id="BlackHole32"></div>
=== Problem 19: free-falling source ===
The source is falling freely without initial velocity from radius $r_0$; it flies by the stationary detector at the moment of emission.
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<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
It is educative to derive the Doppler effect from analogous considerations. Let us consider an emitter (detector) moving radially with physical velocity $v$ (see. (\ref{Schw-v(u)})). From the normalizing condition
\begin{equation}\label{Schw-rad-u0}
1=u^{\mu}u_{\mu}=g_{00}(u^{0})^{2}+g_{11}(u^{1})^{2}=
g_{00}(u^0)^{2}(1-\beta^{2})=\gamma^{-2} h (u^0)^{2},\end{equation}
thus
\begin{equation}\label{Schw-rad-u^mu}
u^{\mu}=\gamma\Big(
\frac{1}{\sqrt{g_{00}}},
\frac{\beta}{\sqrt{-g_{11}}},0,0\Big)=
\gamma\Big(\frac{1}{\sqrt{h}},
\beta\sqrt{h},0,0\Big).\end{equation}
For light
\[0=k^{\mu}k_{\mu}=g_{00}(k^0)^{2}+g_{11}(k^{1})^{2}
\quad\Rightarrow\quad
k^{1}=\pm k^{0}\sqrt{-g_{00}/g_{11}}=\pm hk^{0},\]
therefor the observer registers (or the detector emits) light with frequency
\[\omega_{em}\equiv\omega_{\beta}=
g_{\mu}u^{\mu}k^{\nu}=
g_{00}u^{0}k^{0}+g_{11}u^{1}k^{1}=
h\cdot \frac{\gamma}{\sqrt{h}}k^{0}\pm
\frac{1}{h}\cdot\gamma\beta\sqrt{h}\cdot hk^{0}=
\sqrt{h}\gamma k^{0}(1\pm\beta).\]
Choosing the sign "$+$" here, which corresponds to the emitter moving towards the horizon and light travaling outwards, we obtain the relativistic Dopper effect
\begin{equation}\label{RelatDoppler-radial}
\omega_{\beta}=
\frac{\omega_{0}}{\sqrt{h}}\cdot\gamma(1+\beta)=
\omega_{stat}\cdot\gamma(1+\beta).\end{equation}
Taking into account that in this case the emitter is moving and detector is at rest, so that $\omega_{em}=\omega_{\beta}$, $\omega_{det}=\omega_{stat}$, we get
\begin{equation}\label{Schw-RadialDoppler}
\omega_{det}\equiv \omega_{stat}=
\omega_{em}\frac{\gamma^{-1}}{1+\beta}=
\omega_{em}\sqrt{\frac{1-\beta}{1+\beta}}.\end{equation} </p>
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<div id="BlackHole33"></div>

=== Problem 20: adding the two effects ===
The source is freely falling the same way, while the detector is stationary at $r_{det}>r_{em}$.
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<p style="text-align: left;">
In this case we just need to take into account both the gravitational redshift (\ref{Schw-RedShift}) and the Doppler effect (\ref{Schw-RadialDoppler}). On substituting $\gamma=\sqrt{h_{0}/h_{em}}$, we get
\begin{equation}\label{Schw-Rad-FullRedshift}
\omega_{det}=\omega_{stat}(r_{em})
\sqrt{\frac{h_{em}}{h_{det}}}=
\omega_{em}\frac{\gamma^{-1}}{1+\beta}
\sqrt{\frac{h_{em}}{h_{det}}}=
\frac{\omega_{em}}{1+\beta}\cdot
\frac{h_{em}}{\sqrt{h_{0}h_{det}}}\end{equation}
In the limit $r_{em}\equiv r\sim r_{g}\ll r_{det}$ we obtain $\beta\approx 1$, thus
\[\omega_{det}\approx
\omega_{em}\cdot \frac{h_{em}}{2}=
\omega_{em}\cdot {\textstyle\frac{1}{2}}
\Big(1-\frac{r_{g}}{r_{em}}\Big)\to 0.\]</p>
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<div id="BlackHole34"></div>
=== Problem 21: intensity ===
The source is falling freely and emitting continuously photons with constant frequency, the detector is stationary far away from the horizon $r_{det}\gg r_{g}$. How does the detected light's intensity depend on $r_{em}$ at the moment of emission? On the proper time of detector?
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<p style="text-align: left;">
Repeating the argument of [[Schwarzschild black hole#BlackHole30|this problem]], for the proper time interval between detection of consecutive infinitely close signals (\ref{Schw-dtL(dr)}) we have
\[dt_{\Lambda}=
-\frac{dr}{h}\Big(\frac{1}{v}+\frac{r_{g}}{r}\Big),\]
where $dr$ is the corresponding displacement of the emitter, related with its proper time interval as (see \ref{ShwRad-tau})
\[dr=-\gamma v h\, d\tau.\]
Then using $\gamma=h^{-1/2}$ from (\ref{SchwRad-V}), we get
\begin{equation}\label{Schw-dtL(dTau)}
dt_{\Lambda}=\gamma
\Big(1+v\frac{r_{g}}{r}\Big) d\tau=
h^{-1/2}\Big(1+v\frac{r_{g}}{r}\Big) d\tau.\end{equation}
Intensity of the emitted and detected light is proportional to the photons' frequency and inversely proportional to the proper time interval during which a given number $N$ of photons are being emitted/detected. So, combining (\ref{Schw-Rad-FullRedshift}) and (\ref{Schw-dtL(dTau)}), we get
\begin{equation}\label{SchwRad-I}
\frac{I_{det}}{I_{em}}=
\frac{\omega_{det}}{\omega_{em}}\cdot
\frac{d\tau}{dt_{\Lambda}}=
\frac{1}{1+v}
\frac{h_{em}}{\sqrt{h_{0}h_{det}}}\cdot
\frac{\sqrt{h_{em}}}{1+v\frac{r_{g}}{r}}.\end{equation}
In the limit $h_{0},h_{det}\to 1$ and $r\to r_{g}$, so that $v\to1$, we have
\[\frac{I_{det}}{I_{em}}\approx
\frac{1}{4}h_{em}^{3/2}=
\frac{1}{4}\Big(1-\frac{r_{g}}{r}\Big)^{3/2}.\]
Substituting $r(t_\Lambda)$ from (\ref{Schw-r(tL)}), we finally obtain
\begin{equation}\label{Schw-I(tL)}
I_{det}(t_\Lambda)\approx I_{em}\cdot
\exp\Big\{-\frac{3\;t_\Lambda}{4\;r_g}\Big\}.\end{equation} </p>
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==Orbital motion, effective potential==
Due to high symmetry of the Schwarzschild metric, a particle's worldline is completely determined by the normalizing condition $u^{\mu}u_{\mu}=\epsilon$, where $\epsilon=1$ for a massive particle and $\epsilon=0$ for a massless one, plus two conservation laws---of energy and angular momentum.

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=== Problem 22: impact parameter ===
Show that the ratio of specific energy to specific angular momentum of a particle equals to $r_{g}/b$, where $b$ is the impact parameter at infinity (for unbounded motion).
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<p style="text-align: left;">
For $r\gg r_{g}$ we'll have $h\approx 1$ so if we count the angle from $\pi/2$ the impact parameter is $b=r\cos\varphi$. If we denote the length element of the trajectory by $dl$, then the ratio between the angular momentum and energy is
\[\frac{L}{E}
=\frac{r^2\cdot d\varphi/d\lambda}
{h\cdot dt/d\lambda}
=\frac{r^2 d\varphi}{hdt}\approx
\frac{r^2 (dl\cos\varphi\, /r)}{dl/v}=
r\cdot r\cos\varphi=vb.\]
In the ultrarelativistic case
\[L=Eb.\]</p>
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=== Problem 23: geodesic equations and effective potential ===
Derive the geodesics' equations; bring the equation for $r(\lambda)$ to the form
\[\frac{1}{2}\Big(\frac{dr}{d\lambda}\Big)^{2}
+V_{\epsilon}(r)=\varepsilon,\]
where $V_{\epsilon}(r)$ is a function conventionally termed as effective potential.
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<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The first equation is $\epsilon=u^{\mu}u_{\mu}$. For a massive particle $\epsilon=1$ and we can also choose the parameter $\lambda$ as the natural parameter of the geodesic $d\lambda=ds$. For a massless particle $\epsilon=0$ and the parameter is arbitrary. In the Schwarzschild metric therefore
\begin{equation}\label{Schw-Orbit1}
h\Big(\frac{dt}{d\lambda}\Big)^{2}-
\frac{1}{h}\Big(\frac{dr}{d\lambda}\Big)^{2}-
r^{2}\Big(\frac{d\varphi}{d\lambda}\Big)^{2}=
\epsilon.\end{equation}
The two integrals of motion (\ref{SchwInt-E1},\ref{SchwInt-L1}) are
\begin{equation}\label{Schw-Orbit-Integrals}
E=h\frac{dt}{d\lambda};\qquad
L=r^{2}\frac{d\varphi}{d\lambda}.\end{equation}
Substituting this into (\ref{Schw-Orbit1}) and rearranging terms, we derive the equation for $r(\lambda)$ in the form
\begin{equation}\label{Schw-V}
\frac{1}{2}\Big(\frac{dr}{d\lambda}\Big)^{2}+
V_{\epsilon}(r)=\varepsilon,\qquad\mbox{where}\quad
V(r)=\frac{h}{2}
\Big(\epsilon+\frac{L^2}{r^2}\Big),
\quad \varepsilon=\frac{E^2}{2}.\end{equation}
This is an analogue of one-dimensional motion of a particle in a potential $V$, with the quantity $\varepsilon$ playing the role of full energy (its analytic solution for the given $V$ is expressed through the integral from the square root of a third degree polynomial in the denominator, which is reduced to elliptic Jacobi functions). Note that $V(r=r_{g})=0$.

Recall that for massive particles $E$ and $L$ with the chosen parametrization are the energy and angular momentum per unit mass (see (\ref{SchwInt-E2},\ref{SchwInt-L2})).</p>
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=== Problem 24: bound and unbound motion ===
Plot and investigate the function $V(r)$. Find the radii of circular orbits and analyze their stability; find the regions of parameters $(E,L)$ corresponding to bound and unbound motion, fall into the black hole. Consider the cases of a) massless, b) massive particles.
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<p style="text-align: left;">
Let us first consider $\epsilon=0$.</p>

[[File:BHfig-SchwPhaL.png|center|thumb|400px|The effective potential for a massless particle $V(r/r_{g})$. The position of the maximum, that corresponds to the photon sphere, is at $r/r_{g}=\tfrac{3}{2}$ and does not depend on $L$]]

<p style="text-align: left;"> For massless particles the parameter $\lambda$ is arbitrary, so we will fix it by demanding that far from the horizon, where the metric is asymptotically Euclidean, holds $u^{0}\equiv dt/d\lambda=1$. Then, first, $E=1$, and second, at $r\gg r_{g}$
\[0=u^{\mu}u_{\mu}=(u^{0})^{2}-u^{\alpha}u_{\alpha}
\quad\Rightarrow\quad
\frac{dl}{d\lambda}=1.\]
Here $dl$ is the line element of the (asymptotically) flat space, and for such parametrization we obtain that $L=b$, where $b$ is the impact parameter of the ray at infinity:
\begin{equation}\label{Schw-orb-NullInts}
u^{0}|_{r\to\infty}=1,\;\Rightarrow\quad
E=1;\quad L=b.\end{equation}
Then
\[V_{\epsilon=0}=b^{2}\frac{h}{2r^{2}}=
\frac{b^{2}}{2}
\Big(\frac{1}{r^2}-\frac{r_{g}}{r^3}\Big),
\quad \varepsilon=\frac{1}{2}.\]
At small $r/r_{g}$ the effective potential behaves as $\sim (-r^{-3})$, while at large $r/r_{g}$ as $\sim r^{2}$, and reaches its maximum at
\[r_{max}=\frac{3}{2}r_{g},\quad
V_{max}=V(r_{max})=
\frac{2}{27}\Big(\frac{b}{r_g}\Big)^2.\]
As this is a maximum, the corresponding circular orbit on the so-called "photon sphere" is unstable. For $\varepsilon>V_{max}$ all the trajectories from one side of it escape to infinity, and the ones from the other side fall on the center (more accurately, they fall on the horizon, as we do not yet consider the motion beyond this point). Rewriting the inequality in terms of $b$, we have
\[b<b_{m}\equiv\frac{3\sqrt{3}}{2}\;r_{g}.\]
At larger impact parameters there is a "turning point" for motion from infinity, in which $dr/d\lambda=0$. In the region $r<r_{\max}$ these values of $b$, which do not have the meaning of the impact parameter in this case, correspond to finite orbits falling on the center.</p>
<p style="text-align: left;">
For massive particles it is convenient to express $V$ in terms of dimensionless quantities $\xi=r/r_{g}$ and $l=L/r_{g}$:
\[V_{\epsilon=1}-\frac{1}{2}=-\frac{1}{2\xi}+
\frac{l^2}{2\xi^2}-\frac{l^2}{2\xi^3}.\]
The first term is the Newtonian potential energy, the second one is the centrifugal energy, and only the third term is absent in the Newtonian theory and is unique to General Relativity. It changes the asymptote of $V$ at small $\xi$: $V\sim -\xi^{-3}$ instead of the usual $V\sim\xi^{-2}$.</p>

[[File:BHfig-SchwM1aL.png|center|thumb|400px|Effective potential for a massive particle $V(r/r_{g})$ for $l=\{0, 1.25, \sqrt{3}, 1.85, 2, 2.25, 2.5, 3\}\;\;\;$. There are stable as well as unstable circular orbits. The limiting value $l=\sqrt{3}$ defines the inflection point.]]
[[File:BHfig-SchwM2aL.png|center|thumb|400px|Same figure zoomed in to show the shallow minima]]

<p style="text-align: left;"> The extrema of $V$ are found as the roots of quadratic equation
\[\xi^2-2l^2 \xi +3l^2=0\quad\Rightarrow\quad
\xi_{\pm}=l^{2}\left\{1\pm\sqrt{1-3/l^2}\right\}.\]
Two extrema exist, maximum and minimum, if $l>\sqrt{3}$, i.e. $L>L_{cr}$, where
\[L_{cr}\equiv\sqrt{12}\,GM.\]
The minimum $\xi=\xi_{+}>l^2$ corresponds to a stable circular orbit and non-circular finite motions dangling around it (they are not elliptic). The maximum $\xi=\xi_{-}<l^2$ corresponds to an unstable circular orbit. If $\varepsilon>V(\xi_{-})$, then the motion is infinite with the fall on the center (i.e. at least on the horizon $r=r_g$).

The radius of the stable circular orbit is minimal when the discriminant turns to zero: $l=\sqrt{3}$,
$\xi_{+}=\xi_{-}=l^{2}=3$, and thus \[r_{circ}^{min}=3r_{g}=6GM.\]

For $l<\sqrt{3}$, i.e. $L<L_{cr}$, there are no extrema of $V$ and a particle's motion, either finite ($E<1$) or infinite ($E\geq 1$) always ends with the fall on the center.

In the limit $l\gg 1$, which corresponds to $L\gg r_{g}$,
\[\xi_{-}\approx3/2,\;\xi_{+}\approx2l^2,
\quad\Rightarrow\quad r_{-}=\frac{3}{2}r_{g};\;
r_{+}=\frac{2L^{2}}{r_{g}}=
\frac{L^2}{GM},\]
so the inner unstable orbit tends to the photon sphere, while the outer stable orbit tends to the classical circular one.
</p>
</div>
</div>

<div id="BlackHole38"></div>

=== Problem 25: gravitational cross-section ===
Derive the gravitational capture cross-section for a massless particle; the first correction to it for a massive particle ultra-relativistic at infinity. Find the cross-section for a non-relativistic particle to the first order in $v^2/c^2$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The gravitational capture happens, i.e. a particle moving from infinity falls on the center, if it passes above the effective potential barrier
\[\varepsilon\geq V_{\max}.\]
(a) For a massless particle we have already found the height of the barrier, so the limiting condition is
\[\frac{1}{2}=\frac{2}{27}\Big(\frac{b}{r_g}\Big)^2,\]
and for the capture cross-section we get
\[S_{\gamma}=\pi b^2 =\frac{27\pi}{4}r_{g}^{2}
=27\pi \Big(\frac{GM}{c^2}\Big)^2.\]

For a massive particle $V_{\epsilon=1}$ reaches its maximum in the lesser of the two roots $\xi_{\pm}$:
\[\xi_{\max}^{-1}=\frac{1}{3}\big(1+\sqrt{1-3/l^2}\big).\]
On substituting this into $V(\xi)$, after some transformations we find the maximum:
\[V_{\max}=\frac{1}{27}\Big(
l^2 +9+\frac{(l^2-3)^{3/2}}{l}\Big).\]

(b) Ultrarelativistic case:
\[\gamma\gg1,\quad E=\gamma_{\infty}\gg1,
\quad\varepsilon
=\tfrac{1}{2}\gamma_{\infty}^{2}\gg 1,\]
thus $V_{\max}\gg 1$, which only can be when $l\gg1$. In this limit, in the first order by $l^{-2}$
\[V_{\max}=\frac{2}{27}l^2 +\frac{1}{6},\]
so the limiting condition of capture is
\[l^{2}=\frac{27}{4}\big(
\gamma_{\infty}^{2}-\tfrac{1}{3}\big).\]
Expressing it through the impact parameter $b=L/Ev$, we are led to
\[S_{\gamma\gg1}
=\pi b^2=\pi r_{g}^{2}\frac{l^2}{v^2 E^2}
=\frac{27\pi}{4}r_{g}^{2}\Big(
1+\frac{2}{3\gamma_{\infty}^{2}}\Big).\]

(c) Nonrelativistic case: $E\approx (1+v^2 /2)$, $\varepsilon\approx \tfrac{1}{2}(1+v^2)$, and the limiting condition of capture takes the form
\[1+v^2 =\frac{1}{27}\Big(
l^2 +9+\frac{(l^2-3)^{3/2}}{l}\Big).\]
In the zeroth order by $v^2$ its solution is $l=2$ (note that at $l=\sqrt{3}$ there is no maximum of $V$), while in the first order
\[l^2 =4(1+2v^2).\]
In terms of impact parameter then the cross-section is
\[S_{v\ll 1}=\pi r_{g}^{2}\frac{l^2}{v^2 E^2}
\approx \frac{4\pi r_{g}^{2}}{v^2}(1+v^2).\]</p>
</div>
</div>

<div id="BlackHole39"></div>
=== Problem 26: innermost stable circular orbit ===
Find the minimal radius of stable circular orbit and its parameters. What is the maximum gravitational binding energy of a particle in the Schwarzschild spacetime?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The radius of stable circular orbit is minimal when
\[\xi_{\min}=l_{min}^{2}=3+0.\]
The value of effective potential at $\xi=\xi_{min}$ determines the particle's energy on the corresponding circular orbit (\ref{Schw-V}):
\[\frac{E^2}{2}=\varepsilon=V|_{\epsilon=1}
=\frac{1}{2}\Big(1-\frac{1}{\xi}\Big)
\Big(1+\frac{l^{2}}{\xi^2}\Big)=
\frac{1-\xi_{\min}^{-2}}{2},\]
where in the last equality we plugged in the value $\xi=\xi_{min}=l_{min}^{2}$. Thus
\[E=\sqrt{1-\xi_{min}^{-2}}=\sqrt{\frac{8}{9}}
=\frac{2\sqrt{2}}{3}.\]
The binding energy then is (in the units of $mc^{2}$)
\[1-E=1-\frac{2\sqrt{2}}{3}\approx 0,06.\] </p>
</div>
</div>

==Miscellaneous problems==

<div id="BlackHole40"></div>
=== Problem 27: gravitational lensing ===
Gravitational lensing is the effect of deflection of a light beam's (photon's) trajectory in the gravitational field. Derive the deflection of a photon's trajectory in Schwarzschild metric in the limit $L/r_{g}\gg 1$. Show that it is twice the value for a massive particle with velocity close to $c$ in the Newtonian theory.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
We choose $\lambda$ for massless particles as before and obtain
\[\Big(\frac{dr}{d\lambda}\Big)^{2}+
b^{2}\frac{h}{r^2}=1;\qquad
\frac{d\varphi}{d\lambda}=b\frac{1}{r^2}.\]
Excluding $\lambda$, after integration we obtain $\varphi(r)$. We are interested here in the variation of angle when $r$ goes from infinity to the minimal value, fro which the square root is zero. In the Newtonian theory it is $\pi/2$, so looking for the first correction to this value, we write
\[\Delta\varphi\big|_{\pi/2}=
b\int\limits_{r_{min}}^{\infty}
\frac{dr/r^2} {\sqrt{1-hb^{2}/r^2}}=
\left\|
x=\frac{b}{r},\;\varepsilon=\frac{r_{g}}{r}\ll 1
\right\|=
\int\limits_{0}^{x_{max}}
\frac{dx}{\sqrt{1-x^2+\varepsilon x^3}}.\]

Changing variables as $x^{2}-\varepsilon x^3=y^2$ and using the small parameter $\varepsilon$, we can transform the integral to one that is easy to compute
\[\Delta\varphi\big|_{\pi/2}\approx\int\limits_{0}^{1}
\frac{dy}{\sqrt{1-y^2}}(1+\varepsilon y)=
\frac{\pi}{2}+\varepsilon=
\frac{\pi}{2}+\frac{r_{g}}{b}.\]
The second term is the needed correction, which is half the full correction $\delta\phi$ to the angle's variation when a particle moves from infinity and back to infinity:
\[\delta\varphi\big|_{\pi}=\frac{2r_{g}}{b}=
\frac{4GM}{b c^2}.\]

Now let us calculate the same thing in the Newtonian theory for a fast particle. Integrals of motion are
\begin{align*}
&L=b v_{\infty}=r^{2}\dot{\varphi},\\
&E=U+\frac{\dot{r}^2}{2}+\frac{(r\dot{\varphi})^2}{2}
=\frac{\dot{r}^2}{2}-\frac{r_g}{2r}+\frac{L^2}{2r^2}.
\end{align*}
Then $\varphi(r)$ can be written as
\[\varphi=L\int\frac{dr/r^2}
{\sqrt{2E+\frac{r_g}{r}-\frac{L^2}{r^2}}},\]
and changing variables to $u=r^{-1}$, we obtain for the variation of angle from infinity to the turning point
\[\varphi|_{\pi/2}=\int\limits_{0}^{u_{max}}\frac{du}
{\sqrt{b^{-2}-u^2+\frac{r_g}{L^2}u}}.\]
The last term under the root is a small correction. Let us make another change of variables $u=u'+\varepsilon/2$, where $\varepsilon=r_{g}/L^2 \ll 1$. Then up to terms of higher order by $\varepsilon$ the integral is reduced to
\[\varphi|_{\pi/2}=\int\limits_{-\varepsilon/2}^{b^{-1}}
\frac{du'}{\sqrt{b^{-2}-{u'}^2}}=
\int\limits_{-b\varepsilon/2}^{1}
\frac{d\xi}{\sqrt{1-\xi^2}}
=\frac{\pi}{2}+\frac{b\varepsilon}{2}.\]
Then for motion from infinity to infinity the deflection of the trajectory from a straight line is
\[\delta\varphi|_{\pi}=b\varepsilon=
\frac{b r_g}{L^2}
=\frac{r_g}{b}\frac{1}{v_{\infty}^2}
\to \frac{r_g}{b}.\]
This is exactly half the correct quantity given by GTR.</p>
</div>
</div>

<div id="BlackHole41"></div>
=== Problem 28: generalization of Newtonian potential ===
Show that the $4$-acceleration of a stationary particle in the Schwarzschild metric can be presented in the form
\[a_{\mu}=-\partial_{\mu}\Phi,\quad
\text{where}\quad \Phi=\ln \sqrt{g_{00}}
=\tfrac{1}{2}\ln g_{00}\]
is some generalization of the Newtonian gravitational potential.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a static particle $u^{\mu}=(u^0,0,0,0)$, where $u^0$ is found from the normalizing condition
\[1=u^\mu u_\mu = g_{00}(u^{0})^2\quad
\Rightarrow\quad u^{0}=\frac{1}{\sqrt{g_{00}}}.\]
Then $4$-acceleration is
\[a^\mu=\frac{du^\mu}{ds}
=-\Gamma^{\mu}_{\nu\lambda}u^{\nu}u^{\lambda}
=-\Gamma^{\mu}_{00}(u^0)^{2}\]
and
\[a_{\mu}=-\frac{1}{g_{00}}\Gamma_{\mu\,00}
=-\frac{1}{2g_{00}}(-\partial_{\mu}g_{00})
=\frac{1}{2}\frac{\partial_{\mu}g_{00}}{g_{00}}
=\partial_{\mu}\Phi,\]
where $\Phi=\ln\sqrt{g_{00}}$. </p>
</div>
</div>

<div id="BlackHole42"></div>
=== Problem 29: coordinate-invariant reformulation ===
Let us reformulate the problem in a coordinate-independent manner. Suppose we have an arbitrary stationary metric with timelike Killing vector $\xi^\mu$, and we denote the $4$-velocity of a stationary observer by $u^{\mu}=\xi^{\mu}/V$. What is the $4$-force per unit mass that we need to apply to a test particle in order to make it stay stationary? Show in coordinate-independent way that the answer coincides with $\partial_{\mu}\Phi$ (up to the sign), and rewrite $\Phi$ in coordinate-independent form.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
From the normalizing condition $V^2=\xi^\mu \xi_\mu$. Here we will use a different meaning of acceleration, natural in the context of GTR. On a geodesic the covariant acceleration of a particle $w^\mu = u^\mu \nabla_{\mu} u^\nu$ is zero; if a particle moves not on a geodesic, this means that some (non-gravitational) force (per unit mass) is acting on it, equal to $w^{\mu}=-a^\mu$.

First of all, let us note that, as $\nabla g=0$,
\[\xi^{\mu}\nabla_{\mu}V
=\xi^{\mu}\xi^{\nu}
(\nabla_{\mu}\xi_{\nu}+\nabla_{\nu}\xi_{\mu})=0,\]
and also
\[2V\nabla_{\mu}V=\nabla_{\mu}V^2
=\nabla_{\mu}\xi^{\nu}\xi_{\nu}
=2\xi^{\nu}\nabla_{\mu}\xi_{\nu}.\]
Then using stationarity $u^\mu=\xi_\mu /V$ and the Killing equation, we get
\[ -a^\mu = w^\mu=\frac{1}{V}\xi^{\nu}\nabla_{\nu}
\big(\frac{1}{V}\xi^\mu\big)
=\frac{1}{V^2}\xi_{\nu}\nabla^{\nu}\xi^{\mu}
=-\frac{1}{V^2}\xi_{\nu}\nabla^{\mu}\xi^{\nu}
=-\frac{1}{V}\nabla^{\mu}V=-\nabla^{\mu}\ln V,\]
therefore
\[a^{\mu}=\nabla^{\mu}\Phi,\quad\text{where}
\quad \Phi=\ln V=\tfrac{1}{2}\ln \xi^{\mu}\xi_{\nu}.\]

In the weak field limit $g_{00}\approx 1+\tfrac{2\phi}{c^2}$, so $\Phi\approx \phi/c^2$. $\phi$ is the Newtonian gravitational potential. </p>
</div>
</div>

<div id="BlackHole43"></div>
=== Problem 30: surface gravity ===
Surface gravity $\kappa$ of the Schwarzschild horizon can be defined as acceleration of a stationary particle at the horizon, measured in the proper time of a stationary observer at infinity. Find $\kappa$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The scalar acceleration near the horizon, measured in proper time, is
\begin{align*}
a=&\sqrt{|a^{\mu}a_{\mu}|}
=\sqrt{|g^{\mu\nu}a_{\mu}a_{\nu}|}
=\sqrt{|g^{11}|(a_1)^{1}}=\sqrt{|g^{11}|(\Phi')^2}=\\
&=\sqrt{g_{00}}\;\Phi'(r)
=\sqrt{g_{00}}\;\frac{g_{00}'}{g_{00}}
=\frac{g_{00}'}{2\sqrt{g_{00}}}.
\end{align*}
It tends to infinity at the horizon. The time of remote observer $t$ is related to the proper time as $dt=d\tau\sqrt{g_{00}}$, therefore acceleration measured in time $t$ is
\[a_{\infty}=\sqrt{g_{00}}\;a=\frac{1}{2}g_{00}',\]
and it is finite. At $r=r_{g}$ it gives us the surface gravity
\[\kappa=a_{\infty}\Big|_{r=r_g}=\frac{1}{2r_{g}}
=\frac{c^4}{4MG}.\]
In the last equality we restored the dimensional factor $c^2$. </p>
</div>
</div>

Solving Einstein's equations for a spherically symmetric metric of general form in vacuum (energy-momentum tensor equals to zero), one can reduce the metric to
\[ds^2=f(t)\Big(1-\frac{C}{r}\Big)dt^2
-\Big(1-\frac{C}{r}\Big)^{-1}dr^2-r^2 d\Omega^2,\]
where $C$ is some integration constant, and $f(t)$ an arbitrary function of time $t$.

<div id="BlackHole44"></div>

=== Problem 31: uniqueness in exterior region ===
Suppose all the matter is distributed around the center of symmetry, and its energy-momentum tensor is spherically symmetric, so that the form of $g_{\mu\nu}$ written above is correct. Show that the solution in the exterior region is reduced to the Schwarzschild metric and find the relation between $C$ and the system's mass $M$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
We eliminate the factor $f$ in $g_{00}$ by coordinate transformation $\sqrt{f(t)}dt=dt'$; other components of the metric do not change. In the weak field limit $g_{00}\approx(1+\tfrac{2\varphi}{c^2})$, where $\varphi=-GM/r$ is the Newtonian gravitational potential. Comparing with the asymptote of $g_{00}$ we get
\[C=r_{g}=\frac{2GM}{c^2}.\]</p>
</div>
</div>

<div id="BlackHole45"></div>
=== Problem 32: solution in a spherically symmetric void ===
Let there be a spherically symmetric void $r<r_{0}$ in the spherically symmetric matter distribution. Show that spacetime in the void is flat.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integration constant is determined by the demand of boundedness of $g_{\mu\nu}$ in the region: $C=0$. Then on coordinate transformation $f(t)dt^2=d\tau^2$ we will obtain the flat Minkowskii metric
\[ds^2=d\tau^2-dr^2-r^2 d\Omega^2.\] </p>
</div>
</div>

<div id="BlackHole46"></div>
=== Problem 33: shells ===
Let the matter distribution be spherically symmetric and filling regions $r<r_{0}$ and $r_{1}<r<r_{2}$ ($r_{0}<r_{1}$). Can one affirm, that the solution in the layer of empty space $r_{0}<r<r_{1}$ is also the Schwarzschild metric?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
With such matter distribution the solution in region $r>r_{2}$ is Schwarzschild, as the constant $C$ and function $f(t)$ are fixed in the usual way. In region $r_{0}<r<r_{1}$, however, we do not have the freedom to choose a new time coordinate, as it will be determined by the smooth sewing-up of the metric at the boundaries (in region $r_{1}<r<r_{2}$ the solution is quite different, of course). Therefore $f(t)\neq 1$ in the inner region. </p>
</div>
</div>

<div id="BlackHoleEin1"></div>
=== Problem 34: Einstein equations for spherically symmetric case ===
Consider a static, spherically symmetric spacetime, described by metric
\[ds^{2} =-f(r)dt^{2} +f^{-1} (r)dr^{2} +r^{2} d\Omega ^{2}, \]
and write the Einstein's equations for it.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[(-f)-rf'(r)=-8\pi GPr^{2},\]
where $P=T_{r}^{r} $ is the radial pressure of the matter source. </p>
</div>
</div>

==Different coordinates, maximal extension==
We saw that a particle's proper time of reaching the singularity is finite. However, the Schwarzschild metric has a (removable) coordinate singularity at $r=r_{g}$. In order to eliminate it and analyze the casual structure of the full solution, it is convenient to use other coordinate frames. Everywhere below we transform the coordinates $r$ and $t$, while leaving the angular part unchanged.

<div id="BlackHole47"></div>
=== Problem 35: Rindler metric ===
Make coordinate transformation in the Schwarzschild metric near the horizon $(r-r_{g})\ll r_{g}$ by using physical distance to the horizon as a new radial coordinate instead of $r$, and show that in the new coordinates it reduces near the horizon to the Rindler metric.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The physical distance to the horizon along the "radius" $r$ at $(r-r_{g})\ll r_{g}$ is
\[l=\int\limits_{r_{g}}^{r}\frac{dr}{\sqrt{1-r_{g}/r}}=
\int\limits_{r_{g}}^{r}
\frac{\sqrt{r}dr}{\sqrt{r-r_{g}}}\approx
\sqrt{r_{g}}\int\limits_{0}^{r-r_{g}}
\frac{d\xi}{\sqrt{\xi}}=2\sqrt{r_{g}(r-r_{g})}.\]
Substituting this into the metric, we obtain (restoring the $c$ factors by dimensionality)
\[ds^{2}=l^{2}d\omega^{2}-dl^{2}-r^{2}(l)d\Omega^{2},
\qquad\mbox{where}\quad\omega=\frac{ct}{2r_{g}}.\]
Comparing with the [[Technical_warm-up#Rindler|Rindler metric]], we see that $l\equiv\rho=\frac{c^2}{a}$, and $\tau=t\frac{l}{2r_g}$. Thus the static Schwarzschild metric near the horizon has the same form as Minkowskii metric for a uniformly accelerated observer with acceleration $a=\frac{c^2}{l}$, which tends to infinity at the horizon. Then using the inverse transformation, from Rindler to Minkowskii, we can turn the Schwarzschild metric near the horizon into the flat one, thus showing explicitly that $r=r_g$ is just a removable coordinate singularity.</p>
</div>
</div>

<div id="BlackHole48"></div>

=== Problem 36: tortoise coordinate ===
Derive the Schwarzschild metric in coordinates $t$ and $r^\star=r+r_{g}\ln|r-r_g|$. How do the null geodesics falling to the center look like in $(t,r^\star)$? What range of values of $r^\star$ corresponds to the region $r>r_g$?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Making the change of variables
$(t,r)\to(t,r^\star)$, we get
\[ds^2=\Big(1-\frac{r_g}{r}\Big)
\big(dt^2-(dr^\star)^2\big)-r^2 d\Omega^2,\]
where $r$ should be understood as a function of $r^\star$. This metric is conformally flat$^{*}$ (without its angular part), and with no singularities. This is achieved, however, by pulling the horizon to infinity: $r=r_g$ corresponds to $r^\star =-\infty$.</p>

<p style="text-align: left;">$^{*}$A metric $g_{\mu\nu}$ is said to be conformally flat if it differs from the Minkowskii metric $\eta_{\mu\nu}$ by a factor $g_{\mu\nu}=\omega^{2}(x)\eta_{\mu\nu}$.</p>
</div>
</div>

<div id="BlackHole49"></div>

=== Problem 37: introducing null coordinates ===
Rewrite the metric in coordinates $r$ and $u=t-r^\star$, find the equations of null geodesics and the value of $g=det(g_{\mu\nu})$ at $r=r_{g}$. Likewise in coordinates $r$ and $v=t+r^\star$; in coordinates $(u,v)$. The coordinate frames $(v,r)$ and $(u,r)$ are called the ingoing and outgoing Eddington-Finkelstein coordinates.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In different coordinates
\begin{align*}
ds^2&=\Big(1-\frac{r_g}{r}\Big)dv^2
-\big(dv\,dr+dr\,dv\big)-r^2 d\Omega^2=\\
&=\Big(1-\frac{r_g}{r}\Big)du^2
+\big(du\,dr+dr\,du\big)-r^2 d\Omega^2,\\
&=\frac{1}{2}\Big(1-\frac{r_g}{r}\Big)
\big(du\,dv+dv\,du\big)-r^2 d\Omega^2.
\end{align*}
The null geodesics equations are $v=const$ for the ones falling to the center and $u=const$ for the ones escaping to infinity. Thus, by using the null coordinates $(u,v)$, we rectify the geodesics everywhere including the vicinity of the horizon. In coordinates $(r,v)$ we can analytically extend the solution beyond the horizon. The resulting spacetime region contains all the geodesics corresponding to particles which cross the horizon from the outside in, as well as all the geodesics in the outer region.

In coordinates $(r,u)$ we can likewise extend the solution beyond the horizon, and the resulting spacetime will contain, in addition to all the geodesics in the outer region, all the geodesics corresponding to particles crossing the horizon from the inside out. The two considered analytic extensions extend out metric to "different sides". This will be seen better in Kruskal coordinates.

In coordinates $(r,u)$ and $(r,v)$ the determinant of $g_{\mu\nu}$ is $g=-r^{4}\sin^{2}\theta$, and in coordninates $(u,v)$ we get $g=-\frac{1}{4}(1-r_{g}/r)^{2}r^{4}\sin^{2}\theta$.</p>
</div>
</div>

<div id="BlackHole50"></div>
=== Problem 38: Kruskal-Sekeres metric ===
Rewrite the Schwarzschild metric in coordinates $(u',v')$ and in the Kruskal coordinates $(T,R)$ (Kruskal solution), defined as follows:
\[v'=e^{v/2r_g},\quad u'=-e^{-u/2r_g};\qquad
T=\frac{u'+v'}{2},\quad R=\frac{v'-u'}{2}.\]
What are the equations of null geodesics, surfaces $r=const$ and $t=const$, of the horizon $r=r_{g}$, singularity $r=0$, in the coordinates $(T,R)$? What is the range space of $(T,R)$? Which regions in the Schwarzschild coordinates do the regions $\{\text{I}:\;R>|T|\}$, $\{\text{II}:\;T>|R|\}$, $\{\text{III}:\;R<-|T|\}$ and $\{\text{IV}:\;T<-|R|\}$ correspond to? Which of them are casually connected and which are not? What is the geometry of the spacelike slice $T=const$ and how does it evolve with time $T$?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
After substitution we obtain
\[ds^2=\frac{2r_{g}^3}{r}e^{-r/r_{g}}
(dv'\,du'+du'\,dv')-r^{2}d\Omega^2
=\frac{4r_{g}^3}{r}e^{-r/r_{g}}
(dT^2-dR^2)-r^{2}d\Omega^2.\]
Equations of null geodegics: $u=const$ and $v=const$, or in terms of $(T,R)$: $T=\pm R+const$. The surface $r=const$ is mapped to a hyperbola $T^2-R^2=-2(r-r_g)e^{r/r_g}$; singularity to a hyperbola $T^2-R^2=2r_g$ with ''two'' branches also; the horizon to \emph{two} straight lines $T=\pm R$. A surface $t=const$ is mapped to straight line $T=R\tanh\frac{t}{r_g}$; at $t\to\pm\infty$ it coincides with the horizon. Every point $(T,R)$ here represents a two-sphere.</p>

[[File:BHfig-KruskalCE.png|center|thumb|400px|The Kuskal diagram in coordinates $(T,R)$ and the mapping onto it of the coordinate grid $(t,r)$. Singularity is two hyperbola branches, in the past and future, while the horizon is two straight lines $T=\pm R$. Null geodesics are also straight lines $T\pm R=const$]]

<p style="text-align: left;">The physical region of values $(R,T)$ is given by condition $r>0$ and the region between the two hyperbola branches $T=\pm\sqrt{2r_g+R^2}$, which represent the past and future singularities. Region I is the asymptotically flat space outside of the horizon (we are living here); region II is the analytical extension along timelike geodesics directed inside across the horizon (coordinates $(r,v)$ map regions I and II); region III is the analytic extension along the timelike geodesics directed away from the horizon (coordinates $(r,u)$ map regions I and III); region IV is \emph{another} asymptotically flat spacetime. It is not casually connected to our spacetime region I.

A three-dimensional slice $T=const$ of the entire spacetime for $|T|>\sqrt{2r_g}$ is comprised of two unconnected asymptotically flat regions; for $|T|<\sqrt{2r_g}$ there is a connection between them, the so-called wormhole, or the Einstein-Rosen bridge. This bridge, however, is spacelike: it is seen clearly from the Kruskal diagram that there are no timelike geodesics (or any other timelike curves either) that would correspond to particles coming from one asymptotically flat region into the other.</p>
</div>
</div>

<div id="BlackHole51"></div>
=== Problem 39: Penrose diagram ===
Pass to coordinates
\[v''=\arctan\frac{v'}{\sqrt{r_g}},\quad
u''=\arctan\frac{u'}{\sqrt{r_g}}\]
and draw the spacetime diagram of the Kruskal solution in them.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The coordinates $(v'',u'')$ are needed in order to conformally map the entire Kruskal solution onto a bounded region of parameters $(v'',u'')$,
\[|v''|<\frac{\pi}{2},\quad |u''|<\frac{\pi}{2},
\quad |v''+u''|<\frac{\pi}{2}.\]
This is a square with two opposite angles cut out, rotated by $\pi/4$. Conformality here means that null geodesics, and thus the casual structure, is left invariant under such transformation, and equations of the null geodesics are still $u''=\pm v'' +const$.
[[File:BHfig-KruskalPenroseC2.png|center|thumb|400px|Penrose diagram for the Kruskal solution, with the grid of null geodesics]]</p>
</div>
</div>

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=== Problem 40: more realistic collapse ===
The Kruskal solution describes an eternal black hole. Suppose, for simplicity, that some black hole is formed as a result of radial collapse of a spherically symmetric shell of massless particles. What part of the Kruskal solution will be realized, and what will not be? What is the casual structure of the resulting spacetime?
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<p style="text-align: left;">
Outside of the collapsing shell the spacetime should be the Schwarzschild solution. As the particles are massless, their worldlines on the Kruskal diagram lie on the straight line $T-R=const$, and the exterior region contains a part of region I and a part of region II.</p>

[[File:BHfig-PenroseRealStarC.png|center|thumb|200px|Penrose diagram for the collapse of a massless spherically symmetric shell]]

<p style="text-align: left;">The inner region is just a chunk of Minkowskii spacetime, up to the moment of singularity formation. Sewing together of the tho solutions along the spheres of equal circumference $2\pi r$ gives us the full spacetime realized in the considered idealized case of collapse. We see that singularity and horizon are only in the future.</p>
</div>
</div>

Schwarzschild black hole

2013-05-02T10:01:26Z

Igor:

[[Category:Black Holes|2]]

The spherically symmetric solution of Einstein's equations in vacuum for the spacetime metric has the form$^{*}$
\begin{align}\label{Schw}
ds^{2}=h(r)\,dt^2-h^{-1}(r)\,dr^2-r^2 d\Omega^{2},
&\qquad\mbox{where}\quad
h(r)=1-\frac{r_g}{r};\quad r_{g}=\frac{2GM}{c^{2}};\\
d\Omega^{2}=d\theta^{2}+\sin^{2}\theta\, d\varphi^{2}&\;\text{is the metric of unit sphere.}\nonumber
\end{align}
The Birkhoff's theorem$^{**}$ (1923) states, that this solution is unique up to coordinate transformations. The quantity $r_g$ is called the Schwarzschild radius, or gravitational radius, $M$ is the mass of the central body or black hole.

$^{*}$ K. Schwarzschild, On the gravitational field of a mass point according to Einstein's theory, ''Sitzungsber. Preuss. Akad. Wiss. Phys. Math. Kl.'', p.189 (1916) (there is the translation of the original paper at [http://arxiv.org/abs/physics/9905030v1 arXiv:physics/9905030v1]; please disregard the abstract).

$^{**}$ G.D. Birkhoff, Relativity and Modern Physics, p.253, Harvard University Press, Cambridge (1923);
J.T. Jebsen, "Ark. Mat. Ast. Fys." (Stockholm) 15, nr.18 (1921), see also [http://arxiv.org/abs/physics/0508163 arXiv:physics/0508163v2].

__TOC__

==Simple problems==
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=== Problem 1: local time ===
Find the interval of local time (proper time of stationary observer) at a point $(r,\theta,\varphi)$ in terms of coordinate time $t$, and show that $t$ is the proper time of an observer at infinity. What happens when $r\to r_{g}$?

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<p style="text-align: left;">
The proper time of the stationary observer is $d\tau=ds|_{dr=d\theta=d\varphi=0}$:
\[d\tau=\sqrt{g_{00}}\;dt=\sqrt{1-\frac{r_{g}}{r}}\;dt.\]
At $r\to \infty$ it coincides with $dt$, so the coordinate time $t$ can be interpreted as the proper time of a "remote" observer. At $r\to r_g$ the local time flows slower and asymptotically stops. If one of two twins were to live some time at $r\approx r_g$, he will return to his remote twin having aged less (thought he might have acquired some grey hair due to constant fear of tumbling over the horizon). </p>
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=== Problem 2: measuring distances ===
What is the physical distance between two points with coordinates $(r_{1},\theta,\varphi)$ and $(r_{2},\theta,\varphi)$? Between $(r,\theta,\varphi_{1})$ and $(r,\theta,\varphi_{2})$? How do these distances behave in the limit $r_{1},r\to r_{g}$?
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<p style="text-align: left;">
\[l_{r1-r2}=\int\limits_{r_1}^{r_2}
\frac{dr}{\sqrt{1-\frac{r_g}{r}}};\quad
l_{\varphi1-\varphi_2}=2\pi r|\varphi_1-\varphi_2|;
\quad
l_{\theta1-\theta_2}=2\pi r|\theta_1-\theta_2|.\] </p>
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=== Problem 3: the inner region ===
What would be the answers to the previous two questions for $r<r_g$ and why*? Why the Schwarzschild metric cannot be imagined as a system of "welded" rigid rods in $r<r_g$, as it can be in the external region?
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<p style="text-align: left;">
''This question is given not to be answered but to make one think of the answer. Correct questions and correct answers can be given in terms of a proper coordinate frame, which is regular both in $r>r_g$ and in $r<r_g$. Still, one can say something meaningful as is.''</p>

<p style="text-align: left;">
At $r<r_g$ we have $g_{00}<0$ and $g_{11}>0$, thus the $t$ coordinate is ''spatial'' and $r$ coordinate is ''temporal'' (!):
\[ds^{2}=|h|^{-1}(r)dr^{2}-|h|(r)dt^{2}
-r^{2}d\Omega^{2}.\]
The metric therefore is ''nonstationary'' in this region, depending on the temporal coordinate $r$, but homogeneous, as there is no dependence on spatial coordinates.

Then for an observer "at rest" with respect to this coordinate system we would have $dt=d\theta=d\varphi=0$, and thus
\[d\tau^2=ds^2=-\frac{dr^{2}}{1-r_{g}/r}=
\frac{r dr^{2}}{r_{g}-r}>0,
\quad\Rightarrow\quad
d\tau=\frac{\sqrt{r}dr}{\sqrt{r_{g}-r}}.\]

An observer at rest with respect to the old coordinate system $dr=d\theta=d\varphi=0$, though, does not exist, as it would be $ds^{2}<0$ for him, which corresponds to spacelike geodesics (i.e. particles traveling faster than light).

The last of the two questions cannot be answered without additional assumptions, because ''time'' $t$, which is the spatial coordinate now, in the two points is not given.

The physical distance at $d\theta=d\varphi=dr=0$ is defined as
\[dl^{2}=|h(r)|dt^2.\]
It evidently depends on time $r$.

This very fact that Schwarzschild metric is nonstationary at $r<r_g$, and that a stationary one does not exist in this region, leads to the absence of stationary observers and thus to the impossibility to imagine it "welded" of a system of stiff rods. </p>

<p style="text-align: left;">*This was actually not a very simple problem</p>
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=== Problem 4: acceleration ===
Calculate the acceleration of a test particle with zero velocity.
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<p style="text-align: left;">
If a particle is at rest, its 4-velocity is $u^{\mu}=g_{00}^{-1/2}\delta^{\mu}_{0}$, where the factor is determined from the normalizing condition
\[1=g_{\mu\nu}u^{\mu}u^{\mu}=g_{00}(u^{0})^{2}.\]
Then the $4$-acceleration can be found from the geodesic equation:
\begin{align*}
a^{0}\equiv&\frac{du^0}{ds}=-{\Gamma^{0}}_{00}u^{0}u^{0}\sim
{\Gamma^{0}}_{00}\sim \Gamma_{0,\,00}\sim
(\partial_{0}g_{00}+\partial_{0}g_{00}
-\partial_{0}g_{00})=0;\\
a^{1}\equiv&\frac{du^1}{ds}
=-{\Gamma^{1}}_{00}u^{0}u^{0}
=-g^{11}\Gamma_{1,\,00}\;g_{00}^{-1}
=-(g_{00}g_{11})^{-1}\Gamma_{1,\,00}=\Gamma_{1,00}=
\\ &=\tfrac{1}{2}
(\partial_{0}g_{10}+\partial_{0}g_{10}
-\partial_{1}g_{00})
=-\frac{1}{2}\frac{dg_{00}}{dr}
=-\frac{h'}{2}
=\frac{r_{g}}{2r^2}.
\end{align*}
The scalar acceleration $a$ is then equal to
\[a^{2}=-g_{11}(a^{1})^{2}=\frac{(h')^{2}}{4h}
=-\frac{r_{g}^{2}}{4 r^4}
\Big(1-\frac{r_g}{r}\Big)^{-1}\]
and tends to infinity when we approach the horizon. </p>
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=== Problem 5: Schwarzschild is a vacuum solution ===
Show that Schwarzschild metric is a solution of Einstein's equation in vacuum.
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<p style="text-align: left;">
Straightforward calculation of Christoffel symbols and Ricci tensor yields the vacuum Einstein equation $R_{\mu\nu}=0$. </p>
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==Symmetries and integrals of motion==
For background on Killing vectors see problems [[Equations of General Relativity#equ_oto-kill1|K1]], [[Equations of General Relativity#equ_oto-kill2|K2]], [[Equations of General Relativity#equ_oto-kill3|K3]] of chapter 2.
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=== Problem 6: timelike Killing vector ===
What integral of motion arises due to existence of a timelike Killing vector? Express it through the physical velocity of the particle.
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<p style="text-align: left;">
If a Killing vector field is timelike we can use the coordinate frame in which $K^\mu$ is the unitary vector of the time coordinate (see problem [[Equations of General Relativity#equ_oto-kill2|K2]] of chapter 2.) $K^{\mu}=(1,0,0,0)$, while the spacelike basis vectors are orthogonal to it. Then the integral of motion is energy in the corresponding (stationary) frame $p_{\mu}K^{\mu}=p_{0}\equiv\varepsilon/c$.

Using the result [[Technical_warm-up#u0|u0]], which holds for arbitrary gravitational field, we obtain the expression for energy, which is the integral of motion in a stationary metric
\begin{equation}\label{EnergyStat}
\varepsilon=mc^{2}u_{0}=
mc^{2}\sqrt{g_{00}}\cdot\gamma=
\frac{mc^{2}\sqrt{g_{00}}}
{\sqrt{1-\frac{v^2}{c^2}}}.\end{equation} </p>
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=== Problem 7: Killing vectors of a sphere ===
Derive the Killing vectors for a sphere in Cartesian coordinate system; in spherical coordinates.
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<p style="text-align: left;">
Killing vectors correspond to infinitesimal transformations that leave the metric invariant. Considering a two-dimensional sphere embedded in the tree-dimensional space, we can see that its symmetries are rotations. Three of them are independent: the rotations in each of the three coordinate planes.

Let us consider an infinitesimal rotation in the plane $XY$: $dx=yd\lambda,\; dy=-xd\lambda$. Thus the Killing vector is*
\[K_{1}=x\partial_y-y\partial_x.\]

Using the spherical coordinates
\[x=\sin\theta\cos\varphi;\quad y=\sin\theta\sin\varphi;\quad z=\cos\theta,\]
we obtain $\partial_{x}=-\frac{\sin\varphi}{\sin\theta}\partial_\varphi$, $\partial_y=\frac{\cos\varphi}{\sin\theta}\partial_\varphi$, and therefore $K_{1}=\partial_{\varphi}$.

Considering rotations in plains $XZ$ and $YZ$ in the same way, in the end we obtain
\[\left\{\begin{array}{l}
K_{1}=x\partial_{y}-y\partial_{x}=\partial_{\varphi}\\
K_{2}=z\partial_{x}-x\partial_{z}=
\cos\varphi\;\partial_{\theta}-\cot\theta\sin\varphi\;\partial_\varphi\\
K_{3}=z\partial_{y}-y\partial_{x}=
\sin\varphi\;\partial_{\theta}+\cot\theta\cos\varphi\;\partial_\varphi\;.
\end{array}\right.\] </p>

<p style="text-align: left;">*Hereafter we use the following notation: a 4-vector $A$ is $A=A^{\mu}\partial_{\mu}$, where $\partial_{\mu}\equiv\partial/\partial x_{\mu}$ is the "coordinate" basis, $A^{\mu}$ the coordinates of $A$ in this basis; it is not hard to verify that transformation laws for $A^\mu$ and $\partial_{\mu}$ are adjusted so that $A$ is a quantity that does not depend on a coordinate frame. This also enables us to conveniently recalculate $A^\mu$ when we change the basis. For more detail see e.g. the textbook by Carroll:</p>

<p style="text-align: left;">Carroll S., Spacetime and geometry: an introduction to General Relativity. AW, 2003, ISBN 0805387323</p>
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=== Problem 8: spherical symmetry of Schwarzshild ===
Verify that in coordinates $(t,r,\theta,\varphi)$ vectors
\[ \begin{array}{l}
\Omega^{\mu}=(1,0,0,0),\\
R^{\mu}=(0,0,0,1),\\
S^{\mu}=(0,0,\cos\varphi,-\cot\theta\sin\varphi),\\
T^{\mu}=(0,0,-\sin\varphi,-\cot\theta\cos\varphi)
\end{array}\]
are the Killing vectors of the Schwarzschild metric.
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<p style="text-align: left;">
Vectors $\Omega^\mu$ and $R^\mu$ are Killing vectors because the metric does not depend explicitly either on $t$ or on $\varphi$. The last two correspond to the spherical symmetry (see previous problem). We can also check directly that they obey the Killing equation by evaluating the Christoffel symbols for the Schwarzschild metric. </p>
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=== Problem 9: planar motion ===
Show that existence of Killing vectors $S^\mu$ and $T^\mu$ leads to motion of particles in a plane.
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<p style="text-align: left;">
The integrals of motion corresponding to $S^\mu$ and $T^\mu$ are
\begin{align*}
S=u_{\mu}S^{\mu}=&
\;u_{2}\cos\varphi-u_{3}\cot\theta\sin\varphi;\\
T=u_{\mu}T^{\mu}=&
-u_{2}\sin\varphi-u_{3}\cot\theta\cos\varphi.
\end{align*}
Let us align the coordinate frame in such a way that the initial conditions are
\[\theta|_{t=0}=\pi/2;\quad
u_{2}|_{t=0}\equiv u_{\theta}|_{t=0}=0.\]
Then $S$ and $T$ are zero:
\[u_{2}\cos\varphi=u_{3}\cot\theta\sin\varphi;\qquad
u_{2}\sin\varphi=-u_{3}\cot\theta\cos\varphi.\]
Taking the square and adding the two equations, and also multiplying them, we obtain
\[\left\{\begin{array}{l}
u_{2}^{2}=u_{3}^{2}\cot^{2}\theta,\\
u_{2}^{2}\sin\varphi\cos\varphi=
-u_{3}^{2}\cot^{2}\theta\sin\varphi\cos\varphi
\end{array}\right.\quad\Rightarrow\quad
u_{3}^{2}\sin\varphi\cos\varphi
\cdot\cot^{2}\theta=0.\]
Then either $\varphi=const$, which means that $u_{3}=u_{2}=0$ and the motion is radial, or $\cot^{2}\theta=0$ and the motion takes place in the plane $\theta=\pi/2$. </p>
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=== Problem 10: stability of planar motion ===
Show that the particles' motion in the plane is stable.
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<p style="text-align: left;">
Conservation of $u_{\mu}R^{\mu}$ means that $R\equiv u_{3}\equiv\varphi'=const$. It is not hard to derive from the expressions for $S$ and $T$ that
\begin{align*}
&S\sin\varphi+T\cos\varphi=R\cot\theta,\\
&S\cos\varphi-T\sin\varphi=u_2.
\end{align*}
Taking the square and adding up, we are led to
\[u_2^2 =S^2+T^2+R^2-\frac{R^2}{\sin^2 \theta},\]
then on differentiating we obtain
\[u'_{2}=R^2 \frac{\cos\theta}{\sin^{2}\theta}.\]
Let the trajectory deviate slightly from the plane $\theta=\pi/2$. Then $\theta=\pi/2+\delta\theta$ and
\[(\delta\theta)''=u'_{2}
=R^2 \frac{\cos\theta}{\sin^{2}\theta}
\approx -R^2 \delta\theta,\]
therefore $\theta$ oscillates around the stable point $\pi/2$. </p>
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=== Problem 11: remaining integrals of motion ===
Write down explicitly the conserved quantities $p_{\mu}\Omega^{\mu}$ and $p_{\mu}R^{\mu}$ for movement in the plane $\theta=\pi/2$.
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<p style="text-align: left;">
The first integral of motion is, up to a multiplier, the energy (see Eq. [[Technical_warm-up#EnergyStat|EnergyStat]]):
\begin{equation}\label{SchwInt-E1}
E=\Omega_{\mu}u^{\mu}=
g_{00}\frac{dx^{0}}{d\lambda}=h\frac{dt}{d\lambda}=
\left(1-\frac{r_g}{r}\right)\frac{dt}{d\lambda},\end{equation}
where $\lambda$ is the geodesic parameter: for a massive particle we can always choose the natural parametrization $d\lambda=ds=\gamma^{-1}\sqrt{g_{00}}\,dt$ (see Eq. [[Technical_warm-up#IntervalStaticCase|IntervalStaticCase]]) so that $E$ is energy per unit mass; recovering the multipliers, for true energy we obtain
\begin{equation}\label{SchwInt-E2}
\varepsilon_{m}\equiv mcE=
mc^{2}\sqrt{g_{00}}\,\gamma=
mc^{2}\sqrt{h}\,\gamma=
mc^{2}\sqrt{\frac{1-r_{g}/r}{1-v^{2}/c^{2}}}.\end{equation}

The second integral of motion is the angular momentum (per unit mass also, and the sign is chosen so that in the Newtonian limit we obtain the usual angular momentum)
\begin{equation}\label{SchwInt-L1}
L=-R_{\mu}u^{\mu}=-g_{33}\frac{dx^{3}}{d\lambda}=
r^{2}\sin^{2}\theta\frac{d\varphi}{d\lambda}=
r^{2}\frac{d\varphi}{d\lambda}.\end{equation}
For a massive particle we choose $\lambda=s$, then
$d\lambda=ds=\gamma^{-1}\sqrt{g_{00}}\,dt$ and
\begin{equation}\label{SchwInt-L2}
l_{m}\equiv mL=
mr^{2}\dot{\varphi} \sqrt{g_{00}}\gamma=
m r^{2}\dot{\varphi}
\sqrt{\frac{1-r_{g}/r}{1-v^{2}/c^{2}}}.\end{equation}
Note that conservation of $l_m$ is a generalization to the relativistic case of the second Kepler's law on the sweeping of equal areas per unit time $r^{2}\dot{\varphi}=const$. </p>
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=== Problem 12: work and mass ===
What is the work needed to pull a particle from the horizon to infinity? Will a black hole's mass change if we drop a particle with zero initial velocity from immediate proximity of the horizon?
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<p style="text-align: left;">
At $r\gg r_g$ we'll have $\varepsilon_{m}=mc^{2}$, while near the horizon $\varepsilon_{m}\to 0$. The difference is the work needed to pull the particle away from the horizon to infinity, and it equals the rest mass (energy) of the particle. No, it will not change, because the energy of the falling particle is zero. </p>
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==Radial motion==
Consider a particle's radial motion: $\dot{\varphi}=\dot{\theta}=0$. In this problem one is especially interested in asymptotes of all functions as $r\to r_{g}$.

<p style="text-align: left;">Let us set $c=1$ here and henceforth measure time in the units of length, so that $x^{0}=t$, $\beta=v$, etc., and introduce the notation
\[h(r)\equiv g_{00}(r)=-\frac{1}{g_{11}(r)}=
1-\frac{r_{g}}{r}
\underset{r\to r_g+0}{\longrightarrow}+0.\]</p>

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=== Problem 13: null geodesics ===
Derive the equation for null geodesics (worldlines of massless particles).
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<div class="NavHead">solution</div>
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<p style="text-align: left;">
Equation of null curve is $ds^2=0$. Substituting the metric with $d\theta=d\varphi=0$, we get
\[dt=\pm dr\,\sqrt{-\frac{g_{11}}{g_{00}}}
=\pm\frac{dr}{h}.\]

On integration, we obtain
\begin{equation}\label{Schw-NullGeodesic}
\pm t=r+r_{g}\ln|r-r_{g}|+const.\end{equation}
Due to symmetry it is also a geodesic. This can be verified by evaluating the Christoffel symbols for the Schwarzschild metric and writing down the geodesic equatoin in explicit form. </p>
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=== Problem 14: geodesic motion of massive particle ===
Use energy conservation to derive $v(r)$, $\dot{r}(r)=dr/dt$, $r(t)$ for a massive particle. Initial conditions: $g_{00}|_{\dot{r}=0}=h_{0}$ (the limiting case $h_{0}\to 1$ is especially interesting and simple).
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<div class="NavHead">solution</div>
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<p style="text-align: left;">
In the radial case the physical velocity $v$ is expressed through the $4$-velocity $u^{\mu}=(u^{0},u^{1},0,0)$ as
\begin{equation}\label{Schw-v(u)}
\beta^{2}=\frac{dx^{\alpha}dx_{\alpha}}{d\tau^{2}}=
\frac{-g_{11}(dx^{1})^{2}}{g_{00}(dx^{0})^{2}}=
h^{-2}\Big(\frac{u^1}{u^0}\Big)^2=
\Big(\frac{\dot{r}}{h}\Big)^2,\end{equation}
where dot denotes differentiation by coordinate time $t$, and thus the energy conservation law (see (\ref{SchwInt-E2})) takes the form
\[const=\Big(\frac{\varepsilon}{mc^2}\Big)^{2}=
h\gamma^{2}=\frac{h}{1-\dot{r}^{2}/h^2}.\]

Initial condition fixes the constant*:
\begin{equation}\label{Schw-h0}
\dot{r}\sim v|_{r_{0}}=0\quad
\Rightarrow\quad
\Big(\frac{\varepsilon}{mc^2}\Big)^{2}
=g_{00}(r_{0})=h(r_0)=1-\frac{r_{g}}{r_{0}}
\equiv h_{0},\end{equation}
so for the physical velocity we get
\begin{equation}\label{SchwRad-V}
\gamma^{2}=\frac{h_{0}}{h(r)}
\underset{r\to r_{g}}{\longrightarrow}\infty;
\quad \frac{v}{c}\equiv\beta=
\sqrt{1-\gamma^{-2}}=
\sqrt{1-\frac{h}{h_{0}}}
\underset{r\to r_{g}}{\longrightarrow}1.\end{equation}
It tends to infinity close to the horizon! At the same time, as
$v^{2}=\dot{r}^{2}/h^{2}$ from (\ref{Schw-v(u)}), the coordinate velocity
\begin{equation}\label{Schw-dotR}
\Big|\frac{dr}{dt}\Big|=vh=h \sqrt{1-\frac{h}{h_{0}}}=
\Big(1-\frac{r_{g}}{r}\Big)
\sqrt{1-\frac{1-\frac{r_{g}}{r}}{1-\frac{r_{g}}{r_{0}}}}
\underset{r\to r_{g}}{\thicksim}h\to 0\end{equation}
tends to zero.

After integration we obtain $t(r)$:
\begin{equation}\label{Schw-t(r)}
const\pm t=\int\limits_{r}^{r_{0}}
\frac{dr}{vh}=
\int\limits_{r}^{r_{0}}\frac{rdr}{r-r_{g}}
\bigg[1-\frac{1-r_{g}/r}{1-r_{g}/r_{0}}\bigg]^{-1/2}
\underset{\substack{r\approx r_{g}\\r_{0}\gg r_{g}}}
{\approx}\int\limits_{r}^{r_{0}}
\frac{r\,dr}{r-r_{g}}=r+r_{g}\ln|r-r_{g}|.\end{equation}
As should be expected, the asymptote is the same as for the null geodesics (for $r_{0}\gg r_{g}$).</p>

<p style="text-align: left;">*Note that $r_0$ has the meaning of a turning point only for finite motion; this, however, does not prevent us from using the same notation in the non-finite case. Then $r_0$ is determined from $\varepsilon$ by the same formula and takes negative values, while $h_{0}>1$.</p>
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=== Problem 15: radial motion in terms of proper time ===
Show that the equation of radial motion in terms of proper time of the particle is the same as in the non-relativistic Newtonian theory. Calculate the proper time of the fall from $r=r_0$ to the center. Derive the first correction in $r_{g}/r$ to the Newtonian result. Initial velocity is zero.
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<p style="text-align: left;">
The particle's proper time interval $\tau$ is $d\tau=\gamma^{-1} d\hat{t}$, where $d\hat{t}=\sqrt{h}dt$ is local time. Then
\begin{equation}\label{ShwRad-tau}
\tau=\int dt \gamma^{-1}\sqrt{h}=
\int\limits_{r}^{r_{0}}
\frac{dr}{\gamma v\sqrt{h}}=
\int\limits_{r}^{r_{0}}
\frac{dr}{v \sqrt{h_{0}}}=
\int\limits_{r}^{r_{0}}
\frac{dr}{\sqrt{h_{0}-h}}=
\int\limits_{r}^{r_{0}}
\frac{dr}{\sqrt{\frac{r_{g}}{r_{0}}-
\frac{r_{g}}{r}}}.\end{equation}
On the other hand, in the Newtonian approach from energy conservation we obtain in the same way that
\[(\dot{r})^{2}=2(E-U)=2\;\Big(E+\frac{GM}{r}\Big)
=\frac{r_{g}}{r}-\frac{r_{g}}{r_{0}},\]
which gives us the same equation of motion.

We see that the integral is regular at $r=r_{g}$, is finite and converges even at $r\to 0$. It is not hard to evaluate:
\begin{equation}\label{Schw-TauParametric}
\tau=\frac{r_0}{2}\sqrt{\frac{r_0}{r_g}}
(\eta+\sin\eta),\qquad
r=\frac{r_0}{2}(1+\cos\eta),\qquad \eta\in(0,\pi).
\end{equation}
The proper time of reaching the singularity $r=0$ is therefore
\[\tau_{f}=\frac{\pi}{2}r_{0}
\sqrt{\frac{r_{0}}{r_{g}}}.\]

The integral (\ref{Schw-t(r)}) for $t(r)$ can be rewritten as
\[t(r)=\int \frac{dr}{vh}=\sqrt{h_{0}}
\int\frac{dr}{\sqrt{h_{0}-h}}\cdot \frac{1}{h}=
\underbrace{\sqrt{1-\frac{r_{g}}{r_{0}}}}
_{\text{correction 1}}\,
\underbrace{\int\!\frac{dr}{\sqrt{\frac{r_g}{r}-\frac{r_g}{r_0}}}}_{\tau(r)}
\cdot \underbrace{\frac{1}{1-\frac{r_g}{r}}}
_{\text{correction 2}}.\]
Now we see that the first factor gives the correction taking into account time dilation in the initial point $r_0$, while the second correction accounts for local time dilation and proper time dilation due to acceleration.

If the second factor is close to unity, we can expand it in powers of $r_{g}/r$ and obtain
\[t=\sqrt{1-\frac{r_g}{r_0}}\;\;
\big[\tau+\Delta\tau\big],
\quad\text{where}\quad
\Delta\tau=\int \frac{dr\,r_{g}/r}
{\sqrt{\frac{r_g}{r}-\frac{r_g}{r_0}}}.\]
On integrating, we can express $\Delta\tau$ through $\eta$ (\ref{Schw-TauParametric}), and thus derive the equation of motion in the form analogous to (\ref{Schw-TauParametric}), but with the correction term
\begin{equation}\label{Schw-TauParametric2}
\tau=\frac{r_0}{2}\sqrt{\frac{r_0}{r_g}}\;
\Big([1+\tfrac{r_g}{r_0}]\eta+\sin\eta\Big),\qquad
r=\frac{r_0}{2}(1+\cos\eta).
\end{equation}
Evidently, this approximation works only for $r\gg r_{g}$. </p>
</div>
</div>

<div id="BlackHole29"></div>

=== Problem 16: ultra-relativistic limit ===
Derive the equations of radial motion in the ultra-relativistic limit.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the relativistic limit we should assume $\gamma\gg1$ (near $r_{g}$ this always holds), thus $v\to 1$, and from (\ref{Schw-dotR}) we get
\[\frac{dr}{dt}\approx h\quad
\Rightarrow\quad t=\int \frac{dr}{h}.\]
This is, as should be expected, the null geodesic equation.</p>
</div>
</div>

<div id="BlackHole30"></div>

=== Problem 17: communication from near the black hole ===
A particle (observer) falling into a black hole is emitting photons, which are detected on the same radial line far away from the horizon (i.e. the photons travel from emitter to detector radially). Find $r$, $v$ and $\dot{r}$ as functions of the signal's detection time in the limit $r\to r_g$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The geodesics describing light signals propagating away from the horizon are given by the equation (\ref{Schw-NullGeodesic}) with the plus sign:
\begin{equation}\label{Schw-tLambda}
t=r+r_{g}\ln|r-r_{g}|+t_{\Lambda},\end{equation}
where parameter $t_\Lambda$ parametrizes this family of geodesics and gives the difference in coordinate times of signals' registration at a given $r$. At $r\gg r_g$ this is also the proper time of the remote detector. Then
\begin{equation}\label{Schw-dtL(dr)}
dt_{\Lambda}=dt-\frac{r_{g}dr}{r-r_{g}}=
dt-\frac{r_{g}dr}{rh}=
-\frac{dr}{vh}-\frac{r_{g}dr}{rh}=
-\frac{dr}{h}\Big(\frac{1}{v}+\frac{r_{g}}{r}\Big).\end{equation}
As both $v$ and $r_{g}/r$ tend to unity near the horizon, the integral diverges logarithmically, twice faster than the one for $t(r)$. The meaning of this number is the following: as $r\to r_{g}$, the geodesics of massive and massless particles asymptotically coincide, and it takes the light as much time to "escape" the potential well as it takes the particle to fall in.

Asymptotically we obtain
\begin{equation}\label{Schw-r(tL)}
-t_{\Lambda}\approx 2\int\limits^{r}
\frac{rdr}{r-r_{g}}\approx 2r_{g}\ln|r-r_{g}|
\approx 2t(r)\quad\Rightarrow\quad
r(t_{\Lambda})-r_{g}\sim
r_{g} e^{-t_{\lambda}/2r_{g}};\end{equation}
also $\gamma=h_{0}/h\approx r_{g}/(r-r_{g})$, $v(r)=\sqrt{1-h/h_{0}}\approx 1-(r-r_{g})/2r_{g}$, and $\dot{r}\approx h$, thus
\begin{equation}\label{Schw-v(tL)}
\gamma\sim e^{t_{\Lambda}/2r_{g}};\qquad
(1-v)\sim {\textstyle\frac{1}{2}}
e^{-t_{\Lambda}/2r_{g}};
\qquad \dot{r}\sim e^{-t_{\Lambda}/2r_{g}}.\end{equation}</p>
</div>
</div>

==Blackness of black holes==
A source radiates photons of frequency $\omega_i$, its radial coordinate at the time of emission is $r=r_{em}$. Find the frequency of photons registered by a detector situated at $r=r_{det}$ on the same radial line in different situations described below. By stationary observers here, we mean stationary in the static Schwarzschild metric; "radius" is the radial coordinate $r$.

<div id="BlackHole31"></div>
=== Problem 18: stationary source and detector ===
The source and detector are stationary.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integral of motion that conserves along the light ray due to the timelike Killing vector $\Omega^{\mu}=\partial_{t}$ is energy (up to a factor):
\begin{equation}\label{Schw-OmegaConst}
const\equiv\omega_{0}=
g_{\mu\nu}\Omega^{\mu}k^{\nu}=
g_{00}k^{0}=hk^{0}.\end{equation}
For a static observer $u^{\mu}=(u^{0},0,0,0)$, while
\begin{equation}\label{Schw-u0}
1=u^{\mu}u_{\mu}=g_{\mu\nu}u^{\mu}u^{\nu}=
g_{00}(u^{0})^{2}\quad\Rightarrow\quad
u^{0}=h^{-1/2}.\end{equation}
Therefore the frequency registered by this static observer is
\begin{equation}\label{Schw-omegastat}
\omega_{stat}=g_{\mu\nu}u^{\mu}_{stat}k^{\nu}=
g_{00}u^{0}_{stat}k^{0}=\sqrt{g_{00}}k^{0}=
\sqrt{h}k^{0}=\omega_{0}h^{-1/2}.\end{equation}
Then
\begin{equation}\label{Schw-RedShift0}
\omega_{0}=\omega_{stat}(r)\sqrt{h(r)}=const,\end{equation}
so we obtain
\begin{equation}\label{Schw-RedShift}
\omega_{det}=\omega_{em}
\sqrt{\frac{g_{00}(r_{em})}{g_{00}(r_{det})}}.\end{equation}
This is gravitational redshift.</p>
</div>
</div>

<div id="BlackHole32"></div>
=== Problem 19: free-falling source ===
The source is falling freely without initial velocity from radius $r_0$; it flies by the stationary detector at the moment of emission.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
It is educative to derive the Doppler effect from analogous considerations. Let us consider an emitter (detector) moving radially with physical velocity $v$ (see. (\ref{Schw-v(u)})). From the normalizing condition
\begin{equation}\label{Schw-rad-u0}
1=u^{\mu}u_{\mu}=g_{00}(u^{0})^{2}+g_{11}(u^{1})^{2}=
g_{00}(u^0)^{2}(1-\beta^{2})=\gamma^{-2} h (u^0)^{2},\end{equation}
thus
\begin{equation}\label{Schw-rad-u^mu}
u^{\mu}=\gamma\Big(
\frac{1}{\sqrt{g_{00}}},
\frac{\beta}{\sqrt{-g_{11}}},0,0\Big)=
\gamma\Big(\frac{1}{\sqrt{h}},
\beta\sqrt{h},0,0\Big).\end{equation}
For light
\[0=k^{\mu}k_{\mu}=g_{00}(k^0)^{2}+g_{11}(k^{1})^{2}
\quad\Rightarrow\quad
k^{1}=\pm k^{0}\sqrt{-g_{00}/g_{11}}=\pm hk^{0},\]
therefor the observer registers (or the detector emits) light with frequency
\[\omega_{em}\equiv\omega_{\beta}=
g_{\mu}u^{\mu}k^{\nu}=
g_{00}u^{0}k^{0}+g_{11}u^{1}k^{1}=
h\cdot \frac{\gamma}{\sqrt{h}}k^{0}\pm
\frac{1}{h}\cdot\gamma\beta\sqrt{h}\cdot hk^{0}=
\sqrt{h}\gamma k^{0}(1\pm\beta).\]
Choosing the sign "$+$" here, which corresponds to the emitter moving towards the horizon and light travaling outwards, we obtain the relativistic Dopper effect
\begin{equation}\label{RelatDoppler-radial}
\omega_{\beta}=
\frac{\omega_{0}}{\sqrt{h}}\cdot\gamma(1+\beta)=
\omega_{stat}\cdot\gamma(1+\beta).\end{equation}
Taking into account that in this case the emitter is moving and detector is at rest, so that $\omega_{em}=\omega_{\beta}$, $\omega_{det}=\omega_{stat}$, we get
\begin{equation}\label{Schw-RadialDoppler}
\omega_{det}\equiv \omega_{stat}=
\omega_{em}\frac{\gamma^{-1}}{1+\beta}=
\omega_{em}\sqrt{\frac{1-\beta}{1+\beta}}.\end{equation} </p>
</div>
</div>

<div id="BlackHole33"></div>

=== Problem 20: adding the two effects ===
The source is freely falling the same way, while the detector is stationary at $r_{det}>r_{em}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In this case we just need to take into account both the gravitational redshift (\ref{Schw-RedShift}) and the Doppler effect (\ref{Schw-RadialDoppler}). On substituting $\gamma=\sqrt{h_{0}/h_{em}}$, we get
\begin{equation}\label{Schw-Rad-FullRedshift}
\omega_{det}=\omega_{stat}(r_{em})
\sqrt{\frac{h_{em}}{h_{det}}}=
\omega_{em}\frac{\gamma^{-1}}{1+\beta}
\sqrt{\frac{h_{em}}{h_{det}}}=
\frac{\omega_{em}}{1+\beta}\cdot
\frac{h_{em}}{\sqrt{h_{0}h_{det}}}\end{equation}
In the limit $r_{em}\equiv r\sim r_{g}\ll r_{det}$ we obtain $\beta\approx 1$, thus
\[\omega_{det}\approx
\omega_{em}\cdot \frac{h_{em}}{2}=
\omega_{em}\cdot {\textstyle\frac{1}{2}}
\Big(1-\frac{r_{g}}{r_{em}}\Big)\to 0.\]</p>
</div>
</div>

<div id="BlackHole34"></div>
=== Problem 21: intensity ===
The source is falling freely and emitting continuously photons with constant frequency, the detector is stationary far away from the horizon $r_{det}\gg r_{g}$. How does the detected light's intensity depend on $r_{em}$ at the moment of emission? On the proper time of detector?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Repeating the argument of [[Schwarzschild black hole#BlackHole30|this problem]], for the proper time interval between detection of consecutive infinitely close signals (\ref{Schw-dtL(dr)}) we have
\[dt_{\Lambda}=
-\frac{dr}{h}\Big(\frac{1}{v}+\frac{r_{g}}{r}\Big),\]
where $dr$ is the corresponding displacement of the emitter, related with its proper time interval as (see \ref{ShwRad-tau})
\[dr=-\gamma v h\, d\tau.\]
Then using $\gamma=h^{-1/2}$ from (\ref{SchwRad-V}), we get
\begin{equation}\label{Schw-dtL(dTau)}
dt_{\Lambda}=\gamma
\Big(1+v\frac{r_{g}}{r}\Big) d\tau=
h^{-1/2}\Big(1+v\frac{r_{g}}{r}\Big) d\tau.\end{equation}
Intensity of the emitted and detected light is proportional to the photons' frequency and inversely proportional to the proper time interval during which a given number $N$ of photons are being emitted/detected. So, combining (\ref{Schw-Rad-FullRedshift}) and (\ref{Schw-dtL(dTau)}), we get
\begin{equation}\label{SchwRad-I}
\frac{I_{det}}{I_{em}}=
\frac{\omega_{det}}{\omega_{em}}\cdot
\frac{d\tau}{dt_{\Lambda}}=
\frac{1}{1+v}
\frac{h_{em}}{\sqrt{h_{0}h_{det}}}\cdot
\frac{\sqrt{h_{em}}}{1+v\frac{r_{g}}{r}}.\end{equation}
In the limit $h_{0},h_{det}\to 1$ and $r\to r_{g}$, so that $v\to1$, we have
\[\frac{I_{det}}{I_{em}}\approx
\frac{1}{4}h_{em}^{3/2}=
\frac{1}{4}\Big(1-\frac{r_{g}}{r}\Big)^{3/2}.\]
Substituting $r(t_\Lambda)$ from (\ref{Schw-r(tL)}), we finally obtain
\begin{equation}\label{Schw-I(tL)}
I_{det}(t_\Lambda)\approx I_{em}\cdot
\exp\Big\{-\frac{3\;t_\Lambda}{4\;r_g}\Big\}.\end{equation} </p>
</div>
</div>

==Orbital motion, effective potential==
Due to high symmetry of the Schwarzschild metric, a particle's worldline is completely determined by the normalizing condition $u^{\mu}u_{\mu}=\epsilon$, where $\epsilon=1$ for a massive particle and $\epsilon=0$ for a massless one, plus two conservation laws---of energy and angular momentum.

<div id="BlackHole35"></div>
=== Problem 22: impact parameter ===
Show that the ratio of specific energy to specific angular momentum of a particle equals to $r_{g}/b$, where $b$ is the impact parameter at infinity (for unbounded motion).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For $r\gg r_{g}$ we'll have $h\approx 1$ so if we count the angle from $\pi/2$ the impact parameter is $b=r\cos\varphi$. If we denote the length element of the trajectory by $dl$, then the ratio between the angular momentum and energy is
\[\frac{L}{E}
=\frac{r^2\cdot d\varphi/d\lambda}
{h\cdot dt/d\lambda}
=\frac{r^2 d\varphi}{hdt}\approx
\frac{r^2 (dl\cos\varphi\, /r)}{dl/v}=
r\cdot r\cos\varphi=vb.\]
In the ultrarelativistic case
\[L=Eb.\]</p>
</div>
</div>

<div id="BlackHole36"></div>

=== Problem 23: geodesic equations and effective potential ===
Derive the geodesics' equations; bring the equation for $r(\lambda)$ to the form
\[\frac{1}{2}\Big(\frac{dr}{d\lambda}\Big)^{2}
+V_{\epsilon}(r)=\varepsilon,\]
where $V_{\epsilon}(r)$ is a function conventionally termed as effective potential.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The first equation is $\epsilon=u^{\mu}u_{\mu}$. For a massive particle $\epsilon=1$ and we can also choose the parameter $\lambda$ as the natural parameter of the geodesic $d\lambda=ds$. For a massless particle $\epsilon=0$ and the parameter is arbitrary. In the Schwarzschild metric therefore
\begin{equation}\label{Schw-Orbit1}
h\Big(\frac{dt}{d\lambda}\Big)^{2}-
\frac{1}{h}\Big(\frac{dr}{d\lambda}\Big)^{2}-
r^{2}\Big(\frac{d\varphi}{d\lambda}\Big)^{2}=
\epsilon.\end{equation}
The two integrals of motion (\ref{SchwInt-E1},\ref{SchwInt-L1}) are
\begin{equation}\label{Schw-Orbit-Integrals}
E=h\frac{dt}{d\lambda};\qquad
L=r^{2}\frac{d\varphi}{d\lambda}.\end{equation}
Substituting this into (\ref{Schw-Orbit1}) and rearranging terms, we derive the equation for $r(\lambda)$ in the form
\begin{equation}\label{Schw-V}
\frac{1}{2}\Big(\frac{dr}{d\lambda}\Big)^{2}+
V_{\epsilon}(r)=\varepsilon,\qquad\mbox{where}\quad
V(r)=\frac{h}{2}
\Big(\epsilon+\frac{L^2}{r^2}\Big),
\quad \varepsilon=\frac{E^2}{2}.\end{equation}
This is an analogue of one-dimensional motion of a particle in a potential $V$, with the quantity $\varepsilon$ playing the role of full energy (its analytic solution for the given $V$ is expressed through the integral from the square root of a third degree polynomial in the denominator, which is reduced to elliptic Jacobi functions). Note that $V(r=r_{g})=0$.

Recall that for massive particles $E$ and $L$ with the chosen parametrization are the energy and angular momentum per unit mass (see (\ref{SchwInt-E2},\ref{SchwInt-L2})).</p>
</div>
</div>

<div id="BlackHole37"></div>

=== Problem 24: bound and unbound motion ===
Plot and investigate the function $V(r)$. Find the radii of circular orbits and analyze their stability; find the regions of parameters $(E,L)$ corresponding to bound and unbound motion, fall into the black hole. Consider the cases of a) massless, b) massive particles.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us first consider $\epsilon=0$.</p>

[[File:BHfig-SchwPhaL.png|center|thumb|400px|The effective potential for a massless particle $V(r/r_{g})$. The position of the maximum, that corresponds to the photon sphere, is at $r/r_{g}=\tfrac{3}{2}$ and does not depend on $L$]]

<p style="text-align: left;"> For massless particles the parameter $\lambda$ is arbitrary, so we will fix it by demanding that far from the horizon, where the metric is asymptotically Euclidean, holds $u^{0}\equiv dt/d\lambda=1$. Then, first, $E=1$, and second, at $r\gg r_{g}$
\[0=u^{\mu}u_{\mu}=(u^{0})^{2}-u^{\alpha}u_{\alpha}
\quad\Rightarrow\quad
\frac{dl}{d\lambda}=1.\]
Here $dl$ is the line element of the (asymptotically) flat space, and for such parametrization we obtain that $L=b$, where $b$ is the impact parameter of the ray at infinity:
\begin{equation}\label{Schw-orb-NullInts}
u^{0}|_{r\to\infty}=1,\;\Rightarrow\quad
E=1;\quad L=b.\end{equation}
Then
\[V_{\epsilon=0}=b^{2}\frac{h}{2r^{2}}=
\frac{b^{2}}{2}
\Big(\frac{1}{r^2}-\frac{r_{g}}{r^3}\Big),
\quad \varepsilon=\frac{1}{2}.\]
At small $r/r_{g}$ the effective potential behaves as $\sim (-r^{-3})$, while at large $r/r_{g}$ as $\sim r^{2}$, and reaches its maximum at
\[r_{max}=\frac{3}{2}r_{g},\quad
V_{max}=V(r_{max})=
\frac{2}{27}\Big(\frac{b}{r_g}\Big)^2.\]
As this is a maximum, the corresponding circular orbit on the so-called "photon sphere" is unstable. For $\varepsilon>V_{max}$ all the trajectories from one side of it escape to infinity, and the ones from the other side fall on the center (more accurately, they fall on the horizon, as we do not yet consider the motion beyond this point). Rewriting the inequality in terms of $b$, we have
\[b<b_{m}\equiv\frac{3\sqrt{3}}{2}\;r_{g}.\]
At larger impact parameters there is a "turning point" for motion from infinity, in which $dr/d\lambda=0$. In the region $r<r_{\max}$ these values of $b$, which do not have the meaning of the impact parameter in this case, correspond to finite orbits falling on the center.</p>
<p style="text-align: left;">
For massive particles it is convenient to express $V$ in terms of dimensionless quantities $\xi=r/r_{g}$ and $l=L/r_{g}$:
\[V_{\epsilon=1}-\frac{1}{2}=-\frac{1}{2\xi}+
\frac{l^2}{2\xi^2}-\frac{l^2}{2\xi^3}.\]
The first term is the Newtonian potential energy, the second one is the centrifugal energy, and only the third term is absent in the Newtonian theory and is unique to General Relativity. It changes the asymptote of $V$ at small $\xi$: $V\sim -\xi^{-3}$ instead of the usual $V\sim\xi^{-2}$.</p>

[[File:BHfig-SchwM1aL.png|center|thumb|400px|Effective potential for a massive particle $V(r/r_{g})$ for $l=\{0, 1.25, \sqrt{3}, 1.85, 2, 2.25, 2.5, 3\}\;\;\;$. There are stable as well as unstable circular orbits. The limiting value $l=\sqrt{3}$ defines the inflection point.]]
[[File:BHfig-SchwM2aL.png|center|thumb|400px|Same figure zoomed in to show the shallow minima]]

<p style="text-align: left;"> The extrema of $V$ are found as the roots of quadratic equation
\[\xi^2-2l^2 \xi +3l^2=0\quad\Rightarrow\quad
\xi_{\pm}=l^{2}\left\{1\pm\sqrt{1-3/l^2}\right\}.\]
Two extrema exist, maximum and minimum, if $l>\sqrt{3}$, i.e. $L>L_{cr}$, where
\[L_{cr}\equiv\sqrt{12}\,GM.\]
The minimum $\xi=\xi_{+}>l^2$ corresponds to a stable circular orbit and non-circular finite motions dangling around it (they are not elliptic). The maximum $\xi=\xi_{-}<l^2$ corresponds to an unstable circular orbit. If $\varepsilon>V(\xi_{-})$, then the motion is infinite with the fall on the center (i.e. at least on the horizon $r=r_g$).

The radius of the stable circular orbit is minimal when the discriminant turns to zero: $l=\sqrt{3}$,
$\xi_{+}=\xi_{-}=l^{2}=3$, and thus \[r_{circ}^{min}=3r_{g}=6GM.\]

For $l<\sqrt{3}$, i.e. $L<L_{cr}$, there are no extrema of $V$ and a particle's motion, either finite ($E<1$) or infinite ($E\geq 1$) always ends with the fall on the center.

In the limit $l\gg 1$, which corresponds to $L\gg r_{g}$,
\[\xi_{-}\approx3/2,\;\xi_{+}\approx2l^2,
\quad\Rightarrow\quad r_{-}=\frac{3}{2}r_{g};\;
r_{+}=\frac{2L^{2}}{r_{g}}=
\frac{L^2}{GM},\]
so the inner unstable orbit tends to the photon sphere, while the outer stable orbit tends to the classical circular one.
</p>
</div>
</div>

<div id="BlackHole38"></div>

=== Problem 25: gravitational cross-section ===
Derive the gravitational capture cross-section for a massless particle; the first correction to it for a massive particle ultra-relativistic at infinity. Find the cross-section for a non-relativistic particle to the first order in $v^2/c^2$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The gravitational capture happens, i.e. a particle moving from infinity falls on the center, if it passes above the effective potential barrier
\[\varepsilon\geq V_{\max}.\]
(a) For a massless particle we have already found the height of the barrier, so the limiting condition is
\[\frac{1}{2}=\frac{2}{27}\Big(\frac{b}{r_g}\Big)^2,\]
and for the capture cross-section we get
\[S_{\gamma}=\pi b^2 =\frac{27\pi}{4}r_{g}^{2}
=27\pi \Big(\frac{GM}{c^2}\Big)^2.\]

For a massive particle $V_{\epsilon=1}$ reaches its maximum in the lesser of the two roots $\xi_{\pm}$:
\[\xi_{\max}^{-1}=\frac{1}{3}\big(1+\sqrt{1-3/l^2}\big).\]
On substituting this into $V(\xi)$, after some transformations we find the maximum:
\[V_{\max}=\frac{1}{27}\Big(
l^2 +9+\frac{(l^2-3)^{3/2}}{l}\Big).\]

(b) Ultrarelativistic case:
\[\gamma\gg1,\quad E=\gamma_{\infty}\gg1,
\quad\varepsilon
=\tfrac{1}{2}\gamma_{\infty}^{2}\gg 1,\]
thus $V_{\max}\gg 1$, which only can be when $l\gg1$. In this limit, in the first order by $l^{-2}$
\[V_{\max}=\frac{2}{27}l^2 +\frac{1}{6},\]
so the limiting condition of capture is
\[l^{2}=\frac{27}{4}\big(
\gamma_{\infty}^{2}-\tfrac{1}{3}\big).\]
Expressing it through the impact parameter $b=L/Ev$, we are led to
\[S_{\gamma\gg1}
=\pi b^2=\pi r_{g}^{2}\frac{l^2}{v^2 E^2}
=\frac{27\pi}{4}r_{g}^{2}\Big(
1+\frac{2}{3\gamma_{\infty}^{2}}\Big).\]

(c) Nonrelativistic case: $E\approx (1+v^2 /2)$, $\varepsilon\approx \tfrac{1}{2}(1+v^2)$, and the limiting condition of capture takes the form
\[1+v^2 =\frac{1}{27}\Big(
l^2 +9+\frac{(l^2-3)^{3/2}}{l}\Big).\]
In the zeroth order by $v^2$ its solution is $l=2$ (note that at $l=\sqrt{3}$ there is no maximum of $V$), while in the first order
\[l^2 =4(1+2v^2).\]
In terms of impact parameter then the cross-section is
\[S_{v\ll 1}=\pi r_{g}^{2}\frac{l^2}{v^2 E^2}
\approx \frac{4\pi r_{g}^{2}}{v^2}(1+v^2).\]</p>
</div>
</div>

<div id="BlackHole39"></div>
=== Problem 26: innermost stable circular orbit ===
Find the minimal radius of stable circular orbit and its parameters. What is the maximum gravitational binding energy of a particle in the Schwarzschild spacetime?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The radius of stable circular orbit is minimal when
\[\xi_{\min}=l_{min}^{2}=3+0.\]
The value of effective potential at $\xi=\xi_{min}$ determines the particle's energy on the corresponding circular orbit (\ref{Schw-V}):
\[\frac{E^2}{2}=\varepsilon=V|_{\epsilon=1}
=\frac{1}{2}\Big(1-\frac{1}{\xi}\Big)
\Big(1+\frac{l^{2}}{\xi^2}\Big)=
\frac{1-\xi_{\min}^{-2}}{2},\]
where in the last equality we plugged in the value $\xi=\xi_{min}=l_{min}^{2}$. Thus
\[E=\sqrt{1-\xi_{min}^{-2}}=\sqrt{\frac{8}{9}}
=\frac{2\sqrt{2}}{3}.\]
The binding energy then is (in the units of $mc^{2}$)
\[1-E=1-\frac{2\sqrt{2}}{3}\approx 0,06.\] </p>
</div>
</div>

==Miscellaneous problems==

<div id="BlackHole40"></div>
=== Problem 27: gravitational lensing ===
Gravitational lensing is the effect of deflection of a light beam's (photon's) trajectory in the gravitational field. Derive the deflection of a photon's trajectory in Schwarzschild metric in the limit $L/r_{g}\gg 1$. Show that it is twice the value for a massive particle with velocity close to $c$ in the Newtonian theory.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
We choose $\lambda$ for massless particles as before and obtain
\[\Big(\frac{dr}{d\lambda}\Big)^{2}+
b^{2}\frac{h}{r^2}=1;\qquad
\frac{d\varphi}{d\lambda}=b\frac{1}{r^2}.\]
Excluding $\lambda$, after integration we obtain $\varphi(r)$. We are interested here in the variation of angle when $r$ goes from infinity to the minimal value, fro which the square root is zero. In the Newtonian theory it is $\pi/2$, so looking for the first correction to this value, we write
\[\Delta\varphi\big|_{\pi/2}=
b\int\limits_{r_{min}}^{\infty}
\frac{dr/r^2} {\sqrt{1-hb^{2}/r^2}}=
\left\|
x=\frac{b}{r},\;\varepsilon=\frac{r_{g}}{r}\ll 1
\right\|=
\int\limits_{0}^{x_{max}}
\frac{dx}{\sqrt{1-x^2+\varepsilon x^3}}.\]

Changing variables as $x^{2}-\varepsilon x^3=y^2$ and using the small parameter $\varepsilon$, we can transform the integral to one that is easy to compute
\[\Delta\varphi\big|_{\pi/2}\approx\int\limits_{0}^{1}
\frac{dy}{\sqrt{1-y^2}}(1+\varepsilon y)=
\frac{\pi}{2}+\varepsilon=
\frac{\pi}{2}+\frac{r_{g}}{b}.\]
The second term is the needed correction, which is half the full correction $\delta\phi$ to the angle's variation when a particle moves from infinity and back to infinity:
\[\delta\varphi\big|_{\pi}=\frac{2r_{g}}{b}=
\frac{4GM}{b c^2}.\]

Now let us calculate the same thing in the Newtonian theory for a fast particle. Integrals of motion are
\begin{align*}
&L=b v_{\infty}=r^{2}\dot{\varphi},\\
&E=U+\frac{\dot{r}^2}{2}+\frac{(r\dot{\varphi})^2}{2}
=\frac{\dot{r}^2}{2}-\frac{r_g}{2r}+\frac{L^2}{2r^2}.
\end{align*}
Then $\varphi(r)$ can be written as
\[\varphi=L\int\frac{dr/r^2}
{\sqrt{2E+\frac{r_g}{r}-\frac{L^2}{r^2}}},\]
and changing variables to $u=r^{-1}$, we obtain for the variation of angle from infinity to the turning point
\[\varphi|_{\pi/2}=\int\limits_{0}^{u_{max}}\frac{du}
{\sqrt{b^{-2}-u^2+\frac{r_g}{L^2}u}}.\]
The last term under the root is a small correction. Let us make another change of variables $u=u'+\varepsilon/2$, where $\varepsilon=r_{g}/L^2 \ll 1$. Then up to terms of higher order by $\varepsilon$ the integral is reduced to
\[\varphi|_{\pi/2}=\int\limits_{-\varepsilon/2}^{b^{-1}}
\frac{du'}{\sqrt{b^{-2}-{u'}^2}}=
\int\limits_{-b\varepsilon/2}^{1}
\frac{d\xi}{\sqrt{1-\xi^2}}
=\frac{\pi}{2}+\frac{b\varepsilon}{2}.\]
Then for motion from infinity to infinity the deflection of the trajectory from a straight line is
\[\delta\varphi|_{\pi}=b\varepsilon=
\frac{b r_g}{L^2}
=\frac{r_g}{b}\frac{1}{v_{\infty}^2}
\to \frac{r_g}{b}.\]
This is exactly half the correct quantity given by GTR.</p>
</div>
</div>

<div id="BlackHole41"></div>
=== Problem 28: generalization of Newtonian potential ===
Show that the $4$-acceleration of a stationary particle in the Schwarzschild metric can be presented in the form
\[a_{\mu}=-\partial_{\mu}\Phi,\quad
\text{where}\quad \Phi=\ln \sqrt{g_{00}}
=\tfrac{1}{2}\ln g_{00}\]
is some generalization of the Newtonian gravitational potential.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a static particle $u^{\mu}=(u^0,0,0,0)$, where $u^0$ is found from the normalizing condition
\[1=u^\mu u_\mu = g_{00}(u^{0})^2\quad
\Rightarrow\quad u^{0}=\frac{1}{\sqrt{g_{00}}}.\]
Then $4$-acceleration is
\[a^\mu=\frac{du^\mu}{ds}
=-\Gamma^{\mu}_{\nu\lambda}u^{\nu}u^{\lambda}
=-\Gamma^{\mu}_{00}(u^0)^{2}\]
and
\[a_{\mu}=-\frac{1}{g_{00}}\Gamma_{\mu\,00}
=-\frac{1}{2g_{00}}(-\partial_{\mu}g_{00})
=\frac{1}{2}\frac{\partial_{\mu}g_{00}}{g_{00}}
=\partial_{\mu}\Phi,\]
where $\Phi=\ln\sqrt{g_{00}}$. </p>
</div>
</div>

<div id="BlackHole42"></div>
=== Problem 29: coordinate-invariant reformulation ===
Let us reformulate the problem in a coordinate-independent manner. Suppose we have an arbitrary stationary metric with timelike Killing vector $\xi^\mu$, and we denote the $4$-velocity of a stationary observer by $u^{\mu}=\xi^{\mu}/V$. What is the $4$-force per unit mass that we need to apply to a test particle in order to make it stay stationary? Show in coordinate-independent way that the answer coincides with $\partial_{\mu}\Phi$ (up to the sign), and rewrite $\Phi$ in coordinate-independent form.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
From the normalizing condition $V^2=\xi^\mu \xi_\mu$. Here we will use a different meaning of acceleration, natural in the context of GTR. On a geodesic the covariant acceleration of a particle $w^\mu = u^\mu \nabla_{\mu} u^\nu$ is zero; if a particle moves not on a geodesic, this means that some (non-gravitational) force (per unit mass) is acting on it, equal to $w^{\mu}=-a^\mu$.

First of all, let us note that, as $\nabla g=0$,
\[\xi^{\mu}\nabla_{\mu}V
=\xi^{\mu}\xi^{\nu}
(\nabla_{\mu}\xi_{\nu}+\nabla_{\nu}\xi_{\mu})=0,\]
and also
\[2V\nabla_{\mu}V=\nabla_{\mu}V^2
=\nabla_{\mu}\xi^{\nu}\xi_{\nu}
=2\xi^{\nu}\nabla_{\mu}\xi_{\nu}.\]
Then using stationarity $u^\mu=\xi_\mu /V$ and the Killing equation, we get
\[ -a^\mu = w^\mu=\frac{1}{V}\xi^{\nu}\nabla_{\nu}
\big(\frac{1}{V}\xi^\mu\big)
=\frac{1}{V^2}\xi_{\nu}\nabla^{\nu}\xi^{\mu}
=-\frac{1}{V^2}\xi_{\nu}\nabla^{\mu}\xi^{\nu}
=-\frac{1}{V}\nabla^{\mu}V=-\nabla^{\mu}\ln V,\]
therefore
\[a^{\mu}=\nabla^{\mu}\Phi,\quad\text{where}
\quad \Phi=\ln V=\tfrac{1}{2}\ln \xi^{\mu}\xi_{\nu}.\]

In the weak field limit $g_{00}\approx 1+\tfrac{2\phi}{c^2}$, so $\Phi\approx \phi/c^2$. $\phi$ is the Newtonian gravitational potential. </p>
</div>
</div>

<div id="BlackHole43"></div>
=== Problem 30: surface gravity ===
Surface gravity $\kappa$ of the Schwarzschild horizon can be defined as acceleration of a stationary particle at the horizon, measured in the proper time of a stationary observer at infinity. Find $\kappa$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The scalar acceleration near the horizon, measured in proper time, is
\begin{align*}
a=&\sqrt{|a^{\mu}a_{\mu}|}
=\sqrt{|g^{\mu\nu}a_{\mu}a_{\nu}|}
=\sqrt{|g^{11}|(a_1)^{1}}=\sqrt{|g^{11}|(\Phi')^2}=\\
&=\sqrt{g_{00}}\;\Phi'(r)
=\sqrt{g_{00}}\;\frac{g_{00}'}{g_{00}}
=\frac{g_{00}'}{2\sqrt{g_{00}}}.
\end{align*}
It tends to infinity at the horizon. The time of remote observer $t$ is related to the proper time as $dt=d\tau\sqrt{g_{00}}$, therefore acceleration measured in time $t$ is
\[a_{\infty}=\sqrt{g_{00}}\;a=\frac{1}{2}g_{00}',\]
and it is finite. At $r=r_{g}$ it gives us the surface gravity
\[\kappa=a_{\infty}\Big|_{r=r_g}=\frac{1}{2r_{g}}
=\frac{c^4}{4MG}.\]
In the last equality we restored the dimensional factor $c^2$. </p>
</div>
</div>

Solving Einstein's equations for a spherically symmetric metric of general form in vacuum (energy-momentum tensor equals to zero), one can reduce the metric to
\[ds^2=f(t)\Big(1-\frac{C}{r}\Big)dt^2
-\Big(1-\frac{C}{r}\Big)^{-1}dr^2-r^2 d\Omega^2,\]
where $C$ is some integration constant, and $f(t)$ an arbitrary function of time $t$.

<div id="BlackHole44"></div>

=== Problem 31: uniqueness in exterior region ===
Suppose all the matter is distributed around the center of symmetry, and its energy-momentum tensor is spherically symmetric, so that the form of $g_{\mu\nu}$ written above is correct. Show that the solution in the exterior region is reduced to the Schwarzschild metric and find the relation between $C$ and the system's mass $M$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
We eliminate the factor $f$ in $g_{00}$ by coordinate transformation $\sqrt{f(t)}dt=dt'$; other components of the metric do not change. In the weak field limit $g_{00}\approx(1+\tfrac{2\varphi}{c^2})$, where $\varphi=-GM/r$ is the Newtonian gravitational potential. Comparing with the asymptote of $g_{00}$ we get
\[C=r_{g}=\frac{2GM}{c^2}.\]</p>
</div>
</div>

<div id="BlackHole45"></div>
=== Problem 32: solution in a spherically symmetric void ===
Let there be a spherically symmetric void $r<r_{0}$ in the spherically symmetric matter distribution. Show that spacetime in the void is flat.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integration constant is determined by the demand of boundedness of $g_{\mu\nu}$ in the region: $C=0$. Then on coordinate transformation $f(t)dt^2=d\tau^2$ we will obtain the flat Minkowskii metric
\[ds^2=d\tau^2-dr^2-r^2 d\Omega^2.\] </p>
</div>
</div>

<div id="BlackHole46"></div>
=== Problem 33: shells ===
Let the matter distribution be spherically symmetric and filling regions $r<r_{0}$ and $r_{1}<r<r_{2}$ ($r_{0}<r_{1}$). Can one affirm, that the solution in the layer of empty space $r_{0}<r<r_{1}$ is also the Schwarzschild metric?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
With such matter distribution the solution in region $r>r_{2}$ is Schwarzschild, as the constant $C$ and function $f(t)$ are fixed in the usual way. In region $r_{0}<r<r_{1}$, however, we do not have the freedom to choose a new time coordinate, as it will be determined by the smooth sewing-up of the metric at the boundaries (in region $r_{1}<r<r_{2}$ the solution is quite different, of course). Therefore $f(t)\neq 1$ in the inner region. </p>
</div>
</div>

<div id="BlackHoleEin1"></div>
=== Problem 34: Einstein equations for spherically symmetric case ===
Consider a static, spherically symmetric spacetime, described by metric
\[ds^{2} =-f(r)dt^{2} +f^{-1} (r)dr^{2} +r^{2} d\Omega ^{2}, \]
and write the Einstein's equations for it.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[(-f)-rf'(r)=-8\pi GPr^{2},\]
where $P=T_{r}^{r} $ is the radial pressure of the matter source. </p>
</div>
</div>

==Different coordinates, maximal extension==
We saw that a particle's proper time of reaching the singularity is finite. However, the Schwarzschild metric has a (removable) coordinate singularity at $r=r_{g}$. In order to eliminate it and analyze the casual structure of the full solution, it is convenient to use other coordinate frames. Everywhere below we transform the coordinates $r$ and $t$, while leaving the angular part unchanged.

<div id="BlackHole47"></div>
=== Problem 35: Rindler metric ===
Make coordinate transformation in the Schwarzschild metric near the horizon $(r-r_{g})\ll r_{g}$ by using physical distance to the horizon as a new radial coordinate instead of $r$, and show that in the new coordinates it reduces near the horizon to the Rindler metric.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The physical distance to the horizon along the "radius" $r$ at $(r-r_{g})\ll r_{g}$ is
\[l=\int\limits_{r_{g}}^{r}\frac{dr}{\sqrt{1-r_{g}/r}}=
\int\limits_{r_{g}}^{r}
\frac{\sqrt{r}dr}{\sqrt{r-r_{g}}}\approx
\sqrt{r_{g}}\int\limits_{0}^{r-r_{g}}
\frac{d\xi}{\sqrt{\xi}}=2\sqrt{r_{g}(r-r_{g})}.\]
Substituting this into the metric, we obtain (restoring the $c$ factors by dimensionality)
\[ds^{2}=l^{2}d\omega^{2}-dl^{2}-r^{2}(l)d\Omega^{2},
\qquad\mbox{where}\quad\omega=\frac{ct}{2r_{g}}.\]
Comparing with the [[Technical_warm-up#Rindler|Rindler metric]], we see that $l\equiv\rho=\frac{c^2}{a}$, and $\tau=t\frac{l}{2r_g}$. Thus the static Schwarzschild metric near the horizon has the same form as Minkowskii metric for a uniformly accelerated observer with acceleration $a=\frac{c^2}{l}$, which tends to infinity at the horizon. Then using the inverse transformation, from Rindler to Minkowskii, we can turn the Schwarzschild metric near the horizon into the flat one, thus showing explicitly that $r=r_g$ is just a removable coordinate singularity.</p>
</div>
</div>

<div id="BlackHole48"></div>

=== Problem 36: tortoise coordinate ===
Derive the Schwarzschild metric in coordinates $t$ and $r^\star=r+r_{g}\ln|r-r_g|$. How do the null geodesics falling to the center look like in $(t,r^\star)$? What range of values of $r^\star$ corresponds to the region $r>r_g$?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Making the change of variables
$(t,r)\to(t,r^\star)$, we get
\[ds^2=\Big(1-\frac{r_g}{r}\Big)
\big(dt^2-(dr^\star)^2\big)-r^2 d\Omega^2,\]
where $r$ should be understood as a function of $r^\star$. This metric is conformally flat$^{*}$ (without its angular part), and with no singularities. This is achieved, however, by pulling the horizon to infinity: $r=r_g$ corresponds to $r^\star =-\infty$.</p>

<p style="text-align: left;">$^{*}$A metric $g_{\mu\nu}$ is said to be conformally flat if it differs from the Minkowskii metric $\eta_{\mu\nu}$ by a factor $g_{\mu\nu}=\omega^{2}(x)\eta_{\mu\nu}$.</p>
</div>
</div>

<div id="BlackHole49"></div>

=== Problem 37: introducing null coordinates ===
Rewrite the metric in coordinates $r$ and $u=t-r^\star$, find the equations of null geodesics and the value of $g=det(g_{\mu\nu})$ at $r=r_{g}$. Likewise in coordinates $r$ and $v=t+r^\star$; in coordinates $(u,v)$. The coordinate frames $(v,r)$ and $(u,r)$ are called the ingoing and outgoing Eddington-Finkelstein coordinates.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In different coordinates
\begin{align*}
ds^2&=\Big(1-\frac{r_g}{r}\Big)dv^2
-\big(dv\,dr+dr\,dv\big)-r^2 d\Omega^2=\\
&=\Big(1-\frac{r_g}{r}\Big)du^2
+\big(du\,dr+dr\,du\big)-r^2 d\Omega^2,\\
&=\frac{1}{2}\Big(1-\frac{r_g}{r}\Big)
\big(du\,dv+dv\,du\big)-r^2 d\Omega^2.
\end{align*}
The null geodesics equations are $v=const$ for the ones falling to the center and $u=const$ for the ones escaping to infinity. Thus, by using the null coordinates $(u,v)$, we rectify the geodesics everywhere including the vicinity of the horizon. In coordinates $(r,v)$ we can analytically extend the solution beyond the horizon. The resulting spacetime region contains all the geodesics corresponding to particles which cross the horizon from the outside in, as well as all the geodesics in the outer region.

In coordinates $(r,u)$ we can likewise extend the solution beyond the horizon, and the resulting spacetime will contain, in addition to all the geodesics in the outer region, all the geodesics corresponding to particles crossing the horizon from the inside out. The two considered analytic extensions extend out metric to "different sides". This will be seen better in Kruskal coordinates.

In coordinates $(r,u)$ and $(r,v)$ the determinant of $g_{\mu\nu}$ is $g=-r^{4}\sin^{2}\theta$, and in coordninates $(u,v)$ we get $g=-\frac{1}{4}(1-r_{g}/r)^{2}r^{4}\sin^{2}\theta$.</p>
</div>
</div>

<div id="BlackHole50"></div>
=== Problem 38: Kruskal-Sekeres metric ===
Rewrite the Schwarzschild metric in coordinates $(u',v')$ and in the Kruskal coordinates $(T,R)$ (Kruskal solution), defined as follows:
\[v'=e^{v/2r_g},\quad u'=-e^{-u/2r_g};\qquad
T=\frac{u'+v'}{2},\quad R=\frac{v'-u'}{2}.\]
What are the equations of null geodesics, surfaces $r=const$ and $t=const$, of the horizon $r=r_{g}$, singularity $r=0$, in the coordinates $(T,R)$? What is the range space of $(T,R)$? Which regions in the Schwarzschild coordinates do the regions $\{\text{I}:\;R>|T|\}$, $\{\text{II}:\;T>|R|\}$, $\{\text{III}:\;R<-|T|\}$ and $\{\text{IV}:\;T<-|R|\}$ correspond to? Which of them are casually connected and which are not? What is the geometry of the spacelike slice $T=const$ and how does it evolve with time $T$?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
After substitution we obtain
\[ds^2=\frac{2r_{g}^3}{r}e^{-r/r_{g}}
(dv'\,du'+du'\,dv')-r^{2}d\Omega^2
=\frac{4r_{g}^3}{r}e^{-r/r_{g}}
(dT^2-dR^2)-r^{2}d\Omega^2.\]
Equations of null geodegics: $u=const$ and $v=const$, or in terms of $(T,R)$: $T=\pm R+const$. The surface $r=const$ is mapped to a hyperbola $T^2-R^2=-2(r-r_g)e^{r/r_g}$; singularity to a hyperbola $T^2-R^2=2r_g$ with ''two'' branches also; the horizon to \emph{two} straight lines $T=\pm R$. A surface $t=const$ is mapped to straight line $T=R\tanh\frac{t}{r_g}$; at $t\to\pm\infty$ it coincides with the horizon. Every point $(T,R)$ here represents a two-sphere.</p>

[[File:BHfig-KruskalCE.png|center|thumb|400px|The Kuskal diagram in coordinates $(T,R)$ and the mapping onto it of the coordinate grid $(t,r)$. Singularity is two hyperbola branches, in the past and future, while the horizon is two straight lines $T=\pm R$. Null geodesics are also straight lines $T\pm R=const$]]

<p style="text-align: left;">The physical region of values $(R,T)$ is given by condition $r>0$ and the region between the two hyperbola branches $T=\pm\sqrt{2r_g+R^2}$, which represent the past and future singularities. Region I is the asymptotically flat space outside of the horizon (we are living here); region II is the analytical extension along timelike geodesics directed inside across the horizon (coordinates $(r,v)$ map regions I and II); region III is the analytic extension along the timelike geodesics directed away from the horizon (coordinates $(r,u)$ map regions I and III); region IV is \emph{another} asymptotically flat spacetime. It is not casually connected to our spacetime region I.

A three-dimensional slice $T=const$ of the entire spacetime for $|T|>\sqrt{2r_g}$ is comprised of two unconnected asymptotically flat regions; for $|T|<\sqrt{2r_g}$ there is a connection between them, the so-called wormhole, or the Einstein-Rosen bridge. This bridge, however, is spacelike: it is seen clearly from the Kruskal diagram that there are no timelike geodesics (or any other timelike curves either) that would correspond to particles coming from one asymptotically flat region into the other.</p>
</div>
</div>

<div id="BlackHole51"></div>
=== Problem 39: Penrose diagram ===
Pass to coordinates
\[v''=\arctan\frac{v'}{\sqrt{r_g}},\quad
u''=\arctan\frac{u'}{\sqrt{r_g}}\]
and draw the spacetime diagram of the Kruskal solution in them.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The coordinates $(v'',u'')$ are needed in order to conformally map the entire Kruskal solution onto a bounded region of parameters $(v'',u'')$,
\[|v''|<\frac{\pi}{2},\quad |u''|<\frac{\pi}{2},
\quad |v''+u''|<\frac{\pi}{2}.\]
This is a square with two opposite angles cut out, rotated by $\pi/4$. Conformality here means that null geodesics, and thus the casual structure, is left invariant under such transformation, and equations of the null geodesics are still $u''=\pm v'' +const$.
[[File:BHfig-KruskalPenroseC2.png|center|thumb|400px|Penrose diagram for the Kruskal solution, with the grid of null geodesics]]</p>
</div>
</div>

<div id="BlackHole52"></div>
=== Problem 40: more realistic collapse ===
The Kruskal solution describes an eternal black hole. Suppose, for simplicity, that some black hole is formed as a result of radial collapse of a spherically symmetric shell of massless particles. What part of the Kruskal solution will be realized, and what will not be? What is the casual structure of the resulting spacetime?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Outside of the collapsing shell the spacetime should be the Schwarzschild solution. As the particles are massless, their worldlines on the Kruskal diagram lie on the straight line $T-R=const$, and the exterior region contains a part of region I and a part of region II.</p>

[[File:BHfig-PenroseRealStarC.png|center|thumb|200px|Penrose diagram for the collapse of a massless spherically symmetric shell]]

<p style="text-align: left;">The inner region is just a chunk of Minkowskii spacetime, up to the moment of singularity formation. Sewing together of the tho solutions along the spheres of equal circumference $2\pi r$ gives us the full spacetime realized in the considered idealized case of collapse. We see that singularity and horizon are only in the future.</p>
</div>
</div>

Particles' motion in general black hole spacetimes

2013-05-02T06:26:34Z

Igor:

[[Category:Black Holes|4]]
__TOC__

In this section we use the $(-+++)$ signature, Greek letters for spacetime indices and Latin letters for spatial indices.

==Frames, time intervals and distances==

In the next several problems we again consider the procedure of measuring time and space intervals by different observers, but in a different, more formal and powerful approach.

<div id="OZ01"></div>
=== Problem 1===

Let a particle move with the four-velocity $U^{\mu }$. It can be viewed as
some observer carrying a frame attached to him. Locally, it defines the
hypersurface orthogonal to it. Show that
\begin{equation}
h_{\mu \nu }=g_{\mu \nu }+U_{\mu }U_{\nu } \label{h}
\end{equation}
is (i) the projection operator onto this hypersurface, and at the same time
(ii) the induced metric of the hypersurface. This means that (i) for any
vector projected at this hypersurface by means of $h^\mu_\nu$, only the components orthogonal to $U^{\mu}$ survive, (ii) the repeated application of the projection operation leaves the vector within the hypersurface unchanged. In other words, $h_{\mu \nu}$ satisfies
\begin{align}
&h^{\mu}_{\nu}U^{\nu }=0 ; \label{1} \\
&h^{\mu}_{\nu }h^{\nu}_{\lambda}=h^{\mu }_{\lambda}. \label{2}
\end{align}

<div id="OZ021"></div>
=== Problem 2===

Let us consider a particle moving with the four-velocity $U^{\mu }$. The
interval $ds^{2}$ between two close events is defined in terms of
differentials of coordinates,
\begin{equation}
ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }.
\end{equation}

For given $dx^{\mu }$, what is the value of the proper time $d\tau _{obs}$ between the corresponding events measured by this observer? How can one define
locally the notions of simultaneity and proper distance $dl$ for the observer in terms of its four-velocity and the corresponding projection operator $h^\mu_\nu$ ? How is the interval $ds^{2}$ related to $d\tau _{obs}$ and $dl$?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In order to obtain $d\tau _{obs}$, one should project $dx^{\mu}$ onto the
four-velocity:
\begin{equation}
d\tau _{obs}=-dx^{\mu }U_{\mu }.\label{tau}
\end{equation}
Then, locally, two events are simultaneous when
\begin{equation}
d\tau _{obs}=0. \label{sim}
\end{equation}

The proper distance
\begin{equation}
dl^{2}=h_{\mu \nu }dx^{\mu }dx^{\nu } \label{dl}.
\end{equation}

Then (\ref{h}) implies that
\begin{equation}
ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }=dl^{2}-d\tau _{obs}^{2} \label{ds}.
\end{equation}
</p></div>
</div>

<div id="OZ03"></div>
=== Problem 3===

Let our observer measure the velocity of some other particle passing in its
immediate vicinity. Relate the interval to $d\tau _{obs}$ and the particle's
velocity $w$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The velocity can be defined as
\begin{equation}
w^{2}=\left( \frac{dl}{d\tau _{obs}}\right) ^{2}.
\end{equation}
Then, it follows from (\ref{dl}), (\ref{ds}) that
\begin{equation}
ds^{2}=d\tau _{obs}^{2}(1-w^{2}). \label{sw}
\end{equation}
</p></div>
</div>

<div id="OZ04"></div>
=== Problem 4===

Analyze the formulas derived in the previous three problems applied to the case of flat spacetime (Minkovskii space) and compare them to
the known formulas of special relativity.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In special relativity four-velocity is expressed through velocity (spatial components are denoted by Latin letters)
\begin{equation}
v^{i}=\frac{dx^{i}}{dt},\qquad i=1,2,3\label{vxt}
\end{equation}
and Lorentz factor
\begin{equation}
\gamma =(1-v^{2})^{-1/2}
\end{equation}
as follows:
\begin{align}
U^{\mu}&=\gamma (+1,v^i),\\
U_{\mu }&=\gamma (-1,v^i).
\end{align}

Let us compare the original frame (with coordinates $t,x^i$) and the one
comoving with the observer (with coordinates $t^{\prime }=\tau _{obs}$ and $x^{i\prime}$.) Then using (\ref{vxt}) we get
\begin{equation}
ds^{2}=-dt^{\prime 2}=-dt^{2}+\delta _{ij}dx^{i}dx^{j}=-dt^{2}(1-v^{2}).
\end{equation}

Eqs. (\ref{tau}), (\ref{vxt}) then give us
\begin{equation}
d\tau _{obs}\equiv dt^{\prime }
=\frac{dt-v_{i}dx^{i}}{\sqrt{1-v^{2}}}
=dt\sqrt{1-v^{2}},
\end{equation}
which coincides with the standard Lorentz transformation. Here, $dt$ and $dx^{i} $ are the time and coordinate differences between close events
measured in the laboratory frame. The quantities with primes refer to the comoving frame. Then, $\tau _{obs}=t^{\prime }$ has the meaning of
proper time, as in the case when $dx^{i\prime }=0$, and thus $dl^{\prime}=0$, from (\ref{tau}) we get $ds^{2}=-dt^{\prime 2}$.
</p></div>
</div>

<div id="OZ05"></div>
=== Problem 5===

Consider an observer being at rest with respect to a given coordinate frame:
$x^{i}=const$ ($i=1,2,3$). Find $h_{\mu \nu }$, $d\tau _{obs}$, the
condition of simultaneity and $dl^{2}$ for this case. Show that the
corresponding formulas are equivalent to eqs. (84.6), (84.7) of Landau and Lifshitz [1],
where they are derived in a different way.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

If the observer does not move, its four-velocity is
\begin{equation}
U^{\mu }=(U^{0},0,0,0).
\end{equation}
From the normalization condition one finds
\begin{equation}
g_{00}\left( U^{0}\right) ^{2}=-1,
\end{equation}
whence
\begin{equation}
U^{\mu }=\frac{1}{\sqrt{-g_{00}}}(1,0,0,0).\label{st}
\end{equation}

The covariant components $U_{\mu }=g_{\mu \nu }U^{\nu }$ then are equal to
\begin{equation}
U_{0}=-\sqrt{-g_{00}},\qquad U_{i}=\frac{g_{i 0}}{\sqrt{-g_{00}}}.
\label{uco}
\end{equation}
By substitution into (\ref{h}), one finds
\begin{align}
h_{00}&=0,\quad h_{0i}=0,\label{h0}\\
h_{ij}&=g_{ij}-\frac{g_{0i}g_{0j}}{g_{00}}, \label{hi}
\end{align}
which coincides (up to the choice of the overall signature) with the spatial metric $\gamma_{ij}$ as defined in eq. (84.7) of [1]. Then, $dl^{2}=h_{ij}dx^{i}dx^{j}$ turns into eq. (84.6).

The condition of simultaneity (\ref{sim}) with (\ref{uco}) taken into account reads
\begin{equation}
dt=-dx^{i}\frac{g_{0i}}{g_{00}}
\end{equation}
which coincides with eq. (84.14) of [1].
</p></div>
</div>
[1] Landau L.D., Lifshitz E.M. Vol. 2. The classical theory of fields, 4ed., Butterworth-Heinemann, 1994; ISBN 0-7506-2768-9.

<div id="OZ06"></div>
=== Problem 6===

Consider two events at the same point of space but at different values of
time. Find the relation between $dx^{\mu}$ and $d\tau _{obs}$ for such an
observer.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In this case, $dx^{i}=0$, and we obtain from (\ref{tau}) and (\ref{uco}) that
\begin{equation}
d\tau _{obs}=\sqrt{-g_{00}}\;dt
\end{equation}
which coincides with eq. 84.1 of Landau and Lifshitz$^*$.</p>

<p style="text-align: left;">$^*$ Landau L.D., Lifshitz E.M. Vol. 2. The classical theory of fields. 4ed., Butterworth-Heinemann, 1994; ISBN 0-7506-2768-9.
</p></div>
</div>

==Fiducial observers==

<div id="OZ07"></div>
=== Problem 7===

Consider an observer with
\begin{equation}
U_{\mu }=-N\delta _{\mu }^{0}=-N(1,0,0,0)\text{.} \label{uz}
\end{equation}
We call it a fiducial observer (FidO) in accordance with [1]. This
notion is applied in [1] mainly to static or axially symmetric rotating
black holes. In the latter case it is usually called the ZAMO (zero angular
momentum observer). We will use FidO in a more general context.

Show that a FidO's world-line is orthogonal to hypersurfaces of constant time $t=const$.

[1] Black Holes: The Membrane Paradigm. Edited by Kip S. Thorne, Richard H. Price, Douglas A. Macdonald. Yale University Press New Haven and London, 1986; ISBN 978-030-003-769-2 .

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

A vector $n_\mu$ is called normal to a hypersurface $\Sigma$ if it is orthogonal to any displacement $dx^\mu $ within this hypersurface: $n_\mu dx^\mu =0$. For hypersurfaces of constant time $dx^{\mu}=(0,dx^i)$, so the normal vector is $n_\mu =(n_0 ,0)\sim \partial_t $, with $n_i =0$. Thus, according to (\ref{uz}), $n_{\mu}$ and $U_{\mu}$ are parallel according to (\ref{uz}) and $U_\mu$ is also normal to $t=const$.
</p></div>
</div>

<div id="OZ08"></div>
=== Problem 8===
Find the explicit form of the metric coefficients in terms of the components
of the FidO's four-velocity. Analyze the specific case of axially symmetric metric in
coordinates $(t,\phi ,r,\theta )$ with $g_{0i}=g_{t\phi }\delta _{i}^{\phi }$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

By definition, $U_{i}=0$. Then, the normalization condition gives us
\begin{equation}
U^{\mu }=\frac{1}{N}(1,N^{i}) \label{zamo}
\end{equation}
with some $N^{i}$. It has the simple physical meaning of the velocity of the
observer with respect to the coordinate frame $\{t,x^{i}\}$ defined
as
\begin{equation}
V^{i}\equiv \frac{dx^{i}}{dt}\Big| _{obs}=\frac{U^{i}}{U^{0}}=N^{i}.
\label{vn}
\end{equation}

It follows from (\ref{tau}) that
\begin{equation}
d\tau _{obs}=Ndt .
\end{equation}
Using that
\begin{equation}
0=U_{i}=\frac{g_{i 0}}{N}+\frac{g_{ij}N^{j}}{N},
\end{equation}
we find
\begin{equation}
g_{0i}=-g_{ij}N^{j}.
\end{equation}

In a similar way,
\begin{equation}
-N=U_{0}=\frac{g_{00}}{N}+\frac{g_{0i}N^{i}}{N},
\end{equation}
whence
\begin{equation}
g_{00}=-N^{2}+g_{0i}N^{i}N.
\end{equation}

Collecting all this, we obtain that the metric can be written in the form
\begin{equation}
ds^{2}=-dt^{2}N^{2}+g_{ij}(dx^{i}-N^{i}dt)(dx^{j}-N^{j}dt) \label{N},
\end{equation}
and calculating $h_{\mu \nu }$ we find
\begin{equation}
h_{00}=g_{00}+N^{2},\qquad h_{0i}=g_{0i},\qquad h_{ij}=g_{ij}.
\end{equation}

In the case of a stationary axially symmetric metric
\begin{equation}
ds^{2}
=-dt^{2}N^{2}
+g_{\phi \phi }(d\phi -\omega dt)^{2}
+g_{rr}dr^{2}+g_{\theta \theta }d\theta ^{2}.
\end{equation}
we have
\begin{align}
g_{00}&=-N^{2}+g_{\phi \phi }\omega ^{2}, \label{00m}\\
g_{0\phi }&=-g_{\phi \phi }\omega, \label{03}
\end{align}
and therefore
\begin{align}
h_{00}=\omega ^{2}g_{\phi \phi },
&\qquad h_{0\phi }=g_{0\phi },
\qquad h_{\phi \phi }=g_{\phi \phi };\\
& N^{\phi }=\omega ,\qquad N^{r}=N^{\theta }=0.\label{vf}
\end{align}
</p></div>
</div>

<div id="OZ09"></div>
=== Problem 9===

Consider a stationary metric with the time-like Killing vector field $\xi^\mu =(1,0,0,0)$. Relate the energy $E$ of a particle with four-velocity $u^\mu$ as measured at
infinity by a stationary observer to that measured by a local observer with 4-velocity $U^\mu$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The energy as measured at infinity by a stationary observer is the integral of motion
\begin{equation}
E=-mu_{\mu }\xi ^{\mu }=-mu_{0}.
\end{equation}

On another hand, the energy measured by a local observer is
\begin{equation}
E_{rel}=-mu_{\mu }U^{\mu }\equiv m\gamma =-m(u_{0}U^{0}+u_{i}U^{i}),
\label{rel}
\end{equation}
whence
\begin{equation}
E-m\,u_{i}V^{i}=\frac{E_{rel}}{U^0},\label{e}
\end{equation}
where $V^{i}=U^i / U^0$ is the local observer's velocity in the given frame (\ref{vn}).
</p></div>
</div>

<div id="OZ10"></div>
=== Problem 10===

Express $E_{rel}$ and $E$ in terms of the relative velocity $w$ between a
particle and the observer (i.e. velocity of the particle in the frame of the observer and vice versa).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Using (\ref{tau}), (\ref{rel}) we find that the Lorentz factor is
\begin{equation}
\gamma =-u_{\mu }U^{\mu }
=-\frac{U^{\mu }dx_{\mu }}{ds}=-\frac{d\tau _{obs}}{ds}.
\end{equation}

It follows from (\ref{sw}) that
\begin{equation}
\gamma =\frac{1}{\sqrt{1-w^{2}}},\qquad
E_{rel}=\frac{m}{\sqrt{1-w^{2}}}.
\end{equation}

Then, we find from (\ref{e}) that
\begin{equation}
E-p_{i}V^{i}=\frac{m}{\sqrt{1-w^{2}}}\frac{1}{U_{0}}, \label{ep}
\end{equation}
where $p^{\mu }=mu^{\mu }$ is the particle's four-momentum.
</p></div>
</div>

<div id="OZ11"></div>
=== Problem 11===

Show that in the flat spacetime eq. (\ref{ep}) is reduced to the usual formula
of the Lorentz transformation.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In this case, $U^{0}=(1-V^{2})^{-1/2}=\gamma $, where $V$ is the absolute value of the velocity of the observer (frame 0) with respect to the stationary laboratory frame. Then the energy of a particle in frame 0 is
\begin{equation}
E_{(0)}=\gamma ( E-\mathbf{p}\cdot\mathbf{V}),
\end{equation}
This is the standard formula for energy transformation.
</p></div>
</div>

<div id="OZ12"></div>
=== Problem 12===

Find the expression for $E$ for the case of a static observer ($U^{i}=0$).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Using (\ref{st}) and (\ref{e}) we find that the kinetic term in (\ref{e})
vanishes and
\begin{equation}
E=E_{rel.}\sqrt{-g_{00}}=\frac{m}{\sqrt{1-w^{2}}},
\end{equation}
which coincides with eq. (88.9) of Landau and Lifshitz$^*$.</p>

<p style="text-align: left;">$^*$Landau L.D., Lifshitz E.M. Vol. 2. The classical theory of fields, 4ed., Butterworth-Heinemann, 1994; ISBN 0-7506-2768-9.
</p></div>
</div>

<div id="OZ13"></div>
=== Problem 13===

Find the expression for $E$ for the case of the ZAMO observer and, in particular, in case of axially symmetric metric.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\begin{equation}
E-mu_{i}V^{i}=E_{rel} N=\frac{mN}{\sqrt{1-w^{2}}}.
\end{equation}

In the axially symmetric case, the angular momentum of a particle $mu_{\phi}=L$ is conserved, and according to (\ref{vf}), $V^{\phi }=\omega$. Then,
\begin{equation}
E-\omega L=E_{rel}N=\frac{mN}{\sqrt{1-w^{2}}}.
\end{equation}
</p></div>
</div>

==Collision of particles: general relationships==

<div id="OZ14"></div>
=== Problem 14===
Let two particles collide. Define the energy in the center of mass (CM)
frame $E_{c.m.}$ at the point of collision and relate it to $E_{rel}$ and
the Lorentz factor of relative motion of the two particles.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In general, the momenta and velocities of two particles are defined at
different points and this prevents one from constructing well-defined
quantities. However, as collision occurs at one point, both particles have
characteristics defined just in this point. This enables one to construct
the total momentum $P^{\mu }=p_{1}^{\mu }+p_{2}^{\mu }$ at the point of
collision (the subscript $a=1,2$ enumerates the two particles).

A particle's mass is
\begin{equation}
m^{2}=-p^{\mu }p_{\mu },
\end{equation}
and likewise the full energy (mass) of two particles at the point of collision, which is also their energy in the center of mass frame, is found from
\begin{equation}
E_{c.m.}^{2}=-P^{\mu }P_{\mu }.\label{cm}
\end{equation}
It is is the direct generalization of the corresponding formula of special relativity. Eq (\ref{cm}) implies
\begin{equation}
E_{c.m.}^{2}=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma ,
\end{equation}
where
\begin{equation}
\gamma =-u_{1}^{\mu }u_{2\mu }
\end{equation}
is the Lorentz factor of the two particles' relative motion. It can be expressed in terms of energy $E_{rel}(2,1)$ of particle 1 with respect to particle 2 and vice versa
\begin{equation}
\gamma=\frac{E_{rel.}(1,2)}{m_{1}}=\frac{E_{rel.}(2,1)}{m_{2}}.
\end{equation}
</p></div>
</div>

<div id="OZ1"></div>
=== Problem 15===

Let us consider a collision of particles 1 and 2 viewed from the frame
attached to some other particle 0. How are different Lorentz factors related
to each other? Analyze the case when the laboratory frame coincides with
that of particle 0.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Let us introduce notation
\begin{equation*}
\gamma (0,1)=\gamma _{1}, \quad
\gamma (0,2)=\gamma _{2}, \quad
\gamma (1,2)=\gamma , \quad
u_{0}^{\mu }=U^{\mu}.
\end{equation*}

Each particle's four velocity can be decomposed into components parallel and orthogonal to $U^\mu$ as follows
\begin{equation}
u_{1}^{\mu }=a_{1}U^{\mu }+b_1 n_{1}^{\mu}, \label{ab}
\end{equation}
where $n^{\mu }$ is a unit space-like vector orthogonal to $U^{\mu }$. It is easy to
find from (\ref{ab}) that
\begin{equation}
a_{1}=-u_{1}^{\mu }U_{\mu }=\gamma _{1},
\end{equation}
and from the normalization conditions for $u_{1}^{\mu }$ and $U^{\mu }$ then we see that $b_1$ as introduced in (\ref{ab}) is the velocity of particle 1 in the frame of particle 0:
\begin{equation}
b_{1}^{2}=\frac{1}{\gamma _{1}^{2}-1 },
\end{equation}
so in terms of velocity of particle 1 in the frame of particle 0
\begin{equation}
w_1 =(1-\gamma_1^{-2})^{1/2}=\frac{b_1}{\gamma},
\end{equation}
the decomposition takes form
\begin{equation}
u_{1}^{\mu }=\gamma_{1}(U^{\mu }+w_1 n_{1}^{\mu}).
\end{equation}

Writing down for particle 2 the equation analogous to (\ref{ab}) and
calculating the scalar product $u_{1}^{\mu }u_{2\mu }$, one obtains
\begin{align}
\gamma &=\gamma _{1}\gamma _{2}
-b_1 b_2 (n_{1}^{\mu }n_{2\mu})\\
&=\gamma _{1}\gamma _{2}\big(1-\varepsilon\; w_{1}w_{2}\big),
\label{ga}
\end{align}
where
\begin{equation}
\varepsilon =(n_{1}^{\mu }n_{2\mu }),\qquad |\varepsilon|\leq 1 ,
\end{equation}
or in terms of velocities
\begin{equation}
\frac{1}{\sqrt{1-w^{2}}}
=\frac{1}{\sqrt{1-w_{1}^{2}}}
\frac{1}{\sqrt{1-w_{2}^{2}}}
(1-\varepsilon w_{1}w_{2}). \label{gw}
\end{equation}

In case particle 0 is at rest in the laboratory frame $U^{\mu }=(1,0)$, we have $n_{a}^{\mu}=(0,\vec{n}_{a})$, $a=1,2$, and
\begin{equation}
u_{a}^{\mu }=\gamma _{a}(1,w_{a}\vec{n}_{a}),\qquad \varepsilon =(\vec{n}_{1}
\vec{n}_{2}).
\end{equation}
Then, eqs. (\ref{ga}), (\ref{gw}) turn into those listed in problem 1.3 of Lightman et al $^*$.</p>

<p style="text-align: left;">$^*$ A. P. Lightman, W. H. Press, R. H. Price, and S. A. Teukolsky, Problem book in Relativity and Gravitation (Princeton University Press, Princeton, New Jersey, 1975); ISBN 069-108-160-3.
</p></div>
</div>

<div id="OZ16"></div>
=== Problem 16===

When can $\gamma $ as a function of $\gamma _{1}$ and $\gamma _{2}$ grow
unbounded? How can the answer be interpreted in terms of relative velocities?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
1) If $\gamma _{1}$ and $\gamma _{2}$ are finite, it follows from (\ref{ga})
that $\gamma $ is also finite. It means that if $w_{1}$ and $w_{2}$ are
finite, the relative velocity $w$ of particles 1 and 2 is also finite (in
the sense that it is separated from $c$).</p>

<p style="text-align: left;">
2) Let $\gamma _{1}\rightarrow \infty $ but $\gamma _{2}$ remain finite.
Then,
\begin{equation}
\gamma \approx \gamma _{1}\big( \gamma _{2}-\varepsilon \sqrt{\gamma _{2}^{2}-1}\;\big)
\end{equation}
grows unbound irrespective of $\varepsilon $. It means that the relative
velocity of particles, one of which moves with some finite speed and the
other one almost with the speed of light is always close to the speed of
light.</p>

<p style="text-align: left;">
3) Let $\gamma _{1}\rightarrow \infty $, $\gamma _{2}\rightarrow \infty $.</p>

<p style="text-align: left;">
3a) If $\varepsilon \neq +1$,
\begin{equation}
\gamma \approx \gamma _{1}\gamma _{2}(1-\varepsilon )\rightarrow \infty ,
\qquad w\rightarrow 1.
\end{equation}</p>

<p style="text-align: left;">
3b) Let $\varepsilon =+1$. Then
\begin{equation}
\gamma \approx \frac{1}{2} \Big(\frac{\gamma _{1}}{\gamma _{2}} +\frac{\gamma _{2}}{\gamma _{1}}\Big).
\end{equation}

We see that $\gamma $ remains finite if $\gamma _{1}/\gamma _{2}$ does so.
Otherwise, $\gamma \rightarrow \infty $. The resulting relative velocity
depends on the relative rate with which $w_{1}$ and $w_{2}$ approach the
speed of light.

Thus in case 3, motion in different directions always gives $\gamma
\rightarrow \infty $, whereas motion in the same direction may or may not
lead to divergent $\gamma $.
</p></div>
</div>

<div id="OZ1"></div>
=== Problem 17===

A tetrad basis, or the orthonormal tetrad, is the set of four unit vectors $h_{(a)}^\mu$ (subscripts in parenthesis $a=0,1,2,3$ enumerate these vectors), of which one, $h_{(0)}^\mu$, is timelike, and three vectors $h_{(i)}^\mu$ ($i=1,2,3$) are spacelike, so that
\begin{equation}
g_{\mu\nu}h_{(a)}^\mu h_{(b)}^\nu =\eta_{ab},\qquad a,b=0,1,2,3.
\end{equation}
A vector's tetrad components are
\begin{equation}
u_{(a)}=u_\mu h_{(a)}^\mu,\qquad u^{(b)}=\eta^{ab}u_{(b)}.
\end{equation}

Define the local three-velocities with the help of the tetrad basis attached to the observer, which
would generalize the corresponding formulas of special relativity.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In special relativity,
\begin{equation}
u^{\mu }=\frac{dx^{\mu }}{ds}
=\gamma (1,v^{i}),\qquad
\gamma =\frac{1}{\sqrt{1-v^{2}}}.
\end{equation}
Thus $v^{i}=u^{i}\sqrt{1-v^{2}}$, and using (\ref{sw}), we get
\begin{equation}
v^{i}=\frac{dx^{i}}{d\tau _{obs}}.\label{vi}
\end{equation}

Let us attach a tetrad to some observer with four-velocity $U^\mu$. This means simply that we choose the first, time-like, tetrad vector to be along its world-line
\begin{equation}
h_{(0)}^\mu =U^\mu .
\end{equation}
Then, the natural definition for velocity in the frame of this observer, taking into account (\ref{tau}), will be
\begin{equation}
v^{(i)}=\frac{h_{\mu }^{(i)}dx^{\mu}}{-h^{\mu}_{(0)}dx_{\mu }}.
\end{equation}
It can be rewritten as
\begin{equation}
v^{(i)}=\frac{u^{\mu }h_{\mu }^{(i)}}{u^{\mu }h^{(0)}_\mu}.
\end{equation}

The tetrad components of the projection operator (\ref{h}) are
\begin{equation}
h_{(ab)}=\eta_{ab}+\eta_{0a}\eta_{0b},
\end{equation}
from which it is easy to check explicitly that the two definitions, through $h_{\mu\nu}$ and through $h^\mu_{(a)}$, are equivalent.
</p></div>
</div>

<div id="OZ18"></div>
=== Problem 18===

Derive the analogues of formulas (\ref{e}), (\ref{ep}) for massless particles (photons). Analyze the cases of static and ZAMO observers.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In this case, $m=0$, the velocity of a photon is always 1, and the notion of
a frame comoving with the photon does not have sense. Nonetheless, some
formulas retain their validity with $E$ replaced by the frequency $\nu$ in
accordance with the fact that for photons $E=\hbar\nu$ (we use $\nu=2\pi/\text{(period)}$ for the angular frequency here, so as not to confuse it with the metric function $\omega$).

Let us consider a photon with wave four-vector $k_{\mu }$ in a gravitational
field. In the case of static or stationary metric, $\nu _{0}=-k_{\mu }\xi^{\mu }$ is conserved along the trajectory, where $\xi ^{\mu }$ is the
Killing vector. If it corresponds to time translations, $\nu$ has the
meaning of frequency. It is what is measured by a remote observer (if the
flat infinity exists). In parallel to derivation of (\ref{e}), one can obtain
\begin{equation}
\nu _{0}-k_{i}V^{i}=\frac{\nu }{U^{0}}, \label{nu}
\end{equation}
where $\nu =-k_{\mu }U^{\mu }$, and $U^{\mu }$ is the velocity of the
observer.

For the static observer $V^{i}=0$, $U^{0}=\frac{1}{\sqrt{-g_{00}}}$, and
\begin{equation}
\nu_{0}=\nu \sqrt{-g_{00}}. \label{0g}
\end{equation}

For the ZAMO observer, $V^{\phi }=\omega $, $k_{\phi }=L$ is the angular
momentum which is conserved in the axially symmetric case (in units of $\hbar$), $U^{0}=\frac{1}{N}$ (see eq. (\ref{zamo})). Then
\begin{equation}
\nu _{0}-\omega L=\nu N.
\end{equation}
</p></div>
</div>

<div id="OZ19"></div>
=== Problem 19===

The ergosphere is a surface defined by equation $g_{00}=0$. Show that it is
the surface of infinite redshift for an (almost) static observer.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

From (\ref{0g}) it follows that in the limit $g_{00}\rightarrow 0$ the
frequency $\nu _{0}\rightarrow 0$ for any finite $\nu $. However, for a
nonstatic observer this is not true.
</p></div>
</div>

<div id="OZ20"></div>
=== Problem 20===

Consider an observer orbiting with a constant angular velocity $\Omega $ in
the equatorial plane of the axially symmetric back hole. Analyze what
happens to redshift when the angular velocity approaches the minimum or
maximum values $\Omega _{\pm }$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Eq. (\ref{nu}) reads in this case
\begin{equation}
\nu _{0}-\Omega L=\frac{\nu }{U^{0}}\text{,}
\end{equation}
where $U^{0}$ obeys the normalization condition
\begin{equation}
Y\equiv g_{00}+2g_{0\phi }\Omega +g_{\phi \phi }\Omega ^{2}=-\frac{1}{\left(
U^{0}\right) ^{2}}\text{.}
\end{equation}

It can be rewritten in the form
\begin{align}
Y=&g_{\phi \phi }(\Omega ^{2}-2\omega \Omega +\frac{g_{00}}{g_{\phi \phi }})
=g_{\phi \phi }(\Omega -\Omega _{+})(\Omega -\Omega _{-});\\
&\Omega _{\pm }=\omega \pm \frac{N}{\sqrt{g_{\phi \phi }}},
\end{align}
Then,
\begin{equation}
\nu _{0}-\Omega L=\nu \sqrt{g_{\phi \phi }(\Omega _{+}-\Omega )(\Omega
-\Omega _{-})}\text{.}
\end{equation}

When $g_{00}\rightarrow 0$, $\Omega _{-}\approx \frac{g_{00}}{2\omega }
\rightarrow 0$.

Thus, if, say, $\Omega \rightarrow \Omega _{-}$, $\nu _{0}\rightarrow \Omega
_{-}L$. If $L=0$, then $\nu _{0}\rightarrow 0$.
</p></div>
</div>

<div id="OZ21"></div>
=== Problem 21===

Let two massive particle 1 and 2 collide. Express the energy of each particle
in the centre of mass (CM) frame in terms of their relative Lorentz factor $\gamma (1,2)$. Analyze the limiting cases of ultra-relativistic $\gamma
(1,2)\rightarrow \infty $ and non-relativistic $\gamma (1,2)\approx 1$ collisions.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The energy of particle 1 in the CM frame is
\begin{equation}
\left( E_{1}\right) _{c.m.}=m_{1}\gamma (1,CM) \label{e1}
\end{equation}
where
\begin{equation}
\gamma (1,CM)=-u_{1\mu }U_{c.m.}^{\mu }
\end{equation}
is its Lorentz factor in the same frame, which has four-velocity
\begin{equation}
U_{c.m.}^{\mu }=\frac{P^{\mu }}{\mu}.
\end{equation}
Here $P^\mu$ is the total momentum and $\mu \equiv E_{c.m.}$ is the energy in the center of
mass frame introduced in (\ref{cm})
\begin{align}
P^{\mu }&=m_{1}u_{1}^{\mu }+m_{2}u_{2}^{\mu };\\
\mu ^{2}&=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma (1,2).\label{mu}
\end{align}

Then
\begin{align}
\gamma (1,CM)&=\frac{m_{1}+m_{2}\gamma (1,2)}{\mu };\label{g1}\\
( E_{1}) _{c.m.}
&=m_{1}\gamma (1,CM).\label{e1cm}
\end{align}

Likewise, for the second particle
\begin{align}
\gamma (2,CM)&=\frac{m_{2}+m_{1}\gamma (1,2)}{\mu };\\
( E_{2}) _{c.m.}&=m_{2}\gamma (2,CM).\label{e2cm}
\end{align}

It follows from (\ref{e1cm}), (\ref{e2cm}) that
\begin{equation}
( E_{1}) _{c.m.}+( E_{2}) _{c.m.}=\mu ,
\end{equation}
as it should be.

In the ultra-relativistic limit $\gamma (1,2)\rightarrow \infty $
\begin{equation}
\gamma (1,CM)\approx
\frac{\mu }{2m_{1}}\approx \sqrt{\frac{m_{2}}{2m_{1}}\gamma (1,2)}.
\end{equation}

In the opposite limit of small relative velocity $w(1,2)\ll 1$ we obtain $\gamma (1,2)\approx 1+\frac{w^{2}}{2}$ and
\begin{align}
&\mu \approx (m_{1}+m_{2})+\frac{m_{1}m_{2}}{2(m_{1}+m_{2})}w^{2};\\
&w(1,CM)\approx w\frac{m^{2}}{(m_{1}+m_{2})},
\end{align}
which agrees with formulas of nonrelativistic mechanics -- see Ch. 3, Sec. 13 of Landau and Lifshitz$^*$.</p>

<p style="text-align: left;">$^*$ Landau L.D., Lifshitz E.M. Vol. 2. The classical theory of fields, 4ed., Butterworth-Heinemann, 1994; ISBN 0-7506-2768-9.
</p></div>
</div>

<div id="OZ22"></div>
=== Problem 22===

For a stationary observer in a stationary space-time the quantity $\alpha =(U^0)^{-1}$ is the redshifting factor: if this observer emits a ptoton with frequency $\omega_{em}$, it is detected at infinity by another stationary observer with frequency $\omega_{det}=\alpha \omega_{em}$. For a generic observer this interpretation is invalid, however, $\alpha =ds/dt$ still determines the time dilation for this observer, and thus can still be called the same way. Express the redshifting factor of the center of mass frame $\alpha _{c.m.}$ through the redshifting factors of the colliding particles $\alpha_{1}$ and $\alpha_2$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Starting from the definition,
\begin{equation}
\frac{1}{\alpha _{c.m.}}
=\left( U^{0}\right) _{c.m.}
=\frac{m_{1}U_{1}^{0}+m_{2}U_{2}^{0}}{\mu }
=\frac{\frac{m_{1}}{\alpha _{1}}+\frac{m_{2}}{\alpha _{2}}}{\mu },
\end{equation}
whence
\begin{equation}
\alpha _{c.m.}=\frac{\mu \alpha _{1}\alpha _{2}}{m_{1}\alpha
_{2}+m_{2}\alpha _{1}}.
\end{equation}

If case $\alpha _{2}\ll \frac{m_{2}}{m_{1}}\alpha _{1}$, we get
\begin{equation}
\alpha _{c.m.}\approx \frac{\mu \alpha _{2}}{m_{2}}.
\end{equation}
</p></div>
</div>

<div id="OZ23"></div>
=== Problem 23===

Relate the energy of a particle at infinity $E_{1}$, its energy at the point of collision in the C.M. frame $(E_{1})_{c.m.}$ and $\mu$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

From (\ref{e})
we know that
\begin{equation}
E_{1}-mu_{1i}v_{2}^{i}=\alpha _{2}E_{rel}, \label{e2}
\end{equation}
where $v^{i}$ is the velocity of particle 2 in the stationary frame, $\alpha _{2}=\frac{1}{U_{2}^{0}}$.

A particle's energy in the center of mass frame can be found from (\ref{e1cm})
\begin{equation}
(E_{1})_{c.m.}\mu =m_{1}^{2}+m_{2}E_{rel}(1,2),
\end{equation}
where $E_{rel}(1,2)=m_{1}\gamma (1,2)$, so in terms of $(E_1)_{c.m.}$ we get
\begin{align}
E_{1}-m_{1}u_{1i}v_{2}^{i}
&=\frac{\alpha_{2}(\mu E_{1})_{c.m.} -m_{1}^{2}\alpha _{2}}{m_{2}};\\
E_{2}-m_{2}u_{2i}v_{1}^{i}
&=\frac{\alpha _{1}(\mu E_{2})_{c.m.} -m_{2}^{2}\alpha _{1}}{m_{1}}.
\end{align}

On another hand, applying (\ref{e}) to particle 1 and the center of mass frame, we obtain
\begin{align}
E_{1}-m_{1}u_{1i}V_{c.m.}^{i}
&=\alpha _{c.m.}(E_{1}) _{c.m.}; \label{1cm}\\
E_{2}-m_{2}u_{2i}V_{c.m.}^{i}
&=\alpha _{c.m.}(E_{2}) _{c.m.}.\label{2cm}
\end{align}

The sum of (\ref{1cm}) and (\ref{2cm}) is the total conserved energy $E=E_1 +E_2$:
\begin{equation}
E-\alpha _{c.m.}P_{i}U_{c.m.}^{i}=\alpha _{c.m.}\mu .
\end{equation}
Taking into account that $\frac{\alpha _{c.m.}}{\mu }=\frac{1}{P^{0}}$, this is equivelent to
\begin{equation}
EP^{0}-P_{i}P^{i}=-P^{\mu }P_{\mu }=\mu^{2},
\end{equation}
as it should be.
</p></div>
</div>

<div id="OZ24"></div>
=== Problem 24===

Solve the same problem when both particles are massless (photons). Write
down formulas for the ZAMO observer and for the C.M. frame.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Let $k^\mu_1$ and $k^\mu_2$ be the wave-vectors of two photons. Then vector
\begin{equation}
K^{\mu }=\left( k^{\mu }\right) _{1}+\left( k^{\mu }\right) _{2} \label{K}
\end{equation}
is ''time-like'' (unless the two photons are collinear, but then they would not collide). Dividing by the energy $\mu$ in the center of mass frame
\begin{equation}
\mu ^{2}=-K^{\mu }K_{\mu }=-2k_{1\mu }k^{2\mu },
\end{equation}
we obtain the frame's (time-like) four-velocity
\begin{equation}
U_{c.m.}^{\mu }=\frac{K^{\mu }}{\mu }.
\end{equation}
There is no Lorentz factor or relative velocity for two photons.

The contraction
\begin{equation}
\tilde{\gamma}(1,obs)=-k_{\mu } U^{\mu }
\end{equation}
has the meaning of the frequency of a photon with four-vector $k^\mu$ as measured by the observer with four-velocity $U^\mu$.

Then frequency measured in the CM frame is
\begin{equation}
\left( \nu _{1}\right) _{c.m.}=-k_{1\,\mu} U_{c.m.}^{\mu },
\end{equation}
and taking into account (\ref{mu}) and (\ref{K}), we obtain
\begin{equation}
(\nu _{1}) _{c.m.}
=-k_{1\,\mu}U_{c.m.}^{\mu }
=-\frac{k_{1\,\mu }k_{2}^{\mu }}{\mu }=\frac{\mu }{2},
\end{equation}
as for photons $k_{i\,\mu}k_i^{\mu}=0$. This is quite natural, since for photons the energy is equal to the absolute value of the momentum,
so in the CM frame the energies of both photons should be equal.

Likewise,
\begin{equation}
( \nu _{2}) _{c.m.}
=-k_{2\,\mu}U_{c.m.}^{\mu }
=\frac{\mu }{2}=(\nu_{1}) _{c.m.},
\end{equation}
and
\begin{equation}
\mu =( \nu _{1}) _{c.m.}+( \nu _{2}) _{c.m.}
=2(\nu _{1}) _{c.m.}=2(\nu _{2}) _{c.m.}
\end{equation}

If $\xi$ is the Killing vector field responsible for time translation, the conserved frequency of a photon, equal to the frequency measured at infinity by a stationary observer, is
\begin{equation}
\nu _{\infty }=-k_{\mu }\xi ^{\mu }=-k_{0}.
\end{equation}
In some arbitrary frame it is
\begin{equation}
\nu _{loc}=-k_{\mu }U^{\mu }=-k_{0}U^{0}-k_{i}U^{i}.
\end{equation}

Then
\begin{equation}
\nu _{\infty }-k_{\alpha}V^{\alpha}=\alpha \nu _{loc},\qquad
V^{i}=\frac{U^{i}}{U^{0}},\qquad
\alpha =\frac{1}{U^{0}},\label{obs}
\end{equation}
the same as for the case of massive particle. The difference is only that for photons we
cannot express $\nu _{loc}$ in terms of mass and velocity as $\frac{m}{\sqrt{1-w^{2}}}$.</p>

<p style="text-align: left;">
ZAMO frame
\begin{equation}
\nu _{\infty }-\omega L=N\nu _{loc}.
\end{equation}

CM frame
\begin{align}
&\nu _{\infty }-k_{i}V^{i}=\alpha _{c.m.}\frac{\mu }{2},\\
&\alpha _{c.m.}=\frac{1}{U^{0}}=\frac{\mu }{k_{1}^{0}+k_{2}^{0}}.
\end{align}
</p></div>
</div>

Universe in Problems:About

2013-05-01T18:54:00Z

Igor:

Authors of the Universe in Problems project are Yu. L. Bolotin, V. A. Cherkaskiy, G. I. Ivashkevych, O. A. Lemets, I. V. Tanatarov, D. A. Yerokhin, O.B. Zaslavskii.

The arxiv version without solutions can be found at [http://arxiv.org/abs/0904.0382 arXiv:0904.0382]

All questions and comments please send to this email: universeinproblems@gmail.com.

Universe in Problems:About

2013-05-01T18:53:29Z

Igor:

Authors of the Universe in Problems project are Yu. L. Bolotin, V. A. Cherkaskiy, G. I. Ivashkevych, O. A. Lemets, I. V. Tanatarov, D. A. Yerokhin, O.B. Zaslavskii.

The arxiv version without solutions can be found at [http://arxiv.org/abs/0904.0382 arxiv.org]

All questions and comments please send to this email: universeinproblems@gmail.com.

Particles' motion in general black hole spacetimes

2013-05-01T18:50:36Z

Igor:

[[Category:Black Holes|4]]
__TOC__

In this section we use the $(-+++)$ signature, Greek letters for spacetime indices and Latin letters for spatial indices.

Particles' motion in general black hole spacetimes

2013-05-01T08:40:27Z

Igor:

[[Category:Black Holes|4]]
__TOC__

In this section we use the $(-+++)$ signature, Greek letters for spacetime indices and Latin letters for spatial indices.

==Frames, time intervals and distances==

In the next several problems we again consider the procedure of measuring time and space intervals by different observers, but in a different, more formal and powerful approach.

<div id="OZ01"></div>
=== Problem 1===

Let a particle move with the four-velocity $U^{\mu }$. It can be viewed as
some observer carrying a frame attached to him. Locally, it defines the
hypersurface orthogonal to it. Show that
\begin{equation}
h_{\mu \nu }=g_{\mu \nu }+U_{\mu }U_{\nu } \label{h}
\end{equation}
is (i) the projection operator onto this hypersurface, and at the same time
(ii) the induced metric of the hypersurface. This means that (i) for any
vector projected at this hypersurface by means of $h^\mu_\nu$, only the components orthogonal to $U^{\mu}$ survive, (ii) the repeated application of the projection operation leaves the vector within the hypersurface unchanged. In other words, $h_{\mu \nu}$ satisfies
\begin{align}
&h^{\mu}_{\nu}U^{\nu }=0 ; \label{1} \\
&h^{\mu}_{\nu }h^{\nu}_{\lambda}=h^{\mu }_{\lambda}. \label{2}
\end{align}

<div id="OZ021"></div>
=== Problem 2===

Let us consider a particle moving with the four-velocity $U^{\mu }$. The
interval $ds^{2}$ between two close events is defined in terms of
differentials of coordinates,
\begin{equation}
ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }.
\end{equation}

For given $dx^{\mu }$, what is the value of the proper time $d\tau _{obs}$ between the corresponding events measured by this observer? How can one define
locally the notions of simultaneity and proper distance $dl$ for the observer in terms of its four-velocity and the corresponding projection operator $h^\mu_\nu$ ? How is the interval $ds^{2}$ related to $d\tau _{obs}$ and $dl$?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In order to obtain $d\tau _{obs}$, one should project $dx^{\mu}$ onto the
four-velocity:
\begin{equation}
d\tau _{obs}=-dx^{\mu }U_{\mu }.\label{tau}
\end{equation}
Then, locally, two events are simultaneous when
\begin{equation}
d\tau _{obs}=0. \label{sim}
\end{equation}

The proper distance
\begin{equation}
dl^{2}=h_{\mu \nu }dx^{\mu }dx^{\nu } \label{dl}.
\end{equation}

Then (\ref{h}) implies that
\begin{equation}
ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }=dl^{2}-d\tau _{obs}^{2} \label{ds}.
\end{equation}
</p></div>
</div>

<div id="OZ03"></div>
=== Problem 3===

Let our observer measure the velocity of some other particle passing in its
immediate vicinity. Relate the interval to $d\tau _{obs}$ and the particle's
velocity $w$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The velocity can be defined as
\begin{equation}
w^{2}=\left( \frac{dl}{d\tau _{obs}}\right) ^{2}.
\end{equation}
Then, it follows from (\ref{dl}), (\ref{ds}) that
\begin{equation}
ds^{2}=d\tau _{obs}^{2}(1-w^{2}). \label{sw}
\end{equation}
</p></div>
</div>

<div id="OZ04"></div>
=== Problem 4===

Analyze the formulas derived in the previous three problems applied to the case of flat spacetime (Minkovskii space) and compare them to
the known formulas of special relativity.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In special relativity four-velocity is expressed through velocity (spatial components are denoted by Latin letters)
\begin{equation}
v^{i}=\frac{dx^{i}}{dt},\qquad i=1,2,3\label{vxt}
\end{equation}
and Lorentz factor
\begin{equation}
\gamma =(1-v^{2})^{-1/2}
\end{equation}
as follows:
\begin{align}
U^{\mu}&=\gamma (+1,v^i),\\
U_{\mu }&=\gamma (-1,v^i).
\end{align}

Let us compare the original frame (with coordinates $t,x^i$) and the one
comoving with the observer (with coordinates $t^{\prime }=\tau _{obs}$ and $x^{i\prime}$.) Then using (\ref{vxt}) we get
\begin{equation}
ds^{2}=-dt^{\prime 2}=-dt^{2}+\delta _{ij}dx^{i}dx^{j}=-dt^{2}(1-v^{2}).
\end{equation}

Eqs. (\ref{tau}), (\ref{vxt}) then give us
\begin{equation}
d\tau _{obs}\equiv dt^{\prime }
=\frac{dt-v_{i}dx^{i}}{\sqrt{1-v^{2}}}
=dt\sqrt{1-v^{2}},
\end{equation}
which coincides with the standard Lorentz transformation. Here, $dt$ and $dx^{i} $ are the time and coordinate differences between close events
measured in the laboratory frame. The quantities with primes refer to the comoving frame. Then, $\tau _{obs}=t^{\prime }$ has the meaning of
proper time, as in the case when $dx^{i\prime }=0$, and thus $dl^{\prime}=0$, from (\ref{tau}) we get $ds^{2}=-dt^{\prime 2}$.
</p></div>
</div>

<div id="OZ05"></div>
=== Problem 5===

Consider an observer being at rest with respect to a given coordinate frame:
$x^{i}=const$ ($i=1,2,3$). Find $h_{\mu \nu }$, $d\tau _{obs}$, the
condition of simultaneity and $dl^{2}$ for this case. Show that the
corresponding formulas are equivalent to eqs. (84.6), (84.7) of \cite{lan},
where they are derived in a different way.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

If the observer does not move, its four-velocity is
\begin{equation}
U^{\mu }=(U^{0},0,0,0).
\end{equation}
From the normalization condition one finds
\begin{equation}
g_{00}\left( U^{0}\right) ^{2}=-1,
\end{equation}
whence
\begin{equation}
U^{\mu }=\frac{1}{\sqrt{-g_{00}}}(1,0,0,0).\label{st}
\end{equation}

The covariant components $U_{\mu }=g_{\mu \nu }U^{\nu }$ then are equal to
\begin{equation}
U_{0}=-\sqrt{-g_{00}},\qquad U_{i}=\frac{g_{i 0}}{\sqrt{-g_{00}}}.
\label{uco}
\end{equation}
By substitution into (\ref{h}), one finds
\begin{align}
h_{00}&=0,\quad h_{0i}=0,\label{h0}\\
h_{ij}&=g_{ij}-\frac{g_{0i}g_{0j}}{g_{00}}, \label{hi}
\end{align}
which coincides (up to the choice of the overall signature) with the spatial metric $\gamma_{ij}$ as defined in eq. (84.7) of \cite{lan}. Then, $dl^{2}=h_{ij}dx^{i}dx^{j}$ turns into eq. (84.6).

The condition of simultaneity (\ref{sim}) with (\ref{uco}) taken into account reads
\begin{equation}
dt=-dx^{i}\frac{g_{0i}}{g_{00}}
\end{equation}
which coincides with eq. (84.14) of \cite{lan}.
</p></div>
</div>

<div id="OZ06"></div>
=== Problem 6===

Consider two events at the same point of space but at different values of
time. Find the relation between $dx^{\mu}$ and $d\tau _{obs}$ for such an
observer.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In this case, $dx^{i}=0$, and we obtain from (\ref{tau}) and (\ref{uco}) that
\begin{equation}
d\tau _{obs}=\sqrt{-g_{00}}\;dt
\end{equation}
which coincides with eq. 84.1 of \cite{lan}.
</p></div>
</div>

==Fiducial observers==

<div id="OZ07"></div>
=== Problem 7===

Consider an observer with
\begin{equation}
U_{\mu }=-N\delta _{\mu }^{0}=-N(1,0,0,0)\text{.} \label{uz}
\end{equation}
We call it a fiducial observer (FidO) in accordance with \cite{mb}. This
notion is applied in \cite{mb} mainly to static or axially symmetric rotating
black holes. In the latter case it is usually called the ZAMO (zero angular
momentum observer). We will use FidO in a more general context.

Show that a FidO's world-line is orthogonal to hypersurfaces of constant time $t=const$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

A vector $n_\mu$ is called normal to a hypersurface $\Sigma$ if it is orthogonal to any displacement $dx^\mu $ within this hypersurface: $n_\mu dx^\mu =0$. For hypersurfaces of constant time $dx^{\mu}=(0,dx^i)$, so the normal vector is $n_\mu =(n_0 ,0)\sim \partial_t $, with $n_i =0$. Thus, according to (\ref{uz}), $n_{\mu}$ and $U_{\mu}$ are parallel according to (\ref{uz}) and $U_\mu$ is also normal to $t=const$.
</p></div>
</div>

<div id="OZ08"></div>
=== Problem 8===
Find the explicit form of the metric coefficients in terms of the components
of the FidO's four-velocity. Analyze the specific case of axially symmetric metric in
coordinates $(t,\phi ,r,\theta )$ with $g_{0i}=g_{t\phi }\delta _{i}^{\phi }$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

By definition, $U_{i}=0$. Then, the normalization condition gives us
\begin{equation}
U^{\mu }=\frac{1}{N}(1,N^{i}) \label{zamo}
\end{equation}
with some $N^{i}$. It has the simple physical meaning of the velocity of the
observer with respect to the coordinate frame $\{t,x^{i}\}$ defined
as
\begin{equation}
V^{i}\equiv \frac{dx^{i}}{dt}\Big| _{obs}=\frac{U^{i}}{U^{0}}=N^{i}.
\label{vn}
\end{equation}

It follows from (\ref{tau}) that
\begin{equation}
d\tau _{obs}=Ndt .
\end{equation}
Using that
\begin{equation}
0=U_{i}=\frac{g_{i 0}}{N}+\frac{g_{ij}N^{j}}{N},
\end{equation}%
we find
\begin{equation}
g_{0i}=-g_{ij}N^{j}.
\end{equation}

In a similar way,
\begin{equation}
-N=U_{0}=\frac{g_{00}}{N}+\frac{g_{0i}N^{i}}{N},
\end{equation}
whence
\begin{equation}
g_{00}=-N^{2}+g_{0i}N^{i}N.
\end{equation}

Collecting all this, we obtain that the metric can be written in the form
\begin{equation}
ds^{2}=-dt^{2}N^{2}+g_{ij}(dx^{i}-N^{i}dt)(dx^{j}-N^{j}dt) \label{N},
\end{equation}
and calculating $h_{\mu \nu }$ we find
\begin{equation}
h_{00}=g_{00}+N^{2},\qquad h_{0i}=g_{0i},\qquad h_{ij}=g_{ij}.
\end{equation}

In the case of a stationary axially symmetric metric
\begin{equation}
ds^{2}
=-dt^{2}N^{2}
+g_{\phi \phi }(d\phi -\omega dt)^{2}
+g_{rr}dr^{2}+g_{\theta \theta }d\theta ^{2}.
\end{equation}
we have
\begin{align}
g_{00}&=-N^{2}+g_{\phi \phi }\omega ^{2}, \label{00m}\\
g_{0\phi }&=-g_{\phi \phi }\omega, \label{03}
\end{align}
and therefore
\begin{align}
h_{00}=\omega ^{2}g_{\phi \phi },
&\qquad h_{0\phi }=g_{0\phi },
\qquad h_{\phi \phi }=g_{\phi \phi };\\
& N^{\phi }=\omega ,\qquad N^{r}=N^{\theta }=0.\label{vf}
\end{align}
</p></div>
</div>

<div id="OZ09"></div>
=== Problem 9===

Consider a stationary metric with the time-like Killing vector field $\xi^\mu =(1,0,0,0)$. Relate the energy $E$ of a particle with four-velocity $u^\mu$ as measured at
infinity by a stationary observer to that measured by a local observer with 4-velocity $U^\mu$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The energy as measured at infinity by a stationary observer is the integral of motion
\begin{equation}
E=-mu_{\mu }\xi ^{\mu }=-mu_{0}.
\end{equation}

On another hand, the energy measured by a local observer is
\begin{equation}
E_{rel}=-mu_{\mu }U^{\mu }\equiv m\gamma =-m(u_{0}U^{0}+u_{i}U^{i}),
\label{rel}
\end{equation}
whence
\begin{equation}
E-m\,u_{i}V^{i}=\frac{E_{rel}}{U^0},\label{e}
\end{equation}
where $V^{i}=U^i / U^0$ is the local observer's velocity in the given frame (\ref{vn}).
</p></div>
</div>

<div id="OZ10"></div>
=== Problem 10===

Express $E_{rel}$ and $E$ in terms of the relative velocity $w$ between a
particle and the observer (i.e. velocity of the particle in the frame of the observer and vice versa).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Using (\ref{tau}), (\ref{rel}) we find that the Lorentz factor is
\begin{equation}
\gamma =-u_{\mu }U^{\mu }
=-\frac{U^{\mu }dx_{\mu }}{ds}=-\frac{d\tau _{obs}}{ds}.
\end{equation}

It follows from (\ref{sw}) that
\begin{equation}
\gamma =\frac{1}{\sqrt{1-w^{2}}},\qquad
E_{rel}=\frac{m}{\sqrt{1-w^{2}}}.
\end{equation}

Then, we find from (\ref{e}) that
\begin{equation}
E-p_{i}V^{i}=\frac{m}{\sqrt{1-w^{2}}}\frac{1}{U_{0}}, \label{ep}
\end{equation}
where $p^{\mu }=mu^{\mu }$ is the particle's four-momentum.
</p></div>
</div>

<div id="OZ11"></div>
=== Problem 11===

Show that in the flat spacetime eq. (\ref{ep}) is reduced to the usual formula
of the Lorentz transformation.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In this case, $U^{0}=(1-V^{2})^{-1/2}=\gamma $, where $V$ is the absolute value of the velocity of the observer (frame 0) with respect to the stationary laboratory frame. Then the energy of a particle in frame 0 is
\begin{equation}
E_{(0)}=\gamma ( E-\mathbf{p}\cdot\mathbf{V}),
\end{equation}
This is the standard formula for energy transformation.
</p></div>
</div>

<div id="OZ12"></div>
=== Problem 12===

Find the expression for $E$ for the case of a static observer ($U^{i}=0$).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Using (\ref{st}) and (\ref{e}) we find that the kinetic term in (\ref{e})
vanishes and
\begin{equation}
E=E_{rel.}\sqrt{-g_{00}}=\frac{m}{\sqrt{1-w^{2}}},
\end{equation}
which coincides with eq. (88.9) of \cite{lan}.
</p></div>
</div>

<div id="OZ13"></div>
=== Problem 13===

Find the expression for $E$ for the case of the ZAMO observer and, in particular, in case of axially symmetric metric.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\begin{equation}
E-mu_{i}V^{i}=E_{rel} N=\frac{mN}{\sqrt{1-w^{2}}}.
\end{equation}

In the axially symmetric case, the angular momentum of a particle $mu_{\phi}=L$ is conserved, and according to (\ref{vf}), $V^{\phi }=\omega$. Then,
\begin{equation}
E-\omega L=E_{rel}N=\frac{mN}{\sqrt{1-w^{2}}}.
\end{equation}
</p></div>
</div>

==Collision of particles: general relationships==

<div id="OZ14"></div>
=== Problem 14===
Let two particles collide. Define the energy in the center of mass (CM)
frame $E_{c.m.}$ at the point of collision and relate it to $E_{rel}$ and
the Lorentz factor of relative motion of the two particles.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In general, the momenta and velocities of two particles are defined at
different points and this prevents one from constructing well-defined
quantities. However, as collision occurs at one point, both particles have
characteristics defined just in this point. This enables one to construct
the total momentum $P^{\mu }=p_{1}^{\mu }+p_{2}^{\mu }$ at the point of
collision (the subscript $a=1,2$ enumerates the two particles).

A particle's mass is
\begin{equation}
m^{2}=-p^{\mu }p_{\mu },
\end{equation}
and likewise the full energy (mass) of two particles at the point of collision, which is also their energy in the center of mass frame, is found from
\begin{equation}
E_{c.m.}^{2}=-P^{\mu }P_{\mu }.\label{cm}
\end{equation}
It is is the direct generalization of the corresponding formula of special relativity. Eq (\ref{cm}) implies
\begin{equation}
E_{c.m.}^{2}=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma ,
\end{equation}
where
\begin{equation}
\gamma =-u_{1}^{\mu }u_{2\mu }
\end{equation}
is the Lorentz factor of the two particles' relative motion. It can be expressed in terms of energy $E_{rel}(2,1)$ of particle 1 with respect to particle 2 and vice versa
\begin{equation}
\gamma=\frac{E_{rel.}(1,2)}{m_{1}}=\frac{E_{rel.}(2,1)}{m_{2}}.
\end{equation}
</p></div>
</div>

<div id="OZ1"></div>
=== Problem 15===

Let us consider a collision of particles 1 and 2 viewed from the frame
attached to some other particle 0. How are different Lorentz factors related
to each other? Analyze the case when the laboratory frame coincides with
that of particle 0.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Let us introduce notation
\begin{equation*}
\gamma (0,1)=\gamma _{1}, \quad
\gamma (0,2)=\gamma _{2}, \quad
\gamma (1,2)=\gamma , \quad
u_{0}^{\mu }=U^{\mu}.
\end{equation*}

Each particle's four velocity can be decomposed into components parallel and orthogonal to $U^\mu$ as follows
\begin{equation}
u_{1}^{\mu }=a_{1}U^{\mu }+b_1 n_{1}^{\mu}, \label{ab}
\end{equation}
where $n^{\mu }$ is a unit space-like vector orthogonal to $U^{\mu }$. It is easy to
find from (\ref{ab}) that
\begin{equation}
a_{1}=-u_{1}^{\mu }U_{\mu }=\gamma _{1},
\end{equation}
and from the normalization conditions for $u_{1}^{\mu }$ and $U^{\mu }$ then we see that $b_1$ as introduced in (\ref{ab}) is the velocity of particle 1 in the frame of particle 0:
\begin{equation}
b_{1}^{2}=\frac{1}{\gamma _{1}^{2}-1 },
\end{equation}
so in terms of velocity of particle 1 in the frame of particle 0
\begin{equation}
w_1 =(1-\gamma_1^{-2})^{1/2}=\frac{b_1}{\gamma},
\end{equation}
the decomposition takes form
\begin{equation}
u_{1}^{\mu }=\gamma_{1}(U^{\mu }+w_1 n_{1}^{\mu}).
\end{equation}

Writing down for particle 2 the equation analogous to (\ref{ab}) and
calculating the scalar product $u_{1}^{\mu }u_{2\mu }$, one obtains
\begin{align}
\gamma &=\gamma _{1}\gamma _{2}
-b_1 b_2 (n_{1}^{\mu }n_{2\mu})\\
&=\gamma _{1}\gamma _{2}\big(1-\varepsilon\; w_{1}w_{2}\big),
\label{ga}
\end{align}
where
\begin{equation}
\varepsilon =(n_{1}^{\mu }n_{2\mu }),\qquad |\varepsilon|\leq 1 ,
\end{equation}
or in terms of velocities
\begin{equation}
\frac{1}{\sqrt{1-w^{2}}}
=\frac{1}{\sqrt{1-w_{1}^{2}}}
\frac{1}{\sqrt{1-w_{2}^{2}}}
(1-\varepsilon w_{1}w_{2}). \label{gw}
\end{equation}

In case particle 0 is at rest in the laboratory frame $U^{\mu }=(1,0)$, we have $n_{a}^{\mu}=(0,\vec{n}_{a})$, $a=1,2$, and
\begin{equation}
u_{a}^{\mu }=\gamma _{a}(1,w_{a}\vec{n}_{a}),\qquad \varepsilon =(\vec{n}_{1}
\vec{n}_{2}).
\end{equation}
Then, eqs. (\ref{ga}), (\ref{gw}) turn into those listed in problem 1.3 of \cite{li}.
</p></div>
</div>

<div id="OZ16"></div>
=== Problem 16===

When can $\gamma $ as a function of $\gamma _{1}$ and $\gamma _{2}$ grow
unbounded? How can the answer be interpreted in terms of relative velocities?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

1) If $\gamma _{1}$ and $\gamma _{2}$ are finite, it follows from (\ref{ga})
that $\gamma $ is also finite. It means that if $w_{1}$ and $w_{2}$ are
finite, the relative velocity $w$ of particles 1 and 2 is also finite (in
the sense that it is separated from $c$).

2) Let $\gamma _{1}\rightarrow \infty $ but $\gamma _{2}$ remain finite.
Then,
\begin{equation}
\gamma \approx \gamma _{1}\big( \gamma _{2}-\varepsilon \sqrt{\gamma _{2}^{2}-1}\;\big)
\end{equation}
grows unbound irrespective of $\varepsilon $. It means that the relative
velocity of particles, one of which moves with some finite speed and the
other one almost with the speed of light is always close to the speed of
light.

3) Let $\gamma _{1}\rightarrow \infty $, $\gamma _{2}\rightarrow \infty $.

3a) If $\varepsilon \neq +1$,
\begin{equation}
\gamma \approx \gamma _{1}\gamma _{2}(1-\varepsilon )\rightarrow \infty ,
\qquad w\rightarrow 1.
\end{equation}

3b) Let $\varepsilon =+1$. Then
\begin{equation}
\gamma \approx \frac{1}{2} \Big(\frac{\gamma _{1}}{\gamma _{2}} +\frac{\gamma _{2}}{\gamma _{1}}\Big).
\end{equation}

We see that $\gamma $ remains finite if $\gamma _{1}/\gamma _{2}$ does so.
Otherwise, $\gamma \rightarrow \infty $. The resulting relative velocity
depends on the relative rate with which $w_{1}$ and $w_{2}$ approach the
speed of light.

Thus in case 3, motion in different directions always gives $\gamma
\rightarrow \infty $, whereas motion in the same direction may or may not
lead to divergent $\gamma $.
</p></div>
</div>

<div id="OZ1"></div>
=== Problem 17===

A tetrad basis, or the orthonormal tetrad, is the set of four unit vectors $h_{(a)}^\mu$ (subscripts in parenthesis $a=0,1,2,3$ enumerate these vectors), of which one, $h_{(0)}^\mu$, is timelike, and three vectors $h_{(i)}^\mu$ ($i=1,2,3$) are spacelike, so that
\begin{equation}
g_{\mu\nu}h_{(a)}^\mu h_{(b)}^\nu =\eta_{ab},\qquad a,b=0,1,2,3.
\end{equation}
A vector's tetrad components are
\begin{equation}
u_{(a)}=u_\mu h_{(a)}^\mu,\qquad u^{(b)}=\eta^{ab}u_{(b)}.
\end{equation}

Define the local three-velocities with the help of the tetrad basis attached to the observer, which
would generalize the corresponding formulas of special relativity.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In special relativity,
\begin{equation}
u^{\mu }=\frac{dx^{\mu }}{ds}
=\gamma (1,v^{i}),\qquad
\gamma =\frac{1}{\sqrt{1-v^{2}}}.
\end{equation}
Thus $v^{i}=u^{i}\sqrt{1-v^{2}}$, and using (\ref{sw}), we get
\begin{equation}
v^{i}=\frac{dx^{i}}{d\tau _{obs}}.\label{vi}
\end{equation}

Let us attach a tetrad to some observer with four-velocity $U^\mu$. This means simply that we choose the first, time-like, tetrad vector to be along its world-line
\begin{equation}
h_{(0)}^\mu =U^\mu .
\end{equation}
Then, the natural definition for velocity in the frame of this observer, taking into account (\ref{tau}), will be
\begin{equation}
v^{(i)}=\frac{h_{\mu }^{(i)}dx^{\mu}}{-h^{\mu}_{(0)}dx_{\mu }}.
\end{equation}
It can be rewritten as
\begin{equation}
v^{(i)}=\frac{u^{\mu }h_{\mu }^{(i)}}{u^{\mu }h^{(0)}_\mu}.
\end{equation}

The tetrad components of the projection operator (\ref{h}) are
\begin{equation}
h_{(ab)}=\eta_{ab}+\eta_{0a}\eta_{0b},
\end{equation}
from which it is easy to check explicitly that the two definitions, through $h_{\mu\nu}$ and through $h^\mu_{(a)}$, are equivalent.
</p></div>
</div>

<div id="OZ18"></div>
=== Problem 18===

Derive the analogues of formulas (\ref{e}), (\ref{ep}) for massless particles (photons). Analyze the cases of static and ZAMO observers.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In this case, $m=0$, the velocity of a photon is always 1, and the notion of
a frame comoving with the photon does not have sense. Nonetheless, some
formulas retain their validity with $E$ replaced by the frequency $\nu$ in
accordance with the fact that for photons $E=\hbar\nu$ (we use $\nu=2\pi/\text{(period)}$ for the angular frequency here, so as not to confuse it with the metric function $\omega$).

Let us consider a photon with wave four-vector $k_{\mu }$ in a gravitational
field. In the case of static or stationary metric, $\nu _{0}=-k_{\mu }\xi^{\mu }$ is conserved along the trajectory, where $\xi ^{\mu }$ is the
Killing vector. If it corresponds to time translations, $\nu$ has the
meaning of frequency. It is what is measured by a remote observer (if the
flat infinity exists). In parallel to derivation of (\ref{e}), one can obtain
\begin{equation}
\nu _{0}-k_{i}V^{i}=\frac{\nu }{U^{0}}, \label{nu}
\end{equation}
where $\nu =-k_{\mu }U^{\mu }$, and $U^{\mu }$ is the velocity of the
observer.

For the static observer $V^{i}=0$, $U^{0}=\frac{1}{\sqrt{-g_{00}}}$, and
\begin{equation}
\nu_{0}=\nu \sqrt{-g_{00}}. \label{0g}
\end{equation}

For the ZAMO observer, $V^{\phi }=\omega $, $k_{\phi }=L$ is the angular
momentum which is conserved in the axially symmetric case (in units of $\hbar$), $U^{0}=\frac{1}{N}$ (see eq. (\ref{zamo})). Then
\begin{equation}
\nu _{0}-\omega L=\nu N.
\end{equation}
</p></div>
</div>

<div id="OZ19"></div>
=== Problem 19===

The ergosphere is a surface defined by equation $g_{00}=0$. Show that it is
the surface of infinite redshift for an (almost) static observer.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

From (\ref{0g}) it follows that in the limit $g_{00}\rightarrow 0$ the
frequency $\nu _{0}\rightarrow 0$ for any finite $\nu $. However, for a
nonstatic observer this is not true.
</p></div>
</div>

<div id="OZ20"></div>
=== Problem 20===

Consider an observer orbiting with a constant angular velocity $\Omega $ in
the equatorial plane of the axially symmetric back hole. Analyze what
happens to redshift when the angular velocity approaches the minimum or
maximum values $\Omega _{\pm }$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Eq. (\ref{nu}) reads in this case
\begin{equation}
\nu _{0}-\Omega L=\frac{\nu }{U^{0}}\text{,}
\end{equation}
where $U^{0}$ obeys the normalization condition
\begin{equation}
Y\equiv g_{00}+2g_{0\phi }\Omega +g_{\phi \phi }\Omega ^{2}=-\frac{1}{\left(
U^{0}\right) ^{2}}\text{.}
\end{equation}

It can be rewritten in the form
\begin{align}
Y=&g_{\phi \phi }(\Omega ^{2}-2\omega \Omega +\frac{g_{00}}{g_{\phi \phi }})
=g_{\phi \phi }(\Omega -\Omega _{+})(\Omega -\Omega _{-});\\
&\Omega _{\pm }=\omega \pm \frac{N}{\sqrt{g_{\phi \phi }}},
\end{align}
Then,
\begin{equation}
\nu _{0}-\Omega L=\nu \sqrt{g_{\phi \phi }(\Omega _{+}-\Omega )(\Omega
-\Omega _{-})}\text{.}
\end{equation}

When $g_{00}\rightarrow 0$, $\Omega _{-}\approx \frac{g_{00}}{2\omega }
\rightarrow 0$.

Thus, if, say, $\Omega \rightarrow \Omega _{-}$, $\nu _{0}\rightarrow \Omega
_{-}L$. If $L=0$, then $\nu _{0}\rightarrow 0$.
</p></div>
</div>

<div id="OZ21"></div>
=== Problem 21===

Let two massive particle 1 and 2 collide. Express the energy of each particle
in the centre of mass (CM) frame in terms of their relative Lorentz factor $\gamma (1,2)$. Analyze the limiting cases of ultra-relativistic $\gamma
(1,2)\rightarrow \infty $ and non-relativistic $\gamma (1,2)\approx 1$ collisions.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The energy of particle 1 in the CM frame is
\begin{equation}
\left( E_{1}\right) _{c.m.}=m_{1}\gamma (1,CM) \label{e1}
\end{equation}
where
\begin{equation}
\gamma (1,CM)=-u_{1\mu }U_{c.m.}^{\mu }
\end{equation}
is its Lorentz factor in the same frame, which has four-velocity
\begin{equation}
U_{c.m.}^{\mu }=\frac{P^{\mu }}{\mu}.
\end{equation}
Here $P^\mu$ is the total momentum and $\mu \equiv E_{c.m.}$ is the energy in the center of
mass frame introduced in (\ref{cm})
\begin{align}
P^{\mu }&=m_{1}u_{1}^{\mu }+m_{2}u_{2}^{\mu };\\
\mu ^{2}&=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma (1,2).\label{mu}
\end{align}

Then
\begin{align}
\gamma (1,CM)&=\frac{m_{1}+m_{2}\gamma (1,2)}{\mu };\label{g1}\\
( E_{1}) _{c.m.}
&=m_{1}\gamma (1,CM).\label{e1cm}
\end{align}

Likewise, for the second particle
\begin{align}
\gamma (2,CM)&=\frac{m_{2}+m_{1}\gamma (1,2)}{\mu };\\
( E_{2}) _{c.m.}&=m_{2}\gamma (2,CM).\label{e2cm}
\end{align}

It follows from (\ref{e1cm}), (\ref{e2cm}) that
\begin{equation}
( E_{1}) _{c.m.}+( E_{2}) _{c.m.}=\mu ,
\end{equation}
as it should be.

In the ultra-relativistic limit $\gamma (1,2)\rightarrow \infty $
\begin{equation}
\gamma (1,CM)\approx
\frac{\mu }{2m_{1}}\approx \sqrt{\frac{m_{2}}{2m_{1}}\gamma (1,2)}.
\end{equation}

In the opposite limit of small relative velocity $w(1,2)\ll 1$ we obtain $\gamma (1,2)\approx 1+\frac{w^{2}}{2}$ and
\begin{align}
&\mu \approx (m_{1}+m_{2})+\frac{m_{1}m_{2}}{2(m_{1}+m_{2})}w^{2};\\
&w(1,CM)\approx w\frac{m^{2}}{(m_{1}+m_{2})},
\end{align}
which agrees with formulas of nonrelativistic mechanics -- see Ch. 3, Sec. 13 of \cite{lan}.
</p></div>
</div>

<div id="OZ22"></div>
=== Problem 22===

For a stationary observer in a stationary space-time the quantity $\alpha =(U^0)^{-1}$ is the redshifting factor: if this observer emits a ptoton with frequency $\omega_{em}$, it is detected at infinity by another stationary observer with frequency $\omega_{det}=\alpha \omega_{em}$. For a generic observer this interpretation is invalid, however, $\alpha =ds/dt$ still determines the time dilation for this observer, and thus can still be called the same way. Express the redshifting factor of the center of mass frame $\alpha _{c.m.}$ through the redshifting factors of the colliding particles $\alpha_{1}$ and $\alpha_2$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Starting from the definition,
\begin{equation}
\frac{1}{\alpha _{c.m.}}
=\left( U^{0}\right) _{c.m.}
=\frac{m_{1}U_{1}^{0}+m_{2}U_{2}^{0}}{\mu }
=\frac{\frac{m_{1}}{\alpha _{1}}+\frac{m_{2}}{\alpha _{2}}}{\mu },
\end{equation}
whence
\begin{equation}
\alpha _{c.m.}=\frac{\mu \alpha _{1}\alpha _{2}}{m_{1}\alpha
_{2}+m_{2}\alpha _{1}}.
\end{equation}

If case $\alpha _{2}\ll \frac{m_{2}}{m_{1}}\alpha _{1}$, we get
\begin{equation}
\alpha _{c.m.}\approx \frac{\mu \alpha _{2}}{m_{2}}.
\end{equation}
</p></div>
</div>

<div id="OZ23"></div>
=== Problem 23===

Relate the energy of a particle at infinity $E_{1}$, its energy at the point of collision in the C.M. frame $(E_{1})_{c.m.}$ and $\mu$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

From (\ref{e})
we know that
\begin{equation}
E_{1}-mu_{1i}v_{2}^{i}=\alpha _{2}E_{rel}, \label{e2}
\end{equation}
where $v^{i}$ is the velocity of particle 2 in the stationary frame, $\alpha _{2}=\frac{1}{U_{2}^{0}}$.

A particle's energy in the center of mass frame can be found from (\ref{e1cm})
\begin{equation}
(E_{1})_{c.m.}\mu =m_{1}^{2}+m_{2}E_{rel}(1,2),
\end{equation}
where $E_{rel}(1,2)=m_{1}\gamma (1,2)$, so in terms of $(E_1)_{c.m.}$ we get
\begin{align}
E_{1}-m_{1}u_{1i}v_{2}^{i}
&=\frac{\alpha_{2}(\mu E_{1})_{c.m.} -m_{1}^{2}\alpha _{2}}{m_{2}};\\
E_{2}-m_{2}u_{2i}v_{1}^{i}
&=\frac{\alpha _{1}(\mu E_{2})_{c.m.} -m_{2}^{2}\alpha _{1}}{m_{1}}.
\end{align}

On another hand, applying (\ref{e}) to particle 1 and the center of mass frame, we obtain
\begin{align}
E_{1}-m_{1}u_{1i}V_{c.m.}^{i}
&=\alpha _{c.m.}(E_{1}) _{c.m.}; \label{1cm}\\
E_{2}-m_{2}u_{2i}V_{c.m.}^{i}
&=\alpha _{c.m.}(E_{2}) _{c.m.}.\label{2cm}
\end{align}

The sum of (\ref{1cm}) and (\ref{2cm}) is the total conserved energy $E=E_1 +E_2$:
\begin{equation}
E-\alpha _{c.m.}P_{i}U_{c.m.}^{i}=\alpha _{c.m.}\mu .
\end{equation}
Taking into account that $\frac{\alpha _{c.m.}}{\mu }=\frac{1}{P^{0}}$, this is equivelent to
\begin{equation}
EP^{0}-P_{i}P^{i}=-P^{\mu }P_{\mu }=\mu^{2},
\end{equation}
as it should be.
</p></div>
</div>

<div id="OZ24"></div>
=== Problem 24===

Solve the same problem when both particles are massless (photons). Write
down formulas for the ZAMO observer and for the C.M. frame.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Let $k^\mu_1$ and $k^\mu_2$ be the wave-vectors of two photons. Then vector
\begin{equation}
K^{\mu }=\left( k^{\mu }\right) _{1}+\left( k^{\mu }\right) _{2} \label{K}
\end{equation}
is \emph{time-like} (unless the two photons are collinear, but then they would not collide). Dividing by the energy $\mu$ in the center of mass frame
\begin{equation}
\mu ^{2}=-K^{\mu }K_{\mu }=-2k_{1\mu }k^{2\mu },
\end{equation}
we obtain the frame's (time-like) four-velocity
\begin{equation}
U_{c.m.}^{\mu }=\frac{K^{\mu }}{\mu }.
\end{equation}
There is no Lorentz factor or relative velocity for two photons.

The contraction
\begin{equation}
\tilde{\gamma}(1,obs)=-k_{\mu } U^{\mu }
\end{equation}
has the meaning of the frequency of a photon with four-vector $k^\mu$ as measured by the observer with four-velocity $U^\mu$.

Then frequency measured in the CM frame is
\begin{equation}
\left( \nu _{1}\right) _{c.m.}=-k_{1\,\mu} U_{c.m.}^{\mu },
\end{equation}
and taking into account (\ref{mu}) and (\ref{K}), we obtain
\begin{equation}
(\nu _{1}) _{c.m.}
=-k_{1\,\mu}U_{c.m.}^{\mu }
=-\frac{k_{1\,\mu }k_{2}^{\mu }}{\mu }=\frac{\mu }{2},
\end{equation}
as for photons $k_{i\,\mu}k_i^{\mu}=0$. This is quite natural, since for photons the energy is equal to the absolute value of the momentum,
so in the CM frame the energies of both photons should be equal.

Likewise,
\begin{equation}
( \nu _{2}) _{c.m.}
=-k_{2\,\mu}U_{c.m.}^{\mu }
=\frac{\mu }{2}=(\nu_{1}) _{c.m.},
\end{equation}
and
\begin{equation}
\mu =( \nu _{1}) _{c.m.}+( \nu _{2}) _{c.m.}
=2(\nu _{1}) _{c.m.}=2(\nu _{2}) _{c.m.}
\end{equation}

If $\xi$ is the Killing vector field responsible for time translation, the conserved frequency of a photon, equal to the frequency measured at infinity by a stationary observer, is
\begin{equation}
\nu _{\infty }=-k_{\mu }\xi ^{\mu }=-k_{0}.
\end{equation}
In some arbitrary frame it is
\begin{equation}
\nu _{loc}=-k_{\mu }U^{\mu }=-k_{0}U^{0}-k_{i}U^{i}.
\end{equation}

Then
\begin{equation}
\nu _{\infty }-k_{\alpha}V^{\alpha}=\alpha \nu _{loc},\qquad
V^{i}=\frac{U^{i}}{U^{0}},\qquad
\alpha =\frac{1}{U^{0}},\label{obs}
\end{equation}
the same as for the case of massive particle. The difference is only that for photons we
cannot express $\nu _{loc}$ in terms of mass and velocity as $\frac{m}{\sqrt{1-w^{2}}}$.

ZAMO frame
\begin{equation}
\nu _{\infty }-\omega L=N\nu _{loc}.
\end{equation}

CM frame
\begin{align}
&\nu _{\infty }-k_{i}V^{i}=\alpha _{c.m.}\frac{\mu }{2},\\
&\alpha _{c.m.}=\frac{1}{U^{0}}=\frac{\mu }{k_{1}^{0}+k_{2}^{0}}.
\end{align}
</p></div>
</div>

Main Page

2013-05-01T08:35:26Z

Igor:

= Dynamics of the Universe in Problems =

<p align="center">'''The total number of problems on this website.'''</p>
[[File:Numbers.png‎|center|164px]]

=== About ===

There are [https://www.google.com.ua/search?q=inurl%3Acosmology thousands] of websites with names that contain the term “cosmology”. Many of them are devoted to discussion of fundamental questions: whether there is life on Mars, what was there when there was nothing and the like. Our aim is much more modest. We present here an online living version of our book of problems on cosmology.

=== Why problems, which and for whom?===
The only way to rise above the “popular” level in any science is to master its alphabet, that is, to learn to solve problems, even if most simple at the beginning. To our best knowledge, there are no problem books on cosmology yet, that would include its spectacular recent achievements. Of course, most of excellent modern textbooks on cosmology include problems. However, a reader, exhausted by high theory, may often be thwarted by the lack of time and strength to solve them.

You can familiarize oneself with the modern portrayal of the dynamics of the Universe, including the latest achievements of cosmology, by solving problems from the start. The perseverant will traverse the thorny path of cosmology’s evolution from the traditional Big Bang model to the presently universally accepted Standard Cosmological Model. It is based on the revolutionary discovery of accelerated expansion of the Universe, made at the very end of the last century. The attempts to explain such an unexpected observation led to fundamental reconstruction of the view on the evolution of the Universe that was formed in the XX century.

One only has to be acquainted with the basics of Special and General Relativities and the theory of elementary particles in order to start solving problems. We want to show that even this minimal knowledge is sufficient to be able to solve a very wide range of cosmological problems, and moreover, to understand the essence of difficulties that are encountered by modern cosmology.

No generation of Homo sapiens could withstand the temptation to claim that the true understanding of the nature of the Universe is (almost?) within its grasp. The current generation is not an exception, of course, naturally assuming it has better reason to believe so than ever before. We hope to give our reader the means to evaluate objectively whether this is true or not by himself.

=== Second life online===
We published the first edition of the problem book a couple of years ago, in Russian. However, a printed book, even not taking into account the language, has its obvious drawbacks in itself. The audience is limited by the number of copies printed. The inevitable misprints and errors cannot be corrected – the errors in classical problem books on physics are known to be sometimes corrected in the fifteenth edition. But the main problem that almost immediately manifested itself was that the content of the book with pretentious title “Modern Cosmology” was getting out of date by leaps and bounds. New sections and subsections demanded to be created and populated with problems. The online living problem book is the solution we have come to.

At the moment we have about 1500 problems, almost all of them with solutions, and we hope to bring them online by the fall of 2012. We tried to make references to the authors of problems whenever possible and would be happy to add the references that might still be missing. Some problems have already acquired the status of folklore and their authorship is hard to determine. We welcome all comments, suggestions and corrections to the formulations of problems and their solutions, but most valuable to us would be new problems suggested by the community.

=== Good luck===
Finally, we hope that after getting to grip with the problems our reader not only preserves his interest to cosmology, but tries to make the next step: read original papers. If this transition is overcome seamlessly, we will have achieved our goal.

Welcome!

----

= CONTENTS =
1. [[:Category:Cosmo warm-up|'''Cosmo warm-up''']]

1.1 [[Astronomy "before the Common Era"|Astronomy "before the Common Era"]]

1.2 [[Quantities large and small|Quantities large and small]]

1.3 [[Geometric warm-up|Geometric warm-up]]

1.4 [[Astrophysical warm-up|Astrophysical warm-up]]

1.5 [[Planck scales and fundamental constants|Planck scales and fundamental constants $c$, $G$, $\hbar$]]

1.6 [[Gravity|Gravity]]

1.7 [[Forest for the trees|Forest for the trees]]

1.8 [[Life on Mars?|Life on Mars?]]

1.8.1 [[Life on Mars?#Fine-tuning of the Universe|Fine-tuning of the Universe]]

1.9 [[Thermo warm-up|Thermo warm-up]]

2. [[:Category:Dynamics of the Expanding Universe|'''Dynamics of the Expanding Universe''']]

2.1 [[Homogeneous Universe|Homogeneous and isotropic Universe, Hubble's Law]]

2.2 [[Equations of General Relativity|Equations of General Relativity]]

2.3 [[Friedman-Lemaitre-Robertson-Walker (FLRW) metric|Friedman-Lemaître-Robertson-Walker (FLRW) metric]]

2.4 [[Expanding Universe: ordinarity, difficulties and paradoxes|Expanding Universe: ordinarity, difficulties and paradoxes]]

2.4.1 [[Expanding_Universe:_ordinarity,_difficulties_and_paradoxes#Warm-up|Warm-up]]

2.4.2 [[Expanding_Universe:_ordinarity,_difficulties_and_paradoxes#The_tethered_galaxy_problem|The tethered galaxy problem]]

2.4.3 [[Expanding_Universe:_ordinarity,_difficulties_and_paradoxes#Cosmological_redshift|Cosmological redshift]]

2.5 [[Friedman equations|Friedman equations]]

2.6 [[Newtonian cosmology|Newtonian cosmology]]

2.7 [[Energy balance in an expanding Universe|Energy balance in an expanding Universe]]

2.8 [[Cosmography|Cosmography]]

2.9 [[Light and distances|Light and distances]]

3. [[:Category:Dynamics of the Universe in the Big Bang Model|'''Dynamics of the Universe in the Big Bang Model''']]

3.1 [[General questions|General questions]]

3.2 [[Solutions of Friedman equations in the Big Bang model|Solutions of Friedman equations in the Big Bang model]]

3.3 [[Polytropic equation of state|Polytropic equation of state]]

3.4 [[The role of curvature in the dynamics of the Universe|The role of curvature in the dynamics of the Universe]]

3.5 [[The Milne Universe|The Milne Universe]]

3.6 [[Cosmological horizons|Cosmological horizons]]

3.7 [[Energy conditions and the Raychaudhuri equation|Energy conditions and the Raychaudhuri equation]]

3.6.1 [[Energy_conditions_and_the_Raychaudhuri_equation#Energy conditions|Energy conditions]]

3.6.2 [[Energy_conditions_and_the_Raychaudhuri_equation#Raychaudhuri equation|Raychaudhuri equation]]

3.6.2 [[Energy_conditions_and_the_Raychaudhuri_equation#Sudden Future Singularities|Sudden future singularities]]

3.8 [[Influence of cosmological expansion on local systems|Influence of cosmological expansion on local systems]]

3.9 [[Dynamics of the Universe in terms of redshift and conformal time|Dynamics of the Universe in terms of redshift and conformal time]]

4. [[:Category:Black Holes|'''Black Holes''']]

4.1 [[Technical warm-up|Technical warm-up]]

4.1.1 [[Technical_warm-up#Uniformly_accelerated_observer,_Rindler_metric|Uniformly accelerated observer, Rindler metric]]

4.1.2 [[Technical_warm-up#Metric_in_curved_space-time|Metric in curved spacetime]]

4.2 [[Schwarzschild black hole|Schwarzschild black hole]]

4.2.1 [[Schwarzschild_black_hole#Simple_problems|Simple problems]]

4.2.2 [[Schwarzschild_black_hole#Symmetries_and_integrals_of_motion|Symmetries and integrals of motion]]

4.2.3 [[Schwarzschild_black_hole#Radial_motion|Radial motion]]

4.2.4 [[Schwarzschild_black_hole#Blackness_of_black_holes|Blackness of black holes]]

4.2.5 [[Schwarzschild_black_hole#Orbital_motion,_effective_potential|Orbital motion, effective potential]]

4.2.6 [[Schwarzschild_black_hole#Miscellaneous_problems|Miscellaneous problems]]

4.2.7 [[Schwarzschild_black_hole#Different_coordinates,_maximal_extension|Different coordinates, maximal extension]]

4.3 [[Kerr black hole|Kerr black hole]]

4.3.1 [[Kerr_black_hole#General_axially_symmetric_metric|General axially symmetric metric]]

4.3.2 [[Kerr_black_hole#Limiting_cases|Limiting cases]]

4.3.3 [[Kerr_black_hole#Horizons_and_singularity|Horizons and singularity]]

4.3.4 [[Kerr_black_hole#Stationary_limit|Stationary limit]]

4.3.5 [[Kerr_black_hole#Ergosphere_and_the_Penrose_process|Ergosphere and the Penrose process]]

4.3.6 [[Kerr_black_hole#Integrals_of_motion|Integrals of motion]]

4.3.7 [[Kerr_black_hole#The_laws_of_mechanics_of_black_holes|The laws of mechanics of black holes]]

4.3.8 [[Kerr_black_hole#Particles'_motion_in_the_equatorial_plane|Particles' motion in the equatorial plane]]

4.4 [[Particles' motion in general black hole spacetimes|Particles' motion in general black hole spacetimes]]

4.4.1 [[Particles' motion in general black hole spacetimes#Frames,_time_intervals_and_distances|Frames, time intervals and distances]]

4.4.2 [[Particles' motion in general black hole spacetimes#Fiducial_observers|Fiducial observers]]

4.4.3 [[Particles' motion in general black hole spacetimes#Collision_of_particles:_general_relationships|Collision of particles: general relationships]]

4.5 [[Astrophysical black holes|Astrophysical black holes]]

4.5.1 [[Astrophysical black holes#Preliminary|Preliminary]]

4.5.2 [[Astrophysical black holes#Quantum_effects|Quantum effects]]

5. [[:Category:Cosmic Microwave Background (CMB)|'''Cosmic Microwave Background (CMB)''']]

5.1 [[Thermodynamics of Black-Body Radiation|Thermodynamics of Black-Body Radiation]]

5.2 [[Time Evolution of CMB|Time Evolution of CMB]]

5.3 [[Statistical properties of CMB|Statistical properties of CMB]]

5.4 [[Primary anisotropy of CMB|Primary anisotropy of CMB]]

5.5 [[CMB interaction with other components|CMB interaction with other components]]

5.6 [[Extras|Extras]]

6. [[:Category:Thermodynamics of Universe|'''Thermodynamics of Universe''']]

6.1 [[Thermodynamical Properties of Elementary Particles|Thermodynamical Properties of Elementary Particles]]

6.2 [[Thermodynamics of Non-Relativistic Gas|Thermodynamics of Non-Relativistic Gas]]

6.3 [[Entropy of Expanding Universe|Entropy of Expanding Universe]]

6.4 [[Connection between Temperature and Redshift|Connection between Temperature and Redshift]]

6.5 [[Peculiarities of Thermodynamics in Early Universe|Peculiarities of Thermodynamics in Early Universe]]

6.6 [[The Saha equation|The Saha equation]]

6.7 [[Primary Nucleosynthesis|Primary Nucleosynthesis]]

6.8 [[6_Extras|Extras]]

9. [[:Category:Dark Energy|'''Dark Energy''']]

9.1 [[The Cosmological Constant|The Cosmological Constant]]

9.2 [[Geometry and Destiny|Geometry and Destiny]]

9.3 [[Time-dependent Cosmological Constant|Time-dependent Cosmological Constant]]

9.4 [[Static Einstein's Universe|Static Einstein's Universe]]

9.5 [[Dynamical Forms of Dark Energy|Dynamical Forms of Dark Energy]]

9.5.1 [[Dynamical Forms of Dark Energy#The Quintessence|The Quintessence]]

9.5.2 [[Dynamical Forms of Dark Energy#The K-essence|The K-essence]]

9.5.3 [[Dynamical Forms of Dark Energy#Phantom Energy|Phantom Energy]]

9.5.4 [[Dynamical Forms of Dark Energy#Disintegration of Bound Structures|Disintegration of Bound Structures]]

9.5.5 [[Dynamical Forms of Dark Energy#Big Rip, Pseudo Rip, Little Rip|Big Rip, Pseudo Rip, Little Rip]]

9.5.6 [[Dynamical Forms of Dark Energy#The Statefinder|The Statefinder]]

9.5.7 [[Dynamical Forms of Dark Energy#Crossing the Phantom Divide|Crossing the Phantom Divide]]

9.6 [[Lost and Found|Lost and Found]]

10. [[:Category:Dark Matter|'''Dark Matter''']]

10.1 [[Observational Evidence of the Dark matter Existence|Observational Evidence of the Dark matter Existence]]

10.2 [[ANONIMOUS|ANONIMOUS]]

10.3 [[Dark Matter Halo|Dark Matter Halo]]

10.4 [[Candidates for Dark Matter Particles|Candidates for Dark Matter Particles]]

10.4.1 [[Candidates for Dark Matter Particles#Standard Model Particles as Dark Matter Candidates|Standard Model Particles as Dark Matter Candidates]]

10.4.2 [[Candidates for Dark Matter Particles#Supersymmetric Candidate Particles|Supersymmetric Candidate Particles]]

10.5 [[Dark Matter Detection|Dark Matter Detection]]

10.6 [[The Dark Matter in the Solar System|The Dark Matter in the Solar System]]

10.7 [[The Dark Stars|The Dark Stars]]

10.8 [[Interactions in the Dark Sector|Interactions in the Dark Sector]]

10.8.1 [[Interactions in the Dark Sector#General Analysis|General Analysis]]

10.8.2 [[Interactions in the Dark Sector#Linear Models|Linear Models]]

10.8.3 [[Interactions in the Dark Sector#Non-linear Models|Non-linear Models]]

10.8.4 [[Interactions in the Dark Sector#The Chaplygin Gas|The Chaplygin Gas]]

10.8.5 [[Interactions in the Dark Sector#Universe as the Dynamical System|Universe as the Dynamical System]]

11. [[:Category:Standard Cosmological Model|'''Standard Cosmological Model''']]

11.1 [[Characteristic Parameters and Scales|Characteristic Parameters and Scales]]

11.2 [[Evolution of Universe|Evolution of Universe]]

12. [[:Category:Weak field limit and gravitational waves|'''Weak field limit and gravitational waves''']]

12.1 [[Motivation and symmetries / Introduction|Motivation and symmetries / Introduction]]

12.2 [[Linearized Einstein equations|Linearized Einstein equations]]

12.3 [[Gauge transformations and degrees of freedom|Gauge transformations and degrees of freedom]]

12.4 [[Wave equation|Wave equation]]

12.5 [[Transverse traceless gauge|Transverse traceless gauge]]

12.6 [[Gravitational waves and matter|Gravitational waves and matter]]

12.7 [[Energy of gravitational waves|Energy of gravitational waves]]

12.8 [[Binary systems|Binary systems]]

12.9 [[Gravitational Waves: scale of the phenomenon|Gravitational Waves: scale of the phenomenon]]

12.10 [[Generation and detection of gravitational waves|Generation and detection of gravitational waves]]

13. [[:Category:Observational Cosmology|'''Observational cosmology''']]

13.1 [[Point gravitational lenses|Point gravitational lenses]]

13.2 [[Microlensing and Weak Lensing|Microlensing and weak lensing]]

14. [[:Category:Holographic Universe|'''Holographic Universe''']]

14.1 [[Information, Entropy and Holographic Screen|Information, Entropy and Holographic Screen]]

Particles' motion in general black hole spacetimes

2013-05-01T08:27:39Z

Igor: moved Help:Editing to Particles' motion in general black hole spacetimes: !

[[Category:Black Holes|4]]
__TOC__

In this section we use the $(-+++)$ signature, Greek letters for spacetime indices and Latin letters for spatial indices.

==Frames, time intervals and distances==

In the next several problems we again consider the procedure of measuring time and space intervals by different observers, but in a different, more formal and powerful approach.

<div id="OZ01"></div>
=== Problem 1===

Let a particle move with the four-velocity $U^{\mu }$. It can be viewed as
some observer carrying a frame attached to him. Locally, it defines the
hypersurface orthogonal to it. Show that
\begin{equation}
h_{\mu \nu }=g_{\mu \nu }+U_{\mu }U_{\nu } \label{h}
\end{equation}
is (i) the projection operator onto this hypersurface, and at the same time
(ii) the induced metric of the hypersurface. This means that (i) for any
vector projected at this hypersurface by means of $h^\mu_\nu$, only the components orthogonal to $U^{\mu}$ survive, (ii) the repeated application of the projection operation leaves the vector within the hypersurface unchanged. In other words, $h_{\mu \nu}$ satisfies
\begin{align}
&h^{\mu}_{\nu}U^{\nu }=0 ; \label{1} \\
&h^{\mu}_{\nu }h^{\nu}_{\lambda}=h^{\mu }_{\lambda}. \label{2}
\end{align}

<div id="OZ021"></div>
=== Problem 2===

Let us consider a particle moving with the four-velocity $U^{\mu }$. The
interval $ds^{2}$ between two close events is defined in terms of
differentials of coordinates,%
\begin{equation}
ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }.
\end{equation}

For given $dx^{\mu }$, what is the value of the proper time $d\tau _{obs}$ between the corresponding events measured by this observer? How can one define
locally the notions of simultaneity and proper distance $dl$ for the observer in terms of its four-velocity and the corresponding projection operator $h^\mu_\nu$ ? How is the interval $ds^{2}$ related to $d\tau _{obs}$ and $dl$?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In order to obtain $d\tau _{obs}$, one should project $dx^{\mu}$ onto the
four-velocity:
\begin{equation}
d\tau _{obs}=-dx^{\mu }U_{\mu }.\label{tau}
\end{equation}%
Then, locally, two events are simultaneous when
\begin{equation}
d\tau _{obs}=0. \label{sim}
\end{equation}

The proper distance
\begin{equation}
dl^{2}=h_{\mu \nu }dx^{\mu }dx^{\nu } \label{dl}.
\end{equation}

Then (\ref{h}) implies that
\begin{equation}
ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }=dl^{2}-d\tau _{obs}^{2} \label{ds}.
\end{equation}
</p></div>
</div>

<div id="OZ03"></div>
=== Problem 3===

Let our observer measure the velocity of some other particle passing in its
immediate vicinity. Relate the interval to $d\tau _{obs}$ and the particle's
velocity $w$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The velocity can be defined as
\begin{equation}
w^{2}=\left( \frac{dl}{d\tau _{obs}}\right) ^{2}.
\end{equation}%
Then, it follows from (\ref{dl}), (\ref{ds}) that
\begin{equation}
ds^{2}=d\tau _{obs}^{2}(1-w^{2}). \label{sw}
\end{equation}
</p></div>
</div>

<div id="OZ04"></div>
=== Problem 4===

Analyze the formulas derived in the previous three problems applied to the case of flat spacetime (Minkovskii space) and compare them to
the known formulas of special relativity.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In special relativity four-velocity is expressed through velocity (spatial components are denoted by Latin letters)
\begin{equation}
v^{i}=\frac{dx^{i}}{dt},\qquad i=1,2,3\label{vxt}
\end{equation}
and Lorentz factor
\begin{equation}
\gamma =(1-v^{2})^{-1/2}
\end{equation}
as follows:
\begin{align}
U^{\mu}&=\gamma (+1,v^i),\\
U_{\mu }&=\gamma (-1,v^i).
\end{align}

Let us compare the original frame (with coordinates $t,x^i$) and the one
comoving with the observer (with coordinates $t^{\prime }=\tau _{obs}$ and $x^{i\prime}$.) Then using (\ref{vxt}) we get
\begin{equation}
ds^{2}=-dt^{\prime 2}=-dt^{2}+\delta _{ij}dx^{i}dx^{j}=-dt^{2}(1-v^{2}).
\end{equation}

Eqs. (\ref{tau}), (\ref{vxt}) then give us
\begin{equation}
d\tau _{obs}\equiv dt^{\prime }
=\frac{dt-v_{i}dx^{i}}{\sqrt{1-v^{2}}}
=dt\sqrt{1-v^{2}},
\end{equation}%
which coincides with the standard Lorentz transformation. Here, $dt$ and $dx^{i} $ are the time and coordinate differences between close events
measured in the laboratory frame. The quantities with primes refer to the comoving frame. Then, $\tau _{obs}=t^{\prime }$ has the meaning of
proper time, as in the case when $dx^{i\prime }=0$, and thus $dl^{\prime}=0$, from (\ref{tau}) we get $ds^{2}=-dt^{\prime 2}$.
</p></div>
</div>

<div id="OZ05"></div>
=== Problem 5===

Consider an observer being at rest with respect to a given coordinate frame:
$x^{i}=const$ ($i=1,2,3$). Find $h_{\mu \nu }$, $d\tau _{obs}$, the
condition of simultaneity and $dl^{2}$ for this case. Show that the
corresponding formulas are equivalent to eqs. (84.6), (84.7) of \cite{lan},
where they are derived in a different way.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

If the observer does not move, its four-velocity is
\begin{equation}
U^{\mu }=(U^{0},0,0,0).
\end{equation}%
From the normalization condition one finds
\begin{equation}
g_{00}\left( U^{0}\right) ^{2}=-1,
\end{equation}%
whence
\begin{equation}
U^{\mu }=\frac{1}{\sqrt{-g_{00}}}(1,0,0,0).\label{st}
\end{equation}

The covariant components $U_{\mu }=g_{\mu \nu }U^{\nu }$ then are equal to
\begin{equation}
U_{0}=-\sqrt{-g_{00}},\qquad U_{i}=\frac{g_{i 0}}{\sqrt{-g_{00}}}.
\label{uco}
\end{equation}%
By substitution into (\ref{h}), one finds
\begin{align}
h_{00}&=0,\quad h_{0i}=0,\label{h0}\\
h_{ij}&=g_{ij}-\frac{g_{0i}g_{0j}}{g_{00}}, \label{hi}
\end{align}
which coincides (up to the choice of the overall signature) with the spatial metric $\gamma_{ij}$ as defined in eq. (84.7) of \cite{lan}. Then, $dl^{2}=h_{ij}dx^{i}dx^{j}$ turns into eq. (84.6).

The condition of simultaneity (\ref{sim}) with (\ref{uco}) taken into account reads
\begin{equation}
dt=-dx^{i}\frac{g_{0i}}{g_{00}}
\end{equation}%
which coincides with eq. (84.14) of \cite{lan}.
</p></div>
</div>

<div id="OZ06"></div>
=== Problem 6===

Consider two events at the same point of space but at different values of
time. Find the relation between $dx^{\mu}$ and $d\tau _{obs}$ for such an
observer.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In this case, $dx^{i}=0$, and we obtain from (\ref{tau}) and (\ref{uco}) that
\begin{equation}
d\tau _{obs}=\sqrt{-g_{00}}\;dt
\end{equation}%
which coincides with eq. 84.1 of \cite{lan}.
</p></div>
</div>

==Fiducial observers==

<div id="OZ07"></div>
=== Problem 7===

Consider an observer with
\begin{equation}
U_{\mu }=-N\delta _{\mu }^{0}=-N(1,0,0,0)\text{.} \label{uz}
\end{equation}%
We call it a fiducial observer (FidO) in accordance with \cite{mb}. This
notion is applied in \cite{mb} mainly to static or axially symmetric rotating
black holes. In the latter case it is usually called the ZAMO (zero angular
momentum observer). We will use FidO in a more general context.

Show that a FidO's world-line is orthogonal to hypersurfaces of constant time $t=const$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

A vector $n_\mu$ is called normal to a hypersurface $\Sigma$ if it is orthogonal to any displacement $dx^\mu $ within this hypersurface: $n_\mu dx^\mu =0$. For hypersurfaces of constant time $dx^{\mu}=(0,dx^i)$, so the normal vector is $n_\mu =(n_0 ,0)\sim \partial_t $, with $n_i =0$. Thus, according to (\ref{uz}), $n_{\mu}$ and $U_{\mu}$ are parallel according to (\ref{uz}) and $U_\mu$ is also normal to $t=const$.
</p></div>
</div>

<div id="OZ08"></div>
=== Problem 8===
Find the explicit form of the metric coefficients in terms of the components
of the FidO's four-velocity. Analyze the specific case of axially symmetric metric in
coordinates $(t,\phi ,r,\theta )$ with $g_{0i}=g_{t\phi }\delta _{i}^{\phi }$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

By definition, $U_{i}=0$. Then, the normalization condition gives us%
\begin{equation}
U^{\mu }=\frac{1}{N}(1,N^{i}) \label{zamo}
\end{equation}%
with some $N^{i}$. It has the simple physical meaning of the velocity of the
observer with respect to the coordinate frame $\{t,x^{i}\}$ defined
as
\begin{equation}
V^{i}\equiv \frac{dx^{i}}{dt}\Big| _{obs}=\frac{U^{i}}{U^{0}}=N^{i}.
\label{vn}
\end{equation}

It follows from (\ref{tau}) that%
\begin{equation}
d\tau _{obs}=Ndt .
\end{equation}%
Using that
\begin{equation}
0=U_{i}=\frac{g_{i 0}}{N}+\frac{g_{ij}N^{j}}{N},
\end{equation}%
we find
\begin{equation}
g_{0i}=-g_{ij}N^{j}.
\end{equation}

In a similar way,%
\begin{equation}
-N=U_{0}=\frac{g_{00}}{N}+\frac{g_{0i}N^{i}}{N},
\end{equation}%
whence%
\begin{equation}
g_{00}=-N^{2}+g_{0i}N^{i}N.
\end{equation}

Collecting all this, we obtain that the metric can be written in the form
\begin{equation}
ds^{2}=-dt^{2}N^{2}+g_{ij}(dx^{i}-N^{i}dt)(dx^{j}-N^{j}dt) \label{N},
\end{equation}
and calculating $h_{\mu \nu }$ we find
\begin{equation}
h_{00}=g_{00}+N^{2},\qquad h_{0i}=g_{0i},\qquad h_{ij}=g_{ij}.
\end{equation}

In the case of a stationary axially symmetric metric
\begin{equation}
ds^{2}
=-dt^{2}N^{2}
+g_{\phi \phi }(d\phi -\omega dt)^{2}
+g_{rr}dr^{2}+g_{\theta \theta }d\theta ^{2}.
\end{equation}%
we have
\begin{align}
g_{00}&=-N^{2}+g_{\phi \phi }\omega ^{2}, \label{00m}\\
g_{0\phi }&=-g_{\phi \phi }\omega, \label{03}
\end{align}
and therefore
\begin{align}
h_{00}=\omega ^{2}g_{\phi \phi },
&\qquad h_{0\phi }=g_{0\phi },
\qquad h_{\phi \phi }=g_{\phi \phi };\\
& N^{\phi }=\omega ,\qquad N^{r}=N^{\theta }=0.\label{vf}
\end{align}
</p></div>
</div>

<div id="OZ09"></div>
=== Problem 9===

Consider a stationary metric with the time-like Killing vector field $\xi^\mu =(1,0,0,0)$. Relate the energy $E$ of a particle with four-velocity $u^\mu$ as measured at
infinity by a stationary observer to that measured by a local observer with 4-velocity $U^\mu$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The energy as measured at infinity by a stationary observer is the integral of motion
\begin{equation}
E=-mu_{\mu }\xi ^{\mu }=-mu_{0}.
\end{equation}

On another hand, the energy measured by a local observer is
\begin{equation}
E_{rel}=-mu_{\mu }U^{\mu }\equiv m\gamma =-m(u_{0}U^{0}+u_{i}U^{i}),
\label{rel}
\end{equation}
whence
\begin{equation}
E-m\,u_{i}V^{i}=\frac{E_{rel}}{U^0},\label{e}
\end{equation}%
where $V^{i}=U^i / U^0$ is the local observer's velocity in the given frame (\ref{vn}).
</p></div>
</div>

<div id="OZ10"></div>
=== Problem 10===

Express $E_{rel}$ and $E$ in terms of the relative velocity $w$ between a
particle and the observer (i.e. velocity of the particle in the frame of the observer and vice versa).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Using (\ref{tau}), (\ref{rel}) we find that the Lorentz factor is
\begin{equation}
\gamma =-u_{\mu }U^{\mu }
=-\frac{U^{\mu }dx_{\mu }}{ds}=-\frac{d\tau _{obs}}{ds}.
\end{equation}

It follows from (\ref{sw}) that
\begin{equation}
\gamma =\frac{1}{\sqrt{1-w^{2}}},\qquad
E_{rel}=\frac{m}{\sqrt{1-w^{2}}}.
\end{equation}

Then, we find from (\ref{e}) that
\begin{equation}
E-p_{i}V^{i}=\frac{m}{\sqrt{1-w^{2}}}\frac{1}{U_{0}}, \label{ep}
\end{equation}
where $p^{\mu }=mu^{\mu }$ is the particle's four-momentum.
</p></div>
</div>

<div id="OZ11"></div>
=== Problem 11===

Show that in the flat spacetime eq. (\ref{ep}) is reduced to the usual formula
of the Lorentz transformation.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In this case, $U^{0}=(1-V^{2})^{-1/2}=\gamma $, where $V$ is the absolute value of the velocity of the observer (frame 0) with respect to the stationary laboratory frame. Then the energy of a particle in frame 0 is
\begin{equation}
E_{(0)}=\gamma ( E-\mathbf{p}\cdot\mathbf{V}),
\end{equation}%
This is the standard formula for energy transformation.
</p></div>
</div>

<div id="OZ12"></div>
=== Problem 12===

Find the expression for $E$ for the case of a static observer ($U^{i}=0$).

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Using (\ref{st}) and (\ref{e}) we find that the kinetic term in (\ref{e})
vanishes and
\begin{equation}
E=E_{rel.}\sqrt{-g_{00}}=\frac{m}{\sqrt{1-w^{2}}},
\end{equation}%
which coincides with eq. (88.9) of \cite{lan}.
</p></div>
</div>

<div id="OZ13"></div>
=== Problem 13===

Find the expression for $E$ for the case of the ZAMO observer and, in particular, in case of axially symmetric metric.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\begin{equation}
E-mu_{i}V^{i}=E_{rel} N=\frac{mN}{\sqrt{1-w^{2}}}.
\end{equation}

In the axially symmetric case, the angular momentum of a particle $mu_{\phi}=L$ is conserved, and according to (\ref{vf}), $V^{\phi }=\omega$. Then,
\begin{equation}
E-\omega L=E_{rel}N=\frac{mN}{\sqrt{1-w^{2}}}.
\end{equation}
</p></div>
</div>

==Collision of particles: general relationships==

<div id="OZ14"></div>
=== Problem 14===
Let two particles collide. Define the energy in the center of mass (CM)
frame $E_{c.m.}$ at the point of collision and relate it to $E_{rel}$ and
the Lorentz factor of relative motion of the two particles.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In general, the momenta and velocities of two particles are defined at
different points and this prevents one from constructing well-defined
quantities. However, as collision occurs at one point, both particles have
characteristics defined just in this point. This enables one to construct
the total momentum $P^{\mu }=p_{1}^{\mu }+p_{2}^{\mu }$ at the point of
collision (the subscript $a=1,2$ enumerates the two particles).

A particle's mass is
\begin{equation}
m^{2}=-p^{\mu }p_{\mu },
\end{equation}
and likewise the full energy (mass) of two particles at the point of collision, which is also their energy in the center of mass frame, is found from
\begin{equation}
E_{c.m.}^{2}=-P^{\mu }P_{\mu }.\label{cm}
\end{equation}%
It is is the direct generalization of the corresponding formula of special relativity. Eq (\ref{cm}) implies
\begin{equation}
E_{c.m.}^{2}=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma ,
\end{equation}%
where
\begin{equation}
\gamma =-u_{1}^{\mu }u_{2\mu }
\end{equation}
is the Lorentz factor of the two particles' relative motion. It can be expressed in terms of energy $E_{rel}(2,1)$ of particle 1 with respect to particle 2 and vice versa
\begin{equation}
\gamma=\frac{E_{rel.}(1,2)}{m_{1}}=\frac{E_{rel.}(2,1)}{m_{2}}.
\end{equation}
</p></div>
</div>

<div id="OZ1"></div>
=== Problem 15===

Let us consider a collision of particles 1 and 2 viewed from the frame
attached to some other particle 0. How are different Lorentz factors related
to each other? Analyze the case when the laboratory frame coincides with
that of particle 0.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Let us introduce notation
\begin{equation*}
\gamma (0,1)=\gamma _{1}, \quad
\gamma (0,2)=\gamma _{2}, \quad
\gamma (1,2)=\gamma , \quad
u_{0}^{\mu }=U^{\mu}.
\end{equation*}

Each particle's four velocity can be decomposed into components parallel and orthogonal to $U^\mu$ as follows
\begin{equation}
u_{1}^{\mu }=a_{1}U^{\mu }+b_1 n_{1}^{\mu}, \label{ab}
\end{equation}%
where $n^{\mu }$ is a unit space-like vector orthogonal to $U^{\mu }$. It is easy to
find from (\ref{ab}) that%
\begin{equation}
a_{1}=-u_{1}^{\mu }U_{\mu }=\gamma _{1},
\end{equation}%
and from the normalization conditions for $u_{1}^{\mu }$ and $U^{\mu }$ then we see that %$b_1$ as introduced in (\ref{ab}) is the velocity of particle 1 in the frame of particle 0:
\begin{equation}
b_{1}^{2}=\frac{1}{\gamma _{1}^{2}-1 },
\end{equation}
so in terms of velocity of particle 1 in the frame of particle 0
\begin{equation}
w_1 =(1-\gamma_1^{-2})^{1/2}=\frac{b_1}{\gamma},
\end{equation}
the decomposition takes form
\begin{equation}
u_{1}^{\mu }=\gamma_{1}(U^{\mu }+w_1 n_{1}^{\mu}).
\end{equation}

Writing down for particle 2 the equation analogous to (\ref{ab}) and
calculating the scalar product $u_{1}^{\mu }u_{2\mu }$, one obtains
\begin{align}
\gamma &=\gamma _{1}\gamma _{2}
-b_1 b_2 (n_{1}^{\mu }n_{2\mu})\\
&=\gamma _{1}\gamma _{2}\big(1-\varepsilon\; w_{1}w_{2}\big),
\label{ga}
\end{align}
where
\begin{equation}
\varepsilon =(n_{1}^{\mu }n_{2\mu }),\qquad |\varepsilon|\leq 1 ,
\end{equation}
or in terms of velocities
\begin{equation}
\frac{1}{\sqrt{1-w^{2}}}
=\frac{1}{\sqrt{1-w_{1}^{2}}}
\frac{1}{\sqrt{1-w_{2}^{2}}}
(1-\varepsilon w_{1}w_{2}). \label{gw}
\end{equation}

In case particle 0 is at rest in the laboratory frame $U^{\mu }=(1,0)$, we have $n_{a}^{\mu}=(0,\vec{n}_{a})$, $a=1,2$, and
\begin{equation}
u_{a}^{\mu }=\gamma _{a}(1,w_{a}\vec{n}_{a}),\qquad \varepsilon =(\vec{n}_{1}%
\vec{n}_{2}).
\end{equation}%
Then, eqs. (\ref{ga}), (\ref{gw}) turn into those listed in problem 1.3 of \cite{li}.
</p></div>
</div>

<div id="OZ16"></div>
=== Problem 16===

When can $\gamma $ as a function of $\gamma _{1}$ and $\gamma _{2}$ grow
unbounded? How can the answer be interpreted in terms of relative velocities?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

1) If $\gamma _{1}$ and $\gamma _{2}$ are finite, it follows from (\ref{ga})
that $\gamma $ is also finite. It means that if $w_{1}$ and $w_{2}$ are
finite, the relative velocity $w$ of particles 1 and 2 is also finite (in
the sense that it is separated from $c$).

2) Let $\gamma _{1}\rightarrow \infty $ but $\gamma _{2}$ remain finite.
Then,%
\begin{equation}
\gamma \approx \gamma _{1}\big( \gamma _{2}-\varepsilon \sqrt{\gamma _{2}^{2}-1}\;\big)
\end{equation}%
grows unbound irrespective of $\varepsilon $. It means that the relative
velocity of particles, one of which moves with some finite speed and the
other one almost with the speed of light is always close to the speed of
light.

3) Let $\gamma _{1}\rightarrow \infty $, $\gamma _{2}\rightarrow \infty $.

3a) If $\varepsilon \neq +1$,
\begin{equation}
\gamma \approx \gamma _{1}\gamma _{2}(1-\varepsilon )\rightarrow \infty ,
\qquad w\rightarrow 1.
\end{equation}

3b) Let $\varepsilon =+1$. Then
\begin{equation}
\gamma \approx \frac{1}{2} \Big(\frac{\gamma _{1}}{\gamma _{2}} +\frac{%
\gamma _{2}}{\gamma _{1}}\Big).
\end{equation}

We see that $\gamma $ remains finite if $\gamma _{1}/\gamma _{2}$ does so.
Otherwise, $\gamma \rightarrow \infty $. The resulting relative velocity
depends on the relative rate with which $w_{1}$ and $w_{2}$ approach the
speed of light.

Thus in case 3, motion in different directions always gives $\gamma
\rightarrow \infty $, whereas motion in the same direction may or may not
lead to divergent $\gamma $.
</p></div>
</div>

<div id="OZ1"></div>
=== Problem 17===

A tetrad basis, or the orthonormal tetrad, is the set of four unit vectors $h_{(a)}^\mu$ (subscripts in parenthesis $a=0,1,2,3$ enumerate these vectors), of which one, $h_{(0)}^\mu$, is timelike, and three vectors $h_{(i)}^\mu$ ($i=1,2,3$) are spacelike, so that
\begin{equation}
g_{\mu\nu}h_{(a)}^\mu h_{(b)}^\nu =\eta_{ab},\qquad a,b=0,1,2,3.
\end{equation}
A vector's tetrad components are
\begin{equation}
u_{(a)}=u_\mu h_{(a)}^\mu,\qquad u^{(b)}=\eta^{ab}u_{(b)}.
\end{equation}

Define the local three-velocities with the help of the tetrad basis attached to the observer, which
would generalize the corresponding formulas of special relativity.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In special relativity,%
\begin{equation}
u^{\mu }=\frac{dx^{\mu }}{ds}
=\gamma (1,v^{i}),\qquad
\gamma =\frac{1}{\sqrt{1-v^{2}}}.
\end{equation}%
Thus $v^{i}=u^{i}\sqrt{1-v^{2}}$, and using (\ref{sw}), we get
\begin{equation}
v^{i}=\frac{dx^{i}}{d\tau _{obs}}.\label{vi}
\end{equation}

Let us attach a tetrad to some observer with four-velocity $U^\mu$. This means simply that we choose the first, time-like, tetrad vector to be along its world-line
\begin{equation}
h_{(0)}^\mu =U^\mu .
\end{equation}
Then, the natural definition for velocity in the frame of this observer, taking into account (\ref{tau}), will be
\begin{equation}
v^{(i)}=\frac{h_{\mu }^{(i)}dx^{\mu}}{-h^{\mu}_{(0)}dx_{\mu }}.
\end{equation}%
It can be rewritten as
\begin{equation}
v^{(i)}=\frac{u^{\mu }h_{\mu }^{(i)}}{u^{\mu }h^{(0)}_\mu}.
\end{equation}

The tetrad components of the projection operator (\ref{h}) are
\begin{equation}
h_{(ab)}=\eta_{ab}+\eta_{0a}\eta_{0b},
\end{equation}
from which it is easy to check explicitly that the two definitions, through $h_{\mu\nu}$ and through $h^\mu_{(a)}$, are equivalent.
</p></div>
</div>

<div id="OZ18"></div>
=== Problem 18===

Derive the analogues of formulas (\ref{e}), (\ref{ep}) for massless particles (photons). Analyze the cases of static and ZAMO observers.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In this case, $m=0$, the velocity of a photon is always 1, and the notion of
a frame comoving with the photon does not have sense. Nonetheless, some
formulas retain their validity with $E$ replaced by the frequency $\nu$ in
accordance with the fact that for photons $E=\hbar\nu$ (we use $\nu=2\pi/\text{(period)}$ for the angular frequency here, so as not to confuse it with the metric function $\omega$).

Let us consider a photon with wave four-vector $k_{\mu }$ in a gravitational
field. In the case of static or stationary metric, $\nu _{0}=-k_{\mu }\xi^{\mu }$ is conserved along the trajectory, where $\xi ^{\mu }$ is the
Killing vector. If it corresponds to time translations, $\nu$ has the
meaning of frequency. It is what is measured by a remote observer (if the
flat infinity exists). In parallel to derivation of (\ref{e}), one can obtain
\begin{equation}
\nu _{0}-k_{i}V^{i}=\frac{\nu }{U^{0}}, \label{nu}
\end{equation}%
where $\nu =-k_{\mu }U^{\mu }$, and $U^{\mu }$ is the velocity of the
observer.

For the static observer $V^{i}=0$, $U^{0}=\frac{1}{\sqrt{-g_{00}}}$, and
\begin{equation}
\nu_{0}=\nu \sqrt{-g_{00}}. \label{0g}
\end{equation}

For the ZAMO observer, $V^{\phi }=\omega $, $k_{\phi }=L$ is the angular
momentum which is conserved in the axially symmetric case (in units of $\hbar$), $U^{0}=\frac{1}{N}$ (see eq. (\ref{zamo})). Then
\begin{equation}
\nu _{0}-\omega L=\nu N.
\end{equation}
</p></div>
</div>

<div id="OZ19"></div>
=== Problem 19===

The ergosphere is a surface defined by equation $g_{00}=0$. Show that it is
the surface of infinite redshift for an (almost) static observer.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

From (\ref{0g}) it follows that in the limit $g_{00}\rightarrow 0$ the
frequency $\nu _{0}\rightarrow 0$ for any finite $\nu $. However, for a
nonstatic observer this is not true.
</p></div>
</div>

<div id="OZ20"></div>
=== Problem 20===

Consider an observer orbiting with a constant angular velocity $\Omega $ in
the equatorial plane of the axially symmetric back hole. Analyze what
happens to redshift when the angular velocity approaches the minimum or
maximum values $\Omega _{\pm }$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Eq. (\ref{nu}) reads in this case%
\begin{equation}
\nu _{0}-\Omega L=\frac{\nu }{U^{0}}\text{,}
\end{equation}%
where $U^{0}$ obeys the normalization condition%
\begin{equation}
Y\equiv g_{00}+2g_{0\phi }\Omega +g_{\phi \phi }\Omega ^{2}=-\frac{1}{\left(
U^{0}\right) ^{2}}\text{.}
\end{equation}

It can be rewritten in the form%
\begin{align}
Y=&g_{\phi \phi }(\Omega ^{2}-2\omega \Omega +\frac{g_{00}}{g_{\phi \phi }})
=g_{\phi \phi }(\Omega -\Omega _{+})(\Omega -\Omega _{-});\\
&\Omega _{\pm }=\omega \pm \frac{N}{\sqrt{g_{\phi \phi }}},
\end{align}
Then,%
\begin{equation}
\nu _{0}-\Omega L=\nu \sqrt{g_{\phi \phi }(\Omega _{+}-\Omega )(\Omega
-\Omega _{-})}\text{.}
\end{equation}

When $g_{00}\rightarrow 0$, $\Omega _{-}\approx \frac{g_{00}}{2\omega }%
\rightarrow 0$.

Thus, if, say, $\Omega \rightarrow \Omega _{-}$, $\nu _{0}\rightarrow \Omega
_{-}L$. If $L=0$, then $\nu _{0}\rightarrow 0$.
</p></div>
</div>

<div id="OZ21"></div>
=== Problem 21===

Let two massive particle 1 and 2 collide. Express the energy of each particle
in the centre of mass (CM) frame in terms of their relative Lorentz factor $\gamma (1,2)$. Analyze the limiting cases of ultra-relativistic $\gamma
(1,2)\rightarrow \infty $ and non-relativistic $\gamma (1,2)\approx 1$ collisions.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The energy of particle 1 in the CM frame is
\begin{equation}
\left( E_{1}\right) _{c.m.}=m_{1}\gamma (1,CM) \label{e1}
\end{equation}%
where%
\begin{equation}
\gamma (1,CM)=-u_{1\mu }U_{c.m.}^{\mu }
\end{equation}%
is its Lorentz factor in the same frame, which has four-velocity
\begin{equation}
U_{c.m.}^{\mu }=\frac{P^{\mu }}{\mu}.
\end{equation}
Here $P^\mu$ is the total momentum and $\mu \equiv E_{c.m.}$ is the energy in the center of
mass frame introduced in (\ref{cm})%problem 14
\begin{align}
P^{\mu }&=m_{1}u_{1}^{\mu }+m_{2}u_{2}^{\mu };\\
\mu ^{2}&=m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\gamma (1,2).\label{mu}
\end{align}
%\begin{equation}
%\gamma (1,2)=-u_{1\mu }u_{2}^{\mu }.
%\end{equation}%
Then
\begin{align}
\gamma (1,CM)&=\frac{m_{1}+m_{2}\gamma (1,2)}{\mu };\label{g1}\\
( E_{1}) _{c.m.}
&=m_{1}\gamma (1,CM).\label{e1cm}
\end{align}

Likewise, for the second particle
\begin{align}
\gamma (2,CM)&=\frac{m_{2}+m_{1}\gamma (1,2)}{\mu };\\
( E_{2}) _{c.m.}&=m_{2}\gamma (2,CM).\label{e2cm}
\end{align}

It follows from (\ref{e1cm}), (\ref{e2cm}) that%
\begin{equation}
( E_{1}) _{c.m.}+( E_{2}) _{c.m.}=\mu ,
\end{equation}
as it should be.

In the ultra-relativistic limit $\gamma (1,2)\rightarrow \infty $
\begin{equation}
\gamma (1,CM)\approx
\frac{\mu }{2m_{1}}\approx \sqrt{\frac{m_{2}}{2m_{1}}\gamma (1,2)}.
\end{equation}

In the opposite limit of small relative velocity $w(1,2)\ll 1$ we obtain $\gamma (1,2)\approx 1+\frac{w^{2}}{2}$ and
\begin{align}
&\mu \approx (m_{1}+m_{2})+\frac{m_{1}m_{2}}{2(m_{1}+m_{2})}w^{2};\\
&w(1,CM)\approx w\frac{m^{2}}{(m_{1}+m_{2})},
\end{align}
which agrees with formulas of nonrelativistic mechanics -- see Ch. 3, Sec. 13 of \cite{lan}.
</p></div>
</div>

<div id="OZ22"></div>
=== Problem 22===

For a stationary observer in a stationary space-time the quantity $\alpha =(U^0)^{-1}$ is the redshifting factor: if this observer emits a ptoton with frequency $\omega_{em}$, it is detected at infinity by another stationary observer with frequency $\omega_{det}=\alpha \omega_{em}$. For a generic observer this interpretation is invalid, however, $\alpha =ds/dt$ still determines the time dilation for this observer, and thus can still be called the same way. Express the redshifting factor of the center of mass frame $\alpha _{c.m.}$ through the redshifting factors of the colliding particles $\alpha_{1}$ and $\alpha_2$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Starting from the definition,
\begin{equation}
\frac{1}{\alpha _{c.m.}}
=\left( U^{0}\right) _{c.m.}
=\frac{m_{1}U_{1}^{0}+m_{2}U_{2}^{0}}{\mu }
=\frac{\frac{m_{1}}{\alpha _{1}}+\frac{m_{2}}{\alpha _{2}}}{\mu },
\end{equation}%
whence%
\begin{equation}
\alpha _{c.m.}=\frac{\mu \alpha _{1}\alpha _{2}}{m_{1}\alpha
_{2}+m_{2}\alpha _{1}}.
\end{equation}

If case $\alpha _{2}\ll \frac{m_{2}}{m_{1}}\alpha _{1}$, we get
\begin{equation}
\alpha _{c.m.}\approx \frac{\mu \alpha _{2}}{m_{2}}.
\end{equation}
</p></div>
</div>

<div id="OZ23"></div>
=== Problem 23===

Relate the energy of a particle at infinity $E_{1}$, its energy at the point of collision in the C.M. frame $(E_{1})_{c.m.}$ and $\mu$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

From (\ref{e})
we know that
\begin{equation}
E_{1}-mu_{1i}v_{2}^{i}=\alpha _{2}E_{rel}, \label{e2}
\end{equation}%
where $v^{i}$ is the velocity of particle 2 in the stationary frame, $\alpha _{2}=\frac{1}{U_{2}^{0}}$.

A particle's energy in the center of mass frame can be found from (\ref{e1cm})
\begin{equation}
(E_{1})_{c.m.}\mu =m_{1}^{2}+m_{2}E_{rel}(1,2),
\end{equation}%
where $E_{rel}(1,2)=m_{1}\gamma (1,2)$, so in terms of $(E_1)_{c.m.}$ we get
\begin{align}
E_{1}-m_{1}u_{1i}v_{2}^{i}
&=\frac{\alpha_{2}(\mu E_{1})_{c.m.} -m_{1}^{2}\alpha _{2}}{m_{2}};\\
E_{2}-m_{2}u_{2i}v_{1}^{i}
&=\frac{\alpha _{1}(\mu E_{2})_{c.m.} -m_{2}^{2}\alpha _{1}}{m_{1}}.
\end{align}

On another hand, applying (\ref{e}) to particle 1 and the center of mass frame, we obtain
\begin{align}
E_{1}-m_{1}u_{1i}V_{c.m.}^{i}
&=\alpha _{c.m.}(E_{1}) _{c.m.}; \label{1cm}\\
E_{2}-m_{2}u_{2i}V_{c.m.}^{i}
&=\alpha _{c.m.}(E_{2}) _{c.m.}.\label{2cm}
\end{align}

The sum of (\ref{1cm}) and (\ref{2cm}) is the total conserved energy $E=E_1 +E_2$:
\begin{equation}
E-\alpha _{c.m.}P_{i}U_{c.m.}^{i}=\alpha _{c.m.}\mu .
\end{equation}%
Taking into account that $\frac{\alpha _{c.m.}}{\mu }=\frac{1}{P^{0}}$, this is equivelent to
\begin{equation}
EP^{0}-P_{i}P^{i}=-P^{\mu }P_{\mu }=\mu^{2},
\end{equation}%
as it should be.
</p></div>
</div>

<div id="OZ24"></div>
=== Problem 24===

Solve the same problem when both particles are massless (photons). Write
down formulas for the ZAMO observer and for the C.M. frame.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Let $k^\mu_1$ and $k^\mu_2$ be the wave-vectors of two photons. Then vector
\begin{equation}
K^{\mu }=\left( k^{\mu }\right) _{1}+\left( k^{\mu }\right) _{2} \label{K}
\end{equation}
is \emph{time-like} (unless the two photons are collinear, but then they would not collide). Dividing by the energy $\mu$ in the center of mass frame
\begin{equation}
\mu ^{2}=-K^{\mu }K_{\mu }=-2k_{1\mu }k^{2\mu },
\end{equation}
we obtain the frame's (time-like) four-velocity
\begin{equation}
U_{c.m.}^{\mu }=\frac{K^{\mu }}{\mu }.
\end{equation}%
There is no Lorentz factor or relative velocity for two photons.

The contraction
\begin{equation}
\tilde{\gamma}(1,obs)=-k_{\mu } U^{\mu }
\end{equation}
has the meaning of the frequency of a photon with four-vector $k^\mu$ as measured by the observer with four-velocity $U^\mu$.

Then frequency measured in the CM frame is
\begin{equation}
\left( \nu _{1}\right) _{c.m.}=-k_{1\,\mu} U_{c.m.}^{\mu },
\end{equation}
and taking into account (\ref{mu}) and (\ref{K}), we obtain
\begin{equation}
(\nu _{1}) _{c.m.}
=-k_{1\,\mu}U_{c.m.}^{\mu }
=-\frac{k_{1\,\mu }k_{2}^{\mu }}{\mu }=\frac{\mu }{2},
\end{equation}%
as for photons $k_{i\,\mu}k_i^{\mu}=0$. This is quite natural, since for photons the energy is equal to the absolute value of the momentum,
so in the CM frame the energies of both photons should be equal.

Likewise,
\begin{equation}
( \nu _{2}) _{c.m.}
=-k_{2\,\mu}U_{c.m.}^{\mu }
=\frac{\mu }{2}=(\nu_{1}) _{c.m.},
\end{equation}
and
\begin{equation}
\mu =( \nu _{1}) _{c.m.}+( \nu _{2}) _{c.m.}
=2(\nu _{1}) _{c.m.}=2(\nu _{2}) _{c.m.}
\end{equation}

If $\xi$ is the Killing vector field responsible for time translation, the conserved frequency of a photon, equal to the frequency measured at infinity by a stationary observer, is
\begin{equation}
\nu _{\infty }=-k_{\mu }\xi ^{\mu }=-k_{0}.
\end{equation}
In some arbitrary frame it is
\begin{equation}
\nu _{loc}=-k_{\mu }U^{\mu }=-k_{0}U^{0}-k_{i}U^{i}.
\end{equation}

Then
\begin{equation}
\nu _{\infty }-k_{\alpha}V^{\alpha}=\alpha \nu _{loc},\qquad
V^{i}=\frac{U^{i}}{U^{0}},\qquad
\alpha =\frac{1}{U^{0}},\label{obs}
\end{equation}%
the same as for the case of massive particle. The difference is only that for photons we
cannot express $\nu _{loc}$ in terms of mass and velocity as $\frac{m}{\sqrt{1-w^{2}}}$.

ZAMO frame%
\begin{equation}
\nu _{\infty }-\omega L=N\nu _{loc}.
\end{equation}

CM frame
\begin{align}
&\nu _{\infty }-k_{i}V^{i}=\alpha _{c.m.}\frac{\mu }{2},\\
&\alpha _{c.m.}=\frac{1}{U^{0}}=\frac{\mu }{k_{1}^{0}+k_{2}^{0}}.
\end{align}
</p></div>
</div>

Help:Editing

2013-05-01T08:27:39Z

Igor: moved Help:Editing to Particles' motion in general black hole spacetimes: !

#REDIRECT [[Particles' motion in general black hole spacetimes]]