Cosmo All: Created page with "TOC
New from March-2015

'''Problem 1'''

2015-04-06T22:18:40Z

Created page with "TOC <p style= "color: #999;font-size: 14px">New from March-2015</p> <div id="2501_01"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p st..."

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<p style= "color: #999;font-size: 14px">New from March-2015</p>

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'''Problem 1'''
<p style= "color: #999;font-size: 11px">problem id: 2501_01</p>
Show that the luminosity distance can be in general presented in the following form
\[d_L=\frac{1+z}{H_0\sqrt{\Omega_{k0}}}\sinh\left({H_0\sqrt{\Omega_{k0}}}\int\limits_0^z\frac{dz}{H(z)}\right),\]
where $\Omega_{k0}$ is relative contribution of the space curvature.
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<p style="text-align: left;">Use the relation
\[\rho_{curv}=-\frac3{8\pi G}\frac{k}{R^2},\]
where $R$ is the (comoving) radius of curvature of the open or closed Universe, to find
\[\Omega_{k0}=-\frac{k}{RH_0}.\]
It is easy to see that the expression
\[d_L=\frac{1+z}{H_0\sqrt{\Omega_{k0}}}\sinh\left({H_0\sqrt{\Omega_{k0}}}\int\limits_0^z\frac{dz}{H(z)}\right)\]
reproduces the commonly known expression for the luminosity distance
\[
d_L(z)=(1+z)\left\{
\begin{array}{lr}
R\sinh\left[\frac1{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & open\\
H_0^{-1}\int\limits_0^z\frac{dz'}{E(z')}, & flat\\
R\sin\left[\frac1{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & closed
\end{array}
\right.\]</p>
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'''Problem 2'''
<p style= "color: #999;font-size: 11px">problem id: 2501_02</p>
Give physical interpretation of the conservation equation \[\dot{\rho}+3(\rho+p)H=0.\]
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<p style="text-align: left;">The energy density has time dependence determined by the conservation equation
\[\dot{\rho}=-3H\rho-3Hp.\]
The two terms determine behavior of the homogeneous fluid which contains the energy in a dynamic Universe. The $H$ term provides the "friction", while the density term tracks the reduction in density due to the volume increase during expansion and the pressure term tracks the reduction in pressure energy during expansion.
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'''Problem 3'''
<p style= "color: #999;font-size: 11px">problem id: 2501_03</p>
Show that for both case of matter and radiation domination, the acceleration $\ddot a$ slows as the scale factor grows.
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<p style="text-align: left;">From the second Friedmann equation
\[\frac{\ddot a}{a}=-\frac{4\pi G}{3}(\rho+3p)\]
clearly shows that both for matter and for radiation domination, the acceleration is negative; an expanding Universe dominated by the energy density of matter or radiation will decelerate. The acceleration in a matter dominated phase goes as $a^{-2}$ while in a radiation dominated phase it goes as $a^{-3}$. In either case the deceleration is large when $a(t)$ is small and then slows as the scale factor grows.
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'''Problem 4'''
<p style= "color: #999;font-size: 11px">problem id: 2501_04</p>
Obtain relation between the cosmological and conformal time for the Universe dominated by matter, radiation and cosmological constant respectively.}
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<p style="text-align: left;">\begin{tabular}{|l|l|l|l|l|}
\hline
& $\rho(a)$ & $a(t)$ & $\eta$ & $H$\\
\hline
matter & $1/a^3$ & $t^{2/3}$ & $t^{1/3}$ & $(2/3)/t$\\
\hline
radiation & $1/a^4$ & $t^{1/2}$ & $t^{1/2}$ & $(1/2)/t$\\
\hline
$\Lambda$ & constant & $e^{Ht}$ & $C(e^{Ht}-1)/(aH)$ & constant\\
\hline
\end{tabular}</p>
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'''Problem 5'''
<p style= "color: #999;font-size: 11px">problem id: 2501_05</p>
Show that for the power-paw expansion $a(t)\propto t^\alpha$ with $\alpha<1$ (decelerated expansion) the Hubble radius grows faster than the Universe expands.
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<p style="text-align: left;">Indeed, in the case of the power-paw expansion with $\alpha<1$ (decelerated expansion) the Hubble radius grows faster than the Universe expands: $R_H=H^{-1}\propto t$ while $a(t)\propto t^\alpha$. In power law situations the Hubble radius has expansion velocity greater than the light speed \[\frac{d}{dt}\left(\frac1H\right)=\frac1\alpha.\] This behavior is true only for a decelerating Universe composed of matter and radiation. This is not a physical velocity, violating special relativity, but the velocity of expansion of the metric itself.
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'''Problem 6'''
<p style= "color: #999;font-size: 11px">problem id: 2501_06</p>
Why the cosmological constant cannot be used as a source for inflation in the inflation model?
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<p style="text-align: left;">The cosmological constant do provide the accelerated expansion of Universe needed to realize the inflation: $a(t)\propto e^{Ht}$, $q=-1$. However, in approximation of the cosmological constant, $H$ is constant for all time. Therefore a dynamical mechanism for the limited time of inflation is needed. The physical mechanism for the existence of an approximately constant value of $H$ which lasts for a limited time is given by a scalar field. For a large initial potential energy of scalar field the state equation parameter $w\approx-1$ and the scalar field in process of the "slow roll" imitates the cosmological constant for sufficiently long period of time to solve the flatness and causal problems. Then due to shape of the potential the scalar field exits from the slow-roll regime, oscillates about its' potential minimum decaying into less massive particles insuring that inflation time is finite.</p>
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'''Problem 7'''
<p style= "color: #999;font-size: 11px">problem id: 2501_07</p>
Derive the following formula \[\Delta v\equiv\frac{\Delta z}{1+z}=H_0\Delta t_o \left[1-\frac{E(z)}{1+z}\right].\]
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'''Problem 9'''
<p style= "color: #999;font-size: 11px">problem id: 2501_09</p>
Show that inflation ends when the parameter \[\varepsilon\equiv\frac{M_P^2}{16\pi}\left(\frac{dV}{d\varphi}\frac1V\right)^2=1.\]
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<p style="text-align: left;">Using definition of the parameter $\varepsilon$ one finds
\[\varepsilon=-\frac{\dot H}{H^2}.\]
In the slow-roll approximation
\[H^2=\frac{8\pi}{3M_P^2V},\]
\[3H\dot\varphi=-\frac{dV}{d\varphi}.\]
Therefore
\[\frac{\ddot a}a=H^2(1-\varepsilon).\]
The condition $\varepsilon=1$ is equivalent to $\ddot a=0$. When the value $\varepsilon=1$ is reached due to variation of the potential shape the Universe exits the regime of the accelerated expansion (inflation). Around tend, the inflaton field(s) typically begin oscillating around the minimum of the potential.
(see Inflation and the Higgs Scalar, DANIEL GREEN ([http://arxiv.org/abs/1412.2107 1412.2107]).)</p>
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'''Problem 10'''
<p style= "color: #999;font-size: 11px">problem id: 2501_10</p>
How does the number of e-folds $N$ depend on the slow-roll parameter $\varepsilon$?
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<p style="text-align: left;">\[N\propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi\frac{V}{dV/d\varphi} \propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi/\sqrt\varepsilon.\]
The number of $e$-folds depends on how fast the field is $f$-decreasing. The number of e-folds, $N$ , is inversely proportional to the square root of the slow roll parameter $\varepsilon$ or proportional to the inverse fractional change of the potential with the field, $V/(dV/d\varphi)$.</p>
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'''Problem 11'''
<p style= "color: #999;font-size: 11px">problem id: 2501_11</p>
The exponential increase in $a(t)$ (during the inflation) drastically reduces the temperature since $Ta$ is a constant. After the field disappears, the Universe will need to re-heat to the high temperatures needed to create the light nuclei whose relative abundance is predicted by BB cosmology and is a major success of the BB model.
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'''Problem 17'''
<p style= "color: #999;font-size: 11px">problem id: 2501_17</p>
Suppose that $dq/dt=f(q)$. Find the Hubble parameter in terms of $q$.
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<p style="text-align: left;">\[1+q=\frac{d}{dt}\left(\frac1H\right),\quad \frac{dq}{dt}=f(q)\to dt=\frac{dq}{f(q)},\]
\[\frac1{H(q)}=\int\frac{1+q}{f(q)}dq.\] (see S. Carloni et al., A new approach to reconstruction methods in f(R) gravity, arXiv:1005.1840)
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'''Problem 18'''
<p style= "color: #999;font-size: 11px">problem id: 2501_18</p>
Show that derivative w.r.t. the cosmic time can be related to that w.r.t. redshift as follows: \[\frac1{f(t)}\frac{df(t)}{dt}=-(1+z)H(z)\frac1{f(z)}\frac{df(z)}{dz}.\]
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=New in Observational Cosmology=

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'''Problem 1'''
<p style= "color: #999;font-size: 11px">problem id: 2501_01o</p>
Obtain relations between velocity of cosmological expansion and redshift.
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<p style="text-align: left;">The exact velocity-proper distance relation (Hubble law) reads
\[V_{exp}=HR.\]
For the observer on the Earth living at the time $t=t_0$ the observed redshift serves as a measure of distance at this epoch. For photons moving along the comoving coordinate $r$, $cdt=\pm adr$. Consequently
\[a_0dr=\frac{cdz}{H(z)}.\]
Integration over the redshift gives the proper distance-redshift relation
\[R(t_0,z)\equiv R(z)=\int\limits_0^z\frac{cdz'}{H(z')}.\]
Using the $R(z)$ relation one gets the exact velocity-redshift relation
\[V_{exp}(z)=H_0 R(z)=H_0\int\limits_0^z\frac{cdz'}{H(z')}.\]
where the expansion velocity is taken for an object with the redshift $z$ observed at the time $t=t_0$.</p>
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'''Problem 2'''
<p style= "color: #999;font-size: 11px">problem id: 2501_02o</p>
Why the Linear Distance-Redshift Law in Near Space?
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<p style="text-align: left;">Consider a nearby galaxy and the change of the scale factor during the small time interval $dt$ when the light travels from this galaxy to the observer:
\[a(t_{obs})=a(t_{em})+\left.\frac{da}{dt}\right|_{t=t_{em}}dt.\]
When the distance is short and the expansion speed much less than the speed of light, then the distance $R$ measured by the astronomer using any of the available ways is practically equal to the metric distance $a(t)r$ and the time interval can be approximated as $dt\approx a(t)r/c=R/c$. From this it follows that
\[a(t_{obs})/a(t_{em})=1+z=1+\frac{\dot a}a \frac R c.\]
and finally
\[cz=HR.\]</p>
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'''Problem 3'''
<p style= "color: #999;font-size: 11px">problem id: 2501_03o</p>
Find the exact relativistic Doppler velocity-redshift relation.
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<p style="text-align: left;">\[(1+z)_{Dop}=\sqrt{\frac{c+V}{c-V}}.\]
The exact relativistic Doppler velocity-redshift relation is
\[V_{Dop}(z)=c\frac{2z+z^2}{2+2z+z^2}.\]
For $z\to\infty$, the velocity $V_{Dop}(z)\to c$ corresponding to the limit $V_{Dop}le c$.

\item \label{2501_04o} {\bf Show, that $V_{Dop}$ and $V_{exp}$ give the same results only in the first order of $V/c$.}

\item \label{2501_05o} {\bf Find $V_{exp}(z)$ for the three cosmological models: Einstein-de Sitter, Milne and de Sitter.}

For $\Omega_m=1$, $\Omega_\Lambda=0$: \[V_{exp}(z)=H_0 R(z)=H_0\int\limits_0^z\frac{cdz'}{H(z')}=\frac{2(\sqrt{1+z}-1)}{\sqrt{1+z}}c.\]

For $\Omega_m=0$, $\Omega_\Lambda=0$: \[V_{exp}(z)=\frac{z(1+z/2)}{(1+z)}c.\]

For $\Omega_m=0$, $\Omega_\Lambda=1$: \[V_{exp}(z)=zc.\]</p>
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=NEW 2=

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'''Problem 1'''
<p style= "color: #999;font-size: 11px">problem id: new2015_3</p>
The lookback time is defined as the difference between the present day age of the Universe and its age at redshift $z$, i.e. the difference between the age of the Universe at observation $t_0$ and the age of the Universe, $t$, when the photons were emitted. Find the lookback time for the Universe filled with non-relativistic matter, radiation and a component with state equation $p=w(z)\rho$.
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<p style="text-align: left;">From the definitions of redshift $1+z=1/a$ we have
\[\frac{dz}{dt}=-\frac{dot a}{a^2}=-H(z)(1+z)\]
or
\[dt=-\frac{dt}{H(z)(1+z)}.\]
The lookback time is defined as
\[t_0-t=\int\limits_t^{t_0}dt=\int\limits_0^z\frac{dz'}{H(z')(1+z')}=\frac1{H_0}\int\limits_0^z\frac{dz'}{E(z')(1+z')}\]
where
\[E(z)=\sqrt{\Omega_r(1+z)^4+\Omega_m(1+z)^3+\Omega_k(1+z)^2+\Omega_w\exp\left( 3\int\limits_0^zdz'\frac{1+w(z')}{1+z'}\right)}.\]
From the definition of lookback time it is clear that the cosmological time or the time back to the Big Bang, is given by
\[t(z)=\int\limits_z^\infty\frac{dz'}{H(z')(1+z')}.\]
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'''Problem 2'''
<p style= "color: #999;font-size: 11px">problem id: new2015_4</p>
Find the solutions corrected by LQC Friedmann equations for a matter dominated Universe.
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<p style="text-align: left;">Solving the corrected first Friedmann equation and the conservation equation for a matter dominated Universe
\[H^2=\frac\rho3\left(1-\frac\rho{\rho_c}\right),\quad \dot\rho+3H\rho=0;\]
one obtains the following quantities
\[a(t)=\left(\frac34\rho_ct^2+1\right)^{1/3},\quad \rho(t)=\frac{\rho_c}{\frac34\rho_ct^2+1},\quad H(t)=\frac{\frac12\rho_ct}{\frac34\rho_ct^2+1}.\]
For small values of the energy density ($\rho\ll\rho_c$) the solution of standard Friedmann equations is recovered.</p>
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'''Problem 3'''
<p style= "color: #999;font-size: 11px">problem id: new2015_5</p>
Show that in the case of the flat Friedmann metric, the third power of the scale factor $\varphi=a^3$ satisfies the equation
\[\frac{d^2\varphi}{dt^2}=\frac32(\rho-p)\varphi,\quad 8\pi G=1.\]
Check validity of this equation for different cosmological components: non-relativistic matter, cosmological constant, a component with the state equation $p=w\rho$.
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<p style="text-align: left;">\[\frac{d^2a^3}{dt^2}=3\left(2H^2+\frac{\ddot a}a\right)a^3=3\left[\frac23\rho-\frac16(\rho+3p)\right]a^3=\frac32(\rho-p)\varphi.\]</p>
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$f(R)$ gravity theory is built by direct generalization of the Einstein-Hilbert action with the substitution $R\to f(R)$. The new action is
\[S=\frac1{2\kappa}\int d^4x\sqrt{-g}f(R)+S_m(g_{\mu\nu},\psi);\quad \kappa\equiv8\pi G.\]
Here $\psi$ is general notion for the matter fields. The chosen generalization contains function $f(R)$ which depends solely on the Ricci scalar $R$, but it does not include other invariants such as $R_{\mu\nu}R^{\mu\nu}$. The reason of that is the following: the action $f(R)$ is sufficiently general to reflect basic features of gravity, and at the same time it is simple enough so that the calculations present no technical difficulty. The function $f(R)$ must satisfy the stability conditions
\[f'(R)>0,\quad f''(R)>0.\]

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'''Problem 1'''
<p style= "color: #999;font-size: 11px">problem id: f_r_1</p>
Obtain field equations for the $f(R)$ gravity.
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<p style="text-align: left;">\[f'(R)R_{\mu\nu}-\frac12f(R)g_{\mu\nu}-\left(\nabla_\mu\nabla_\nu-g_{\mu\nu}\square\right)f'(R)=\kappa T_{\mu\nu}.\]
Here \[T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g^{\mu\nu}},\]
$\nabla_\mu$ is covariant derivative associated with the Levi-Civita connection of the metric, and $\square=\nabla_\mu\nabla^\mu$.</p>
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'''Problem 2'''
<p style= "color: #999;font-size: 11px">problem id: f_r_2</p>
Obtain equation relating the scalar curvature $R$ with trace of the stress-energy tensor.
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<p style="text-align: left;">\begin{equation}\label{f_r_2_e_1}
f'(R)R-2f(R)+3\square f'(R)=\kappa T,
\end{equation}
where $T=g^{\mu\nu}T_{\mu\nu}$. $R$ and $T$ are related in differential rather in algebraic way like $R=-kT$ in general relativity. Note that equality $T=0$ does not imply that $R$ equals to zero or a constant.</p>
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'''Problem 3'''
<p style= "color: #999;font-size: 11px">problem id: f_r_3</p>
Solutions with $R=const$ are called the maximally symmetric. Show that in the case $R=0$, $T_{\mu\nu}=0$ the maximally symmetric solution is the Minkowski space, and in the case $R=const\equiv C$, $T_{\mu\nu}=0$ the maximally symmetric solution coincides with the de- Sitter or anti-de Sitter depending on sign of $C$.
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<p style="text-align: left;">For $R=const$ and $T_{\mu\nu}=0$ the equation \ref{f_r_2_e_1} obtained in the previous problem takes on the form \[f'(R)R-2f(R)=0.\]
For given function $f(R)$ this equation represents an algebraic equation for $R$. If $R=0$ is a root of this equation then it immediately follows from the field equations that $R_{\mu\nu}=0$ and the maximally symmetric solution is the Minkowski space-time. If the equation has a root $R=const=C$ then \[R_{\mu\nu}=\frac C4 g_{\mu\nu}\] and the maximally symmetric solution corresponds to the de- Sitter or anti-de Sitter (or to the cosmological constant in usual General Relativity) depending on sign of $C$.</p>
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'''Problem 4'''
<p style= "color: #999;font-size: 11px">problem id: f_r_4</p>
Use the FLRW metrics and the stress-energy tensor for an ideal liquid to obtain an analogue of the Friedmann equations for the $f(R)$ cosmology.
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<p style="text-align: left;">\[H^2=\frac\kappa{3f'}\left[\rho+\frac12(Rf'-f)-3H\dot Rf''\right];\]
\[2\dot H+H^2=-\frac\kappa{f'}\left[p+\dot R^2f'''+2H\dot Rf''+\ddot Rf''+\frac12(f-Rf')\right].\]</p>
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'''Problem 5'''
<p style= "color: #999;font-size: 11px">problem id: f_r_5</p>
Show that introduction of effective energy density \[\rho_{eff}=\frac{Rf'-f}{2f'}-\frac{3H\dot Rf''}{f'}\] and effective pressure \[p_{eff}=\frac{\dot R^2f'''+2H\dot Rf''+\ddot Rf''+\frac12(f-Rf')}{f'}\] allows to represent the equations obtained in the previous problem in form of the standard Friedmann equations
\begin{align}
\nonumber H^2&=\frac\kappa3\rho_{eff};\\
\nonumber \frac{\ddot a}a&=-\frac\kappa6\left(\rho_{eff}+3p_{eff}\right).
\end{align}
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'''Problem 6'''
<p style= "color: #999;font-size: 11px">problem id: f_r_6</p>
What condition must the function $f(R)$ satisfy to in order to make \[w_{eff}\equiv\frac{p_{eff}}{\rho_{eff}}=-1?\]
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<p style="text-align: left;">To reproduce (imitate) the de Sitter equation of state (i.e. the cosmological constant) $w_{eff}=-1$ the following condition is required \[\frac{f'''}{f''}=\frac{\dot RH-\ddot R}{\dot R^2}.\]
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'''Problem '''
<p style= "color: #999;font-size: 11px">problem id: </p>

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'''Problem '''
<p style= "color: #999;font-size: 11px">problem id: </p>

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

New from march - Revision history

Cosmo All at 22:37, 6 April 2015