Difference between revisions of "Kerr black hole"

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(The laws of mechanics of black holes)
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==Particles' motion in the equatorial plane==
 
==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. ===
 +
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. ===
 +
What is the angular velocity of a particle with zero energy?
 +
<div class="NavFrame collapsed">
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  <div class="NavHead">solution</div>
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  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">
 +
For $\theta=\pi/2$ and $\rho^2=r^2$ we get
 +
\[\label{PenroseAngVelocity2}
 +
\Omega_{P}=\frac{\frac{2\mu r}{\rho^2}-1}
 +
{\frac{2\mu r}{\rho^2}a\sin^2 \theta}=
 +
\frac{1-\frac{\rho^2}{2\mu r}}{a\sin^2 \theta}=
 +
\Big\rwavy \rho^2=r^2,\; \sin\theta=1\Big\lwavy
 +
=\frac{1-\frac{r}{2\mu}}{a}.\]
 +
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. ===
 +
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">
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  <div class="NavHead">solution</div>
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  <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. ===
 +
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. ===
 +
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 moments 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. ===
 +
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. ===
 +
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.
 +
\begin{figure}[!hbt]\center
 +
\includegraphics*[width=0.65\textwidth]{BHfig-Kerr-XiMin}
 +
\parbox{0.7\textwidth}{\caption{''Radius'' of minimal 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.}}
 +
\end{figure}
 +
 +
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 the stability condition holds when $\xi>9$, and for the ``$-$'' sign 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 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>

Revision as of 11:25, 23 July 2012

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).

Contents

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 $\theta$.

Problem 1: preliminary algebra

Find the components of metric tensor $g_{\mu\nu}$ and its inverse $g^{\mu\nu}$.

Problem 2: integrals of motion

Write down the integrals of motion corresponding to Killing vectors $\boldsymbol{\xi}_{t}=\partial_t$ and $\boldsymbol{\xi}_{\varphi}=\partial_\varphi$.

Problem 3: Zero Angular Momentum Observer/particle

Find the coordinate angular velocity $\Omega=\tfrac{d\varphi}{dt}$ of a particle with zero angular momentum $u_{\mu}(\partial_{\varphi})^{\mu}=0$.

Problem 4: some more simple algebra

Calculate $A,B,C,D,\omega$ for the Kerr metric.

Limiting cases

Problem 5: Schwarzshild limit

Show that in the limit $a\to 0$ the Kerr metric turns into Schwarzschild with $r_{g}=2\mu$.

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 $r=0$?

Problem 7: weak field rotation effect

Write the Kerr metric in the limit $a/r \to 0$ up to linear terms.

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"

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.

Problem 9: null surfaces in Kerr metric

Find the surfaces $g^{rr}=0$ for the Kerr metric. Are they spheres?

Problem 10: horizon area

Calculate surface areas of the outer and inner horizons.

Problem 11: black holes and naked singularities

What values of $a$ lead to existence of horizons?

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.

Problem 12: $r=0$ is not a point.

Derive the internal metric of the surface $r=0$ in Kerr solution.

Problem 13: circular singularity

Show that the set of points $\rho=0$ is a circle. How it it situated relative to the inner horizon?

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 the problems on blackness of Schwarzshild black hole). In the general case the two do not necessarily have to coincide.

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?

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?

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?

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?

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.

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 ((7)) and ((8)) do not hold, as they were derived in assumption that $g_{00}>0$.

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?

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?

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 $B$.

Integrals of motion

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?

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 problem on negative energy.

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 $\tilde{\Omega}\in(\Omega_{-},\Omega_{+})$.

The laws of mechanics of black holes

If a Killing vector is null on some null hypersurface $\Sigma$, $\Sigma$ is called a Killing horizon.

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.

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 corresponding problem. Calculate $\kappa$ for particles with zero angular momentum in the Kerr metric. What is it for the critical black hole, with $a=\mu$?

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.

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?

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.

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.

Problem 30.

Put down explicit expressions for the metric components and parameters $A,B,C,D,\omega$

Problem 31.

What is the angular velocity of a particle with zero energy?

Problem 32.

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.

Problem 33.

Integrate the equations of motion for null geodesics with $L=aE$, investigate the asymptotes close to the horizons, limits $a\to 0$ and $a\to \mu$.

Problem 34.

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 $a\to\mu$.

Problem 35.

Find $L^2$ and $E^2$ as functions of radii for circular geodesics of the massive particles.

Problem 36.

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 $a/\mu\to 0,1$.