# Difference between revisions of "Kerr black hole"

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Let us introduce notation | 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 | 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} | \begin{equation}\label{Kerr-ABCD} |

## Latest revision as of 17:38, 8 December 2013

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

- 1 General axially symmetric metric
- 2 Limiting cases
- 3 Horizons and singularity
- 4 Stationary limit
- 5 Ergosphere and the Penrose process
- 6 Integrals of motion
- 7 The laws of mechanics of black holes
- 8 Particles' motion in the equatorial plane
- 8.1 Problem 30: preparatory algebra
- 8.2 Problem 31: zero energy particles
- 8.3 Problem 32: geodesics and effective potential
- 8.4 Problem 33: principal null geodesics
- 8.5 Problem 34: innermost stable circular orbits, massless case
- 8.6 Problem 35: circular orbits for massive particles
- 8.7 Problem 36: innermost stable circular orbits, massive case

## 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 $g^{\mu\nu}$.

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}

### Problem 2: integrals of motion

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

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 \[E=mc^{2}u_{t};\qquad L=-m u_\varphi.\]

### Problem 3: Zero Angular Momentum Observer/particle

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

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 $\omega(r,\theta)$.

### Problem 4: some more simple algebra

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

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}

## Limiting cases

### Problem 5: Schwarzshild limit

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

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.

### 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 $r=0$?

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.

### Problem 7: weak field rotation effect

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

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}

## 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.

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 \[0=g^{\mu\nu}\delta^{r}_{\mu}\delta^{r}_{\nu} =g^{rr}.\]

### Problem 9: null surfaces in Kerr metric

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

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.

### Problem 10: horizon area

Calculate surface areas of the outer and inner horizons.

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}

### Problem 11: black holes and naked singularities

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

Solutions of $\Delta=0$ exist when \[a<m.\]

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.

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 $\eta=a\sin\theta$.

### Problem 13: circular singularity

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

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.

## 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?

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.

### 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?

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 $\Omega\pm=\omega$.

### 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?

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}

### 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?

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.

## 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 $g_{00}>0$.

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.

### 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?

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.

### 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?

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.

### 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 $B$.

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.

## 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?

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.

### 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.

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.

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: \[l_{m}<-m\sqrt{\frac{A}{\Omega_{+}\Omega_{-}}}= -m\sqrt{A\frac{\omega}{\Omega_P}}.\]

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

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 \[L<\frac{E}{\Omega_{H}}.\]

## 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.

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 $K^{\mu}K_{\mu}\to 0$.

### 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 $a=\mu$?

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 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.

### 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.

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 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}

### 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?

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.

### 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.

\[\delta\alpha=\delta\Big(\frac{J}{\mu^2}\Big) =\frac{1}{\mu^3} \Big[\mu \delta J-2J \delta\mu\Big],\] 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.

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: preparatory algebra

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

\[g_{\mu\nu}=\begin{pmatrix} 1-\frac{2\mu}{r}&0&0&\frac{2\mu a}{r}\\ 0&-\frac{r^2}{\Delta}&0&0\\ 9&0&-r^2&0\\ \frac{2\mu a}{r}&0&0& -\frac{\Sigma^2}{r^2} \end{pmatrix},\quad g^{\mu\nu}=\begin{pmatrix} \frac{\Sigma^2}{r^2\Delta} &0&0&\frac{2\mu a}{\Delta r}\\ 0&-\frac{\Delta}{r^2}&0&0\\ 9&0&-\frac{1}{r^2}&0\\ \frac{2\mu a}{\Delta r}&0&0& -\frac{1}{\Delta}\Big(1-\frac{2\mu}{r}\Big) \end{pmatrix}\] 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.

### Problem 31: zero energy particles

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

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 \[\Omega_{P}|_{r_+} =\frac{1}{a}\Big(1-\frac{r_+}{2\mu}\Big) =\frac{1}{a}\frac{2\mu-r_+}{2\mu} =\frac{r_-}{2\mu a}=\Omega_H.\]

### 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.

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}

### 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 $a\to \mu$.

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 $(r-r_{+})^{-1}$.

### 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 $a\to\mu$.

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 ($\xi_2$).

### Problem 35: circular orbits for massive particles

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

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}

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

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.

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.