Kerr black hole
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 $g^{\mu\nu}$.
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 \[\Sigma^{2}=(r^2+a^2)^{2}-a^{2}\Delta \sin^{2}\theta,\] 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 problem \ref{BlackHole70}.
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.
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.
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 problem Schwarzschild_black_hole#BlackHole43. 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 problem \ref{BlackHole74} and the previous one \ref{BlackHole76}) \[(u^{t})^2=\frac{1}{A} =\Big(\frac{\rho^2 \Delta}{\Sigma^2}\Big)^{-1},\quad \text{thus}\quad \partial_{\mu}\ln u^{t}=-\tfrac{1}{2} \frac{\partial_{\mu} A}{A}.\] 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.
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 (\ref{BlackHole75}), 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.
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.
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.