Difference between revisions of "Kerr black hole"

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}

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

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 $\omega(r,\theta)$.

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}

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.

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.

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}

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

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.

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}

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

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 $\eta=a\sin\theta$.

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.

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.

• This is a schematic view of the Kerr solution in section $t,\theta=const$. Solid lines denote ergosurfaces, the dashed ones are horizons. The thin circle shows the value of $a$, equal to its radius.
• 

  a=0.5 (slowly rotating)

• 

  a=0.97 (generic astrophysical)

• 

  a=1.0 (extremal)

• 

  a=1.1 (naked singularity)

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 $\Omega\pm=\omega$.

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}

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.

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.

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.

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.

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.

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.

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.

$\Omega_{\pm}(\xi)$ and $\omega(\xi)$ in the equatorial plane on the Kerr solution for $\alpha=a/\mu=0.9375$. A particle's energy in the ergosphere is negative if $\Omega\in(\Omega_{-},\Omega_{P})$; $\Omega_{P}(\xi)$ in the equatorial plane is a straight line.

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

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

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

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.

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}

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.

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