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 \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}

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 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.

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}

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.

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

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

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}

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

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 ($\xi_2$).

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}

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.

" Radius of innermost stable circular orbit as function of $\alpha=a/\mu$: the upper curve for $L<0$, the lower one for $L>0$. The horizon is shown by the dashed line."

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.