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) \cite{Kerr63}. In the Boyer-Lindquist coordinates \cite{BoyerLindquist67} 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 $r\to\infty$.
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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.
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.
Write down the integrals of motion corresponding to Killing vectors $\partial_t$ and $\partial_\varphi$.
A particle's integrals of motion are \begin{equation}\label{AxiSimm-Integrals} \bm{u}\cdot\bm{\xi}_t=u_{t}; \quad \bm{u}\cdot\bm{\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.
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.
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}