Difference between revisions of "Kerr black hole"
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==Limiting cases== | ==Limiting cases== | ||
− | + | <div id="label"></div> | |
+ | === Problem 5: Schwarzshild limit === | ||
+ | Show that in the limit $a\to 0$ the Kerr metric turns into Schwarzschild with $r_{g}=2\mu$. | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;"> | ||
+ | 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. </p> | ||
+ | </div> | ||
+ | </div> | ||
+ | |||
+ | <div id="label"></div> | ||
+ | === 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 $r=0$? | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;"> | ||
+ | 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{где}\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. </p> | ||
+ | </div> | ||
+ | </div> | ||
+ | |||
+ | <div id="label"></div> | ||
+ | === Problem 7: weak field rotation effect === | ||
+ | Write the Kerr metric in the limit $a/r \to 0$ up to linear terms. | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;"> | ||
+ | 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} </p> | ||
+ | </div> | ||
+ | </div> | ||
+ | |||
+ | |||
+ | |||
+ | |||
==Horizons and singularity== | ==Horizons and singularity== | ||
Revision as of 22:41, 18 July 2012
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 $\partial_t$ and $\partial_\varphi$.
A particle's integrals of motion are \begin{equation}\label{AxiSimm-Integrals} \mathbf{u}\cdot\mathbf{\xi}_t=u_{t}; \quad \mathbf{u}\cdot\mathbf{\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{где}\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}