Difference between revisions of "Kerr black hole"
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</div> | </div> | ||
+ | ==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" | ||
+ | <div id="label"></div> | ||
+ | === 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. | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;"> | ||
+ | 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 | ||
+ | \[0=g^{\mu\nu}\delta^{r}_{\mu}\delta^{r}_{\nu} | ||
+ | =g^{rr}.\] </p> | ||
+ | </div> | ||
+ | </div> | ||
+ | <div id="label"></div> | ||
+ | === Problem 9: null surfaces in Kerr metric === | ||
+ | Find the surfaces $g^{rr}=0$ for the Kerr metric. Are they spheres? | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;"> | ||
+ | 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{где}\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. </p> | ||
+ | </div> | ||
+ | </div> | ||
+ | |||
+ | <div id="label"></div> | ||
+ | === Problem 10: horizon area === | ||
+ | Calculate surface areas of the outer and inner horizons. | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;"> | ||
+ | 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} </p> | ||
+ | </div> | ||
+ | </div> | ||
+ | |||
+ | <div id="label"></div> | ||
+ | === Problem 11: black holes and naked singularities === | ||
+ | What values of $a$ lead to existence of horizons? | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;"> | ||
+ | Solutions of $\Delta=0$ exist when | ||
+ | \[a<m.\] </p> | ||
+ | </div> | ||
+ | </div> | ||
+ | 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. | ||
+ | |||
+ | <div id="label"></div> | ||
+ | === Problem 12: surface $r=0$. === | ||
+ | Derive the internal metric of the surface $r=0$ in Kerr solution. | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;"> | ||
+ | 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\lwavy a\sin\theta=\eta\Big\rwavy= | ||
+ | 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 $\eta=a\sin\theta$. </p> | ||
+ | </div> | ||
+ | </div> | ||
+ | |||
+ | <div id="label"></div> | ||
+ | === Problem 13: circular singularity === | ||
+ | Show that the set of points $\rho=0$ is a circle. How it it situated relative to the inner horizon? | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;"> | ||
+ | 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. </p> | ||
+ | </div> | ||
+ | </div>. | ||
==Stationary limit== | ==Stationary limit== |
Revision as of 22:49, 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}
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{где}\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: surface $r=0$.
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\lwavy a\sin\theta=\eta\Big\rwavy= 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.