Finding Black Holes 2
- Spherically Symmetric Case
For the spherically symmetric case, \(f\) is constant. Thus
\[ \begin{equation}
D_as^a=\frac{1}{\sqrt{\gamma}}\partial_r (\sqrt{\gamma}s^r)
\end{equation}
\]
Because the spatial line element is written in a diagonal form,as
\[ \begin{equation}
\gamma_{ij}=diag(\gamma_{rr},\gamma_{\theta\theta},\gamma_{\theta\theta}\sin{\theta}^2)
\end{equation}
\]
then \(D_as^a+K_{ab}s^as^b-K=0\) can be write as
\[\begin{equation}
\boxed{ \partial_r(log \gamma_{\theta\theta})-2\sqrt{\gamma_{rr}}K^\theta_\theta =0}
\end{equation}
\]
Indeed, \(A(r) =4\pi \gamma_{\theta\theta}(r)\) denotes the surface area of a radius, \(r\), and using the equation of \(\gamma_{\theta\theta}\) in the form
\[-2\alpha K^\theta_\theta=(\partial_t-\beta^r\partial_r)log \gamma_{\theta\theta}
\]
the equation (3) is written as
\[ \begin{equation}
( \partial_t+(\alpha \gamma^{-1/2}-\beta^r)\partial_r)A(r)=0, or, k^a\nabla_aA=0
\end{equation}
\]
Thus, the apparent horizon in spherical symmetry may be defined as the surface where the local variation rate of its area along outgoing light rays is zero.
- Check equation \(\partial_r(log \gamma_{\theta\theta})-2\sqrt{\gamma_{rr}}K^\theta_\theta =0\):
consider a nonrotating black hole in isotropic coordinates
\[ \begin{align*}
\gamma_{rr}&=\varphi^4\\
\gamma_{\theta\theta} &=\varphi^4 r^2\\
K_{ab} &=0
\end{align*}
\]
the expansion equation (3) is:
\[ \begin{align}
2\frac{\varphi_{,r}}{\varphi}+\frac{1}{2r} &=0\\
\Rightarrow -\frac{M}{r^2}(1+\frac{M}{2r})^{-1}+\frac{1}{2r}&=0\\
\Rightarrow r &=\frac{M}{2}
\end{align}
\]
