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:
    \[ \begin{align} 2\frac{\varphi_{,r}}{\varphi}+\frac{1}{2r} &=0\\ \Rightarrow -\frac{M}{r^2}(1+\frac{M}{2r})^{-1}+\frac{1}{2r}&=0\\ \Rightarrow r &=\frac{M}{2} \end{align} \]

Guess you like

Origin www.cnblogs.com/yuewen-chen/p/11537350.html