Finding Black Holes 2

Posted 2022-11-07 yuewen-chen

• Spherically Symmetric Case
  
  For the spherically symmetric case, \(f\) is constant. Thus
  \[ \beginequation D_as^a=\frac1\sqrt\gamma\partial_r (\sqrt\gammas^r) \endequation \]
  Because the spatial line element is written in a diagonal form,as
  \[ \beginequation \gamma_ij=diag(\gamma_rr,\gamma_\theta\theta,\gamma_\theta\theta\sin\theta^2) \endequation \]
  then \(D_as^a+K_abs^as^b-K=0\) can be write as
  \[ \beginequation \boxed \partial_r(log \gamma_\theta\theta)-2\sqrt\gamma_rrK^\theta_\theta =0 \endequation \]
  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
  \[ \beginequation ( \partial_t+(\alpha \gamma^-1/2-\beta^r)\partial_r)A(r)=0, or, k^a\nabla_aA=0 \endequation \]
  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 \(\partial_r(log \gamma_\theta\theta)-2\sqrt\gamma_rrK^\theta_\theta =0\):
  
  consider a nonrotating black hole in isotropic coordinates
  \[ \beginalign* \gamma_rr&=\varphi^4\ \gamma_\theta\theta &=\varphi^4 r^2\ K_ab &=0 \endalign* \]
  the expansion equation (3) is:
  \[ \beginalign 2\frac\varphi_,r\varphi+\frac12r &=0\ \Rightarrow -\fracMr^2(1+\fracM2r)^-1+\frac12r&=0\ \Rightarrow r &=\fracM2 \endalign \]