Why the Place that Jang'S Equation Blow Up is Apparent Horiozn?

• Why the Place that Jang'S Equation Blow Up is Apparent Horiozn?
  
  Let's consider a simple example:
  Let us assume that the three metric is \[ds^2=g_{rr}dr^2+R^2(r)d\Omega\]
  extrinsic curvature is \[K^{ab}=n^an^bK_L+(g^{ab}-n^an^b)K_R\]
  where \(K_L\) and \(K_R\) are two scalar and \(n^a\) is the outward pointing unit normal to the surface of constant \(R\).
  The Spherical symmetry Jang's equation is
  \[ \begin{equation} \frac{\sqrt{g^{rr}}}{R^2}( \frac{R^2 f_r\sqrt{g^{rr}}}{\sqrt{1+f_r^2 g^{rr}}})_{,r} +(2K_R+K_L)-K_L\frac{f_r^2 g^{rr}}{1+f_r^2 g^{rr}}=0 \end{equation} \]
  The Apparent Horizon condition(expansion equation) is
  \[ \begin{equation} \boxed{ \sqrt{g^{rr}}R_{,r}+RK_R=0} \end{equation} \]
  If Jang's equation blow up, \(\Longleftrightarrow\)
  \[ \begin{equation} \displaystyle \frac{f_r\sqrt{g^{rr}}}{\sqrt{1+f_r^2 g^{rr}}}=1,\text{at the Apparent Horiozn} \end{equation} \]
  ,then Jang's equation reduce to
  \[ \begin{align} \frac{\sqrt{g^{rr}}}{R^2}( R^2)_{,r}+2K_R &=0\\ \Longrightarrow \sqrt{g^{rr}} R_{,r}+RK_{R} &=0 \end{align} \]

转载于:https://www.cnblogs.com/yuewen-chen/p/11542732.html