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 $$ds2=g_{rr}dr2+R^2(r)d\Omega$$
extrinsic curvature is $$K{ab}=nanbK_L+(g-nanb)K_R$$
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
\[ \begin{align}
\frac{\sqrt{g^{rr}}}{R^2}( R^2)_{,r}+2K_R &=0\\
\Longrightarrow \sqrt{g^{rr}} R_{,r}+RK_{R} &=0
\end{align}
\]
