I faced "any apparent horizon is a minimal surface", but I don't know how I can relate a physical concept (apparent horizon) to pure mathematical concept (minimal surface). How can I prove it?
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2$\begingroup$ Where did you encounter this statement? $\endgroup$– SivaNov 24, 2013 at 2:43
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2$\begingroup$ The statement definitely needs a clarification. Minimal surface has mean curvature of zero, while apparent horizon can be curved. $\endgroup$– user23660Nov 24, 2013 at 5:29
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I found the answer:
The apparent horizon $\mathcal{H}$ is defined as the outer boundary of the region of $\Sigma$ (a hypersurface of spacetime with induced metric $h_{ab}$ and extrinsic curvature $K_{ab}$) which contains trapped or marginally trapped surfaces. $\mathcal{H}$ itself must be a marginally trapped surface, and thus it satisfies $$k+K^{ab}(h_{ab}-n_an_b)=0$$
where $k$ is the trace of the extrinsic curvature of $H$ as a submanifold of $\Sigma$ and $n^a$ is the unit outward normal to $H$ on $\Sigma$.
Now, in the case of time symmetric data (the extrinsic curvature $K_{ab}$ of $\Sigma$ has been set to zero) by above equation one can conclude $k=0$. Then minimal surface and apparent horizon coincide in the case of time symmetric data.