# Is any apparent horizon a minimal surface?

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?

• Where did you encounter this statement? – Siva Nov 24 '13 at 2:43
• The statement definitely needs a clarification. Minimal surface has mean curvature of zero, while apparent horizon can be curved. – user23660 Nov 24 '13 at 5:29

## 1 Answer

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.