# Why apparent horizon area provides a lower bound for event horizon area?

In Penrose's original formulation of Penrose Inequality, i.e. in Naked Singularity (1973), he speculated a thin sell matter collapsing at the speed of light, forming an ingoing null hypersurface N. Consider outgoing null hypersurface H. Along the line, he said

"Assuming cosmic censorship, we deduce the existence of an absolute-event horizon E intersecting H in a two-surface $$S_1$$, the surface $$S_1$$ lying farther out along the generators of H than $$S_0$$. Furthermore, the total surface area $$A_1$$ of $$S_1$$ is not less than the area $$A_0$$ of $$S_0$$."

where $$S_0$$ is apparent horizon.

My question is how could one deduce that area of event horizon is no less than area of apparent horizon simply from the fact that event horizon lies outside of apparent horizon? For example, the event horizon of the Schwarzschild black hole at $$r=2m$$ is known to be minimal, possessing a smaller area than those both outside and inside of it. To compare area, we are assuming knowledge of metric, right? But outside apparent horizon $$S_0$$ in Penrose's set-up, the spacetime is not Minkowski. What am I missing? Thanks in advance!

In Penrose's setup, the spacetime interior to the null hypersurface $$H$$ is always Minkowski.Since the area of a cross-section $$S$$ of $$H$$ has to be equal when measured from the interior and exterior metrics, when can always use the Minkowski metric to measure the area of $$S$$. In particular, if $$S_1$$ lies further out on $$H$$ than $$S_0$$, it must also have a greater area.
• Thanks! I guess I mis-pictured in my mind that $S_1$ lies off null hyperplane H. A quick question: do people always assume the scenario of collapsing null shells when talking about Penrose inequality? If not, what argument would replace this part invoking Minkowski metric? Thanks! Jan 25 at 0:12