Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 52 / 1 / c / Solution

The physical argument for the Penrose inequality combines weak cosmic censorship conjecture, the dominant energy condition, and relaxation to a stationary black hole. Work in geometrized units. Let $M_f$ and $A_f$ be the final Kerr black hole mass and horizon area. Positive energy radiated to infinity gives $E_{\mathrm{ADM}}\ge M_f$, where $E_{\mathrm{ADM}}$ is the initial ADM energy. For a Kerr black hole with $a_f=J_f/M_f$,If the initial apparent horizon obeys the necessary apparent-horizon area comparison with the enclosing event horizon, and Hawking's area theorem applies during the evolution, thenConsequently the anticipated answer, under those additional hypotheses, isThe bound is saturated by a nonrotating Schwarzschild black hole with no energy loss. Rotation or outgoing radiation makes the argument's inequalities stricter.

There is an essential qualification: inclusion inside an event horizon does not by itself compare areas. An arbitrary apparent horizon on general, non-time-symmetric initial data need not satisfy the displayed apparent-horizon area comparison; the unqualified version with its area is not universally true, even with the dominant energy condition. On time-symmetric data the relevant outermost minimal surface is an outer area-minimizing surface, as used in the Riemannian Penrose inequality, with nonnegative scalar curvature. In more general formulations an appropriate enclosing-area quantity is needed. The physical expectation is conditional on this area comparison, as well as on censorship, predictability, settling, and the energy assumptions; the mere presence of a trapped surface does not supply every step.