The stationary Killing field becomes null where
$$
g_{tt}=-\left(1-\frac{2Mr}{\Sigma}\right)=0.
$$
Solving $r^2-2Mr+a^2\cos^2\theta=0$ gives the stationary-limit surfaces
$$
r_E^\pm(\theta)=M\pm\sqrt{M^2-a^2\cos^2\theta}.
$$
The exterior <Kerr ergoregion> is
$$
\boxed{r_+<r<r_E^+(\theta)},
\qquad
r_+=M+\sqrt{M^2-a^2}.
$$
The event horizon is the constant-radius surface $\Delta=0$, whereas the outer stationary limit lies outside it except at the poles, where the two meet.

Inside the ergoregion $\partial_t$ is spacelike, so future-directed particles can have negative conserved stationary Killing energy while still following causal trajectories. Sending such a particle through the horizon permits another particle to escape with increased positive energy. This is the <Penrose process>, which extracts the black hole's rotational energy.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 311 / 4 / b / Solution

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The Penrose process extracts rotational energy by allowing a fragment with negative stationary Killing energy to enter the black hole while another fragment escapes with more energy than the original body carried.


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 Incoming links: Penrose process