Relativity and gravitational theory Black holes Proposition

Penrose process

Builds on Kerr metric Ergosphere Killing vector field Event horizon

Statement

Use geometric units $G = c = 1$ . Let $(M, g)$ be the Kerr metric of mass $M$ and angular momentum $J = aM$ with $0 < \abs{a} < M$ . Let $\xi = \partial_t$ and $\eta = \partial_\phi$ be its Killing vector fields in Boyer–Lindquist coordinates. For a particle with future-directed causal four-momentum $p$ , its conserved energy is $E = -g(\xi, p)$ and its conserved angular momentum is $L = g(\eta, p)$ . Let $\Omega_H = a / (r_+^2 + a^2)$ with $r_+ = M + \sqrt{M^2 - a^2}$ be the angular velocity of the event horizon.

At each point of the ergosphere there are future-directed timelike four-momenta $p$ with $E < 0$ .
Suppose a particle of momentum $p_0$ inside the ergosphere splits into two particles, $p_0 = p_1 + p_2$ , with $E_2 < 0$ . Then the energy $E_1 = E_0 - E_2$ of the first particle exceeds $E_0$ .
If the second particle crosses the event horizon, then $E_2 - \Omega_H L_2 \ge 0$ . So $\Omega_H L_2 \le E_2 < 0$ : the second particle carries angular momentum against the rotation of the hole. The changes $\delta M = E_2$ and $\delta J = L_2$ of the black hole satisfy $\delta M \ge \Omega_H\, \delta J$ .

Remarks

Part 1 holds because $\xi$ is spacelike in the ergosphere, so the timelike cone contains vectors with $g(\xi, p) > 0$ . Part 2 is conservation of $E$ , which follows from conservation of four-momentum and the fact that $E$ is linear in $p$ . For part 3, the vector field $\chi = \xi + \Omega_H \eta$ is a future-directed null normal of the horizon, and $-g(\chi, p) \ge 0$ for any future-directed causal $p$ .

The inequality $\delta M \ge \Omega_H \delta J$ is equivalent to the statement that the irreducible mass $M_{\mathrm{irr}} = \sqrt{A / 16\pi}$ does not decrease, where $A$ is the horizon area. Hence at most $M - M_{\mathrm{irr}}$ can be extracted. For an extremal black hole, $a = M$ , one has $M_{\mathrm{irr}} = M/\sqrt{2}$ , so at most about $29\%$ of the mass can be extracted. For the Schwarzschild metric ( $a = 0$ ) the ergosphere is empty and the process is impossible.

Going deeper

For a student

The Penrose process shows that the mass of a rotating black hole is not all locked away. The bound $\delta M \ge \Omega_H \delta J$ says, in this simple setting, that the area of the horizon cannot decrease, which the Hawking area theorem later proved in general. It leads to the split of the mass into an irreducible part and a rotational part, and to the laws of black hole mechanics. The wave analogue of the process is superradiance, in which waves scattered by a rotating black hole come back amplified.

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Tag	`TW2587J15Z`
URI	`m2://TW2587J15Z`
Kind	statement, proposition
Writing	W0 stub
MSC	83C57, 83C40
Status	This copy of the site has no build data. Run `m2 status TW2587J15Z`.
About	Ergosphere, Kerr metric
Uses	Kerr metric, Ergosphere, Killing vector field, Event horizon
Same as	Wikidata Q616172 Wikipedia Penrose process
Changed	3 November 2026