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For a future-directed causal four-momentum $p$ crossing a future Killing horizon with generator $\xi =t+{\Omega }_H\phi$, the Lorentzian inner product obeys $p\cdot \xi \le 0$. With $E=-p\cdot t$ and $L=p\cdot \phi$, this is $E\ge {\Omega }_{H}L$. A massive particle has strict inequality. The same bound follows in the nonextremal quasistationary regime from the Physical-process first law for a rotating black hole and the second law of black-hole mechanics. Negative Killing energy in a Penrose process is allowed, but its angular momentum must satisfy this bound.

Let $p^a$ be the future-directed causal four-momentum of the fragment captured by the Kerr black hole. In signature $(-+++)$ its conserved Killing energy and axial angular momentum areThe future-directed null generator of the Killing horizon is ${\chi }^a=k^a+{\Omega }_H{m}^a$, where ${\Omega }_H$ is the Kerr horizon angular velocity. At a crossing of the future event horizon, the scalar product of a future-directed causal vector with a future-directed null vector is nonpositive. For example, in a local orthonormal frame, $\chi =\alpha (1,\mathbf{n})$ with $|\mathbf{n}|=1$ and $\alpha >0$, while $p^0\ge |\mathbf{p}|$; hence $p\cdot \chi =\alpha (-p^0+\mathbf{p}\cdot \mathbf{n})\le 0$. ThereforeThis is a local causal restriction, so it applies also to an extremal Kerr black hole. A massive fragment crossing a regular future event horizon has a strict inequality; equality is the reversible limiting case, with causal four-momentum approaching the generator direction.

The Penrose process exploits the fact that $k^a$ becomes spacelike in the Kerr ergoregion, allowing future-directed fragments with negative Killing energy. For a positive choice of Kerr horizon angular velocity, a captured negative-energy fragment must have sufficiently negative angular momentum to satisfy the inequality. Conservation of energy in the splitting then gives $E_{\mathrm{escape}}=E_{\mathrm{initial}}-E_{\mathrm{capture}}>E_{\mathrm{initial}}$: the escaping fragment extracts rotational energy and the captured fragment reduces the black hole's mass and angular momentum.

For the independent thermodynamic argument, take a subextremal Kerr black hole and a small capture process covered by the Physical-process first law of black-hole mechanics. To first order, $\delta M=E$, $\delta J=L$, soThe second law of black-hole mechanics, under its classical null energy condition and regularity hypotheses, gives $\delta A\ge 0$; since $\kappa >0$, this reproduces $E\ge {\Omega }_{H}L$. The ideal reversible limit has $\delta A\to 0$. One should not infer a general finite-capture equality for an extremal black hole by simply substituting $\kappa =0$: the linearized nonextremal process proof and its differentiability assumptions must first be controlled. The causal proof already establishes the required inequality there without this issue.

For a Kerr black hole, the future horizon generator is $\xi =t+{\Omega }_H\phi$. A future-directed causal four-momentum $p$ and a future-directed null vector $\xi$ satisfy $g(p,\xi )\le 0$ in the $(-+++)$ convention. The conserved quantities from the Killing vector field are $E=-p\cdot t$ and $L=p\cdot \phi$. ThereforeFor a massive particle the inequality is strict. Equality for a null particle requires momentum parallel to the horizon generator and describes the limiting tangent case. In a Penrose process the absorbed particle may have $E<0$; for ${\Omega }_H>0$ this requires sufficiently negative angular momentum.

In the nonextremal test-particle regime, the Physical-process first law for a rotating black hole identifies $\delta M=E$ and $\delta J=L$, soby the second law of black-hole mechanics. The usual hypotheses of that area law, including the null energy condition and the required global regularity conditions, are understood. This thermodynamic derivation is first order in the capture process; the causal horizon flux inequality for causal matter also applies to the extremal horizon without dividing by $\kappa$.

The stationary Killing field becomes null whereSolving $r^2-2Mr+a^2{\cos }^2\theta =0$ gives the stationary-limit surfacesThe exterior Kerr ergoregion isThe 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.