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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.

A Killing horizon must be a null hypersurface whose normal is a Killing vector field. The outer stationary-limit surface instead has $s=r-r_e(\theta )=0$ and, in Boyer-Lindquist coordinates,On it, $r_e^2-2M{r}_e+a^2{\cos }^2\theta =0$, so $\Delta (r_e)=a^2{\sin }^2\theta >0$ away from the rotation axis. Its normal is spacelike and the surface timelike. Thus the Kerr stationary-limit surface is generically timelike and is not a Killing horizon, despite $K^2=0$ there. For example its stationary null vector is not normal, since $g(K,{\partial }_{\varphi })=-2Mar{\sin }^2\theta /\Sigma \ne 0$ away from the axis. The actual outer Killing horizon is $r=r_{+}=M+\sqrt{M^2-a^2}$, generated by $K+{\Omega }_H\Phi$, with Kerr horizon angular velocity ${\Omega }_H=a/(r_{+}^2+a^2)$. The two surfaces meet on the axis, which does not make the entire stationary-limit surface null.

Axisymmetry preserves the horizon, so the axial Killing vector field $m$ is tangent to it, so its null normal $\xi =k+\Omega m$ obeys $\xi \cdot m=0$ and ${\xi }^2=0$. Therefore $k\cdot m=-\Omega m^2$ andThus $\boxed{k^2={\Omega }^2{m}^2}$ there. A generator has $dt/d\alpha =1$, $d\phi /d\alpha =\Omega$, making $\Omega =d\phi /dt$ its angular velocity relative to the nonrotating stationary frame at infinity. For Kerr it is the Kerr horizon angular velocity $a/(r_{+}^2+a^2)$; the argument itself is not Kerr-specific.

The inverse Kerr metric in ingoing Kerr coordinates gives the raised normal to a surface of constant $r$:On $r=r_{+}$, $\Delta =0$, so the normal is null and tangent to the horizon. It reduces toA constant linear combination of the two Killing vector fields is again a Killing vector field. Therefore $k+{\Omega }_{H}m$ is normal to this null hypersurface, establishing that it is a Killing horizon, with Kerr horizon angular velocityThe second equality uses $\Delta (r_{+})=0$. This normal calculation proves hypersurface orthogonality on the horizon, rather than merely proving that the proposed vector happens to have zero norm there.

A generator of the Killing horizon is an orbit of $k+{\Omega }_{H}m$. In ingoing Kerr coordinates it has constant $r=r_{+}$ and $\theta$, with $d\chi /dv={\Omega }_H$. Since the coordinate shifts depend only on $r$, this is also $d\varphi /dt={\Omega }_H$ in the limiting Boyer-Lindquist coordinates description.

Thus ${\Omega }_H$ is the angular velocity of the horizon relative to the nonrotating stationary frame at infinity. The stationary Killing vector field $k$ is normalized to unit time translation there, while the axial Killing vector field has $2\pi$-periodic orbits. This normalization makes the Kerr horizon angular velocity physically definite. It describes the rotation of the null generators and the dragging of inertial frames, not a material solid surface rotating through space. In the Schwarzschild black hole limit $a=0$, ${\Omega }_H=0$.