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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 $\kappa$.