Let $k^a=(d/d\lambda)^a$ be an affine horizon generator, normalized so that the horizon Killing field is $\xi^a=\kappa\lambda k^a$. To first order in the small total injected energy, expansion and shear-squared terms are second order, and the <Null Raychaudhuri equation> on the horizon becomes
$$
\frac{d\theta}{d\lambda}=-8\pi T_{ab}k^ak^b.
$$
The flux of Killing energy through all the shells is
$$
\Delta M
=\int_{\mathcal H^+}T_{ab}\xi^ak^b\,d\lambda\,dA
=\kappa\int\lambda T_{ab}k^ak^b\,d\lambda\,dA.
$$
Substitute the linearized focusing equation and integrate by parts. Stationarity before and after the process makes the endpoint term $\lambda\theta$ vanish, while $dA/d\lambda=\theta A$ to first order. Therefore
$$
\Delta M
=\frac{\kappa}{8\pi}\int\theta\,d\lambda\,dA
=\frac{\kappa}{8\pi}\Delta A.
$$
Using the <Hawking temperature> $T_H=\kappa/(2\pi)$ and <Bekenstein-Hawking entropy> $S_{BH}=A/4$ gives the <Physical-process first law of black-hole mechanics>
$$
\boxed{\Delta M=\frac{\kappa}{8\pi}\Delta A
=T_H\Delta S_{BH}}.
$$
The derivation is linear in the stress tensor, so separated shells simply contribute additively.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 311 / 2 / b / Solution

{c}
{title2=$T_H$}
{wiki}

In units $G=\hbar=c=k_B=1$, a stationary black hole with surface gravity $\kappa$ has Hawking temperature $T_H=\kappa/(2\pi)$.


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 Incoming links: Hawking temperature