Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 52 / 4 / b / Solution

The laws of black-hole mechanics initially relate geometric quantities in a way resembling thermodynamics. Hawking radiation supplies a physical temperature: in units $c=1$, while displaying $G,\hslash ,k_B$,The Hawking temperature $T_H$ is measured with the stationary time normalized at infinity. The radiation's thermal occupation factor is physically observable; propagation to infinity also introduces greybody factors, so the distant spectrum need not be a perfect blackbody spectrum at every frequency.

The Zeroth law of black-hole mechanics says that surface gravity $\kappa$ is constant on a stationary Killing horizon under its standard hypotheses. Through the Hawking temperature, this becomes uniform equilibrium temperature. The First law of black-hole mechanics isUsing $T_H$ identifies its area term as $T_H\delta S_{\mathrm{BH}}$. Integrating $\delta S_{\mathrm{BH}}=k_B\delta A/(4G\hslash )$ gives the Bekenstein-Hawking entropy, up to an additive constant. Thus $M$ is the energy, the angular and charge terms are work terms, and the geometrical law is the ordinary thermodynamic first law with a fixed entropy normalization.

The second law of black-hole mechanics, or Hawking's area theorem, gives nondecreasing area in the classical setting with the requisite energy and predictability assumptions. It corresponds to increasing Bekenstein-Hawking entropy. Semiclassical black-hole evaporation can decrease the area: the classical null energy condition need not hold for the quantum expectation of the stress-energy tensor. The appropriate extension is the generalized second law, that $S_{\mathrm{BH}}+S_{\mathrm{outside}}$ does not decrease, with the exterior entropy and its renormalization treated consistently. Hawking's temperature identification motivates this law; thermality alone does not prove every form of it.

The third law of black-hole mechanics is the unattainability, by an admissible finite physical process, of zero surface gravity. With Hawking temperature it becomes unattainability of absolute zero. It is not the assertion that an extremal black hole has vanishing entropy: its area can remain nonzero when $\kappa =0$.

For a Schwarzschild black hole, $\kappa =1/(4GM)$ and $A=16\pi G^2{M}^2$, giving $T_H=\hslash /(8\pi k_{B}GM)$ and $S_{\mathrm{BH}}=4\pi k_{B}G{M}^2/\hslash$. Their product satisfies $T_{H}d{S}_{\mathrm{BH}}=dM$, explicitly checking the first law. Quantum radiation turns the temperature and entropy in the mechanical analogy into physical thermodynamic quantities.