Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 69 / 1 / Solution

Use $G=\hslash =c_{\mathrm{light}}=k_B=1$ and normalize the stationary time at infinity. For a regular stationary, asymptotically flat electrovacuum black hole, the Kerr-Newman metric hasInitially assume $M>0$ and $d>0$, so the outer Killing horizon is nondegenerate. Its surface gravity, angular velocity, electric potential and area areThese are horizon quantities for the generator $\chi ={\partial }_t+{\Omega }_H{\partial }_{\varphi }$.

The laws of black-hole mechanics provide the first reason to regard $\kappa$ and area as thermodynamic variables. The Zeroth law of black-hole mechanics makes $\kappa$ constant on an equilibrium horizon. The first law for the Kerr-Newman family readsFor example, write the horizon relation as $M=(r_{+}^2+a^2+Q^2)/(2r_{+})$ and use $J=Ma$; differentiating eliminates $d{r}_{+}$ and $da$ to give this law. The last two terms are rotational and electric work. The classical second law of black-hole mechanics says that area cannot decrease under the null energy condition and appropriate global horizon assumptions. The third law of black-hole mechanics concerns unattainability of a zero-surface-gravity regular horizon by a finite physical process.

This analogy alone does not determine the temperature scale: if entropy were $\eta A_H$, the first law would only give $T=\kappa /(8\pi \eta )$. The decisive input is quantum field theory in the collapsing or stationary background. In a collapse vacuum that is regular for freely falling observers, the late outgoing retarded time $u$ and a regular affine null coordinate $U$ satisfy the Hawking exponential ray mapAn outgoing mode $e^{-i\omega u}$ therefore behaves as $(-U/C{)}^{i\omega /\kappa }$. Continuing this power through the horizon changes its amplitude by the factor $e^{-\pi \omega /\kappa }$. Its positive- and negative-frequency decomposition consequently has the thermal ratio of Hawking Bogoliubov coefficientsCombining that ratio with bosonic normalization $|\alpha {|}^2-|\beta {|}^2=1$ gives a Planck occupation $1/(e^{2\pi \omega /\kappa }-1)$. The fermionic normalization instead gives the corresponding Fermi factor. Thus Hawking radiation fixesFor rotating charged modes, the horizon energy is $\tilde{\omega }=\omega -m_{\varphi }{\Omega }_H-q{\Phi }_H$, so the emission spectrum contains the same temperature with angular-momentum and charge chemical potentials. Exterior scattering supplies greybody factors; it changes the received flux, not the horizon temperature. Superradiant bosonic modes require the usual signed absorption factor, and a globally regular thermal bath need not exist throughout an asymptotically flat rotating exterior. The collapse-state emission argument is the relevant one.

A complementary check is the Euclidean black-hole regularity condition. The local corotating nondegenerate horizon geometry is Rindler-like:Smoothness at $\rho =0$ requires $\tau$ to have period $2\pi /\kappa$. Imaginary-time periodicity is precisely inverse temperature, agreeing with the radiation calculation. Matching the resulting temperature to the first law fixes the Bekenstein-Hawking entropy to $S=A_H/4$. The generalized second law then uses $S_{\mathrm{outside}}+A_H/4$: the classical area theorem alone does not apply to the negative-energy quantum flux responsible for evaporation.

Substitution gives the Kerr-Newman horizon temperatureIt reduces to $1/(8\pi M)$ for a Schwarzschild black hole. The nonextremal limit towards $d=0$ gives zero temperature. At exact extremality the Euclidean horizon is degenerate, so the elementary conical-period argument does not itself fix a period; zero temperature here is the limiting semiclassical result. The reasoning assumes ordinary Einstein–Maxwell dynamics, a regular horizon, the stated normalization at infinity and a regime where quantum fields on a slowly evolving classical geometry are a useful approximation. The mechanical laws, radiation spectrum and Euclidean regularity agree under these assumptions, which is substantially stronger evidence than the classical analogy alone.