Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 311 / 2

Let $k$ generate the Killing horizon and let $\ell =fk$ be an affinely parametrized generator wherever $k\ne 0$. Choose an auxiliary null vector $n$ with $n\cdot \ell =-1$ and the screen-space projector $q_{ab}=g_{ab}+{\ell }_a{n}_b+n_a{\ell }_b$. The optical tensor is the screen projection $B_{ab}=q_a{}^c{q}_b{}^d{\nabla }_d{\ell }_c$.

Since the projector annihilates $k$, the symmetric part isby the Killing equation. Its trace is the null expansion and its trace-free symmetric part is the null shear, so both vanish. Also $\ell$ is normal to the null hypersurface. Frobenius theorem gives ${\ell }_{[a}{\nabla }_b{\ell }_{c]}=0$, which implies $q_a{}^c{q}_b{}^d{\nabla }_{[d}{\ell }_{c]}=0$: the null twist vanishes. These arguments use derivatives tangent to the horizon and hence apply to any smooth null geodesic congruence containing its generators, independently of its extension off the horizon. At a regular bifurcation surface, use the smooth affine generator and continuity. ThusThis is the vanishing of the optical scalars on a Killing horizon.

The Physical-process first law for a rotating black hole applies to a small perturbation of a stationary nonextremal vacuum black hole by infalling neutral matter, followed by relaxation to stationarity. Its first-order form isHere the background horizon generator is $\xi =t+{\Omega }_H\phi$, normalized by the unit time translation at infinity, and its surface gravity is constant, $\kappa >0$, as in the Zeroth law of black-hole mechanics. The formula neglects terms quadratic in the perturbation. Choose the process sufficiently weak that the horizon generators used below have no large caustics, and assume appropriate early and late stationary limits and convergent fluxes.

On the background Killing horizon, choose an affine generator $\ell =d/d\lambda$ such that $\xi =\kappa \lambda \ell$; the bifurcation or early stationary limit corresponds to $\lambda =0$. The background null expansion and null shear vanish. The null twist is zero because the perturbed event horizon is a null hypersurface. Linearizing the Null Raychaudhuri equation and using the Einstein field equations givesThe expansion squared and shear squared are second order, and the metric-trace part of the Einstein equations vanishes in this null contraction. Future stationarity sets $\delta \theta \to 0$. To first order $\delta \theta$ is the derivative of the fractional area change, so, using the background area element $d{A}_0$,The integration by parts uses $[\lambda \delta \theta {]}_0^{\infty }=0$, guaranteed by the stated endpoint conditions.

The conserved stress-energy current from a Killing vector gives the absorbed mass and angular momentum, with signsConsequentlywhich proves the law. Any other escaping flux must be accounted for separately when identifying these absorbed charges with the changes in the black hole. Electric-charge variations would require the corresponding potential-charge term; they are excluded in this version. The convention $16\pi G=1$ in Question 1 is not needed here.

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 $\kappa$.