Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 311 / 2 / e

Differentiate $E=-m{K}_a{U}^a+q\Phi$ along the worldline. The symmetric contraction $U^a{U}^b{\nabla }_a{K}_b$ vanishes by the Killing equation. Using the Lorentz force and $K^a{F}_{ab}={\nabla }_b\Phi$, the remaining force term cancels $q{U}^a{\nabla }_a\Phi$. Hence

For the displayed gauge field,which already vanishes at infinity. A static future-directed particle has $U=K/\sqrt{f}$, soIf $qQ<0$, the first term tends to zero at $r_{+}$ while the second remains negative. Lowering the particle sufficiently close to the horizon therefore produces $E<0$. Dropping it into the hole reduces the black-hole mass by $|E|$, while the external agent receives the corresponding positive work: this is charged-particle black-hole energy extraction.

In the reversible limit the particle is released arbitrarily close to the horizon, soHolding $r_{+}$ fixed and using $Q^2=r_{+}^{d-3}{r}_{-}^{d-3}$,Discharging to $Q=0$ therefore extracts at mostHawking's area theorem forbids the horizon area, and hence $r_{+}$, from decreasing. The smallest possible final neutral mass is consequently $r_{+}^{d-3}/2$, giving exactly the same upper bound.