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

Every metric coefficient and the gauge potential are independent of $t$. Thereforeso $K={\partial }_t$ is a Killing vector field.

Define the ingoing coordinateThen $dt=dv-dr/f$ andThis is the higher-dimensional analogue of Ingoing Eddington-Finkelstein coordinates. Its metric and inverse are regular at $r=r_{+}$, including when the zero is degenerate. A gauge transformation removes the singular radial term in the transformed potential, leaving the regular representative $A=-Q{r}^{-(d-3)}dv$. These expressions analytically extend the spacetime through the future outer horizon.

The normal to $r=r_{+}$ is $dr$, whose squared norm is $g^{rr}=f(r_{+})=0$, so the surface is a null hypersurface. In ingoing coordinates $K={\partial }_v$, and on the horizonThus $K$ is both tangent and normal there, making the surface a Killing horizon. For a static metric of this form, the surface gravity is $\kappa =f^{\prime }(r_{+})/2$. With $n=d-3$,

Curvature invariants and the electromagnetic invariant diverge at $r=0$, so it is a genuine curvature singularity, not a coordinate singularity. The causal structures are the higher-dimensional analogues of the familiar four-dimensional diagrams:

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.