Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 66 / 1 / b / ii / Solution

Take the first centre at the origin. If some listed centres coincide, first combine their poles; let $M$ be the total residue at the origin, so $M=M_1$ when the centres are distinct. The contributions from centres away from the origin are real analytic there. Their constant term iswhere $F$ is analytic in $r$ and in local angular charts. This follows by substituting $\mathbf{x}=r\mathbf{n}(\theta ,\varphi )$ in the ordinary Taylor expansions of the regular contributions. The constant $h$ is independent of angle, which makes the hinted coordinate differential integrable.

Put $A=M+hr$, $Q=rH=A+r^{2}F$, and defineinitially for $r>0$. Substitution into the metric givesThe apparently singular radial coefficient is actuallysince $Q-A=r^{2}F$. All coefficients therefore extend analytically to negative and positive $r$ near zero, with $Q(0)=M>0$. At the horizon $g_{vv}=0$, $g_{vr}=1$, $g_{rr}=4MF(0,\theta ,\varphi )$ and the angular metric is $M^{2}d{\Omega }_2^2$. The $v,r$ determinant is $-1$ both before and at the extension, so the metric remains nondegenerate and Lorentzian. Angular coordinate singularities are handled by ordinary sphere charts. This explicitly constructs the analytic extension of a Majumdar–Papapetrou horizon.

Inverting the radial block gives $g^{rr}=r^2/Q^2$, which vanishes at $r=0$. Thus the hypersurface is null. Its normal agrees there with the metric dual of $\xi ={\partial }_v$: ${\xi }_{a}d{x}^a=dr$ on the horizon. The metric is independent of $v$, so $\xi$ is a Killing vector field and the null hypersurface is a Killing horizon.

The surface gravity obeys ${\nabla }_a({\xi }^2)=-2\kappa {\xi }_a$ on a Killing horizon. Here ${\xi }^2=-r^2/Q^2$ has a double zero and its entire differential vanishes at $r=0$, while ${\xi }_a$ is the nonzero normal. ConsequentlyThe horizon is degenerate. Its cross-sectional area is $4\pi M^2$, independent of the centres away from the origin. The essential cancellation would fail with an arbitrary constant in place of the displayed regular part $h$.