Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 56 1 f Solution Created 2026-10-03 Updated 2026-10-14
Set . Since , the integrand defining the tortoise coordinate has a simple pole:where is analytic near . Define the Kruskal–Szekeres coordinates in the right exterior by and . ThenThe right-hand side has positive nonzero derivative at . Continuing this relation analytically, rather than retaining an absolute value, makes an analytic function of through both signs. The Kruskal extension across a simple static horizon has metricIts radial coefficient tends to , finite and nonzero, while . Therefore both null surfaces and are regular horizon branches, intersecting at a smooth two-sphere .
The stationary Killing vector field becomesIt is null, tangent and normal on each horizon branch and vanishes on their intersection. This proves that is a bifurcate Killing horizon, with bifurcation surface of area . The exponential definitions specify one exterior; allowing all signs of supplies the remaining quadrants of the extension.