Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 1 / d / Solution

The normal to a surface of constant $r$ is $n_a=(dr{)}_a$, and the inverse ingoing metric givesHence $r=r_{+}$ is a null hypersurface. LetThis constant linear combination of Killing vector fields is Killing. Its squared norm iswhich vanishes at $r_{+}$. More strongly, lowering its index in the ingoing metric gives ${\xi }_a=(dr{)}_a$ on the horizon. It is therefore the null normal and generator, so the surface is a Killing horizon.

Using ${\nabla }_a({\xi }^2)=-2\kappa {\xi }_a$ on a Killing horizon, the squared term contributes no first derivative at $r_{+}$ andDifferentiating $f$ at its simple outer root givesso the surface gravity isIt vanishes in the extremal case $r_{+}=r_{-}$.