Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 56 / 2 / c

The conserved Killing energy of a geodesic tangent $p^a$ is $E=-K_a{p}^a$, where $K={\partial }_t$. Conservation follows from $p^a{\nabla }_a{p}^b=0$ and the Killing equation. For the Kerr metric, $K^2=g_{tt}=-(1-2Mr/\Sigma )$. Outside the outer Kerr ergoregion, $K$ is timelike. Its orthogonal complement is positive definite, so a nonzero causal vector cannot have $K\cdot p=0$. Equivalently, if both are future-directed, $-K\cdot p>0$. HenceThis excludes the open exterior of the stationary-limit surface; on the surface itself a zero-energy null tangent can be proportional to the null stationary vector.

A Killing horizon must be a null hypersurface whose normal is a Killing vector field. The outer stationary-limit surface instead has $s=r-r_e(\theta )=0$ and, in Boyer-Lindquist coordinates,On it, $r_e^2-2M{r}_e+a^2{\cos }^2\theta =0$, so $\Delta (r_e)=a^2{\sin }^2\theta >0$ away from the rotation axis. Its normal is spacelike and the surface timelike. Thus the Kerr stationary-limit surface is generically timelike and is not a Killing horizon, despite $K^2=0$ there. For example its stationary null vector is not normal, since $g(K,{\partial }_{\varphi })=-2Mar{\sin }^2\theta /\Sigma \ne 0$ away from the axis. The actual outer Killing horizon is $r=r_{+}=M+\sqrt{M^2-a^2}$, generated by $K+{\Omega }_H\Phi$, with Kerr horizon angular velocity ${\Omega }_H=a/(r_{+}^2+a^2)$. The two surfaces meet on the axis, which does not make the entire stationary-limit surface null.