# Raychaudhuri equation for black holes

Since the Raychaudhuri equation is defined only for timeline and null geodesic congruences, is it valid to use his equation to describe the null generators of the event horizon of a black hole?

Assuming that it is valid, if the black hole is stationary, the expansion parameter, $$\theta=k^a_{;a}-\kappa$$,will be zero. Here $$k$$ is a Killing vector that is timelike outside the black hole (must exist because the spacetime is stationary), while $$\kappa$$ is the surface gravity, defined by, $$k^ak^b_{;a}=\kappa k^b$$. But since $$k$$ is a Killing vector, its trace will be zero and hence the surface gravity of all stationary black holes will also turn out to be zero, which cannot be true. Can someone point out my mistake?

• The trace of $\nabla_ak_b$ is zero, but $\theta\ne\nabla_ak^a$. – Drake Marquis Jun 23 at 12:06
• In Poisson's book that's what he has said. For non-affine paramterizations, $\theta$ equals the divergence of $k$ minus $\kappa$, the surface gravity. It's the last problem of section 2.6. – Sounak Sinha Jun 23 at 12:10
• Why is $\theta=0$, for a stationary black hole? – Drake Marquis Jun 23 at 12:52
• The key here is that when Poisson stated $\theta=0$ for a stationary black hole, he referred to the affinely parameterized geodesic congruences. In that case, $\nabla_ak_b$ is no longer antisymmetric, which compensates $\kappa$. – Drake Marquis Jun 23 at 14:16
• There is a subtilty in applying Raychaudhuri equation to the null generators of some null hypersurface. The relation $\theta=\nabla_ak^a-\kappa$ applies to the congruence define all over (some open subset of) the spacetime. In the case of the null generators $\xi^a$ of the horizon, the null $\xi^a$ is defined only on the horizon. In order to use that relation, you have to extend $\xi^a$ to some null congruence near the horizon. Although the expansion is independent of the extension, $\nabla_ak^a$ depends on it. So $\theta=\nabla_ak^a-\kappa$ cannot be used in this case. – Drake Marquis Jun 26 at 11:51