 Solution

A Killing horizon is a null hypersurface with a Killing vector field $\xi$ as its normal and generator. Write $\xi =d/d\alpha$ along its orbits. These are null geodesics, but their Killing parameter need not be affine:The coefficient is the surface gravity for the chosen normalization; nondegeneracy means $\kappa \ne 0$. If $\lambda$ is affine and $q=d\lambda /d\alpha$, then $\xi =qd/d\lambda$ and $d(\log q)/d\alpha =\kappa$. For constant $\kappa$, $\lambda ={\lambda }_0+C{e}^{\kappa \alpha }$.

The Killing equation yields ${\partial }_{\mu }({\xi }^2)=2{\xi }^{\nu }{\nabla }_{\mu }{\xi }_{\nu }=-2{\xi }^{\nu }{\nabla }_{\nu }{\xi }_{\mu }$. On the horizon, $\boxed{{\partial }_{\mu }{\xi }^2=-2\kappa {\xi }_{\mu }}$. Multiplying $\xi$ by a positive constant multiplies $\kappa$ by that constant; physical temperature therefore needs a stated time normalization.