Past exam of the general relativity II course of the University of Oxford / 2017 / 2 / b / iii

The Killing equation impliesHere $K^2=-{\lambda }^2$, so in the regular coordinatesThis vector vanishes at $\lambda =0$. The Killing vector field $K={\partial }_v$ is nonzero there, so comparison with the definition of surface gravity yieldsHence this is a degenerate Killing horizon. The argument uses the vanishing gradient of $K^2$ on the horizon; the condition $K^2=0$ by itself would not force zero surface gravity.