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

A nondegenerate Lorentzian metric requires $M\ne 0$; take $M>0$ as the radius convention. The original coordinates describe $\lambda >0$, with $\lambda =0$ a candidate boundary rather than an ordinary point of that chart. The change of coordinate givesSubstituting and cancelling the two terms proportional to $d{\lambda }^2/{\lambda }^2$ producesThe $(v,\lambda )$ block of this metric tensor has determinant $-M^2$ and inverse metricAll coefficients are smooth through $\lambda =0$, and the block has one negative and one positive eigenvalue. Together with the positive round two-sphere factor this is a regular Lorentzian metric. It extends to a neighborhood with either sign of $\lambda$. The usual polar-coordinate problem at the poles of the two-sphere is removed by separate angular charts. Therefore $\lambda =0$ is only a coordinate singularity of the original chart. This is the degenerate-horizon patch of Bertotti-Robinson spacetime.

On the extension in part (i), the hypersurface $\mathcal{N}$ has nonzero normal covector $n_a={\partial }_a\lambda$. Its squared norm isThis vanishes on $\mathcal{N}$, which is therefore a null hypersurface. Raising the index givesSince ${\partial }_t={\partial }_v$ at fixed $\lambda$, the original Killing vector field becomes the smooth nonzero field $K={\partial }_v$. The transformed metric tensor is independent of $v$, so $K$ remains a Killing vector field, andThus $K$ is both tangent and normal to $\mathcal{N}$. This is exactly a Killing horizon:The Killing vector field itself does not vanish there; a vanishing squared norm means nullness, not a zero vector.

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.