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

Since the real conformal factor $\Lambda$ is nowhere zero, ${\Lambda }^2>0$ at every point. For any tangent vector,The sign is unchanged. With metric signature $(-,+,+,+)$, negative, zero and positive values therefore describe the same timelike vectors, null vectors and spacelike vectors for both metric tensors. In particular the two null cones coincide.

Choose the same time orientation for the two Lorentzian metrics. A future-directed causal curve or timelike curve for one has the same character for the other. Hence the causal future, causal past and chronological future of every point agree:This is the preservation of causal structure by a smooth, nonvanishing conformal transformation. It does not assert preservation of proper time or of affine parameter lengths.

Put $\omega =\log |\Lambda |$, which is smooth because $\Lambda$ is nowhere zero. The conformal change of the Levi-Civita connection isThis follows directly by differentiating ${\tilde{g}}_{ab}=e^{2\omega }{g}_{ab}$ and using ${\tilde{g}}^{ab}=e^{-2\omega }{g}^{ab}$ in the Christoffel symbol formula.

Let $V=dx/d\tau$ be a geodesic tangent for $g$, with $\tau$ an affine parameter. Since ${\nabla }_{V}V=0$,For a null geodesic the second term vanishes. The remaining acceleration is tangent to the same curve and can be removed by changing its parameter. Explicitly set $\tilde{V}=fV$, with $f=d\tau /d\tilde{\tau }$. ThenIt vanishes when $f=C^{-1}{\Lambda }^{-2}$, orThis affine parameter change under conformal rescaling shows that conformal preservation of null geodesic paths means preservation of unparametrized curves; the original affine parameter usually changes.

For a non-null geodesic, the component of ${\mathrm{grad}}_g\omega$ orthogonal to $V$ generally survives. A change of parameter can remove only the tangential component of acceleration. As a concrete example use Minkowski spacetime,The curve $(t,x,y,z)=(\tau ,0,0,0)$ is a timelike geodesic for $g$, but along itThis acceleration is not parallel to ${\partial }_t$, so the curve is not a geodesic of $\tilde{g}$ under any reparametrization. Special non-null curves may still be preserved, for example when $\Lambda$ is constant; nullness is the general guarantee.

The product rule for the Lie derivative and the Killing equation giveConsequently $K$ is a Conformal Killing vector field for the rescaled metric tensor, with smooth conformal Killing factorIt remains an actual Killing vector field precisely when $K(\Lambda )=0$: the conformal factor must be constant along its integral curves of a vector field.

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.