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

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.