Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 36D (General Relativity)

An affinely parameterized geodesic satisfies ${\ddot{x}}^a+{\Gamma }_{bc}^a{\dot{x}}^b{\dot{x}}^c=0$. With a general regular parameter $\lambda$ it satisfieswhere, if $s$ is affine, $f=s^{\prime \prime }/s^{\prime }$. Put $u=\log \Omega$. Substituting ${\tilde{g}}_{ab}=e^{2u}{g}_{ab}$ and ${\tilde{g}}^{ab}=e^{-2u}{g}^{ab}$ into the Christoffel symbol formula givesFor an affinely parameterized null tangent $k$, contracting the difference with $k^b{k}^c$ gives $2k(u)k^a$, since $g(k,k)=0$. The same curve therefore satisfies a nonaffine geodesic equation in the conformal metric. It becomes affine there after choosing $d\tilde{\lambda }/d\lambda =C{\Omega }^2$, obtained by setting $d\log (d\tilde{\lambda }/d\lambda )/d\lambda =2u^{\prime }$. Conformal transformations preserve unparameterized null geodesics.

In two dimensions each antisymmetric index pair of the Riemann curvature tensor has only one possible independent component, the pair 01. Pair exchange symmetry then leaves just $R_{0101}$. Contracting with the inverse metric givesChoose $u$ solving $\square u=R/2$, using the granted solvability, and let $\Omega =e^u>0$. The stated conformal transformation formula gives $\tilde{R}=0$. The one-component result makes the full curvature zero. A flat Lorentzian metric is locally isometric to Minkowski spacetime: parallel transport on a simply connected small neighbourhood supplies a parallel orthonormal frame, whose closed dual coframe integrates to flat coordinates. This is a local conclusion; vanishing curvature does not supply global Minkowski coordinates on an arbitrary topology.

In local two-dimensional Minkowski coordinates $U=T-X$, $V=T+X$, a regular smooth null curve obeys $U^{\prime }{V}^{\prime }=0$. Its nonzero continuous tangent stays on one of the two null directions locally, hence the curve lies on $U=\mathrm{constant}$ or $V=\mathrm{constant}$, a straight null line. It is a reparameterized null geodesic. Conformal invariance proves that every regular null curve in the original two-dimensional metric is a null geodesic, without requiring global flat coordinates.

For the displayed metric put $u=\log |t|$. When $t>0$, $d{s}^2=t(-d{u}^2+d{\theta }^2)$; when $t<0$, $d{s}^2=|t|(d{u}^2-d{\theta }^2)$. In both cases the two null families areIf $\theta$ is angular, the endpoints $0,2\pi$ are identified and these wind repeatedly around the cylinder as $t\to 0$ and as $|t|\to \infty$. To find the original affine behaviour, the Killing momentum $\ell =t\dot{\theta }$ is constant and the null condition gives ${\dot{t}}^2={\ell }^2$. A nontrivial null curve has $\ell \ne 0$, so $t$ is affine-linear: it reaches the excluded boundary $t=0$ in finite affine time despite infinitely many windings, and reaches $|t|=\infty$ only in infinite affine time.

The metric is already locally flat. For $t>0$, set $T=2\sqrt{t}\cosh (\theta /2)$, $X=2\sqrt{t}\sinh (\theta /2)$; for $t<0$ interchange the hyperbolic sine and cosine with $2\sqrt{|t|}$. Both give $d{s}^2=-d{T}^2+d{X}^2$. The angular identification acts as a Lorentz boost, so these are not global single-valued Minkowski coordinates. Constant-$t$ circles are spacelike for $t>0$ and timelike for $t<0$; the negative branch therefore has closed timelike curves, though its null curves continue to change $t$. If the angular endpoints are not identified, the formulas describe the corresponding unwrapped strips instead. This exhibits concretely the local/global qualification in the earlier request.