Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 54 / 2 / b

In the two null coordinates the inverse metric has $g^{UV}=g^{VU}=-f^{-1}$ and $g^{UU}=g^{VV}=0$. ThereforeBoth tangents are null and future-directed, since $f>0$ and both coordinate directions are future-directed. To prove the affine parameter property, use the null-gradient affine-geodesic identity. The one-form $k_a=-{\nabla }_{a}U$ is closed, so ${\nabla }_a{k}_b={\nabla }_b{k}_a$. ConsequentlyThe same argument applies to ${\ell }_a=-{\nabla }_{a}V$. Thus both vector fields satisfy the geodesic equation without a nonaffinity term, and generate affinely parametrized null geodesic congruences. The coordinate vectors alone need not be affine; the factors $f^{-1}$ are essential.

For the spherical null expansions in double-null coordinates, the cross-sectional area of a symmetry sphere is $A=4\pi r^2$. For the two affine null geodesic congruences, the definition of null expansion therefore givesThere is no extra derivative of $f$ in these formulas: the null expansion differentiates the transverse area, not the normalization of the tangent. Equivalently, the projected metric is $q_{AB}=r^2{\gamma }_{AB}$ and its optical tensor is $(k(r)/r)q_{AB}$, or $(\ell (r)/r)q_{AB}$ for the other congruence. It is purely trace, so both null shears vanish.

The null twist vanishes because the covector tangents are gradients of $-U$ and $-V$, and hence their exterior derivatives vanish. Each congruence is orthogonal to a family of null hypersurfaces. This also identifies the Frobenius theorem reason for zero null twist.

Use the Null Raychaudhuri equation for $\ell =f^{-1}{\partial }_U$. Its null shear and null twist vanish, and $R_{ab}{\ell }^a{\ell }^b=f^{-2}{R}_{UU}$. Substitution of ${\theta }_{\ell }=2r_{,U}/(fr)$ givesThe quadratic terms cancel. Thus the double-null focusing identity isHere the last equality uses the Einstein field equations in units $G=1$; terms proportional to the metric vanish in this null component. The inequality follows from the null energy condition.

For any point of the future domain of dependence $D^{+}(\Sigma )$, the past continuation of its $\ell$ generator must meet the Cauchy hypersurface $\Sigma$: every past-inextendible causal curve from that point meets $\Sigma$. Along its future continuation, $U$ increases, and $f^{-1}{r}_{,U}$ is nonincreasing by the focusing identity. Its initial value is negative, so it stays negative. Since $f>0$, $r_{,U}<0$ throughout the regular part of $D^{+}(\Sigma )$. The argument is used only where the smooth double-null chart has $r>0$; it does not extend a congruence beyond a singular endpoint.

A spherical trapped surface has both future null expansions negative. Thus the initial sphere has $r_{,U}<0$ and $r_{,V}<0$. The result of the previous part already gives $r_{,U}<0$ at every regular future sphere in $D^{+}(\Sigma )$.

Apply the same Null Raychaudhuri equation to $k=f^{-1}{\partial }_V$. The exchanged-coordinate double-null focusing identity isAlong $U=U_0$, increasing $V$ is future-directed. ThereforeBoth null expansions remain negative, proving persistence of spherical trapping along this future null direction. As usual, the conclusion concerns every $V_1>V_0$ for which the sphere exists in the regular chart in $D^{+}(\Sigma )$; the inequality does not assert existence of such spheres beyond a focusing singularity.