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

A null geodesic congruence is a smooth family of null geodesics filling a region without intersections there, with one tangent direction assigned smoothly at every point. Choose an affine parameter $\lambda$ and tangent $U^a$, so $U^a{U}_a=0$ and $U^b{\nabla }_b{U}^a=0$. Choose a second null vector $N^a$ with $U\cdot N=-1$. The screen-space projector isIt annihilates $U,N$ and restricts to a positive-definite metric on the two-dimensional transverse screen. Project the covariant derivative of the tangent to obtain the optical tensorIts decomposition defines the three optical quantities:The null expansion is the fractional rate of change of a transverse area, $\theta =A^{-1}dA/d\lambda$. The null shear changes a beam's shape without changing its area to first order. The null twist is its antisymmetric transverse deformation. The factor $1/2$ uses four spacetime dimensions.

For an affinely parametrized null geodesic congruence, put $B^a{}_b={\nabla }_b{U}^a$. Differentiating along a ray and commuting covariant derivatives givesusing $[{\nabla }_c,{\nabla }_b]V^a=R^a{}_{dcb}{V}^d$. The geodesic acceleration term vanishes. Since $U$ is null and affine, $B^a{}_b{U}^b=0$ and $U_a{B}^a{}_b=0$. Its trace is therefore the screen trace $\theta$, and the trace of its square equals that of the optical tensor squared. Taking the trace, with $R_{db}=R^a{}_{dab}$, yieldsNow decompose the optical tensor into null expansion, null shear and null twist. The symmetric trace-free part is orthogonal to the trace and to the antisymmetric part, while ${\hat{\omega }}_{ab}{\hat{\omega }}^{ba}=-{\hat{\omega }}_{ab}{\hat{\omega }}^{ab}$. HenceSubstitution proves the Null Raychaudhuri equation:The squares use the positive-definite screen metric. Thus null shear contributes to focusing, whereas nonzero null twist has the opposite sign.

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.