Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 56 / 3

A null geodesic congruence is a smooth family of null geodesics filling a region, with one curve through each point before caustics form. Choose an affine parameter $\lambda$ and tangent $U^a$, so $U^2=0$ and ${\nabla }_{U}U=0$. Choose a null partner $N^a$ with $U\cdot N=-1$, transported parallel to $U$ along each generator. The screen-space projectorprojects onto the positive-definite two-dimensional plane transverse to $U,N$. Define the optical tensor and its decomposition bywhereThese are respectively the null expansion, null shear and null twist, the latter also called rotation. In the Null Raychaudhuri equation, ${\hat{\sigma }}^2={\hat{\sigma }}_{ab}{\hat{\sigma }}^{ab}$ and ${\hat{\omega }}^2={\hat{\omega }}_{ab}{\hat{\omega }}^{ab}$ use the screen metric. The trace convention for $\theta$ is important: it is not half the trace. For an affine null tangent it also equals ${\nabla }_a{U}^a$.

The Frobenius theorem says that hypersurface orthogonality of the nonzero covector $U_a$ is equivalent to $U_{[a}{\nabla }_b{U}_{c]}=0$, or locally $U_a=h{\nabla }_{a}S$ for some nonzero $h$. This immediately makes the screen projection of ${\nabla }_{[b}{U}_{a]}$ zero, since its antisymmetric part is a wedge product with $\nabla S$, which is proportional to $U$. Thus hypersurface orthogonality gives zero null twist.

For the converse, put $B_{ab}={\nabla }_b{U}_a$. Nullness gives $U^a{B}_{ab}=\frac{1}{2}{\nabla }_b(U^2)=0$, and the affine geodesic equation gives $B_{ab}{U}^b=0$. Resolve both tensor slots in the basis consisting of $U,N$ and the screen. These two identities exclude covector factors of $N$. Consequentlyfor some covector $q$. If $\hat{\omega }=0$, wedging this identity with $U$ makes $U_{[a}{\nabla }_b{U}_{c]}=0$. Applying Frobenius theorem completes the proof:This is the null twist vanishes exactly for hypersurface-orthogonal geodesics criterion; the geodesic assumption ensures that the screen criterion implies the full Frobenius condition. The resulting hypersurfaces are null, and the congruence follows their null generators.

Transport two connecting vectors between neighbouring rays so that they commute with $U$, and project them onto the screen. Their infinitesimal transverse lengths and angles evolve according to the symmetric part of the optical tensor. The trace part gives uniform fractional dilation, while the trace-free null shear stretches one transverse direction and contracts the other, changing a circular beam to an ellipse without changing its area to first order. With zero null twist, there is no antisymmetric rotation of the beam.

If $q_{AB}$ is the metric on an infinitesimal transverse element, then $d{q}_{AB}/d\lambda =2{\hat{B}}_{(AB)}$. Its area $A$ therefore obeysThus positive null expansion means a beam's cross-sectional area increases, and negative expansion means it decreases; null shear records the directional distortion rather than this overall area change.

Write the Schwarzschild metric in ingoing Eddington-Finkelstein coordinates as $d{s}^2=-fd{v}^2+2dvdr+r^{2}d{\Omega }^2$, with $f=1-2M/r$. Ingoing radial null geodesics have $v,\theta ,\varphi$ constant. Since ${\Gamma }^a{}_{rr}=0$, choose their future affine tangent $U=-E{\partial }_r$, where $E>0$ is constant along a ray; a common normalization sets $E=1$. The parallel null partner for ingoing Schwarzschild rays isThe radial conditions give $U\cdot N=-E{N}^v=-1$ and $N^2=-f/E^2+2(E^{-1})(f/(2E))=0$. Parallel transport can be checked explicitly: ${\Gamma }^v{}_{rv}=0$ and ${\Gamma }^r{}_{rv}=-f^{\prime }/2$, sowith angular components also zero. Thus ${\nabla }_{U}N=0$. The field is regular at $r=2M$ and throughout the regular advanced chart $r>0$; no vector field can be defined at the curvature-singular endpoint. If the affine normalization differs between rays, the same formula works for $E$ constant along each ray, since ${\partial }_{r}E=0$.

For the radial null pair above, the screen-space projector selects the angular tangent plane, with metric $q_{AB}=r^2{\gamma }_{AB}$. The ingoing covector is $U_a=-E(dv{)}_a$, and ${\Gamma }^v{}_{AB}=-r{\gamma }_{AB}$. Hence the optical tensor isIt is purely symmetric and proportional to the screen metric, givingFor $E=1$, the null expansion is $-2/r$. The area check $A\propto r^2$ and $dr/d\lambda =-E$ gives $A^{\prime }/A=-2E/r$, in agreement with the spherical null expansion in ingoing coordinates. In vacuum, $d\theta /d\lambda =-2E^2/r^2=-{\theta }^2/2$ also satisfies the Null Raychaudhuri equation; the ingoing beam focuses at the singularity rather than remaining future complete.

The second law of black-hole mechanics states that the total area of a classical future event horizon cannot decrease toward the future, under the appropriate causal and energy assumptions. Use the null convergence condition $R_{ab}{U}^a{U}^b\ge 0$; in general relativity the Einstein field equations imply it from the null energy condition. Assume, as allowed here, that the horizon's null generators remain on the horizon and are complete to the future. The horizon is an achronal boundary, and its smooth generators are hypersurface-orthogonal, so their null twist is zero. The Null Raychaudhuri equation therefore givesIf some smooth horizon point had ${\theta }_0<0$, then while the optical description remains regular, $\theta$ stays negative andThusforcing focusing no later than ${\lambda }_0+2/|{\theta }_0|$.

The assumed geometrical result is that this focusing produces a conjugate point to a spacelike surface, namely a focal point relative to a small spacelike horizon cross-section through the original point. A null geodesic beyond such a focal point ceases to generate an achronal boundary: a timelike variation connects the cross-section to points just beyond it. Since the cross-section and the generator lie on the achronal horizon, this is impossible. Future completeness ensures that a singular endpoint cannot terminate the generator before the finite focusing parameter. This future-complete horizon focusing proof of the area theorem rules out ${\theta }_0<0$.

Therefore $\theta \ge 0$ on smooth horizon pieces, and the transverse-area identity givesIntegrating over generators proves area nondecrease between later cross-sections. At crease sets, new generators may enter the horizon from the past, adding area; horizon generators do not leave it toward the future under the stated assumptions. Hence the statement also applies to total horizon area through black-hole mergers. Hawking's area theorem is classical: quantum stress-energy can violate the null energy condition, so it does not forbid a shrinking evaporating black hole.