Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 2

Locally write the null hypersurface as a level set $u=0$ and its normal as $n_a={\nabla }_{a}u$. Since a null normal is also tangent, $n^a$ lies within the hypersurface. The symmetry of the Levi-Civita connection givesThe scalar $n^2$ vanishes on the hypersurface, so its gradient there is normal and hence proportional to $n_a$:This is the nonaffine geodesic equation. A rescaling of $n$ removes $\kappa$, proving that the null hypersurface normal generates null geodesics lying in the hypersurface.

Let $U^a$ be the affine null tangent and choose another null vector $N^a$ with $U\cdot N=-1$. The screen-space projectorprojects onto the $(D-2)$-dimensional transverse space. The optical tensor isIts irreducible decomposition definesthe null expansion,the null shear, andthe null twist, also called rotation.

Affine geodesic evolution and the Ricci identity give the screen-projected optical equationTaking the screen trace yieldsThe optical decomposition and the antisymmetry of $\hat{\omega }$ implyTherefore the $D$-dimensional Null Raychaudhuri equation isand the requested constant is $\boxed{A=D-2}$.

For generators of a null hypersurface, the Frobenius theorem gives ${\hat{\omega }}_{ab}=0$. The null energy condition and the Einstein field equations imply $R_{ab}{U}^a{U}^b\ge 0$, while the shear norm is nonnegative. With $A=D-2$, Raychaudhuri's equation givesAs long as $\theta <0$ remains finite,If $\theta (0)={\theta }_0<0$, integration givesThe right-hand side reaches zero at $\lambda =A/|{\theta }_0|$, which a finite negative $\theta$ cannot cross. Hence the null focusing theorem forcesprovided the generator extends that far.

Suppose, for contradiction, that every future null geodesic orthogonal to the compact trapped surface $T$ extends beyond $L=A/|{\theta }_0|$. Both families begin with expansion at most ${\theta }_0<0$, so part d gives a point conjugate to $T$ on every generator by affine length $L$. A generator of the achronal boundary ${\dot{J}}^{+}(T)$ cannot remain on that boundary beyond its first conjugate point, because afterward it can be deformed to a timelike curve from $T$.

The two bundles of initial null directions over compact $T$, restricted to $0\le \lambda \le L$, form a compact set. Their image under the geodesic exponential map contains all of ${\dot{J}}^{+}(T)$, so this achronal boundary is compact. Project it along a complete timelike flow onto the noncompact Cauchy hypersurface $\Sigma$. The projection is both open and closed in the connected Cauchy surface, hence would be all of $\Sigma$; compactness of the source would then make $\Sigma$ compact, a contradiction.

Therefore at least one orthogonal future null geodesic cannot extend to affine length $L$. Its maximal future development is future-inextendible with total affine lengthwhich is the incompleteness conclusion of the Penrose singularity theorem.