Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 311 / 2

Choose an auxiliary null vector $N^a$ with $U\cdot N=-1$. The screen-space projector isand the optical tensor is ${\hat{B}}_{ab}=P_a{}^c{P}_b{}^d{B}_{cd}$ for $B_{ab}={\nabla }_b{U}_a$. Because the screen is two-dimensional, its irreducible decomposition iswhereThese are respectively the null expansion, null shear, and null twist, also called rotation.

Affine geodesic motion gives $U^c{\nabla }_c{U}_a=0$. Commuting the two covariant derivatives and applying the product rule yieldsThis proves the required transport identity. The curvature term is symmetric after screen projection, so taking the antisymmetric screen part and inserting the optical decomposition gives the null-twist propagation equationIn particular, an initially twist-free congruence remains twist-free.

Choose a spacelike two-dimensional cross-section $S$ of the null hypersurface $\mathcal{N}$ and coordinates $y^i$ on $S$. Extend $y^i$ along the null generators, and choose their parameter $\lambda$ to be affine, so $U={\partial }_{\lambda }$ on $\mathcal{N}$. At every point choose the other null normal $L$ with $g(U,L)=1$, shoot out the affinely parametrized null geodesic with tangent $L$, and call its affine parameter $r$. Transport $(\lambda ,y^i)$ along those transverse geodesics.

This construction gives Gaussian null coordinates. The hypersurface is $r=0$, and the coordinate conditions implyBecause $\lambda$ is affine on the generators, ${\partial }_r{g}_{\lambda \lambda }{|}_{r=0}=0$. Smoothness then factors the remaining components as $g_{\lambda \lambda }=r^{2}F$ and $g_{\lambda i}=r{h}_i$, giving

On $r=0$, the screen metric is $h_{ij}$ and $U={\partial }_{\lambda }$. Therefore the null expansion isThe derivative formula for the determinant givesand henceThus $\theta$ is the fractional rate of change of an infinitesimal transverse area carried along the generators: positive expansion enlarges the beam and negative expansion focuses it.