Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 311 / 1 / a

Because $U$ is a null tangent, metric compatibility givesBecause $U$ is tangent to an affinely parametrized null geodesic congruence,by the geodesic equation. Thus $B$ has no component along $U$ in either index.

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With $U^2=N^2=0$ and $U\cdot N=-1$, define the screen-space projectorDirect contraction gives $P^a{}_b{U}^b=P^a{}_b{N}^b=0$, while a vector orthogonal to both $U$ and $N$ is unchanged. Moreover $P^a{}_c{P}^c{}_b=P^a{}_b$ and $P^a{}_a=2$, so its image is precisely the two-dimensional transverse space.

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Decompose the optical tensor on the two-dimensional screen space asThe null expansion, null shear, and null twist are respectivelyThe shear is symmetric and trace-free, while the twist is antisymmetric.

Solved by gpt-5.6-sol high.