Differentiate $B_{ab}=\nabla_bU_a$ along an affinely parametrized generator. Commuting <covariant derivatives> and using $\nabla_UU=0$ gives the optical evolution equation
$$
U^c\nabla_cB_{ab}
=-B_{ac}B^c{}_b-R_{acbd}U^cU^d.
$$
Taking its screen trace and using
$$
\widehat B_{ab}\widehat B^{ba}
=\frac12\theta^2
+\widehat\sigma_{ab}\widehat\sigma^{ab}
-\widehat\omega_{ab}\widehat\omega^{ab}
$$
produces the <Null Raychaudhuri equation>
$$
\frac{d\theta}{d\lambda}
=-\frac12\theta^2
-\widehat\sigma_{ab}\widehat\sigma^{ab}
+\widehat\omega_{ab}\widehat\omega^{ab}
-R_{ab}U^aU^b.
$$
The generators lie in a <null hypersurface>, so they are hypersurface orthogonal. The <Frobenius theorem> therefore gives $\widehat\omega_{ab}=0$, leaving
$$
\boxed{
\frac{d\theta}{d\lambda}
=-\frac12\theta^2
-\widehat\sigma_{ab}\widehat\sigma^{ab}
-R_{ab}U^aU^b}.
$$

Solved by gpt-5.6-sol high.

Solution

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

{wiki}

A null hypersurface has a degenerate induced metric. Its normal covector raises to a null vector tangent to the hypersurface, whose integral curves are its null generators.


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 Incoming links: Null hypersurface