Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 311 / 2 / b / Solution

Parallel transport the auxiliary null vector $N$ and a screen basis along each generator of the null geodesic congruence. This keeps the screen-space projector fixed under the corresponding transported derivative. Use the curvature convention $[{\nabla }_c,{\nabla }_b]V_a=-R^d{}_{acb}{V}_d$.

With $B_{ab}={\nabla }_b{U}_a$, differentiation of the affine geodesic equation givesHere the first term comes from commuting the derivative past $U^c$, and the second from the curvature commutator. The constraints $B_{ab}{U}^a=B_{ab}{U}^b=0$ follow respectively from the null normalization and affine geodesic equation. Projecting onto the screen therefore yields the optical matrix equationThe trace of the tidal matrix is $R_{ab}{U}^a{U}^b$; extra terms from replacing the screen trace by the spacetime trace vanish by curvature antisymmetry. Taking the trace givesDecompose the optical tensor as $\hat{B}=\frac{1}{2}\theta I+\hat{\sigma }+\hat{\omega }$. The null shear is symmetric and trace-free and the null twist is antisymmetric. Their cross traces vanish, andThe negative sign in the last term is the identity $\mathrm{tr}({\hat{\omega }}^2)=-{\sum }_{I,J}{\hat{\omega }}_{IJ}^2$ on the positive-definite screen. Substitution proves the four-dimensional Null Raychaudhuri equation:The coefficient is $1/(D-2)$ in $D$ spacetime dimensions; the factor $1/2$ here belongs to the four-dimensional congruence in this question, not the five-dimensional geometry of the next question.