Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 54 / 2 / a

A null geodesic congruence is a smooth family of null geodesics filling a region without intersections there, with one tangent direction assigned smoothly at every point. Choose an affine parameter $\lambda$ and tangent $U^a$, so $U^a{U}_a=0$ and $U^b{\nabla }_b{U}^a=0$. Choose a second null vector $N^a$ with $U\cdot N=-1$. The screen-space projector isIt annihilates $U,N$ and restricts to a positive-definite metric on the two-dimensional transverse screen. Project the covariant derivative of the tangent to obtain the optical tensorIts decomposition defines the three optical quantities:The null expansion is the fractional rate of change of a transverse area, $\theta =A^{-1}dA/d\lambda$. The null shear changes a beam's shape without changing its area to first order. The null twist is its antisymmetric transverse deformation. The factor $1/2$ uses four spacetime dimensions.

For an affinely parametrized null geodesic congruence, put $B^a{}_b={\nabla }_b{U}^a$. Differentiating along a ray and commuting covariant derivatives givesusing $[{\nabla }_c,{\nabla }_b]V^a=R^a{}_{dcb}{V}^d$. The geodesic acceleration term vanishes. Since $U$ is null and affine, $B^a{}_b{U}^b=0$ and $U_a{B}^a{}_b=0$. Its trace is therefore the screen trace $\theta$, and the trace of its square equals that of the optical tensor squared. Taking the trace, with $R_{db}=R^a{}_{dab}$, yieldsNow decompose the optical tensor into null expansion, null shear and null twist. The symmetric trace-free part is orthogonal to the trace and to the antisymmetric part, while ${\hat{\omega }}_{ab}{\hat{\omega }}^{ba}=-{\hat{\omega }}_{ab}{\hat{\omega }}^{ab}$. HenceSubstitution proves the Null Raychaudhuri equation:The squares use the positive-definite screen metric. Thus null shear contributes to focusing, whereas nonzero null twist has the opposite sign.