Contracting the <Einstein field equations> with the null tangent eliminates the trace term and gives
$$
R_{ab}U^aU^b=8\pi T_{ab}U^aU^b\geq0
$$
by the <null energy condition>. The screen metric is positive definite, so $\widehat\sigma_{ab}\widehat\sigma^{ab}\geq0$. The <Null Raychaudhuri equation> consequently implies
$$
\frac{d\theta}{d\lambda}\leq-\frac12\theta^2.
$$
While $\theta<0$ this is equivalent to
$$
\frac{d}{d\lambda}\left(\frac1\theta\right)
\geq\frac12.
$$
If $	heta(\lambda_0)=\theta_0<0$, integration gives
$$
\frac1{\theta(\lambda)}
\geq\frac1{\theta_0}+\frac{\lambda-\lambda_0}{2}.
$$
The right-hand side reaches zero after affine distance $2/|\theta_0|$. A finite negative expansion cannot pass through this value, so $	heta\to-\infty$ no later than that point. This is the <null focusing theorem>.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-311/1/c/solution