Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 311 / 3

Because $t_a={\nabla }_{a}t$ is a gradient, ${\nabla }_a{t}_b={\nabla }_b{t}_a$. ThereforeThus $t^a$ is an affinely parametrized timelike geodesic vector field, and because $t$ is constant on $S$, it emanates orthogonally from $S$.

With $c=-{\nabla }_a{t}^a$, commute covariant derivatives and use the geodesic equation:This is the hypersurface-orthogonal Raychaudhuri equation written in terms of positive convergence rather than expansion.

The strong energy condition and Einstein's equation imply $R_{ab}{t}^a{t}^b\ge 0$. Since the congruence is hypersurface orthogonal, its vorticity vanishes. The tensor in the hint is the trace-free spatial shear, soBecause ${\nabla }_a{t}^a=-c$, part b gives

Let $\tau$ be proper time along a geodesic. While $c$ is finite and positive,ThusThe right side reaches zero at $\tau =3/c_0$, so the convergence must diverge no later than that proper time.

Suppose instead that every future-directed normal timelike geodesic were complete. Choose a point more than $3/c_0$ proper time to the future of $S$. Global hyperbolicity supplies a longest timelike curve from $S$ to that point; it is a geodesic orthogonal to $S$ and has no focal point before its endpoint. But part d makes the convergence of every such normal congruence diverge within proper time $3/c_0$. Nearby geodesics then intersect, producing a focal point after which the geodesic cannot maximize proper time. This contradiction proves that at least one timelike geodesic has finite maximal proper time, so $(M,g)$ has timelike geodesic incompleteness.