Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 3 / d / ii

Table of contents
- Solution ii

Solution

In vacuum, neglecting

A_{mn} A^{mn}

leaves

\frac{d K}{d t} = \frac{1}{3} K^{2} .

Separating variables and imposing

K (t_{0}) = K_{0} > 0

gives

- \frac{1}{K ( t )} + \frac{1}{K _{0}} = \frac{t - t _{0}}{3},

hence

K (t) = \frac{K _{0}}{1 - \frac{K _{0}}{3} ( t - t _{0} )} .

The denominator vanishes at

t_{\infty} = t_{0} + \frac{3}{K _{0}} .

This finite-time blow-up of mean curvature in geodesic slicing is another direct expression of the gauge's singularity problem.

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