 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-311/3/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

Let

t^{a}

exhibit the stable causality of

g

, so

g - t_{a} t_{b}

r^{a} = \frac{t ^{a}}{2}, g_{ab}^{'} = g_{ab} - r_{a} r_{b} = g_{ab} - \frac{1}{2} t_{a} t_{b} .

r^{a}

remains timelike for

g^{'}

. Widening the cones of

g^{'}

once more using

r

gives

g_{ab}^{'} - r_{a} r_{b} = g_{ab} - t_{a} t_{b},

which is causal by assumption. Hence

g - rr is itself stably causal .

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