Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 309 / 1 / d

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- Solution d

Solution

Use proper time

s

, so

k = - 1

. The first integral becomes

\overset{r}{˙}^{2} = e^{2} - 1 - r^{2} = A^{2} - r^{2}, A = e^{2} - 1 .

For an outward geodesic starting at the origin,

r (s) = A sin s

until it next reaches

r = 0

. This occurs at

T = π .

Thus every radial timelike geodesic in anti-de Sitter spacetime through the origin returns after the same proper time; restoring anti-de Sitter radius

a

gives

T = πa

.

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