Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 2 / c

Solution

γ_{ij} = diag (\frac{( r + M ) ^{2}}{r ^{2}}, (r + M)^{2}, (r + M)^{2} sin^{2} θ) .

β_{r} = g_{t r} = M / r

. Raising its index with

γ^{rr} = r^{2} / (r + M)^{2}

β^{r} = \frac{M r}{( r + M ) ^{2}}, β^{θ} = β^{ϕ} = 0.

Now

β_{i} β^{i} = \frac{M ^{2}}{( r + M ) ^{2}}

and

g_{tt} = - α^{2} + β_{i} β^{i} = - (r - M) / (r + M)

. The positive lapse function is consequently

α = \frac{r}{r + M} .

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