Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 38C / c / i

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- Solution i

Solution

The Schwarzschild exterior is vacuum, so

R_{μν} = 0, R = 0,

and consequently

C_{α β γ δ} = R_{α β γ δ}

. Put

f (r) = 1 - \frac{2 M}{r} .

The relevant Christoffel symbols are

Γ^{t}_{t r} = \frac{f ^{'}}{2 f}, Γ^{r}_{rr} = - \frac{f ^{'}}{2 f}, Γ^{r}_{tt} = \frac{1}{2} f f^{'} .

With the convention

R^{α}_{β γ δ} = \partial_{γ} Γ_{β δ}^{α} - \partial_{δ} Γ_{β γ}^{α} + \dots,

one finds

R_{t r t r} = \frac{1}{2} f^{''} (r) .

Since

f^{''} = - 4 M / r^{3}

,

C_{t r t r} = - \frac{2 M}{r ^{3}} (r ≫ 2 M) .

This is the leading Radial Weyl curvature of Schwarzschild spacetime. The opposite convention for the Riemann tensor reverses the sign but not the physical tidal magnitude.

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