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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 37D / b / Solution

Let

L (x, \overset{x}{˙}) = g_{μν} \overset{x}{˙}^{μ} \overset{x}{˙}^{ν} = \frac{d s}{d λ} .

The Euler-Lagrange equation gives

\frac{d}{d λ} (\frac{g _{β ν} x ˙ ^{ν}}{L}) - \frac{1}{2 L} \partial_{β} g_{μν} \overset{x}{˙}^{μ} \overset{x}{˙}^{ν} = 0.

Since

\overset{x}{˙}^{μ} = L d x^{μ} / d s

and

d / d λ = L d / d s

, this becomes

\frac{d}{d s} (g_{β ν} \frac{d x ^{ν}}{d s}) - \frac{1}{2} \partial_{β} g_{μν} \frac{d x ^{μ}}{d s} \frac{d x ^{ν}}{d s} = 0.

Expanding the derivative, multiplying by the inverse metric, and using the Christoffel symbol

Γ_{μν}^{α} = \frac{1}{2} g^{α β} (\partial_{μ} g_{β ν} + \partial_{ν} g_{β μ} - \partial_{β} g_{μν})

gives the geodesic equation

\frac{d ^{2} x ^{α}}{d s ^{2}} + Γ_{μν}^{α} \frac{d x ^{μ}}{d s} \frac{d x ^{ν}}{d s} = 0 .

Solved by gpt-5.6-sol high.

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