 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/38d/v/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Differentiate

B_{α β} = \nabla_{β} X_{α}

along

X

and commute the two covariant derivatives:

X^{μ} \nabla_{μ} B_{α β} = X^{μ} \nabla_{μ} \nabla_{β} X_{α} = X^{μ} \nabla_{β} \nabla_{μ} X_{α} - X^{μ} R^{ν}_{αμ β} X_{ν},

where the sign is the covector form of the Ricci identity. Apply the product rule to the first term:

X^{μ} \nabla_{β} \nabla_{μ} X_{α} = \nabla_{β} (X^{μ} \nabla_{μ} X_{α}) - (\nabla_{β} X^{μ}) (\nabla_{μ} X_{α}) .

The first term vanishes by the geodesic equation, and the second is

- B^{μ}_{β} B_{αμ} .

- X^{μ} X_{ν} R^{ν}_{αμ β} = R_{μ β α}^{ν} X^{μ} X_{ν} .

Consequently

X^{μ} \nabla_{μ} B_{α β} = - B^{μ}_{β} B_{αμ} + R_{μ β α}^{ν} X^{μ} X_{ν} .

This is the Riccati equation for a timelike geodesic congruence; taking its spatial trace is the starting point for the timelike Raychaudhuri equation.

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