Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 2 / b

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- Solution b

Solution

Define the acceleration of the normal congruence by

a_{σ} = n^{ρ} \nabla_{ρ} n_{σ} .

Because

S^{σγ}

is spatial, differentiating

n_{σ} S^{σγ} = 0

gives

n_{σ} \nabla_{ρ} S^{σγ} = - (\nabla_{ρ} n_{σ}) S^{σγ} .

Insert

δ^{ρ}_{σ} = ⊥^{ρ}_{σ} - n^{ρ} n_{σ}

into the contracted derivative and project the free index:

⊥^{γ}_{α} \nabla_{μ} S^{μ}_{γ} = ⊥^{γ}_{α} ⊥^{ρ}_{σ} \nabla_{ρ} S^{σ}_{γ} - ⊥^{γ}_{α} n^{ρ} n_{σ} \nabla_{ρ} S^{σ}_{γ} = D_{μ} S^{μ}_{α} + a_{σ} S^{σ}_{α} .

Rearranging proves

D_{μ} S^{μ}_{α} = ⊥^{γ}_{α} \nabla_{μ} S^{μ}_{γ} - a_{σ} S^{σ}_{α} .

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