 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-53/3/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

δ g_{00} = 2 a^{2} ψ

δ g_{0 i} = - a^{2} B_{i}

, and

δ g_{ij} = 2 a^{2} (ϕ δ_{ij} - E_{ij})

. Apply part (i). The time-time component changes by

- 2 a^{2} (T^{'} + H T)

, and the time-space component changes by

a^{2} (L_{i}^{'} - \partial_{i} T)

. Hence

ψ = ψ - T^{'} - H T, B_{i} = B_{i} + \partial_{i} T - L_{i}^{'} .

For the spatial components,

δ g_{ij} - δ g_{ij} = 2 a^{2} [H T δ_{ij} + \partial_{(i} L_{j)}] .

Taking the trace and the symmetric trace-free part separately gives

ϕ = ϕ + H T + \frac{1}{3} \partial_{i} L^{i}, E_{ij} = E_{ij} - \partial_{⟨ i} L_{j ⟩},

where

\partial_{⟨ i} L_{j ⟩} = \frac{1}{2} (\partial_{i} L_{j} + \partial_{j} L_{i}) - δ_{ij} \partial_{k} L^{k} /3

. This explicitly accounts for the trace term in

ϕ

; it must not be absorbed into the trace-free shear.

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