 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-62/3/b/solution

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

- g = a^{4}

g^{00} = - a^{- 2}

and

g^{ij} = a^{- 2} δ^{ij}

- (a^{2} δ ϕ^{'})^{'} + a^{2} \nabla^{2} δ ϕ = 0.

k

, the Laplacian contributes

- k^{2}

, where

k = ∣ k ∣

. Thus

δ ϕ_{k}^{''} + 2 \frac{a ^{'}}{a} δ ϕ_{k}^{'} + k^{2} δ ϕ_{k} = 0.

Since

a^{'} / a = - 1/ τ

, this yields

δ ϕ_{k}^{''} - \frac{2}{τ} δ ϕ_{k}^{'} = - k^{2} δ ϕ_{k} .

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