 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-59/2/c/solution

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

Let

B = H Φ + Φ^{'}

δ ϕ = 2 B / ϕ_{0}^{'}

. Differentiating it and using the background field equation gives

ϕ_{0}^{'} δ ϕ^{'} = 2 [B^{'} - \frac{ϕ _{0}^{''}}{ϕ _{0}^{'}} B] = 2 B^{'} + 4 H B .

Insert this into the energy constraint, whose right-hand side in the PDF contains

- ϕ_{0}^{'2} Φ

2 (- 3 H B + \nabla^{2} Φ) = - ϕ_{0}^{'2} Φ + 2 B^{'} + 4 H B .

Rearranging and expanding

B

gives

Φ^{''} + 6 H Φ^{'} - \nabla^{2} Φ + (H^{'} + 5 H^{2} - \frac{1}{2} ϕ_{0}^{'2}) Φ = 0.

Here

ϕ_{0}^{'2} /2 = 3 H^{2}

and

H^{'} = - 2 H^{2}

Φ

cancels, proving

Φ^{''} + 6 \frac{a ^{'}}{a} Φ^{'} - \nabla^{2} Φ = 0.

This also identifies the derivative lost from the background-field square in the TeX transcription.

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