 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-310/3/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

- k^{2} Φ = 4 π G a^{2} \overset{ρ}{ˉ}_{m} Δ_{m} .

Because nonrelativistic matter has

\overset{ρ}{ˉ}_{m} \propto a^{- 3}

, this implies

Φ \propto Δ_{m} / a

. Write

Φ = C Δ_{m} / a

and use

H = a^{'} / a

. Direct differentiation in the stated potential equation gives

Δ_{m}^{''} + H Δ_{m}^{'} + (H^{'} - H^{2}) Δ_{m} = 0.

The supplied Friedmann equation identity therefore yields

Δ_{m}^{''} + H Δ_{m}^{'} - 4 π G a^{2} \overset{ρ}{ˉ}_{m} Δ_{m} = 0.

Since

d / d τ = a d / d t

\ddot{Δ}_{m} + 2 H \dot{Δ}_{m} - 4 π G \overset{ρ}{ˉ}_{m} Δ_{m} = 0 .

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