 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/14a/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

\dot{ϕ}^{2} ≪ V (ϕ), ∣ \ddot{ϕ} ∣ ≪ 3 H ∣ \dot{ϕ} ∣.

The first condition makes the energy density potential dominated, and the second neglects inertial acceleration relative to Hubble friction. The scalar and Friedmann equations reduce to

3 H \dot{ϕ} ≃ - V^{'} (ϕ), H^{2} ≃ \frac{8 π G}{3 c ^{2}} V (ϕ) .

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