 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/15b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

2 H \dot{H} = \frac{1}{3 M _{Pl}^{2}} (\dot{ϕ} \ddot{ϕ} + c^{2} V^{'} (ϕ) \dot{ϕ}) .

\ddot{ϕ} + c^{2} V^{'} = - 3 H \dot{ϕ}

, hence

2 H \dot{H} = - \frac{H ϕ ˙ ^{2}}{M _{Pl}^{2}} .

By continuity this also holds through isolated zeros of

H

\dot{H} = - \frac{ϕ ˙ ^{2}}{2 M _{Pl}^{2}} .

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