 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/9e/a/solution

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

ϵ = \frac{M _{Pl}^{2}}{2} (\frac{V ^{'}}{V})^{2} .

\dot{ϕ} = - \frac{V ^{'}}{3 H}, H^{2} = \frac{V}{3 M _{Pl}^{2}} .

Therefore

\frac{ϕ ˙ ^{2}}{2 V} = \frac{V ^{'2}}{18 H ^{2} V} = \frac{M _{Pl}^{2} V ^{'2}}{6 V ^{2}} = \frac{ϵ}{3} .

Thus

ϵ ≪ 1

implies

\dot{ϕ}^{2} ≪ V

, consistently with potential-energy domination.

The same equations give

\frac{H}{ϕ ˙} = - \frac{3 H ^{2}}{V ^{'}} = - \frac{1}{M _{Pl}^{2}} \frac{V}{V ^{'}} .

On the branch

V^{'} / V > 0

ϵ

turns this into

\frac{H}{ϕ ˙} = - \frac{1}{M _{Pl}^{2}} \frac{V}{V ^{'}} = - \frac{1}{2 M _{Pl}} \frac{1}{ϵ} .

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