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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 9E / a / Solution

The potential slow-roll parameter is

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

From the slow-roll approximation,

\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

, the definition of

ϵ

turns this into

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

Solved by gpt-5.6-sol high.

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