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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/9d/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 9D / i / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-29

Under the slow-roll approximation,

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

For

V = m^{2} ϕ^{2} /2

and the branch

ϕ > 0

A = \frac{4 π G}{3 c ^{2}} .

Then

H ≃ A m ϕ

and

\dot{ϕ} ≃ - \frac{m}{3 A} = - \frac{m c}{12 π G} .

With

ϕ (0) = ϕ_{0}

and

a (0) = a_{0}

,

ϕ (t) = ϕ_{0} - \frac{m c}{12 π G} t,

and integrating

\overset{a}{˙} / a = H

gives

a (t) = a_{0} exp [\frac{2 π G}{c ^{2}} (ϕ_{0}^{2} - ϕ (t)^{2})] .

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