 Solution

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

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

For

V = λ ϕ^{4} /4

and positive

ϕ

, the slow-roll Friedmann equation gives

H ≃ ϕ^{2} \frac{2 π G λ}{3 c ^{2}} .

The scalar equation then reduces to

\dot{ϕ} ≃ - c \frac{λ}{6 π G} ϕ .

With

ϕ (t_{0}) = ϕ_{0}

ϕ (t) = ϕ_{0} exp [- c \frac{λ}{6 π G} (t - t_{0})] .

Furthermore,

\frac{d l o g a}{d ϕ} = \frac{H}{ϕ ˙} ≃ - \frac{3 H ^{2}}{V ^{'}} = - \frac{2 π G}{c ^{2}} ϕ .

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

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