 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/15b/d/solution

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

For the quartic potential,

\frac{V}{V ^{'}} = \frac{ϕ}{4} .

Integrating the result of part (c) along the trajectory from

ϕ_{i}

ϕ_{f}

gives

N ≃ - 8 π G \int_{ϕ_{i}}^{ϕ_{f}} \frac{ϕ}{4} d ϕ = π G (ϕ_{i}^{2} - ϕ_{f}^{2}) .

Because the negative field rolls toward zero,

ϕ_{i} < ϕ_{f} < 0

and

∣ ϕ_{i} ∣ > ∣ ϕ_{f} ∣

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