 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-310/2/a/solution

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

V^{'} = - m^{2} ϕ, V^{''} = - m^{2} .

ϵ_{V} = \frac{M _{Pl}^{2} m ^{4} ϕ ^{2}}{2 ( Λ ^{4} - \frac{1}{2} m ^{2} ϕ ^{2} ) ^{2}} ≃ \frac{M _{Pl}^{2} m ^{4} ϕ ^{2}}{2 Λ ^{8}}

and

η_{V} = - \frac{M _{Pl}^{2} m ^{2}}{Λ ^{4} - \frac{1}{2} m ^{2} ϕ ^{2}} ≃ - \frac{M _{Pl}^{2} m ^{2}}{Λ ^{4}} .

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