 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-55/5/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

H^{2} ≃ V / (3 M_{Pl}^{2})

and

\dot{Φ} ≃ - V^{'} / (3 H)

. Substituting in the stated slow-roll curvature power spectrum yields

P_{R} = \frac{H ^{4}}{4 π ^{2} Φ ˙ ^{2}} ≃ \frac{9 H ^{6}}{4 π ^{2} V ^{'2}} = \frac{V ^{3}}{12 π ^{2} M _{Pl}^{6} V ^{'2}} .

ϵ_{V} = M_{Pl}^{2} V^{'2} / (2 V^{2})

P_{R} ≃ \frac{V}{24 π ^{2} M _{Pl}^{4} ϵ _{V}} = \frac{8}{3 ϵ _{V}} (\frac{V ^{1/4}}{8 π M _{Pl}})^{4} .

All quantities are evaluated at horizon exit. The two forms have the same normalization because

(8 π)^{4} = 64 π^{2}

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