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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-70/1/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 70 / 1 / c / Solution by

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

At leading order the scalar curvature power and total tensor power are

P_{S} = \frac{V}{24 π ^{2} M ^{4} ϵ _{V}}, P_{T} = \frac{2 V}{3 π ^{2} M ^{4}}, ϵ_{V} = \frac{M ^{2}}{2} (\frac{V ^{'}}{V})^{2}, η_{V} = \frac{M ^{2} V ^{''}}{V} .

Applying the preceding

d / d ln k

relation gives the scalar spectral index

n - 1 = - 6 ϵ_{V} + 2 η_{V}

and the tensor spectral index

n_{T} = - 2 ϵ_{V}

. For this hilltop,

η_{V} ≃ - \frac{36 M ^{2} ϕ ^{2}}{σ ^{4}} = - \frac{3}{2 N}, ϵ_{V} ≃ \frac{72 M ^{2} ϕ ^{6}}{σ ^{8}} = \frac{σ ^{4}}{192 M ^{4} N ^{3}} .

Thus, fifty e-folds before the end,

n ≃ 1 - \frac{3}{50} = 0.94, n_{T} ≃ - \frac{σ ^{4}}{96 M ^{4} ( 50 ) ^{3}} = - 5.26 \times 1 0^{- 5} (\frac{σ}{m _{pl}})^{4} .

The tiny

- 6 ϵ_{V}

contribution and finite-endpoint corrections have been omitted in the quoted leading scalar tilt. The tensor tilt is nearly zero but slightly negative; treating

H

as exactly constant in its spectrum would incorrectly discard its leading nonzero value.

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