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 Incoming links: E-fold count for the natural-inflation cosine potential

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 9E / c / Solution Created 2026-09-24 Updated 2026-09-25

For the natural inflation potential, write

x = ϕ / f

. Then

V^{'} = - \frac{V _{0}}{f} sin x

and the half-angle identity gives

\frac{V ^{'}}{V} = \frac{1}{f} \frac{s i n x}{1 + c o s x} = \frac{1}{f} tan \frac{x}{2} .

Therefore, on the stated rolling branch,

ϵ (ϕ) = \frac{M _{Pl}}{2 f} tan \frac{ϕ}{2 f} .

Substitution into part (b) gives

N = \frac{f}{M _{Pl}^{2}} \int_{ϕ_{f}}^{ϕ_{i}} cot \frac{ϕ}{2 f} d ϕ .

Since

\int cot \frac{ϕ}{2 f} d ϕ = 2 f lo g sin \frac{ϕ}{2 f},

the e-fold count for the natural-inflation cosine potential is

N = \frac{2 f ^{2}}{M _{Pl}^{2}} [lo g (sin \frac{ϕ _{i}}{2 f}) - lo g (sin \frac{ϕ _{f}}{2 f})] .

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