Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 310 / 4 / b

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- Solution b

Solution

The right-hand side of the slow-roll curvature power spectrum must be evaluated separately for each mode at cosmological horizon exit,

k = a H

. Taking logarithms gives

lo g Δ_{R}^{2} = 2 lo g H - lo g ϵ + constant .

For

N = lo g a

,

\frac{d l o g H}{d N} = - ϵ, \frac{d l o g ϵ}{d N} = η, \frac{d l o g k}{d N} = 1 - ϵ .

Hence

n_{s} - 1 = \frac{- 2 ϵ - η}{1 - ϵ} = - 2 ϵ - η + O (ϵ^{2}, ϵη) .

This is the scalar spectral index in Hubble slow-roll parameters.

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