Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 310 / 4 / c

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- Solution c

Solution

At cosmological horizon exit,

d lo g k / d N = 1 - ϵ ≃ 1

. Differentiating the slow-roll curvature power spectrum and using the Hubble slow-roll parameters gives

n_{s} - 1 = \frac{d l o g Δ _{R}^{2}}{d l o g k} = - 2 ϵ - η .

The supplied relations imply

ϵ = \frac{M _{Pl}^{2}}{2} (\frac{V ^{'}}{V})^{2}, η = 4 ϵ - 2 M_{Pl}^{2} \frac{V ^{''}}{V} .

Consequently the scalar spectral index is

n_{s} - 1 = - 3 M_{Pl}^{2} (\frac{V ^{'}}{V})^{2} + 2 M_{Pl}^{2} \frac{V ^{''}}{V} .

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