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

A perturbation $\delta \varphi$ shifts the time at which the rolling inflaton reaches a fixed value by $\delta t\simeq \delta \varphi /\dot{\varphi }$. The corresponding perturbation in local expansion is $\delta N=H\delta t$, which is the comoving curvature perturbation up to the stated sign convention:ThereforeUsing ${\dot{\varphi }}^2=2ϵH^2{M}_{\mathrm{Pl}}^2$ gives the slow-roll curvature power spectrumFor each $k$, the slowly varying quantities on the right are evaluated at cosmological horizon exit, $k=aH$.