 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-53/4/iv/solution

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

Now use proper-time dots and

H = \overset{a}{˙} / a

. Differentiate the slow-roll relation

H^{2} ≃ V / (3 M_{Pl}^{2})

, then substitute

\dot{\overset{ˉ}{Φ}} ≃ - V_{, Φ} / (3 H)

2 H \dot{H} ≃ \frac{V _{, Φ} Φ ˉ ˙}{3 M _{Pl}^{2}}, - \frac{H ˙}{H ^{2}} ≃ \frac{V _{, Φ}^{2}}{18 M _{Pl}^{2} H ^{4}} ≃ \frac{M _{Pl}^{2}}{2} (\frac{V _{, Φ}}{V})^{2} = ϵ_{V} .

Thus the Hubble slow-roll parameter

ϵ_{H}

agrees with

ϵ_{V}

to this order. Since

P_{ψ} \propto H_{k}^{2}

and

k = a H

at horizon exit,

\frac{d l n P _{ψ}}{d t} = 2 \frac{H ˙}{H}, \frac{d l n k}{d t} = H + \frac{H ˙}{H} = H (1 - ϵ_{H}) .

n_{t} = - \frac{2 ϵ _{H}}{1 - ϵ _{H}} ≃ - 2 ϵ_{V} .

The final equality is the requested first-order slow-roll approximation; corrections beyond that order also affect the horizon-exit spectrum itself.

 New to topics? Read the docs here!