 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/15b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Since

H = \overset{a}{˙} / a

, differentiation gives the Hubble parameter identity

\dot{H} = \frac{a ¨}{a} - H^{2} .

Subtracting the given Friedmann equation from the Raychaudhuri equation therefore yields

\dot{H} = - \frac{8 π G}{3} (\dot{ϕ}^{2} - V) - \frac{8 π G}{3} (\frac{1}{2} \dot{ϕ}^{2} + V) = - 4 π G \dot{ϕ}^{2} .

On the other hand, differentiating the Friedmann equation gives

2 H \dot{H} = \frac{8 π G}{3} \dot{ϕ} (\ddot{ϕ} + V^{'} (ϕ)) .

Substitute

\dot{H} = - 4 π G \dot{ϕ}^{2}

and divide by

(8 π G /3) \dot{ϕ}

, which is permitted by the assumption

\dot{ϕ} \neq = 0

. The result is the homogeneous inflaton equation of motion

\ddot{ϕ} + 3 H \dot{ϕ} + V^{'} (ϕ) = 0 .

The middle term is Hubble friction.

 New to topics? Read the docs here!