 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/9b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

H^{2} = 8 π Gρ /3

\overset{ρ}{˙} = - 3 H (ρ + P)

2 H \dot{H} = \frac{8 π G}{3} \overset{ρ}{˙} = - 8 π G H (ρ + P),

\dot{H} = - 4 π G (ρ + P)

. Since

H = \overset{a}{˙} / a

\frac{a ¨}{a} = \dot{H} + H^{2} = - 4 π G (ρ + P) + \frac{8 π G}{3} ρ .

\frac{a ¨}{a} = - \frac{4 π G}{3} (ρ + 3 P) .

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