 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/9b/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

P = wρ

\frac{d ρ}{ρ} = - 3 (1 + w) \frac{d a}{a} .

Since

a (t_{0}) = 1

ρ = ρ_{0} a^{- 3 (1 + w)}, ρ_{0} = ρ (t_{0}),

When

Q = 0

, the expanding branch satisfies

\overset{a}{˙} = \frac{8 π G ρ _{0}}{3 c ^{2}} a^{- (1 + 3 w) /2} = H_{0} a^{- (1 + 3 w) /2},

where

H_{0} = \overset{a}{˙} (t_{0}) > 0

. Because

w > - 1

, integration gives

a (t)^{3 (1 + w) /2} = 1 + \frac{3}{2} (1 + w) H_{0} (t - t_{0}),

a (t) = [1 + \frac{3}{2} (1 + w) H_{0} (t - t_{0})]^{2/ [3 (1 + w)]} .

It vanishes at

t_{*} = t_{0} - \frac{2}{3 ( 1 + w ) H _{0}} < t_{0},

∣ w ∣ < 1

ensures

1 + w > 0

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