 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-3/14b/c/solution

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

Put

a = α a

with

α = \frac{c}{R}, Γ = \frac{8 π G ρ _{0} R ^{4}}{3 c ^{6}} .

Then

(\frac{a ˙}{a})^{2} = \frac{Γ}{a ^{4}} - \frac{1}{a ^{2}} + \frac{Λ}{3} .

For

y = a^{2}

this becomes

\overset{y}{˙}^{2} = 4Γ - 4 y + \frac{4Λ}{3} y^{2} .

The expanding big-bang branch has

y (0) = 0, \overset{y}{˙} (0) = 2 Γ .

Differentiation gives

\overset{y}{¨} = \frac{4Λ}{3} y - 2.

Thus, with

ω = 4Λ/3

y (t) = \frac{3}{2Λ} [1 - cosh (ω t) + λ sinh (ω t)],

where the initial derivative gives

λ = 2 \frac{ΛΓ}{3} .

For

λ > 1

a

expands forever and becomes asymptotically exponential. For

λ < 1

, it reaches a maximum and recollapses to zero. At

λ = 1

y

increases monotonically to

3/ (2Λ)

, the limiting static radius.

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