 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/9b/solution

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

δ m

on the surface of a homogeneous sphere, its Newtonian mechanical energy is

E = \frac{1}{2} δ m \dot{R}^{2} - \frac{GM δ m}{R} .

The spatially flat case corresponds to zero specific energy. Using

M = \frac{4 π}{3} ρ R^{3}

and dividing by

δ m R^{2} /2

gives

(\frac{R ˙}{R})^{2} = \frac{8 π G}{3} ρ .

Since

R

is proportional to the scale factor

a

H^{2} = (\frac{a ˙}{a})^{2} = \frac{8 π G}{3} ρ .

Now take

a (t_{0}) = 1

and

ρ = ρ_{0} a^{- 4}

. Put

C = \frac{8 π G ρ _{0}}{3} .

The contracting branch of the Friedmann equation is

\frac{a ˙}{a} = - C a^{- 2}, \overset{a}{˙} = - \frac{C}{a} .

Integrating

a d a = - C d t

from

t_{0}

gives

\frac{1}{2} (a (t)^{2} - 1) = - C (t - t_{0}) .

Therefore

a (t) = 1 - 2 C (t - t_{0})

on the contracting interval, reaching

a = 0

t = t_{0} + 1/ (2 C)

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