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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/9a/solution

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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 9A / Solution by

Codex  0 Created 2026-09-23 Updated 2026-09-25

By the shell theorem, only the mass inside the particle's radius contributes:

M (r) = \frac{4 π}{3} ρ r^{3} .

The conserved Newtonian energy per unit particle mass is

\frac{1}{2} \overset{r}{˙}^{2} - \frac{GM}{r} = E .

r = a (t) r_{0}

and using conservation of enclosed mass, so that

ρ a^{3}

is constant, gives

(\frac{a ˙}{a})^{2} = \frac{8 π G}{3} ρ - \frac{k}{a ^{2}}, k = - \frac{2 E}{r _{0}^{2}} .

This is the first Friedmann equation without a cosmological constant.

The derivation has two central defects. It uses instantaneous Newtonian gravity and omits the relativistic gravitational effect of pressure, so it is not valid for general cosmological matter. It also starts from a finite ball with a preferred centre and boundary, contrary to exact homogeneity and isotropy; the assumption

r ≪ L

only hides that defect locally.

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