 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/14d/a/solution

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

Since

P = P (ρ)

and

ρ = \overset{ρ}{ˉ} (1 + δ)

\frac{c ^{2}}{ρ} \nabla P = \frac{c ^{2}}{ρ} \frac{d P}{d ρ} \nabla ρ ≃ c_{s}^{2} \nabla δ .

To first order, continuity and the divergence of Euler give

\dot{δ} + θ = 0, \dot{θ} + 2 \frac{a ˙}{a} θ = - \frac{4 π G}{c ^{2}} \overset{ρ}{ˉ} δ - c_{s}^{2} \nabla^{2} δ .

Eliminating

θ

and using

\nabla \mapsto i k / a

yields

\ddot{δ} + 2 \frac{a ˙}{a} \dot{δ} - (\frac{4 π G}{c ^{2}} \overset{ρ}{ˉ} - \frac{c _{s}^{2} k ^{2}}{a ^{2}}) δ = 0.

The second term is Hubble damping, gravity drives growth, and pressure produces oscillatory restoration. Their balance defines

k_{J}^{2} = \frac{4 π G ρ ˉ a ^{2}}{c ^{2} c _{s}^{2}}, λ_{J} = \frac{2 πa}{k _{J}} .

Modes longer than the Jeans length are gravitationally unstable; shorter modes are pressure supported.

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