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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 9B / c / Solution

Insert

ρ = \overset{ρ}{ˉ} + ϵ δ ρ, v = ϵ δ v

into the continuity equation. Its zeroth-order part is the homogeneous cosmological continuity equation

\dot{\overset{ρ}{ˉ}} + 3 H \overset{ρ}{ˉ} = 0.

Since

\overset{ρ}{ˉ}

is spatially homogeneous, the terms of first order in

ϵ

give

\frac{\partial δ ρ}{\partial t} + 3 Hδ ρ + \frac{ρ ˉ}{a} \nabla \cdot δ v = 0.

On the other hand, the quotient rule and the background equation imply

\frac{\partial}{\partial t} (\frac{δ ρ}{ρ ˉ}) = \frac{1}{ρ ˉ} \frac{\partial δ ρ}{\partial t} - \frac{δ ρ}{ρ ˉ ^{2}} \dot{\overset{ρ}{ˉ}} = \frac{1}{ρ ˉ} \frac{\partial δ ρ}{\partial t} + 3 H \frac{δ ρ}{ρ ˉ} .

Substitution of the first-order equation yields the linearized cosmological continuity equation

\frac{\partial}{\partial t} (\frac{δ ρ}{ρ ˉ}) = - \frac{1}{a} \nabla \cdot δ v .

The ratio

δ ρ / \overset{ρ}{ˉ}

is the density contrast.

Solved by gpt-5.6-sol high.

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