Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 59 / 2 / d

Solution

Φ_{k}^{''} + 6 H Φ_{k}^{'} + k^{2} Φ_{k} = 0

Φ_{k} = a^{- 3} u_{k}

. Direct differentiation removes the first derivative and gives

u_{k}^{''} + [k^{2} - 3 H^{'} - 9 H^{2}] u_{k} = u_{k}^{''} + [k^{2} - \frac{3}{4 τ ^{2}}] u_{k} = 0.

For

∣ k τ ∣ ≫ 1

, the inverse-square term is negligible, and the positive incoming-frequency branch has

u_{k} \sim A_{k} e^{- ik τ}

. Therefore

Φ_{k} \sim A_{k} a^{- 3} e^{- ik τ} .

δ ϕ_{k} = \frac{2 ( Φ _{k}^{'} + H Φ _{k} )}{ϕ _{0}^{'}} \sim \frac{2 ik A _{k}}{C} a^{- 1} e^{- ik τ},

where

ϕ_{0}^{'} = - C / a^{2}

. Thus the scalar perturbation has precisely the required

a^{- 1}

behavior. Conversely, the two constraints imply the potential equation in part (c); reconstructing

δ ϕ

by the momentum constraint therefore satisfies the energy constraint too. This checks compatibility beyond simply comparing powers of

a

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