Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 312 / 3 / i

Table of contents
- Solution i

Solution

Let

θ = \nabla \cdot v

. Vanishing vorticity implies in Fourier space

v (q) = - i \frac{q}{q ^{2}} θ (q) .

Fourier transforming

\nabla \cdot (δ v)

in the continuity equation gives

δ^{'} (k) + θ (k) = - \int_{q_{1}, q_{2}} δ_{D} (k - q_{1} - q_{2}) α (q_{1}, q_{2}) θ (q_{1}) δ (q_{2}),

with the alpha mode-coupling kernel

α (q_{1}, q_{2}) = \frac{( q _{1} + q _{2} ) \cdot q _{1}}{q _{1}^{2}} .

Taking the divergence of

(v \cdot \nabla) v

and symmetrizing its two velocity arguments gives

θ^{'} (k) + H θ (k) + \frac{3}{2} H^{2} Ω_{m} δ (k) = - \int_{q_{1}, q_{2}} δ_{D} (k - q_{1} - q_{2}) β (q_{1}, q_{2}) θ (q_{1}) θ (q_{2}),

where the beta mode-coupling kernel is

β (q_{1}, q_{2}) = \frac{∣ q _{1} + q _{2} ∣ ^{2} ( q _{1} \cdot q _{2} )}{2 q _{1}^{2} q _{2}^{2}} .

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