Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 2 / b

Table of contents
- Solution b

Solution

Linearize the inviscid momentum equation in the uniformly rotating frame. The Coriolis acceleration is

2 Ω \times δ u

, and the equilibrium pressure gradient cancels gravity. For perturbations proportional to

e^{i k_{x} x - iω t}

, the horizontal components are

- iω ρ δ u_{x} - 2Ω ρ δ u_{y} = - i k_{x} δ p,

- iω ρ δ u_{y} + 2Ω ρ δ u_{x} = 0.

The vertical component is

- iω ρ δ u_{z} = - g δ ρ - \frac{d δ p}{d z} .

Linearizing mass conservation,

\partial_{t} δ ρ + \nabla \cdot (ρ δ u) = 0,

gives

- iω δ ρ + δ u_{z} \frac{d ρ}{d z} = - ρ (i k_{x} δ u_{x} + \frac{d δ u _{z}}{d z}) .

Finally, linearizing the adiabatic pressure equation gives

- iω δ p + δ u_{z} \frac{d p}{d z} = - γ p (i k_{x} δ u_{x} + \frac{d δ u _{z}}{d z}) .

These are the stated five equations. Self-gravity contributes no perturbation because it is neglected, and the equilibrium centrifugal term has already been absorbed or omitted.

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