 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-64/3/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Use perturbations

v, b, p, ϑ

D_{0} = \partial_{t} - \frac{3}{2} Ω x \partial_{y} .

This is advection by the background Keplerian shearing sheet. Subtracting the equilibrium before retaining first-order terms gives the nine linearized equations:

D_{0} v_{x} - 2Ω v_{y} D_{0} v_{y} + \frac{1}{2} Ω v_{x} D_{0} v_{z} D_{0} b_{x} D_{0} b_{y} D_{0} b_{z} D_{0} ϑ \partial_{x} v_{x} + \partial_{y} v_{y} + \partial_{z} v_{z} \partial_{x} b_{x} + \partial_{y} b_{y} + \partial_{z} b_{z} = - ρ_{0}^{- 1} \partial_{x} p - N^{2} ϑ + \frac{B _{0}}{4 π ρ _{0}} \partial_{z} b_{x}, = - ρ_{0}^{- 1} \partial_{y} p + \frac{B _{0}}{4 π ρ _{0}} \partial_{z} b_{y}, = - ρ_{0}^{- 1} \partial_{z} p + \frac{B _{0}}{4 π ρ _{0}} \partial_{z} b_{z}, = B_{0} \partial_{z} v_{x}, = - \frac{3}{2} Ω b_{x} + B_{0} \partial_{z} v_{y}, = B_{0} \partial_{z} v_{z}, = v_{x}, = 0, = 0.

The azimuthal coefficient

Ω/2

combines the Coriolis acceleration with perturbation advection of the background shear. The

- 3Ω b_{x} /2

term is the winding of radial magnetic field by that shear. The tidal force cancels when the equations at a fixed position are subtracted; it does not add a separate Eulerian perturbation force. Magnetic pressure is already included in

p

, leaving only linear magnetic tension. In this normalization

ϑ

has dimensions of length because

D_{0} ϑ = v_{x}

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