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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-75/2/d/solution

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Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 75 / 2 / d / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Linearization removes the Poisson bracket from

\nabla_{∥}

. For Fourier perturbations proportional to

e^{i k \cdot x - iω t}

, the two reduced equations become

- iω Ψ = i v_{A}^{2} d_{i} k_{∥} b, - iωb = i d_{i} k_{∥} k_{⊥}^{2} Ψ.

(1)

Eliminating either amplitude gives

ω^{2} = v_{A}^{2} d_{i}^{2} k_{∥}^{2} k_{⊥}^{2}, ω = \pm k_{∥} v_{A} d_{i} k_{⊥} .

(2)

Since

k = (k_{⊥}^{2} + k_{∥}^{2})^{1/2} = k_{⊥} [1 + O (ϵ^{2})]

, this is the specified dispersion relation to the accuracy of the anisotropic reduction. The reduced equations cannot retain the subleading correction replacing

k_{⊥}

by the full

k

The full electron magnetohydrodynamics equation checks that correction directly. Its linearization is

\partial_{t} δ B = - α B_{0} \partial_{z} \nabla \times δ B

. Hence

- iω δ B = α B_{0} k_{∥} k \times δ B .

(3)

Applying the cross-product operator twice and using

k \cdot δ B = 0

gives

ω^{2} = (α B_{0})^{2} k_{∥}^{2} k^{2} = v_{A}^{2} d_{i}^{2} k_{∥}^{2} k^{2} .

(4)

Thus the full model gives the stated relation exactly and the reduced model gives its leading anisotropic limit, with all signs and normalization factors consistent.

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