 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-75/1/3/solution

Solving the polarization quadratic gives all four signed roots:

ω_{h, s} = \frac{q}{2} (hk d_{i} + s 4 + k^{2} d_{i}^{2}), h, s = \pm 1.

A sign-independent description is the pair of positive frequency magnitudes

Ω_{fast, slow} = ∣ k_{∥} ∣ v_{A} [1 + \frac{k ^{2} d _{i}^{2}}{4} \pm \frac{k d _{i}}{2}] .

For wavelengths long compared with the ion skin depth,

k d_{i} ≪ 1

, both magnitudes approach

∣ k_{∥} ∣ v_{A}

. The Hall correction is then small and the usual Alfvén wave dispersion

ω^{2} = k_{∥}^{2} v_{A}^{2}

is recovered. In an incompressible system the two transverse polarizations include the shear and pseudo-Alfvén wave sectors; away from this limit Hall coupling selects circular helicity combinations. A small parallel wavenumber alone does not remove the relative Hall splitting: the expansion parameter is the full

k d_{i}

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