Solution

Linearize about zero bulk flow and the uniform guide field. Put

b = δ B / 4 π ρ

, and take plane waves proportional to

e^{i (k \cdot x - ω t)}

. Let

k_{∥} = k \cdot \hat{z}

k = ∣ k ∣

q = k_{∥} v_{A}

, and

d_{i} = \frac{c}{ω _{p i}}, v_{A} d_{i} = \frac{c B _{0}}{4 π e n} .

Both

u

and

b

k

. Projecting momentum onto that plane eliminates total pressure. The remaining magnetic-tension term gives

- iω u = i q b, ω u = - q b .

In the induction equation, the Hall contribution is

- c (B_{0} \partial_{z}) (\nabla \times δ B) / (4 π e n)

; its two derivative factors of

i

in the Fourier mode produce a positive coefficient multiplying

k \times b

. Thus

- iω b = i q u + q d_{i} k \times b, ω b = - q u + i q d_{i} k \times b .

Eliminating

u

gives

(ω^{2} - q^{2}) b = iω q d_{i} k \times b .

On the transverse plane,

(i k \times)^{2} = k^{2}

, so applying that operator again proves the incompressible Hall-MHD wave dispersion:

(ω^{2} - k_{∥}^{2} v_{A}^{2})^{2} = ω^{2} k_{∥}^{2} v_{A}^{2} k^{2} \frac{c ^{2}}{ω _{p i}^{2}} .

i k \times b_{h} = hk b_{h}

h = \pm 1

, each polarization obeys

ω^{2} - h q k d_{i} ω - q^{2} = 0

. For

k_{∥} = 0

the projected linear restoring terms vanish and these modes have zero frequency; the propagating calculation assumes

k_{∥} \neq = 0

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