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For field amplitudes of order the guide field, the ion Lorentz-force response over time $\tau$ is $\Delta u/v_A=O(v_A\tau /l)\ll 1$. The electron-current velocity is of order $v_A{d}_i/l$, larger than a bulk velocity of order $v_A$. Thus the leading Hall magnetohydrodynamics induction equation becomes the closed electron magnetohydrodynamics equation, while ions remain fixed to leading order on this fast scale. The first nonzero ion response need not have zero acceleration. This limit retains the fast branch of incompressible Hall-MHD wave dispersion, not the slow branch; very oblique modes must separately satisfy the fast-time ordering.

Linearize about zero bulk flow and the uniform guide field. Put $\mathbf{b}=\delta \mathbf{B}/\sqrt{4\pi \rho }$, and take plane waves proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$. Let $k_{\parallel }=\mathbf{k}\cdot \hat{\mathbf{z}}$, $k=|\mathbf{k}|$, $q=k_{\parallel }{v}_A$, andBoth $\mathbf{u}$ and $\mathbf{b}$ are perpendicular to $\mathbf{k}$. Projecting momentum onto that plane eliminates total pressure. The remaining magnetic-tension term givesIn the induction equation, the Hall contribution is $-c(B_0{\partial }_z)(\nabla \times \delta \mathbf{B})/(4\pi en)$; its two derivative factors of $i$ in the Fourier mode produce a positive coefficient multiplying $\mathbf{k}\times \mathbf{b}$. ThusEliminating $\mathbf{u}$ givesOn the transverse plane, $(i\mathbf{k}\times {)}^2=k^2$, so applying that operator again proves the incompressible Hall-MHD wave dispersion:Equivalently, in a helicity decomposition of a transverse Fourier mode with $i\mathbf{k}\times {\mathbf{b}}_h=hk{\mathbf{b}}_h$, $h=\pm 1$, each polarization obeys ${\omega }^2-hqk{d}_i\omega -q^2=0$. For $k_{\parallel }=0$ the projected linear restoring terms vanish and these modes have zero frequency; the propagating calculation assumes $k_{\parallel }\ne 0$.