Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 75 / 1

Use Gaussian units, uniform number density $n$, and ion mass density $\rho =n{m}_i$. Here $e>0$ is the magnitude of the elementary charge, as fixed by the printed current relation. Neglect electron inertia and displacement current. ThenFreezing the magnetic flux into this electron velocity, while retaining the ion momentum equation, gives the closed Hall magnetohydrodynamics systemThe pressure enforces the incompressible flow constraint; taking the divergence of momentum determines it up to the usual boundary-dependent constant. The induction equation preserves the solenoidal magnetic-field constraint. One can absorb $B^2/(8\pi )$ into magnetohydrodynamic total pressure, replacing the Lorentz term by magnetic tension. Keeping ion inertia here is important: electron magnetohydrodynamics with fixed ions will be a later limit, not the complete two-field system.

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 $k_{\parallel }\ne 0$.

Solving the polarization quadratic gives all four signed roots:A sign-independent description is the pair of positive frequency magnitudesFor wavelengths long compared with the ion skin depth, $k{d}_i\ll 1$, both magnitudes approach $|k_{\parallel }|v_A$. The Hall correction is then small and the usual Alfvén wave dispersion ${\omega }^2=k_{\parallel }^2{v}_A^2$ 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 $k{d}_i$.

When $k{d}_i\gg 1$, expand the square root in the exact frequencies. The fast branch haswhile the slow branch has ${\Omega }_{\mathrm{slow}}=|q|[(k{d}_i{)}^{-1}+O((k{d}_i{)}^{-3})]$. HenceThe fast dispersive wave is the mode sought in the question's subsequent fast-time limit. In a strongly oblique wave, $k\simeq k_{\perp }$, so its frequency scales as $|\omega |\simeq |k_{\parallel }|v_A{k}_{\perp }{d}_i$. The slow branch instead tends to $(|k_{\parallel }|/k){\Omega }_i$, where the ion cyclotron frequency is ${\Omega }_i=v_A/d_i=e{B}_0/(m_{i}c)$.

There is a terminology qualification: in the pressure-free incompressible Hall equations used here, the fast branch is conventionally a whistler wave. The question calls the requested small-scale waves kinetic Alfvén waves, but a general kinetic-Alfvén dispersion includes pressure/kinetic ordering absent from these equations. The boxed dispersion is the result for the printed model; no extra pressure response or kinetic term has been silently added.

Write $ϵ=l/d_i\ll 1$ and $\delta =v_A\tau /l\ll 1$. For magnetic amplitudes of order $B_0$, ion acceleration is of order $v_A^2/l$. Thus its change over the fast time isThe ions are consequently fixed to leading order on this timescale. Their first nonzero response is driven by the momentum equation, with negligible self-advection for an initially resting ion flow; one need not assert that the dimensional Lorentz acceleration literally vanishes. In fast-time, Alfvén-velocity dimensionless variables, the ion equation reads ${\partial }_{\hat{t}}\hat{\mathbf{u}}=O(\delta )$, giving ${\partial }_{\hat{t}}\hat{\mathbf{u}}=0$ at leading order.

Meanwhile the electron-current velocity is $j/(en)\sim v_A{d}_i/l=v_A/ϵ$, larger than a bulk flow of order $v_A$. The ion-advection term in induction is smaller than Hall induction by $O(ϵ)$ even if that slow flow is retained. The leading magnetic equation is thereforewith stationary ions at leading order. This is the fast-time electron-MHD limit of Hall magnetohydrodynamics and is precisely electron magnetohydrodynamics.

The already obtained fast-wave eigenvector has $u/b=|q/\omega |\sim 1/(k{d}_i)\ll 1$, confirming that ion motion cannot affect its leading frequency. The closed magnetic equation therefore suffices to obtain that branch without using momentum to close it. Its timescale is ${\tau }_w\sim 1/(v_A{d}_i|k_{\parallel }|k)$; compatibility with ${\tau }_w\ll l/v_A$ requires $|k_{\parallel }|d_i\gg 1$ when $l\sim k^{-1}$. Arbitrarily small $k_{\parallel }$ can violate this fast-time assumption even though $k{d}_i\gg 1$. The slow branch does not obey the stated ordering and is not retained by the fixed-ion reduction.