 Solution

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.