Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 75 / 1

Let ${\rho }_i={\rho }_n=\rho$ and $\mu ={\mu }_{in}={\mu }_{ni}$. Use normal modes proportional to $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$, so $\mathrm{Im}\omega <0$ means damping. The linear displacements satisfyFor a nonzero-frequency perturbation, linearizing the ideal magnetohydrodynamic induction equation givesProject both momentum equations with $\mathsf{P}=I-\mathbf{k}{\mathbf{k}}^T/k^2$. This removes the pressure gradients. The magnetic tension is already transverse to $\mathbf{k}$, and, with $a=k_{\parallel }{v}_A$, the projected equations areHere $v_A=B_0/\sqrt{4\pi \rho }$ uses the ion density, as stipulated. For either of the two independent transverse polarizations, the equal-density ion-neutral Alfvén dispersion relation follows fromIts determinant is $\omega [{\omega }^3+2i\mu {\omega }^2-a^2\omega -i\mu a^2]$. Removing the static neutral-displacement factor givesThe removed factor is a time-independent neutral displacement with no velocity or magnetic perturbation, a relabeling of the homogeneous neutral fluid rather than a fourth physical wave. Direct elimination in the first-order system for $(u_i,u_n,\delta B)$ gives exactly the same cubic. This also handles the zero-frequency limits without introducing a displacement-label mode.

For the strong-collision ion-neutral Alfvén modes, put $a=k_{\parallel }{v}_A$ and assume $0<|a|/\mu \ll 1$. Rewrite the cubic asThe two slow roots obey $\omega =O(a)$, so the first approximation is ${\omega }_0^2=a^2/2$. Set $\omega ={\omega }_0+\delta \omega$ and retain the next order, of size $a^3$:ConsequentlyThese are weakly damped Alfvén waves. Their phase speed is $v_A/\sqrt{2}=B_0/\sqrt{4\pi ({\rho }_i+{\rho }_n)}$, since collisions make the neutral mass participate in the oscillation. Their ambipolar damping rate is ${\Gamma }_A=a^2/(8\mu )$, with ${\Gamma }_A/|\mathrm{Re}{\omega }_{\pm }|\ll 1$.

For the remaining root, at $a=0$ the cubic is ${\omega }^2(\omega +2i\mu )$, so start from ${\omega }_{d0}=-2i\mu$. Denote the cubic by $F(\omega ,a)$. Its value and derivative at this root areThe first correction is therefore $\delta {\omega }_d=i{a}^2/(4\mu )$, givingThis branch is purely damped. Indeed, the real growth-rate polynomial for $s=-i\omega$ is $s^3+2\mu s^2+a^{2}s+\mu a^2$, and its simple real root near $-2\mu$ stays real for sufficiently small real $a^2$. It represents rapid decay of ion-neutral relative motion. In the $a\to 0$ limit, the other two frequencies tend to zero; all three branches are accounted for.

The neutral displacement equation gives, for each nonzero-frequency root,The vector displacements are parallel as complex polarization vectors, with this scalar amplitude and phase ratio.

For the rapidly damped root, $\mu -i{\omega }_d=-\mu +a^2/(4\mu )+O(a^4/{\mu }^3)$, and henceThe two fluids move almost equally and oppositely; collisions remove the counterflow at a rate close to $2\mu$.

For either weakly damped Alfvén wave, expand the ratio around unity:Equivalently, the exact slippage relation is ${\mathbf{\xi }}_i-{\mathbf{\xi }}_n=-(i\omega /\mu ){\mathbf{\xi }}_n$. Thus the relative slippage has size $|a|/(\sqrt{2}\mu )\ll 1$ and is primarily a small phase difference. The fluids almost share a common displacement, but magnetic tension acts directly on the ions and drag accelerates the neutrals. This imperfect locking causes ambipolar damping.

The sign of the damping is also checked by the perturbation energy balance. With periodic or vanishing-flux boundaries,It follows by dotting each linear momentum equation with its velocity and the induction equation with $\delta \mathbf{B}/(4\pi )$; magnetic-tension work cancels magnetic-energy change, pressure contributes only boundary flux, and the two drag terms combine into the negative square. Even though spatial diffusion was neglected, ion-neutral friction still dissipates the wave energy.