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Ambipolar damping is attenuation of magnetic disturbances by friction between the magnetically coupled charged fluid and a neutral fluid. In a strongly collisional mixture, the fluids nearly move together, but their small relative velocity dissipates energy. With drag coefficient ${\rho }_i{\mu }_{in}={\rho }_n{\mu }_{ni}$, the frictional energy loss is $-{\rho }_i{\mu }_{in}\int |{\mathbf{u}}_i-{\mathbf{u}}_n{|}^{2}dV$. Neglecting other dissipative processes does not remove this collisional loss.

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.

For $|a|\ll \mu$, the equal-density ion-neutral cubic has a rapidly damped mode ${\omega }_d=-2i\mu +i{a}^2/(4\mu )+O(a^4/{\mu }^3)$ and two propagating modes ${\omega }_{\pm }=\pm a/\sqrt{2}-i{a}^2/(8\mu )+O(a^3/{\mu }^2)$. The reduced phase speed uses the total density ${\rho }_i+{\rho }_n=2{\rho }_i$. The exact displacement ratio is ${\xi }_n/{\xi }_i=\mu /(\mu -i\omega )$. It is nearly $-1$ for the rapidly damped counterflow and nearly $1\pm ia/(\sqrt{2}\mu )$ for the propagating modes. Thus the latter have small ion-neutral slippage of relative size $|a|/\mu$, responsible for ambipolar damping.