Categories: Physics, Plasma physics, Plasma waves.

Alfvén waves

In the magnetohydrodynamic description of a plasma, we split the velocity u\vb{u}, electric current J\vb{J}, magnetic field B\vb{B} and electric field E\vb{E} like so, into a constant uniform equilibrium (subscript 00) and a small unknown perturbation (subscript 11):

u=u0+u1J=J0+J1B=B0+B1E=E0+E1\begin{aligned} \vb{u} = \vb{u}_0 + \vb{u}_1 \qquad \vb{J} = \vb{J}_0 + \vb{J}_1 \qquad \vb{B} = \vb{B}_0 + \vb{B}_1 \qquad \vb{E} = \vb{E}_0 + \vb{E}_1 \end{aligned}

Inserting this decomposition into the ideal form of the generalized Ohm’s law and keeping only terms that are first-order in the perturbation, we get:

0=(E0+E1)+(u0+u1)×(B0+B1)=E1+u1×B0\begin{aligned} 0 &= (\vb{E}_0 + \vb{E}_1) + (\vb{u}_0 + \vb{u}_1) \cross (\vb{B}_0 + \vb{B}_1) \\ &= \vb{E}_1 + \vb{u}_1 \cross \vb{B}_0 \end{aligned}

We do this for the momentum equation too, assuming that J0 ⁣= ⁣0\vb{J}_0 \!=\! 0 (to be justified later). Note that the temperature is set to zero, such that the pressure vanishes:

ρu1t=J1×B0\begin{aligned} \rho \pdv{\vb{u}_1}{t} = \vb{J}_1 \cross \vb{B}_0 \end{aligned}

Where ρ\rho is the uniform equilibrium density. We would like an equation for J1\vb{J}_1, which is provided by the magnetohydrodynamic form of Ampère’s law:

×B1=μ0J1    J1=1μ0×B1\begin{aligned} \nabla \cross \vb{B}_1 = \mu_0 \vb{J}_1 \qquad \implies \quad \vb{J}_1 = \frac{1}{\mu_0} \nabla \cross \vb{B}_1 \end{aligned}

Substituting this into the above momentum equation, and differentiating with respect to tt:

ρ2u1t2=1μ0((×B1t)×B0)\begin{aligned} \rho \pdvn{2}{\vb{u}_1}{t} = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \pdv{\vb{B}_1}{t} \Big) \cross \vb{B}_0 \bigg) \end{aligned}

For which we can use Faraday’s law to rewrite B1/t\ipdv{\vb{B}_1}{t}, incorporating Ohm’s law too:

B1t=×E1=×(u1×B0)\begin{aligned} \pdv{\vb{B}_1}{t} = - \nabla \cross \vb{E}_1 = \nabla \cross (\vb{u}_1 \cross \vb{B}_0) \end{aligned}

Inserting this back into the momentum equation for u1\vb{u}_1 thus yields its final form:

ρ2u1t2=1μ0((×(×(u1×B0)))×B0)\begin{aligned} \rho \pdvn{2}{\vb{u}_1}{t} = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vb{B}_0) \big) \Big) \cross \vb{B}_0 \bigg) \end{aligned}

Suppose the magnetic field is pointing in zz-direction, i.e. B0=B0e^z\vb{B}_0 = B_0 \vu{e}_z. Then Faraday’s law justifies our earlier assumption that J0=0\vb{J}_0 = 0, and the equation can be written as:

2u1t2=vA2((×(×(u1×e^z)))×e^z)\begin{aligned} \pdvn{2}{\vb{u}_1}{t} = v_A^2 \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg) \end{aligned}

Where we have defined the so-called Alfvén velocity vAv_A to be given by:

vAB02μ0ρ\begin{aligned} \boxed{ v_A \equiv \sqrt{\frac{B_0^2}{\mu_0 \rho}} } \end{aligned}

Now, consider the following plane-wave ansatz for u1\vb{u}_1, with wavevector k\vb{k} and frequency ω\omega:

u1(r,t)=u1exp(ikriωt)\begin{aligned} \vb{u}_1(\vb{r}, t) &= \vb{u}_1 \exp(i \vb{k} \cdot \vb{r} - i \omega t) \end{aligned}

Inserting this into the above differential equation for u1\vb{u}_1 leads to:

ω2u1=vA2((k×(k×(u1×e^z)))×e^z)\begin{aligned} \omega^2 \vb{u}_1 = v_A^2 \bigg( \Big( \vb{k} \cross \big( \vb{k} \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg) \end{aligned}

To evaluate this, we rotate our coordinate system around the zz-axis such that k=(0,k,k)\vb{k} = (0, k_\perp, k_\parallel), i.e. the wavevector’s xx-component is zero. Calculating the cross products:

ω2u1=vA2(([0kk]×([0kk]×([u1xu1yu1z]×[001])))×[001])=vA2(([0kk]×([0kk]×[u1yu1x0]))×[001])=vA2(([0kk]×[ku1xku1yku1y])×[001])=vA2([(k2 ⁣+k2)u1yk2u1xkku1x]×[001])=vA2[k2u1x(k2 ⁣+k2)u1y0]\begin{aligned} \omega^2 \vb{u}_1 &= v_A^2 \bigg( \Big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix} \cross \big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix} \cross ( \begin{bmatrix} u_{1x} \\ u_{1y} \\ u_{1z} \end{bmatrix} \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} ) \big) \Big) \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg) \\ &= v_A^2 \bigg( \Big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix} \cross \big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix} \cross \begin{bmatrix} u_{1y} \\ -u_{1x} \\ 0 \end{bmatrix} \big) \Big) \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg) \\ &= v_A^2 \bigg( \Big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix} \cross \begin{bmatrix} k_\parallel u_{1x} \\ k_\parallel u_{1y} \\ -k_\perp u_{1y} \end{bmatrix} \Big) \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg) \\ &= v_A^2 \bigg( \begin{bmatrix} -(k_\perp^2 \!+ k_\parallel^2) u_{1y} \\ k_\parallel^2 u_{1x} \\ -k_\perp k_\parallel u_{1x} \end{bmatrix} \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg) \\ &= v_A^2 \begin{bmatrix} k_\parallel^2 u_{1x} \\ (k_\perp^2 \!+ k_\parallel^2) u_{1y} \\ 0 \end{bmatrix} \end{aligned}

We rewrite this equation in matrix form, using that k2 ⁣+k2=k2k2k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2:

[ω2vA2k2000ω2vA2k2000ω2]u1=0\begin{aligned} \begin{bmatrix} \omega^2 - v_A^2 k_\parallel^2 & 0 & 0 \\ 0 & \omega^2 - v_A^2 k^2 & 0 \\ 0 & 0 & \omega^2 \end{bmatrix} \vb{u}_1 = 0 \end{aligned}

This has the form of an eigenvalue problem for ω2\omega^2, meaning we must find non-trivial solutions, where we cannot simply choose the components of u1\vb{u}_1 to satisfy the equation. To achieve this, we demand that the matrix’ determinant is zero:

(ω2vA2k2)(ω2vA2k2)ω2=0\begin{aligned} \big(\omega^2 - v_A^2 k_\parallel^2\big) \: \big(\omega^2 - v_A^2 k^2\big) \: \omega^2 = 0 \end{aligned}

This equation has three solutions for ω2\omega^2, one for each of its three factors being zero. The simplest case ω2=0\omega^2 = 0 is of no interest to us, because we are looking for waves.

The first interesting case is ω2=vA2k2\omega^2 = v_A^2 k_\parallel^2, yielding the following dispersion relation:

ω=±vAk\begin{aligned} \boxed{ \omega = \pm v_A k_\parallel } \end{aligned}

The resulting waves are called shear Alfvén waves. From the eigenvalue problem, we see that in this case u1=(u1x,0,0)\vb{u}_1 = (u_{1x}, 0, 0), meaning u1k=0\vb{u}_1 \cdot \vb{k} = 0: these waves are transverse. The phase velocity vpv_p and group velocity vgv_g are as follows, where θ\theta is the angle between k\vb{k} and B0\vb{B}_0:

vp=ωk=vAkk=vAcos(θ)vg=ωk=vA\begin{aligned} v_p = \frac{|\omega|}{k} = v_A \frac{k_\parallel}{k} = v_A \cos(\theta) \qquad \qquad v_g = \pdv{|\omega|}{k} = v_A \end{aligned}

The other interesting case is ω2=vA2k2\omega^2 = v_A^2 k^2, which leads to so-called compressional Alfvén waves, with the simple dispersion relation:

ω=±vAk\begin{aligned} \boxed{ \omega = \pm v_A k } \end{aligned}

Looking at the eigenvalue problem reveals that u1=(0,u1y,0)\vb{u}_1 = (0, u_{1y}, 0), meaning u1k=u1yk\vb{u}_1 \cdot \vb{k} = u_{1y} k_\perp, so these waves are not necessarily transverse, nor longitudinal (since kk_\parallel is free). The phase velocity vpv_p and group velocity vgv_g are given by:

vp=ωk=vAvg=ωk=vA\begin{aligned} v_p = \frac{|\omega|}{k} = v_A \qquad \qquad v_g = \pdv{|\omega|}{k} = v_A \end{aligned}

The mechanism behind both of these oscillations is magnetic tension: the waves are “ripples” in the field lines, which get straightened out by Faraday’s law, but the ions’ inertia causes them to overshoot and form ripples again.


  1. M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.