Categories: Physics, Plasma physics, Plasma waves.

# Alfvén waves

In the magnetohydrodynamic description of a plasma, we split the velocity $\vb{u}$, electric current $\vb{J}$, magnetic field $\vb{B}$ and electric field $\vb{E}$ like so, into a constant uniform equilibrium (subscript $0$) and a small unknown perturbation (subscript $1$):

\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:

\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 $\vb{J}_0 \!=\! 0$ (to be justified later). Note that the temperature is set to zero, such that the pressure vanishes:

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

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

\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 $t$:

\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 $\ipdv{\vb{B}_1}{t}$, incorporating Ohm’s law too:

\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 $\vb{u}_1$ thus yields its final form:

\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 $z$-direction, i.e. $\vb{B}_0 = B_0 \vu{e}_z$. Then Faraday’s law justifies our earlier assumption that $\vb{J}_0 = 0$, and the equation can be written as:

\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 $v_A$ to be given by:

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

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

\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 $\vb{u}_1$ leads to:

\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 $z$-axis such that $\vb{k} = (0, k_\perp, k_\parallel)$, i.e. the wavevector’s $x$-component is zero. Calculating the cross products:

\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 $k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$:

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

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

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

\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 $\vb{u}_1 = (u_{1x}, 0, 0)$, meaning $\vb{u}_1 \cdot \vb{k} = 0$: these waves are transverse. The phase velocity $v_p$ and group velocity $v_g$ are as follows, where $\theta$ is the angle between $\vb{k}$ and $\vb{B}_0$:

\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 $\omega^2 = v_A^2 k^2$, which leads to so-called compressional Alfvén waves, with the simple dispersion relation:

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

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

\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.