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.

## References

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 M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.