path: root/source/know/concept/alfven-waves/index.md

---
title: "Alfvén waves"
sort_title: "Alfven waves" # sic
date: 2022-01-31
categories:
- Physics
- Plasma physics
- Plasma waves
layout: "concept"
---

In the [magnetohydrodynamic](/know/concept/magnetohydrodynamics/) description of a plasma,
we split the velocity $$\vb{u}$$, electric current $$\vb{J}$$,
[magnetic field](/know/concept/magnetic-field/) $$\vb{B}$$
and [electric field](/know/concept/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.