From 43c5b696aaf421dec7aee967002999d9145da35e Mon Sep 17 00:00:00 2001
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Date: Tue, 1 Feb 2022 11:23:35 +0100
Subject: Expand knowledge base

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+---
+title: "Alfvén waves"
+firstLetter: "A"
+publishDate: 2022-01-31
+categories:
+- Physics
+- Plasma physics
+- Plasma waves
+
+date: 2022-01-30T19:26:33+01:00
+draft: false
+markup: pandoc
+---
+
+# Alfvén waves
+
+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 momentum equation,
+and differentiating with respect to $t$:
+
+$$\begin{aligned}
+    \rho \pdv[2]{\vb{u}_1}{t}
+    = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \pdv{\vb{B}_1}{t} \Big) \cross \vb{B}_0 \bigg)
+\end{aligned}$$
+
+For which we can use Faraday's law to rewrite $\pdv*{\vb{B}_1}{t}$,
+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 into the momentum equation for $\vb{u}_1$
+thus yields its final form:
+
+$$\begin{aligned}
+    \rho \pdv[2]{\vb{u}_1}{t}
+    = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vb{B}_0) \big) \Big) \cross \vb{B}_0 \bigg)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \pdv[2]{\vb{u}_1}{t}
+    = v_A^2 \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg)
+\end{aligned}$$
+
+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.
+
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