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diff --git a/content/know/concept/alfven-waves/index.pdc b/content/know/concept/alfven-waves/index.pdc new file mode 100644 index 0000000..ba87bee --- /dev/null +++ b/content/know/concept/alfven-waves/index.pdc @@ -0,0 +1,249 @@ +--- +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. + |