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diff --git a/source/know/concept/alfven-waves/index.md b/source/know/concept/alfven-waves/index.md index c560c67..31576f3 100644 --- a/source/know/concept/alfven-waves/index.md +++ b/source/know/concept/alfven-waves/index.md @@ -10,11 +10,11 @@ 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 $1$): +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 $$1$$): $$\begin{aligned} \vb{u} @@ -41,7 +41,7 @@ $$\begin{aligned} \end{aligned}$$ We do this for the momentum equation too, -assuming that $\vb{J}_0 \!=\! 0$ (to be justified later). +assuming that $$\vb{J}_0 \!=\! 0$$ (to be justified later). Note that the temperature is set to zero, such that the pressure vanishes: $$\begin{aligned} @@ -49,8 +49,8 @@ $$\begin{aligned} = \vb{J}_1 \cross \vb{B}_0 \end{aligned}$$ -Where $\rho$ is the uniform equilibrium density. -We would like an equation for $\vb{J}_1$, +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: $$\begin{aligned} @@ -62,14 +62,14 @@ $$\begin{aligned} \end{aligned}$$ Substituting this into the momentum equation, -and differentiating with respect to $t$: +and differentiating with respect to $$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}$$ -For which we can use Faraday's law to rewrite $\ipdv{\vb{B}_1}{t}$, +For which we can use Faraday's law to rewrite $$\ipdv{\vb{B}_1}{t}$$, incorporating Ohm's law too: $$\begin{aligned} @@ -78,7 +78,7 @@ $$\begin{aligned} = \nabla \cross (\vb{u}_1 \cross \vb{B}_0) \end{aligned}$$ -Inserting this into the momentum equation for $\vb{u}_1$ +Inserting this into the momentum equation for $$\vb{u}_1$$ thus yields its final form: $$\begin{aligned} @@ -86,9 +86,9 @@ $$\begin{aligned} = \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}$$ -Suppose the magnetic field is pointing in $z$-direction, -i.e. $\vb{B}_0 = B_0 \vu{e}_z$. -Then Faraday's law justifies our earlier assumption that $\vb{J}_0 = 0$, +Suppose the magnetic field is pointing in $$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} @@ -96,7 +96,7 @@ $$\begin{aligned} = v_A^2 \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg) \end{aligned}$$ -Where we have defined the so-called **Alfvén velocity** $v_A$ to be given by: +Where we have defined the so-called **Alfvén velocity** $$v_A$$ to be given by: $$\begin{aligned} \boxed{ @@ -105,24 +105,24 @@ $$\begin{aligned} } \end{aligned}$$ -Now, consider the following plane-wave ansatz for $\vb{u}_1$, -with wavevector $\vb{k}$ and frequency $\omega$: +Now, consider the following plane-wave ansatz for $$\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}$$ -Inserting this into the above differential equation for $\vb{u}_1$ leads to: +Inserting this into the above differential equation for $$\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}$$ -To evaluate this, we rotate our coordinate system around the $z$-axis -such that $\vb{k} = (0, k_\perp, k_\parallel)$, -i.e. the wavevector's $x$-component is zero. +To evaluate this, we rotate our coordinate system around the $$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} @@ -149,7 +149,7 @@ $$\begin{aligned} \end{aligned}$$ We rewrite this equation in matrix form, -using that $k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$: +using that $$k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$$: $$\begin{aligned} \begin{bmatrix} @@ -161,9 +161,9 @@ $$\begin{aligned} = 0 \end{aligned}$$ -This has the form of an eigenvalue problem for $\omega^2$, +This has the form of an eigenvalue problem for $$\omega^2$$, meaning we must find non-trivial solutions, -where we cannot simply choose the components of $\vb{u}_1$ to satisfy the equation. +where we cannot simply choose the components of $$\vb{u}_1$$ to satisfy the equation. To achieve this, we demand that the matrix' determinant is zero: $$\begin{aligned} @@ -171,12 +171,12 @@ $$\begin{aligned} = 0 \end{aligned}$$ -This equation has three solutions for $\omega^2$, +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, +The simplest case $$\omega^2 = 0$$ is of no interest to us, because we are looking for waves. -The first interesting case is $\omega^2 = v_A^2 k_\parallel^2$, +The first interesting case is $$\omega^2 = v_A^2 k_\parallel^2$$, yielding the following dispersion relation: $$\begin{aligned} @@ -188,10 +188,10 @@ $$\begin{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 $\vb{u}_1 \cdot \vb{k} = 0$: +$$\vb{u}_1 = (u_{1x}, 0, 0)$$, 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 $\vb{B}_0$: +The phase velocity $$v_p$$ and group velocity $$v_g$$ are as follows, +where $$\theta$$ is the angle between $$\vb{k}$$ and $$\vb{B}_0$$: $$\begin{aligned} v_p @@ -204,7 +204,7 @@ $$\begin{aligned} = v_A \end{aligned}$$ -The other interesting case is $\omega^2 = v_A^2 k^2$, +The other interesting case is $$\omega^2 = v_A^2 k^2$$, which leads to so-called **compressional Alfvén waves**, with the simple dispersion relation: @@ -215,10 +215,10 @@ $$\begin{aligned} } \end{aligned}$$ -Looking at the eigenvalue problem reveals that $\vb{u}_1 = (0, u_{1y}, 0)$, -meaning $\vb{u}_1 \cdot \vb{k} = u_{1y} k_\perp$, -so these waves are not necessarily transverse, nor longitudinal (since $k_\parallel$ is free). -The phase velocity $v_p$ and group velocity $v_g$ are given by: +Looking at the eigenvalue problem reveals that $$\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 |