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\newcommand\E{\mathbf{E}}
% Matter velocity
\newcommand\uvec{\mathbf{u}}
\newcommand\rvec{\mathbf{r}}
\newcommand\J{\mathbf{J}}
% Area
\newcommand\s{\mathbf{s}}
\newcommand\n{\mathbf{n}}
% Frequency
\newcommand\w{\omega}
% Wave vector
\newcommand\kvec{\mathbf{k}}
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\newcommand{\zunit}{\hat{z}}
\newcommand{\const}{\mathit{const}}
\renewcommand{\Re}{\mathrm{Re}}
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\title{Magneto-hydro-dynamic (MHD) waves}
\author{NGL}
\begin{document}
\maketitle

\section{MHD equations}
\subsection{Equations}
In non-relativistic magnetohydrodynamics (MHD), the following
equations are used.
First, let us consider hydrodynamic equations:
\begin{enumerate}
\item
Continuity equation
\begin{equation}
\dot{\rho}+\diverg(\rho\uvec)=0
\end{equation}
\item
Adiabatic state equation
\begin{equation}
\left(\pderiv{}{t}+\uvec\cdot\nabla\right)(p\rho^{-\gamma})=0
\end{equation}
For homogeneous media, we can use just $p\rho^{-\gamma}=\const$.
\item
The Euler's equation (Newton's Second law per unit volume):
\begin{equation}
\rho\left(\dot{\uvec}+(\uvec\cdot\nabla)\uvec\right)=
-\grad p+\J\times\B
\end{equation}
The last term is the Lorentz force, caused by presence of current $\J$
in the magnetic field $\B$.
\end{enumerate}

Now, let us consider electrodynamic equations.
\begin{enumerate}
\item
Ohm's law:
\begin{equation}
\J=\sigma\E'=\sigma(\E+\uvec\times\B)
\label{eq:cond}
\end{equation}
where $\E'$ is the electric field in the system of the medium moving
at non-relativistic speed $u\ll c$. The conductivity is assumed to be high
enough so that $E\sim uB$.
\item
Maxwell's equation \#1. We assume that the medium is tenuous enough so
that $\varepsilon=1$, $\mu=1$:
\[
\curl\B=\frac{1}{c^2}\dot{\E}+\mu_0\J
\]
The first term is of the order $B(u/c)(v/c)/L$ and is neglected
compared to $B/L$, where $L$ is the typical length scale and $v\sim
L/T$ is the typical wave speed, with $T$ being the time scale. Thus,
\begin{equation}
\curl\B=\mu_0\J
\label{eq:curlB}
\end{equation}
\item
Maxwell's equation \#2:
\begin{equation}
\dot{\B}=-\curl\E
\label{eq:curlE}
\end{equation}
\end{enumerate}

\subsection{Summary}
From  (\ref{eq:cond}) and (\ref{eq:curlB}), $\J=\curl\B/\mu_0$ and
\begin{equation}
\E=\frac{\J}{\sigma}-\uvec\times\B=\eta\curl\B
-\uvec\times\B
\label{eq:E}
\end{equation}
where we introduced $\eta=(\mu_0\sigma)^{-1}$, the magnetic field
diffusion coefficient.  After substituting this into (\ref{eq:curlE})
and collecting all together:
\begin{eqnarray}
\dot{\rho}+\diverg(\rho\uvec)&=&0
\label{eq:continuity}
\\
p\rho^{-\gamma}&=&\const
\label{eq:state}
\\
\rho\left(\dot{\uvec}+(\uvec\cdot\nabla)\uvec\right)&=&
-\grad p+\frac{1}{\mu_0}(\curl\B)\times\B
\label{eq:euler}
\\
\dot{\B}&=&\curl(\uvec\times\B)+\eta\nabla^2\B
\label{eq:frozen}
\end{eqnarray}
where we used
$\curl(\curl\B)=\grad(\diverg\B)-\nabla^2\B$.

\subsection{Frozen-in $\B$}
For infinite conductivity, $\eta=0$. Equation (\ref{eq:frozen}), which
now takes the form
\[
\dot{\B}=\curl(\uvec\times\B)
\]
describes a frozen-in magnetic field, so that the magnetic field lines
can be visualized to be moving together with the matter. This is
proven by considering a magnetic flux
\[
\Phi=\int_S \B\cdot d\s
\]
through an area $S$ enclosed by contour $C$ that is transformed
together with the moving matter and showing that it is constant in
time. Consider its change in time $\Delta t$:
\[
\Delta\Phi=\Delta t\int_C \curl(\uvec\times\B)\cdot d\s
+ \int_{S'=\Delta C} \B\cdot d\s
\]
The first term can be transformed using
\[
\int_S \curl(\uvec\times\B)\cdot d\s=
\oint_C (\uvec\times\B)\cdot d\lvec=
\oint_C \B\cdot(d\lvec\times\uvec)
\]
In the second term, the integration is over the surface that is
created by the ``slit'' between $C$ at $t$ and the transformed contour
at $t+\Delta t$. This term also can be transformed into integration
over $C$ using the area element $d\s=(\uvec\Delta t)\times d\lvec$ and
then cancels the first term.

\subsection{Energy flow}
Using (\ref{eq:E}), the Poynting vector is
\[
\mathbf{S}=\E\times\mathbf{H}=\eta\frac{1}{\mu_0}(\curl\B)\times\B
+\frac{1}{\mu_0}\uvec_\perp B^2
\]
where we used $\uvec_\perp B^2=\B\times(\uvec\times\B)=
\uvec B^2-\B(\uvec\cdot\B)$.

\section{Linearized MHD equations}
Consider an infinite uniform system, in which all physical quantities
consist of a large part, denoted with index~$0$, which is constant
(both in time and space), plus a small variable part (with index~$1$),
e.g. $\B=\B_0+\B_1$. The speed has only the small part
$\uvec=\uvec_1$. We also assume infinite conductivity ($\eta=0$), so
that the waves are not attenuated. From the state equation
(\ref{eq:state}) we get
\[
p_1=\rho_1\gamma\frac{p_0}{\rho_0}= c_s^2\rho_1
\]
where $c_s$ is the speed of sound. We use this to substitute $p_1$
everywhere in terms of $\rho_1$. The MHD equations
(\ref{eq:continuity},\ref{eq:euler}--\ref{eq:frozen}) become after
linearization
\begin{eqnarray}
\dot{\rho_1}+\rho_0\diverg\uvec&=&0 \\
\dot{\B_1}&=&\curl(\uvec\times\B_0) \\
\rho_0\dot{\uvec}&=&-c_s^2\grad\rho_1+\frac{1}{\mu_0}(\curl\B_1)\times\B_0
\end{eqnarray}

\subsection{Sound}
Disregard all electrodynamics. The linearized equations are
\begin{eqnarray*}
\dot{\rho_1}+\rho_0\diverg\uvec&=&0
\\
\rho_0\dot{\uvec}&=&-c_s^2\grad\rho_1
\end{eqnarray*}
Taking the time derivative of the first equation, we get the wave
equation
\[
\ddot{\rho_1}-c_s^2\nabla^2\rho_1=0
\]

\section{Plane waves}
The plane waves are described by the harmonic time and space
dependence
\[
\rho_1=\Re(\tilde{\rho}e^{-i\w t+i\kvec\cdot\rvec})
\]
etc. We use a tilde to denote the complex amplitude.  The differential
operators become (when they act on $\tilde{\rho}$ etc.)
$\nabla\rightarrow i\kvec$ and $\partial/\partial t\rightarrow -i\w$.
The equations for complex amplitudes become
\begin{eqnarray}
\w\tilde{\rho}&=&\rho_0\kvec\cdot\tilde{\uvec} \\
\w\tilde{\B}&=&-\kvec\times(\tilde{\uvec}\times\B_0) \\
\w\rho_0\tilde{\uvec}&=&c_s^2\kvec\tilde{\rho}-
\frac{1}{\mu_0}(\kvec\times\tilde{\B})\times\B_0
\end{eqnarray}
Denote $\n=\kvec/\w$:
\begin{eqnarray}
\tilde{\rho}&=&\rho_0\n\cdot\tilde{\uvec}\\
\tilde{\B}&=&-\n\times(\tilde{\uvec}\times\B_0) \\
\rho_0\tilde{\uvec}&=&c_s^2\n\tilde{\rho}-
\frac{1}{\mu_0}(\n\times\tilde{\B})\times\B_0
\end{eqnarray}
Substituting the first and second equation into the third, we get
\[
\tilde{\uvec}=c_s^2\n(\n\cdot\tilde{\uvec})+
\frac{1}{\rho_0\mu_0}(\n\times(\n\times(\tilde{\uvec}\times\B_0)))\times\B_0
\]
Taking $\B_0$ along the $z$-axis, and introducing the Alfv\'en speed
$v_A=B_0/\sqrt{\rho_0\mu_0}$, and also dropping the tilde over
$\uvec$, we have
\begin{equation}
\uvec=c_s^2\n(\n\cdot\uvec)+
v_A^2\zunit\times(\n\times(\n\times(\zunit\times\uvec)))
\label{eq:u}
\end{equation}
The cross-products can be expanded to yield
\begin{equation}
\uvec=(c_s^2+v_A^2)\n(\n\cdot\uvec)+
v_A^2\left(
\uvec n_z^2
-\zunit n_z(\n\cdot\uvec)
-\n n_z u_z\right)
\label{eq:uexp}
\end{equation}
where $n_z=\zunit\cdot\n$ and $u_z=\zunit\cdot\uvec$.

\subsection{Alfv\'en waves}
Consider $\uvec\perp\n$, so that
$\n\cdot\uvec=0$. Then (\ref{eq:u}) becomes
\[
\uvec=v_A^2\zunit\times(\n\times(\n\times(\zunit\times\uvec)))
\]
We see that we also have automatically $\uvec\perp\zunit$. Thus, the
displacements are perpendicular to the plane containing $\kvec$ and
$\B_0$. There is no density perturbation because
$\tilde{\rho}=\rho_0\n\cdot\uvec=0$. From the second version
(\ref{eq:uexp}) we get (taking into account that $u_z=0$)
\[
\uvec=v_A^2\uvec n_z^2
\]
from where the dispersion equation of Alfv\'en waves immediately
follows:
\[
1=v_A^2n_z^2
\]
or
\[
k_z=\pm\w/v_A
\]
Note that Alfv\'en wave, being a purely transverse (shear) wave, does
not depend on the sound speed $c_s$ at all.

\subsection{Other waves}
Let us split (\ref{eq:uexp}) into parts $\parallel\zunit$ and
$\perp\zunit$, denoting $\uvec=u_z\zunit+\uvec_\perp$ and
$\n=n_z\zunit+\n_\perp$.
Notice that another way to write (\ref{eq:uexp}) is
\[
\uvec-v_A^2n_z^2\uvec_\perp=c_s^2\n(\n\cdot\uvec)+
v_A^2\n_\perp(\n_\perp\cdot\uvec_\perp)
\]
The component of $\uvec$ which is perpendicular to both $\zunit$ and
$\n$ is the Alfv\'en wave, and can be decoupled from other
modes. ({\bf NOTE:} One should check the logical validity of this
statement.) Therefore, now let us assume that $\uvec$ is in the plane
of $\B_0$ and $\kvec$, so that $\uvec_\perp\parallel\n_\perp$.  Thus,
$\n\cdot\uvec=n_zu_z+\n_\perp\cdot\uvec_\perp=n_zu_z+n_\perp u_\perp$,
and we can again rewrite the equation for $\uvec$ as
\[
\uvec-v_A^2n^2\uvec_\perp=c_s^2\n(\n\cdot\uvec)
\]

The parallel part is
\begin{equation}
(1-c_s^2n_z^2)u_z=c_s^2n_z n_\perp u_\perp
\label{eq:pardisp}
\end{equation}
The perpendicular part is
\begin{equation}
(1-v_A^2n^2-c_s^2n_\perp^2)u_\perp=c_s^2 n_\perp n_zu_z
\label{eq:perpdisp}
\end{equation}

Substituting equations into each other we find that nontrivial
solution is possible only when
\[
(1-c_s^2n_z^2)(1-v_A^2n^2-c_s^2n_\perp^2)=c_s^4n_\perp^2 n_z^2
\]
or
\begin{equation}
c_s^2v_A^2n^4\cos^2\theta-(c_s^2+v_A^2)n^2+1=0
\label{eq:nmaeq}
\end{equation}
where we used $n_z=n\cos\theta$.  The solution of this equation gives
the two remaining modes. Usually this equation is solved in terms of
the phase velocity $v=1/n$, the equation for which reads
\[
v^4-(c_s^2+v_A^2)v^2+c_s^2v_A^2\cos^2\theta=0
\]
with the solution
\begin{equation}
v_{1,2}^2=\frac{1}{2}\left(
c_s^2+v_A^2\pm\sqrt{c_s^4+v_A^4-2c_s^2v_A^2\cos2\theta}
\right)
\label{eq:vma}
\end{equation}
In the limiting case $c_s\ll v_A$ we get $v_1\approx v_A$ (fast magnetoacoustic
wave) and $v_2\approx c_s\cos\theta$ (slow). In the opposite case ($c_s\gg v_A$),
we have $v_1\approx c_s$ (acoustic wave) and the slow wave is very similar
to Alfv\'en wave, $v_2\approx v_A\cos\theta$.


\subsection{Polarization of plasma displacement of the magnetoacoustic waves}
Using (\ref{eq:perpdisp}), (\ref{eq:vma}), $1/n^2=v^2$ and
$n_\perp=n\sin\theta$, $n_z=n\cos\theta$, we get
\[
 \left(\frac{u_z}{u_\perp}\right)_{1,2}=\frac{1/n^2-c_s^2v_A^2\sin^2\theta}
{c_s^2\cos\theta\sin\theta}
\]
\begin{equation}
\left(\frac{u_z}{u_\perp}\right)_{1,2}=
\frac{c_s^2\cos2\theta-v_A^2\pm\sqrt{c_s^4+v_A^4-2c_s^2v_A^2\cos2\theta}}
{c_s^2\sin2\theta}
\label{eq:mapol}
\end{equation}
From here, we immediately see that
\begin{equation}
\left(\frac{u_z}{u_\perp}\right)_1\left(\frac{u_z}{u_\perp}\right)_2=-1
\label{eq:acperp}
\end{equation}
which means that the $\uvec_1\perp\uvec_2$. Another expression for polarization
can be obtained by solving (\ref{eq:nmaeq}) and substituting into (\ref{eq:pardisp}),
which leads to
\[
 \left(\frac{u_z}{u_\perp}\right)_{1,2}=-\tan\theta\frac{
c_s^2+v_A^2\mp\sqrt{c_s^4+v_A^4-2c_s^2v_A^2\cos2\theta}}{
c_s^2-v_A^2\mp\sqrt{c_s^4+v_A^4-2c_s^2v_A^2\cos2\theta}}
\]
Using (\ref{eq:acperp}), we get formulas useful for $c_s>v_A$:
\begin{eqnarray}
 \left(\frac{u_z}{u_\perp}\right)_1=\cot\theta\frac{
c_s^2-v_A^2+\sqrt{c_s^4+v_A^4-2c_s^2v_A^2\cos2\theta}}{
c_s^2+v_A^2+\sqrt{c_s^4+v_A^4-2c_s^2v_A^2\cos2\theta}}
\\
 \left(\frac{u_z}{u_\perp}\right)_2=-\tan\theta\frac{
c_s^2+v_A^2+\sqrt{c_s^4+v_A^4-2c_s^2v_A^2\cos2\theta}}{
c_s^2-v_A^2+\sqrt{c_s^4+v_A^4-2c_s^2v_A^2\cos2\theta}}
\end{eqnarray}

This gives some insight into polarization in the limiting case
$c_s\gg v_A$. We have $(u_z/u_\perp)_1\approx\cot\theta$, $(u_z/u_\perp)_2\approx-\tan\theta$,
that is, $\uvec$ is approximately
parallel to $\kvec$ for the fast (acoustic) wave, and perpendicular for the slow wave.
The phase velocity of the slow wave is very similar to that of
Alfv\'en wave, as obtained in the end of the previous Section.

Using (\ref{eq:mapol}), in the opposite limiting case ($c_s\ll v_A$) and intermediate
range of angles (not too close to zero or $\pi/2$), we find that for the fast magnetoacoustic
wave ($v_1\approx v_A$) $\uvec_1\perp\zunit$ and for the slow ($v_2\approx c_s\cos\theta$)
$\uvec_2\parallel\zunit$.

Last remark: the polarization of the field of the wave $\E$, $\B_1$ is found from
$\E=-\uvec\times\B_0$ and $\B_1=\n\times\E$.

\end{document}
