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%TCIDATA{Created=Tue Aug 20 14:31:10 1996}
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\markright{{\sf Propagation in the Ionosphere and Magnetosphere}}
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\begin{document}


\section{7.6\quad Propagation in the Ionosphere and Magnetosphere.}

\subsection{7.6.1}

\quad \thinspace \textbf{a.} For an ionized plasma, the dielectric constant $%
\varepsilon =1-\omega _p^2/\omega ^2$ can be derived from the simple model
of $N$ \textsl{free} electrons of mass $m,$ charge $e.$ Each electron's
position is described by $-m\omega ^2\mathbf{x=}$ $e\mathbf{E,}$ the current
is $\mathbf{J=}$ $Ne\mathbf{v=}-i\omega Ne\mathbf{x=}$ $i\left( Ne^2/m\omega
\right) \mathbf{E,}$ so that $\sigma =iNe^2\mathbf{/}m\omega $ and $%
\varepsilon =1+4\pi i\sigma /\omega $ gives the desired result, with $\omega
_p^2$ $=4\pi Ne^2/m.$

\smallskip\ 

\textbf{b}. The relation between $k$ and $\omega ,$ $k^{2}$ $=\varepsilon
\omega ^{2}/c^{2}$, gives 
\begin{equation}
k^{2}=\frac{\omega ^{2}-\omega _{p}^{2}}{c^{2}}\,\,\,\,\,\,\,\,\text{or}%
\,\,\,\,\,\,\,\omega ^{2}=\,\,\omega _{p}^{2}+k^{2}c^{2}.  \label{1}
\end{equation}
Considered as a relation for $\hbar \omega $ as a function of $\hbar k,$ (%
\ref{1}) gives the energy-momentum relation for a new kind of
``quasiparticle'', the \textsl{transverse plasmon.}

\smallskip\ 

\textbf{c. }The plasmon dispersion curve in the figure shows once again (see
7.5.7d) that propagation is not possible at frequencies below $\omega _{p},$
which acts as a cut-off frequency. It is also interesting to note that
plasmons have the same energy-momentum relation as relativistic particles of
mass $\hbar \omega _{p}/c^{2}$. For $kc\ll $ $\omega _{p}$, they behave like
newtonian particles of this mass. For $kc\gg $ $\omega _{p}$, they behave
like photons. \FRAME{dtbpFU}{3.4272in}{2.0003in}{0pt}{\Qcb{Plot of $\omega
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\newpage

\subsection{7.6.2}

\quad \thinspace \textbf{a.} Wave propagation in the ionosphere is further
complicated by the presence of the slowly varying magnetic field of the
earth, which for the present purposes will be assumed to have the locally
constant value $\mathbf{B}_0$. Then the medium is anisotropic (see 7.5.1
c(i)), which implies that birefringence occurs at least in certain
directions of propagation.

\smallskip\ 

\textbf{b.} It is clear on physical grounds that the response will be
different if $\mathbf{E}$ is parallel to $\mathbf{B}_0$ than it is for $%
\mathbf{E}$ perpendicular to $\mathbf{B}_0.$ In fact, for $\mathbf{E}$ $/\!/$
$\mathbf{B}_0$ it is as if there were no $\mathbf{B}_0$, while for $\mathbf{E%
}$ $\perp \mathbf{B}_0$ the two circularly polarized components have
different response functions, but are independent of each other. Thus the
three directions $\mathbf{\epsilon }_{\pm }=\mathbf{\epsilon }_1\pm i\mathbf{%
\epsilon }_2$ and $\mathbf{\epsilon }_3,$ with $\mathbf{\epsilon }_1,\mathbf{%
\epsilon }_2,\mathbf{\epsilon }_3,$ mutually orthogonal and $\mathbf{%
\epsilon }_3$ $/\!/\mathbf{\ B}_0,$ diagonalize the dielectric tensor $%
\varepsilon _{pq}.$

\smallskip\ 

\textbf{c}. In the simple, independent electron model, we have 
\begin{equation}
m\frac{d\mathbf{v}}{dt}+\frac{e}{c}\mathbf{B}_{0}\times \mathbf{v=}e\mathbf{%
E.}  \label{2}
\end{equation}
If $\mathbf{E=}E_{0}(\mathbf{\epsilon }_{1}\pm i\mathbf{\epsilon }_{2})\exp
(-i\omega t),$ we find 
\[
\mathbf{v=-}\frac{ie\mathbf{\mathbf{E}}}{m(\omega \mp \omega _{B})}\,,
\]
with $\omega _{B}=\left| e\right| B_{0}/mc,$ the cyclotron frequency. One
can understand the frequency shift by changing to a frame rotating with
frequency $\pm \omega _{B}$. Setting $Ne\mathbf{v=}$ $\sigma \mathbf{E}$ and 
$\varepsilon =1+4\pi i\sigma /\omega ,$ one has 
\begin{equation}
\varepsilon _{\pm }=1-\frac{\omega _{p}^{2}}{\omega (\omega \mp \omega _{B})}%
\,.  \label{3}
\end{equation}

\newpage

\subsection{7.6.3}

\quad \thinspace \textbf{a.} For $\mathbf{k}$ parallel to $\mathbf{B}_{0}$,
the relation between $k$ and $\omega $ for left-circularly polarized $\left(
+\right) $ waves is not too different from the case $\mathbf{B}_{0}=0;$ but
for right-circularly polarized $\left( -\right) $ waves, the relation 
\begin{equation}
c^{2}k^{2}=\omega \left( \omega -\frac{\omega _{p}^{2}}{\omega -\omega _{B}}%
\right)   \label{4}
\end{equation}
is qualitatively different for $\omega <\omega _{B},$ where $k^{2}$ is
positive. For $\omega \ll \omega _{B}$ and $\ll \omega _{p}^{2}/\omega _{B},$
we have $c^{2}k^{2}=\omega \omega _{p}^{2}/\omega _{B}.$ The energy-momentum
relation is quadratic, as for a non-relativistic particle in quantum
mechanics. This wave is called a \textsl{whistler} in radio reception and a 
\textsl{helicon} in solid state plasma physics.

\smallskip\ 

\textbf{b. }Whistlers are generated by thunderstorms and travel down the
magnetic field lines to the antipodal point on the surface of the earth.

\smallskip\ 

\textbf{c.} For $\omega \gg \omega _{B},\,\,\,\varepsilon _{+}$ and $%
\,\varepsilon _{-}$ are both negative up to a frequency that is essentially $%
\omega _{p}$ modified by the presence of $B_{0}.$ If the electron density $N$
varies slowly, $k$ will change smoothly and remain positive as long as the
local plasma frequency is below $\omega .$ The physics is the same as for a
quantum mechanical particle moving in a slowly varying potential. If the
density is such that the local $\omega _{p}$ is above $\omega ,$ a turning
point is reached and the wave is reflected.

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