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Cold-Plasma Dielectric Permittivity

In a collisionless plasma, the linearized cold-plasma equations are written [see Eqs. (408)-(411)]:
$\displaystyle m_i n\,\frac{\partial {\bf V}}{\partial t}$ $\textstyle =$ $\displaystyle {\bf j}\times{\bf B}_0,$ (461)
$\displaystyle {\bf E}$ $\textstyle =$ $\displaystyle - {\bf V}\times{\bf B}_0 +\frac{{\bf j}\times{\bf B}_0}{ne}
+\frac{m_e}{ne^2}\frac{\partial{\bf j}}{\partial t}.$ (462)

Substitution of plane wave solutions of the type (450) into the above equations yields
$\displaystyle -{\rm i}\,\omega\,m_i n\,{\bf V}$ $\textstyle =$ $\displaystyle {\bf j}\times{\bf B}_0,$ (463)
$\displaystyle {\bf E}$ $\textstyle =$ $\displaystyle - {\bf V}\times{\bf B}_0 +\frac{{\bf j}\times{\bf B}_0}{ne}
-{\rm i}\,\omega\,\frac{m_e}{ne^2}\,{\bf j}.$ (464)

$\displaystyle { \Pi}_e$ $\textstyle =$ $\displaystyle \sqrt{\frac{n\,e^2}{\epsilon_0\,m_e}},$ (465)
$\displaystyle { \Pi}_i$ $\textstyle =$ $\displaystyle \sqrt{\frac{n\,e^2}{\epsilon_0\,m_i}},$ (466)
$\displaystyle {\Omega}_e$ $\textstyle =$ $\displaystyle -\frac{e\,B_0}{m_e},$ (467)
$\displaystyle {\Omega}_i$ $\textstyle =$ $\displaystyle \frac{e\,B_0}{m_i},$ (468)

be the electron plasma frequency, the ion plasma frequency, the electron cyclotron frequency, and the ion cyclotron frequency, respectively. The ``plasma frequency,'' $\omega_p$, mentioned in Sect. 1, is identical to the electron plasma frequency, ${ \Pi}_e$. Eliminating the fluid velocity ${\bf V}$ between Eqs. (463) and (464), and making use of the above definitions, we obtain
{\rm i}\,\omega\,\epsilon_0\,{\bf E} =
\frac{ \omega^2\,{\bf...
...\bf b} + {\Omega}_e\,{\Omega}_i\,{\bf j}_\perp}{{\Pi}_e^{~2}}.
\end{displaymath} (469)

The parallel component of the above equation is readily solved to give

j_\parallel = \frac{{\Pi}_e^{~2}}{\omega^2}\,({\rm i}\,\omega\,\epsilon_0\,
\end{displaymath} (470)

In solving for ${\bf j}_\perp$, it is helpful to define the vectors:
$\displaystyle {\bf e}_+$ $\textstyle =$ $\displaystyle \frac{ {\bf e}_1 +{\rm i}\,{\bf e}_2}{\sqrt{2}},$ (471)
$\displaystyle {\bf e}_-$ $\textstyle =$ $\displaystyle \frac{ {\bf e}_1 - {\rm i}\,{\bf e}_2}{\sqrt{2}}.$ (472)

Here, $({\bf e}_1, {\bf e}_2, {\bf b})$ are a set of mutually orthogonal, right-handed unit vectors. Note that
{\bf b}\times {\bf e}_\pm = \mp {\rm i}\,{\bf e}_\pm.
\end{displaymath} (473)

It is easily demonstrated that
j_\pm = \frac{{\Pi}_e^{~2}}
{\omega^2 \pm \omega\,{\Omega}_e
+ {\Omega}_e\,{\Omega}_i}\,{\rm i}\,\omega\,\epsilon_0\,E_\pm,
\end{displaymath} (474)

where $j_\pm = {\bf j}\cdot {\bf e}_\pm$, etc.

The conductivity tensor is diagonal in the basis $({\bf e}_+, {\bf e}_-,{\bf b})$. Its elements are given by the coefficients of $E_\pm$ and $E_\parallel$ in Eqs. (474) and (470), respectively. Thus, the dielectric permittivity (457) takes the form

{\bf K}_{\rm circ} = \left(\begin{array}{ccc}
\end{displaymath} (475)

$\displaystyle R$ $\textstyle \simeq$ $\displaystyle 1 - \frac{{\Pi}_e^{~2}}
{\omega^2+\omega\,{\Omega}_e + {\Omega}_e\,{\Omega}_i},$ (476)
$\displaystyle L$ $\textstyle \simeq$ $\displaystyle 1 - \frac{{\Pi}_e^{~2}}
{\Omega}_e\,{\Omega}_i },$ (477)
$\displaystyle P$ $\textstyle \simeq$ $\displaystyle 1- \frac{{\Pi}_e^{~2}}{\omega^2}.$ (478)

Here, $R$ and $L$ represent the permittivities for right- and left-handed circularly polarized waves, respectively. The permittivity parallel to the magnetic field, $P$, is identical to that of an unmagnetized plasma.

In fact, the above expressions are only approximate, because the small mass-ratio ordering $m_e/m_i\ll 1$ has already been folded into the cold-plasma equations. The exact expressions, which are most easily obtained by solving the individual charged particle equations of motion, and then summing to obtain the fluid response, are:

$\displaystyle R$ $\textstyle =$ $\displaystyle 1 - \frac{{\Pi}_e^{~2}}{\omega^2}
\!\left(\frac{\omega}{\omega + ...
\left(\frac{\omega}{\omega + {\Omega}_i}\right),$ (479)
$\displaystyle L$ $\textstyle =$ $\displaystyle 1 - \frac{{\Pi}_e^{~2}}{\omega^2}\!
\left(\frac{\omega}{\omega -{...
\left(\frac{\omega}{\omega -{\Omega}_i}\right),$ (480)
$\displaystyle P$ $\textstyle =$ $\displaystyle 1 - \frac{{\Pi}_e^{~2}}{\omega^2}-\frac{{\Pi}_i^{~2}}{\omega^2}.$ (481)

Equations (476)-(478) and (479)-(481) are equivalent in the limit $m_e/m_i\rightarrow 0$. Note that Eqs. (479)-(481) generalize in a fairly obvious manner in plasmas consisting of more than two particle species.

In order to obtain the actual dielectric permittivity, it is necessary to transform back to the Cartesian basis $({\bf e}_1, {\bf e}_2, {\bf b})$. Let ${\bf b} \equiv {\bf e}_3$, for ease of notation. It follows that the components of an arbitrary vector ${\bf W}$ in the Cartesian basis are related to the components in the ``circular'' basis via

\left(\!\begin{array}{c} W_1 \\ W_2 \\ W_3 \end{array}\! \ri...
...ft(\!\begin{array}{c} W_+ \\ W_- \\ W_3 \end{array}\! \right),
\end{displaymath} (482)

where the unitary matrix ${\bf U}$ is written
{\bf U} = \frac{1}{\sqrt{2}}\left(\begin{array}{ccc}
{\rm i} & -{\rm i}&0\\
\end{displaymath} (483)

The dielectric permittivity in the Cartesian basis is then
{\bf K} = {\bf U}\, {\bf K}_{\rm circ}\, {\bf U}^\dag .
\end{displaymath} (484)

We obtain
{\bf K} = \left(\begin{array}{ccc}
S&-{\rm i}\,D&0\\
{\rm i}\,D & S&0\\
\end{displaymath} (485)

S =\frac{R+L}{2},
\end{displaymath} (486)

D = \frac{R-L}{2},
\end{displaymath} (487)

represent the sum and difference of the right- and left-handed dielectric permittivities, respectively.

next up previous
Next: Cold-Plasma Dispersion Relation Up: Waves in Cold Plasmas Previous: Plane Waves in a
Richard Fitzpatrick 2011-03-31