The cold-plasma dielectric permittivity

In a collisionless plasma, the linearized cold-plasma equations are written [see Eqs. (408)-(411)]:

$$m_i n\,\frac{\partial {\bf V}}{\partial t} = {\bf j}\times{\bf B}_0,$$ (437)

$${\bf E} = - {\bf V}\times{\bf B}_0 +\frac{{\bf j}\times{\bf B}_0}{ne} +\frac{m_e}{ne^2}\frac{\partial{\bf j}}{\partial t}.$$ (438)

Substitution of plane wave solutions of the type (426) into the above equations yields

$$-{\rm i}\,\omega\,m_i n\,{\bf V} = {\bf j}\times{\bf B}_0,$$ (439)

$${\bf E} = - {\bf V}\times{\bf B}_0 +\frac{{\bf j}\times{\bf B}_0}{ne} -{\rm i}\,\omega\,\frac{m_e}{ne^2}\,{\bf j}.$$ (440)

Let

$${ \Pi}_e = \sqrt{\frac{n\,e^2}{\epsilon_0\,m_e}},$$ (441)

$${ \Pi}_i = \sqrt{\frac{n\,e^2}{\epsilon_0\,m_i}},$$ (442)

$${\Omega}_e = -\frac{e\,B_0}{m_e},$$ (443)

$${\Omega}_i = \frac{e\,B_0}{m_i},$$ (444)

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. (439) and (440), and making use of the above definitions, we obtain

$${\rm i}\,\omega\,\epsilon_0\,{\bf E} = \frac{ \omega^2\,{\bf... ...\bf b} + {\Omega}_e\,{\Omega}_i\,{\bf j}_\perp}{{\Pi}_e^{~2}}.$$ (445)

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

$$j_\parallel = \frac{{\Pi}_e^{~2}}{\omega^2}\,({\rm i}\,\omega\,\epsilon_0\, E_\parallel).$$ (446)

In solving for ${\bf j}_\perp$, it is helpful to define the vectors:

$${\bf e}_+ = \frac{ {\bf e}_1 +{\rm i}\,{\bf e}_2}{\sqrt{2}},$$ (447)

$${\bf e}_- = \frac{ -{\rm i}\,{\bf e}_1 - {\bf e}_2}{\sqrt{2}}.$$ (448)

Here, $({\bf e}_1, {\bf e}_2, {\bf b})$ are a set of mutually orthogonal, right-handed unit vectors. Note that

$${\bf e}_\pm \times {\bf b} = \pm {\rm i}\,{\bf e}_\pm.$$ (449)

It is easily demonstrated that

$$j_\pm = \frac{{\Pi}_e^{~2}} {\omega^2 \pm \omega\,{\Omega}_e + {\Omega}_e\,{\Omega}_i}\,{\rm i}\,\omega\,\epsilon_0\,E_\pm.$$ (450)

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. (450) and (446), respectively. Thus, the dielectric permittivity (433) takes the form

$${\bf K}_{\rm circ} = \left(\begin{array}{ccc} R&0&0\\ 0&L&0\\ 0&0&P\end{array}\right),$$ (451)

where

$$R \simeq 1 - \frac{{\Pi}_e^{~2}} {\omega^2+\omega\,{\Omega}_e + {\Omega}_e\,{\Omega}_i},$$ (452)

$$L \simeq 1 - \frac{{\Pi}_e^{~2}} {\omega^2-\omega\,{\Omega}_e+ {\Omega}_e\,{\Omega}_i },$$ (453)

$$P \simeq 1- \frac{{\Pi}_e^{~2}}{\omega^2}.$$ (454)

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:

$$R = 1 - \frac{{\Pi}_e^{~2}}{\omega^2} \!\left(\frac{\omega}{\omega + ... ...frac{{\Pi}_i^{~2}}{\omega^2}\! \left(\frac{\omega}{\omega + {\Omega}_i}\right),$$ (455)

$$L = 1 - \frac{{\Pi}_e^{~2}}{\omega^2}\! \left(\frac{\omega}{\omega -{... ...\frac{{\Pi}_i^{~2}}{\omega^2}\! \left(\frac{\omega}{\omega -{\Omega}_i}\right),$$ (456)

$$P = 1 - \frac{{\Pi}_e^{~2}}{\omega^2}-\frac{{\Pi}_i^{~2}}{\omega^2}.$$ (457)

Equations (452)-(454) and (455)-(457) are equivalent in the limit $m_e/m_i\rightarrow 0$. Note that Eqs. (455)-(457) 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

$$\left(\!\begin{array}{c} W_1 \\ W_2 \\ W_3 \end{array}\! \ri... ...ft(\!\begin{array}{c} W_+ \\ W_- \\ W_3 \end{array}\! \right),$$ (458)

where the unitary matrix ${\bf U}$ is written

$${\bf U} = \frac{1}{\sqrt{2}}\left(\begin{array}{ccc} 1&-{\rm i}&0\\ {\rm i} & -1&0\\ 0&0&\sqrt{2}\end{array}\right).$$ (459)

The dielectric permittivity in the Cartesian basis is then

$${\bf K} = {\bf U}\, {\bf K}_{\rm circ}\, {\bf U}^\dag .$$ (460)

We obtain

$${\bf K} = \left(\begin{array}{ccc} S&-{\rm i}\,D&0\\ {\rm i}\,D & S&0\\ 0&0&P\end{array}\right),$$ (461)

where

$$S =\frac{R+L}{2},$$ (462)

and

$$D = \frac{R-L}{2},$$ (463)

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