Cold-Plasma Dielectric Permittivity

In a collisionless plasma, the linearized cold-plasma equations are written [see Equations (4.225)–(4.227)],

$$m_i \,n_e\,\frac{\partial {\bf V}}{\partial t} = {\bf j}\times{\bf B}_0,$$ (5.12)

$${\bf E} = - {\bf V}\times{\bf B}_0 +\frac{{\bf j}\times{\bf B}_0}{n_e\,e} +\frac{m_e}{n_e\,e^2}\frac{\partial{\bf j}}{\partial t},$$ (5.13)

where $n_e$ is the equilibrium electron number density. Substitution of plane-wave solutions of the type (5.1) into the previous equations yields

$$-{\rm i}\,\omega\,m_i \,n_e\,{\bf V} = {\bf j}\times{\bf B}_0,$$ (5.14)

$${\bf E} = - {\bf V}\times{\bf B}_0 +\frac{{\bf j}\times{\bf B}_0}{n_e\,e} -{\rm i}\,\frac{\omega\,m_e}{n_e\,e^2}\,{\bf j}.$$ (5.15)

Let

$${ {\mit\Pi}}_e = \sqrt{\frac{n_e\,e^2}{\epsilon_0\,m_e}},$$ (5.16)

$${ {\mit\Pi}}_i = \sqrt{\frac{n_e\,e^2}{\epsilon_0\,m_i}},$$ (5.17)

$${{\mit\Omega}}_e = -\frac{e\,B_0}{m_e},$$ (5.18)

$${{\mit\Omega}}_i = \frac{e\,B_0}{m_i},$$ (5.19)

be the electron plasma frequency, the ion plasma frequency, the electron cyclotron frequency, and the ion cyclotron frequency, respectively. Eliminating the fluid velocity, ${\bf V}$, between Equations (5.14) and (5.15), and making use of the previous definitions, we obtain

$${\rm i}\,\omega\,\epsilon_0\,{\bf E} = \frac{ \omega^2\,{\bf j} -... ..._e\,{{\mit\Omega}}_i\,{\bf b}\times ({\bf j}\times{\bf b})}{{{\mit\Pi}}_e^{2}},$$ (5.20)

where ${\bf b} = {\bf B}_0/B_0$.

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

$$j_\parallel = {\rm i}\,\omega\,\epsilon_0\,\frac{{{\mit\Pi}}_e^{2}}{\omega^2}\, E_\parallel,$$ (5.21)

where $j_\parallel \equiv {\bf j}\cdot{\bf b}$, et cetera. In solving for ${\bf j}_\perp\equiv {\bf j} - j_\parallel\,{\bf b}$, it is helpful to define the vectors

$${\bf e}_+ = \frac{ {\bf e}_1 +{\rm i}\,{\bf e}_2}{\sqrt{2}},$$ (5.22)

$${\bf e}_- = \frac{ {\bf e}_1 - {\rm i}\,{\bf e}_2}{\sqrt{2}}.$$ (5.23)

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

$${\bf e}_\pm\times{\bf b} = \pm {\rm i}\,{\bf e}_\pm,$$ (5.24)

$${\bf b}\times ({\bf e}_\pm\times{\bf b}) = {\bf e}_\pm.$$ (5.25)

It follows that

$$j_\pm = {\rm i}\,\omega\,\epsilon_0\left(\frac{{{\mit\Pi}}_e^{2}}... ...\pm \omega\,{{\mit\Omega}}_e + {{\mit\Omega}}_e\,{{\mit\Omega}}_i}\right)E_\pm,$$ (5.26)

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

The conductivity tensor is diagonal in the “circular” basis $({\bf e}_+, {\bf e}_-,{\bf b})$. In fact, its elements are the coefficients of $E_\pm$ and $E_\parallel$ in Equations (5.26) and (5.21), respectively. Thus, the dielectric permittivity tensor, defined in Equation (5.8), takes the form

$${\bf K}_{\rm circ} = \left(\begin{array}{ccc} R,&0,&0\\ 0,&L,&0\\ 0,&0,&P\end{array}\right),$$ (5.27)

where

$$R \simeq 1 - \frac{{{\mit\Pi}}_e^{2}} {\omega^2+\omega\,{{\mit\Omega}}_e + {{\mit\Omega}}_e\,{{\mit\Omega}}_i},$$ (5.28)

$$L \simeq 1 - \frac{{{\mit\Pi}}_e^{2}} {\omega^2-\omega\,{{\mit\Omega}}_e+ {{\mit\Omega}}_e\,{{\mit\Omega}}_i },$$ (5.29)

$$P \simeq 1- \frac{{{\mit\Pi}}_e^{2}}{\omega^2}.$$ (5.30)

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.

The previous expressions are only approximate because the small mass-ratio ordering $m_e/m_i\ll 1$ has already been incorporated 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

$$R = 1 - \frac{{{\mit\Pi}}_e^{2}}{\omega^2} \!\left(\frac{\omega}{\o... ...\Pi}}_i^{2}}{\omega^2}\! \left(\frac{\omega}{\omega + {{\mit\Omega}}_i}\right),$$ (5.31)

$$L = 1 - \frac{{{\mit\Pi}}_e^{2}}{\omega^2}\! \left(\frac{\omega}{\o... ...t\Pi}}_i^{2}}{\omega^2}\! \left(\frac{\omega}{\omega -{{\mit\Omega}}_i}\right),$$ (5.32)

$$P = 1 - \frac{{{\mit\Pi}}_e^{2}}{\omega^2}-\frac{{{\mit\Pi}}_i^{2}}{\omega^2}.$$ (5.33)

Equations (5.28)–(5.30) and (5.31)–(5.33) are equivalent in the limit $m_e/m_i\rightarrow 0$. Furthermore, Equations (5.31)–(5.33) generalize in a fairly obvious manner to plasmas consisting of more than two particle species.

In order to obtain the standard expression for dielectric permittivity tensor, it is necessary to transform 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 ${\bf a}$ in the Cartesian basis are related to the components in the “circular” basis via

$$\left(\!\begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array}\! \right) = {\bf U} \left(\!\begin{array}{c} a_+ \\ a_- \\ a_3 \end{array}\! \right),$$ (5.34)

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

$${\bf U} = \frac{1}{\sqrt{2}}\left(\begin{array}{ccc} 1,&1,&0\\ {\rm i}, & -{\rm i},&0\\ 0,&0,&\sqrt{2}\end{array}\right).$$ (5.35)

The dielectric permittivity in the Cartesian basis is then

$${\bf K} = {\bf U}\, {\bf K}_{\rm circ}\, {\bf U}^\dag .$$ (5.36)

We obtain

$${\bf K} = \left(\begin{array}{ccc} S,&-{\rm i}\,D,&0\\ {\rm i}\,D, & S,&0\\ 0,&0,&P\end{array}\right),$$ (5.37)

where

$$S =\frac{R+L}{2},$$ (5.38)

and

$$D = \frac{R-L}{2},$$ (5.39)

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