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Low Frequency Behavior

Near $\omega = 0$ the qualitative behavior depends upon whether or not there is a ``resonance'' there. If there is, then $\epsilon(\omega\approx 0)$ can begin with a complex component that attenuates the propagation of EM energy in a (nearly static) applied electric field. This (as we shall see) accurately describes conduction and resistance. If there isn't, then $\epsilon $ is nearly all real and the material is a dielectric insulator.

Suppose there are both ``free'' electrons (counted by $f_f$) that are ``resonant'' at zero frequency, and ``bound'' electrons (counted by $f_b$). Then if we start out with:

$\displaystyle \epsilon(\omega)$ $\textstyle =$ $\displaystyle \epsilon_0\left (1 + \frac{N e^2}{m}
\sum_i \frac{f_i}{(\omega_i^2 - \omega^2 - i \omega \gamma_i)}\right)$  
  $\textstyle =$ $\displaystyle \epsilon_0\left (1 + \frac{N e^2}{m}
\sum_b \frac{f_b}{(\omega_b^2 - \omega^2 - i \omega \gamma_b)}\right)$  
    $\displaystyle \hspace{1cm} + \frac{N e^2}{m}
\sum_f \frac{f_f}{(- \omega^2 - i \omega \gamma_f)}$  
  $\textstyle =$ $\displaystyle \epsilon_b + i \epsilon_0 \frac{N e^2 f_f}{m \omega
(\gamma_0 - i \omega)}$ (9.117)

where $\epsilon_b$ is now only the contribution from all the ``bound'' dipoles.

We can understand this from

\begin{displaymath}
\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$H$} = \mbox{\boldmath$J$} + \frac{d\mbox{\boldmath$D$}}{dt}
\end{displaymath} (9.118)

(Maxwell/Ampere's Law). Let's first of all think of this in terms of a plain old static current, sustained according to Ohm's Law:
\begin{displaymath}
\mbox{\boldmath$J$} = \sigma \mbox{\boldmath$E$}.
\end{displaymath} (9.119)

If we assume a harmonic time dependence and a ``normal'' dielectric constant $\epsilon_b$, we get:

$\displaystyle \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$H$}$ $\textstyle =$ $\displaystyle \left(\sigma - i\omega \epsilon_b \right)\mbox{\boldmath$E$}$  
  $\textstyle =$ $\displaystyle -i \omega\left ( \epsilon_b + i \frac {\sigma}{\omega} \right ) \mbox{\boldmath$E$}.$ (9.120)

On the other hand, we can instead set the static current to zero and consider all ``currents'' present to be the result of the polarization response $\mbox{\boldmath$D$}$ to the field $\mbox{\boldmath$E$}$. In this case:

$\displaystyle \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$H$}$ $\textstyle =$ $\displaystyle - i\omega \epsilon \mbox{\boldmath$E$}$  
  $\textstyle =$ $\displaystyle -i \omega\left ( \epsilon_b +
i \epsilon_0 \frac{N e^2}{m} \frac{f_f}{(\gamma_0 - i \omega)} \right)
\mbox{\boldmath$E$}$ (9.121)

Equating the two latter terms in the brackets and simplifying, we obtain the following relation for the conductivity:

\begin{displaymath}
\sigma = \epsilon_0  \frac{n_f e^2}{m}\frac{1}{(\gamma_0 - i \omega)}.
\end{displaymath} (9.122)

This is the Drude Model with $n_f = f_f N$ the number of ``free'' electrons per unit volume. It is primarily useful for the insight that it gives us concerning the ``conductivity'' being closely related to the zero-frequency complex part of the permittivity. Note that at $\omega = 0$ it is purely real, as it should be, recovering the usual Ohm's Law.

We conclude that the distinction between dielectrics and conductors is a matter of perspective away from the purely static case. Away from the static case, ``conductivity'' is simply a feature of resonant amplitudes. It is a matter of taste whether a description is better made in terms of dielectric constants and conductivity or complex dielectric.


next up previous contents
Next: High Frequency Limit; Plasma Up: Dispersion Previous: Attenuation by a complex   Contents
Robert G. Brown 2007-12-28