I've been reading the chapter on plasmon polaritons and polarons, and I have come to a halt trying to understand this part of the chapter:

The dielectric function of the free electron gas follows from (6) and (7): $$ \text{(CGS)}\quad \epsilon(\omega) = 1-\frac{4\pi ne^2}{m\omega^2}; \qquad \text{(SI)}\quad \epsilon(\omega) = 1-\frac{ne^2}{\epsilon_0m\omega^2}. \tag{8} $$ The plasma frequency $\omega_p$ is defined by the relation $$ \text{(CGS)}\quad \omega_p^2=4\pi ne^2/m; \qquad \text{(SI)}\quad \omega_p^2 = ne^2/\epsilon_0m. \tag{9} $$ A plasma is a medium with equal concentration of positive and negative charges, of which at least one charge type is mobile. In a solid the negative charges of the conduction electrons are balanced by an equal concentration of positive charge of the ion cores. We write the dielectric function (8) as $$ \epsilon(\omega) = 1-\frac{\omega_p^2}{\omega^2},\tag{10}$$ plotted in Fig. 1.

If the positive ion core background has a dielectric constant labeled $\epsilon(\infty)$ essentially constant up to frequencies well above $\omega_p$, then (8) becomes $$ \epsilon(\omega) = \epsilon(\infty)-4\pi ne^2/m\omega^2 = \epsilon(\infty)\left[1-\bar\omega_p^2/\omega^2\right],\tag{11} $$ where $\bar\omega_p$ is defined as $$ \bar\omega_p^2 = 4\pi ne^2/\epsilon(\infty)m. \tag{12} $$ Notice that $\epsilon=0$ at $\omega=\bar\omega_p$.

I just need to know how we arrive at equation no. 11 and equation no. 12. In particular, what does the $\infty$ in $\epsilon(\infty)$ signify?

  • $\begingroup$ Use (10) to evaluate $\epsilon(\infty)$ and combine that with (8)+(9) and rearrange to get (11)+(12). And $\epsilon(\infty)$ represents the dielectric constant in the limit of high frequencies (it converges). $\endgroup$
    – lemon
    Oct 11, 2016 at 9:12
  • $\begingroup$ umm.. I know that part but in equation 10 we had 1- (value) but in 11 we have $\epsilon(\infty)$ - (value). Does equation 10 assume that $\epsilon(\infty)$ is 1? Thanks for pointing out that $\infty$ stands for high frequency limit. :) $\endgroup$ Oct 11, 2016 at 9:18
  • $\begingroup$ Yes, stick $\omega=\infty$ into (10) and what do you get? $\endgroup$
    – lemon
    Oct 11, 2016 at 9:30
  • $\begingroup$ @lemon No, that's pretty much completely wrong. Please have a closer look. $\endgroup$ Oct 11, 2016 at 9:50

1 Answer 1


Here $\epsilon(\infty)$ denotes the dielectric constant of the ion background. The excerpt might be clearer if modified a bit:

If the positive ion core background has a dielectric constant, labeled $\epsilon(\infty)$, which is essentially constant up to frequencies well above $\omega_p$, then ...

In this framework you are splitting the system as a plasma that consists of fixed ions and mobile conduction electrons. However, it's important to realize that the background of fixed ions still contains electrons: in copper, for example, the $3d$ electrons are mobile, but the core electrons - in the $1s$, $2s$, $2p$, $3s$ and $3p$ shells - are still pretty much bound to the nucleus. These are the "ionic cores" of the excerpt - the nucleus plus the inner electrons.

More importantly, these ionic cores can still be polarized by an external field, since the inner electrons can be displaced slightly with respect to the nucleus. This adds an additional term to the polarizability and therefore to the global dielectric constant.

In addition, Kittel is making the assumption that the frequency dependence of said ionic cores is relatively flat in the region of interest. This is a reasonable approximation: this polarizability does depend on frequency, but only in regimes where you can 'talk' to the excitations of this core, which tend to be at higher energies than the valence electron.

This means, then, that you have some additional term in the permittivity, which you don't know but which is reasonably flat, and this means that you can replace the background of $1$ with some suitable constant, which Kittel labels $\epsilon(\infty)$. Everything after that (including e.g. equation 12) is just pushing symbols around.

As to why Kittel chose $\epsilon(\infty)$ as the symbol to use, though - you'd have to ask Kittel.

  • $\begingroup$ I think he uses that symbol just because once you assume $\epsilon(\omega)=A(1-\omega_p^2/\omega^2)$ where $A$ is some constant you get $\epsilon(\omega \to \infty)=A$ $\endgroup$
    – valerio
    Oct 11, 2016 at 10:05
  • $\begingroup$ @valerio92 That's of limited use, though - the assumption that the background permittivity is flat only works for low frequencies, so getting $A$ from $\omega\to\infty$ only works in the vaguest, most symbolic of senses. $\endgroup$ Oct 11, 2016 at 10:10
  • $\begingroup$ Kittel does use the phrase "$\epsilon(\infty)$ is essentially constant upto frequencies well above $\omega_p$". Doesn't that mean that he's using the "constant" nature of background from low frequency and going up to high frequencies? The high frequencies needn't be so high but they're sufficiently higher than $\omega_p$ to be considered $\infty$ wrt $\omega_p$ $\endgroup$ Oct 11, 2016 at 10:16
  • $\begingroup$ @Sad_lab_rat This is all terminology - there's nothing to see there, really. Your argument works, but it's only notation (and, in any case $\infty$ is actual infinity unless you preface it with "except you shouldn't go too high", which Kittel doesn't do). The physics is the only thing that actually matters. $\endgroup$ Oct 11, 2016 at 10:18
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    $\begingroup$ @RobinEkman Perhaps I wasn't clear earlier: The reasons for the notation do not matter. Only the physics does, and the notation is not the physics. $\endgroup$ Oct 11, 2016 at 12:40

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