# What is the "free charge" in Langmuir oscillations for T>0?

• A
Gold Member
Summary:
Cold Langmuir waves have dielectric constant given by ##\epsilon (\omega) = \epsilon_0 ( 1 - \frac{\omega_{pe}^2}{\omega^2})##. This implies the existence of a free charge ##\rho_f = \nabla \cdot (\epsilon \vec{E})##. In this picture, the electrons' and ions' displacement are considered part of the bound charge. So what is this unaccounted free charge in reality?
I did a homework problem in plasma physics recently, and got the right answer (I already submitted the assignment, that's why I didn't put this in the homework subforum), but I had to introduce a new charge density term that doesn't seem to actually exist (but it's zero at T=0). The problem was to derive the dielectric constant ##\epsilon(\omega)## for cold Langmuir waves (as a function of ##\omega##, so the answer isn't automatically 0) such that ##\nabla \cdot (\epsilon \vec{E}) = 0 ##.

I took the approach of assuming a non-zero free charge ("free charge" here is defined in the sense of free vs bound charges, as discussed in Griffiths Ch 4 and Jackson Ch 4). I took the bound charge to be the entire charge distribution in the plasma: $$\rho_b = e(n_i - n_e)$$ where ##n_i## is the density of ions (assumed to be singly-ionized) and ##n_e## is the density of electrons. So my macroscopic Poisson equation looks like: $$\nabla \cdot (\epsilon \vec{E}) = \rho_f$$

I then use the following perturbative approximation for small amplitude longitudinal waves in the plasma: $$n_i = n_0$$ $$n_e = n_0 + n_1$$ $$\vec{u} = u_1 \hat{k}$$ $$\vec{E} = E_1 \hat{k}$$
where ##\vec{u}## is the fluid flow velocity, and where all of the above quantities with subscript "1" (not 0) are written with exponential notation: $$\chi_1 = \tilde{\chi}_1 e^{i(\vec{k} \cdot \vec{x} - \omega t)}$$ where ##\chi## is any of the above quantities and ##\tilde{\chi}_1## is the complex amplitude of that quantity. So, for example, ##n_e = n_0 + \tilde{n}_1 e^{i(\vec{k} \cdot \vec{x} - \omega t)}##. Lastly, because these are small-oscillation waves, the perturbation scale is given by ##\frac{|\tilde{n}_1|}{n_0}##, and I only take up to first order.

With all that notation out of the way, the way my solution proceeded was to use my macroscopic Poisson equation ##\nabla \cdot \vec{E} = \frac{\rho_f}{\epsilon}## and combining it with the microscopic Poisson equation ##\nabla \cdot \vec{E} = \frac{e}{\epsilon_0} (n_i - n_e) + \frac{\rho_f}{\epsilon_0} ##, and plugging in the wave expansions to get $$\rho_f = \frac{\epsilon_r}{\epsilon_r - 1} en_1$$ where ##\epsilon_r = \epsilon / \epsilon_0##.

From there, I use the adiabatic equation of state, continuity equation, and Cauchy momentum equation to this massive system of equations for ##\epsilon_r## and I successfully derive the well-known result $$\epsilon_r = 1 - \frac{\omega_{pe}^2}{\omega^2}$$ Since for Langmuir waves at T=0 we know that ##\omega = \omega_{pe}##, it follows that this free charge I made up is 0 at T=0. How am I supposed to interpret it at non-zero temperature? It has the form of a longitudinal wave that is larger in magnitude at 180deg out of phase with the bound charge density (makes sense for dieletric type behavior). But why does there exist a charge density *in addition to* the plasma itself? Did I make a bad ansatz?

Gold Member
Alright, I did some more thinking about this and noticed even at finite temperature, ##\rho_f = 0##.

I derived the dispersion relationship while taking ##\rho_f## to be non-zero, and ended up with: $$\omega^2 = \gamma v_{therm}^2 + \frac{\omega_{pe}^2}{1-\epsilon_r}$$

By comparison with the Bohm-Gross dispersion relationship $$\omega^2 = \gamma v_{therm}^2 + \omega_{pe}^2$$
we can conclude that ##\epsilon_r = 0## and therefore that ##\rho_f = \nabla \cdot (\epsilon \vec{E}) = 0##

This settles it that the solution I came up with where I introduced a free charge is just a hack and there's no deeper significance to it.