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

AI Thread Summary
The discussion revolves around the concept of "free charge" in the context of Langmuir oscillations at temperatures above absolute zero. The original poster derived the dielectric constant for cold Langmuir waves, introducing a free charge density that appeared to be zero at T=0 but questioned its validity at higher temperatures. They concluded that even at finite temperatures, the free charge density remains zero, aligning with the Bohm-Gross dispersion relationship. Ultimately, the introduction of free charge was deemed unnecessary, revealing it as a mere workaround without deeper significance. The findings highlight the complexities of plasma physics and the behavior of charge densities in oscillatory systems.
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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?
 
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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.
 
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