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

• Twigg
In summary, the conversation discusses the problem of deriving the dielectric constant for cold Langmuir waves and the use of a perturbative approximation for small amplitude longitudinal waves in a plasma. The solution derived introduces a non-zero free charge, but further analysis shows that the free charge is actually zero even at finite temperature. This solution is deemed to be a hack with no deeper significance.
Twigg
Gold Member
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?

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.

## 1. What is the definition of "free charge" in Langmuir oscillations for T>0?

The term "free charge" refers to the excess charge carriers present in a material that are not bound to any specific atom or molecule. In Langmuir oscillations, these free charges are responsible for the oscillatory behavior of the material's electric field.

## 2. How does the presence of free charge affect Langmuir oscillations at high temperatures (T>0)?

At high temperatures, the free charge in a material increases due to the thermal excitation of electrons. This increase in free charge leads to a stronger electric field and a higher frequency of Langmuir oscillations.

## 3. Can Langmuir oscillations occur in materials with no free charge at T>0?

No, Langmuir oscillations require the presence of free charge in order to occur. Without free charge, there would be no excess charge carriers to cause the oscillatory behavior in the electric field.

## 4. How do Langmuir oscillations at T>0 differ from those at T=0?

At T=0, there is no thermal excitation of electrons, so there is no increase in free charge. This results in a lower frequency of Langmuir oscillations compared to those at T>0, where the thermal excitation of electrons increases the free charge and leads to a higher frequency.

## 5. What practical applications are there for understanding Langmuir oscillations at T>0?

Langmuir oscillations at T>0 have many practical applications, including in plasma physics, semiconductor devices, and particle accelerators. Understanding the behavior of free charge and its effects on Langmuir oscillations is crucial for the development of these technologies.

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