Dispersion relation for electron plasma waves at large and small wavelengths

Question

I am currently reading F.Chen's Introduction to Plasma Physics and Controlled Fusion

On page 83, he derives the dispersion relation for the electron plasma wave: $$ \omega^2 = \omega^2_p + \frac{3}{2}k^2v_{\text{th}}^2 $$

where $v_{\text{th}} = \sqrt{\frac{2k_B T_e}{m_e}}$ represents the thermal velocity. Note that $k_B$ is the Boltzmann constant and $k$ is the wave number. We can then derive the expression for the group velocity by: $$ v_g = \frac{d\omega}{dk} = \frac{3}{2} \frac{k}{\omega} v_{\text{th}}^2 $$

This is the velocity in which information is carried by the electron plasma wave.

He then proceeds to state the following:

(a) At large $k$ (small $\lambda$), information travels essentially at the thermal velocity

(b) At small $k$ (large $\lambda$), information travels more slowly than $v_{\text{th}}$ even though the phase velocity $v_\phi = \frac{\omega}{k}$ is greater than $v_{th}$. This is because the density gradient is small at large $\lambda$ and thermal motions carry very little net momentum into the adjacent layers.

For (a), how does he arrive at the conclusion that: $$ v_{g} = \frac{3}{2} \frac{k}{\omega} v_{\text{th}}^2 \approx v_{th} $$

for large $k$?

For (b), I do not really understand the argument regarding the density gradient and require some explanation, if $v_\phi = \frac{\omega}{k}$ is large because $k$ is large, can't the conclusion that $v_g << v_{th}$ be deduced from: $$ v_{g} = \frac{3}{2} \frac{k}{\omega} v_{\text{th}}^2 << v_{th} $$

without referring to the density gradient?

Answer 1

(a):

$$v_g=\frac 3 2 \frac k{\omega}v^2_{\rm th}$$

with

$$\frac k{\omega}=\frac k{\sqrt{\omega_p^2+ \frac 3 2 k^2v^2_{\rm th}}}\rightarrow \frac k {\frac 3 2 k v_{\rm th}}=\frac 1 {\frac 3 2 v_{\rm th}}$$

where the arrow means $\omega_p \ll \omega$.

Substitution shows: $v_g = v_{\rm th}$.

Answer 2

First, keep the derivative in terms of the wavenumber, $k$, and the thermal speed, $v_{th}$ to see that the group speed is given by: $$ \frac{ \partial \omega }{ \partial k } = \frac{ 3 \ k \ v_{th}^{2} }{ 2 \sqrt{ \omega_{pe}^{2} + \tfrac{ 3 }{ 2 } k^{2} \ v_{th}^{2} } } \tag{0} $$

(a) At large $k$ (small $\lambda$), information travels essentially at the thermal velocity

Take the limit of Equation 0 above as $k \rightarrow \infty$ and you will get a result proportional to $v_{th}$.

(b) At small $k$ (large $\lambda$), information travels more slowly than $v_{th}$ even though the phase velocity $v_{\phi} = \tfrac{ \omega }{ k }$ is greater than $v_{th}$. This is because the density gradient is small at large $\lambda$ and thermal motions carry very little net momentum into the adjacent layers.

Take the limit of Equation 0 above as $k \rightarrow 0$ and you will get a result that asymptotically approaches zero.

For (b), I do not really understand the argument regarding the density gradient and require some explanation...

It's a relative statement. When the wavelength is large, the density oscillation of the Langmuir waves are spread out over large distances, thus the density gradients will be small (i.e., since gradients are the change in some quantity over distance). It's another way of saying that the change in density is gradual relative to some other physically relevant parameter (e.g., gyroradius).

...if $v_{\phi} = \tfrac{ \omega }{ k }$ is large because $k$ is large...

I think you have this backwards. In the limit of large $k$ for constant $\omega$, the phase speed will go to zero.

...can't the conclusion that $v_{g} \ll v_{th}$ be deduced from... without referring to the density gradient?

Okay, I think you meant to say small $k$ here but no matter. Physically, the limit of large $k$ (small $\lambda$) is just saying that the wavelengths asymptotically approach some lower boundary. In a plasma, the smallest physically meaningful wavelength is roughly $2 \pi \ \lambda_{De}$, where $\lambda_{De}$ is the Debye length. Generally Langmuir wave wavelengths approach the electron skin depth, which tends to be much much larger than the Debye length in most plasmas. Since Langmuir waves are really just longitudinal thermal osciallations, it makes physical sense that information would not exceed the local thermal speed of the electrons.

In the opposite limit (i.e., small $k$, large $\lambda$), the electrons can only oscillate so fast and so far before they would decouple to the local ions, i.e., you can't arbitrarily pull electrons out of a plasma without affecting the ions and the local quasi-neutral system. Eventually the electric fields will always do work to eliminate themselves. Even so, if the wavelength gets extremely large, the electrons have to cover larger distances per unit time than the local thermal speed would allow thus preventing the net transfer of matter/energy. The end result is a mode with a finite phase speed but zero group speed.