Langmuir wave could also be treated as electron acoustic/sound wave. In cold plasma it is just Langmuir oscillation that could not spread, in warm plasma it could propagate as a kind of longitudinal wave.

Langmuir oscillation

The physical model of Langmuir oscillation and wave is based on the assumption that the movement of ion can be neglected. Which means in the double fluid equation, only the equation for electrons are kept. ($v_i=0, p_i=0$) Compare with ion sound wave, could we treat Langmuir wave as “electron sound/acoustic wave” ? Langmuir oscillation and wave applies to the condition of plasma without external magnetic field.

double fluid equations

Langmuir oscillation treats the plasma as a double fluid. We can make some changes to the two fluid equations that describes the Langmuir oscillation.

$\begin{array}{l l} \frac{\partial n_{\alpha}}{\partial t}+\nabla\cdot (n_{\alpha}\vec{u_{\alpha}})=0 & (a)\\ \frac{\partial \vec{u_{\alpha}}}{\partial t}+(\vec{u_{\alpha}}\cdot\nabla)\vec{u_{\alpha}}-\frac{q_{\alpha}}{m_{\alpha}}(\vec{E}+\vec{u_{\alpha}}\times\vec{B})+\frac{\nabla p_{\alpha}}{n_{\alpha}m_{\alpha}}=\frac{\vec{F_{\alpha}}}{n_{\alpha}m_{\alpha}} & (b)\\ \vec{J}=\sum_{\alpha} n_{\alpha}q_{\alpha}\vec{u_{\alpha}} & (c)\\ \rho_{q}=\sum_{\alpha}n_{\alpha}q_{\alpha} & (d) \end{array}$

one dimensional case

In order to make the matters simple, we only consider the 1 dimensional condition along the x axis. And suppose the plasma are formed by electron and proton.

from \eqref{double_fluid}a, we get equation: $\frac{\partial n_{\alpha}}{\partial t}+\frac{\partial}{\partial x}(n_{\alpha}u_{\alpha})=0$,

from \eqref{double_fluid}b, we get the equation: $n_e m_e(\frac{\partial u_e}{\partial t}+u_e\frac{\partial u_e}{\partial x})+e n_e E_x+\frac{\partial p_e}{\partial x}=0$ further we can get $n_e m_e(\frac{\partial u_e}{\partial t}+u_e\frac{\partial u_e}{\partial x})=-\frac{\partial p_e}{\partial x}-en_e E_x$, among them $F_x=0, \vec{u_x}\times\vec{B}=0$, this should be the simplification of this case.

from \eqref{double_fluid}c and Ampere’s law, we can get the relation that: $\frac{\partial E_x}{\partial x}=\frac{e(n_i-n_e)}{\epsilon_0}$. Combine together forms the equations to describe the Langmuir oscillation.

$\begin{array}{l} \frac{\partial n_{\alpha}}{\partial t}+\frac{\partial}{\partial x}(n_{\alpha}u_{\alpha})=0\\ m_e n_e(\frac{\partial u_e}{\partial t}+u_e\frac{\partial u_e}{\partial x})=-\frac{\partial p_e}{\partial x}-en_e E_x\\ \frac{\partial E_x}{\partial x}=\frac{e(n_i-n_e)}{\epsilon_0} \end{array}$

Langmuir oscillation in simple case

Based on the one dimensional case, we make an assumption that: $n_e=n_0+n_1$ and pressure is constant $p_e=constant$, where $n_0=n_i$, thus the above equation is simplified to:

$\begin{array}{l} \frac{\partial n_1}{\partial t}+n_0\frac{\partial u_e}{\partial x}=0\\ m_e \frac{\partial u_e}{\partial t}=-eE_1\\ \frac{\partial E_1}{\partial x}=\frac{e n_1}{\epsilon_0} \end{array}$

eliminate $E_1$ and $u_e$ from the above equations and combine them as a whole equation, we get wave function for Langmuir oscillation:

$\frac{\partial^2 n_1}{\partial t^2}+\frac{e^2 n_0}{m_e\epsilon_0}n_1=0$

Suppose the density fluctuation has the form: $e^{-i\omega}$, then the above equation becomes: $(-i\omega)^2 n_1+\frac{e^2 n_e}{m_e \epsilon_0}n_1=0$, then we get the oscillation frequency as: $\omega_{pe}^2=\frac{e^2 n_e}{m_e \epsilon_0}$, which is also called the plasma oscillation frequency. We can also see $\omega$ is not depend on wave vector $k$, thus, the group velocity is: $v_g=\frac{d\omega}{dk}=0$. Phase velocity $v_p=\frac{\omega}{k}$ do not have any meaning.

Langmuir wave in hot plasma

The Languire oscillation in cold plasma can not propagate, since its group velocity equals to zero. But in the hot plasma condition, due to the existence of pressure. There exist Languire wave that can propagate in plasma. In hot plasma $\frac{\partial p_e}{\partial x}\neq 0$. Thus the equations becomes:

$\begin{array}{l} \frac{\partial n_1}{\partial t}+n_0\frac{\partial u_e}{\partial x}=0\\ m_e n_0 \frac{\partial u_e}{\partial t}=-\frac{\partial p_e}{\partial x}-e n_0 E_1\\ \frac{\partial E_1}{\partial x}=\frac{e n_1}{\epsilon_0} \end{array}$

But only the three equations are not enough, we should add a equations of static to make the equations close. Also we assume the state is adiatic $\frac{\partial{T_1}}{\partial x}=0$. and equation of static is $p_e n_e^{\gamma}=constant$ and $p_e=n_e T_e$. From the 2 equations we can get: $\frac{\partial p_e}{\partial x}=\gamma Te\frac{\partial n_1}{\partial x}$
Combine this relation and \refeq{Languir_wave_eq}, we can get: $-m_e\frac{\partial^2 n_1}{\partial t^2}=-\gamma T_e\frac{\partial^2 n_1}{\partial x^2}+\frac{e^2 n_0}{\epsilon_0}n_1$. Further we can get the dispersion relation of Langmuir wave:

$\omega^2=\frac{e^2 n_0}{m_e \epsilon_0}+\frac{\gamma T_e} {m_e}k^2=\omega^2_{pe}+\frac{\gamma T_e}{m_e}k^2$

From which we can see $\omega$ is a function of wave vector k, thus this wave is propagating with: $v_g=\frac{d\omega}{dk}=\sqrt{\frac{\gamma T_e}{m_e}}=v_s > 0$

How could electrostatic perturbation exist alone?

Why for the Langmuir wave, there is only electric perturbation, no magnetic perturbation presents?
To explain this we will start from basic Maxwell equation.

$\nabla\times\vec{E} = -\frac{\partial \vec{B}}{\partial t}\\ \nabla\times\vec{B} = \mu_0\vec{j} + \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$

We make a replacement of the differential operator by applying the perturbation in this form: $~e^{i(\vec{k}\cdot\vec{r}-\omega t)}$, where we have the relation: $\nabla \sim \vec{ik}$, $\frac{\partial}{\partial t} \sim -i\omega$. Thus the above two equation becomes:

$\begin{array}{l} \vec{k}\times\vec{E_1} = -\omega\vec{B_1}\\ \vec{k}\times\vec{B_1} = \mu_0\vec{j_1} - \mu_0\epsilon_0\omega\vec{E_1} \end{array}$

From equation we can see that if $\vec{k}\parallel\vec{E_1}$, then we could have $\vec{B_1} = 0$. Which means even we have zero magnetic perturbation ($\vec{B_1} = 0$), there could be non-zero electrostatic perturbation $\vec{E_1} \neq 0$. But there is one key condition for the non-zero electrostatic perturbation to exist. There should be plasma.

Why do we need plasma here?
You can understand it by looking at the second equation.

$\vec{k}\times\vec{B_1} = \mu_0\vec{j_1} - \mu_0\epsilon_0\omega\vec{E_1}$

If the magnetic perturbation is zero ($\vec{B_1} = 0$), then we have $\vec{k}\times\vec{E_1} = 0$ (Since $\vec{k}\parallel\vec{E_1}$, there must be $\vec{k}\perp\vec{B_1}$. It is obvious that only when $\vec{B_1} = 0$, will the term $\vec{k}\times\vec{B_1}$ be zero). Which lead to the equation: $\mu_0\vec{j_1} - \mu_0\epsilon_0\omega\vec{E_1} = 0$, where means if you wish non-zero electrostatic perturbation $\vec{E_1} \neq 0$, then there should be non-zero real plasma current $\vec{j_1} \neq 0$ to balance it. Otherwise the magnetic perturbation will not be 0. So electrostatic perturbation/wave could only exist in plasma. In the vacuum, where we always have $\vec{j_1} = 0$, only electromagnetic wave could exist.