Edu

Plasma - is a state of matter, when it consists of mobile charged particles: ions and electrons. Its unique properties are realted with that fact that it's a gas of mobile chrages.

The mass of the electron $m=9.10 \cdot 10^{-31}~\text{kg}$, the elementary charge $e=1.60 \cdot 10^{-19}~\text{C}$.

Electrons are are much lighter than ions, so the electronic contribution is the main one in the most of processes. Let $n_0$ be the concentration of the ions ($m^{-3}$ units) and $n$ be the concentration of the electrons. Let's consider that all ions are similar and that their charge is $+e$.

A1 If we assume that the velocity of the electrons at a certain point is $\vec{v}$, what is the current density $\vec{j}_t$ at that point?

The macroscopic electrodynamic properties of the matter are described by its relative permittivity $\varepsilon$ and conductivity $\sigma$. More precisely, if there is an electric field $\vec{E}$ in this matter there is also the polarization density $\vec{P}$ and the conduction current $\vec{j}$:
$\vec{P} = \varepsilon_0 (\varepsilon-1) \vec{E}, \quad \vec{j} = \sigma \vec{E}.$

A2 How to express the total current $\vec{j}_t$ from the conduction current $\vec{j}$ and the time derivative of the polarization density $\partial \vec{P}/\partial t$.

We can redefine the relative permittivity $\varepsilon$ in such a way that the polarization density also compounds the conduction current and $\vec{j}_t = \partial \vec{P}/\partial t$. Note, that the relative permittivity becomes a complex number.

A3 Show that for the harmonic external field $\tilde{E} e^{-i\omega t}$ with the angular frequency $\omega$, the redefinition of the $\varepsilon$ is
$\tilde{\varepsilon} = \varepsilon + \frac{i \sigma}{\omega \varepsilon_0}.$

We can find the relative permittivity of a plasma if we can describe the motion of the electrons in it. The electrons interact with the external field $\tilde{E} e^{-i\omega t}$ and the electric field of the other particles. Also, a plasma is electroneutral at fairly large distances and fields of individual particles is shielded.

Particulary, when the plasma is in equilibrium without the external field the average charge density is zero. According to Maxwell's equation $\operatorname{div} \vec{E}=\rho/\varepsilon_0$ the electric field is zero everywhere. However, every single electron has it's own electric field $\vec{E} = \pm e\vec{r}/\left( 4 \pi \varepsilon_0 r^3 \right)$, so only the total electric field is zero by averaging over finite volume with a characteristic radius $\lambda_D$, called the Debye length. In the other words, at the distances of the order of $\lambda_D$, the electric field of the particular charge is shielded by the others.

Since the magnetic field generated by the system is always negligible in the presence of the electric interaction we can choose the potential $\Phi(\vec{r})$ of the electric field. Thus, the substitution $\vec{E} = - \operatorname{grad} \Phi$ in Maxwell's equation gives us
$-\Delta \Phi = \frac{e}{\varepsilon_0} \left(n_0 - n \right).$
Also, the value of the $\Phi(\vec{r})$ is proportional to the potential energy at that particular point and a thermodynamic equilibrium of particles is described by the Boltzmann distribution
$n(\vec{r}) = \langle n \rangle \exp \left( \frac{e \Phi}{k_\text{B}T} \right), \quad n_0(\vec{r}) = \langle n_0 \rangle \exp\left( -\frac{e \Phi}{k_\text{B}T} \right),$
where $\langle n \rangle = \langle n_0 \rangle$ are the average concentrations of electrons and ions in the plasma, so the constant in the potential is chosen so that if $V \langle n \rangle = \int n(\vec{r}) \, dV$.

Finally, we have a nonlinear differential equation for $\Phi$:
$\Delta \Phi = \frac{e \langle n \rangle}{\varepsilon_0} \left[ \exp \left( \frac{e \Phi}{k_\text{B}T} \right) - \exp \left(-\frac{e \Phi}{k_\text{B}T} \right) \right] \tag{1}$

Let's consider how the electric field of the charge $q$ behaves on the distances where the thermal energy is much larger than the potential electric energy, i.e. $e \Phi \ll k_\text{B} T$. In this approach the equation (1) becomes linear and has a spherically symmetric solution corresponding to the shielded electric field of charge $q$:
$\Phi(r) = \frac{q}{4\pi \varepsilon_0 r} e^{-r/\lambda_D}.$

A4 Using that for a spherically symmetric function, the Lapplacian could be written as
$\Delta \Phi = \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d \Phi}{dr} \right),$
calculate the $\lambda_D$. Express your answer in terms of $\langle n \rangle$ and $T$. Estimate the value of $\lambda_D$ in the gas discharge, where$\langle n \rangle = 10^{16}~\text{м}^{-3}$, $T=10^4~\text{K}$.

A5 Estimate the average number of particles $N_\text{loc}$ whose field is not shielded at the given point in the plasma.

If $N_\text{loc} \gg1$, the electric field at each point is the sum of numerous fields that cancel each other out, so the only field that interacting with the particular electron is the external field.

A6 Analyze the motion of the electron in the plasma in the external field $\tilde{E}e^{-i\omega t}$ and find the relative permittivity $\tilde{\varepsilon}$ of the plasma.

The plasma has the frequency of self-oscillations: the oscillations that could exist without a periodic external field. Such a oscillations are called the Langmuir waves.

A7 Find the circular frequency $\omega_p$ of the plasma self-oscillations.