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# 11: Plasma physics

• • Johan Wevers

## Introduction

The degree of ionization $$\alpha$$ of a plasma is defined by: $$\displaystyle\alpha=\frac{n_{\rm e}}{n_{\rm e}+n_{\rm 0}}$$ where $$n_{\rm e}$$ is the electron density and $$n_{\rm 0}$$ the density of the neutrals. If a plasma also contains negative charged ions $$\alpha$$ is not well defined.

The probability that a test particle collides with another is given by $$dP=n\sigma dx$$ where $$\sigma$$ is the cross section. The collision frequency $$\nu_{\rm c}=1/\tau_{\rm c}=n\sigma v$$. The mean free path is given by $$\lambda_{\rm v}=1/n\sigma$$. The rate coefficient $$K$$ is defined by $$K=\left\langle \sigma v \right\rangle$$. The number of collisions per unit of time and volume between particles of kind 1 and 2 is given by $$n_1n_2\left\langle \sigma v \right\rangle=Kn_1n_2$$.

The potential energy of an electron is given by:

$V(r)=\frac{-e}{4\pi\varepsilon_0r}\exp\left(-\frac{r}{\lambda_{\rm D}}\right) ~~~\mbox{with}~~~ \lambda_{\rm D}=\sqrt{\frac{\varepsilon_0kT_{\rm e}T_{\rm i}}{e^2(n_{\rm e}T_{\rm i}+n_{\rm i}T_{\rm e})}}\approx \sqrt{\frac{\varepsilon_0kT_{\rm e}}{n_{\rm e}e^2}}$

because charge is shielded in a plasma. Here, $$\lambda_{\rm D}$$ is the Debye length. For distances $$<\lambda_{\rm D}$$ the plasma cannot be assumed to be quasi-neutral. Deviations of charge neutrality by thermal motion are compensated by oscillations with frequency

$\omega_{\rm pe}=\sqrt{\frac{n_{\rm e}e^2}{m_{\rm e}\varepsilon_0}}$

The distance of closest approach when two equally charged particles collide for a deviation of $$\pi/2$$ is $$2b_0=e^2/(4\pi\varepsilon_0 \frac{1}{2} mv^2)$$. A “neat” plasma is defined as a plasma for which: $$b_0<n_{\rm e}^{-1/3}\ll\lambda_{\rm D}\ll L_{\rm p}$$ holds. Here $$L_{\rm p}:=|n_{\rm e}/\bigtriangledown n_{\rm e}|$$ is the gradient length of the plasma.

## Transport

Relaxation times are defined as $$\tau=1/\nu_{\rm c}$$. Starting with $$\sigma_{\rm m}=4\pi b_0^2\ln(\Lambda_{\rm C})$$ and with $$\frac{1}{2} mv^2=kT$$ it can be found that:

$\tau_{\rm m}=\frac{4\pi\varepsilon_0^2m^2v^3}{ne^4\ln(\Lambda_{\rm C})}= \frac{8\sqrt{2}\pi\varepsilon_0^2\sqrt{m}(kT)^{3/2}}{ne^4\ln(\Lambda_{\rm C})}$

For momentum transfer between electrons and ions for a Maxwellian velocity distribution:

$\tau_{\rm ee}=\frac{6\pi\sqrt{3}\varepsilon_0^2\sqrt{m_{\rm e}}(kT_{\rm e})^{3/2}}{n_{\rm e}e^4\ln(\Lambda_{\rm C})}\approx\tau_{\rm ei}~~,~~ \tau_{\rm ii}=\frac{6\pi\sqrt{3}\varepsilon_0^2\sqrt{m_{\rm i}}(kT_{\rm i})^{3/2}}{n_{\rm i}e^4\ln(\Lambda_{\rm C})}$

The energy relaxation times for identical particles are equal to the momentum relaxation times. Because for e-i collisions the energy transfer is only $$\sim2m_{\rm e}/m_{\rm i}$$ this is a slow process. Approximately: $$\tau_{\rm ee}:\tau_{\rm ei}:\tau_{\rm ie}:\tau_{\rm ie}^E=1:1:\sqrt{m_{\rm i}/m_{\rm e}}:m_{\rm i}/m_{\rm e}$$.

The relaxation for e-o interaction is much more complicated. For $$T>10$$ eV approximately: $$\sigma_{\rm eo}=10^{-17}v_{\rm e}^{-2/5}$$, for lower energies this can be a factor of ten lower.

The resistivity $$\eta=E/J$$ of a plasma is given by:

$\eta=\frac{n_{\rm e}e^2}{m_{\rm e}\nu_{\rm ei}}=\frac{e^2\sqrt{m_{\rm e}}\ln(\Lambda_{\rm C})}{6\pi\sqrt{3}\varepsilon_0^2(kT_{\rm e})^{3/2}}$

The diffusion coefficient $$D$$ is defined by means of the flux $$\Gamma$$ by $$\vec{\Gamma}=n\vec{v}_{\rm diff}=-D\nabla n$$. The equation of continuity is $$\partial_tn+\nabla(nv_{\rm diff})=0\Rightarrow\partial_tn=D\nabla^2n$$. One finds that $$D=\frac{1}{3}\lambda_{\rm v}v$$. A rough estimate gives $$\tau_{\rm D}=L_{\rm p}/D=L_{\rm p}^2\tau_{\rm c}/\lambda_{\rm v}^2$$. For magnetized plasmas $$\lambda_{\rm v}$$ must be replaced with the cyclotron radius. In electrical fields also $$\vec{J}=ne\mu\vec{E}=e(n_{\rm e}\mu_{\rm e}+n_{\rm i}\mu_{\rm i})\vec{E}$$ with $$\mu=e/m\nu_{\rm c}$$ the mobility of the particles. The Einstein ratio is:

$\frac{D}{\mu}=\frac{kT}{e}$

Because a plasma is electrically neutral electrons and ions are strongly coupled and they don’t diffuse independently. The coefficient of ambipolar diffusion $$D_{\rm amb}$$ is defined by $$\vec{\Gamma}=\vec{\Gamma}_{\rm i}=\vec{\Gamma}_{\rm e}=-D_{\rm amb}\nabla n_{\rm e,i}$$. From this it follows that

$D_{\rm amb}=\frac{kT_{\rm e}/e-kT_{\rm i}/e}{1/\mu_{\rm e}-1/\mu_{\rm i}}\approx\frac{kT_{\rm e}\mu_{\rm i}}{e}$

In an external magnetic field $$B_0$$ particles will move in spiral orbits with cyclotron radius $$\rho=mv/eB_0$$ and with cyclotron frequency $$\Omega=B_0e/m$$. The helical orbit is perturbed by collisions. A plasma is called magnetized if $$\lambda_{\rm v}>\rho_{\rm e,i}$$. Therefore electrons are magnetized if

$\frac{\rho_{\rm e}}{\lambda_{\rm ee}}=\frac{\sqrt{m_{\rm e}}e^3n_{\rm e}\ln(\Lambda_{\rm C})}{6\pi\sqrt{3}\varepsilon_0^2(kT_{\rm e})^{3/2}B_0}<1$

Magnetization of only the electrons is sufficient to reasonably confine the plasma because they are coupled to the ions by charge neutrality. In case of magnetic confinement: $$\nabla p=\vec{J}\times\vec{B}$$. Combined with the two stationary Maxwell equations for the $$B$$-field these form the ideal magneto-hydrodynamic equations. For a uniform $$B$$-field: $$p=nkT=B^2/2\mu_0$$.

If both magnetic and electric fields are present electrons and ions will move in the same direction. If $$\vec{E}=E_r\vec{e}_r+E_z\vec{e}_z$$ and $$\vec{B}=B_z\vec{e}_z$$ the $$\vec{E}\times\vec{B}$$ drift results in a velocity $$\vec{u}=(\vec{E}\times\vec{B}\,)/B^2$$ and the velocity in the $$r,\varphi$$ plane is $$\dot{r}(r,\varphi,t)=\vec{u}+\dot{\vec{\rho}}(t)$$.

## Elastic collisions

### General

The scattering angle of a particle interacting with another particle, as shown in the figure below is:

$\chi=\pi-2b\int\limits_{r_a}^\infty \frac{dr}{r^2\sqrt1-\frac{b^2}{r^2}-\frac{W(r)}{E_0}}$

Particles with an impact parameter between $$b$$ and $$b+db$$, moving through a ring with $$d\sigma=2\pi bdb$$ leave the scattering area at a solid angle $$d\Omega=2\pi\sin(\chi)d\chi$$. The differential cross section is then defined as:

$I(\Omega)=\left|\frac{d\sigma}{d\Omega}\right|=\frac{b}{\sin(\chi)}\frac{\partial b}{\partial \chi}$ Figure $$\PageIndex{1}$$

For a potential energy $$W(r)=kr^{-n}$$ it follows that: $$I(\Omega,v)\sim v^{-4/n}$$.

For low energies, $$\cal O$$(1 eV), $$\sigma$$ has a Ramsauer minimum. It arises from the interference of matter waves behind the object. $$I(\Omega)$$ for angles $$0<\chi<\lambda/4$$ is larger than the classical value.

### The Coulomb interaction

For the Coulomb interaction: $$2b_0=q_1q_2/2\pi\varepsilon_0mv_0^2$$, so $$W(r)=2b_0/r$$. This gives $$b=b_0\cot( \frac{1}{2} \chi)$$ and

$I(\Omega=\frac{b}{\sin(\chi)}\frac{\partial b}{\partial \chi}=\frac{b_0^2}{4\sin^2( \frac{1}{2} \chi)}$

Because the influence of a particle vanishes at $$r=\lambda_{\rm D}$$ then: $$\sigma=\pi(\lambda_{\rm D}^2-b_0^2)$$. Because $$dp=d(mv)=mv_0(1-\cos\chi)$$ the cross section for momentum transfer $$\sigma_{\rm m}$$ is given by:

$\sigma_{\rm m}=\int(1-\cos\chi)I(\Omega)d\Omega=4\pi b_0^2\ln\left(\frac{1}{\sin( \frac{1}{2} \chi_{\rm min})}\right)= 4\pi b_0^2\ln\left(\frac{\lambda_{\rm D}}{b_0}\right):=4\pi b_0^2\ln(\Lambda_{\rm C}) \sim \frac{\ln(v^4)}{v^4}$

where $$\ln(\Lambda_{\rm C})$$ is the Coulomb-logarithm. For this quantity: $$\Lambda_{\rm C}=\lambda_{\rm D}/b_0=9n(\lambda_{\rm D})$$.

### The induced dipole interaction

The induced dipole interaction, with $$\vec{p}=\alpha\vec{E}$$, gives a potential $$V$$ and an energy $$W$$ in a dipole field:

$V(r)=\frac{\vec{p}\cdot\vec{e}_r}{4\pi\varepsilon_0r^2}~~,~~ W(r)=-\frac{|e|p}{8\pi\varepsilon_0 r^2}=-\frac{\alpha e^2}{2(4\pi\varepsilon_0)^2r^4}$

with $$\displaystyle b_a=\sqrt{\frac{2e^2\alpha}{(4\pi\varepsilon_0)^2 \frac{1}{2} mv_0^2}}$$ and therefore: $$\displaystyle\chi=\pi-2b\int\limits_{r_a}^\infty \frac{dr}{r^2\sqrt1-\frac{b^2}{r^2}+\frac{b_a^4}{4r^4}}$$

If $$b\geq b_a$$ the charge would hit the atom. Repulsive nuclear forces prevent this from happening. If the scattering angle is a large compared to  $$2\pi$$ it is called capture. The cross section for capture $$\sigma_{\rm orb}=\pi b_a^2$$ is called the Langevin limit, and is a lower limit estimate for the total cross section.

### The centre of mass system

If collisions of two particles with masses $$m_1$$ and $$m_2$$ which scatter in the centre of mass system by an angle $$\chi$$ result in scattering at an angle $$\theta$$ in the laboratory system:

$\tan(\theta)=\frac{m_2\sin(\chi)}{m_1+m_2\cos(\chi)}$

The energy loss $$\Delta E$$ of the incoming particle is given by:

$\frac{\Delta E}{E}=\frac{ \frac{1}{2} m_2v_2^2}{\mbox{\frac{1}{2}}m_1v_1^2}=\frac{2m_1m_2}{(m_1+m_2)^2}(1-\cos(\chi))$

### Scattering of light

Scattering of light by free electrons is called Thomson scattering. The scattering is free from collective effects if $$k\lambda_{\rm D}\ll1$$. The cross section $$\sigma=6.65\cdot10^{-29}$$m$$^2$$ and

$\frac{\Delta f}{f}=\frac{2v}{c}\sin( \frac{1}{2} \chi)$

This gives for the scattered energy $$E_{\rm scat}\sim n\lambda_0^4/(\lambda^2-\lambda_0^2)^2$$ with $$n$$ the density. If $$\lambda\gg\lambda_0$$ it is called Rayleigh scattering. Thomson scattering is a limit of Compton scattering, which is given by $$\lambda'-\lambda=\lambda_{\rm C}(1-\cos\chi)$$ with $$\lambda_{\rm C}=h/mc$$ and cannot be used if relativistic effects become important.

## Thermodynamic equilibrium and reversibility

Planck’s radiation law and the Maxwell velocity distribution hold for a plasma in equilibrium:

$\rho(\nu,T)d\nu=\frac{8\pi h\nu^3}{c^3}\frac{1}{\exp(h\nu/kT)-1}d\nu~~,~~ N(E,T)dE=\frac{2\pi n}{(\pi kT)^{3/2}}\sqrt{E}\exp\left(-\frac{E}{kT}\right)dE$

Detailed balancing” means that the number of reactions in one direction equals the number of reactions in the opposite direction because both processes have equal probability if one corrects for the volume of the phase space available. For the reaction

$\sum_{\rm forward}{\rm X}_{\rm forward}~\lower.2ex\hbox{\leftarrow}\kern-2.3ex\raise.7ex\hbox{\rightarrow}~\sum_{\rm back}{\rm X}_{\rm back}$

Given microscopic reversibility in a plasma at equilibrium :

$\prod_{\rm forward}\hat{\eta}_{\rm forward}=\prod_{\rm back}\hat{\eta}_{\rm back}$

If the velocity distribution is Maxwellian, this gives:

$\hat{\eta}_{x}=\frac{n_x}{g_x}\frac{h^3}{(2\pi m_xkT)^{3/2}}{\rm e}^{-E_{\rm kin}/kT}$

where $$g$$ is the statistical weight of the state and $$n/g:=\eta$$. For ground state electrons $$g=2$$, for excited states usually $$g=2j+1=2n^2$$.

With this one finds for the Boltzmann balance, $${\rm X}_p+{\rm e}^-~\rightleftharpoons ~{\rm X}_1+{\rm e}^-+(E_{1p})$$

$\frac{n_p^{\rm B}}{n_1}=\frac{g_p}{g_1}\exp\left(\frac{E_p-E_1}{kT_{\rm e}}\right)$

And for the Saha balance, $${\rm X}_p+{\rm e}^-+(E_{pi})~\rightleftharpoons ~{\rm X}_1^++2{\rm e}^-$$:

$\frac{n_p^{\rm S}}{g_p}=\frac{n_1^+}{g_1^+}\frac{n_{\rm e}}{g_{\rm e}} \frac{h^3}{(2\pi m_{\rm e}kT_{\rm e})^{3/2}}\exp\left(\frac{E_{pi}}{kT_{\rm e}}\right)$

Because the number of particles on the left-hand side and right-hand side of the equation is different, a factor $$g/V_{\rm e}$$ remains. This factor causes the Saha-jump.

From microscopic reversibility one can derive that for the rate coefficients $$K(p,q,T):=\left\langle \sigma v \right\rangle_{pq}$$ and:

$K(q,p,T)=\frac{g_p}{g_q}K(p,q,T)\exp\left(\frac{\Delta E_{pq}}{kT}\right)$

## Inelastic collisions

### Types of collisions

The kinetic energy of a system can be split into the motion of the centre of mass and motion relative to the centre of mass. The energy relative to the centre of mass system is available for reactions. This energy is given by

$E=\frac{m_1m_2(v_1-v_2)^2}{2(m_1+m_2)}$

Some types of inelastic collisions important for plasma physics are:

1. Excitation:  $${\rm A}_p+{\rm e}^- \rightleftharpoons {\rm A}_q+{\rm e}^-$$
2. Decay: $${\rm A}_q\rightleftharpoons {\rm A}_p+hf$$
3. Ionisation and 3-particles recombination: $${\rm A}_p+{\rm e}^-\rightleftharpoons {\rm A}^++2{\rm e}^-$$
4. Radiative recombination: $${\rm A}^++{\rm e}^-\rightleftharpoons {\rm A}_p+hf$$
5. Stimulated emission: $${\rm A}_q+hf\rightarrow{\rm A}_p+2hf$$
6. Associative ionisation: $$\rm A^{**}+B\rightleftharpoons ~AB^++e^-$$
7. Penning ionisation: b.v. $$\rm Ne^*+Ar~\rightleftharpoons ~Ar^++Ne+e^-$$
8. Charge transfer: $$\rm A^++B~\rightleftharpoons ~A+ B^+$$
9. Resonant charge transfer: $$\rm A^++A~\rightleftharpoons ~A+A^+$$

### Cross sections

Collisions between an electron and an atom can be approximated by a collision between an electron and one of the electrons of that atom. This results in

$\frac{d\sigma}{d(\Delta E)}=\frac{\pi Z^2 e^4}{(4\pi\varepsilon_0)^2E(\Delta E)^2}$

Then it follows for the transition $$p\rightarrow q$$: $$\displaystyle \sigma_{pq}(E)=\frac{\pi Z^2e^4\Delta E_{q,q+1}}{(4\pi\varepsilon_0)^2E(\Delta E)_{pq}^2}$$

For ionization from state $$p$$ to a good approximation it holds that: $$\displaystyle \sigma_p=4\pi a_0^2 Ry\left(\frac{1}{E_p}-\frac{1}{E}\right)\ln\left(\frac{1.25\beta E}{E_p}\right)$$

For resonant charge transfer: $$\displaystyle\sigma_{\rm ex}=\frac{A[1-B\ln(E)]^2}{1+CE^{3.3}}$$

$\underbrace{n_pA_{pq}}_{\rm emission}+ \underbrace{n_pB_{pq}\rho(\nu,T)}_{\rm stimulated~emission}= \underbrace{n_qB_{qp}\rho(\nu,T)}_{\rm absorption}$

Here, $$A_{pq}$$ is the matrix element of the transition $$p\rightarrow q$$, and is given by:

$A_{pq}=\frac{8\pi^2e^2\nu^3|r_{pq}|^2}{3\hbar\varepsilon_0c^3}~~\mbox{with}~~ r_{pq}=\langle\psi_p|\vec{r}\,|\psi_q\rangle$

For hydrogenic atoms: $$A_p=1.58\cdot10^8Z^4p^{-4.5}$$, with $$A_p=1/\tau_p=\sum\limits_qA_{pq}$$. The intensity $$I$$ of a line is given by $$I_{pq}=hfA_{pq}n_p/4\pi$$. The Einstein coefficients $$B$$ are given by:

$B_{pq}=\frac{c^3A_{pq}}{8\pi h\nu^3}~~\mbox{and}~~\frac{B_{pq}}{B_{qp}}=\frac{g_q}{g_p}$

A spectral line is broadened by several mechanisms:

1. Because the states have a finite life time. The natural life time of a state $$p$$ is given by $$\tau_p=1/\sum\limits_q A_{pq}$$. From the uncertainty relation then follows: $$\Delta(h\nu)\cdot\tau_p=\frac{1}{2} \hbar$$, this gives
$\Delta\nu=\frac{1}{4\pi\tau_p}=\frac{\sum\limits_qA_{pq}}{4\pi}$
The natural line width is usually $$\ll$$ than the broadening due to the following two mechanisms:
2. Doppler broadening is caused by the thermal motion of the particles:
$\frac{\Delta\lambda}{\lambda}=\frac{2}{c}\sqrt{\frac{2\ln(2)kT_{\rm i}}{m_{\rm i}}}$
This broadening results in a Gaussian line profile: $$k_\nu=k_0\exp(-[2\sqrt{\ln 2}(\nu-\nu_0)/\Delta\nu_D]^2)$$, with $$k$$ the coefficient of absorption or emission.
3. Stark broadening is caused by the electric field of the electrons:
$\Delta\lambda_{1/2}=\left[\frac{n_{\rm e}}{C(n_{\rm e},T_{\rm e})}\right]^{2/3}$
for the H-$$\beta$$ line: $$C(n_{\rm e},T_{\rm e})\approx3\cdot10^{14}$$Å$$^{-3/2}$$cm$$^{-3}$$.

The natural broadening and the Stark broadening result in a Lorentz profile of a spectral line: $$k_\nu=\frac{1}{2} k_0\Delta\nu_L/ [( \frac{1}{2} \Delta\nu_L)^2+(\nu-\nu_0)^2]$$. The total line shape is a convolution of the Gauss- and Lorentz profile and is called a Voigt profile.

The number of transitions $$p\rightarrow q$$ is given by $$n_pB_{pq}\rho$$ and by $$n_pn_{hf}\left\langle \sigma_{\rm a} c \right\rangle=n_p(\rho d\nu/h\nu)\sigma_{\rm a}c$$ where $$d\nu$$ is the line width. Then the cross section of absorption processes follows: $$\sigma_{\rm a}=B_{pq}h\nu/cd\nu$$.

The background radiation in a plasma originates from two processes:

1. Free-Bound radiation, originating from radiative recombination. The emission is given by:
$\varepsilon_{fb}=\frac{C_1}{\lambda^2}\frac{z_{\rm i}n_{\rm i}n_{\rm e}}{\sqrt{kT_{\rm e}}}\left[1-\exp\left(-\frac{hc}{\lambda kT_{\rm e}}\right)\right]\xi_{fb}(\lambda,T_{\rm e})$
with $$C_1=1.63\cdot10^{-43}$$ Wm$$^4$$K$$^{1/2}$$sr$$^{-1}$$ and $$\xi$$, the Biberman factor.
2. Free-free radiation, originating from the acceleration of particles in the EM-field of other particles:
$\varepsilon_{ff}=\frac{C_1}{\lambda^2}\frac{z_{\rm i}n_{\rm i}n_{\rm e}}{\sqrt{kT_{\rm e}}}\exp\left(-\frac{hc}{\lambda kT_{\rm e}}\right)\xi_{ff}(\lambda,T_{\rm e})$

## The Boltzmann transport equation

It is assumed that there exists a distribution function $$F$$ for the plasma so that

$F(\vec{r},\vec{v},t)=F_r(\vec{r},t)\cdot F_v(\vec{v},t)=F_1(x,t)F_2(y,t)F_3(z,t)F_4(v_x,t)F_5(v_y,t)F_6(v_z,t)$

Then the BTE is: $$\displaystyle \frac{dF}{dt}=\frac{\partial F}{\partial t}+\nabla_r\cdot(F\vec{v}\,)+\nabla_v\cdot(F\vec{a}\,)=\left(\frac{\partial F}{\partial t}\right)_{\rm coll-rad}$$

Assuming that $$v$$ does not depend on $$r$$ and $$a_i$$ does not depend on $$v_i$$, then $$\nabla_r\cdot(F\vec{v}\,)=\vec{v}\cdot\nabla F$$ and $$\nabla_v\cdot(F\vec{a}\,)=\vec{a}\cdot\nabla_vF$$ holds. This is also true in magnetic fields because $$\partial a_i/\partial x_i=0$$. The velocity is separated into a thermal velocity $$\vec{v}_{\rm t}$$ and a drift velocity $$\vec{w}$$. The total density is given by $$n=\int Fd\vec{v}$$ and $$\int\vec{v}Fd\vec{v}=n\vec{w}$$.

The balance equations can be derived by means of the moment method:

1. Mass balance: $$\displaystyle\int({\rm BTE})d\vec{v}\Rightarrow\frac{\partial n}{\partial t}+\nabla\cdot(n\vec{w})=\left(\frac{\partial n}{\partial t}\right)_{\rm cr}$$
2. Momentum balance: $$\displaystyle\int({\rm BTE})m\vec{v}d\vec{v}\Rightarrow mn\frac{d\vec{w}}{dt}+\nabla\mbox{\sfd T}'+\nabla p=mn\left\langle \vec{a}~ \right\rangle+\vec{R}$$
3. Energy balance: $$\displaystyle\int({\rm BTE})mv^2d\vec{v}\Rightarrow\frac{3}{2}\frac{dp}{dt}+\frac{5}{2}p\nabla\cdot\vec{w}+\nabla\cdot\vec{q}=Q$$

Here, $$\left\langle \vec{a}~ \right\rangle=e/m(\vec{E}+\vec{w}\times\vec{B}\,)$$ is the average acceleration, $$\vec{q}=\frac{1}{2} nm\left\langle \vec{v}_{\rm t}^{~2}\vec{v}_{\rm t} \right\rangle$$ the heat flow, $$\displaystyle Q=\int\frac{mv_{\rm t}^2}{r}\left(\frac{\partial F}{\partial t}\right)_{\rm cr}d\vec{v}$$ the source term for energy production, $$\vec{R}$$ is a friction term and $$p=nkT$$ the pressure.

A thermodynamic derivation gives for the total pressure: $$\displaystyle p=nkT=\sum_ip_i-\frac{e^2(n_{\rm e}+z_{\rm i}n_{\rm i})}{24\pi\varepsilon_0\lambda_{\rm D}}$$

The electrical conductance in a plasma follows from the momentum balance, if $$w_{\rm e}\gg w_{\rm i}$$:

$\eta\vec{J}=\vec{E}-\frac{\vec{J}\times\vec{B}+\nabla p_{\rm e}}{en_{\rm e}}$

In a plasma where only elastic e-a collisions are important the equilibrium energy distribution function is the Druyvesteyn distribution:

$N(E)dE=Cn_{\rm e}\left(\frac{E}{E_0}\right)^{3/2}\exp\left[-\frac{3m_{\rm e}}{m_0}\left(\frac{E}{E_0}\right)^2\right]dE$

with $$E_0=eE\lambda_{\rm v}=eE/n\sigma$$.

These models are first-moment equations for excited states. One assumes the Quasi-steady-state solution is valid, where $$\forall_{p>1}[(\partial n_p/\partial t=0)\wedge(\nabla\cdot(n_p\vec{w}_p)=0)]$$. This results in:

$\left(\frac{\partial n_{p>1}}{\partial t}\right)_{\rm cr}=0~~,~~ \frac{\partial n_1}{\partial t}+\nabla\cdot(n_1\vec{w}_1)=\left(\frac{\partial n_1}{\partial t}\right)_{\rm cr}~~,~~ \frac{\partial n_{\rm i}}{\partial t}+\nabla\cdot(n_{\rm i}\vec{w}_{\rm i})=\left(\frac{\partial n_{\rm i}}{\partial t}\right)_{\rm cr}$

with solutions $$n_p=r_p^0n_p^{\rm S}+r_p^1n_p^{\rm B}=b_pn_p^{\rm S}$$. Further for all collision-dominated levels with $$\delta b_p:=b_p-1=b_0p_{\rm eff}^{-x}$$ with $$p_{\rm eff}=\sqrt{Ry/E_{p{\rm i}}}$$ and $$5\leq x\leq6$$. For systems in ESP, where only collisional (de)excitation between levels $$p$$ and $$p\pm1$$ is taken into account  $$x=6$$. Even in plasma’s far from equilibrium the excited levels will eventually reach ESP, so from a certain level up the level densities can be calculated.

To find the population densities of the lower levels in the stationary case one has to start with a macroscopic equilibrium:

$\mbox{Number of populating processes of level}~p~=~ \mbox{Number of depopulating processes of level}~p~,$

When this is expanded it becomes:

$\underbrace{n_{\rm e}\sum_{q<p}n_qK_{qp}}_{\rm coll.~excit.}+ \underbrace{n_{\rm e}\sum_{q>p}n_qK_{qp}}_{\rm coll.~deexcit.}+ \underbrace{\sum_{q>p}n_qA_{qp}}_{\rm rad.~deex.~to}+ \underbrace{n_{\rm e}^2n_{\rm i}K_{+p}}_{\rm coll.~recomb.}+ \underbrace{n_{\rm e}n_{\rm i}\alpha_{\rm rad}}_{\rm rad.~recomb}=$

$\underbrace{n_{\rm e}n_p\sum_{q<p}K_{pq}}_{\rm coll.~deexcit.}+ \underbrace{n_{\rm e}n_p\sum_{q>p}K_{pq}}_{\rm coll.~excit.}+ \underbrace{n_p\sum_{q<p}A_{pq}}_{\rm rad.~deex.~from}+ \underbrace{n_{\rm e}n_pK_{p+}}_{\rm coll.~ion.}$

## Waves in plasma’s

Interaction of electromagnetic waves in plasma’s results in scattering and absorption of energy. For electromagnetic waves with complex wave number $$k=\omega(n+i\kappa)/c$$ in one dimension one finds: $$E_x=E_0{\rm e}^{-\kappa\omega x/c}\cos[\omega(t-nx/c)]$$. The refractive index $$n$$ is given by:

$n=c\frac{k}{\omega}=\frac{c}{v_{\rm f}}=\sqrt{1-\frac{\omega_{\rm p}^2}{\omega^2}}$

For disturbances in the $$z$$-direction in a cold, homogeneous, magnetized plasma: $$\vec{B}=B_0\vec{e}_z+\vec{\hat{B}}{\rm e}^{i(kz-\omega t)}$$ and $$n=n_0+\hat{n}{\rm e}^{i(kz-\omega t)}$$ (external $$E$$ fields are screened) it follows, with the definitions $$\alpha=\omega_{\rm p}/\omega$$ and $$\beta=\Omega/\omega$$ and $$\omega_{\rm p}^2=\omega_{\rm pi}^2+\omega_{\rm pe}^2$$:

$\vec{J}=\vec{\vec{\sigma}}\vec{E}~~,\mbox{with}~~ \vec{\vec{\sigma}}=i\varepsilon_0\omega\sum_s\alpha_s^2 \left(\begin{array}{ccc} \displaystyle\frac{1}{1-\beta_s^2}&\displaystyle\frac{-i\beta_s}{1-\beta_s^2}&0\\ \displaystyle\frac{i\beta_s}{1-\beta_s^2}&\displaystyle\frac{1}{1-\beta_s^2}&0\\ 0&0&1 \end{array}\right)$

where the sum is taken over particle species $$s$$. The dielectric tensor $$\cal E$$, with property:

$\vec{k}\cdot \left( \vec{\vec{\cal E}}\cdot \vec{\textrm{E}} \right )=0$

is given by $$\vec{\vec{\cal E}} =\vec{\vec{I}}-\vec{\vec{\sigma}}/i \cal E_0 \omega$$

With the definitions $$\displaystyle S=1-\sum_s\frac{\alpha_s^2}{1-\beta_s^2}~~,~~ D=\sum_s\frac{\alpha_s^2\beta_s}{1-\beta_s^2}~~,~~ P=1-\sum_s\alpha_s^2$$

it follows that:

$\vec{\vec{\cal E}}=\left(\begin{array}{ccc}S&- iD&0\\iD&S&0\\0&0&P\end{array}\right)$

The eigenvalues of this Hermitian matrix are $$R=S+D$$, $$L=S-D$$, $$\lambda_3=P$$, with eigenvectors $$\vec{e}_{\rm r}=\frac{1}{2} \sqrt{2}(1,i,0)$$, $$\vec{e}_{\rm l}=\frac{1}{2} \sqrt{2}(1,-i,0)$$ and $$\vec{e}_{\rm 3}=(0,0,1)$$. $$\vec{e}_{\rm r}$$ is connected with a right rotating field for which $$iE_x/E_y=1$$ and $$\vec{e}_{\rm l}$$ is connected with a left rotating field for which $$iE_x/E_y=-1$$. When $$k$$ makes an angle $$\theta$$ with $$\vec{B}$$ one finds:

$\tan^2(\theta)=\frac{P(n^2-R)(n^2-L)}{S(n^2-RL/S)(n^2-P)}$

where $$n$$ is the refractive index. From this the following solutions can be obtained:

#### A. (\theta = 0 \): Transmission in the z-direction

1. $$P=0$$: $$E_x=E_y=0$$. This describes a longitudinal linearly polarized wave.
2. $$n^2=L$$: a left, circularly polarized wave.
3. $$n^2=R$$: a right, circularly polarized wave.

#### B. $$\theta=\pi/2$$: transmission $$\perp$$ the $$B$$-field.

1. $$n^2=P$$: the ordinary mode: $$E_x=E_y=0$$. This is a transverse linearly polarized wave.
2. $$n^2=RL/S$$: the extraordinary mode: $$iE_x/E_y=-D/S$$, an elliptically polarized wave.

Resonance frequencies are frequencies for which $$n^2\rightarrow\infty$$, so $$v_{\rm f}=0$$. For these: $$\tan(\theta)=-P/S$$. For $$R\rightarrow\infty$$ this gives the electron cyclotron resonance frequency $$\omega=\Omega_{\rm e}$$, for $$L\rightarrow\infty$$ the ion cyclotron resonance frequency $$\omega=\Omega_{\rm i}$$ and for $$S=0$$ for the extraordinary mode:

$\alpha^2\left(1-\frac{m_{\rm i}}{m_{\rm e}}\frac{\Omega_{\rm i}^2}{\omega^2}\right)= \left(1-\frac{m_{\rm i}^2}{m_{\rm e}^2}\frac{\Omega_{\rm i}^2}{\omega^2}\right) \left(1-\frac{\Omega_{\rm i}^2}{\omega^2}\right)$

Cut-off frequencies are frequencies for which $$n^2=0$$, so $$v_{\rm f}\rightarrow\infty$$. For these: $$P=0$$ or $$R=0$$ or $$L=0$$.

In the case that $$\beta^2\gg1$$ one finds Alfvén waves propagating parallel to the field lines. With the Alfvén velocity

$v_{\rm A}=\frac{\Omega_{\rm e}\Omega_{\rm i}}{\omega_{\rm pe}^2+\omega_{\rm pi}^2}c^2$

follows: $$n=\sqrt{1+c/v_{\rm A}}$$, and in case $$v_{\rm A}\ll c$$ then: $$\omega=kv_{\rm A}$$.

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