Kinetic Description of a Plasma

Abstract

This chapter is devoted to the Boltzmann transport equation. The Boltzmann equation has been initially established for a dilute gas of neutral molecules and it provides the standard kinetic approach to describe the microscopic evolution of a gas to equilibrium. The Boltzmann equation can be applied to the description of a medium in which the dominant interactions are of short-range type so that it is suitable to describe the evolution of a plasma determined by electron-molecule and ion-molecule collisions. The Boltzmann equation fails however when applied to long-range Coulomb interactions. The collisional terms for both elastic and inelastic electron-molecule collisions are then consistently derived. The chapter ends with the establishment of the equations for the moments of the Boltzmann equation, i.e. the fluid equations, for electron-neutral interactions.

Appendices

Appendices

1.1 A.2.1 Liouville Relation dq dp = dq ^′ dp ^′

Let us consider a transformation to pass from the variables q(t) and p(t), describing a system in a given instant t, to the variables $q^{{\prime}}(t + dt)$ and $p^{{\prime}}(t + dt)$, describing the same system at the instant t + dt. The evolutions in time of q and p are given by Hamilton’s equations

$$\displaystyle{ \dot{q}\; =\; \frac{\partial H} {\partial p} \;\;\;\;\mbox{ and}\;\;\;\;\dot{p}\; =\; -\;\frac{\partial H} {\partial q}, }$$

(2.180)

so that we may write

$$\displaystyle\begin{array}{rcl} & & q^{{\prime}}\; =\; q\; +\;\dot{ q}\;\;dt\; =\; q\; +\; \frac{\partial H} {\partial p} \;\;dt{}\end{array}$$

(2.181)

$$\displaystyle\begin{array}{rcl} & & p^{{\prime}}\; =\; p\; +\;\dot{ p}\;\;dt\; =\; p\; -\;\frac{\partial H} {\partial q} \;\;dt.{}\end{array}$$

(2.182)

Differentiating $q^{{\prime}}$ and $p^{{\prime}}$ in terms of q and p we obtain

$$\displaystyle\begin{array}{rcl} & & dq^{{\prime}}\; =\; dq\; +\; \frac{\partial } {\partial q}\left (\frac{\partial H} {\partial p} \;\;dt\right )dq\; +\; \frac{\partial } {\partial p}\left (\frac{\partial H} {\partial p} \;\;dt\right )dp{}\end{array}$$

(2.183)

$$\displaystyle\begin{array}{rcl} & & dp^{{\prime}}\; =\; dp\; +\; \frac{\partial } {\partial q}\left (-\;\frac{\partial H} {\partial q} \;\;dt\right )dq\; +\; \frac{\partial } {\partial p}\left (-\;\frac{\partial H} {\partial q} \;\;dt\right )dp,{}\end{array}$$

(2.184)

so that we may write in accordance with Jacobi’s theorem

$$\displaystyle{ dq^{{\prime}}dp^{{\prime}}\; =\; J\;dq\;dp, }$$

(2.185)

where J is the determinant of the transformation

$$\displaystyle{ J\; = \left \vert \begin{array}{cc} 1 + \frac{\partial ^{2}H} {\partial q\partial p}\;dt& \frac{\partial ^{2}H} {\partial p^{2}} \;dt\\ & \\ -\;\frac{\partial ^{2}H} {\partial q^{2}} \;dt &1 - \frac{\partial ^{2}H} {\partial p\partial q}\;dt\\ \end{array} \right \vert }$$

(2.186)

and therefore

$$\displaystyle\begin{array}{rcl} J& =& 1\; + \left ( \frac{\partial ^{2}H} {\partial q\partial p}\; -\; \frac{\partial ^{2}H} {\partial p\partial q}\right )dt\; +\; \mathcal{O}(dt^{2}) \\ & =& 1\; +\; \mathcal{O}(dt^{2}). {}\end{array}$$

(2.187)

Neglecting the terms of $\mathcal{O}(dt^{2})$ order, we have J = 1, so that

$$\displaystyle{ dq^{{\prime}}dp^{{\prime}}\; =\; dq\;dp. }$$

(2.188)

This proof is equivalent to calculate the Poisson brackets of variables $q^{{\prime}}$ and $p^{{\prime}}$ with respect to q and p

$$\displaystyle{ [q^{{\prime}},p^{{\prime}}]_{ qp}\; =\; \frac{\partial (q^{{\prime}},p^{{\prime}})} {\partial (q,p)} \; =\; \frac{\partial q^{{\prime}}} {\partial q} \;\frac{\partial p^{{\prime}}} {\partial p} \; -\;\frac{\partial q^{{\prime}}} {\partial p} \;\frac{\partial p^{{\prime}}} {\partial q} }$$

(2.189)

and verifying thus that

$$\displaystyle{ [q^{{\prime}},p^{{\prime}}]_{ qp}\; =\; 1. }$$

(2.190)

This shows that the transformation $(q,p) \rightarrow (q^{{\prime}},p^{{\prime}})$ is canonical and the condition $dq^{{\prime}}dp^{{\prime}} = dq\;dp$ is satisfied (see e.g. Goldstein 1980).

1.2 A.2.2 Demonstration of V = V ^′ and |v| = |v ^′ | in Sect. 2.3.1

From equation (2.81) we may write before and after a given collision, respectively

$$\displaystyle\begin{array}{rcl} & & (m + M)\;\mathbf{V}\; =\; m\;\mathbf{v_{e}}\; +\; M\;\mathbf{v_{o}}{}\end{array}$$

(2.191)

$$\displaystyle\begin{array}{rcl} & & (m + M)\;\mathbf{V^{{\prime}}}\; =\; m\;\mathbf{v_{ e}^{{\prime}}}\; +\; M\;\mathbf{v_{ o}^{{\prime}}}.{}\end{array}$$

(2.192)

The right-hand side members of these two expressions are equal due to momentum conservation, so that the left-hand side members are also equal, and hence $\mathbf{V} = \mathbf{V^{{\prime}}}$.

On the other hand, from equations (2.82) and (2.83), we may write

$$\displaystyle\begin{array}{rcl} & & v_{e}^{\;2}\; =\; V ^{2} + \left ( \frac{M} {m + M}\right )^{2}v^{2} +\; 2\; \frac{M} {m + M}\;(\mathbf{V}.\;\mathbf{v}){}\end{array}$$

(2.193)

$$\displaystyle\begin{array}{rcl} & & v_{o}^{\;2}\; =\; V ^{2} + \left ( \frac{m} {m + M}\right )^{2}v^{2} -\; 2\; \frac{m} {m + M}\;(\mathbf{V}.\;\mathbf{v}){}\end{array}$$

(2.194)

and from equations (2.84) and (2.85), with $\mathbf{V} = \mathbf{V^{{\prime}}}$, we get similar expressions for $v_{e}^{{\prime}\;2}$ and $v_{o}^{{\prime}\;2}$. Substituting these expressions into the equation for energy conservation

$$\displaystyle{ m\;v_{e}^{\;2} +\; M\;v_{ 0}^{\;2}\; =\; m\;v_{ e}^{{\prime}\;2} +\; M\;v_{ 0}^{{\prime}\;2}, }$$

(2.195)

we rapidly obtain $v = v^{{\prime}}$.

1.3 A.2.3 Demonstration of Equation (2.91)

Let us consider the triangle v _o, V and $\mathbf{v_{e}^{{\prime}}}$ shown in Fig. 2.9 obtained from Fig. 2.5. We easily obtain

$$\displaystyle\begin{array}{rcl} \vert \mathbf{v_{e}^{{\prime}}}-\mathbf{V}\vert ^{2}& =& \vert \mathbf{V} -\mathbf{v_{ o}}\vert ^{2}\;\sin ^{2}\chi _{ 0}\; +\; \left (\vert \mathbf{v_{e}^{{\prime}}}-\mathbf{v_{ o}}\vert \;-\;\vert \mathbf{V} -\mathbf{v_{o}}\vert \;\cos \chi _{0}\right )^{2} \\ & =& \vert \mathbf{v_{e}^{{\prime}}}-\mathbf{v_{ o}}\vert ^{2}\; +\; \vert \mathbf{V} -\mathbf{v_{ o}}\vert ^{2}\; -\; 2\;\vert \mathbf{v_{ e}^{{\prime}}}-\mathbf{v_{ o}}\vert \;\vert \mathbf{V} -\mathbf{v_{o}}\vert \;\cos \chi _{0}.{}\end{array}$$

(2.196)

Fig. 2.9

Triangle obtained from Fig. 2.5, in which χ ₀ is the scattering angle in the laboratory system

Substituting equations (2.84) and (2.85), with $\mathbf{V^{{\prime}}} = \mathbf{V}$ and $\vert \mathbf{v^{{\prime}}}\vert = \vert \mathbf{v}\vert$, we may write

$$\displaystyle{ \left ( \frac{M} {m+M}\right )^{2}\vert \mathbf{v}\vert ^{2}\;=\;\vert \mathbf{v_{ e}^{{\prime}}}-\mathbf{v_{ o}}\vert ^{2}\; +\; \left ( \frac{m} {m+M}\right )^{2}\vert \mathbf{v}\vert ^{2}\;-\;2\;\vert \mathbf{v_{ e}^{{\prime}}}-\mathbf{v_{ o}}\vert \left ( \frac{m} {m + M}\right )\vert \mathbf{v}\vert \;\cos \chi _{0}. }$$

(2.197)

Inserting now (2.88), we still have

$$\displaystyle{ \cos \chi _{0}\; =\; \frac{m + M\;\cos \chi } {\sqrt{m^{2 } + M^{2 } + 2\;mM\;\cos \chi }}, }$$

(2.198)

from which we obtain equation (2.91).

Exercises

Exercise 2.1:

Determine the electron rate coefficients of the inelastic electron-molecule processes for excitation of a given state j from state i, with u _ij denoting the threshold-energy:

• (a) $\sigma _{ij}(u) = 0$, for u < u _ij , and $\sigma _{ij}(u) = a_{ij} =\mathrm{ const}$, for u > u _ij ;

• (b) $\sigma _{ij}(u) = 0$, for u ≤ u _ij , and $\sigma _{ij}(u) = a_{ij}\;(u - u_{ij})$, for u ≥ u _ij ,

in the case of electrons with a Maxwellian velocity distribution at temperature T _e .

Resolution:

For electrons following a Maxwellian velocity distribution at temperature T _e , the isotropic component of the velocity distribution is

$$\displaystyle{f_{e}^{0}(v_{ e})\; =\; n_{e}\left ( \frac{m} {2\pi k_{B}T_{e}}\right )^{3/2}\exp \left (-\; \frac{mv_{e}^{\;2}} {2k_{B}T_{e}}\right ),}$$

obeying to the normalization condition

$$\displaystyle{\int _{0}^{\infty }f_{ e}^{0}(v_{ e})\;4\pi v_{e}^{\;2}\;dv_{ e}\; =\; n_{e},}$$

whereas the electron rate coefficient (2.132) is given by

$$\displaystyle{C_{ij}\; =\;<\!\! v_{e}\;\sigma _{ij}(v_{e})\!\!>\;=\; \frac{1} {n_{e}}\int _{v_{ij}}^{\infty }v_{ e}\;\sigma _{ij}(v_{e})\;f_{e}^{0}(v_{ e})\;4\pi v_{e}^{\;2}\;dv_{ e},}$$

being $v_{ij} = \sqrt{2u_{ij } /m}$.

Rewriting these expressions in terms of the electron energy $u = \frac{1} {2}mv_{e}^{\;2}$, we have

$$\displaystyle{f_{e}^{0}(u)\; =\; n_{ e}\left ( \frac{m} {2\pi k_{B}T_{e}}\right )^{3/2}\exp \left (-\; \frac{u} {k_{B}T_{e}}\right )}$$

and

$$\displaystyle{C_{ij}\; =\; \frac{1} {n_{e}}\; \frac{8\pi } {m^{2}}\int _{u_{ij}}^{\infty }u\;\sigma _{ ij}(u)\;f_{e}^{0}(u)\;du,}$$

from which we obtain the following expressions for the two cases proposed:

• (a)
  $$\displaystyle{C_{ij}\, =\; \sqrt{\frac{8k_{B } T_{e } } {\pi m}} \;a_{ij}\left (1 +\; \frac{u_{ij}} {k_{B}T_{e}}\right )\exp \left (-\; \frac{u_{ij}} {k_{B}T_{e}}\right )\;;}$$
• (b)
  $$\displaystyle{C_{0j}\, =\; \sqrt{\frac{8k_{B } T_{e } } {\pi m}} \;2k_{B}T_{e}\;a_{ij}\left (1 +\; \frac{u_{ij}} {2k_{B}T_{e}}\right )\exp \left (-\; \frac{u_{ij}} {k_{B}T_{e}}\right ).}$$

Exercise 2.2:

Using the Klein-Rosseland relation (2.140) on the cross sections of Exercise 2.1 show that the electron rate coefficients of direct and reverse processes obey the equation (2.135).

Resolution:

Using the relation (2.140) we obtain the reverse cross sections of Exercise 2.1: (a) $\sigma _{ji}(u) = a_{ij}\;((u + u_{ij})/u)$, for u > 0; (b) $\sigma _{ji}(u) = a_{ij}\;(u + u_{ij})$, for u ≥ 0. Calculating then the electron rate coefficient of the superelastic process

$$\displaystyle{C_{ji}\; =\; \frac{1} {n_{e}}\; \frac{8\pi } {m^{2}}\int _{0}^{\infty }u\;\sigma _{ ji}(u)\;f_{e}^{0}(u)\;du,}$$

we obtain

$$\displaystyle{\frac{C_{ji}} {C_{ij}}\; =\;\exp \left ( \frac{u_{ij}} {k_{B}T_{e}}\right )}$$

in both situations.

Exercise 2.3:

The cross sections for electron impact excitation of a dipole-allowed transition at high energies is of the form (Massey and Burhop 1969)

$$\displaystyle{\sigma _{ij}(u)\; =\; \frac{a_{ij}} {u} \;\ln \left (\!b_{ij}\; \frac{u} {u_{ij}}\!\right ),}$$

being $u = \frac{1} {2}mv_{e}^{\;2}$ the electron energy, u _ij the threshold-energy, and a _ij and b _ij two constants dependent of the gas. Determine the expression for the corresponding electron rate coefficient in the case of a Maxwellian velocity distribution at temperature T _e .

Resolution:

In the case of a Maxwellian velocity distribution, the expression for the rate coefficient is as follows

$$\displaystyle{C_{ij}\; =\; \sqrt{ \frac{8} {\pi m}}\; \frac{1} {(k_{B}T_{e})^{3/2}}\int _{u_{ij}}^{\infty }u\;\sigma _{ ij}(u)\;\exp \left (-\; \frac{u} {k_{B}T_{e}}\right )\;du,}$$

so that assuming, as in Exercise 2.1-(ii), a linear dependence $\sigma _{ij}(u) = c_{ij}\;(u - u_{ij})$ at low energies up to the energy u ^∗ > u _ij , we may write

$$\displaystyle\begin{array}{rcl} C_{ij}& =& \sqrt{\frac{8k_{B } T_{e } } {\pi m}} \;2k_{B}T_{e}\;c_{ij}\;\Bigg\{\!\!\left (1 + \frac{u_{ij}} {2k_{B}T_{e}}\right )\exp \left (-\; \frac{u_{ij}} {k_{B}T_{e}}\right ) -\left (1 + \frac{u^{{\ast}}} {2k_{B}T_{e}}\right ) {}\\ & & \exp \left (-\; \frac{u^{{\ast}}} {k_{B}T_{e}}\right )\!\!\Bigg\} +\; \sqrt{ \frac{8} {\pi m}}\; \frac{a_{ij}} {(k_{B}T_{e})^{3/2}}\int _{u^{{\ast}}}^{\infty }\ln \left (b_{ ij}\; \frac{u} {u_{ij}}\right )\exp \left (-\; \frac{u} {k_{B}T_{e}}\right )\;du. {}\\ \end{array}$$

Making the substitution $\eta = b_{ij}\;u/u_{ij}$, we still obtain for the high-energy term

$$\displaystyle{C_{ij}\; =\; \ldots \ldots \ldots.. +\; \sqrt{ \frac{8} {\pi m}}\; \frac{a_{ij}} {(k_{B}T_{e})^{3/2}}\;\frac{u_{ij}} {b_{ij}} \int _{\eta ^{{\ast}}}^{\infty }\ln \eta \;\exp \left (-\; \frac{u_{ij}} {b_{ij}\;k_{B}T_{e}}\;\eta \right )\;d\eta,}$$

which presents the following primitive for the integrand function with $a\,=\,u_{ij}/$ $(b_{ij}\;k_{B}T_{e})$

$$\displaystyle\begin{array}{rcl} & & \int \ln \eta \;\exp (-\;a\eta )\;d\eta \; =\; -\;\frac{1} {a}\;\ln \eta \;\exp (-\;a\eta ) +\; \frac{1} {a}\int \frac{1} {\eta } \;\exp (-\;a\eta )\;d\eta {}\\ & & =\; -\;\frac{1} {a}\;\ln \eta \;\exp (-\;a\eta ) +\; \frac{1} {a}\;\Bigg(\ln \eta \;-\;a\eta \; +\; \frac{(a\eta )^{2}} {2.\;2!} \; -\;\frac{(a\eta )^{3}} {3.\;3!} \; + \ldots \ldots..\Bigg). {}\\ \end{array}$$

Exercise 2.4:

In a plasma column of radius R and infinite length the gas temperature presents a parabolic radial profile

$$\displaystyle{T_{o}(r)\; =\; (T_{o}(0) - T_{W})\left (1 -\; \frac{r^{2}} {R^{2}}\right ) +\; T_{W},}$$

being T _o (0) and T _W  = T _o (R) the gas temperatures at the axis and at the wall, respectively. Neglecting the radial dependence of the thermal conductivity coefficient through the gas temperature, determine the power loss by thermal conduction, per volume unity.

Resolution:

Using an analogous expression for molecules as indicated in Sect. 2.4.3, the power loss by thermal conduction by volume unity is $P_{cond} = -\;k_{T}\;\triangledown ^{2}T_{o}$, being k _T the thermal conductivity coefficient. We obtain therefore

$$\displaystyle{P_{cond}\; =\; -\;k_{T}\;\frac{1} {r}\; \frac{d} {dr}\left (r\;\frac{dT_{o}} {dr} \right ) =\; \frac{4} {R^{2}}\;k_{T}\left (T_{o}(0) - T_{W}\right ).}$$

On the other hand, the radially averaged temperature is

$$\displaystyle{\overline{T}\; =\; \frac{1} {R^{2}}\int _{0}^{R}T_{ o}(r)\;2r\;dr\; =\; \frac{1} {2}\;\left (T_{o}(0) + T_{W}\right ),}$$

so that the gas temperature at the axis of the column is $T_{o}(0) = 2\overline{T} - T_{W}$. Using this latter relation in the expression above, we obtain at the end

$$\displaystyle{P_{cond}\; =\; \frac{8} {R^{2}}\;k_{T}\left (\overline{T} - T_{W}\right ).}$$