1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
|
---
title: "Boltzmann relation"
firstLetter: "B"
publishDate: 2021-10-18
categories:
- Physics
- Plasma physics
date: 2021-10-18T15:25:39+02:00
draft: false
markup: pandoc
---
# Boltzmann relation
In a plasma where the ions and electrons are both in thermal equilibrium,
and in the absence of short-lived induced electromagnetic fields,
their densities $n_i$ and $n_e$ can be predicted.
By definition, a particle in an [electric field](/know/concept/electric-field/) $\vb{E}$
experiences a [Lorentz force](/know/concept/lorentz-force/) $\vb{F}_e$.
This corresponds to a force density $\vb{f}_e$,
such that $\vb{F}_e = \vb{f}_e \dd{V}$.
For the electrons, we thus have:
$$\begin{aligned}
\vb{f}_e
= q_e n_e \vb{E}
= - q_e n_e \nabla \phi
\end{aligned}$$
Meanwhile, if we treat the electrons as a gas
obeying the ideal gas law $p_e = k_B T_e n_e$,
then the pressure $p_e$ leads to another force density $\vb{f}_p$:
$$\begin{aligned}
\vb{f}_p
= - \nabla p_e
= - k_B T_e \nabla n_e
\end{aligned}$$
At equilibrium, we demand that $\vb{f}_e = \vb{f}_p$,
and isolate this equation for $\nabla n_e$, yielding:
$$\begin{aligned}
k_B T_e \nabla n_e
= q_e n_e \nabla \phi
\quad \implies \quad
\nabla n_e
= \frac{q_e \nabla \phi}{k_B T_e} n_e
= \nabla \bigg( \frac{q_e \phi}{k_B T_e} \bigg) n_e
\end{aligned}$$
This equation is straightforward to integrate,
leading to the following expression for $n_e$,
known as the **Boltzmann relation**,
due to its resemblance to the statistical Boltzmann distribution
(see [canonical ensemble](/know/concept/canonical-ensemble/)):
$$\begin{aligned}
\boxed{
n_e(\vb{r})
= n_{e0} \exp\!\bigg( \!-\! \frac{q_e \phi(\vb{r})}{k_B T_e} \bigg)
}
\end{aligned}$$
Where the linearity factor $n_{e0}$ represents
the electron density for $\phi = 0$.
We can do the same for ions instead of electrons,
leading to the following ion density $n_i$:
$$\begin{aligned}
\boxed{
n_i(\vb{r})
= n_{i0} \exp\!\bigg( \!-\! \frac{q_i \phi(\vb{r})}{k_B T_i} \bigg)
}
\end{aligned}$$
However, due to their larger mass,
ions are much slower to respond to fluctuations in the above equilibrium.
Consequently, after a perturbation,
the ions spend much more time in a transient non-equilibrium state
than the electrons, so this formula for $n_i$ is only valid
if the perturbation is sufficiently slow,
allowing the ions to keep up.
Usually, electrons do not suffer the same issue,
thanks to their small mass and fast response.
## References
1. P.M. Bellan,
*Fundamentals of plasma physics*,
1st edition, Cambridge.
2. M. Salewski, A.H. Nielsen,
*Plasma physics: lecture notes*,
2021, unpublished.
|
