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---
title: "Two-fluid equations"
sort_title: "Two-fluid equations"
date: 2021-10-19
categories:
- Physics
- Plasma physics
layout: "concept"
---
The **two-fluid model** describes a plasma as two separate but overlapping fluids,
one for ions and one for electrons.
Instead of tracking individual particles,
it gives the dynamics of fluid elements $$\dd{V}$$ (i.e. small "blobs").
These blobs are assumed to be much larger than
the [Debye length](/know/concept/debye-length/),
such that electromagnetic interactions between nearby blobs can be ignored.
From Newton's second law, we know that the velocity $$\vb{v}$$
of a particle with mass $$m$$ and charge $$q$$ is as follows,
when subjected only to the [Lorentz force](/know/concept/lorentz-force/):
$$\begin{aligned}
m \dv{\vb{v}}{t}
= q (\vb{E} + \vb{v} \cross \vb{B})
\end{aligned}$$
From here, the derivation is similar to that of the
[Navier-Stokes equations](/know/concept/navier-stokes-equations/).
We replace $$\idv{}{t}$$ with a
[material derivative](/know/concept/material-derivative/) $$\mathrm{D}/\mathrm{D}t$$,
and define $$\vb{u}$$ as the blob's center-of-mass velocity:
$$\begin{aligned}
m n \frac{\mathrm{D} \vb{u}}{\mathrm{D} t}
= q n (\vb{E} + \vb{u} \cross \vb{B})
\end{aligned}$$
Where we have multiplied by the number density $$n$$ of the particles.
Due to particle collisions in the fluid,
stresses become important. Therefore, we include
the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{P}$$,
leading to the following two equations:
$$\begin{aligned}
m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
&= q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) + \nabla \cdot \hat{P}_i{}^\top
\\
m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
&= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) + \nabla \cdot \hat{P}_e{}^\top
\end{aligned}$$
Where the subscripts $$i$$ and $$e$$ refer to ions and electrons, respectively.
Finally, we also account for momentum transfer between ions and electrons
due to [Rutherford scattering](/know/concept/rutherford-scattering/),
leading to these **two-fluid momentum equations**:
$$\begin{aligned}
\boxed{
\begin{aligned}
m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
&= q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) + \nabla \cdot \hat{P}_i{}^\top - f_{ie} m_i n_i (\vb{u}_i - \vb{u}_e)
\\
m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
&= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) + \nabla \cdot \hat{P}_e{}^\top - f_{ei} m_e n_e (\vb{u}_e - \vb{u}_i)
\end{aligned}
}
\end{aligned}$$
Where $$f_{ie}$$ is the mean frequency at which an ion collides with electrons,
and vice versa for $$f_{ei}$$.
For simplicity, we assume that the plasma is isotropic
and that shear stresses are negligible,
in which case the stress term can be replaced
by the gradient $$- \nabla p$$ of a scalar pressure $$p$$:
$$\begin{aligned}
m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
&= q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) - \nabla p_i - f_{ie} m_i n_i (\vb{u}_i - \vb{u}_e)
\\
m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
&= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) - \nabla p_e - f_{ei} m_e n_e (\vb{u}_e - \vb{u}_i)
\end{aligned}$$
Next, we demand that matter is conserved.
In other words, the rate at which particles enter/leave a volume $$V$$
must be equal to the flux through the enclosing surface $$S$$:
$$\begin{aligned}
0
&= \pdv{}{t}\int_V n \dd{V} + \oint_S n \vb{u} \cdot \dd{\vb{S}}
= \int_V \Big( \pdv{n}{t} + \nabla \cdot (n \vb{u}) \Big) \dd{V}
\end{aligned}$$
Where we have used the divergence theorem.
Since $$V$$ is arbitrary, we can remove the integrals,
leading to the following **continuity equations**:
$$\begin{aligned}
\boxed{
\pdv{n_i}{t} + \nabla \cdot (n_i \vb{u}_i)
= 0
\qquad \quad
\pdv{n_e}{t} + \nabla \cdot (n_e \vb{u}_e)
= 0
}
\end{aligned}$$
These are 8 equations (2 scalar continuity, 2 vector momentum),
but 16 unknowns $$\vb{u}_i$$, $$\vb{u}_e$$, $$\vb{E}$$, $$\vb{B}$$, $$n_i$$, $$n_e$$, $$p_i$$ and $$p_e$$.
We would like to close this system, so we need 8 more.
An obvious choice is [Maxwell's equations](/know/concept/maxwells-equations/),
in particular Faraday's and Ampère's law
(since Gauss' laws are redundant; see the article on Maxwell's equations):
$$\begin{aligned}
\boxed{
\nabla \cross \vb{E} = - \pdv{\vb{B}}{t}
\qquad \quad
\nabla \cross \vb{B} = \mu_0 \Big( n_i q_i \vb{u}_i + n_e q_e \vb{u}_e + \varepsilon_0 \pdv{\vb{E}}{t} \Big)
}
\end{aligned}$$
Now we have 14 equations, so we need 2 more, for the pressures $$p_i$$ and $$p_e$$.
This turns out to be the thermodynamic **equation of state**:
for quasistatic, reversible, adiabatic compression
of a gas with constant heat capacity (i.e. a *calorically perfect* gas),
it turns out that:
$$\begin{aligned}
\frac{\mathrm{D}}{\mathrm{D} t} \big( p V^\gamma \big) = 0
\qquad \quad
\gamma
\equiv \frac{C_P}{C_V}
= \frac{N + 2}{N}
\end{aligned}$$
Where $$\gamma$$ is the *heat capacity ratio*,
and can be calculated from the number of degrees of freedom $$N$$
of each particle in the gas.
In a fully ionized plasma, $$N = 3$$.
The density $$n \propto 1/V$$,
so since $$p V^\gamma$$ is constant in time,
for some constant $$C$$:
$$\begin{aligned}
\frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{p}{n^\gamma} \Big) = 0
\quad \implies \quad
p = C n^\gamma
\end{aligned}$$
In the two-fluid model, we thus have the following two equations of state,
giving us a set of 16 equations for 16 unknowns:
$$\begin{aligned}
\boxed{
\frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{p_i}{n_i^\gamma} \Big)
= 0
\qquad \quad
\frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{p_e}{n_e^\gamma} \Big)
= 0
}
\end{aligned}$$
Note that from the relation $$p = C n^\gamma$$,
we can calculate the $$\nabla p$$ term in the momentum equation,
using simple differentiation and the ideal gas law:
$$\begin{aligned}
p = C n^\gamma
\quad \implies \quad
\nabla p
= \gamma \frac{C n^{\gamma}}{n} \nabla n
= \gamma p \frac{\nabla n}{n}
= \gamma k_B T \nabla n
\end{aligned}$$
Note that the ideal gas law was not used immediately,
to allow for $$\gamma \neq 1$$.
## Fluid drifts
The momentum equations reduce to the following
if we assume the flow is steady $$\ipdv{\vb{u}}{t} = 0$$,
and neglect electron-ion momentum transfer on the right:
$$\begin{aligned}
m_i n_i (\vb{u}_i \cdot \nabla) \vb{u}_i
&\approx q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) - \nabla p_i
\\
m_e n_e (\vb{u}_e \cdot \nabla) \vb{u}_e
&\approx q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) - \nabla p_e
\end{aligned}$$
We take the cross product with $$\vb{B}$$,
which leaves only the component $$\vb{u}_\perp$$ of $$\vb{u}$$
perpendicular to $$\vb{B}$$ in the Lorentz term:
$$\begin{aligned}
0
&= q n (\vb{E} + \vb{u}_\perp \cross \vb{B}) \cross \vb{B} - \nabla p \cross \vb{B} - m n \big( (\vb{u} \cdot \nabla) \vb{u} \big) \cross \vb{B}
\\
&= q n (\vb{E} \cross \vb{B} - \vb{u}_\perp B^2) - \nabla p \cross \vb{B} - m n \big( (\vb{u} \cdot \nabla) \vb{u} \big) \cross \vb{B}
\end{aligned}$$
Isolating for $$\vb{u}_\perp$$ tells us
that the fluids drifts perpendicularly to $$\vb{B}$$,
with velocity $$\vb{u}_\perp$$:
$$\begin{aligned}
\vb{u}_\perp
= \frac{\vb{E} \cross \vb{B}}{B^2} - \frac{\nabla p \cross \vb{B}}{q n B^2}
- \frac{m \big( (\vb{u} \cdot \nabla) \vb{u} \big) \cross \vb{B}}{q B^2}
\end{aligned}$$
The last term is often neglected,
which turns out to be a valid approximation if $$\vb{E} = 0$$,
or if $$\vb{E}$$ is parallel to $$\nabla p$$.
The first term is the familiar $$\vb{E} \cross \vb{B}$$ drift $$\vb{v}_E$$
from [guiding center theory](/know/concept/guiding-center-theory/),
and the second term is called the **diamagnetic drift** $$\vb{v}_D$$:
$$\begin{aligned}
\boxed{
\vb{v}_E
= \frac{\vb{E} \cross \vb{B}}{B^2}
}
\qquad \quad
\boxed{
\vb{v}_D
= - \frac{\nabla p \cross \vb{B}}{q n B^2}
}
\end{aligned}$$
It is called *diamagnetic* because
it creates a current that induces
a magnetic field opposite to the original $$\vb{B}$$.
In a quasi-neutral plasma $$q_e n_e = - q_i n_i$$,
the current density $$\vb{J}$$ is given by:
$$\begin{aligned}
\vb{J}
= q_e n_e (\vb{v}_{De} - \vb{v}_{Di})
= q_e n_e \Big( \frac{\nabla p_i \cross \vb{B}}{q_i n_i B^2} - \frac{\nabla p_e \cross \vb{B}}{q_e n_e B^2} \Big)
= \frac{\vb{B} \cross \nabla (p_i + p_e)}{B^2}
\end{aligned}$$
Using the ideal gas law $$p = k_B T n$$,
this can be rewritten as follows:
$$\begin{aligned}
\vb{J}
= k_B \frac{\vb{B} \cross \nabla (T_i n_i + T_e n_e)}{B^2}
\end{aligned}$$
Curiously, $$\vb{v}_D$$ does not involve any net movement of particles,
because a pressure gradient does not necessarily cause particles to move.
Instead, there is a higher density of gyration paths
in the high-pressure region,
so that the particle flux through a reference plane is higher.
This causes the fluid elements to drift,
but not the guiding centers.
## References
1. F.F. Chen,
*Introduction to plasma physics and controlled fusion*,
3rd edition, Springer.
2. M. Salewski, A.H. Nielsen,
*Plasma physics: lecture notes*,
2021, unpublished.
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