---
title: "Two-fluid equations"
firstLetter: "T"
publishDate: 2021-10-19
categories:
- Physics
- Plasma physics

date: 2021-10-18T10:12:20+02:00
draft: false
markup: pandoc
---

# Two-fluid equations

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 $\dv*{t}$ 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**:

$$\begin{aligned}
    \boxed{
        \pdv{n_i}{t} + \nabla \cdot (n_i \vb{u}_i)
        = 0
        \qquad \quad
        \pdv{n_e}{t} + \nabla \cdot (n_e \vb{u}_e)
        = 0
    }
\end{aligned}$$

Currently, we have 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 in fact 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 $\pdv*{\vb{u}}{t} = 0$,
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.