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\begin{document}
{\LARGE A Derivation of the One-fluid Plasma Model
from the Multi-fluid Model.}\\
by Alec Johnson, November 28, 2006

\section{Definition of quantities.}

\nl $s$ = species index; typically $s\in\{e,i\}$, \\
        where $e$ denotes electron and $i$ denotes ion.

\p All quantities with a subscript $s$ are specific
  to the species $s$.

\pskip
\nl $q_s$ = charge of a particle
\nl $m_s$ = mass of a particle
\nl $n_s$ = particle number density
\def\fvel{\mathbf{u}}
\def\mfvel{u}
\nl $\fvel_s$ = average particle velocity
\nl $\sigma_s := n_s q_s$ = charge density 
  (Some call this $\rho_{qs}$.)
\nl $\rho_s := n_s m_s$ = mass density
  (Some call this $\rho_{ms}$.)
\def\cur{\mathbf{J}}
\nl $\cur_s := \sigma_s \fvel_s$ = charge flux = current density \\
  (Some call this $\cur_{qs}$.)
\nl $(\rho_s \fvel_s)$ = mass flux = momentum density \\
  (Some call this $\cur_{ms}$.)
\def\Pressure{\mathbb{P}}
%\def\Pressure{\diaunder[\undertilde|P]}
\nl $\Pressure_s$ = pressure tensor
\nl $p_s$ = scalar pressure
\def\energy{\varepsilon}
\nl $\energy_s$ = gas-dynamic energy per volume
\nl $T_s$ = temperature
\nl $S_s$ = entropy
\def\drag{\mathbf{R}}
\nl $\drag_s$ = drag force (per volume) on species $s$
  due to collisions with other species.
\def\heatflux{\mathbf{q}}
\nl $\heatflux$ = heat flux vector
\nl $Q_s$ = rate of heat transfer from other species
  to species $s$ (due to collisions).
\nl $K_s$ = rate of energy transfer from other species
  to species $s$ due to collisions \\
  $=\drag_s\dotp\fvel_s+Q_s$.

\p Quantities without a subscript apply to the plasma
as a whole.

\pskip
\nl $\sigma := \sum_s \sigma_s$ = net charge density
\nl $\rho := \sum \rho_s$ = net mass density
\nl $\cur := \sum \cur_s$ = net current density
\nl $(\rho \fvel) := \sum \rho_s \fvel_s$ = net momentum density \\
  (i.e. $\fvel := \frac{\sum \rho_s \fvel_s}{\rho}$.)
\nl $\Pressure := \sum_s \Pressure_s$ = pressure tensor
\def\efield{\mathbf{E}}

\pskip
\nl $\efield$ = electric field
\def\bfield{\mathbf{B}}
\nl $\bfield$ = magnetic field

\pskip
\nl $V_s$ = some volume element convected by $\fvel_s$.
\def\unitnorm{\hat{n}}
\nl Let $\int := \int_{V_s}$ and let $\oint := \int_{\partial V_s}$.
\def\deltabar{{\mathchar'26\mkern-9mu \delta}}
\def\dc{\deltabar_t} % conservative derivative
\def\Dt{\partial_t} %{\frac{\partial}{\partial t}}
\newcommand{\Dx}[1]{\ensuremath{\partial_{#1}}}
\def\div{\nabla\dotp}
\def\curl{\nabla\times}
\nl Let $d_t^s := \Dt + \fvel_s\dotp\nabla$
  denote the \emph{convective derivative}. 
\nl Let $\dc^s := \alpha \mapsto
      (\Dt \alpha + \nabla\dotp(\fvel_s\alpha))$
  denote the \emph{conservative derivative}.

\section{Conservation of mass.}

\nl We assume that particles are conserved.
This means that the number
of particles of species $s$ in a volume convected
by $\fvel_s$ remains constant:
\nl $d_t\int_{V_s} n_s = 0$ 
  $\iff \dc^s(n_s)=0$ \\
  $\iff$ \framebox{$\Dt(n_s)+\div(\fvel_s n_s)=0$}
\nl Multiplying by $q_s$ gives charge conservation: \\
  $\Dt(\sigma_s)+\div(\cur_s)=0$
\nl Multiplying by $m_s$ gives mass conservation: \\
  $\Dt(\rho_s)+\div(\rho_s \fvel_s)=0$

\pskip
{\bf Deriving one-fluid equation.}

\np If we sum over all species $s$, we get
  equations for conservation of net charge and total mass:
\nl $\Dt \sigma +\div \cur =0$
\nl \framebox{$\Dt \rho +\div(\rho \fvel)=0$}

\section{Conservation of momentum.}

To write conservation of momentum, we must identify
the sources of forces.  The fundamental forces are
electromagnetic force, nuclear force, and gravitational force.
We are ignoring gravity.

\p
We model the net force on a species as the sum of a macroscopic 
electromagnetic field (averaged over a region roughly the size
of a Debye sphere) plus a pressure force (due to collisions
with particles of the same species, which cancel everywhere
in a convected test volume except at the boundary) plus a
resistive drag force due to collisions with other species:

\pskip
\nl $d_t \int m_s n_s\fvel_s
   = - \oint \unitnorm\dotp\Pressure_s
     + \int(q_s n_s(\efield+\fvel_s\times\bfield))
     + \int \drag_s$
{\small
\nl \framebox{$\Dt(\rho_s\fvel_s) + \div(\rho_s\fvel_s\fvel_s)
   = - \div\Pressure_s
     + \sigma_s\efield+\cur_s\times\bfield
     + \drag_s$}
}

\pskip
{\bf Deriving one-fluid momentum equation.}

\p We sum over all species.  Since total momentum
is conserved for any collision process, $\sum_s \drag_s=0$.
For the linear terms the sum effectively replaces each
species quantity with the corresponding net quantity.

\def\diffvel{\mathbf{w}}
\def\mdiffvel{w}
\p
\nl We handle the nonlinear term in the standard manner of
  statistical mechanics, by absorbing the nonlinearity
  into the pressure tensor.  We define:
  \\ \framebox{$\diffvel_s := \fvel_s-\fvel$}
\nl $\diffvel_s$ is the average velocity of species $s$ relative to
  the net velocity $\fvel$, also known as the \emph{diffusive velocity}
  for species $s$.
%\begin{multicols}{2}
\nl Then $\sum_s \rho_s\diffvel_s
       = \sum_s \rho_s(\fvel_s-\fvel)
       = \sum_s \rho_s\fvel_s - \sum_s \rho_s\fvel
       = \rho\fvel - \rho\fvel
       = 0.$
%\end{multicols}
\nl So $\sum_s \rho_s\fvel_s\fvel_s
       = \sum_s \rho_s(\fvel+\diffvel_s)(\fvel+\diffvel_s)
   \\  = \rho\fvel\fvel
         + \cancel{\sum_s \rho_s\fvel\diffvel_s}
         + \cancel{\sum_s \rho_s\diffvel_s\fvel}
         + \sum_s \rho_s\diffvel_s\diffvel_s
         $
\nl Define $\Pressure_{0s}:=\Pressure_s + \rho_s\diffvel_s\diffvel_s$
\nl Define \framebox{$\Pressure_{0}:=\sum_s \Pressure_{0s}
    = \Pressure + \sum_s \rho_s\diffvel_s\diffvel_s$}

\p
\nl So the net (i.e. one-fluid) momentum equation is:
\nl \framebox{$\Dt(\rho\fvel) + \div(\rho\fvel\fvel)
   = - \div\Pressure_0
     + \sigma\efield+\cur\times\bfield $}

\section{Conservation of energy.}

The total energy of the system is the energy
of the electromagnetic field plus the (kinetic)
energy in each species.  (Again we are ignoring gravity.)
The kinetic energy $\energy_s$ of species $s$ is the sum of the
macroscopic kinetic energy $\half\rho_s\mfvel_s^2$
plus the thermal energy per unit volume, $\rho_s e_s$,
where $e_s$ is the thermal energy per unit mass: \\
\framebox{$\energy_s = \rho_s e_s + \half\rho_s\mfvel_s^2$}

\p
The gas-dynamic energy of the species $s$ within
the convected test volume
$V_s$ is changed as a result of work performed by
electromagnetic and pressure forces, heat flow
within the species, and interaction (collisions)
with other species.
Energy balance for species $s$ is thus:
\nl $d_t\int\energy_s =
   -\oint(\unitnorm\dotp \Pressure_s\dotp\fvel_s)
  + -\oint\unitnorm\dotp(\heatflux_s) \\
  + \int \fvel_s\dotp(n_s q_s(\efield+\cancel{\fvel_s\times\bfield}))
  + \int K_s
  $.
\nl Recall that $K_s$ denotes the rate of transfer of energy
    from other species to $s$ due to collisions.

\nl Cast as a differential equation, this says:
%\framebox{
%\begin{tabular}{rcl}
%    $\Dt\energy_s + \div(\fvel_s\energy_s)$
%    & = &
%    $ - \div(\Pressure_s\dotp\fvel_s)
%      + -\div \heatflux_s $
%    \\
%    & & $ + \cur_s\dotp\efield + K_s$
%\end{tabular}
%}
\nl
%{\small
% \framebox{
  $\Dt\energy_s + \div(\fvel_s\energy_s)=
   -\div(\Pressure_s\dotp\fvel_s)
  + -\div \heatflux_s 
  + \cur_s\dotp\efield+ K_s $
% }
%}
\subsection{Deriving one-fluid energy equation.}

\nl We again sum over all species,
as we did for the one-fluid momentum balance.
Since total kinetic energy is conserved for any collisional process,
$\sum_s K_s=0$.
We again deal with nonlinear terms using $\fvel_s = \fvel + \diffvel_s$.
In deriving the one-fluid momentum equation we linearized by absorbing
the nonlinearity from the inertial term into the definition of pressure.
Here we linearize by absorbing anything we don't
want into the definition of the heat flux.
\nl The equation we want is:
\nl $\Dt\energy_s + \div(\fvel\energy_s)=
   -\div(\Pressure_{0s}\dotp\fvel)
   -\div \heatflux_{0s}
   + \cur_s\dotp\efield + K_s
  $
(Summing over all species gives this same equation without
the $s$ subscripts, which is the desired 1-fluid equation.)

\nl Comparing this with the actual conservation of energy equation
for species $s$ tells how we need to define $\heatflux_{0s}$.
\nl Taking the difference of the two equations and summing
over species gives:
\nl $\div \sum_s \left(\diffvel_s\energy_s
   +\Pressure_s\dotp\fvel_s-\Pressure_{0s}\dotp\fvel
   +\heatflux_s - \heatflux_{0s}\right) = 0$ \\
   \hbox{(1-fluid heat flux requirement)}
\nl There is a great deal of freedom here in choosing how to work
  backwards to a definition of $\heatflux_{0s}$.  Of course
  we could just throw out the divergence and the sum in the
  previous equation and satisfy the book-keeping requirement
  of the exact 1-fluid equation, but we seek a definition of
  $\heatflux_{0s}$ that minimizes the size of the terms appearing
  in $\heatflux_{0s}-\heatflux_s$.  The most obvious requirement
  to satisfy is to express the heat flux requirement purely in
  terms of relative velocities, eliminating all references to
  absolute velocity.  You can throw out the divergence at this
  point, since I don't see any way to use the divergence
  to make anything cancel.
\nl $\sum_s \big(\diffvel_s\energy_s
   +(\Pressure_s-\Pressure_{0s})\dotp\fvel
   + \Pressure_s\dotp\diffvel_s
   +\heatflux_s - \heatflux_{0s}\big) = 0$
\nl Using $\sum_s\rho_s\diffvel_s=0$
  and $\Pressure_{0s}-\Pressure_{s}=\rho_s\diffvel_s\diffvel_s$, get:
\nl $\sum_s \big(
   \diffvel_s\rho_s
     (e_s+\cancel{\half\mfvel^2}+\cancel{\diffvel_s\dotp\fvel}
       +\half\mdiffvel_s^2) \\
   - \cancel{\rho_s\diffvel_s\diffvel_s\dotp\fvel}
   + \Pressure_s\dotp\diffvel_s
   +\heatflux_s - \heatflux_{0s}\big) = 0$

\nl So we can make the definition \\
  \framebox{
    $\heatflux_{0s}:=\heatflux_s+\diffvel_s\rho_s(e_s+\half\mdiffvel_s^2)
      + \Pressure_s\dotp\diffvel_s$ } 
\nl This says that the contribution of each species to the effective
  heat flux is its own heat flux plus its relative energy flux
  plus the relative work performed by its internal pressure.
  (The reason no interactions with other species are incorporated
  here is that in the average they cancel.)

\np So the net (i.e. one-fluid) energy equation is:
\nl \framebox{$\Dt\energy + \div(\fvel\energy)=
   -\div(\Pressure_{0}\dotp\fvel)
   -\div \heatflux_{0}
   + \cur\dotp\efield
  $}
\nl Recall here that $\energy$, $\Pressure_0$, $\heatflux_0$,
  and $\cur$ are sums of the corresponding quantities for
  each species.

\np Hopefully $\heatflux_0-\heatflux
  := \sum_s\diffvel_s\rho_s(e_s+\half\mdiffvel_s^2)+\Pressure_s\dotp\diffvel_s$
  \\ and $\Pressure_{0}-\Pressure = \sum_s\rho_s\diffvel_s\diffvel_s$
  are small and well-behaved.

%The total gas-dynamic energy is: \\
%$\energy := \sum_s \energy_s = 
%             \sum_s\frac{p}{\gamma-1} + \half\rho\mfvel^2$
%
%The total energy is always conserved, so it is conserved across
%a shock.  The total energy is the sum of the gas-dynamic
%energy and the energy of the electromagnetic field.  Since the
%electromagnetic field is continuous, it follows that the total
%gas-dynamic energy is conserved across a shock.

%\section{Conservation of electromagnetic field}
%
%These conservation laws are coupled with Maxwell's
%equations of electromagnetism.
%These are most commonly expressed as:
%\begin{itemize}
%\item $\div \efield = \frac{\sigma}{\epsilon_0}$ \hfill (Gauss's law)
%\item $\div \bfield = 0$ \hfill (no monopoles)
%\item $\curl\efield = -\Dt\bfield$ \hfill (Faraday's law)
%\item $\curl\bfield = \mu_0\cur+\mu_0\epsilon_0\Dt\efield$
%  \hfill (extended Ampere's law)
%\end{itemize}
%
%In the classical limit, the force of the electromagnetic field
%on a particle is given by the Lorentz force law:
%\def\vel{\mathbf{v}}
%$$\mathbf{F} = q(\efield+\vel\times\bfield)$$
%where $\vel$ is the force on a particle.
%
%\p The last two equations can be read as evolution equations
%for $\bfield$ and $\efield$.
%The first two equations are constraints that continue to
%hold if they hold at time 0.
%Taking the curl of the evolution equations and setting
%the current to 0 show that the speed of light $c$ satisfies
%$c^2\mu_0\epsilon_0=1$.
%
%\p We can write the evolution equations as conservation laws
%with current as a source term for electromagnetic field:
%\begin{gather*}
%  \Dt \begin{bmatrix}
%    \bfield \\ \efield
%    \end{bmatrix} + \curl
%        \begin{bmatrix}
%    \efield \\ -c^2\bfield
%    \end{bmatrix} = 
%    \begin{bmatrix}
%     0 \\ -\frac{1}{\epsilon_0}\cur
%    \end{bmatrix}
%\end{gather*}
%Remark: a curl, like any spatial differential operator,
%can be viewed as a divergence: \\
%$\underline{\nabla}\times\underline{v}
%=\Dx{j}\mathbf{e}_i\epsilon_{ijk}v_k
%=-\underline{\nabla}\cdot(\underline{\underline{\underline{\epsilon}}}\dotp\underline{v})$.
%
%\subsection{Nondimensionalized electrodynamics}
%
%By rescaling we can write Maxwell's equations
%in nondimensional form:
%\begin{itemize}
%\item $\div \efield = \Big(\frac{\sigma}{\epsilon_0}\Big)$
%\item $\div (c\bfield) = 0$
%\item $\curl\efield = -\partial_{(ct)}(c\bfield)$
%\item $\curl(c\bfield)
%  = \Big(\frac{\cur}{c\epsilon_0}\Big)+\partial_{(ct)}\efield$
%\end{itemize}
%
%Under this rescaling the Lorentz force law becomes:
%$$\mathbf{F} = q\Big(\efield+\Big(\frac{\vel}{c}\Big)\times(c\bfield)\Big)$$
%
%The quantities in parentheses represent rescaled charge density,
%current, magnetic field, time, and velocity.

%\newpage
%
%\section{Constitutive relations for species balance laws}
%
%\subsection{Constitutive relations for ideal gas.}
%
%\subsubsection{Pressure}
%
%\def\idtwo{\mathbf{I}}
%\def\viscstress{\tau}
%
%We can write the pressure tensor as a sum
%of a component $p\idtwo$ that is
%independent of the rate of deformation
%minus a viscous stress tensor $\viscstress$.
%
%% For an isotropic fluid this viscous stress tensor
%% is an isotropic linear function of the even part
%% of the rate of deformation tensor.  This allows
%% us to write:
%% \nl \framebox{$\viscstress = (\lambda\div\fvel)\idtwo
%%   + \mu(\nabla\fvel+(\nabla\fvel)^T)$}
%
%For an ideal gas the viscous stress tensor is zero, and the
%pressure is jointly proportional to density and temperature:
%$p=\rho R T$.  The temperature is directly related to the
%energy per mass: $e=c_v T$.  So the relationship between energy
%and pressure is simply: $\rho e=\frac{c_v}{R}p$.
%
%\subsubsection{Heat Flux}
%
%For an ideal gas there is no heat flux:
%$\heatflux_s=0$
%
%\subsubsection{Energy}
%
%For an ideal gas, $\rho_s e_s=\frac{p_s}{\gamma_s-1}$.
%So for an ideal gas, \\
%\framebox{$\energy_s = \frac{p_s}{\gamma_s-1} + \half\rho_s\mfvel_s^2$}


%\section{Summary of Ideal Plasma}
%
%\def\deltabar{{\mathchar'26\mkern-9mu \delta}}
%\def\dcs{\deltabar_t} % conservative derivative
%\def\dcv{d_t} % convected derivative
%
%\subsection{Two-fluid equations}
%
%The two-fluid equations are simply the gas-dynamic
%equations for each species with electromagnetic
%source terms.
%
%\begin{itemize}
%\item $\Dt \rho_s + \div(\rho_s \fvel_s)=0$
%\item $\dcs^s(\rho_s \fvel_s)
%   = -\nabla p_s + (\sigma_s\efield+\cur_s\times\bfield) + \drag_s$
%\item $\dcs^s\energy_s = -\div(p_s\fvel_s)+\cur_s\dotp\efield
%         + \fvel_s\dotp\drag_s$
%\end{itemize}
%Note here that $\rho_s=m_s n_s$ and $\cur_s=q_s \fvel_s$.
%For simplicity we will assume $\drag_s=0$.
%In the absence of shocks we can replace energy conservation
%with entropy conservation:
%$$\dcv^s S_s=0$, where $S_s=\ln(p_s \rho_s^{-\gamma})$$


%\section{Resume of balance laws}

\end{document}