Dimensional equations
Here are the following equations for deriving the MHD model.
(1)\[
\div \vb{B} = 0
\]
(2)\[
\pdv{B}{t} = - \curl{\vb{E}}
\]
(3)\[
\curl \vb{B} = \mu_0 \vb{j}
\]
(4)\[
\vb{j} = \sigma \left( \vb{E} + \vb{u} \cross \vb{B} - \frac{1}{ne} \vb{j} \cross \vb{B} + \nu \laplacian \vb{j} \right)
\]
(5)\[
\pdv{\rho}{t} + \div{\rho \vb{u}} = 0
\]
(6)\[
\pdv{t}(\rho \vb{u}) + \div(\rho \vb{u} \vb{u}) = -\grad P + \vb{j} \cross \vb{B}
\]
(7)\[
\pdv{t}(\rho e_t) + \div(\rho \vb{u} e_t) = - \div(P \vb{u}) + \vb{j} \cdot \vb{E}
\]
With some manipulations, and by defining the total pressure \(P* = P + \flatfrac{\vb{B}^2}{2 \mu_0}\),
the total energy per unit volume \(E = \rho e_t + \flatfrac{\vb{B}^2}{2 \mu_0}\),
and the Poynting vector \(\vb*{\Pi} = \frac{\vb{E} \cross \vb{B}}{\mu_0}\) ,
we can obtain the following system of equations:
(8)\[\begin{split}
\begin{aligned}
& \pdv{\rho}{t} = -\div{\rho \vb{u}} \\
& \pdv{t}(\rho \vb{u}) = - \div[ \rho \vb{u}\vb{u} + P^* \vb{I} - \frac{\vb{B} \vb{B}}{\mu_0}] \\
& \pdv{E}{t} = - \div [ ( \rho e_t + P) \vb{u} + \vb*{\Pi}] = - \div[ \left(E + P^* - \frac{\vb{B}^2}{\mu_0} \right) \vb{u} + \vb*{\Pi} ] \\
& \pdv{B}{t} = - \curl \vb{E} %= - \curl[ \eta \vb{j} - \vb{u} \cross \vb{B} ]
\end{aligned}
\end{split}\]
where the electric field is expressed thanks to Ohm’s law as:
\[
\vb{E} = \eta \vb{j} - \vb{u} \cross \vb{B} + \frac{1}{n e} \vb{j} \cross \vb{B} - \nu \laplacian \vb{j}
\]
The system is closed with a polytropic equation of state:
(9)\[
P \rho^{-\gamma} = \mathrm{constant} \implies \rho e = \frac{P}{\gamma - 1}
\]
Dimensionless equations
Let us introduce the following reference quantities:
A reference magnetic field \(B_0\)
A reference plasma density \(n_0\)
The mass of the proton \(m_p\)
The Alfven velocity \(v_{0}\):
\[
v_{0} = \frac{B_0}{\sqrt{m_p n_0 \mu_0}}
\]
\[
\omega_0 = \frac{e B_0}{m_p}
\]
From these reference quantities, we can deduce the other reference quantities
\[
\delta_0 = \frac{v_0}{\omega_0} = \frac{1}{e} \sqrt{\frac{m_p}{n_0 \mu_0}}
\]
\[
\rho_0 = m_p n_0
\]
\[
P_0 = \rho_0 v_0^2 = \frac{B_0^2}{\mu_0}
\]
\[
E_0 = V_0 B_0 = \frac{B_0^2}{\sqrt{m_p n_0 \mu_0}}
\]
\[
j_0 = \frac{B_0}{\delta_0 \mu_0} = e B_0 \sqrt{\frac{n_0 }{m_p \mu_0}}
\]
\[
\eta_0 = \mu_0 v_0 \delta_0 = \frac{B_0 }{e n_0}
\]
\[
\nu_0 = \mu_0 v_0 \delta_0^3 = \frac{B_0 m_p}{e^3 n_0^2 \mu_0 }
\]
By defining non-dimensional fields as \(\tilde{\phi} = \frac{\phi}{\phi_0}\), we can obtain the following equations:
\[\begin{split}
\begin{aligned}
& \pdv{\tilde{\rho}}{\tilde{t}} = -\tilde{\grad} \cdot \left( \tilde{\rho} \tilde{\vb{u}} \right) \\
& \pdv{\tilde{t}}(\tilde{\rho} \tilde{\vb{u}}) = - \tilde{\grad} \cdot \left[ \tilde{\rho} \tilde{\vb{u}}\tilde{\vb{u}} + \tilde{P}^* \vb{I} - \tilde{\vb{B}} \tilde{\vb{B}} \right] \\
& \pdv{\tilde{E}}{\tilde{t}} = - \tilde{\grad} \cdot \left[ ( \tilde{\rho} \tilde{e}_t + \tilde{P}) \tilde{\vb{u}} + \tilde{\vb*{\Pi}}\right] = - \tilde{\grad} \cdot \left[ \left(\tilde{E} + \tilde{P}^* - \tilde{\vb{B}}^2 \right) \tilde{\vb{u}} + \tilde{\vb*{\Pi}} \right] \\
& \pdv{\tilde{\vb{B}}}{\tilde{t}} = - \tilde{\grad} \cross \vb{\tilde{E}} %= - \curl[ \eta \vb{j} - \tilde{\vb{u}} \cross \tilde{\vb{B}} ]
\end{aligned}
\end{split}\]
with
\[\begin{split}
\begin{aligned}
& \tilde{\vb*{\Pi}} = \tilde{\vb{E}} \cross \tilde{\vb{B}} \\
& \tilde{\vb{E}} = \tilde{\eta} \tilde{\vb{j}} - \tilde{\vb{u}} \cross \tilde{\vb{B}} + \frac{1}{\tilde{n}} \tilde{\vb{j}} \cross \tilde{\vb{B}} - \tilde{\nu} \tilde{\nabla}^2 \tilde{\vb{j}}
\end{aligned}
\end{split}\]
In the following, equations will be considered under their non-dimensional form, with the tildes dropped for readability.
Splitting of the magnetic field
To numerically impose an external magnetic field, the magnetic field is decomposed as
\[
\vb{B} = \vb{B}_0 + \vb{B}_1
\]
where \(\vb{B}_0\) is the imposed external field and \(\vb{B}_1\) is the difference between the total and external magnetic fields.
Inserting this splitting in the definition of the total energy \(E\), one can define the “reduced” total energy \(E_1\):
\[
E = \underbrace{\rho e_t + \frac{\vb{B}_1^2}{2}}_{E_1} + \frac{\vb{B}_0^2}{2} + \vb{B}_0 \cdot \vb{B}_1
\]
The derivative of \(E\) is related to \(E_1\)’s one following
\[\begin{split}
\begin{aligned}
\pdv{E}{t} & = \pdv{E_1}{t} + \vb{B}_0 \cdot \pdv{\vb{B}_0}{t} + \vb{B}_0 \cdot \pdv{\vb{B}_1}{t} + \pdv{\vb{B}_0}{t} \cdot \vb{B}_1 \\
& = \pdv{E_1}{t} + \vb{B}_0 \cdot \pdv*{\vb{B}}{t} + \pdv{\vb{B}_0}{t} \cdot \vb{B}_1
\end{aligned}
\end{split}\]
Moreover the term \(\vb{B}_0 \cdot \pdv*{\vb{B}}{t}\) can be expanded using the induction equation as:
\[\begin{split}
\begin{aligned}
\vb{B}_0 \cdot \pdv*{\vb{B}}{t} &= - \vb{B}_0 \cdot \left( \curl \vb{E} \right) \\
&= - \div(\vb{E} \cross \vb{B}_0) - \vb{E} \cdot \underbrace{ \left( \curl \vb{B}_0\right)}_{ = \vb{j}_0 = \vb{0}}
\end{aligned}
\end{split}\]
The rotational of \(\vb{B}_0\) is zero because it is external; the currents generating it are located outside of the computational domain.
Therefore, the equation for energy and magnetic field can be written in terms of \(E_1\) and \(B_1\) as follows
\[\begin{split}
\begin{aligned}
& \pdv{E_1}{t} = - \grad \cdot \left[ ( \rho e_t + P) \vb{u} + \vb{E} \cross \vb{B}_1 \right] - \pdv{\vb{B}_0}{t} \cdot \vb{B}_1 % = - \grad \cdot \left[ \left(E_1 + P_1^* - \vb{B}_1^2 \right) \vb{u} + \vb{E} \cross \vb{B}_1 \right]
\\
& \pdv{\vb{B}_1}{t} = - \grad \cross \vb{E} - \pdv{\vb{B}_0}{t}
\end{aligned}
\end{split}\]