path: root/source/know/concept/magnetohydrodynamics/index.md

---
title: "Magnetohydrodynamics"
sort_title: "Magnetohydrodynamics"
date: 2021-10-21
categories:
- Physics
- Plasma physics
- Electromagnetism
layout: "concept"
---

**Magnetohydrodynamics** (MHD) describes the dynamics
of fluids that are electrically conductive.
Notably, it is often suitable to describe plasmas,
and can be regarded as a special case of the
[two-fluid model](/know/concept/two-fluid-equations/);
we will derive it as such,
but the results are not specific to plasmas.

In the two-fluid model, we described the plasma as two separate fluids,
but in MHD we treat it as a single conductive fluid.
The macroscopic pressure $$p$$
and electric current density $$\vb{J}$$ are:

$$\begin{aligned}
    p
    &= p_i + p_e
    \\
    \vb{J}
    &= q_i n_i \vb{u}_i + q_e n_e \vb{u}_e
\end{aligned}$$

Meanwhile, the macroscopic mass density $$\rho$$
and center-of-mass flow velocity $$\vb{u}$$ are as follows,
although the ions dominate both due to their large mass,
so $$\rho \approx m_i n_i$$ and $$\vb{u} \approx \vb{u}_i$$:

$$\begin{aligned}
    \rho
    &= m_i n_i + m_e n_e
    \\
    \vb{u}
    &= \frac{1}{\rho} \Big( m_i n_i \vb{u}_i + m_e n_e \vb{u}_e \Big)
\end{aligned}$$

With these quantities in mind,
we add up the two-fluid continuity equations,
multiplied by their respective particles' masses:

$$\begin{aligned}
    0
    &= m_i \pdv{n_i}{t} + m_e \pdv{n_e}{t} + m_i \nabla \cdot (n_i \vb{u}_i) + m_e \nabla \cdot (n_e \vb{u}_e)
\end{aligned}$$

After some straightforward rearranging,
we arrive at the single-fluid continuity relation:

$$\begin{aligned}
    \boxed{
        \pdv{\rho}{t} + \nabla \cdot (\rho \vb{u})
        = 0
    }
\end{aligned}$$

Next, consider the two-fluid momentum equations
for the ions and electrons, respectively:

$$\begin{aligned}
    m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
    &= q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) - \nabla p_i - f_{ie} m_i n_i (\vb{u}_i - \vb{u}_e)
    \\
    m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
    &= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) - \nabla p_e - f_{ei} m_e n_e (\vb{u}_e - \vb{u}_i)
\end{aligned}$$

We will assume that electrons' inertia
is negligible compared to the Lorentz force.
Let $$\tau_\mathrm{char}$$ be the characteristic timescale of the plasma's dynamics
(i.e. nothing notable happens in times shorter than $$\tau_\mathrm{char}$$),
then this assumption can be written as:

$$\begin{aligned}
    1
    \gg \frac{\big| m_e n_e \mathrm{D} \vb{u}_e / \mathrm{D} t \big|}{\big| q_e n_e \vb{u}_e \cross \vb{B} \big|}
    \sim \frac{m_e n_e |\vb{u}_e| / \tau_\mathrm{char}}{q_e n_e |\vb{u}_e| |\vb{B}|}
    = \frac{m_e}{q_e |\vb{B}| \tau_\mathrm{char}}
    = \frac{1}{\omega_{ce} \tau_\mathrm{char}}
\end{aligned}$$

Where we have recognized the cyclotron frequency $$\omega_c$$
(see [Lorentz force](/know/concept/lorentz-force/)).
In other words, our assumption is equivalent to
the electron gyration period $$2 \pi / \omega_{ce}$$
being small compared to the macroscopic timescale $$\tau_\mathrm{char}$$.
We can thus ignore the left-hand side of the electron momentum equation, leaving:

$$\begin{aligned}
    m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
    &= q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) - \nabla p_i - f_{ie} m_i n_i (\vb{u}_i - \vb{u}_e)
    \\
    0
    &= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) - \nabla p_e - f_{ei} m_e n_e (\vb{u}_e - \vb{u}_i)
\end{aligned}$$

We add up these momentum equations,
recognizing the pressure $$p$$ and current $$\vb{J}$$:

$$\begin{aligned}
    m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
    &= (q_i n_i + q_e n_e) \vb{E} + \vb{J} \cross \vb{B} - \nabla p
    - f_{ie} m_i n_i (\vb{u}_i \!-\! \vb{u}_e) - f_{ei} m_e n_e (\vb{u}_e \!-\! \vb{u}_i)
    \\
    &= (q_i n_i + q_e n_e) \vb{E} + \vb{J} \cross \vb{B} - \nabla p
\end{aligned}$$

Where we have used $$f_{ie} m_i n_i = f_{ei} m_e n_e$$
because momentum is conserved by the underlying
[Rutherford scattering](/know/concept/rutherford-scattering/) process,
which is [elastic](/know/concept/elastic-collision/).
In other words, the momentum given by ions to electrons
is equal to the momentum received by electrons from ions.

Since the two-fluid model assumes that
the [Debye length](/know/concept/debye-length/) $$\lambda_D$$
is small compared to a "blob" $$\dd{V}$$ of the fluid,
we can invoke quasi-neutrality $$q_i n_i + q_e n_e = 0$$.
Using that $$\rho \approx m_i n_i$$ and $$\vb{u} \approx \vb{u}_i$$,
we thus arrive at the **momentum equation**:

$$\begin{aligned}
    \boxed{
        \rho \frac{\mathrm{D} \vb{u}}{\mathrm{D} t}
        = \vb{J} \cross \vb{B} - \nabla p
    }
\end{aligned}$$

However, we found this by combining two equations into one,
so some information was implicitly lost;
we need a second one to keep our system of equations complete.
Therefore we return to the electrons' momentum equation,
after a bit of rearranging:

$$\begin{aligned}
    \vb{E} + \vb{u}_e \cross \vb{B} - \frac{\nabla p_e}{q_e n_e}
    = \frac{f_{ei} m_e}{q_e} (\vb{u}_e - \vb{u}_i)
\end{aligned}$$

Again using quasi-neutrality $$q_i n_i = - q_e n_e$$,
the current density $$\vb{J} = q_e n_e (\vb{u}_e \!-\! \vb{u}_i)$$,
so:

$$\begin{aligned}
    \vb{E} + \vb{u}_e \cross \vb{B} - \frac{\nabla p_e}{q_e n_e}
    = \eta \vb{J}
    \qquad \qquad
    \eta
    \equiv \frac{f_{ei} m_e}{n_e q_e^2}
\end{aligned}$$

Where $$\eta$$ is the electrical resistivity of the plasma,
see [Spitzer resistivity](/know/concept/spitzer-resistivity/)
for more information and a rough estimate of its value in a plasma.

Now, using that $$\vb{u} \approx \vb{u}_i$$,
we add $$(\vb{u} \!-\! \vb{u}_i) \cross \vb{B} \approx 0$$ to the equation,
and insert $$\vb{J}$$ again:

$$\begin{aligned}
    \eta \vb{J}
    &= \vb{E} + \vb{u} \cross \vb{B} + (\vb{u}_e - \vb{u}_i) \cross \vb{B} - \frac{\nabla p_e}{q_e n_e}
    \\
    &= \vb{E} + \vb{u} \cross \vb{B} + \frac{\vb{J} \cross \vb{B}}{q_e n_e} - \frac{\nabla p_e}{q_e n_e}
\end{aligned}$$

Next, we want to get rid of the pressure term.
To do so, we take the curl of the equation:

$$\begin{aligned}
    \nabla \cross (\eta \vb{J})
    = - \pdv{\vb{B}}{t} + \nabla \cross (\vb{u} \cross \vb{B}) + \nabla \cross \frac{\vb{J} \cross \vb{B}}{q_e n_e}
    - \nabla \cross \frac{\nabla p_e}{q_e n_e}
\end{aligned}$$

Where we have used [Faraday's law](/know/concept/maxwells-equations/).
This is the **induction equation**,
and is used to compute $$\vb{B}$$.
The pressure term can be rewritten using the ideal gas law $$p_e = k_B T_e n_e$$:

$$\begin{aligned}
    \nabla \cross \frac{\nabla p_e}{q_e n_e}
    &= \frac{k_B}{q_e} \nabla \cross \frac{\nabla (n_e T_e)}{n_e}
    \\
    &= \frac{k_B}{q_e} \nabla \cross \Big( \nabla T_e + T_e \frac{\nabla n_e}{n_e} \Big)
\end{aligned}$$

The curl of a gradient is always zero,
and we notice that $$\nabla n_e / n_e = \nabla\! \ln(n_e)$$.
Then we use the vector identity $$\nabla \cross (f \nabla g) = \nabla f \cross \nabla g$$ to get:

$$\begin{aligned}
    \nabla \cross \frac{\nabla p_e}{q_e n_e}
    &= \frac{k_B}{q_e} \nabla \cross \big( T_e \: \nabla\! \ln(n_e) \big)
    \\
    &= \frac{k_B}{q_e} \big( \nabla T_e \cross \nabla\! \ln(n_e) \big)
    \\
    &= \frac{k_B}{q_e n_e} \big( \nabla T_e \cross \nabla n_e \big)
\end{aligned}$$

It is reasonable to assume that $$\nabla T_e$$ and $$\nabla n_e$$
point in roughly the same direction,
in which case the pressure term can be neglected.
Consequently, $$p_e$$ has no effect on the dynamics of $$\vb{B}$$,
so we argue that it can also be dropped
from the original equation (before taking the curl):

$$\begin{aligned}
    \boxed{
        \vb{E} + \vb{u} \cross \vb{B} + \frac{\vb{J} \cross \vb{B}}{q_e n_e}
        = \eta \vb{J}
    }
\end{aligned}$$

This is known as the **generalized Ohm's law**,
since it contains the relation $$\vb{E} = \eta \vb{J}$$.

Next, consider [Ampère's law](/know/concept/maxwells-equations/),
where we would like to neglect the last term:

$$\begin{aligned}
    \nabla \cross \vb{B}
    = \mu_0 \vb{J} + \frac{1}{c^2} \pdv{\vb{E}}{t}
\end{aligned}$$

From Faraday's law, we can obtain a scale estimate for $$\vb{E}$$.
Recall that $$\tau_\mathrm{char}$$ is the characteristic timescale of the plasma,
and let $$\lambda_\mathrm{char} \gg \lambda_D$$ be its characteristic length scale:

$$\begin{aligned}
    \nabla \cross \vb{E}
    = - \pdv{\vb{B}}{t}
    \qquad \implies \qquad
    |\vb{E}|
    \sim \frac{\lambda_\mathrm{char}}{\tau_\mathrm{char}} |\vb{B}|
\end{aligned}$$

From this, we find that we can neglect the last term in Ampère's law
as long as the characteristic velocity $$v_\mathrm{char}$$ is tiny compared to $$c$$,
i.e. the plasma must be non-relativistic:

$$\begin{aligned}
    1
    \gg \frac{\big| (\ipdv{\vb{E}}{t}) / c^2 \big|}{\big| \nabla \cross \vb{B} \big|}
    \sim \frac{|\vb{E}| / \tau_\mathrm{char}}{|\vb{B}| c^2 / \lambda_\mathrm{char}}
    \sim \frac{|\vb{B}| \lambda_\mathrm{char}^2 / \tau_\mathrm{char}^2}{|\vb{B}| c^2}
    = \frac{v_\mathrm{char}^2}{c^2}
\end{aligned}$$

We thus have the following reduced form of Ampère's law,
in addition to Faraday's law:

$$\begin{aligned}
    \boxed{
        \nabla \cross \vb{B}
        = \mu_0 \vb{J}
    }
    \qquad \qquad
    \boxed{
        \nabla \cross \vb{E}
        = - \pdv{\vb{B}}{t}
    }
\end{aligned}$$

Finally, we revisit the thermodynamic equation of state,
for a single fluid this time.
Using the product rule of differentiation yields:

$$\begin{aligned}
    0
    &= \frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{p}{\rho^\gamma} \Big)
    = \frac{\mathrm{D} p}{\mathrm{D} t} \rho^{-\gamma} - p \gamma \rho^{-\gamma - 1} \frac{\mathrm{D} \rho}{\mathrm{D} t}
\end{aligned}$$

The continuity equation allows us to rewrite
the [material derivative](/know/concept/material-derivative/)
$$\mathrm{D} \rho / \mathrm{D} t$$ as follows:

$$\begin{aligned}
    0
    &= \pdv{\rho}{t} + \nabla \cdot (\rho \vb{u})
    \\
    &= \pdv{\rho}{t} + \rho \nabla \cdot \vb{u} + \vb{u} \cdot \nabla \rho
    \\
    &= \rho \nabla \cdot \vb{u} + \frac{\mathrm{D} \rho}{\mathrm{D} t}
\end{aligned}$$

Inserting this into the equation of state
leads us to a differential equation for $$p$$:

$$\begin{aligned}
    0
    = \frac{\mathrm{D} p}{\mathrm{D} t} + p \gamma \frac{1}{\rho} \rho \nabla \cdot \vb{u}
    \quad \implies \quad
    \boxed{
        \frac{\mathrm{D} p}{\mathrm{D} t} = - p \gamma \nabla \cdot \vb{u}
    }
\end{aligned}$$

This closes the set of 14 MHD equations for 14 unknowns.
Originally, the two-fluid model had 16 of each,
but we have merged $$n_i$$ and $$n_e$$ into $$\rho$$,
and $$p_i$$ and $$p_i$$ into $$p$$.



## Ohm's law variants

It is worth discussing the generalized Ohm's law in more detail.
Its full form was:

$$\begin{aligned}
    \vb{E} + \vb{u} \cross \vb{B} + \frac{\vb{J} \cross \vb{B}}{q_e n_e}
    = \eta \vb{J}
\end{aligned}$$

However, most authors neglect some terms:
the full form is used for **Hall MHD**,
where $$\vb{J} \cross \vb{B}$$ is called the **Hall term**.
It can be dropped in any of the following cases:

$$\begin{aligned}
    1
    &\gg \frac{\big| \vb{J} \cross \vb{B} / q_e n_e \big|}{\big| \vb{u} \cross \vb{B} \big|}
    \sim \frac{\rho v_\mathrm{char} / \tau_\mathrm{char}}{v_\mathrm{char} |\vb{B}| q_i n_i}
    \approx \frac{m_i n_i}{|\vb{B}| q_i n_i \tau_\mathrm{char}}
    = \frac{1}{\omega_{ci} \tau_\mathrm{char}}
    \\
    1
    &\gg \frac{\big| \vb{J} \cross \vb{B} / q_e n_e \big|}{\big| \eta \vb{J} \big|}
    \sim \frac{|\vb{J}| |\vb{B}| q_e^2 n_e}{f_{ei} m_e |\vb{J}| q_e n_e}
    = \frac{|\vb{B}| q_e}{f_{ei} m_e}
    = \frac{\omega_{ce}}{f_{ei}}
\end{aligned}$$

Where we have used the MHD momentum equation with $$\nabla p \approx 0$$
to obtain the scale estimate $$|\vb{J} \cross \vb{B}| \sim \rho v_\mathrm{char} / \tau_\mathrm{char}$$.
In other words, if the ion gyration period is short $$\tau_\mathrm{char} \gg \omega_{ci}$$,
and/or if the electron gyration period is long
compared to the electron-ion collision period $$\omega_{ce} \ll f_{ei}$$,
then we are left with this form of Ohm's law, used in **resistive MHD**:

$$\begin{aligned}
    \vb{E} + \vb{u} \cross \vb{B}
    = \eta \vb{J}
\end{aligned}$$

Finally, we can neglect the resistive term $$\eta \vb{J}$$
if the Lorentz force is much larger.
We formalize this condition as follows,
where we have used Ampère's law to find $$|\vb{J}| \sim |\vb{B}| / \mu_0 \lambda_\mathrm{char}$$:

$$\begin{aligned}
    1
    \ll \frac{\big| \vb{u} \cross \vb{B} \big|}{\big| \eta \vb{J} \big|}
    \sim \frac{v_\mathrm{char} |\vb{B}|}{\eta |\vb{J}|}
    \sim \frac{v_\mathrm{char} |\vb{B}|}{\eta |\vb{B}| / \mu_0 \lambda_\mathrm{char}}
    = \mathrm{R_m}
\end{aligned}$$

Where we have defined the **magnetic Reynolds number** $$\mathrm{R_m}$$ as follows,
which is analogous to the fluid [Reynolds number](/know/concept/reynolds-number/) $$\mathrm{Re}$$:

$$\begin{aligned}
    \boxed{
        \mathrm{R_m}
        \equiv \frac{v_\mathrm{char} \lambda_\mathrm{char}}{\eta / \mu_0}
    }
\end{aligned}$$

If $$\mathrm{R_m} \ll 1$$, the plasma is "electrically viscous",
meaning resistivity needs to be accounted for,
whereas if $$\mathrm{R_m} \gg 1$$, the resistivity is negligible,
in which case we have **ideal MHD**:

$$\begin{aligned}
    \boxed{
        \vb{E} + \vb{u} \cross \vb{B}
        = 0
    }
\end{aligned}$$



## References
1.  P.M. Bellan,
    *Fundamentals of plasma physics*,
    1st edition, Cambridge.
2.  M. Salewski, A.H. Nielsen,
    *Plasma physics: lecture notes*,
    2021, unpublished.