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+---
+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:
+
+$$\begin{aligned}
+    p
+    = p_i + p_e
+    \qquad \quad
+    \vb{J}
+    = q_i n_i \vb{u}_i + q_e n_e \vb{u}_e
+\end{aligned}$$
+
+Meanwhile, the macroscopic mass density $\rho$
+and center-of-mass flow velocity $\vb{u}$
+are as follows, although the ions dominate due to their large mass:
+
+$$\begin{aligned}
+    \rho
+    = m_i n_i + m_e n_e
+    \approx m_i n_i
+    \qquad \quad
+    \vb{u}
+    = \frac{1}{\rho} \Big( m_i n_i \vb{u}_i + m_e n_e \vb{u}_e \Big)
+    \approx \vb{u}_i
+\end{aligned}$$
+
+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](/know/concept/lorentz-force/).
+Let $\tau_\mathrm{char}$ be the characteristic timescale of the plasma's dynamics,
+i.e. nothing noticable happens in times shorter than $\tau_\mathrm{char}$,
+then this assumption can be written as:
+
+$$\begin{aligned}
+    1
+    \gg \frac{\big| m_e n_e \mathrm{D} \vb{u}_e / \mathrm{D} t \big|}{\big| q_e n_e \vb{u}_e \cross \vb{B} \big|}
+    \sim \frac{m_e n_e |\vb{u}_e| / \tau_\mathrm{char}}{q_e n_e |\vb{u}_e| |\vb{B}|}
+    = \frac{m_e}{q_e |\vb{B}| \tau_\mathrm{char}}
+    = \frac{1}{\omega_{ce} \tau_\mathrm{char}}
+    \ll 1
+\end{aligned}$$
+
+Where we have recognized the cyclotron frequency $\omega_c$ (see Lorentz force article).
+In other words, our assumption is equivalent to
+the electron gyration period $2 \pi / \omega_{ce}$
+being small compared to the macroscopic dynamics' timescale $\tau_\mathrm{char}$.
+By construction, 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 momentum equation.
+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:
+
+$$\begin{aligned}
+    \vb{E} + \vb{u}_e \cross \vb{B} - \frac{\nabla p_e}{q_e n_e}
+    = \eta \vb{J}
+    \qquad \quad
+    \eta
+    \equiv \frac{f_{ei} m_e}{n_e q_e^2}
+\end{aligned}$$
+
+Where $\eta$ is the electrical resistivity of the plasma,
+see [Spitzer resistivity](/know/concept/spitzer-resistivity/)
+for more information, and a rough estimate of this quantity for 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.
+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 $\nabla \cross (f \nabla g) = \nabla f \cross \nabla g$,
+leading to:
+
+$$\begin{aligned}
+    \nabla \cross \frac{\nabla p_e}{q_e n_e}
+    = \frac{k_B}{q_e} \nabla \cross \big( T_e \: \nabla\! \ln(n_e) \big)
+    = \frac{k_B}{q_e} \big( \nabla T_e \cross \nabla\! \ln(n_e) \big)
+    = \frac{k_B}{q_e n_e} \big( \nabla T_e \cross \nabla n_e \big)
+\end{aligned}$$
+
+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 be dropped from the original (non-curled) equation too, leaving:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{E} + \vb{u} \cross \vb{B} + \frac{\vb{J} \cross \vb{B}}{q_e n_e}
+        = \eta \vb{J}
+    }
+\end{aligned}$$
+
+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 lengthscale:
+
+$$\begin{aligned}
+    \nabla \cross \vb{E}
+    = - \pdv{\vb{B}}{t}
+    \quad \implies \quad
+    |\vb{E}|
+    \sim \frac{\lambda_\mathrm{char}}{\tau_\mathrm{char}} |\vb{B}|
+\end{aligned}$$
+
+From this, we find when we can neglect
+the last term in Ampère's law:
+the characteristic velocity $v_\mathrm{char}$
+must be tiny compared to $c$,
+i.e. the plasma must be non-relativistic:
+
+$$\begin{aligned}
+    1
+    \gg \frac{\big| (\ipdv{\vb{E}}{t}) / c^2 \big|}{\big| \nabla \cross \vb{B} \big|}
+    \sim \frac{|\vb{E}| / \tau_\mathrm{char}}{|\vb{B}| c^2 / \lambda_\mathrm{char}}
+    \sim \frac{|\vb{B}| \lambda_\mathrm{char}^2 / \tau_\mathrm{char}^2}{|\vb{B}| c^2}
+    = \frac{v_\mathrm{char}^2}{c^2}
+    \ll 1
+\end{aligned}$$
+
+We thus have the following reduced form of Ampère's law,
+in addition to Faraday's law:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cross \vb{B}
+        = \mu_0 \vb{J}
+    }
+    \qquad \quad
+    \boxed{
+        \nabla \cross \vb{E}
+        = - \pdv{\vb{B}}{t}
+    }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \pdv{\rho}{t} + \nabla \cdot (\rho \vb{u})
+    = \pdv{\rho}{t} + \rho \nabla \cdot \vb{u} + \vb{u} \cdot \nabla \rho
+    = \rho \nabla \cdot \vb{u} + \frac{\mathrm{D} \rho}{\mathrm{D} t}
+    = 0
+\end{aligned}$$
+
+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 of its terms:
+this form is used for **Hall MHD**,
+where $\vb{J} \cross \vb{B}$ is called the *Hall term*.
+This term can be dropped in any of the following cases:
+
+$$\begin{gathered}
+    1
+    \gg \frac{\big| \vb{J} \cross \vb{B} / q_e n_e \big|}{\big| \vb{u} \cross \vb{B} \big|}
+    \sim \frac{\rho v_\mathrm{char} / \tau_\mathrm{char}}{v_\mathrm{char} |\vb{B}| q_i n_i}
+    \approx \frac{m_i n_i}{|\vb{B}| q_i n_i \tau_\mathrm{char}}
+    = \frac{1}{\omega_{ci} \tau_\mathrm{char}}
+    \ll 1
+    \\
+    1
+    \gg \frac{\big| \vb{J} \cross \vb{B} / q_e n_e \big|}{\big| \eta \vb{J} \big|}
+    \sim \frac{|\vb{J}| |\vb{B}| q_e^2 n_e}{f_{ei} m_e |\vb{J}| q_e n_e}
+    = \frac{|\vb{B}| q_e}{f_{ei} m_e}
+    = \frac{\omega_{ce}}{f_{ei}}
+    \ll 1
+\end{gathered}$$
+
+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 resisitive 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 $\vb{J} \sim \vb{B} / \mu_0 \lambda_\mathrm{char}$:
+
+$$\begin{aligned}
+    1
+    \ll \frac{\big| \vb{u} \cross \vb{B} \big|}{\big| \eta \vb{J} \big|}
+    \sim \frac{v_\mathrm{char} |\vb{B}|}{\eta \vb{J}}
+    \sim \frac{v_\mathrm{char} |\vb{B}|}{\eta |\vb{B}| / \mu_0 \lambda_\mathrm{char}}
+    = \mathrm{R_m}
+    \gg 1
+\end{aligned}$$
+
+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",
+such that resistivity needs to be accounted for,
+whereas if $\mathrm{R_m} \gg 1$, the resistivity is negligible,
+in which case we have **ideal MHD**:
+
+$$\begin{aligned}
+    \vb{E} + \vb{u} \cross \vb{B}
+    = 0
+\end{aligned}$$
+
+
+
+## References
+1.  P.M. Bellan,
+    *Fundamentals of plasma physics*,
+    1st edition, Cambridge.
+2.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.