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diff --git a/content/know/concept/magnetohydrodynamics/index.pdc b/content/know/concept/magnetohydrodynamics/index.pdc new file mode 100644 index 0000000..6672a74 --- /dev/null +++ b/content/know/concept/magnetohydrodynamics/index.pdc @@ -0,0 +1,404 @@ +--- +title: "Magnetohydrodynamics" +firstLetter: "M" +publishDate: 2021-10-21 +categories: +- Physics +- Plasma physics +- Electromagnetism + +date: 2021-10-20T18:02:54+02:00 +draft: false +markup: pandoc +--- + +# Magnetohydrodynamics + +**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: + +$$\begin{aligned} + 1 + \gg \frac{\big| (\pdv*{\vb{E}}{t}) / c^2 \big|}{\big| \nabla \cross \vb{B} \big|} + \sim \frac{|\vb{E}| / \tau_\mathrm{char}}{|\vb{B}| c^2 / \lambda_\mathrm{char}} + \sim \frac{|\vb{B}| \lambda_\mathrm{char}^2 / \tau_\mathrm{char}^2}{|\vb{B}| c^2} + = \frac{v_\mathrm{char}^2}{c^2} + \ll 1 +\end{aligned}$$ + +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. |