From 075683cdf4588fe16f41d9f7b46b9720b42b2553 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Wed, 17 Jul 2024 10:01:43 +0200 Subject: Improve knowledge base --- source/know/concept/magnetohydrodynamics/index.md | 122 +++++++++++----------- 1 file changed, 61 insertions(+), 61 deletions(-) (limited to 'source/know/concept/magnetohydrodynamics') diff --git a/source/know/concept/magnetohydrodynamics/index.md b/source/know/concept/magnetohydrodynamics/index.md index bcc23f3..4431dfa 100644 --- a/source/know/concept/magnetohydrodynamics/index.md +++ b/source/know/concept/magnetohydrodynamics/index.md @@ -24,24 +24,23 @@ and electric current density $$\vb{J}$$ are: $$\begin{aligned} p - = p_i + p_e - \qquad \quad + &= p_i + p_e + \\ \vb{J} - = q_i n_i \vb{u}_i + q_e n_e \vb{u}_e + &= q_i n_i \vb{u}_i + q_e n_e \vb{u}_e \end{aligned}$$ 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: +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 $$\vb{u} \approx \vb{u}_i$$: $$\begin{aligned} \rho - = m_i n_i + m_e n_e - \approx m_i n_i - \qquad \quad + &= m_i n_i + m_e n_e + \\ \vb{u} - = \frac{1}{\rho} \Big( m_i n_i \vb{u}_i + m_e n_e \vb{u}_e \Big) - \approx \vb{u}_i + &= \frac{1}{\rho} \Big( m_i n_i \vb{u}_i + m_e n_e \vb{u}_e \Big) \end{aligned}$$ With these quantities in mind, @@ -75,9 +74,9 @@ $$\begin{aligned} \end{aligned}$$ 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 $$\tau_\mathrm{char}$$, +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: $$\begin{aligned} @@ -86,15 +85,14 @@ $$\begin{aligned} \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$$ (see Lorentz force article). +Where we have recognized the cyclotron frequency $$\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}$$ -being small compared to the macroscopic dynamics' timescale $$\tau_\mathrm{char}$$. -By construction, we can thus ignore the left-hand side -of the electron momentum equation, leaving: +being small compared to the macroscopic timescale $$\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} @@ -138,8 +136,8 @@ $$\begin{aligned} However, we found this by combining two equations into one, so some information was implicitly lost; -we need a second momentum equation. -Therefore, we return to the electrons' momentum equation, +we need a second one to keep our system of equations complete. +Therefore we return to the electrons' momentum equation, after a bit of rearranging: $$\begin{aligned} @@ -154,14 +152,14 @@ so: $$\begin{aligned} \vb{E} + \vb{u}_e \cross \vb{B} - \frac{\nabla p_e}{q_e n_e} = \eta \vb{J} - \qquad \quad + \qquad \qquad \eta \equiv \frac{f_{ei} m_e}{n_e q_e^2} \end{aligned}$$ Where $$\eta$$ is the electrical resistivity of the plasma, see [Spitzer resistivity](/know/concept/spitzer-resistivity/) -for more information, and a rough estimate of this quantity for a plasma. +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, @@ -183,34 +181,37 @@ $$\begin{aligned} - \nabla \cross \frac{\nabla p_e}{q_e n_e} \end{aligned}$$ -Where we have used Faraday's law. +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 $$p_e = k_B T_e n_e$$: $$\begin{aligned} \nabla \cross \frac{\nabla p_e}{q_e n_e} - = \frac{k_B}{q_e} \nabla \cross \frac{\nabla (n_e T_e)}{n_e} - = \frac{k_B}{q_e} \nabla \cross \Big( \nabla T_e + T_e \frac{\nabla n_e}{n_e} \Big) + &= \frac{k_B}{q_e} \nabla \cross \frac{\nabla (n_e T_e)}{n_e} + \\ + &= \frac{k_B}{q_e} \nabla \cross \Big( \nabla T_e + T_e \frac{\nabla n_e}{n_e} \Big) \end{aligned}$$ 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: +Then we use the vector identity $$\nabla \cross (f \nabla g) = \nabla f \cross \nabla g$$ to get: $$\begin{aligned} \nabla \cross \frac{\nabla p_e}{q_e n_e} - = \frac{k_B}{q_e} \nabla \cross \big( T_e \: \nabla\! \ln(n_e) \big) - = \frac{k_B}{q_e} \big( \nabla T_e \cross \nabla\! \ln(n_e) \big) - = \frac{k_B}{q_e n_e} \big( \nabla T_e \cross \nabla n_e \big) + &= \frac{k_B}{q_e} \nabla \cross \big( T_e \: \nabla\! \ln(n_e) \big) + \\ + &= \frac{k_B}{q_e} \big( \nabla T_e \cross \nabla\! \ln(n_e) \big) + \\ + &= \frac{k_B}{q_e n_e} \big( \nabla T_e \cross \nabla n_e \big) \end{aligned}$$ 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: +so we argue that it can also be dropped +from the original equation (before taking the curl): $$\begin{aligned} \boxed{ @@ -232,20 +233,18 @@ $$\begin{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: +and let $$\lambda_\mathrm{char} \gg \lambda_D$$ be its characteristic length scale: $$\begin{aligned} \nabla \cross \vb{E} = - \pdv{\vb{B}}{t} - \quad \implies \quad + \qquad \implies \qquad |\vb{E}| \sim \frac{\lambda_\mathrm{char}}{\tau_\mathrm{char}} |\vb{B}| \end{aligned}$$ -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 $$c$$, +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: $$\begin{aligned} @@ -254,7 +253,6 @@ $$\begin{aligned} \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}$$ We thus have the following reduced form of Ampère's law, @@ -265,7 +263,7 @@ $$\begin{aligned} \nabla \cross \vb{B} = \mu_0 \vb{J} } - \qquad \quad + \qquad \qquad \boxed{ \nabla \cross \vb{E} = - \pdv{\vb{B}}{t} @@ -287,10 +285,12 @@ the [material derivative](/know/concept/material-derivative/) $$\mathrm{D} \rho / \mathrm{D} t$$ as follows: $$\begin{aligned} - \pdv{\rho}{t} + \nabla \cdot (\rho \vb{u}) - = \pdv{\rho}{t} + \rho \nabla \cdot \vb{u} + \vb{u} \cdot \nabla \rho - = \rho \nabla \cdot \vb{u} + \frac{\mathrm{D} \rho}{\mathrm{D} t} - = 0 + 0 + &= \pdv{\rho}{t} + \nabla \cdot (\rho \vb{u}) + \\ + &= \pdv{\rho}{t} + \rho \nabla \cdot \vb{u} + \vb{u} \cdot \nabla \rho + \\ + &= \rho \nabla \cdot \vb{u} + \frac{\mathrm{D} \rho}{\mathrm{D} t} \end{aligned}$$ Inserting this into the equation of state @@ -311,6 +311,7 @@ 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. @@ -321,29 +322,27 @@ $$\begin{aligned} = \eta \vb{J} \end{aligned}$$ -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: +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: -$$\begin{gathered} +$$\begin{aligned} 1 - \gg \frac{\big| \vb{J} \cross \vb{B} / q_e n_e \big|}{\big| \vb{u} \cross \vb{B} \big|} + &\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|} + &\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}$$ +\end{aligned}$$ Where we have used the MHD momentum equation with $$\nabla p \approx 0$$ -to obtain the scale estimate $$\vb{J} \cross \vb{B} \sim \rho v_\mathrm{char} / \tau_\mathrm{char}$$. +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 $$\omega_{ce} \ll f_{ei}$$, @@ -354,18 +353,17 @@ $$\begin{aligned} = \eta \vb{J} \end{aligned}$$ -Finally, we can neglect the resisitive term $$\eta \vb{J}$$ +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 $$\vb{J} \sim \vb{B} / \mu_0 \lambda_\mathrm{char}$$: +where we have used Ampère's law to find $$|\vb{J}| \sim |\vb{B}| / \mu_0 \lambda_\mathrm{char}$$: $$\begin{aligned} 1 \ll \frac{\big| \vb{u} \cross \vb{B} \big|}{\big| \eta \vb{J} \big|} - \sim \frac{v_\mathrm{char} |\vb{B}|}{\eta \vb{J}} + \sim \frac{v_\mathrm{char} |\vb{B}|}{\eta |\vb{J}|} \sim \frac{v_\mathrm{char} |\vb{B}|}{\eta |\vb{B}| / \mu_0 \lambda_\mathrm{char}} = \mathrm{R_m} - \gg 1 \end{aligned}$$ Where we have defined the **magnetic Reynolds number** $$\mathrm{R_m}$$ as follows, @@ -379,13 +377,15 @@ $$\begin{aligned} \end{aligned}$$ If $$\mathrm{R_m} \ll 1$$, the plasma is "electrically viscous", -such that resistivity needs to be accounted for, +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**: $$\begin{aligned} - \vb{E} + \vb{u} \cross \vb{B} - = 0 + \boxed{ + \vb{E} + \vb{u} \cross \vb{B} + = 0 + } \end{aligned}$$ -- cgit v1.2.3