Changes between Version 1 and Version 2 of MagnetoHydroDynamics


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Timestamp:
12/30/20 23:17:22 (4 years ago)
Author:
Jonathan
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  • MagnetoHydroDynamics

    v1 v2  
     1== Units in Astrobear ==
     2
     3Astrobear uses something like rationalized electromagnetic units (extra factor of $\sqrt{4 \pi}$ in the electric and magnetic fields) - or Lorentz-Heaviside but scaling the $E$ field by $c$ and the charge density $\rho$ (and current $J$ ) by $\frac{1}{c}$
     4
     5|| $E = c E^{LH} = \frac{c }{\sqrt{4\pi}}E^{G}$ ||
     6|| $\rho = \frac{1}{c}\rho^{LH} = \frac{\sqrt{4 \pi} }{c}\rho^{G}$||
     7|| $J = \frac{1}{c} J^{LH} = \frac{\sqrt{4 \pi}}{c} J^{G}$||
     8|| $B = B^{LH} = \frac{1}{\sqrt{4\pi}} B^{G}$ ||
     9
     10Using the approach in the appendix of Jackson, we have
     11|| $k_1 = \frac{c^2}{4 \pi} $ ||
     12|| $k_2 = \frac{1}{4\pi}$ ||
     13|| $k_3 = 1$ ||
     14|| $\alpha = 1$ ||
     15|| $\mu_0 = 1$ ||
     16|| $\epsilon_0 = \frac{1}{c^2}$ ||
     17This allows us to write Maxwell's equations as
     18
     19|| $\nabla \cdot \mathbf{E} = c^2 \rho$ ||
     20|| $\nabla \times \mathbf{B} = \mathbf{J} + \frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t}$ ||
     21|| $\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}$ ||
     22|| $\nabla \cdot \mathbf{B} = 0$ ||
     23
     24as well as
     25|| Lorentz Force Law || $\mathbf{F} = q \left ( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right)$ ||
     26|| Coulomb's Law || $\mathbf{F} = -\frac{c^2}{4 \pi} \frac{q_1 q_2}{r^2}\hat{\mathbf{r}}$ ||
     27
     28
     29Most of the time we don't care about $E$, $\rho$, or $J$, however for the Hall MHD terms, we need to determine the electron charge in our system of units.
     30
     31For our system of equations, the elementary charge is
     32
     33|| $e = 4.80320425 \times 10^{-10} \mbox{statC} = 1.70269157 \times 10^{-9} \mbox{hsu} = 5.67956774 \times 10^{-20} \left [\mbox{g}^{1/2} \mbox{cm}^{1/2} \right]$
     34
     35
     36
     37
     38
     39
     40=== MHD equations ===
     41
     42Using the system of units described above, we can write the following simplified equations
     43
     44
     45|| Induction equation || $\frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \mathbf{E}$ ||
     46|| Ampere's equation (without Maxwell's correction) || $\mathbf{J} = \nabla \times \mathbf{B}$ ||
     47|| Lorentz Force Law || $\frac{\partial \rho \mathbf{v}}{\partial t}=\mathbf{J} \times \mathbf{B}$ ||
     48|| Poynting's theorem (electromagnetic energy) || $\frac{\partial e}{\partial t} = - \nabla \cdot \mathbf{S} - \mathbf{J}\cdot \mathbf{E}  $ ||
     49|| Poynting vector || $\mathbf{S} = \mathbf{E} \times \mathbf{B} $ ||
     50
     51=== Equations for momentum, magnetic energy, and magnetic field ===
     52Subsituting Ampere's equation and the definition of the Poynting vector into the equations for momentum, magnetic energy, and the magnetic field, we arrive at
     53
     54$\frac{\partial \rho \mathbf{v}}{\partial t} = \mathbf{J} \times \mathbf{B} = \left (\nabla \times \mathbf{B} \right) \times \mathbf{B} = \nabla \cdot \left [ \mathbf{BB}-\frac{B^2}{2} \mathbf{I} \right ]$
     55
     56$\frac{\partial e}{\partial t} = -\nabla \cdot \left ( \mathbf{E}  \times \mathbf{B} \right) - \left ( \nabla \times \mathbf{B} \right) \cdot \mathbf{E} $
     57
     58$\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) $
     59
     60== Ohm's Law ==
     61Not the momentum equation only involves the magnetic field $\mathbf{B}$ and is the same regardless of the electric field $\mathbf{E}$.  Note it is also already in a form involving a total divergence making it straightforward to implement in a conservative finite volume scheme.  The energy and induction equation, however, will depend on our choice of $\mathbf{E}$.
     62
     63The $\mathbf{E}$ field can be calculated using a Generalized Ohm's law
     64
     65$\frac{m_e}{n e^2} \frac{\partial J}{\partial t} - \frac{1}{ne}\nabla \cdot \mathbf{P_e} = \mathbf{E} + \mathbf{v} \times \mathbf{B} - \eta\mathbf{J} - \frac{1}{ne} \mathbf{J} \times \mathbf{B} $
     66
     67Under under certain conditions, we can ignore various terms resulting in the following approximations
     68
     69|| Resistive + Hall MHD || $\mathbf{E} =  -\mathbf{v} \times \mathbf{B} + \eta \mathbf{J} +  \frac{1}{ne} \mathbf{J} \times \mathbf{B}$ ||
     70|| Resistive MHD || $ \mathbf{E} = -\mathbf{v} \times \mathbf{B} + \eta \mathbf{J}$ ||
     71|| Ideal MHD || $\mathbf{E} = - \mathbf{v}\times \mathbf{B}$ ||
     72
     73
     74
     75== Ideal MHD ==
     76For ideal MHD we substitute
     77
     78$\mathbf{E} = -\mathbf{v} \times \mathbf{B} $
     79
     80into the equations above and after some vector calculus (including a very messy vector quadruple product with a del operator), manage to write the energy term as a total divergence (and the field as a curl)
     81
     82$\frac{\partial e}{\partial t} = \nabla \cdot \left ( \left (\mathbf{v} \times \mathbf{B} \right ) \times \mathbf{B} \right) + \left ( \nabla \times \mathbf{B} \right) \cdot \left ( \mathbf{v} \times \mathbf{B} \right) = \nabla \cdot \left [ \mathbf{B} \left ( \mathbf{B} \cdot \mathbf{v} \right) - \frac{B^2}{2} \mathbf{v} \right]$
     83
     84$\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) $
     85
     86== Resistive MHD ==
     87
     88For resistive MHD, it will be convenient to define the resistive component of the electric field
     89$ \color{green}{\mathbf{E}_{\scriptscriptstyle R} =  \eta \mathbf{J} = \eta \nabla \times \mathbf{B}}$
     90
     91Then our total electric field is given by
     92
     93$ \mathbf{E} = -\mathbf{v} \times \mathbf{B} \color{green}{+ \mathbf{E}_{\scriptscriptstyle R}}$
     94
     95We then need to add this contribution to the induction equation $\color{green}{-\nabla \times \mathbf{E}_{\scriptscriptstyle R}}$ and the energy equation where it enters both the Poynting flux $-\color{green}{\nabla \cdot \left ( \mathbf{E}_{\scriptscriptstyle R} \times \mathbf{B} \right)}$ and the work done on the the charge distribution $\color{green}{ - \mathbf{J} \cdot \mathbf{E}_{\scriptscriptstyle R}}$. 
     96
     97However the work done on the charge distribution would need to also be added to the thermal energy cancelling any changes in the total energy due to the $\color{green}{\mathbf{J} \cdot \mathbf{E}_{\scriptscriptstyle R}}$ term.  This leaves us with.
     98
     99$\frac{\partial e}{\partial t} = \nabla \cdot \left [ \mathbf{B} \left ( \mathbf{B} \cdot \mathbf{v} \right) - \frac{B^2}{2} \mathbf{v}  \color{green}{- \mathbf{E}_{\scriptscriptstyle R} \times \mathbf{B}} \right] $
     100
     101$\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) \color{green}{ - \nabla \times \mathbf{E}_{\scriptscriptstyle R}} =  \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right)  \color{green}{- \nabla \times \left ( \eta \nabla \times \mathbf{B} \right )} $
     102
     103Note the resistive part of the induction equation
     104
     105$\frac{\partial \mathbf{B}}{\partial t} =  \color{green}{- \nabla \times \left ( \eta \nabla \times \mathbf{B} \right )} $
     106
     107can be expanded as
     108
     109$\frac{\partial \mathbf{B}}{\partial t} =  \color{green}{- \nabla \times \left ( \eta \nabla \times \mathbf{B} \right ) = \eta \nabla^2 \mathbf{B} -  \vec{\nabla} \mathbf{B}  \cdot \nabla \eta} $
     110
     111Which hints to the diffusive nature of the resistive term.  It also involves a second derivative so the time stepping criteria will be $\Delta t \propto \Delta x^2$ and an implicit solve might be worth considering.
     112
     113== Hall+Resistive MHD ==
     114
     115As in resistive MHD, it will be convenient to define the Hall component of the electric field
     116$ \color{red}{\mathbf{E}_{\scriptscriptstyle H} =} \color{red}{  \frac{1}{ne} \mathbf{J} \times \mathbf{B} =  \frac{1}{ne} \left ( \nabla \times \mathbf{B} \right ) \times \mathbf{B}}$
     117
     118Then our total electric field is given by
     119
     120$ \mathbf{E} = -\mathbf{v} \times \mathbf{B} \color{green}{+ \mathbf{E}_{\scriptscriptstyle R}} \color{red}{+\mathbf{E}_{\scriptscriptstyle H}}$
     121
     122Again we need to add the additional electric field component to the energy equation (in both the Poynting Flux and the work on the charge distribution) as well as the induction equation.  Also note in this case the electric field is perpendicular to the current - so it does no work on the charge distribution.
     123
     124$\mathbf{J} \cdot \color{red}{\mathbf{E}_{\scriptscriptstyle H}} = \mathbf{J} \cdot \color{red}{\frac{1}{ne} \mathbf{J} \times \mathbf{B}} = 0$
     125
     126So as in the Resistive case, only the Poynting Flux term needs to be added to the energy equation.
     127
     128$\frac{\partial \rho \mathbf{v}}{\partial t} =  \left (\nabla \times \mathbf{B} \right) \times \mathbf{B} = \nabla \cdot \left [ \mathbf{BB}-\frac{B^2}{2} \mathbf{I} \right ]$
     129
     130$\frac{\partial e}{\partial t} = \nabla \cdot \left [ \mathbf{B} \left ( \mathbf{B} \cdot \mathbf{v} \right) - \frac{B^2}{2} \mathbf{v}  \color{green}{ - \mathbf{E}_{\scriptscriptstyle R} \times \mathbf{B}} \color{red}{ -\mathbf{E}_{\scriptscriptstyle H} \times \mathbf{B}} \right] $
     131
     132
     133$\frac{\partial B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) \color{green}{- \nabla \times \mathbf{E}_{\scriptscriptstyle R}} \color{red}{- \nabla \times \mathbf{E}_{\scriptscriptstyle H}} $
     134
     135
     136
     137
     138
     139
    1140
    2141==== MHD ====