Changes between Version 2 and Version 3 of ResistiveMhd


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Timestamp:
01/04/21 11:30:52 (4 years ago)
Author:
Jonathan
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  • ResistiveMhd

    v2 v3  
    44Astrobear 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 $\frac{1}{c}$
    55
    6 || AstroBEAR || Lorentz-Heaviside || Gaussian ||  Physical Units ||
    7 || $E $ || $ c E^{LH} $ || $ \frac{c }{\sqrt{4\pi}}E^{G}$ || $\mbox{g}^{1/2} \mbox{cm}^{1/2} \mbox{s}^{-2}$ ||
    8 || $\rho $ || $ \frac{1}{c}\rho^{LH} $ || $ \frac{\sqrt{4 \pi} }{c}\rho^{G}$ || $\mbox{g}^{1/2} \mbox{cm}^{-5/2}$ ||
    9 || $J$ || $\frac{1}{c} J^{LH} $ || $ \frac{\sqrt{4 \pi}}{c} J^{G}$ || $\mbox{g}^{1/2} \mbox{cm}^{-3/2} \mbox{s}^{-1}$ ||
    10 || $B$ || $B^{LH}$ || $\frac{1}{\sqrt{4\pi}} B^{G}$ || $\mbox{g}^{1/2} \mbox{cm}^{-1/2} \mbox{s}^{-1}$ ||
     6|| AstroBEAR || Lorentz-Heaviside || EMU || Gaussian ||  Physical Units ||
     7|| $E $ || $ c E^{LH} $ || $ \frac{1}{\sqrt{4\pi}} E^{EMU}$ || $ \frac{c }{\sqrt{4\pi}}E^{G}$ || $\mbox{g}^{1/2} \mbox{cm}^{1/2} \mbox{s}^{-2}$ ||
     8|| $\rho $ || $ \frac{1}{c}\rho^{LH} $ || $ \sqrt{4 \pi}\rho^{EMU}$ || $ \frac{\sqrt{4 \pi} }{c}\rho^{G}$ || $\mbox{g}^{1/2} \mbox{cm}^{-5/2}$ ||
     9|| $J$ || $\frac{1}{c} J^{LH} $ || $ \sqrt{4 \pi} J^{EMU}$|| $ \frac{\sqrt{4 \pi}}{c} J^{G}$ || $\mbox{g}^{1/2} \mbox{cm}^{-3/2} \mbox{s}^{-1}$ ||
     10|| $B$ || $B^{LH}$ || $\frac{1}{\sqrt{4\pi}} B^{EMU}$|| $\frac{1}{\sqrt{4\pi}} B^{G}$ || $\mbox{g}^{1/2} \mbox{cm}^{-1/2} \mbox{s}^{-1}$ ||
    1111
    12 Using the approach in the appendix of Jackson, we have
     12Or - Using the approach in the appendix of Jackson, we have
    1313|| $k_1 = \frac{c^2}{4 \pi} $ ||
    1414|| $k_2 = \frac{1}{4\pi}$ ||
     
    3535|| $e = 4.80320425 \times 10^{-10} \mbox{statC} = 1.70269157 \times 10^{-9} \mbox{hsu} = 5.67956774 \times 10^{-20} \left [\mbox{g}^{1/2} \mbox{cm}^{1/2} \right]$
    3636
    37 Using the system of units described above, we can write the following simplified equations
     37Now for MHD we assume that the plasma is neutral ($\rho = 0$) and that $\mathbf{J} >> \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}$ and instead of tracking the electric field $\mathbf{E}$ through time, we will use Ohm's law to calculate the electric field.
    3838
     39This leaves us with the following set of equations
    3940
    4041|| Induction equation || $\frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \mathbf{E}$ ||
    4142|| Ampere's equation (without Maxwell's correction) || $\mathbf{J} = \nabla \times \mathbf{B}$ ||
    4243|| Lorentz Force Law || $\frac{\partial \rho \mathbf{v}}{\partial t}=\mathbf{J} \times \mathbf{B}$ ||
     44|| Generalized Ohm's Law || $\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} $ ||
     45
     46In addition we have equations for the total electromagnetic energy involving the net poynting flux $\left ( -\nabla \cdot \mathbf{S} \right)$ and work done on the gas $\left (-\mathbf{J} \cdot \mathbf{E} \right) $.
     47
    4348|| Poynting's theorem (electromagnetic energy) || $\frac{\partial e}{\partial t} = - \nabla \cdot \mathbf{S} - \mathbf{J}\cdot \mathbf{E}  $ ||
    4449|| Poynting vector || $\mathbf{S} = \mathbf{E} \times \mathbf{B} $ ||
     
    5863The $\mathbf{E}$ field can be calculated using a Generalized Ohm's law
    5964
    60 $\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} $
     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} $
    6166
    6267Under under certain conditions, we can ignore various terms resulting in the following approximations
     
    6671|| Ideal MHD || $\mathbf{E} = - \mathbf{v}\times \mathbf{B}$ ||
    6772
     73Also - the Biermann Battery term can be included $\left ( \frac{1}{ne} \nabla \cdot \mathbf{P}_e \right)$ which then allows for magnetic fields to be generated without the need for nascient fields.
    6874
    6975
     
    7177For ideal MHD we substitute
    7278
    73 $\mathbf{E} = -\mathbf{v} \times \mathbf{B} $
     79$\mathbf{E} = \mathbf{E_{EMF}} = -\mathbf{v} \times \mathbf{B} $
    7480
    7581into 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)