Changes between Version 2 and Version 3 of ResistiveMhd
- Timestamp:
- 01/04/21 11:30:52 (4 years ago)
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ResistiveMhd
v2 v3 4 4 Astrobear 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 $\frac{1}{c}$ 5 5 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 || $B$ || $B^{LH}$ || $\frac{1}{\sqrt{4\pi}} B^{EMU}$|| $\frac{1}{\sqrt{4\pi}} B^{G}$ || $\mbox{g}^{1/2} \mbox{cm}^{-1/2} \mbox{s}^{-1}$ || 11 11 12 Using the approach in the appendix of Jackson, we have12 Or - Using the approach in the appendix of Jackson, we have 13 13 || $k_1 = \frac{c^2}{4 \pi} $ || 14 14 || $k_2 = \frac{1}{4\pi}$ || … … 35 35 || $e = 4.80320425 \times 10^{-10} \mbox{statC} = 1.70269157 \times 10^{-9} \mbox{hsu} = 5.67956774 \times 10^{-20} \left [\mbox{g}^{1/2} \mbox{cm}^{1/2} \right]$ 36 36 37 Using the system of units described above, we can write the following simplified equations 37 Now 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. 38 38 39 This leaves us with the following set of equations 39 40 40 41 || Induction equation || $\frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \mathbf{E}$ || 41 42 || Ampere's equation (without Maxwell's correction) || $\mathbf{J} = \nabla \times \mathbf{B}$ || 42 43 || 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 46 In 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 43 48 || Poynting's theorem (electromagnetic energy) || $\frac{\partial e}{\partial t} = - \nabla \cdot \mathbf{S} - \mathbf{J}\cdot \mathbf{E} $ || 44 49 || Poynting vector || $\mathbf{S} = \mathbf{E} \times \mathbf{B} $ || … … 58 63 The $\mathbf{E}$ field can be calculated using a Generalized Ohm's law 59 64 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} $ 61 66 62 67 Under under certain conditions, we can ignore various terms resulting in the following approximations … … 66 71 || Ideal MHD || $\mathbf{E} = - \mathbf{v}\times \mathbf{B}$ || 67 72 73 Also - 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. 68 74 69 75 … … 71 77 For ideal MHD we substitute 72 78 73 $\mathbf{E} = -\mathbf{v} \times \mathbf{B} $79 $\mathbf{E} = \mathbf{E_{EMF}} = -\mathbf{v} \times \mathbf{B} $ 74 80 75 81 into 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)