| 1 | == Units in Astrobear == |
| 2 | |
| 3 | 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 $\frac{1}{c}$ |
| 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 | |
| 10 | Using 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 | || $\epsilon_0 = \frac{1}{c^2}$ || |
| 17 | This 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 | |
| 24 | as 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 | |
| 29 | Most 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 | |
| 31 | For 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 | |
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| 38 | |
| 39 | |
| 40 | === MHD equations === |
| 41 | |
| 42 | Using 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 === |
| 52 | Subsituting 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 == |
| 61 | Not 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 | |
| 63 | The $\mathbf{E}$ 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 | |
| 67 | Under 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 == |
| 76 | For ideal MHD we substitute |
| 77 | |
| 78 | $\mathbf{E} = -\mathbf{v} \times \mathbf{B} $ |
| 79 | |
| 80 | 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) |
| 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 | |
| 88 | For 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 | |
| 91 | Then our total electric field is given by |
| 92 | |
| 93 | $ \mathbf{E} = -\mathbf{v} \times \mathbf{B} \color{green}{+ \mathbf{E}_{\scriptscriptstyle R}}$ |
| 94 | |
| 95 | We 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 | |
| 97 | However 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 | |
| 103 | Note 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 | |
| 107 | can 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 | |
| 111 | Which 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 | |
| 115 | As 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 | |
| 118 | Then 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 | |
| 122 | Again 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 | |
| 126 | So 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}} $ |
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