| 1 | | == Generalized Equations == |
| 2 | | We start with a Generalized Ohm's law |
| | 1 | === MHD equations === |
| | 2 | |
| | 3 | |
| | 4 | || Poynting's theorem (electromagnetic energy) || $\frac{\partial e}{\partial t} = - \nabla \cdot \mathbf{S} - \mathbf{J}\cdot \mathbf{E} $ || |
| | 5 | || Poynting vector || $\mathbf{S} = \mathbf{E} \times \mathbf{B} $ || |
| | 6 | || Induction equation || $\frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \mathbf{E}$ || |
| | 7 | || Ampere's equation (without Maxwell's correction) || $\mathbf{J} = \nabla \times \mathbf{B}$ || |
| | 8 | || Lorentz Force Law || $\frac{\partial \rho \mathbf{v}}{\partial t}=\mathbf{J} \times \mathbf{B}$ || |
| | 9 | |
| | 10 | === Equations for momentum, magnetic energy, and magnetic field === |
| | 11 | 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 |
| | 12 | |
| | 13 | $\frac{\partial \rho \mathbf{v}}{\partial t} = \mathbf{J} \times \mathbf{B} = \left (\nabla \times \mathbf{B} \right) \times \mathbf{B} = \nabla \cdot \left [ \mathbf{BB}-\frac{B^2}{2} \mathbf{I} \right ]$ |
| | 14 | |
| | 15 | $\frac{\partial e}{\partial t} = -\nabla \cdot \left ( \mathbf{E} \times \mathbf{B} \right) - \left ( \nabla \times \mathbf{B} \right) \cdot \mathbf{E} $ |
| | 16 | |
| | 17 | $\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) $ |
| | 18 | |
| | 19 | == Ohm's Law == |
| | 20 | 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 $\mathbf{E}$. |
| | 21 | |
| | 22 | The $\mathbf{E}$ field can be calculated using a Generalized Ohm's law |
| 12 | | === Electromagnetic energy === |
| 13 | | We also need Poynting's theorem for the change in electro-magnetic energy |
| 14 | | |
| 15 | | $\frac{\partial e}{\partial t} = - \nabla \cdot \mathbf{S} - \mathbf{J}\cdot \mathbf{E} $ |
| 16 | | |
| 17 | | which is just the net loss of energy due to diverging electromagnetic Poynting flux $\mathbf{S} = \mathbf{E} \times \mathbf{B}$ minus the work done on the charge distribution $\mathbf{J} \cdot \mathbf{E} $ |
| 18 | | |
| 19 | | === Induction equation === |
| 20 | | |
| 21 | | $\frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \mathbf{E}$ |
| 22 | | |
| 23 | | === Ampere's equation === |
| 24 | | (without Maxwell's correction for time varying electric field) |
| 25 | | |
| 26 | | $\mathbf{J} = \nabla \times \mathbf{B}$ and |
| 27 | | |
| 28 | | === Lorentz force law === |
| 29 | | $\frac{\partial \rho \mathbf{v}}{\partial t}=\mathbf{J} \times \mathbf{B}$ |
| 30 | | |
| 31 | | === Final set of equations === |
| 32 | | |
| 33 | | $\frac{\partial \rho \mathbf{v}}{\partial t} = \mathbf{J} \times \mathbf{B} = \left (\nabla \times \mathbf{B} \right) \times \mathbf{B} = \nabla \cdot \left [ \mathbf{BB}-\frac{B^2}{2} \mathbf{I} \right ]$ |
| 34 | | |
| 35 | | $\frac{\partial e}{\partial t} = -\nabla \cdot \left ( \mathbf{E} \times \mathbf{B} \right) - \left ( \nabla \times \mathbf{B} \right) \cdot \mathbf{E} $ |
| 36 | | |
| 37 | | $\frac{\partial B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) $ |
| 45 | | into the equations above and after some vector calculus, manage to write momentum and energy as a total divergence (and the field as a curl) |
| | 39 | 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) |
| | 40 | |
| | 41 | $\frac{\partial e}{\partial t} = \nabla \cdot \left ( \left (\mathbf{v} \times \mathbf{B} \right ) \times \mathbf{B} \right) + \left ( \nabla \times \mathbf{B} \right) \cdot \left ( \mathbf{v} \times \mathbf{B} \right) = \nabla \cdot \left [ \mathbf{B} \left ( \mathbf{B} \cdot \mathbf{v} \right) - \frac{B^2}{2} \mathbf{v} \right]$ |
| | 42 | |
| | 43 | $\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) $ |
| | 44 | |
| | 45 | == Resistive MHD == |
| | 46 | |
| | 47 | For resistive MHD, it will be convenient to define the resistive component of the electric field |
| | 48 | $ \color{green}{\mathbf{E}_{\scriptscriptstyle R} = \eta \mathbf{J} = \eta \nabla \times \mathbf{B}}$ |
| | 49 | |
| | 50 | Then our total electric field is given by |
| | 51 | |
| | 52 | $ \mathbf{E} = -\mathbf{v} \times \mathbf{B} \color{green}{+ \mathbf{E}_{\scriptscriptstyle R}}$ |
| | 53 | |
| | 54 | We then need to add this contribution to the induction equation $\color{green}{-\nabla \times \mathbf{E'}}$ and the energy equation where it enters both the Poynting flux $-\color{green}{\nabla \cdot \left ( \mathbf{E'} \times \mathbf{B} \right)}$ and the work done on the the charge distribution $\color{green}{ - \mathbf{J} \cdot \mathbf{E'}}$. |
| | 55 | |
| | 56 | 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. |
| | 57 | |
| | 58 | $\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] $ |
| | 59 | |
| | 60 | $\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 )} $ |
| | 61 | |
| | 62 | Note the resistive part of the induction equation |
| | 63 | |
| | 64 | $\frac{\partial \mathbf{B}}{\partial t} = \color{green}{- \nabla \times \left ( \eta \nabla \times \mathbf{B} \right )} $ |
| | 65 | |
| | 66 | can be expanded as |
| | 67 | |
| | 68 | $\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} $ |
| | 69 | |
| | 70 | 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. |
| | 71 | |
| | 72 | == Hall+Resistive MHD == |
| | 73 | |
| | 74 | As in resistive MHD, it will be convenient to define the Hall component of the electric field |
| | 75 | $ \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}}$ |
| | 76 | |
| | 77 | Then our total electric field is given by |
| | 78 | |
| | 79 | $ \mathbf{E} = -\mathbf{v} \times \mathbf{B} \color{green}{+ \mathbf{E}_{\scriptscriptstyle R}} \color{red}{+\mathbf{E}_{\scriptscriptstyle H}}$ |
| | 80 | |
| | 81 | 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. |
| | 82 | |
| | 83 | $\mathbf{J} \cdot \color{red}{\mathbf{E}_{\scriptscriptstyle H}} = \mathbf{J} \cdot \color{red}{\frac{1}{ne} \mathbf{J} \times \mathbf{B}} = 0$ |
| | 84 | |
| | 85 | So as in the Resistive case, only the Poynting Flux term needs to be added to the energy equation. |
| 49 | | $\frac{\partial e}{\partial t} = \nabla \cdot \left ( \left (\mathbf{v} \times \mathbf{B} \right ) \times \mathbf{B} \right) + \left ( \nabla \times \mathbf{B} \right) \cdot \left ( \mathbf{v} \times \mathbf{B} \right) = \nabla \cdot \left [ \mathbf{B} \left ( \mathbf{B} \cdot \mathbf{v} \right) - \frac{B^2}{2} \mathbf{v} \right]$ |
| 50 | | |
| 51 | | $\frac{\partial B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) $ |
| 52 | | |
| 53 | | == Resistive MHD == |
| 54 | | |
| 55 | | $ \mathbf{E} = -\mathbf{v} \times \mathbf{B} + \eta \mathbf{J} = -\mathbf{v} \times \mathbf{E} + \eta \nabla \times \mathbf{B}$ |
| 56 | | |
| 57 | | into the equations above and after some vector calculus, manage to write momentum and energy as a total divergence (and the field as a curl) |
| 58 | | |
| 59 | | $\frac{\partial \rho \mathbf{v}}{\partial t} = \left (\nabla \times \mathbf{B} \right) \times \mathbf{B} = \nabla \cdot \left [ \mathbf{BB}-\frac{B^2}{2} \mathbf{I} \right ]$ |
| 60 | | |
| 61 | | $\frac{\partial e}{\partial t} = \nabla \cdot \left [ \mathbf{B} \left ( \mathbf{B} \cdot \mathbf{v} \right) - \frac{B^2}{2} \mathbf{v} - \left ( \eta \nabla \times \mathbf{B} \right ) \times \mathbf{B} \right]$ |
| 62 | | |
| 63 | | $\frac{\partial B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) - \nabla \times \left ( \eta \nabla \times \mathbf{B} \right ) $ |
| 64 | | |
| 65 | | Note the energy equation is missing the additional $\mathbf{J} \cdot \mathbf{\eta \mathbf{J}}$ term which would correspond to the loss in magnetic energy due to ohmic dissipation. However this loss in magnetic energy is balanced by a gain in thermal energy - so those terms cancel in the final energy equation. |
| | 89 | $\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] $ |
| 68 | | == Hall+Resistive MHD == |
| 69 | | |
| 70 | | $\mathbf{E} = -\mathbf{v} \times \mathbf{B} + \eta \mathbf{J} + \frac{1}{ne} \mathbf{J} \times \mathbf{B}$ |
| 71 | | |
| 72 | | |
| 73 | | $\frac{\partial \rho \mathbf{v}}{\partial t} = \left (\nabla \times \mathbf{B} \right) \times \mathbf{B} = \nabla \cdot \left [ \mathbf{BB}-\frac{B^2}{2} \mathbf{I} \right ]$ |
| 74 | | |
| 75 | | $\frac{\partial e}{\partial t} = \nabla \cdot \left [ \mathbf{B} \left ( \mathbf{B} \cdot \mathbf{v} \right) - \frac{B^2}{2} \mathbf{v} - \left ( \eta \nabla \times \mathbf{B} \right ) \times \mathbf{B} - \left ( \frac{1}{ne} \left ( \nabla \times \mathbf{B} \right) \times \mathbf{B} \right) \times \mathbf{B} \right]$ |
| 76 | | |
| 77 | | $\frac{\partial B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) - \nabla \times \left ( \eta \nabla \times \mathbf{B} \right ) - \nabla \times \left ( \frac{1}{ne} \left( \nabla \times \mathbf{B} \right) \times \mathbf{B} \right) $ |
| 78 | | |
| 79 | | |
| 80 | | We can also combine the Hall and Resistive terms |
| 81 | | |
| 82 | | $\frac{\partial \rho \mathbf{v}}{\partial t} = \left (\nabla \times \mathbf{B} \right) \times \mathbf{B} = \nabla \cdot \left [ \mathbf{BB}-\frac{B^2}{2} \mathbf{I} \right ]$ |
| 83 | | |
| 84 | | $\frac{\partial e}{\partial t} = \nabla \cdot \left [ \mathbf{B} \left ( \mathbf{B} \cdot \mathbf{v} \right) - \frac{B^2}{2} \mathbf{v} - \left ( \eta \nabla \times \mathbf{B} + \left ( \frac{1}{ne} \left ( \nabla \times \mathbf{B} \right) \times \mathbf{B} \right) \right ) \times \mathbf{B} \right]$ |
| 85 | | |
| 86 | | $\frac{\partial B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) - \nabla \times \left ( \eta \nabla \times \mathbf{B} + \frac{1}{ne} \left( \nabla \times \mathbf{B} \right) \times \mathbf{B} \right) $ |
| 87 | | |
| 88 | | Or defining $\mathbf{E}' = \eta \nabla \times \mathbf{B} + \frac{1}{ne} \left( \nabla \times \mathbf{B} \right) \times \mathbf{B} $ |
| 89 | | |
| 90 | | $\frac{\partial \rho \mathbf{v}}{\partial t} = \left (\nabla \times \mathbf{B} \right) \times \mathbf{B} = \nabla \cdot \left [ \mathbf{BB}-\frac{B^2}{2} \mathbf{I} \right ]$ |
| 91 | | |
| 92 | | $\frac{\partial e}{\partial t} = \nabla \cdot \left [ \mathbf{B} \left ( \mathbf{B} \cdot \mathbf{v} \right) - \frac{B^2}{2} \mathbf{v} - \mathbf{E}' \times \mathbf{B} \right]$ |
| 93 | | |
| 94 | | $\frac{\partial B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) - \nabla \times \mathbf{E}' $ |
| 95 | | |
| 96 | | |
| 97 | | |
| 98 | | |
| 99 | | This along with |
| 100 | | |
| 101 | | we arrive at |
| 102 | | |
| 103 | | || Ideal MHD || $\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left [ \mathbf{v}\times \mathbf{B} \right ]$ || |
| 104 | | || Resistive MHD || $ \frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left [ \mathbf{v} \times \mathbf{B} - \frac{\mathbf{J}}{\sigma} \right ]$ || |
| 105 | | || Hall MHD || $\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left [ \mathbf{v} \times \mathbf{B} - \frac{1}{ne} \mathbf{J} \times \mathbf{B} \right ]$ || |
| 106 | | |
| 107 | | In addition, Poynting's theorem states that the electromagnetic energy goes as |
| 108 | | |
| 109 | | $\frac{\partial u}{\partial t} = - \nabla \cdot \mathbf{S} - \mathbf{J}\cdot \mathbf{E} $ |
| 110 | | |
| 111 | | |
| 112 | | $\frac{\partial u}{\partial t} = - \nabla \cdot \left (\mathbf{E} \times \mathbf{B}\right) - \mathbf{J}\cdot \mathbf{E} $ |
| 113 | | |
| 114 | | |
| 115 | | |
| 116 | | In Ideal MHD we have $\mathbf{E} = \mathbf{v} \times \mathbf{B}$ |
| 117 | | |
| 118 | | |
| 119 | | |
| 120 | | |
| 121 | | |
| 122 | | |
| 123 | | Now for Hall MHD, we can write the induction equation as |
| 124 | | |
| 125 | | $\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left [ \mathbf{v} \times \mathbf{B} - \mathbf{v}_\mbox{H} \times \mathbf{B} \right ]$ |
| 126 | | where |
| 127 | | |
| 128 | | $ \mathbf{v}_\mbox{H} = \frac{\mathbf{J}}{ne} $ |
| 129 | | |
| 130 | | Note the Hall velocity is always parallel to the current and therefore perpendicular to the field (using Ampere's law without Maxwell's addition ignoring time varying electric fields) and as a result the corresponding energy term involving $\mathbf{v_H} \cdot \mathbf{B} = 0$. |
| 131 | | |
| 132 | | |
| 133 | | Expanding just the Hall term further, we arrive at |
| 134 | | |
| 135 | | $\frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \left [ \frac{1}{ne} \left ( \nabla \times \mathbf{B} \right ) \times \mathbf{B} \right ]$ |
| 136 | | |
| 137 | | which we can compare to the resistive mhd term |
| 138 | | |
| 139 | | $\frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \left [ \left ( \eta \nabla \times \mathbf{B} \right ) \right ]$ |
| 140 | | |
| 141 | | So it is likewise a |
| | 92 | $\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}} $ |
| 147 | | For resistive MHD, we calculate $\xi = \nabla \times \mathbf{B}$ at cell edges for which each component can be calculated by differencing the 4 cell faces that share the edge. |
| | 98 | For resistive MHD, we calculate $\mathbf{E} = \eta \nabla \times \mathbf{B}$ at cell edges for which each component can be calculated by differencing the 4 cell faces that share the edge. |
| | 99 | |
| | 100 | Getting the curl of B at a cell edge is straight forward as it involves difference the 4 adjacent cell faces. For example, the Z component would be given by |
| | 101 | |
| | 102 | $E^z_{i+1/2,j+1/2,k} = \eta_{i+1/2,j+1/2,k} \left ( \frac{B^y_{i+1,j+1/2,k} - B^y_{i,j+1/2,k}}{\Delta x} - \frac{B^x_{i+1/2,j+1,k} - B^x_{i+1/2,j,k}}{\Delta y} \right)$ |
| 165 | 120 | |
| 166 | 121 | $\xi^z_{i+1/2,j+1/2,k} = \frac{B^x_{i+1/2,j,k} + B^x_{i+1/2,j+1,k}}{2} \left ( \frac{b^z_{i+1,j,k} + b^z_{i+1,j+1,k} - b^z_{i,j,k}-b^z_{i,j+1,k}}{2 \Delta x} - \frac{B^x_{i+1/2, j, k+1} + B^x_{i+1/2,j+1,k+1}-B^x_{i+1/2,j,k-1}-B^x_{i+1/2,j+1,k-1}}{4 \Delta z} \right ) $ |
