MHD equations
Poynting's theorem (electromagnetic energy) | \frac{\partial e}{\partial t} = - \nabla \cdot \mathbf{S} - \mathbf{J}\cdot \mathbf{E}
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Poynting vector | \mathbf{S} = \mathbf{E} \times \mathbf{B}
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Induction equation | \frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \mathbf{E}
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Ampere's equation (without Maxwell's correction) | \mathbf{J} = \nabla \times \mathbf{B}
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Lorentz Force Law | \frac{\partial \rho \mathbf{v}}{\partial t}=\mathbf{J} \times \mathbf{B}
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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
\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 ]
\frac{\partial e}{\partial t} = -\nabla \cdot \left ( \mathbf{E} \times \mathbf{B} \right) - \left ( \nabla \times \mathbf{B} \right) \cdot \mathbf{E}
\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right)
Ohm's Law
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}.
The \mathbf{E} field can be calculated using a 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}
Under under certain conditions, we can ignore various terms resulting in the following approximations
Resistive + Hall MHD | \mathbf{E} = -\mathbf{v} \times \mathbf{B} + \eta \mathbf{J} + \frac{1}{ne} \mathbf{J} \times \mathbf{B}
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Resistive MHD | \mathbf{E} = -\mathbf{v} \times \mathbf{B} + \eta \mathbf{J}
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Ideal MHD | \mathbf{E} = - \mathbf{v}\times \mathbf{B}
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Ideal MHD
For ideal MHD we substitute
\mathbf{E} = -\mathbf{v} \times \mathbf{B}
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)
\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]
\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right)
Resistive MHD
For resistive MHD, it will be convenient to define the resistive component of the electric field
\color{green}{\mathbf{E}_{\scriptscriptstyle R} = \eta \mathbf{J} = \eta \nabla \times \mathbf{B}}
Then our total electric field is given by
\mathbf{E} = -\mathbf{v} \times \mathbf{B} \color{green}{+ \mathbf{E}_{\scriptscriptstyle R}}
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}}.
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.
\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]
\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 )}
Note the resistive part of the induction equation
\frac{\partial \mathbf{B}}{\partial t} = \color{green}{- \nabla \times \left ( \eta \nabla \times \mathbf{B} \right )}
can be expanded as
\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}
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.
Hall+Resistive MHD
As in resistive MHD, it will be convenient to define the Hall component of the electric field
\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}}
Then our total electric field is given by
\mathbf{E} = -\mathbf{v} \times \mathbf{B} \color{green}{+ \mathbf{E}_{\scriptscriptstyle R}} \color{red}{+\mathbf{E}_{\scriptscriptstyle H}}
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.
\mathbf{J} \cdot \color{red}{\mathbf{E}_{\scriptscriptstyle H}} = \mathbf{J} \cdot \color{red}{\frac{1}{ne} \mathbf{J} \times \mathbf{B}} = 0
So as in the Resistive case, only the Poynting Flux term needs to be added to the energy equation.
\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 ]
\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]
\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}}
Discretization
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.
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
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)
For Hall MHD we need to calculate \xi = \left (\nabla \times \mathbf{B} \right) \times \mathbf{B} along with n at cell edges.
\xi_i = B_k d_l B_m \epsilon_{ijk} \epsilon_{jlm} = B_k d_l B_m \left(\delta_{kl} \delta_{im} - \delta_{km} \delta_{il} \right )
So we have
\xi_z = B_x \left ( d_x B_z - d_z B_x \right) + B_y \left ( d_y B_z - d_z B_y \right)
So we need
- edge centered estimates of the perpendicular fields (which can be found by averaging the two adjacent faces)
- transverse gradients of the parallel field (which can be found by differencing adjacent face centered averages of the parallel field (found by averaging the adjacent cell centered fields)
- parallel gradients of the transverse field
Using b to represent cell centered values, B to represent face centered values, we have
\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 )
+\frac{B^y_{i,j+1/2,k} + B^y_{i+1,j+1/2,k}}{2} \left ( \frac{b^z_{i,j+1,k} + b^z_{i+1,j+1,k} - b^z_{i,j,k}-b^z_{i+1,j,k}}{2 \Delta y} - \frac{B^y_{i, j+1/2, k+1} + B^y_{i+1,j+1/2,k+1}-B^y_{i,j+1/2,k-1}-B^y_{i+1,j+1/2,k-1}}{4 \Delta z} \right )
Now if we represent edge centered perpendicular B fields
\beta^x_{i+1/2,j+1/2,k} = \frac{B^x_{i+1/2,j,k}+B^x_{i+1/2,j+1,k}}{2}
and face centered parallel fields
\alpha^z_{i,j+1/2,k} = \frac{b^z_{i,j,k}+b^z_{i,j+1,k}}{2}
this simplifies to
\xi^z_{i+1/2,j+1/2,k} = \beta^x_{i+1/2,j+1/2,k} \left ( \frac{\alpha^z_{i+1,j+1/2,k} - \alpha^z_{i,j+1/2,k}}{\Delta x} - \frac{\beta^x_{i+1/2, j+1/2, k+1} -\beta^x_{i+1/2,j+1/2,k-1}}{2 \Delta z} \right )
+\beta^y_{i+1/2,j+1/2,k} \left ( \frac{\alpha^z_{i+1/2,j+1,k} - \alpha^z_{i+1/2,j,k}}{\Delta x} - \frac{\beta^y_{i+1/2, j+1/2, k+1} -\beta^y_{i+1/2,j+1/2,k-1}}{2 \Delta z} \right )
So to calculate 'z' edges we need adjacent face centered z fields and edge centered transverse fields
These fields won't be used for any of the other directional updates - so can be discarded after each dimension