Generalized Equations

We start with 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}

And then under certain conditions, we can ignore various terms yielding various approximations

Resistive + Hall MHD	\mathbf{E} = -\mathbf{v} \times \mathbf{B} + \eta \mathbf{J} + \frac{1}{ne} \mathbf{J} \times \mathbf{B}
Resistive MHD	\mathbf{E} = -\mathbf{v} \times \mathbf{B} + \eta \mathbf{J}
Ideal MHD	\mathbf{E} = - \mathbf{v}\times \mathbf{B}

Electromagnetic energy

We also need Poynting's theorem for the change in electro-magnetic energy

\frac{\partial e}{\partial t} = - \nabla \cdot \mathbf{S} - \mathbf{J}\cdot \mathbf{E}

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}

Induction equation

\frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \mathbf{E}

Ampere's equation

(without Maxwell's correction for time varying electric field)

\mathbf{J} = \nabla \times \mathbf{B} and

Lorentz force law

\frac{\partial \rho \mathbf{v}}{\partial t}=\mathbf{J} \times \mathbf{B}

Final set of equations

\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 B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right)

Ideal MHD

For ideal MHD we substitute

\mathbf{E} = -\mathbf{v} \times \mathbf{B}

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)

\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 ( \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 B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right)

Resistive MHD

\mathbf{E} = -\mathbf{v} \times \mathbf{B} + \eta \mathbf{J} = -\mathbf{v} \times \mathbf{E} + \eta \nabla \times \mathbf{B}

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)

\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} - \left ( \eta \nabla \times \mathbf{B} \right ) \times \mathbf{B} \right]

\frac{\partial B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) - \nabla \times \left ( \eta \nabla \times \mathbf{B} \right )

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.

Hall+Resistive MHD

\mathbf{E} = -\mathbf{v} \times \mathbf{B} + \eta \mathbf{J} + \frac{1}{ne} \mathbf{J} \times \mathbf{B}

\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} - \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]

\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)

We can also combine the Hall and Resistive terms

\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} - \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]

\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)

Or defining \mathbf{E}' = \eta \nabla \times \mathbf{B} + \frac{1}{ne} \left( \nabla \times \mathbf{B} \right) \times \mathbf{B}

\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} - \mathbf{E}' \times \mathbf{B} \right]

\frac{\partial B}{\partial t} = \nabla \times \left ( \mathbf{v} \times \mathbf{B} \right) - \nabla \times \mathbf{E}'

This along with

we arrive at

Ideal MHD	\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left [ \mathbf{v}\times \mathbf{B} \right ]
Resistive MHD	\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left [ \mathbf{v} \times \mathbf{B} - \frac{\mathbf{J}}{\sigma} \right ]
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 ]

In addition, Poynting's theorem states that the electromagnetic energy goes as

\frac{\partial u}{\partial t} = - \nabla \cdot \mathbf{S} - \mathbf{J}\cdot \mathbf{E}

\frac{\partial u}{\partial t} = - \nabla \cdot \left (\mathbf{E} \times \mathbf{B}\right) - \mathbf{J}\cdot \mathbf{E}

In Ideal MHD we have \mathbf{E} = \mathbf{v} \times \mathbf{B}

Now for Hall MHD, we can write the induction equation as

\frac{\partial \mathbf{B}}{\partial t} = \nabla \times \left [ \mathbf{v} \times \mathbf{B} - \mathbf{v}_\mbox{H} \times \mathbf{B} \right ]

where

\mathbf{v}_\mbox{H} = \frac{\mathbf{J}}{ne}

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.

Expanding just the Hall term further, we arrive at

\frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \left [ \frac{1}{ne} \left ( \nabla \times \mathbf{B} \right ) \times \mathbf{B} \right ]

which we can compare to the resistive mhd term

\frac{\partial \mathbf{B}}{\partial t} = -\nabla \times \left [ \left ( \eta \nabla \times \mathbf{B} \right ) \right ]

So it is likewise a

Discretization

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.

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

So steps are

\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