Final MHD equations

The induction equation is derived by eliminating E from Faraday's Law () and Ohm's Law (), using (), () and a vector identity:

$$\frac{\partial {\bf B}}{\partial t} = {\mbox{\boldmath$\nabla... ... \times B}) + \frac{1}{\mu_{0} \sigma} \nabla^{2} {\bf B} \, .$$ (3.43)

E has now been eliminated and $\rho = 0$, so Poisson's equation () does not contribute to the final set of equations. Equation () is effectively a boundary condition, since if $\nabla \cdot$ ${\bf B} =0$ initially, then taking the divergence of () implies that $\nabla \cdot$${\bf B}$ remains zero henceforth.

After making the above approximations, the final set of MHD equations are the induction equation and equations (), and () with $\rho = 0$:

$$\frac{\partial \eta}{\partial t} +{\mbox{\boldmath$\nabla \cdot$ }} (\eta {\bf U}) = 0 \, ,$$ (3.44)

$$\eta \left[\frac{\partial {\bf U}} {\partial t} + ({\bf U} {\... ... U} \right] = - \nabla p + {\bf J \times B} + \eta {\bf g}\, .$$ (3.45)

We now have one scalar and two vector equations in two scalar quantities ($\eta$, p) and two vector quantities (${\bf B}$, ${\bf U}$). We thus require one more scalar equation to close the set of equations. This can either be the energy conservation equation () or, as is commonly adopted, an equation of state for the fluid; in this case the adiabatic equation of state ().