 Solution

Start in an inertial frame with the inviscid magnetohydrodynamic momentum equation,Here $\rho$ is mass density, $P$ is the isotropic fluid pressure, $\mathbf{v}$ is inertial-frame velocity, $\mathbf{J}$ is electric current density, $\mathbf{B}$ is the magnetic field, and $\Phi$ is gravitational potential per unit mass. In the low-frequency approximation to Ampère's law, $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$. The vector calculus identityseparates the Lorentz force density into magnetic tension and a magnetic pressure gradient. Transform the material acceleration to a rigid frame rotating with constant angular velocity ${\mathbf{\Omega }}_0$, using the stated frame transformation. The rotating-frame magnetohydrodynamic momentum equation isThe final terms are the Coriolis force and centrifugal force densities; $\mathbf{u}$ is the velocity relative to the rotating frame, and $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ uses that frame.

This requires a smooth conducting-fluid continuum with nonrelativistic motion, a constant magnetic permeability ${\mu }_0$, the solenoidal magnetic-field constraint, a negligible displacement current and negligible net electric-charge force. Viscous forces and anisotropic fluid stresses are omitted. A time-dependent frame would add an Euler force. Incompressible flow is not needed for this force equation. Finite electrical resistivity does not change its displayed instantaneous magnetic force; the later frozen-field calculation additionally uses ideal magnetohydrodynamics. A differentially rotating fluid does not define a single rigid frame with spatially varying ${\mathbf{\Omega }}_0$: choose a fixed local reference rate and retain the fluid's shear in $\mathbf{u}$.