# Could the Cauchy momentum equation be analogous to ion current?

I'm trying to understand magnetohydrodynamics, in particular the Cauchy momentum equation

$$\rho \left({\frac {\partial }{\partial t}}+\mathbf {v} \cdot \nabla \right)\mathbf {v} =\mathbf {J} \times \mathbf {B} -\nabla p.}$$

Could this be analogous to an ion current? Could the other MHD equations be described in such a way?

Thanks

Additonal MHD questions: Is there a way to calculate the velocity field? Also, the magnetic pressure can be calculated using the kink instability?

Could this be analogous to an ion current?

Pure MHD is a single fluid theory, i.e., ions and electrons are not treated separately. In two-fluid MHD, yes, you do treat ions and electrons separately.

Could the other MHD equations be described in such a way?

I am not entirely sure what you are asking but if it's tied to the above question, then no, not unless you move to adapted versions of MHD that deal with two magnetized fluids, ions and electrons.

Is there a way to calculate the velocity field?

I am not sure I follow. You have the causal dynamical equation of motion written down already. That is the equation for the velocity field, is it not?

Also, the magnetic pressure can be calculated using the kink instability?

The magnetic pressure is derived from the $$\mathbf{J} \times \mathbf{B}$$ term. Since there are no electric fields in pure MHD, you can replace $$\mathbf{J}$$ with $$\nabla \times \mathbf{B}/\mu_{o}$$ then use some vector calculus identities to show that: $$\mathbf{J} \times \mathbf{B} = \frac{ \nabla \left( \mathbf{B} \cdot \mathbf{B} \right) }{ 2 \ \mu_{o} } - \frac{ \left( \mathbf{B} \cdot \nabla \right) \mathbf{B} }{ \mu_{o} } \tag{0}$$ The magnetic pressure is given by the first term on the right-hand side. The second term is referred to as magnetic tension, i.e., it is the term that resists "bending of the field" (fields are continuous constructs and don't physically bend like a metal bar, thus the quotes).