Pressure in Incompressible Euler vs Pressure Incompressible Ideal MHD

Question

We know the incompressible 2D Euler equations are

$$U_t+U\cdot \nabla U=-\nabla P_1$$

where $P_1$ is the hydrodynamic pressure (never mind the incoppressibility condition), while the momentum equation for the 2D incompressible ideal MHD is

$$u_t+u\cdot\nabla u=-\nabla p+B\cdot\nabla B$$

where $p=P_2+\frac{1}{2}|B|^2$ is the total pressure, $P_2$ the hydrodynamic pressure, and $B$ the magnetic field.

It is known that solutions of MHD converge to solutions of Euler as $B\to 0$. My question is, are the hydrodynamic pressures $P_1$ and $P_2$ the same scalar function regardless of there being a magnetic field or not? In other words, if I subtract the two equations, would the $P_1$ and $P_2$ cancel out?

My instinct tells me that the hydrodynamic pressure, which is the pressure exerted by fluid particles on themselves, would be different if there is a magnetic field or not, unless the hydrodynamic pressure is some intrinsic property of the fluid that is invariant under external forces such as a magnetic field.

Thanks!

Answer 1

Using this we get that $$|B|^2=B∙B⇒∇(B∙B)=2B∙∇B+2B×(∇×B)$$ Substituting we get: $$u_t+u∙∇u=-∇P_2+B×(∇×B)$$ Now, I'm not to familiar with magnetism, but the term $B×(∇×B)$ need not be zero. Now a few words about how such an equation is solved. The pair $(u,P_2)$ is a solution determined by the initial condition and the forcing term, in this case that term is $B×(∇×B)$. If $B×(∇×B)=0$ the yes, $P_1=P_2$ but if not then in general $P_2≠P_1$