= Solution
The divergence-free poloidal field can be represented by the <poloidal magnetic flux function>
$$
\mathbf B_p=\nabla\psi\times\nabla\phi
=-\frac1r\mathbf e_\phi\times\nabla\psi.
$$
In a steady axisymmetric <ideal magnetohydrodynamics> flow, the azimuthal component of $\nabla\times(\mathbf u\times\mathbf B)=0$ makes $\mathbf u_p$ parallel to $\mathbf B_p$. Write
$$
\boxed{\rho\mathbf u_p=k\mathbf B_p}.
$$
Mass conservation and $\nabla\cdot\mathbf B=0$ then imply $\mathbf B\cdot\nabla k=0$, so the <magnetohydrodynamic mass loading> $k=k(\psi)$ is constant along each magnetic line.
The poloidal part of $\mathbf u\times\mathbf B$ is
$$
\mathbf u\times\mathbf B
=\frac1r\left(u_\phi-\frac{kB_\phi}{\rho}\right)\nabla\psi.
$$
Its curl vanishes only if its coefficient is a flux function, giving the <field-line angular velocity>
$$
\boxed{\frac{u_\phi}{r}-\frac{kB_\phi}{r\rho}=\omega(\psi)}.
$$
The azimuthal component of momentum conservation is a divergence of matter and magnetic angular-momentum flux. Dividing its field-line constant by the mass loading yields the <magnetohydrodynamic angular-momentum invariant>
$$
\boxed{ru_\phi-\frac{rB_\phi}{\mu_0k}=\ell(\psi)}.
$$
The conservative total-energy equation similarly gives the <magnetohydrodynamic Bernoulli invariant>
$$
\boxed{\frac12|\mathbf u|^2+\Phi+h
-\frac{r\omega B_\phi}{\mu_0k}=\epsilon(\psi)}.
$$
Finally, the <entropy advection equation> and $\mathbf u_p\parallel\mathbf B_p$ imply $s=s(\psi)$. Thus $k,\omega,\ell,\epsilon$, and $s$ are constant along each magnetic field line.
Back to article page
