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The flux representation gives $B_R=-{\psi }_z/R$ and $B_z={\psi }_R/R$. Therefore $\nabla \cdot \mathbf{B}=0$ and $\mathbf{B}\cdot \nabla \psi =0$. With magnetohydrodynamic mass loading $k=k(\psi )$,Thus the steady continuity equation is satisfied.

The cross product in the ideal magnetohydrodynamic induction equation simplifies toHence steady induction requires $\mathbf{B}\cdot \nabla (v/R)=0$. On a regular connected poloidal flux surface, this integrates locally toThis is the field-line angular velocity of an axisymmetric wind. It need not equal the gas angular velocity when $B_{\varphi }\ne 0$, because $u_{\varphi }=R\omega +k{B}_{\varphi }/\rho$. The regular-surface qualification matters: if $\psi$ is constant and the field is purely toroidal, induction alone cannot turn an arbitrary $v(R,z)/R$ into a function of $\psi$.

Put $\mathbf{F}=(\nabla \times \mathbf{B})\times \mathbf{B}$ and $E=|\mathbf{u}{|}^2/2+\Phi$. Dot the pressure-free momentum equation with $\mathbf{u}$. Since $\mathbf{B}\cdot \mathbf{F}=0$, axisymmetry givesThe azimuthal momentum equation in cylindrical coordinates similarly givesSubtracting $\omega$ times the second relation from the first, and using $\mathbf{B}\cdot \nabla \omega =0$, proves that $E-\omega R{u}_{\varphi }$ is constant along a flowing regular poloidal flux surface. For nonzero mass loading its corotating energy invariant of an axisymmetric magnetic wind isThis is the invariant labeled $ϵ$ in the question. It is a corotating mechanical energy, distinct from the total magnetohydrodynamic Bernoulli invariant, which includes electromagnetic energy flux. In standard notation it equals total specific energy minus $\omega$ times the total specific angular-momentum invariant.

Near the footpoint, negligible $k{B}_{\varphi }/\rho$ gives $u_{\varphi }\simeq R\omega$, while the poloidal velocity is parallel to the poloidal magnetic field. The gas approximately corotates and slides along a rotating field line as a bead on a wire, with effective potentialA decreasing effective potential converts corotating potential energy into poloidal kinetic energy. This is the mechanism of magnetocentrifugal acceleration; local launching does not alone guarantee global escape.

For an attractive point mass, take $\Phi =-GM/(R^2+z^2{)}^{1/2}$ and ${\omega }^2=GM/R_0^3$. Parameterize the outward straight line by $R=R_0+s\sin \alpha$, $z=s\cos \alpha$. The first derivative of ${\Phi }_{\mathrm{eff}}$ vanishes at $s=0$ because the footpoint is in circular radial balance. The meridional Hessian there is ${\omega }^2\mathrm{diag}(-3,1)$, soThus the thirty-degree magnetocentrifugal launching criterion in its strict negative-curvature form isAbove this angle the equilibrium is unstable to an outward displacement; below it the near-footpoint potential initially rises. The source's proportionality constant must be negative to describe attraction.

The marginal straight-line launch at thirty degrees has zero quadratic curvature. For precisely the straight line and point-mass potential here, let $x=s/R_0$ and $a=\sin \alpha$. Direct expansion givesAt $\alpha =\pi /6$, this is $-7x^3/16+O(x^4)$, so the outward direction is downhill at cubic order. Consequently the strict inequality is the nondegenerate quadratic criterion; it should not be interpreted as excluding every one-sided nonlinear marginal launch at equality. An exactly resting bead remains an equilibrium until displaced.