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The imposed constraint now fixes $\dot{\varphi }=\omega$, while a torque allows $h=r^2\omega$ to vary as the particle moves radially. The meridional equations become $\ddot{r}=-{\Phi }_r+{\omega }^{2}r$ and $\ddot{z}=-{\Phi }_z$. Hence $\boxed{{\Phi }_2=\Phi -\frac{1}{2}{\omega }^2{r}^2}$ is the appropriate effective potential. One must not substitute a variable h into the fixed-angular-momentum potential and then differentiate as though h were constant.

For a point mass and nonzero $\omega$, the stationary point is $(r_c,0)$ with $r_c=(GM/{\omega }^2{)}^{1/3}$. Its meridional Hessian matrix isIt is a saddle, at ${\Phi }_2=-3GM/(2r_c)$. Near it the contours are hyperbolas, with the saddle contour locally tangent to $z=\pm \sqrt{3}(r-r_c)$, as shown in the right panel. On the equatorial plane the potential tends downward both toward the point mass and at large cylindrical radius; at fixed r it increases as $|z|$ increases. The radial perturbation obeys $\delta \ddot{r}=3{\omega }^2\delta r$, while the vertical perturbation obeys $\delta \ddot{z}=-{\omega }^2\delta z$. Thus the radial motion has growing and decaying exponential solutions and the vertical motion is oscillatory. For $\omega =0$ no such stationary point exists.

This saddle is the basis of magnetocentrifugal acceleration. A sufficiently strong, open poloidal magnetic field can enforce approximate corotation with its disc footpoint while allowing motion along the field. If its outward tangent makes angle $\vartheta$ with the vertical, a small distance l along it has $\delta r=l\sin \vartheta$, $\delta z=l\cos \vartheta$. Its leading potential change at a Keplerian rotation footpoint isThe thirty-degree magnetocentrifugal launching criterion follows: cold local launching has a downhill quadratic direction when $\boxed{\vartheta >30^{\circ }}$ from the vertical. Equality is marginal at quadratic order and depends on higher terms and field-line geometry. The field transmits the torque needed to increase the material's angular momentum; a freely moving particle has no such constraint and instead sees the stable minimum in part (a). The local criterion does not by itself prove global escape or determine the mass loading of a complete magnetohydrodynamic wind.

Matter corotating with the footpoint has fixed angular velocity ${\Omega }_0=(GM/R_0^3{)}^{1/2}$. Its effective potential in that rotating frame is the sum of Newtonian gravitational potential and centrifugal potential,Along an outward-inclined straight magnetic field line, write $R=R_0+s\sin i$ and $z=s\cos i$, where $s$ is distance from the footpoint. The exact potential along it isIts first derivative at $s=0$ is zero by the Keplerian rotation balance. At the footpoint the meridional Hessian is $\mathrm{diag}(-3{\Omega }_0^2,{\Omega }_0^2)$ with zero mixed derivative. HenceWhen $0\le i<30^{\circ }$, the potential initially rises and a cold parcel must overcome a barrier. When $i>30^{\circ }$, its quadratic curvature is negative, so the footpoint is a local maximum along the field and a small outward displacement is accelerated further outward:This is the thirty-degree magnetocentrifugal launching criterion. It describes the lack of a local cold-launch barrier, not a nonzero force on a parcel exactly at rest at $s=0$. At exactly $30^{\circ }$ the quadratic term vanishes; for the specified straight line the next term is $-7{\Omega }_0^2{s}^3/(16R_0)$, as in marginal straight-line launch at thirty degrees. Global wind escape additionally depends on conditions farther along the line.

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.