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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.

Put $\beta =2(q-1)>0$. With zero viscosity and $S=f(r)\Sigma$, the wind-loaded Keplerian disk evolution equation becomesOn an interval with $f>0$, introduce the inward travel-time coordinate and weighted densityPrimes on $x^{\prime }$ and $g^{\prime }$ denote the new variables, not differentiation. To verify the transformation, let $W=f{r}^{2-1/\beta }$ and $w=\beta rf$. Its logarithmic derivative satisfiesSince ${\partial }_{x^{\prime }}=w{\partial }_r$, multiplication of the density equation by W gives exactlyThis establishes inward transport by an angular-momentum-enhanced disk wind, including the amplitude weighting rather than only a characteristic speed.

The general local advection equation solution is $g^{\prime }=F(x^{\prime }+t)$; an initial profile fixes F on the interval swept out by its characteristics. A constant wave phase therefore obeys $d{x}^{\prime }/dt=-1$, orThus the transformed wave propagates inward. The physical surface density is not a rigid profile in r: its weighting accounts for radial compression and loss of mass into the outflow. The same velocity follows independently from the mass flux in part (a), $u_r=\mathcal{F}/(2\pi r\Sigma )=-\beta rf$. If f vanishes, the coordinate change applies only on the neighboring positive-loss intervals; at a loss-free point this wind-driven velocity vanishes.

A magnetocentrifugal acceleration mechanism can produce $q>1$. A poloidal field anchored to the disc transmits a torque to the outflow and enforces approximate corotation below its Alfvén surface. The wind's total specific angular momentum is ${\Omega }_0{r}_A^2$, where $r_A$ is its cylindrical Alfvén radius, compared with the footpoint value $h_0={\Omega }_0{r}_0^2$. Hence $q=(r_A/r_0{)}^2>1$ for a lever arm beyond the footpoint. Magnetic stresses remove the surplus angular momentum and drive accretion without an effective internal viscosity.

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.

For local cold magnetocentrifugal acceleration, the field must guide the gas and enforce approximate corotation with its disk footpoint in the sub-Alfvénic launching region. Let that footpoint be $R=R_f$, and keep its field-line angular velocity ${\Omega }_f=\Omega (R_f)$ constant along the field. In this rotating frame the relevant effective potential isThe footpoint is a critical point because ${\partial }_R\Phi (R_f,0)=R_f{\Omega }_f^2$ and ${\partial }_z\Phi (R_f,0)=0$. The meridional Hessian matrix there isTake distance $\ell$ along a smooth outward magnetic field line whose initial tangent is $(\sin \alpha ,\cos \alpha )$ in the $(R,z)$ plane. Since the gradient of $V$ vanishes at the footpoint, field-line curvature does not contribute to its second derivative. ConsequentlyA negative quadratic coefficient gives a downhill displacement and acceleration without thermal assistance. The magnetocentrifugal launching criterion in a flattened power-law potential is thereforeAngles are measured from the vertical, with $0\le \alpha <\pi /2$. Below the critical angle the footpoint is a local minimum along the field, so a cold particle cannot cross the initial potential barrier. At equality the quadratic test is marginal; higher derivatives and the unspecified field curvature can decide the outcome. The threshold is a local condition for launch, not proof that the wind escapes along an arbitrary global field geometry. The spherical point-mass limit $\lambda =1$, $\beta \to 1$ recovers ${\alpha }_{\mathrm{crit}}=30^{\circ }$, consistent with Ogilvie's discussion of cold disk winds, section 9.9.

Ideal flux freezing anchors a field line in the highly conducting disc, so its field-line angular velocity naturally equals the angular velocity of its footpoint. A circular orbit in the central potential is Keplerian:

Put $r=r_0+\delta r$ and expandThe linear radial term vanishes because $GM/r_0^2={\Omega }_0^2{r}_0$. The radial and vertical second derivatives are respectively $-3{\Omega }_0^2$ and ${\Omega }_0^2$, so

If the field line makes angle $\theta$ from the vertical, then locally $\delta r=z\tan \theta$. Its quadratic potential change iswhich is negative exactly whenThis is the local magnetocentrifugal acceleration launching criterion.