Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 65 / 3

Use a steady axisymmetric flow governed by ideal magnetohydrodynamics and take the Newtonian gravitational potential to be axisymmetric. Work locally on connected regular magnetic flux surfaces with nonzero poloidal flow. As usual for a wind without an imposed toroidal loop voltage, $E_{\varphi }=0$; regularity on an included rotation axis guarantees this condition. It is needed to exclude steady cross-field flow in an annular domain.

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/R$, the poloidal magnetic flux function givesThe poloidal magnetic field is divergence-free. The zero toroidal electric field, with $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, gives ${\mathbf{u}}_p\parallel {\mathbf{B}}_p$. Define $k$ by $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$. Steady mass conservation then givesThus $k=k(\psi )$ is the magnetohydrodynamic mass loading, the mass flux per unit poloidal magnetic flux. Subtracting $(k/\rho )\mathbf{B}$ from the full velocity leaves only an azimuthal component. The cross product isThe steady ideal magnetohydrodynamic induction equation requires $0=\nabla \times (\omega \nabla \psi )=\nabla \omega \times \nabla \psi$. Consequently the field-line angular velocity is $\omega =\omega (\psi )$ and

Neither gas pressure nor axisymmetric gravity exerts an azimuthal torque. Multiplying the azimuthal momentum equation by $R$ combines the cylindrical curvature terms into angular-momentum transport:Insert $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$ and use ${\mathbf{B}}_p\cdot \nabla k=0$. It follows thatso the magnetohydrodynamic angular-momentum invariant isThe two terms account respectively for matter and Maxwell stress tensor transport of angular momentum.

Finally, subtracting $\gamma$ times the logarithmic mass density equation from the logarithmic pressure equation gives $D\ln (p/{\rho }^{\gamma })/Dt=0$. The specific entropy of a fixed-composition ideal gas is a function of $p/{\rho }^{\gamma }$. Steadiness and axisymmetry reduce its advection to ${\mathbf{u}}_p\cdot \nabla s=0$. HenceThese flux-label functions are local on connected regular surfaces; disconnected components with the same numerical label need not share constants without an additional matching condition.

On one poloidal flux surface, the field-line angular velocity $\omega (\psi )$ is constant. In a frame rotating at that rate the relative velocity is $\mathbf{v}=\mathbf{u}-R\omega {\mathbf{e}}_{\varphi }=k\mathbf{B}/\rho$, parallel to the magnetic field. The Lorentz force is perpendicular to this motion, and the Coriolis force also does no work. Centrifugal acceleration comes from the effective potential $-R^2{\omega }^2/2$. Since entropy is constant along the surface, pressure work integrates to specific enthalpy $w$, with $dw=dp/\rho$ along the flow.

Therefore kinetic energy relative to the rotating poloidal flux surface, enthalpy and effective gravitational-plus-centrifugal potential have a constant sum:This is a corotating energy integral, not the laboratory energy alone: Maxwell stress tensor stresses can transfer energy and angular momentum in the laboratory frame. In fact the expression equals $|\mathbf{u}{|}^2/2+w+\Phi -\omega R{u}_{\varphi }$, namely the total magnetohydrodynamic Bernoulli invariant minus $\omega \ell$. Differential rotation between surfaces does not invalidate the argument, because the chosen angular velocity is fixed along the particular surface being followed.

All invariants in this part refer to the selected magnetic field line. Its squared poloidal Alfvén number isAt the Alfvén surface crossing it equals unity, so ${\rho }_a={\mu }_0{k}^2$ and $M_{Ap}^2=1/y$. Subtract the two azimuthal integrals in part (a) to findFinite azimuthal fields and velocity at $y=1$ require the Alfvén-surface regularity condition for an axisymmetric wind $\ell =\omega R_a^2$. HenceHere $v_{\varphi }$ denotes the azimuthal velocity relative to field-line rotation. For gravity of a point mass at $z=0$, $\Phi =-GM/R=-GM/(R_{a}x)$.

Insert these quantities in the corotating energy integral with $w=0$, and divide by the gravitational energy scale $GM/R_a$. This gives the cold radial magnetohydrodynamic wind integral,withBoth are dimensionless. The apparent $0/0$ at $x=y=1$ must be evaluated on the smooth solution branch, not assigned an arbitrary value.

A constant-energy curve obeys $f_x+f_{y}dy/dx=0$. At a nondegenerate magnetosonic point, the vanishing coefficient of the mass density derivative requires simultaneous vanishing of its numerator:The curve must admit a real crossing slope. Expanding about such a point gives $f_{yy}{m}^2+2f_{xy}m+f_{xx}=0$, $m=dy/dx$, so a regular saddle point of a scalar function crossing has $f_{xy}^2-f_{xx}{f}_{yy}>0$. These are the form of the slow and fast regularity conditions in a warm wind, with the enthalpy contribution retained in $f$.

In the strictly cold approximation used to obtain the displayed $f$, the slow speed is zero. There is a cold-limit degeneracy of a slow magnetosonic point: a finite-speed interior slow crossing need not survive. Indeed,so this cold equation has no ordinary sub-Alfvénic slow critical point. A small retained enthalpy is needed to locate that point, which can approach the launching boundary as the sound speed tends to zero. The fast critical point still obeys the simultaneous conditions above wherever the reduced expression is nonsingular.