Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 64 / 2

For the isothermal equation of state, $p=c_s^2\rho$, with constant isothermal sound speed $c_s$. A nonzero steady plane-parallel flow has constant mass flux $\rho u=j$. Choose the positive flow direction for convenience. The continuity equation then gives ${\rho }^{\prime }/\rho =-u^{\prime }/u$, and the steady Euler momentum equation becomesIntegrating and absorbing the reference velocity into the constant yields, for the Mach number $\mathcal{M}=u/c_s$,For the opposite flow direction use $\mathcal{M}=|u|/c_s$; the same relation holds.

A smooth sonic point has $u=c_s$, so its momentum equation requires ${\Phi }^{\prime }=0$. Moreover $f(\mathcal{M})={\mathcal{M}}^2/2-\ln \mathcal{M}$ satisfies $f^{\prime }(1)=0$, $f^{\prime \prime }(1)=2$ and has its minimum $f(1)=1/2$. Thus, if the flow reaches $\mathcal{M}=1$ at $x_{\ast }$,A sonic crossing is therefore possible at a maximum of the potential, withAt a nondegenerate maximum, the expansions give $(\mathcal{M}-1{)}^2=-{\Phi }^{\prime \prime }(x_{\ast })(x-x_{\ast }{)}^2/(2c_s^2)+\cdots$. Choosing either smooth sign gives a crossing slope ${\mathcal{M}}^{\prime }(x_{\ast })=\pm \sqrt{-{\Phi }^{\prime \prime }(x_{\ast })/(2c_s^2)}$, joining a subsonic flow to a supersonic flow, or the reverse. This establishes the plane-parallel isothermal sonic transition at a potential maximum. For a globally defined crossing, the same maximum must also be high enough for the Bernoulli relation to remain real over the intended domain.

Decompose $\mathbf{u}={\mathbf{u}}_p+u_{\varphi }{\mathbf{e}}_{\varphi }$ and $\mathbf{B}={\mathbf{B}}_p+B_{\varphi }{\mathbf{e}}_{\varphi }$. In a steady axisymmetric vector field, $\nabla \times \mathbf{E}=0$ and $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. The radial and vertical components of its curl areTherefore $R{E}_{\varphi }=C$ is a constant. Regularity at the symmetry axis sets $C=0$; equivalently, one imposes no toroidal loop electromotive force, $\oint \mathbf{E}\cdot d\mathbf{l}=2\pi C=0$. Under this usual condition,which implieswhere ${\mathbf{B}}_p\ne 0$. This is the regular-axis alignment of steady poloidal ideal flow. It allows nonzero toroidal components as well.

The axis or zero-loop-voltage condition is necessary for the unqualified conclusion. On an annular domain, $\mathbf{B}=(B_0/R){\mathbf{e}}_R$ and $\mathbf{u}=U_0{\mathbf{e}}_z$ give $\mathbf{E}=-(U_0{B}_0/R){\mathbf{e}}_{\varphi }$ with zero curl, yet the poloidal vectors are perpendicular. This can satisfy the remaining steady equations as well: choose constant mass density ${\rho }_0$, $\Phi =\pi G{\rho }_0{R}^2$ and $p=p_0-\pi G{\rho }_0^2{R}^2>0$ on the annulus. The field has zero curl, radial pressure balances gravity, the flow advects no equilibrium quantity, and ${\nabla }^2\Phi =4\pi G{\rho }_0$. It carries a nonzero loop voltage. Steady axisymmetry alone on an annulus therefore does not exclude such cross-field flow.

Use the aligned poloidal flow from part (i), and define the magnetohydrodynamic mass loading $\eta =\rho \alpha$, so $\rho {\mathbf{u}}_p=\eta {\mathbf{B}}_p$. Axisymmetric mass conservation and $\nabla \cdot {\mathbf{B}}_p=0$ give ${\mathbf{B}}_p\cdot \nabla \eta =0$. Introduce the poloidal magnetic flux function $\Psi$ by ${\mathbf{B}}_p=R^{-1}\nabla \Psi \times {\mathbf{e}}_{\varphi }$, and write $u_{\varphi }=R\Omega$. ThenThe steady induction equation says the curl of this expression is zero. Hence $\nabla {\Omega }_F\times \nabla \Psi =0$, wherethe field-line angular velocity ${\Omega }_F$ is constant on each connected magnetic surface and along its field lines.

To control the difference between matter rotation and field-line rotation, use the azimuthal component of the ideal magnetohydrodynamic momentum equation. The inertial and magnetic terms are respectively $\rho R^{-1}{\mathbf{u}}_p\cdot \nabla (R{u}_{\varphi })$ and $({\mu }_{0}R{)}^{-1}{\mathbf{B}}_p\cdot \nabla (R{B}_{\varphi })$; pressure and gravity have no azimuthal derivative. Multiplying by $R$ and using constant mass loading givesThus the magnetohydrodynamic angular-momentum invariant $L=R^2\Omega -R{B}_{\varphi }/({\mu }_0\eta )$ is constant along the line when $\eta \ne 0$. Let $v_{Ap}=|{\mathbf{B}}_p|/\sqrt{{\mu }_0\rho }$ and define $M_{Ap}^2=|{\mathbf{u}}_p{|}^2/v_{Ap}^2={\mu }_0{\eta }^2/\rho$. Eliminating $B_{\varphi }$ between the two invariants givesFor a highly sub-Alfvénic poloidal flow, $M_{Ap}^2\ll 1$, this is $\Omega \simeq {\Omega }_F$ along a nonsingular field-line segment. This proves sub-Alfvénic isorotation in a steady axisymmetric wind, with the correction explicitly displayed. If the poloidal flow is exactly zero, the induction equation directly gives ${\mathbf{B}}_p\cdot \nabla \Omega =0$, namely Ferraro's law of isorotation. The approximate claim presumes regular invariants: small $M_{Ap}$ alone is not a uniform relative bound if $L/R^2$ is allowed to become arbitrarily large.

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.