Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 70 / 2

Axisymmetry makes all cylindrical components independent of $\varphi$. The divergence-free condition is therefore ${\partial }_R(R{B}_R)+{\partial }_z(R{B}_z)=0$. On a simply connected meridional patch introduce a streamfunction $\psi$ so thatThese definitions satisfy the solenoidal condition identically and represent any such poloidal field locally. Since $\nabla \varphi =e_{\varphi }/R$, they give $B_p=\nabla \psi \times \nabla \varphi$. The toroidal field is independent of this constraint, yielding the full representation $B=B_p+B_{\varphi }{e}_{\varphi }$. Global use of a single flux function assumes the corresponding meridional domain has no obstructing flux topology; a regular field extends to the axis by the usual limiting construction.

The axisymmetric magnetic flux function has a direct physical normalization. The flux through an annulus in a plane of constant $z$ is $\int 2\pi R{B}_{z}dR=2\pi \Delta \psi$. Thus $\psi$ is poloidal magnetic flux divided by $2\pi$, up to an arbitrary additive constant. Also $B\cdot \nabla \psi =0$, including the toroidal part, so its level surfaces are axisymmetric magnetic flux surfaces. Field lines remain on those surfaces even when they wind azimuthally.

Put $I=R{B}_{\varphi }$ and define ${\Delta }_{\ast }\psi ={\psi }_{RR}-R^{-1}{\psi }_R+{\psi }_{zz}$. Direct cylindrical differentiation givesThe azimuthal component of the Lorentz force is proportional to $I_z{\psi }_R-I_R{\psi }_z$. A force-free magnetic field therefore has parallel meridional gradients of $I$ and $\psi$, implying $I=f(\psi )$ on connected regular flux surfaces. ConsequentlyWith this dependence the poloidal part of $\nabla \times B$ is $f^{\prime }(\psi )B_p$. Its full cross product with $B$ isVanishing force therefore requires ${\Delta }_{\ast }\psi +f{f}^{\prime }=0$ wherever the flux gradient is nonzero, extending by regularity through ordinary isolated nulls. Finally, evaluating the three-dimensional cylindrical divergence gives $R^2\nabla \cdot (R^{-2}\nabla \psi )={\Delta }_{\ast }\psi$, soThis derives the Grad-Shafranov equation for a force-free magnetic field. The function $f$ is set by boundary/current information; it is not an additional fixed constant. Its single-valued global form presumes the usual connected flux-surface labeling. The equation also makes $\nabla \times B=f^{\prime }(\psi )B$, explicitly showing that current is parallel to the magnetic field.

For fixed volume define $E_B={\int }_V{B}^2/(2{\mu }_0)dV$ and $J=\nabla \times B/{\mu }_0$. Differentiate and use the ideal magnetohydrodynamic induction equation. The vector identitythen gives the general balanceThe volume term is the work done by the magnetic force on the fluid, with the opposite sign for magnetic energy. Under the force-free assumption of part (b), it vanishes, leaving the requested boundary expression. Without that assumption, or another reason for zero net Lorentz work, a boundary-only balance is not generally valid.

For an axisymmetric boundary its normal has no azimuthal component. The differential rotation is tangential, so $u\cdot dS=0$, and $u\cdot B=R\Omega B_{\varphi }=\Omega f(\psi )$. Therefore ${\dot{E}}_B={\mu }_0^{-1}{\oint }_S\Omega f(\psi )B\cdot dS$. A thin axisymmetric tube between neighboring flux labels carries flux $2\pi d\psi$. Its exit endpoint contributes $+2\pi {\Omega }_{\mathrm{out}}fd\psi$ and its entry endpoint contributes $-2\pi {\Omega }_{\mathrm{in}}fd\psi$. Pair the endpoints, counting each once, to obtainThe endpoint convention fixes the sign. Include each connected tube segment through $V$ once; a field line with no boundary crossing contributes nothing. This is magnetic-energy injection by differential boundary rotation. A common angular velocity at both ends produces no such injection.