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

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.