Force-free, axisymmetric magnetic field in MHD

[ergospherical]

Homework Statement

Show that a force-free, axisymmetric field has ##B_{\phi} = f(\psi)/R## in cylindrical polars, where ##f## is an arbitrary function and ##\psi(R,z) = rA_{\phi}(R,z)## is the poloidal flux function, and find the equation satisfied by ##\psi##.

Relevant Equations

##-\nabla^2 \mathbf{B} = \lambda^2 \mathbf{B}##

Force free: ##\mathbf{J} \times \mathbf{B} \sim (\nabla \times \mathbf{B}) \times \mathbf{B} = 0##
(N.B. MHD applies so ##\epsilon_0|\partial \mathbf{E}/\partial t|/|\mathbf{J}| \ll 1##).

Axisymmetric: can write ##\mathbf{B} = \nabla \psi \times \nabla \phi + B_{\phi} \mathbf{e}_{\phi}##
(##\phi## is the azimuthal coordinate i.e. ##\nabla \phi = \mathbf{e}_{\phi}/r##)

Inserting into ##(\nabla \times \mathbf{B}) \times \mathbf{B} = 0## gives a bit of a mess. Is there an easier route?

 

[pasmith]

If the cross-product of two vectors is zero, then those vectors are scalar multiples of each other. Hence a force-free field satisfies [tex]\nabla \times \mathbf{B} = \alpha\mathbf{B}[/tex] for some scalar field [itex]\alpha[/itex], which (taking the divergence of the above with [itex]\nabla \cdot \mathbf{B} = 0[/itex]) must satisfy [tex]
\mathbf{B} \cdot \nabla \alpha = 0.[/tex] Mestel, Stellar Magnetism (2nd ed) at 58ff gives the equations satisfied by the poloidal and toroidal components of [itex]\mathbf{B}[/itex] in the axisymmetric case.