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 $$\nabla \times \mathbf{B} = \alpha\mathbf{B}$$ for some scalar field $\alpha$, which (taking the divergence of the above with $\nabla \cdot \mathbf{B} = 0$) must satisfy $$\mathbf{B} \cdot \nabla \alpha = 0.$$ Mestel, Stellar Magnetism (2nd ed) at 58ff gives the equations satisfied by the poloidal and toroidal components of $\mathbf{B}$ in the axisymmetric case.