Chandrasekhar–Kendall function

If the force-free magnetic field equation is written as ${\displaystyle \mathrm{\nabla }\times \mathbf{H}=\lambda \mathbf{H}}$, where ${\displaystyle \mathbf{H}}$ is the magnetic field and ${\displaystyle \lambda }$ is the force-free parameter, with the assumption of divergence free field, ${\displaystyle \mathrm{\nabla }\cdot \mathbf{H}=0}$, then the most general solution for the axisymmetric case is

${\displaystyle \mathbf{H}=\frac{1}{\lambda }\mathrm{\nabla }\times (\mathrm{\nabla }\times \psi \hat{\mathbf{n}})+\mathrm{\nabla }\times \psi \hat{\mathbf{n}}}$

where ${\displaystyle \hat{\mathbf{n}}}$ is a unit vector and the scalar function ${\displaystyle \psi }$ satisfies the Helmholtz equation, i.e.,

${\displaystyle {\mathrm{\nabla }}^2\psi +{\lambda }^2\psi =0.}$

The same equation also appears in Beltrami flows from fluid dynamics where, the vorticity vector is parallel to the velocity vector, i.e., ${\displaystyle \mathrm{\nabla }\times \mathbf{v}=\lambda \mathbf{v}}$.

Derivation

Taking curl of the equation ${\displaystyle \mathrm{\nabla }\times \mathbf{H}=\lambda \mathbf{H}}$ and using this same equation, we get

${\displaystyle \mathrm{\nabla }\times (\mathrm{\nabla }\times \mathbf{H})={\lambda }^2\mathbf{H}}$.

In the vector identity ${\displaystyle \mathrm{\nabla }\times \left(\mathrm{\nabla }\times \mathbf{H}\right)=\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{H})-{\mathrm{\nabla }}^2\mathbf{H}}$, we can set ${\displaystyle \mathrm{\nabla }\cdot \mathbf{H}=0}$ since it is solenoidal, which leads to a vector Helmholtz equation,

${\displaystyle {\mathrm{\nabla }}^2\mathbf{H}+{\lambda }^2\mathbf{H}=0}$.

Every solution of above equation is not the solution of original equation, but the converse is true. If ${\displaystyle \psi }$ is a scalar function which satisfies the equation ${\displaystyle {\mathrm{\nabla }}^2\psi +{\lambda }^2\psi =0}$, then the three linearly independent solutions of the vector Helmholtz equation are given by

${\displaystyle \mathbf{L}=\mathrm{\nabla }\psi ,\mathbf{T}=\mathrm{\nabla }\times \psi \hat{\mathbf{n}},\mathbf{S}=\frac{1}{\lambda }\mathrm{\nabla }\times \mathbf{T}}$

where ${\displaystyle \hat{\mathbf{n}}}$ is a fixed unit vector. Since ${\displaystyle \mathrm{\nabla }\times \mathbf{S}=\lambda \mathbf{T}}$, it can be found that ${\displaystyle \mathrm{\nabla }\times (\mathbf{S}+\mathbf{T})=\lambda (\mathbf{S}+\mathbf{T})}$. But this is same as the original equation, therefore ${\displaystyle \mathbf{H}=\mathbf{S}+\mathbf{T}}$, where ${\displaystyle \mathbf{S}}$ is the poloidal field and ${\displaystyle \mathbf{T}}$ is the toroidal field. Thus, substituting ${\displaystyle \mathbf{T}}$ in ${\displaystyle \mathbf{S}}$, we get the most general solution as

${\displaystyle \mathbf{H}=\frac{1}{\lambda }\mathrm{\nabla }\times (\mathrm{\nabla }\times \psi \hat{\mathbf{n}})+\mathrm{\nabla }\times \psi \hat{\mathbf{n}}.}$