# Chandrasekhar–Kendall function

Chandrasekhar–Kendall functions are the eigenfunctions of the curl operator derived by Subrahmanyan Chandrasekhar and P. C. Kendall in 1957 while attempting to solve the force-free magnetic fields. The functions were independently derived by both, and the two decided to publish their findings in the same paper.
If the force-free magnetic field equation is written as $\nabla \times \mathbf {H} =\lambda \mathbf {H}$, where $\mathbf {H}$ is the magnetic field and $\lambda$ is the force-free parameter, with the assumption of divergence free field, $\nabla \cdot \mathbf {H} =0$, then the most general solution for the axisymmetric case is
    $\mathbf {H} ={\frac {1}{\lambda }}\nabla \times (\nabla \times \psi \mathbf {\hat {n}} )+\nabla \times \psi \mathbf {\hat {n}}$
where $\mathbf {\hat {n}}$ is a unit vector and the scalar function $\psi$ satisfies the Helmholtz equation, i.e.,
    $\nabla ^{2}\psi +\lambda ^{2}\psi =0.$
The same equation also appears in Beltrami flows from fluid dynamics where, the vorticity vector is parallel to the velocity vector, i.e., $\nabla \times \mathbf {v} =\lambda \mathbf {v}$.

## Derivation

Taking curl of the equation $\nabla \times \mathbf {H} =\lambda \mathbf {H}$ and using this same equation, we get
    $\nabla \times (\nabla \times \mathbf {H} )=\lambda ^{2}\mathbf {H}$.
In the vector identity $\nabla \times \left(\nabla \times \mathbf {H} \right)=\nabla (\nabla \cdot \mathbf {H} )-\nabla ^{2}\mathbf {H}$, we can set $\nabla \cdot \mathbf {H} =0$ since it is solenoidal, which leads to a vector Helmholtz equation,
    $\nabla ^{2}\mathbf {H} +\lambda ^{2}\mathbf {H} =0$.
Every solution of above equation is not the solution of original equation, but the converse is true. If $\psi$ is a scalar function which satisfies the equation $\nabla ^{2}\psi +\lambda ^{2}\psi =0$, then the three linearly independent solutions of the vector Helmholtz equation are given by
    $\mathbf {L} =\nabla \psi ,\quad \mathbf {T} =\nabla \times \psi \mathbf {\hat {n}} ,\quad \mathbf {S} ={\frac {1}{\lambda }}\nabla \times \mathbf {T}$
where $\mathbf {\hat {n}}$ is a fixed unit vector. Since $\nabla \times \mathbf {S} =\lambda \mathbf {T}$, it can be found that $\nabla \times (\mathbf {S} +\mathbf {T} )=\lambda (\mathbf {S} +\mathbf {T} )$. But this is same as the original equation, therefore $\mathbf {H} =\mathbf {S} +\mathbf {T}$, where $\mathbf {S}$ is the poloidal field and $\mathbf {T}$ is the toroidal field. Thus, substituting $\mathbf {T}$ in $\mathbf {S}$, we get the most general solution as
    $\mathbf {H} ={\frac {1}{\lambda }}\nabla \times (\nabla \times \psi \mathbf {\hat {n}} )+\nabla \times \psi \mathbf {\hat {n}} .$

## Cylindrical polar coordinates

Taking the unit vector in the $z$ direction, i.e., $\mathbf {\hat {n}} =\mathbf {e} _{z}$, with a periodicity $L$ in the $z$ direction with vanishing boundary conditions at $r=a$, the solution is given by
    $\psi =J_{m}(\mu _{j}r)e^{im\theta +ikz},\quad \lambda =\pm (\mu _{j}^{2}+k^{2})^{1/2}$
where $J_{m}$ is the Bessel function, $k=\pm 2\pi n/L,\ n=0,1,2,\ldots$, the integers $m=0,\pm 1,\pm 2,\ldots$ and $\mu _{j}$ is determined by the boundary condition $ak\mu _{j}J_{m}'(\mu _{j}a)+m\lambda J_{m}(\mu _{j}a)=0.$ The eigenvalues for $m=n=0$ has to be dealt separately. Since here $\mathbf {\hat {n}} =\mathbf {e} _{z}$, we can think of $z$ direction to be toroidal and $\theta$ direction to be poloidal, consistent with the convention.