Astronomy:Chandrasekhar–Kendall function

Short description: Axisymmetric eigenfunctions

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 [math]\displaystyle{ \nabla\times\mathbf{H}=\lambda\mathbf{H} }[/math], where [math]\displaystyle{ \mathbf{H} }[/math] is the magnetic field and [math]\displaystyle{ \lambda }[/math] is the force-free parameter, with the assumption of divergence free field, [math]\displaystyle{ \nabla\cdot\mathbf{H}=0 }[/math], then the most general solution for the axisymmetric case is

[math]\displaystyle{ \mathbf{H} = \frac{1}{\lambda}\nabla\times(\nabla\times\psi\mathbf{\hat n}) + \nabla \times \psi \mathbf{\hat n} }[/math]

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

[math]\displaystyle{ \nabla^2\psi + \lambda^2\psi=0. }[/math]

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

Derivation

Taking curl of the equation [math]\displaystyle{ \nabla\times\mathbf{H}=\lambda\mathbf{H} }[/math] and using this same equation, we get

[math]\displaystyle{ \nabla\times(\nabla\times\mathbf{H}) = \lambda^2\mathbf{H} }[/math].

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

[math]\displaystyle{ \nabla^2\mathbf{H}+\lambda^2\mathbf{H}=0 }[/math].

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

[math]\displaystyle{ \mathbf{L} = \nabla\psi,\quad \mathbf{T} = \nabla\times\psi\mathbf{\hat n}, \quad \mathbf{S} = \frac{1}{\lambda}\nabla\times\mathbf{T} }[/math]

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

[math]\displaystyle{ \mathbf{H} = \frac{1}{\lambda}\nabla\times(\nabla\times\psi\mathbf{\hat n}) + \nabla \times \psi \mathbf{\hat n}. }[/math]

Cylindrical polar coordinates

Taking the unit vector in the [math]\displaystyle{ z }[/math] direction, i.e., [math]\displaystyle{ \mathbf{\hat n}=\mathbf{e}_z }[/math], with a periodicity [math]\displaystyle{ L }[/math] in the [math]\displaystyle{ z }[/math] direction with vanishing boundary conditions at [math]\displaystyle{ r=a }[/math], the solution is given by

[math]\displaystyle{ \psi = J_m(\mu_jr)e^{im\theta+ikz}, \quad \lambda =\pm(\mu_j^2+k^2)^{1/2} }[/math]

where [math]\displaystyle{ J_m }[/math] is the Bessel function, [math]\displaystyle{ k=\pm 2\pi n/L, \ n = 0,1,2,\ldots }[/math], the integers [math]\displaystyle{ m =0,\pm 1,\pm 2,\ldots }[/math] and [math]\displaystyle{ \mu_j }[/math] is determined by the boundary condition [math]\displaystyle{ a k\mu_j J_m'(\mu_j a)+m \lambda J_m(\mu_j a) =0. }[/math] The eigenvalues for [math]\displaystyle{ m=n=0 }[/math] has to be dealt separately. Since here [math]\displaystyle{ \mathbf{\hat n}=\mathbf{e}_z }[/math], we can think of [math]\displaystyle{ z }[/math] direction to be toroidal and [math]\displaystyle{ \theta }[/math] direction to be poloidal, consistent with the convention.

See also

• Poloidal–toroidal decomposition
• Woltjer's theorem