Astronomy:Chandrasekhar-Kendall function

Chandrasekhar-Kendall functions are the axisymmetric eigenfunctions of the curl operator, derived by Subrahmanyan Chandrasekhar and P.C. Kendall in 1957, in attempting to solve the force-free magnetic fields. The results were independently derived by both, but were agreed to publish the paper together. If the force-free magnetic field equation is written as $\mathrm{\nabla }\times \mathbf{𝐇}=\lambda \mathbf{𝐇}$ with the assumption of divergence free field ($\mathrm{\nabla }\cdot \mathbf{𝐇}=0$), then the most general solution for axisymmetric case is

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

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

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

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

Derivation

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

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

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

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

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

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

where $\hat{\mathbf{n}}$ is a fixed unit vector. Since $\mathrm{\nabla }\times \mathbf{𝐒}=\lambda \mathbf{𝐓}$, it can be found that $\mathrm{\nabla }\times (\mathbf{𝐒}+\mathbf{𝐓})=\lambda (\mathbf{𝐒}+\mathbf{𝐓})$. 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

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

Cylindrical polar coordinates

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

$\psi =J_m({\mu }_{j}r)e^{im\theta +ikz},\lambda =\pm ({\mu }_j^2+k^2{)}^{1/2}$

where $J_m$ is the Bessel function, $k=\pm 2\pi n/L,n=0,1,2,...$, the integers $m=0,\pm 1,\pm 2,...$ and ${\mu }_j$ is determined by the boundary condition $ak{\mu }_j{J_m}^{\prime }({\mu }_{j}a)+m\lambda J_m({\mu }_{j}a)=0.$ The eigenvalues for $m=n=0$ has to be dealt separately. Since here $\hat{\mathbf{n}}={\mathbf{𝐞}}_z$, we can think of $z$ direction to be toroidal and $\theta$ direction to be poloidal, consistent with the convention.

See also

• Poloidal–toroidal decomposition
• Woltjer's theorem