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.{{Cite journal|last=Chandrasekhar|first=Subrahmanyan|year=1956|title=On force-free magnetic fields|journal=Proceedings of the National Academy of Sciences|volume=42|issue=1|pages=1–5|doi=10.1073/pnas.42.1.1|pmid=16589804|pmc=534220|issn=0027-8424|doi-access=free|language=en}}{{Cite journal|last1=Chandrasekhar|first1=Subrahmanyan|last2=Kendall|first2=P. C.|date=September 1957|title=On Force-Free Magnetic Fields|journal=The Astrophysical Journal|language=en|volume=126|issue=1|pages=1–5 |doi=10.1086/146413|pmid=16589804|issn=0004-637X|bibcode=1957ApJ...126..457C|pmc=534220}} 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{{Cite journal|last1=Montgomery|first1=David|last2=Turner|first2=Leaf|last3=Vahala|first3=George|date=1978|title=Three-dimensional magnetohydrodynamic turbulence in cylindrical geometry|journal=Physics of Fluids|language=en|volume=21|issue=5|pages=757–764|doi=10.1063/1.862295}}{{Cite journal|last=Yoshida|first=Z.|date=1991-07-01|title=Discrete Eigenstates of Plasmas Described by the Chandrasekhar–Kendall Functions|journal=Progress of Theoretical Physics|language=en|volume=86|issue=1|pages=45–55|doi=10.1143/ptp/86.1.45|issn=0033-068X|doi-access=free}}

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

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 $a k\mu_j J_m'(\mu_j a)+m \lambda J_m(\mu_j a) =0.$ 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.

References

Category:Astrophysics

Category:Plasma theory and modeling

Chandrasekhar–Kendall function

Derivation

Cylindrical polar coordinates

See also

References