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It is well know that planewaves are a complete basis for solutions to the wave equation. Let us assume a 2D space, and at fixed temporal frequency, the equation reduces to the Helmholtz equation. In cylindrical coordinates, the most appropriate solutions are the two kinds of Hankel functions, representing outgoing and incoming wave solutions. Actually, the Hankel functions should be multiplied by $e^{i m \theta}$ to produce cylindrical harmonics, which are a complete basis. My question is this:

If cylindrical harmonics are a complete basis, is there a closed form expression relating them to planewaves?

I know that 1st kind Bessel functions $J_m$ have a planewave decomposition by way of the Jacobi-Anger identity. However, a Hankel function's real part is a bessel function while its imaginary part is a 2nd kind Bessel function (Neumann function) with a singularity at the origin. I can't find an analogous expression for expression Neumann functions in terms of planewaves.

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    [Cross-posted to physics.SE](http://physics.stackexchange.com/questions/14888/hankel-function-in-terms-of-planewaves). Please point out when you cross-post; not doing so unnecessarily duplicates efforts.2011-09-20

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I once worked out a method to get the Neumann functions in terms of plane waves by physical reasoning. I think you know that if you take a uniform distribution of sine waves in all different directions in 2d space, in phase at the origin, you get the Bessel functions; the 0th order if they are in phase about the circle, the 1st order if they phase ranges over degrees around the circle, etc. Considering for a moment just the 0th order bessel function, you can map from this circular distribution into a 1-dimensional Fourier transform giving just the radial component of the wave. It's something like sqrt(x^2 - 1) with |x| < 1. (I don't remember what the actual distribution was but it had poles at +/-1.) Then if you take the analytical continuation, that is, the same function greater than 1, you get the Fourier transform of the Neumann function; that is, the radial component of it.

The question is: how do you show how the Neumann function is physically composed of sine waves? This is tricky, but here's how I did it. For electromagnetic waves, there are currents flowing in the boundary of two mediums. I divide space into two halves, on the right a vaccuum and on the left an ultra-slow dielectric (I call it the "dielectric glue") where light virtually crawls. I arrange for sine waves to arrive on the boundary from the left so they create currents on the boundary. Knowing the laws of refraction, I can calculate the currents and arrange by Fourier components so that there is a net resultant of just a single line of current on the boundary. I know that on the vacuum side, this must generate the Neumann function at a given frequency; in the zero-frequency limit this becomes the 1/r magnetic field of a wire with DC current.

The waves that make this field distribution in the vaccuum are the evanescent waves which result from total internal reflection of the ordinary waves inside the dielectric glue. By using the laws of refraction you get the right distribution of ordinary waves to give you the desired current distribution on the boundary. The resulting field in the vaccuum must be physically correct, and it is made up of the evanescent portion of the incident waves from the left hand side.

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    That's the beauty of it! the waves that get through aren't the ordinary plane waves, but the exponentially decaying tails. Those are exactly what you need to build up the Neumann functions.2011-09-23
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Please google before posting. The expansion you're looking for is derived in this article, which I found immediately as one of the first hits of a Google search for "hankel plane waves".

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    Thinking about the angular spectrum a bit more, it seems like it is related to Marty Green's answer about the radial distribution having poles at +/-1. Although the presence of homoegeneous waves in the cited paper does not agree with Marty's refraction argument. I was sort of hoping someone had already done the hard work for me :).2011-09-23