# Plane wave expansion in cylindrical coordinates

I am trying to solve scattering problem in 2D and got to expand the wave function in cylindrical system which comes out to be Hankel function. Can you tell me how to expand the plane wave $\exp(i {\bf k}\cdot {\bf r})$ in terms of Hankel function, please?

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Please take a look at en.wikipedia.org/wiki/Fourier%E2%80%93Bessel_series. The method is always the same, "integrate and use the orthoganility relations" –  Jorge Nov 21 '12 at 16:43

The Hankel functions are not really the most natural way to get a cylindrical coordinates expansion for a plane wave, which is in terms of Bessel functions. (Why is this? it's because the Hankel functions are singular at the origin and plane waves are not.) You can then rephrase it in terms of Hankel functions if necessary. While this is of course an example of a Fourier-Bessel series, it is quite a simple one and it does not call upon any fancy result other than standard Fourier series.

To derive it, consider a plane wave along the $x$ axis, so $\mathbf{k}\cdot\mathbf{r}=kx=kr \cos(\theta)$, where $\theta\in[0,2\pi)$ is the positive angle from the $x$ axis. Then your plane wave, $$e^{i\mathbf{k}\cdot\mathbf{r}}=e^{ikr \cos(\theta)},$$ is a periodic function of $\theta$ (for fixed $r$) and can therefore be expanded as a Fourier series. Thus you can write $$e^{i\mathbf{k}\cdot\mathbf{r}}=e^{ikr \cos(\theta)}=\sum_{n=-\infty}^\infty c_n(kr)e^{in\theta},$$ where the coefficients, of course, depend on $r$.

All you need now, of course, is a good expression for these Fourier coefficients, and here standard Fourier series theory gives an unambiguous answer: they are given by the integral $$c_n(kr)=\frac{1}{2\pi}\int_0^{2\pi} e^{ikr \cos(\theta)}e^{-in\theta}\mathrm{d}\theta.$$ This is Bessel's first integral, (given e.g. in Jackson, p140, or in a similar form in DLMF 10.9.2), and it comes out essentially equal to a regular Bessel function: $$c_n(kr)=\frac{1}{2\pi}\int_0^{2\pi} e^{ikr \cos(\theta)}e^{-in\theta}\mathrm{d}\theta=i^n J_n(kr).$$ (Alternatively, I would usually choose this as a definition for the Bessel function, and get its other properties from it, but that's a matter of taste.)

All you need to do then is put it all together: $$e^{i\mathbf{k}\cdot\mathbf{r}}=\sum_{n=-\infty}^\infty i^n J_n(kr)e^{in\theta}.$$ For a general plane wave, you'll need to displace the angle $\theta$ to $\theta_r-\theta_k$, giving you the necessary factors in $e^{-in\theta_k}$ at each term.

A good question at this point is why did this come out in terms of Bessel functions (which represent standing radial waves), instead of Hankel functions, which describe travelling waves? You can always represent the standing Bessel function as a sum of two travelling Hankel functions, quite easily, as $J_n(kr)= \frac{1}{2} \left[ H_n^{(1)}(kr)+H_n^{(2)}(kr) \right]$. This means that you can express your plane wave as $$e^{i\mathbf{k}\cdot\mathbf{r}}=\sum_{n=-\infty}^\infty \frac{i^n}{2} e^{-in\theta_k} H_n^{(1)}(kr)e^{in\theta}+\sum_{n=-\infty}^\infty \frac{i^n}{2} e^{-in\theta_k} H_n^{(2)}(kr)e^{in\theta}.$$

This more complicated expression gives a plane wave as a superposition of both incoming and outgoing waves! Why? well, the plane wave $e^{ikx}$ is incoming at negative $x$, and it is outgoing at positive $x$, so obviously you need both types to describe it.

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