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I understand that; in cylindrical coordinates, the basic solutions of the Helmholtz equation are of the form Hankel function of integer order times a complex exponential term ($E=H_{n}(kr)e^{in\alpha}$). However, In some literature book page:481, the plane wave expansion in cylindrical functions is expressed as a summation of the of Hankel/Bessel function. My question is, can I consider just one order (say order one of Hankel function) and get the expression of $E=H_{1}(kr)e^{i\alpha}$ or do I need to consider the summation of all the orders of the Hankel function and obtain; $E=\sum_{n} H_{n}(kr)e^{in\alpha}$

Which method is accurate?

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    $\begingroup$ It has always been my understanding that we only take solutions of the first kind since they are finite at the origin while the second kind are divergent. $\endgroup$ – AngusTheMan Jul 6 '15 at 10:22
  • $\begingroup$ @AngusTheMan Thanks for the response. 'kind' and the 'order' are two difefrent things. Here, I consider the 'first kind' (sorry I could not mention it). So that two expressions I need to clarify should be $E=H_{1}^{1}(kr)e^{i\alpha}$ and $E=\sum_{n} H_{n}^{1}(kr)e^{in\alpha}$ $\endgroup$ – QuantumGirl Jul 6 '15 at 10:26
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    $\begingroup$ Ah I see, well now you have me curious too! :) $\endgroup$ – AngusTheMan Jul 6 '15 at 10:29
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In general you need both kinds of functions $H_n^{(1)}(k\,r)\,e^{i\,n\,\phi} = (J_n(k\,r) + i\,Y_n(k\,r))\,e^{i\,n\,\phi}$ and $H_n^{(2)}(k\,r)\,e^{i\,n\,\phi} = (J_n(k\,r) - i\,Y_n(k\,r))\,e^{i\,n\,\phi}$ in any superposition, where $(r\,\phi)$ are the polar co-ordinates and $k$ the wavenumber. If the homogeneous region in question contains the point $r=0$ and you can be sure that there are no singularities at the origin (which is almost always, but see later on), then the superposition weights of the corresponding kinds of Hankel function must be the same: they must arise in pairs of the form $\alpha\,(H_n^{(1)}(k\,r) + H_n^{(2)}(k\,r)))\,e^{i\,n\,\phi} = \alpha\,J_n(k\,r) \,e^{i\,n\,\phi}$ because the Bessel functions of the second kind (Neumann functions) have branch points at the origin (and the spherical Hankel functions of order $n$ , which are the normal ones divided by $\sqrt{k\,r}$ have poles of order $|n|+1$).

If you're working with a field in a homogeneous cylindrical shell which doesn't include $r=0$ then arbitrary superpositions of all the functions are possible.

Now you can just as well say that your solutions are superpositions of the functions $J_n(k\,r)\,e^{i\,n\,\phi}$ and $Y_n(k\,r)\,e^{i\,n\,\phi}$, with the $Y_n$ terms absent in any homogeneous region including $r=0$ and everything will be equivalent. The reason that people work with Hankel functions instead is their neat asymptotic behavior:

\begin{align} H_\alpha^{(1)}(z) &\sim \sqrt{\frac{2}{\pi z}}\exp\left(i\left(z-\frac{\alpha\pi}{2}-\frac{\pi}{4}\right)\right) &&\text{ for } -\pi<\arg z<2\pi \\ H_\alpha^{(2)}(z) &\sim \sqrt{\frac{2}{\pi z}}\exp\left(-i\left(z-\frac{\alpha\pi}{2}-\frac{\pi}{4}\right)\right) && \text{ for } -2\pi<\arg z<\pi \end{align}

and so they decompose an EM field into inward propagating and outward propagating cylindrical waves, so they are especially useful for studying antenna problems with cylindrical symmetry, or the radiation shed from an cylindrically symmetric optical fiber, for example. If you know you've only got radiation going outwards, you use only the $H_n^{(1)}(k\,r)\,e^{i\,n\,\phi}$ functions in the superposition. If you know you've only got radiation coming inwards, you use only the $H_n^{(2)}(k\,r)\,e^{i\,n\,\phi}$ functions (see footnote).

But if you're interested in plane wave expansions, you include the point $r=0$ and there are no singularities in a plane wave, so that only the $J_n$ terms can be present. Indeed, the plane wave expansion follows from the generating function for the sequence $\{J_n\}_{n=-\infty}^\infty$ or the Jacobi-Anger expansion:

$$ e^{i z \cos \theta} = \sum\limits_{n=-\infty}^{\infty} i^n\, J_n(z)\, e^{i n \theta}$$

(the generating function formula is $e^{i \,z\,(t-t^{-1})}=\sum\limits_{n=-\infty}^\infty J_n(z)\,t^n$). Putting $z\,\cos \theta=k\,r\,\cos \theta=\vec{k}\cdot\vec{r}$ then yields the plane wave expansion:

$$e^{i\,\vec{k}\cdot\vec{r}}=\sum\limits_{n=-\infty}^{\infty} i^n\, J_n(k\,r)\, e^{i\, n\, \theta}$$


Footnote: Antenna Problems

Sometimes it is physically reasonable to have the singularity at $r=0$, at least at the scale you consider the problem. If, for example, you model an antenna by a thread current with constant phase along the thread's length, you think of the solution of the Helmholtz equation as $H_n^{(1)}(k\,r)\,e^{i\,n\,\phi} = (J_n(k\,r) + i\,Y_n(k\,r))\,e^{i\,n\,\phi}$ for a radiating antenna and $H_n^{(1)}(k\,r)\,e^{i\,n\,\phi} = (J_n(k\,r)- i\,Y_n(k\,r))\,e^{i\,n\,\phi}$ for a receiving antenna. The singularity arises from the idealization of a thread current, with zero diameter. This kind of calculation you would use to model the antenna's transmission and so forth. But if you need to design the conductor to carry the required current, or if you need to work out the radiation impedance of the system (whose reactive component diverges to infinity for a thread current), then you're going to break the problem up into two regions: one outside the wire modelling the radiation and containing only $H_n^{(1)}(k\,r)\,e^{i\,n\,\phi}$ terms (if a radiating antenna) and one inside the wire, where there are only functions of the kind $J_n$. Actually these latter have a very different, complex wavenumber to account for the conductor, but the singularities are still the same for complex arguments, so there are only $J_n$ terms present.

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  • $\begingroup$ Thank you soo much for the detailed answer. I got your points. I was previously curious about the singularity at $r=0$, and now it is clarified. However, still this thing bothers me. Say, I need to model outward propagating cylindrical waves(E). Is it fair enough to consider only one order (If I take first order) $E=H_{1}^{1}(kr)e^{i\theta}$ or the summation of all the orders $E=\sum_{n} H_{n}^{1}(kr)e^{in\theta}$? Thanks algain $\endgroup$ – QuantumGirl Jul 7 '15 at 5:53

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