Consider electromagnetic cylindrical waves. Cylindrical waves can be derived from the plane waves using energy conservation consideration: since the power must be a constant the amplitude of a cylindrical wave must decrease with $\sqrt{r}$. Therefore a cylindrical wave expression must be

$$\mathbf{E}(r,t)=\frac{\mathbf{E}_0}{\sqrt{r}} \mathrm{sin}(kr-\omega t)$$

The function $\sqrt{r} \mathbf{E}(r,t)$ satisfies one dimensional wave equation

$$\frac{\partial^2\xi}{\partial r^2}-\frac{1}{c^2}\frac{\partial^2\xi}{\partial t^2}=0$$

In complex notation the cylindrical wave becomes $$\mathbf{E}(r,t)=\frac{\mathbf{E}_0}{\sqrt{r}} e^{j(kr-\omega t)}\tag{1}$$

If we call $\xi$ a generic component of $\mathbf{E}$, the three dimensional wave equation is

$$\nabla^2\xi-\frac{1}{c^2}\frac{\partial^2\xi}{\partial t^2}=\square \xi=0$$

The solution in cylindrical coordinates is

$$\xi (r,\phi , z,t) =\sum_{\omega,n,h} R^{0}_{\omega, n, h } H_n\Bigg(r \sqrt{\frac{\omega^2}{c^2}-h^2}\Bigg) e^{j(n\phi +hz-\omega t)} \tag{2}$$

Where $R^{0}_{\omega, n, h }$ is a (complex) constant and $H_n$ is the Hankel function of order $n$.

Under the assumption of cylindrical symmetry of the wave, that is

$$\frac{\partial \xi}{\partial \phi}=0 \,\,\,\,\,\, \mathrm{and} \,\,\,\,\,\, \frac{\partial \xi}{\partial z}=0$$ the asymptotic approximation of $(2)$ (for $r >> \frac{c}{\omega}$) lead to a field that is the same as $(1)$.

My question is: why (under cylindrical symmetry) is $(2)$ equal to $(1)$ only at large distances?

I always thought that $(1)$ gives the expression of a cylindrical wave in all the circumstances. So is $(1)$ "wrong" for small $r$? Or are $(1)$ and $(2)$ describing two different things? If so, what are the differences?

(I have an identical doubt for spherical waves).

  • $\begingroup$ Note that you're mixing conventions in this question: if you use $j$ for $\sqrt{-1}$, then you should write $e^{j(\omega t-kx)}$ as the engineers do; if you must have a positive sign on the $kx$, use $i$ instead. These are engineer vs physicist conventions and you break them at your own risk (it can and will cause you pain down the line), and you shouldn't break them in a place that puts others in a position of being required to use your notation. $\endgroup$ Commented Aug 19, 2017 at 5:07

3 Answers 3


Energy conservation alone is not enough to get exact solution for a cylindrical wave equation. You get the correct asymoptotic solution,

$$\mathbf{E}(r,t)\sim\frac{\mathbf{E}_0}{\sqrt{r}} \mathrm{sin}(kr-\omega t)\;\text{ as }\;r\to\infty,$$

but it's only that — asymptotic as $r\to\infty$, and invalid for $r\to0$.

To see what happens, consider the curvature of the wave front far from the origin and near it. You'll see that far from the origin, indeed, the wavefront is close to being flat, so you can approximate the wave function with a (fading) sinusoid. But near the origin the wavefront is quite curved, and its curvature becomes infinite at the origin. Clearly something must become different there.

It's important to understand that whatever coordinates you choose to solve the wave equation, any solution to it still remains the solution — provided you're interested only in the domain where the solution is nonsingular. So, for example, the function

$$f(x,y,z,t)=J_0\left(k\sqrt{x^2+y^2}\right)e^{i\omega t}$$

still solves the 3D wave equation, as does the function

$$g(x,y,z,t)=\sin(kx)e^{i\omega t}$$

and many others.

What distinguishes the solutions represented in terms of Bessel/Neumann/Hankel functions is their particular behavior on rotation around the origin: such solutions are eigenfunctions of rotation operator.

How do you convert your $\cos$-solution to a Bessel-function one? Since we want the solution to be an eigenfunction of rotation operator, (we'll consider the one invariant under rotation for simplicity), one of the ways is to integrate over all directions. Here's an example for $0\text{th}$ order Bessel function:

$$J_0(r)=\frac1{2\pi}\int\limits_0^{2\pi} \cos(r\cos\varphi)\,\mathrm d\varphi.$$

Here the interference of all the rotated cosines automatically gives you both: fading with $r\to\infty$ to satisfy energy conservation, and changes in wavelength for $r\to0$ to account for "lumping" of the waves near the origin.

  • $\begingroup$ If I may ask, I do not see why energy considerations give an epression for $\mathbf{E}(r,t)$ that is not valid near the axis: what do you mean by the "curvature" of the wavefront, mathematically? It is very reasonable that near the source (e.g. a infinitely long wire) the wave front cannot be described with that simple function but only using energy consideration one gets that expression. $\endgroup$
    – Sørën
    Commented Aug 15, 2017 at 23:06
  • $\begingroup$ @Sørën do you know what curvature is? I mean exactly this. Just consider the difference between directions of propagation in nearby points: far from the origin they are mostly parallel, so your wave is close enough to a flat wave, while near the origin the directions vary quite widely as you move from point to point. $\endgroup$
    – Ruslan
    Commented Aug 16, 2017 at 5:21

Cylindrical coordinates only really make sense in two or more dimensions, where $(x,y)\rightarrow (r,\phi)$ would represent the transformation from Cartesian to cylindrical coordinates. When you have only one dimension, then the transformation $x\rightarrow r$ is somewhat trivial in that it does not really chance anything. So you end up with something that still looks like the Cartesian case. That is why the solution in (1) looks more like the plane wave solution that you would have for the Cartesian case in two or more dimensions.

BTW, in two (or more) dimensions the solution in cylindrical coordinates contains the Bessel functions rather than the Hankel functions, because one would assume that the solution is finite at the origin.

  • $\begingroup$ The running wave is Hankel function anyway. It can't be finite at the origin in general. $\endgroup$
    – Ruslan
    Commented Aug 15, 2017 at 5:40
  • $\begingroup$ @Ruslan: what do you mean by the "running wave"? The solution in three dimensions for a region that contains the origin has to finite at the origin to be physical. Such a solution only contains the Bessel function. An example of that is the modes in an optical fibre. $\endgroup$ Commented Aug 15, 2017 at 7:02
  • $\begingroup$ A running wave may emanate from (or be absorbed by) a cylindrical surface of finite radius, for example. Outside the cylinder the solution will be represented by Hankel functions, and will be finite. Inside the cylinder the solution will depend on what the medium there is and how it interacts with the medium outside. This all depends on boundary conditions (and their placement!), you may not simply discard the second solution of a differential equation because it diverges somewhere. $\endgroup$
    – Ruslan
    Commented Aug 15, 2017 at 9:33

Also: consider the concepts of near and far field. The exact solution contains both. In the far field, only traveling waves exist ($1/r^2$ power for spherical cases) that asymptotically approach plane waves. Near field falls off faster (hence: "near"), moreover there may be phase differences with the driving oscillation--power can go into to field and back to the antenna, for example. I believe the credit-card chip is a near field device, and is hence more secure than RFID--which radiates your information.


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