# Is Huygens principle true for any shape of wavefront?

I have read from a few sources that cylindrical waves propagate leaving a wake behind, differently from spherical and planar waves, which would propagate sharply, ‘cleanly’.

“The wave equation may allow any shape wave front, but Huygens principle does not hold for any shape wave front. For example cylindrical waves do not propagate 'cleanly' without a wake whereas spherical waves and plane waves do.”

Another example is this article I found, although it is a little complex for my understanding and I may be missing something. It reads:

“By way of contrast, the cylindrically spreading pressure pulse depends on time t and the propagation delay, r/c, individually, which means that it does not retain the shape of the source signature as it propagates. Instead, after the propagation delay at time t = r/c, the pulse exhibits an extended tail, or wake, which decays to zero asymptotically as 1/t2.”

Considering they’re all 3-dimensional waves, I find it odd that cylindrically shaped waves would propagate differently from spherical or planar waves. What explains this? Are those comments accurate? Also, do they say the same thing or I’m misunderstanding?

I don’t understand complex math, so any simple answer would be appreciated. Thanks.

• One way to see that the comments are accurate is to note that eigenfunctions of Laplacian, corresponding to planar and spherical waves, i.e. complex exponentials $\exp(i\vec k\vec x)$ and spherical Hankel functions $h^{(1,2)}_n(kr)$ have constant distance between nodes, while the analogous functions for cylindrical waves, i.e. cylindrical Hankel functions $H^{(1,2)}_n(kr)$, have varying distance between nodes, depending on the distance from the origin. This variation leads to shapes of wave packets not being preserved while they propagate, especially near the origin. Oct 3, 2020 at 15:13
• But is this true for any cylindrical wave? This image for example: i.stack.imgur.com/X2G2x.png - All points on the wavefront seem to be at the same distance from the origin. In this case, would there be a wake? Also, why wouldn’t Huygens principle be true for any shape of wave front?
– user137288
Oct 3, 2020 at 15:34
• Yes, this is true for any cylindrical wave: Hankel functions are the single-frequency solutions of the wave equations in cylindrical coordinates, so you can expand any waveform in this basis. Your image is just a drawing, it doesn't show anything quantitatively. Do note though, that the differences from equal-period function are very small, see this chart comparing the Bessel function with its asymptotic form (and the zeros in the table at the bottom): i.stack.imgur.com/qG3JV.png Oct 3, 2020 at 15:56
• And yeah, that wasn't an answer. I'm just commenting your question because this is not an explanation, but rather a way to make sure that the statements you cite are correct. Oct 3, 2020 at 15:57

Consider a line of sources of spherical waves, placed along the $$z$$ axis. For concreteness, let the waves emitted be square pulses. If we take a cross section of the wave field in the $$xOz$$ plane, we'll see the following some time after the (synchronous) emission of the pulses: Notice how, if we look at the $$Ox$$ line, we'll see that there's not just a single pulse visible. Instead there's a train of pulses: wavefronts from different sources arrive at a given point at different times: If we increase density of the sources along $$z$$ axis, we'll get something like the following: At different times the waves will look like this: We've made a cylindrical wave from a line of spherical waves. Compare this arrangement to a continuous line of monochromatic spherical sources, resulting in a cylindrical wave:

$$\frac1\pi\int\limits_{-\infty}^\infty \operatorname{sinc}\left(\sqrt{x^2+y^2}\right)\,\mathrm{d}y=J_0(x).$$

Let's now arrange the sources along a plane $$yOz$$ instead of the line $$Oz$$. We'll get the following result along the $$Ox$$ line: Or, at varying times, This corresponds to a planar wave made of a plane of spherical sources. For monochromatic sources this corresponds to the arrangement*

$$\frac1{2\pi} \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \operatorname{sinc}\left(\sqrt{x^2+y^2+z^2}\right)\,\mathrm{d}y\mathrm{d}z=\cos(x).$$

Notice how we've obtained a perfect (in the limit of continuous distribution of sources) piecewise-constant wave field. If we now make the sources emit a negative pulse so as to cancel this bump, we'll get the exact same shape, just of different sign (and with different propagation distance and amplitude). Adding these two—positive and negative—waves together by the principle of superposition, we'll get a planar pulse with well-defined beginning and ending.

Do you see what we'll get if we attempt to do the same subtraction with the waves from the line of sources? Obviously we'll get a messy wave with two peaks—one positive (outer) and one negative (inner)—and exactly what is predicted by the answers you link to: a wake in the inner region, lack of an ending to the cylindrical wave pulse.

So what we have here is just what this answer at Math.SE says: we have the simple effect of difference between arrival of waves from different sources in even dimensions, which leads to the wake, while this effect nicely cancels out in odd dimensions, resulting in well-formed wave pulses.

*Strictly speaking, this integral diverges. But with the appropriate regularization the equation still holds, see this post for details.

• “wavefronts from different sources arrive at a given point at different times” — I understand that, but why would that be different for a planar wave made of spherical sources? Help me understand that please.
– user137288
Oct 13, 2020 at 15:24
• @RobertoValente a cylindrical wave is a 2-dimensional wave $f(x,y)$, trivially extended to 3rd dimension: $g(x,y,z)=f(x,y)$. Oct 13, 2020 at 15:52
• @RobertoValente we have a function of 2 parameters. When we extend it to larger-dimensional domain, to take 3 parameters, we simply make a new function that ignores its third parameter—the $z$ coordinate. Thus, it's basically identical to the original 2D function, so its properties like presence of wake are also identical to the ones of the 2D wave. Oct 13, 2020 at 16:05
• The placement of individual, discrete, sources along the $z$ coordinate is just an ansatz. After we take the limit of $\Delta z\to0$ (and the length $\ell$ of the line that contains the sources $\ell\to\infty$), all the dependence of the waveform on $z$ will be removed—due to the continuous translational symmetry in this coordinate. Oct 13, 2020 at 22:21
• @RobertoValente as for your next comment, nothing is in principle different for the planar wave. You can still see that, although the sources no longer emit the wave, you still get the pulses from faraway sources, and continue doing so indefinitely, unless the sources emit a negative pulse. But the actual difference is the shape of this compound signal from faraway sources: it's constant in time and space (where it's nonzero after initial rise), unlike in the case when the sources are on the line instead of the plane. Oct 13, 2020 at 22:23