My understanding of Huygens' principle is, that it describes the way how a wave moves: Instead of moving straightforward, it propagates in all directions via producing secondary wavelets. 'Secondary wavelets' and 'wave's propagation' are synonyms.

But I realized that many others assume that waves move straight, and Huygens' principle doesn't say anything about the original wave, just that in addition to the original wave that moves straight, there are also secondary wavelets which are spherical. And these wavelets just continue straight, because Huygens' principle only applies to waves, not wavelets. I wonder if it could be the correct interpretation, because it seems to void most of Huygens' principle. For example, if refraction occurs due to Huygens' principle, refracted waves shouldn't be able to refract again, because they are composed of those wavelets, since the original wave just continues straight. So it's not a 'real' wave, and Huygens' principle doesn't apply to it.

Now to my main question (with my understanding of Huygens' principle): Let's start with the most simple case: propagation of waves in free space. Suppose we have a planar wavefront of troughs followed by a wavefront of crests of the same waves (½ wavelength behind). After one revolution, we have a lot of semicircular secondary wavefronts superimposed on each other all along the initial wavefront. (We should really have full spherical wavefronts, but for now let's ignore backward waves, and focus on 2 dimensions only, considering only the forward semicircles. It's enough complicated as is.) The result is a thick wavefront starting from the location where the wavefront initially was, and ending one wavelength ahead. After another revolution, we have a wavefront twice as thick. The starting point never changes (as long as we ignore backward waves), but the endpoint does, so the wavefront is just growing. The same is true for the wave crests (and everything in between), just their start and end points are a half wavelength behind. So behind the very first part of the waves - where the troughs have no crests to compete with - we should have destructive interference in every wave.

Here is how it should look after one revolution:wavefronts after 1 revolution Whatever the answer should be, one needs to be careful that only straightforward waves should survive. And I'm not concerned specifically about backward waves (and I have no idea why people assumed so), but about every possible direction.

I think that the answer may be that although there are troughs and crests all over, they have different densities, which causes that some parts should survive. But I'm waiting for others to confirm it before diving in the details.

When it comes to refraction, the problem with direction gets worse. If every point of a wavefront produces waves in every direction with and without refraction, then how can waves change direction? What exactly is different after refraction than before?

The question gets even more complicated, when we have refraction and diffraction simultaneously. In that case the side waves don't get canceled (this is the cause for diffraction), so how can refraction have an effect?

Some commenters told me that Huygens' principle is not accurate. It's just an approximation. I wonder if they're right, because I didn't see such a statement anywhere. If they're right, then I'm not interested in understanding Huygens' principle, because I'm interested in the real facts of wave propagation.

One of them told me that the exact truth is Maxwell's equations. But I couldn't find an interpretation of Maxwell's equations which predicts wave propagation, at least not a well explained one.

This is really a fundamental question. If you know of an article or inexpensive ebook etc. that explains this in a way that all my questions will be answered, please give me a link. (in addition to, or without, your answer.) (It's not easy to find. I searched a lot before posting this question.)

  • $\begingroup$ Are you looking for a mathematical explanation? Is calculus OK? $\endgroup$
    – G. Smith
    Sep 22, 2020 at 19:13
  • $\begingroup$ @G.Smith No. I'm looking for a layman explanation. $\endgroup$
    – George Lee
    Sep 23, 2020 at 15:14
  • $\begingroup$ @G.Smith: I see that no answer is coming, so I'm changing my mind. Even if I will not understand, it may be helpful for others. But please try to make it relatively simple, so it will be helpful for me too. (I know some very basic calculus) $\endgroup$
    – George Lee
    Sep 24, 2020 at 15:54
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    $\begingroup$ Huygens did not know anything about the wavelength of light, did not know about superposition or interference of light. He used it to explain (double) refraction. The principle is not fundamental in any way. It is not even really true. $\endgroup$
    – user137289
    Oct 1, 2020 at 20:49
  • $\begingroup$ @Pieter: Then how would you explain slit interference? $\endgroup$
    – George Lee
    Oct 2, 2020 at 17:56

2 Answers 2


Imagine a simple plane wave, moving to the right, in a direction perpendicular to the wavefronts (i.e., to the "isophase planes"). Huygens wavelets emitted at all points on a given wavefront interfere constructively with each other in the forward direction, as you know. They never interfere with the wavelets from another wavefront in the same wave train, because they are moving at the same velocity as the wave train.

You want to know why there is destructive interference in the backward direction. To reach that understanding, you need to introduce something that accounts for the fact that the wave is actually moving.

The Huygens principle as it is usually presented can easily lead to confusion. The normal presentation begins with the assumption that every wave is actually a monochromatic wave train that starts out stationary. If that were true, there would be waves moving in both the forward and backward directions.

So, now re-do Huygens principle, taking time into account. In the backward (left) direction, an emitting point on the forward moving wavefront encounters backward-moving wavelets emitted by the forward-moving wavefront to the right of it, because it has moved to the right. The encounter is slightly too soon to be in step, because of the fact that the backward-moving wavelet travels less than a full wavelength before hitting the advancing left wavefront. Add that up over all the wavelets emitted by all points in the wavefronts ahead of it, and the sum washes out to zero: that is destructive interference, so this version of Huygens principle does not produce a backward-moving wavefront.

Edited 10/1/20

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    $\begingroup$ I agree that by this explanation there would be destructive interference.. but would it prevent backward waves? Because if the previous source emits a circular wave, and moments later, a source ahead emits another wave, the “previous” wave would always be “ahead” in the backward direction, since the velocities of both would be the same. Does that make sense? $\endgroup$
    – user137288
    Sep 30, 2020 at 16:42
  • $\begingroup$ Not quite. Please try again. But first, make some sketches and think some more. $\endgroup$
    – S. McGrew
    Sep 30, 2020 at 17:04
  • $\begingroup$ The point is that in the forward direction there is constructive interference; and in the backward direction there is destructive tnterference. $\endgroup$
    – S. McGrew
    Sep 30, 2020 at 18:30
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    $\begingroup$ Yes, I am familiar with that article. I'm not very comfortable with the author's assignment of a dipole property to each point, though it may be mathematically equivalent. $\endgroup$
    – S. McGrew
    Sep 30, 2020 at 19:00
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    $\begingroup$ When ignoring the circular wavelets (as you - and everyone - did), and focusing only on front and back, I find that after propagating a quarter of a wavelength, there is destructive interference between the front and back waves. But after another quarter, there is constructive interference. This means that the backward wave is not canceled. $\endgroup$
    – George Lee
    Oct 1, 2020 at 18:32

The wave equation has two initial conditions: the initial displacement, and the initial speed of the initial displacement. If the initial speed of the initial displacement is given the appropriate value then the backward wave is canceled. As an ongoing wave propagates this happens 'automatically' so there are no backward waves in an ongoing wave.

See my


Huygens' Principle geometric derivation and elimination of the wake and backward wave, rev2, 3/21/20

especially the appendices, particularly appendix D. {sorry for the math--hope its ok}

Now there is a peer reviewed version (Nature Scientific Reports):


  • $\begingroup$ It's very technical and hard to understand. If you or anyone can explain it in a simple way, it would be very appreciated. $\endgroup$
    – George Lee
    Sep 27, 2020 at 20:48
  • $\begingroup$ I do not know of a simpler explanation--wish I did. $\endgroup$
    – user45664
    Sep 27, 2020 at 21:15
  • $\begingroup$ @user45664 if the “initial displacement” was produced by the moving planar source you mention, while already moving with said velocity, would it also create waves only ahead of its path and not back? Or does this model only work in the context of an already propagating wave? $\endgroup$
    – user137288
    Sep 30, 2020 at 23:20
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    $\begingroup$ Suppose a planar source is moving in some direction, and begins sending waves forward and backwards at the same speed. I’m having a hard time understanding why there wouldn’t be any wave opposite to its direction of movement. What would cancel it? $\endgroup$
    – user137288
    Oct 1, 2020 at 22:06
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    $\begingroup$ The wave sent backwards due to the source motion is negative, canceling the original backward wave. $\endgroup$
    – user45664
    Oct 2, 2020 at 0:27

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