When we were studying mechanical waves like sound waves, and waves on strings in class, we never studied Huygens' principle with these - and nor did we really derive the laws of reflection or refraction with them. We always in any case took them to travel linearly, and did not talk of any wavefronts then. But in studying geometric optics recently, we've begun by proving the laws of reflection and refraction by using Huygens' principle. In doing so we looked at the behaviour of wavefronts - not linear waves - which we had not with sound waves previously. In real life, I understand there must exist mostly wavefronts - there can hardly be any isolated linear wave disturbances, so reflection and refraction occur the way they do.

But if we were to talk of wavefronts of sound, will they still follow Huygens' principle? Alternately, will linear light waves behave like sound waves - for example, will they set up standing waves as well? And is Huygens' principle any more than a simple method of analysis of how wavefronts propagate? It must just a consequence of the wave behaviour - and must be obtainable from the wave theory. Surely it isn't an independent principle?

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    $\begingroup$ Yes, all wavefronts follow this principle. TO be blunt, Huygens proposed it and as a formalism it works perfectly. BTW, his name includes the trailing "s" so watch those apostrophes :-) $\endgroup$ Sep 14, 2015 at 13:43
  • $\begingroup$ see researchgate.net/publication/316994209 for a geometrical derivation $\endgroup$
    – user45664
    May 19, 2017 at 22:15

1 Answer 1


The grounding of Huygens's principle is essentially the observation that the Green's Function for the Helmholtz Wave Equation is the spherical wave source


Since approximately monotonal sound waves also fulfill the Helmholtz equation, the reasoning below, and thus Huygens's principle, applies exactly to them as well.

A simple calculation to show that one can sum up the effect of such sources on a wavefront and get approximately the right answer runs as follows. We look at a semi-infinite region $V$ with boundary $\partial V$. Only part of that boundary - an aperture $A$ - has a significantly nonzero disturbance $\psi(\vec{r})$. We wish to find $\psi$ at some position $\vec{r}_0$ away from the aperture. We consider two functions $\psi(\vec{r})$ and the Green's function $\psi_G(\vec{r}-\vec{r}_0)$ and form the vector field $\psi(\vec{r})\,\nabla \psi_G(\vec{r}-\vec{r}_0)-\psi_G(\vec{r}-\vec{r}_0)\,\nabla \psi(\vec{r})$ and then plug this little beast into Gauss's divergence theorem for a surface comprising (1) the boundary $\partial V$, which, by assumption, for this calculation is the same as simply the aperture $A$ (because the field is by assumption small elsewhere) and (2) a small sphere of radius $\epsilon$ that excises the singularity in $\psi_G(\vec{r}-\vec{r}_0)$ at $\vec{r}_0$. So we are applying the divergence theorem to the part of the volume within $\partial V$ that is outside the little "quarantining" sphere. The divergence theorem yields:

$$\int_{\partial V} (\psi(\vec{r})\,\nabla \psi_G(\vec{r}-\vec{r}_0)-\psi_G(\vec{r}-\vec{r}_0)\,\nabla \psi(\vec{r}))\cdot \hat{n} \,\mathrm{d} S = -4\,\pi\,\psi(\vec{r}_0) + \int_V (\psi(\vec{r})\,\nabla^2 \psi_G(\vec{r}-\vec{r}_0)-\psi_G(\vec{r}-\vec{r}_0)\,\nabla^2 \psi(\vec{r}))\,\mathrm{d} V + O(\epsilon) = -4\,\pi\,\psi(\vec{r}_0) + O(\epsilon)$$

where the $-4\,\pi\,\psi(\vec{r}_0)$ term (recall that $\psi(\vec{r}_0)$ is what we want to find) comes from the surface integral of the little quarantining sphere and the volume integral vanishes because both $\psi$ and $\psi_G(\vec{r}-\vec{r}_0)$ fulfill the Helmholtz equation outside the quarantining sphere. So you can see that what we're left with is that (after we take the limit as $\epsilon\to 0$):

$$\psi(\vec{r}_0) \propto \int_A (\psi(\vec{r})\,\nabla \psi_G(\vec{r}-\vec{r}_0)-\psi_G(\vec{r}-\vec{r}_0)\,\nabla \psi(\vec{r}))\cdot \hat{n} \,\mathrm{d} S \approx \int_A \frac{\exp(i\,k|\vec{r}-\vec{r}_0|)}{|\vec{r}-\vec{r}_0|}(i\,k\,\psi(\vec{r})\,\cos(\theta(\vec{r}))-\nabla \psi(\vec{r})\cdot \hat{n}) \,\mathrm{d} A$$

which, under various approximations, leads to Huygen's principle (witness the first term, which sums up the spherical wave $\exp(i\,k|\vec{r}-\vec{r}_0|)/|\vec{r}-\vec{r}_0|$ weighted by the value of the field $\psi(\vec{r})$ on the aperture. $\cos(\theta(\vec{r}))$ is responsible for the so-called obliquity factor.

Can your explanation please by simplified to high school level? I apologise, but I cannot follow the maths.

Apologies on my part then. It's a good idea to state one's level in the question: your own question is well and thoughtfully put, so you actually come across as someone more advanced than high school. Describe your level in terms of what you know rather than age as we have several teenage physicists on this site who approach graduate level.

Anyhow, you will need to be content with the explanation that Huygens's principle was postulated by Huygens as a "guess" to explain the nature of waves. He understood the linear superposition principle: that the disturbance caused by the sum of waves is the sum of the individual disturbances and thus understood, for example, that a line of point radiators could sum together to yield a plane wavefront. In the 19th century, mathematicians came up with a rigorous description of these thoughts - the Green's function method is essentially the building up of general solutions to linear equations by linear superposition of "fundamental solutions". Lo and behold, if you work out what the "fundamental solution" for the monochromatic wave equation (Helmholtz's equation) is, it turns out to be a spherical wave point emitter, exactly as Huygens had guessed.

So Huygens's principle works for any phenomenon that is described by the Helmholtz equation. This includes light, sound and water waves (in certain approximations). The Helmholtz equation is another form of the D'Alembert wave equation that is valid when the waves are roughly monochromatic or monotonal. Even in modern quantum field theory, where the relevant equations aren't wave equations, certain path integrals are still calculated by Huygens-like methods.

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    $\begingroup$ Can your explanation please by simplified to high school level? I apologise, but I cannot follow the maths. $\endgroup$
    – Charles
    Sep 14, 2015 at 14:48
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    $\begingroup$ the above description is an excellent summation of the origin of the Huygens' principle. It is not on the "high school" level because Maxwell's equations or the Helmholtz' equation are not high school level subjects but you asked for the origins. Huygens' principle is, at some level certainly, is high school level, but to go deeper you need calculus. Good luck! $\endgroup$
    – hyportnex
    Sep 14, 2015 at 21:16
  • $\begingroup$ @Charles Sorry for that. Please see the updates at the end of my answer. $\endgroup$ Sep 14, 2015 at 22:40
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    $\begingroup$ Huygens original explanation of his principle does not use calculus and is interesting and useful: "Treatise on Light", Christiaan Huygens, Forgotten Books, 2013 {there are two 'a's in his first name} $\endgroup$
    – user45664
    Jun 3, 2016 at 21:16
  • $\begingroup$ @Charles may be more knowledgeable than me by now; but, for what it's worth, this is meant to be understandable at high-school level: "Exact derivation of Kirchhoff's integral theorem and diffraction formula using high-school math". The derivation involves rigging secondary sources so as to give the same wave function as the primary source(s) in the region of interest. Therein lies Huygens' principle. $\endgroup$ Feb 8, 2020 at 12:10

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