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I know it has to do with diffraction of sound, as it is a wave, but how exactly does this diffraction occur?

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For the sake of this argument, imagine an infinitely thin barrier (or a thick barrier with a very sharp corner, if you prefer). Think of a molecule right at the edge of the barrier but not touching it. A sound wave approaches. What this means is that the nearest neighbor molecule behind it is approaching. It may hit straight on and accelerate the first molecule forward. But the neighbor molecule might be aligned off-center, so that it hits the first molecule a glancing blow and accelerates it in some other direction. That's the kernel of a wave propagating in a new direction. Imagine that there are $10^{23}$ molecules in the system ... you will end up with diffraction.

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  • $\begingroup$ Where does the barrier come into play? $\endgroup$ Oct 6 '19 at 13:38
  • $\begingroup$ @AaronStevens The barrier is a stand-in for the wall. $\endgroup$
    – garyp
    Oct 6 '19 at 15:46
  • $\begingroup$ Where does the wall come into play? What I mean is that your answer says to imagine a molecule at the edge of the barrier, but then you never explicitly mention the role of the barrier after this. $\endgroup$ Oct 6 '19 at 15:59
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    $\begingroup$ Your answer seems right to me, but diffraction occurs for all kind of waves, doesn’t it? Can Huygens principle always be explained on the molecular level, similar to your answer? Seems hard to believe in the view of the generality of that principle. Or, maybe, Huygens principle indeed only applies in media comprised of molecules? $\endgroup$ Oct 6 '19 at 16:18
  • $\begingroup$ @HartmutBraun Huygens' principle cannot always be explained on a molecular level. For light diffracting from a barrier in a vacuum Huygen's principle can be applied, but there are no molecules to generate the wavelets. I think all diffraction can be explained by arguments similar to Huygens' principle, but there would be differences in the details. $\endgroup$
    – garyp
    Oct 6 '19 at 16:36
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It is not quite clear to me what you are asking, but I would refer to simulations like on PhET: Wave Interference.

There you can switch between sound waves, surface waves on water (ripple tank) or light waves.

There are two important things to realize:

  • The wavelength of sound waves is of the order of 1 meter, comparable to the height of walls.
  • We experience sound level as a logarithmic measure of the intensity, and for example even a tenfold reduction in power reduces the sound level by only 10 dB, for example from 60 dB to 50 dB.
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    $\begingroup$ I think the OP just wants to understand how diffraction of a wave occurs. $\endgroup$ Oct 6 '19 at 11:57
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If the propagation of sound waves and light waves are governed by exactly the same wave equation: $$\frac{\partial^2 f}{\partial t^2}=c^2 \nabla^2f,$$ where $c$ is the propagation speed of the waves and $f$ is the thing that propagates (say, pressure for a sound wave or electric field for a light wave) why can you hear but not see what’s happening on the other side of a wall?

There are two main reasons, the more important of which Pieter already pointed out: the wavelength of sound waves is a million times bigger. Sound in the human auditory range diffracts around a 10-m wall the same way light we can see diffracts around a “wall” with a height of 0.01 mm. (The speed is also different by a factor of about a million, but that doesn't change the spatial characteristics of how the waves behave as they travel from source to you, only how long that trip takes.)

The other, subtler reason is that we process auditory and visual signals very differently.

All that really matters to your brain about the pressure signals coming into your ears is the spectrum—the amount of energy at each frequency. So the signal can be diffracted (or bounce off irregular surfaces or whatever) without affecting your ability to extract information from it. (Incidentally, in many cases scattering is more important than diffraction—when you hear a voice from the other room, much of that sound energy reaches you by bouncing off a wall or two.)

Your eyes, on the other hand, don’t care about the spectrum—color is just some sort of average of all the constituent frequencies—but they do need light rays originating from various positions on some object you'd like to see to be traveling predictably along the lines from those positions. Otherwise no image of the object is formed. Diffraction (or bounces off rough surfaces) messes this up; even if the light still enters your eye, the image that would have been formed on your retina by rays traveling direct from the source is now scrambled, making it impossible for your brain to extract from the light the information it would have gotten from that image.

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    $\begingroup$ This answer was down-voted, but I think that might be a little harsh. This answer seems to address the word "hear" in the subject. We don't know for sure if the OP was interested in the physics of diffraction or diffraction coupled to hearing. This answer might be made more clear (my opinion) by pointing out that I can hear conversations coming from beyond a wall, but I cannot see who is talking. But even then there are issues to deal with: my radio can pick up signals that have diffracted around many obstacles, but I still hear the music perfectly. The argument gets complicated. $\endgroup$
    – garyp
    Oct 6 '19 at 16:50
  • $\begingroup$ That is a nice way to put it—hear the conversation but not see who’s talking. I see no complication whatsoever though: radio waves diffract around things because they have long wavelengths unlike visible light. Your radio is not trying to focus the radio waves into an image—it just responds to the amplitude at a single point (much more similar to the way your ears interpret signal than your eyes). So the diffracted radio waves still carry the information. $\endgroup$
    – Ben51
    Oct 6 '19 at 17:15
  • $\begingroup$ The diffracted light still carries the information, too, but it's harder to make it intelligible. Harder, but not impossible. A hologram captures a diffracted light field, and the reconstruction makes it intelligible. $\endgroup$
    – garyp
    Oct 6 '19 at 17:37
  • $\begingroup$ Good point. But our eyes can’t do that. $\endgroup$
    – Ben51
    Oct 6 '19 at 17:37
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If you can follow this logic you should get a feel for what happens...

Waves that aren't physically confined have a tendency to spread.

Imagine using a stick to agitate water in a tank- you will make waves that travel outwards in a circle, spreading over an increasing area.

If the tank is a long channel, maybe a foot across and say twenty feet long, when the waves try to spread they are constrained by the walls, so their movement is directed along the channel.

Now imagine the channel opens out into a pond at the far end. When the waves arrive there they are no longer constrained by the walls so they open out again.

So returning to your question, when something makes a sound on the other side of a wall, the sound waves try to spread in all directions. The sound that makes it over the wall is like the water wave leaving the channel- it spreads out beyond the wall.

When you consider the effect in more detail you can show (I won't try to do it here) that the extent of the spread when the wave moves out of the channel into the pond depends upon the width of the opening at the end of the channel compared with the wavelength of the wave. When the two are about the same the spread is greatest. That explains why typical audible sounds (which have wavelengths of perhaps a few metres down to a few centimetres) spread around everyday gaps and obstacles.

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An intuitive picture can be obtained using Huygen's construction. Although it is not a quantitative method, it is very useful for intuition. It says that every point on a wavefront acts as a trigger for a secondary spherical wavefront emanating from it. The superposition of these secondary wavefronts results in the whole wavefront moving forward - and so on. With this picture of wave propagation, imagine what happens if a section of wavefront is screened out by a wall -- the secondary spherical wavefronts right on the edge simply radiate into the geometric shadow region. That is diffraction. It is simply a consequence of the tendency of waves to spread.

In fact, a mathematically rigorous version of this argument is precisely the Kirchhoff approximation. It is still an approximation though. A truly rigorous geometric picture of diffraction, one that is derived directly from the wave equation without any additional postulates, does not exist. So based on our current understanding of wave physics, this is probably as far as you can go before getting the answer "solve the wave equation."

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