Sound waves travel with constant speed, but air molecules that transfer action move with different speeds than the ones described by Maxwell distribution.

Why does the sound wave not smear out and dissipate quickly?

UPDATE. It is not just that should ''smear out'', but that it should do it quickly. If we look at distribution of speed of atoms


then we will see that there is no steep peak. There is a lot of atoms that are 30-50 percent quicker or slower that the rest. With length of sound wave measuring in 1-10 meter noticeable smear out should happen in tenths and hundreds of a second = tens of meters. And in ultrasonic frequency a lot quicker.

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    $\begingroup$ It is so good to see new accounts asking so interesting questions. $\endgroup$
    – peterh
    Jul 14, 2016 at 12:37
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    $\begingroup$ My favorite everyday example of this phenomenon is thunder: if you are close to the lightning you hear a sharp crack, but if you are far away you hear a long rumble, since the sound wave does smear out to an extent. $\endgroup$
    – Asher
    Jul 14, 2016 at 13:02
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    $\begingroup$ One thing to contemplate is the difference in length scale between the mean free path of individual air molecules (and the timescale for scattering) versus the wavelength on frequency of a sound wave you can hear. You will find out that the sound wave basically sees a uniform medium to propagate through. $\endgroup$
    – Jon Custer
    Jul 14, 2016 at 14:17
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    $\begingroup$ @Asher : The thunder dispersion you're describing is multipath propagation. $\endgroup$ Jul 14, 2016 at 14:52
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    $\begingroup$ @Asher: A bolt of lightning is kilometers long. So the sound source has different distances to the observer (listener). I'm guessing that as the reason for a long rumble. Smearing would dampen the sharpness of the crack, but I don't instinctively feel that it's enough to make it into a quarter minute rumble. As a citizen of a country that had a civil war, I know that bombs don't smear out into long rumbles. contd... $\endgroup$ Jul 15, 2016 at 10:05

6 Answers 6


It does, but the effects are negligible in the regions we think about.

If you think about a volume of air as a box of atoms bouncing around, you can apply an oscillating pressure gradient across that box and show that it behaves close enough to an ideal wave propagation medium that you can get away with using such an ideal model. The variations you are looking at "smooth out" on a timescale much shorter than the timescale of the sound wave being transmitted. This is a case where the central limit theorem is quite useful - you can basically show that the variance of the statistical medium you are thinking of is sufficiently negligible when occurring over the timescales we think of when we think of sound waves. That's not to say the effects you are thinking about don't occur, just that they are small enough compared to other effects that we can get away with handwaving them away and still have a useful predictive model left over.

The term used for this is "relaxation." The assumption is that the stochastic system you bring up "relaxes" fast enough compared to the behaviors we care about that we don't have to concern ourselves with those details. The random behaviors obscure any information that might have been held in the exact structure of the medium. All that is left is a homogeneous system which, because of the central limit theorem and large number of particles, behaves almost as an ideal wave propagation medium.

This assumption is not always valid. There are times where you need to use a more complete model, which includes the statistical model of the air molecules. One particular case where we have to do this is when dealing with objects that approach the speed of sound. As you approach the speed of sound, the assumption that the stochastic effects are on a short enough timescale that we can ignore them starts to fall apart. The timescale of the events we care about start to get closer to the relaxation time of the stochastic system of particles. Now we have to account for the sorts of effects you are looking at, because they have a substantial effect. Now we start seeing behaviors like shock waves which never appeared at lower speeds.

We also have to start considering more complete models when dealing with very loud sounds. Once a sound gets above 196dB, you cannot use the nice simple ideal wave propagation formulas because the low-pressure side of the wave is so low that you get a 0atm vacuum. Modeling this correctly requires including effects that were not in the simple model we use every day for normal volume sounds at normal speeds.

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    $\begingroup$ I think you nailed it. Silly of me, to forget about central theorem. It would be nice to reformulate it to simple phrase: see central limit theorem, as we sum up a lot of random variables we get very steep peak. $\endgroup$
    – Vashu
    Jul 15, 2016 at 10:25

They do, its just usually negligible in practice. There's also scattering because the particles are not all the same (H2O, N2, O2, etc.)--but that, too, is usually negligible. Its mainly because there are so many particles in a single wave. Consider that the wave must be extremely short before it becomes noticeable (megahertz).

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    $\begingroup$ So what would be the time constant (time for step response to reach 1-1/e) at 20m distance? $\endgroup$
    – Previous
    Jul 15, 2016 at 8:22

I think this may be correct. The individual air molecules are indeed moving about randomly, and oscillating. When a large amount of energy is generated, these random motions are superposed by an alomost uniform behavior, making the propagation of sound possible. This is quite like the free electrons in a metal. Without a potrntial difference, the motion of these electrons are random just like air molecules. However, on the apllication of an emf, these electrons random velocities are superposed , which causes an orderly flow of electrons,just as in the case of air molecules(only they oscillate longitudinally). You can also relate this to heat popagation down a rod.

  • $\begingroup$ The initial distribution may be maxwellian, but if the external agent is powerful enough, this distribution curve gets a high peak for most of the particles. $\endgroup$
    – Lelouch
    Jul 14, 2016 at 13:21

Consider a box full of air in stady state (closed system). If the volume of that box is $22.4~\mathrm l$ there are $6.022\cdot10^{23}$ particles inside that have maxwellian distribution of speeds. Those speeds are randomly oriented and we can defocus from single-particle description to whole body desctiption supposing it is continuous and homogeneous.

Then we can define pressure and temperature as a measure of total energy of all particles in the scope. If the volume is in steady state the temperature and pressure are uniform in space and time but still the individual molecules keep maxwellian distribution and change their [kinetic] energies and directions wildly.

If the sound is introduced into this volume the steady state is broken and the pressure uniformity is disturbed. Let's pretend that the pressure is partially uniform and there is enough particles in these parts that we can introduce local temperature and local pressure values correctly. In the boundary, the particles from high-pressure volume bring higher momentum than the particles from the low-pressure volume in total. There are cases that particles from high-energy tail from low-pressure part transfer their momentum to low-energy tail particles from high-pressure volume, but the probability of such collision is low.

This way the pressure wave, known as a soud, propagate through the volume following the Huygens principle.

Now imagine we have such box in open space, vacuum, and we suddenly remove the walls. The molecules will dissipate quicky and the sound will smear as well. The approximation of the system as a continuum will become invalid.


The speed of sound is not constant. It is a function of the elastic modulus (or its closest analog in the material) and the density of the material. In gases, it noticeably depends on temperature. In solids, the speed of sound depends on the kind of wave; different kinds of wave have different elastic moduli.

For purposes of estimating the speed of sound, the relevant properties (such as pressure, density, temperature, and elastic modulus) are estimated for small volumes that are much larger than the volume of a single atom.


I think the sound wave DOES spread out and dissipates over enough distance. Consider this:

A small stick of dynamite explodes in a flat empty field. The compression wave of the air (which is what cuases sound) moves away from the source of the explosion at about 340m/s and propagates in a hemispherical manner outward.

One second after the explosion, the radius of the compression wave is approximately 340 meters and the "thickness" of the compression wave is so-and-so thick. As the radius of the wave continues to expand, the "thickness' contracts. You can compare this phenomenon to stretching a piece of elastic; as you pull the material, in thins out.

Energy is dissipated from the compression wave as it expands and grows in radius. Eventually the hemispherical wave is so large and "thin" it can no longer be heard by humans, and soon after, the energy is totally absorbed by the atmosphere.

Basically: the energy from the explosion, which creates the sound, is transferred to more and more air molecules as the wave expands. The wave isn't smeared so much as thinning out.

Hope this makes sense and helps!

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    $\begingroup$ So why isn't it smeared out as well? $\endgroup$ Jul 14, 2016 at 16:32
  • $\begingroup$ @immibis Can you clarify what you mean exactly by "smeared out"? Are you referring to the "thickness" of the compression wave? $\endgroup$ Jul 14, 2016 at 18:01
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    $\begingroup$ Isn't this caused by the fact, that intensity reduces with the squared distance? If you fire flash, instead of dynamite, you will observe the same behaviour for light intensity. On the other hand, fire a laser beam and observe its intensity change over the distance. $\endgroup$
    – Crowley
    Jul 15, 2016 at 8:52
  • $\begingroup$ @Garrettfromhp: I think the questioner's expectation, in saying "smear out", it that since the air molecules have a wide range of speeds, that the energy of a sound wave "should" arrive at you over a wide time period, even if it was emitted in a very short time period. So yes, the question is why the thickness of the expanding shell of sound that you're describing, fails to rapidly increase as the shell expands. You describe dissipation perpendicular to the direction of travel, which is fine, but I think the questioner is asking about dissipation in the direction of travel. $\endgroup$ Jul 15, 2016 at 9:53
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    $\begingroup$ @Garrettfromhp I mean exactly what the question means by "smeared out", because you don't seem to have answered the question. $\endgroup$ Jul 15, 2016 at 13:40

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