Is there an upper frequency limit to ultrasound?

Question

Wikipedia has this diagram of the acoustic frequency spectrum:

Is there an upper limit to the frequencies you can transmit through the air? Are they absorbed more and more at higher frequencies, resulting in shorter travel distances? Are there "windows" of absorption and transmission like there are for radio waves?

Update:

This says

An additional problem is created by the strong absorption of ultrasound in air, at least for frequencies higher than 250 kHz. As a consequence, ultrasound of e.g. 1 or 2 MHz can propagate in air over a distance of not more than a few centimeter

Also found some references, but I can't access them: http://scholar.google.com/scholar?q=Atmospheric+absorption+of+sound

Answer 1

The previous answer is qualitatively too generous. The maximum frequency is the mean free path divided by the speed of sound, and it is a gradual thing, defined by greater and greater attenuation as you approach the limit rather than a sharp cutoff, as there is for phonons in a solid.

The mean free path in air is 68nm, and the mean inter-atomic spacing is some tens of nms about 30, while the speed of sound in air is 300 m/s, so that the absolute maximum frequency is about 5 Ghz. This is much less than 100 Ghz.

Ultrasound in fluids and solids is not similarly limited, because fluids are much denser, and solids have periodicity. In fluids, you should be able to go to 1nm wavelength sound waves, perhaps a little shorter, with a speed of sound in the range of 1500 m/s. This gives 1500 Ghz as the cutoff, much, much higher than in air.

In a solid, the phonon frequency is periodic, since phonons are defined by lattice displacements. In this case, the maximum frequency is estimated by twice the inter-atomic distance over the speed of sound. this gives 20,000 Ghz as the limiting phonon frequency, again higher, because the speed of sound in solids can be 2-3 times higher, and (twice) the interatomic spacing is five times smaller than a liquid. So it is safe to put the upper limit of ultrasound in metals at 100,000 Ghz, and then only for small-atom metals. If you look at optical phonon bands, you can get frequencies like this over a wide range of modes.

Attenuation buildup

The speed of sound is determined by the relation of pressure to density. From this alone, Newton derived the propagation of sound in any medium.

To understand what happens as you get to higher frequencies, one must understand the statistical nature of sound in a medium like air or water. Where a sound-wave has high pressure, there is just a tendency for a few more air molecules or water molecules to be present per cubic nm. This tendency is statistical, so it has shot-noise, due to the discrete nature of atoms. When the number of extra air or water molecules in one wavelength gets to be order 1, the shot noise dominates the pressure relationship, and the linear relation between pressure and response is wrecked.

This means that such short-wavelength pressure variations are washed out by thermal fluctuations in a short time or just in any instantiation of a statistical state, and they cannot coherently propagate the pressure long distances, unlike long-wavelength pressure waves. You can see this by attenuation lengths for the ultra-sound-waves. As you go to shorter wavelengths, the attenuation length decreases.

The attenuation coefficient for ultrasound describes how the waves die down over distance. By comparing the attenuation length to the wavelength, one gets an estimate for the maximum frequency at which ultrasound can coherently propagate in a statistical fluid. is less than a wavelength for air-sound-waves of order 70nm, so that these waves die out too quickly to propagate as sound. This is how nature enforces the cutoff for statistical fluids.

Impossibility of Wikipedia attenuation model at high frequencies

The attenuation model on Wikipedia states that the attenuation length (length to e-folding) shrinks inversely as the first power of the frequency. This leads to an attenuation over one wavelength which is independent of the wavelength, and equal to approximately $10^{-4}$ in water.

This model is clearly wrong, since the attenuation must be comparable to the wavelength at the point where the pressure variations have atomic scale shot-noise, meaning at the nm scale and below.

Unfortunately, I was not able to find either free data or a more accurate attenuation model in a quick search. So I leave the answer as is. The coefficient of quadratic dependence of the attenuation on the frequency should lead the attenuation over one wavelength to be order 1 when the wavelength is order 1nm.

Answer 2

I read this just recently, so I though I could quickly find the source. But I can't! I hope I didn't dream it?

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It's quite simple: there can't be frequencies higher than the average frequency with which the air particles collide (yeah, it's ok to treat them as classical points here). At least not in the form of linear pressure waves as we know sound, there are other kinds such as the waves that supernovae push through the galaxies.

In principle, there could also be confined frequencies of absorption like there are for electromagnetic radiation or sound in solids; that would require resonances. But in a gas, these could only be molecular vibrations, which – in the case of small, stable molecules as you have in the air – are very high, above this upper threshold: going after Peter Morgan's estimate, or even a more generous one*, the highest possible frequency is still well below the exitation frequencies of air, which are those you also see in the electromagnetic absorption spectrum (starting at $\approx 3\:\mathrm{THz}$).

_{*extra-generous estimate: the frequency is at most of the order of the average spacing between molecules, divided by the average speed.} $$ \nu_{\mathrm{max}}\approx \frac{\sqrt{\overline{E_{\mathrm{kin}}}/2m}}{\sqrt[3]{m/\rho}} \approx \frac{\sqrt{kT}\sqrt[3]{\rho}}{m^{\frac56}} $$ _{(only order of magnitude), which would be $\approx 100\:\mathrm{GHz}$.}