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More precisely, how small is the potency a listener hears, compared to the potency of the emitter.

I'd like to present a simple and yet reasonable approximation, to a high school audience (I am a teacher)

Intuitively, it should decrease with a square (of distance; because of the expansion of the area the wave covers) and exponentially (because of accumulation losses in the way) .Here, they mention both decreases: . The air absorption component is linear in dB (which, as far as I know, means exponential on energy)

I am just not sure of how to put them together, and if there are other ideas that a first approximation should cover.

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You meant square, not square root, right? Imagine the solution is something like $dE/dV=C \exp(-r/r_0)/r^2$. – Luboš Motl Apr 18 '12 at 4:59
@Pygmalion if losses occur this is no longer correct. In lossless media, just take $r_0\to\infty$ – Tobias Kienzler Apr 18 '12 at 10:53
Related: – Qmechanic Apr 18 '12 at 14:42

edit Sorry, I guess high school students are neither familiar with complex numbers nor interested in actual equations, so my answer below is probably not what you're looking for. I guess it boils down to mentioning spherical waves (there's a nice animation at the Wikipedia article) and the Beer-Lambert law and stating that you can simply multiply both kinds of decay envelopes.

You should also mention that the distance decay is only correct for point sources, while the other extreme, an infinite planar source, does not spatially decay at all. Reality lies in between...

For a point source, you have a spherical wave with complex valued amplitude (for a physical meaning, take the real part only)

$$A(\vec x, t) = A_0\frac{e^{i(\vec k\vec x-\omega t)}}{|\vec x|^2}.$$

The wave vector $\vec k = \vec k_r + i\vec k_i$ is also complex valued to treat absorption (and fulfils a dispersion relation like $\vec k^2 = \omega^2c^2$).

edit2[ Of course, physical reality consists of real numbers only, so this is just a construct to simplify things. Since the complex conjugate of this complex valued amplitude also fulfils the wave equation, and since the superposition is applicable for linear wave theory, you can take the real part of $A$, i.e.

$$A_r(\vec x, t) = \Re A(\vec x,t) = |A_0|e^{-\vec k_i\vec x}\frac{\sin(\vec k_r\vec x-\omega t + \phi)}{|\vec x|^2}$$

(where $\phi$ is the comlex phase of $A_0$)


You are interested in the intensity $I$, which is the squared absolute value of $A$:

$$I/|A_0|^2 = |A|^2/|A_0|^2 \\= e^{i(\vec k\vec x-\omega t)}e^{-i(\vec k^*\vec x-\omega t)} / |\vec x|^2 \\= e^{i((\overbrace{\vec k-\vec k^*}^{=2i\vec k_i})\vec x}/|\vec x|^2 \\ = \frac{e^{-2\vec k_i\vec x}}{r^2}$$

So in total, you get just what Luboš said in his comment

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As far as I know the exponential part is superfluous. – Pygmalion Apr 18 '12 at 9:33
@Pygmalion most certainly not, if you have absorptive media (Lossless media are of course included for $\vec k_i=0$). Just take $k_i=\tfrac12\ln2$. Then the intensity at distances 2,4,8 and 16 is divided by a factor of 4,16,32 and 256 from the $1/r^2$ part but by 4,16,64 and 65.536 from the exponential part. In the far field, the exponential part is what dominates the damping! – Tobias Kienzler Apr 18 '12 at 10:52

It all depends how much you want to model the problem according to realities. Getting exact answers is almost an impossible task. I was working for a while on noise propagation calculation for traffic and industry sources as a business. Things are extremely complicated, you have dozens of formulas to calculate that. In fact very precise EU and state-dependent regulations exist for calculating sound/noise propagation. Eventually, everybody use special software for that and I guess only those who write software are fully aware of all details.

But as a bottom line: absorption due to air itself (which by the way largery depends on humidity and air temperature!) is much smaller than intensity fall due to geometric reasons (I am talking about earthly dimensions). Even when sound propagates through thick vegetation, the additive reduction according to EU-regulated modelling, is few dB per 100 meter. And bottom line, when doing real absorption calculation in air, exponential formula fails. I would gladly provide you exact formulas from calculation regulations, but they are no longer at my disposal.

As it happens, absorption due to presence of soft materials (like for example earth ground) is more important, and reflections are also a much more acute effect. Other effects are well described in the document you cite. So I think it would be fairly right to say for real problems, that intensity reduces $\frac{1}{r^2}$ for point sources (mostly industrial sources) and $\frac{1}{r}$ for linear sources (mostly traffic sources). If you only explain this difference (point and line sources) I think you shall do enough.

Note: The right derivation for point source (without theoretical absorption) is rather simple

$$A(r, t) = A_0\frac{\sin(k r-\omega t)}{r}.$$

$$I_0 = |A(1,t)|^2 = \frac{A_0^2}{2}$$

$$I = |A(r,t)|^2 = \frac{A_0^2}{2 r^2} = \frac{I_0}{r^2}$$

And here is for linear sources

$$A(\rho, t) = A_0\frac{\sin(k \rho-\omega t)}{\sqrt{\rho}}.$$

$$I_0 = |A(1,t)|^2 = \frac{A_0^2}{2}$$

$$I = |A(\rho,t)|^2 = \frac{A_0^2}{2 \rho} = \frac{I_0}{\rho}$$

There is even much simpler explanation for the formulas above. Intensity is power over area

$$I = \frac{P}{A}.$$

In case if there is no absorption, power remains constant, while area changes. In case of the point source, area increases as $A \propto r^2$, while in case of the linear source, area increases as $A \propto \rho$!

PS: In the document you cite it says "Under 'normal' circumstances, atmospheric absorption can be neglected except where long distances or very high frequencies are involved." This is in full agreement with general experience, as this absorption presents a serious effect in case you calculate distrubtion of noise from huge industrial transformers. Long distance effect is on the other hand completely diminished by other effects.

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for the problem of sources you mentioned at the top, it can be more convenient to numerically solve the wave propagation directly instead of treating each source's field individually. Also, you cannot neglect the exponential dampening in lossy media (why should sound not propagate through walls otherwise?) – Tobias Kienzler Apr 18 '12 at 10:55
We are talking about propagation through air. Propagation through the wall is zillion times more complex. Even loses due to propagation through thick vegetation is only few dB per 100 meters!!!!! – Pygmalion Apr 18 '12 at 11:02
true, and of course if one considers walls, one has to take interfaces into account as well, which is beyond the scope of the question – Tobias Kienzler Apr 18 '12 at 14:32
Whatever, if it would be clear that the question is purely theoretical I wouldn't lose time explaining all this stuff. What a waste of time. – Pygmalion Apr 18 '12 at 15:00

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