4
$\begingroup$

In attempting to find a reference for explaining sound propagation, I came across this article in Wikipedia:

Stokes law(sound attenuation)

What I learnt in high school is that sound is propagated by an inverse square law.

My question is, what (if any) is the relationship between Stoke's law and the inverse square law? Is the inverse square law simply an approximation/generalization of Stoke's law? Or do they in fact have nothing to do with each other?

Lastly, if my assumption that the inverse square law is an approximation of Stoke's law, can anyone point me to a reference?

$\endgroup$
1
  • 2
    $\begingroup$ I believe Stoke's law is the energy loss of the sound wave due to scattering and absorption. The inverse square law however applies to the reduction in intensity of the sound due to the wave spreading through space. In reality both factors apply to reduce the intensity of sound. $\endgroup$
    – Kenshin
    Oct 5, 2013 at 12:10

3 Answers 3

4
$\begingroup$

As Chris and leftaroundabout said, these are really independent things.

The $1/r^2$ law comes from the fact that as the energy in the sound wave travels away from its source, it gets spread out over a larger and larger sphere. Since the total energy must remain the same (assuming no dissipation), the amount of energy in each unit of area on the sphere must decrease. The rate it must decrease is $1/r^2$. So this law is really a result of conservation of energy.

The other law says what happens if there is dissipation. Then as the energy travels through the medium (say, air), then some of the energy in the sound will be lost to heat. The way it works is that after a sound wave has traveled a distance $l_d$, the amplitude of the wave decreases by a factor of $e$.

Putting these two together, we expect intensity $I$ as a function of distance to the source $r$ to have the form $I(r) = I_0 R_0^2 \frac{e^{-2r/l_d}}{r^2}.$ Now why didn't your teachers tell you about this. Well if you calculate $l_d$ according to the formula on wikipedia, you get about six miles. So four sounds that you ordinarilly hear, it is not an important effect.

Now as leftaroundabout said, the attenuation law was derived assuming a plane wave, but I'm fairly certain that you will get the functional form I said for the spherical wave. The decay length might be slightly different, but it will just be an $O(1)$ correction.

$\endgroup$
3
$\begingroup$

They do in fact have nothing to do with each other. Inverse-square is a simple conservation-of-energy argument for point sources, i.e. it basically assumes that no absorption takes place anywhere – which is a pretty good approximation for dry air, provided of course there are no solid obstacles anywhere.

Stokes' law is much more specific in that it starts with a plane wave, while inverse-square amounts to a spherical-wave model. Now, any wave looks locally like a plane wave if you zoom in close enough, but Stokes' law only makes sense if you look at an area much larger than the wavelength. Which you can do with a spherical wave, but you need to move out far away from the source, where $r^2$ becomes basically constant1.

For audio applications, Stokes' law is largely irrelevant, because a) you'll hardly get to a proper plane-wave scenario and b) the viscosity of air is so low that the predicted absorption is usually much smaller than any losses at solid boundaries, and to the inverse-square spreading. Where Stokes' law is important is at high frequencies in fluids with significant viscosity, e.g. in medical ultrasonic scans.

Inverse-square itself needs to be handled with care: in lots of applications, reflection plays a significant role, so the spreading is limited and the intensity drops slower than $1/r^2$, or there are solid obstacles between source and observer so the intensity is obviously lower.


1For the purpose of its effect on magnitude only. Of course, the variations in $r$ still act on the phase, where a large offset doesn't play any role.

$\endgroup$
1
$\begingroup$

For the fluctuation analisys of sound in ideal fluid over persistent density in 2 dimensions, we can use Green's Theorem. Let's say $k = dI\;ds$ where $I$ is sound intensity and $s$ is orthogonal path line to the vectorial gradient of the sound emission. In fact it may sound complicated but it is very simple, it is the simple representation that the variation of sound's amplitude follow the variation of the section that goes perpendiculary through. With Green theorem, we can say that the area modulus of the closed loop section is the same as the path's lenght. And so, $dI = k / dA$ where $A$ is the area of the path's loop. We can transpose $dA$ to radial perspective so $dA = r\;dr\;d\theta$, where $r$ is radius and $\theta$ is the angle. Then $$ A = \int_0^r\int_0^{2\pi} r\;dr\;d\theta=\pi r^2 $$ (do you recognise the formula for the area of a circle!). From there, $$ I = \frac{k }{\pi r^2} $$

At this point, $\pi$ can be merged in the constant $k$ so that $I = k / r^2$ where $k$ is a constant relative to some physics proprieties of the fluid medium and so on.

The same logic is applied when we want to get into 3D. The extention of Green theorem (I think this one is more known as Stokes theorem) says that the modulus of the volume of an closed area is the same as the modulus of the area itself. An other way of considering it : Stokes says "hey man, the area is simply the variation of the variation of the looping path orthogonal to the flow that goes perpendiculary through!". Green hear that and says " dude this is the variation of the area I just found!" Then Stock says "Green stupid geinus I know that! In fact, overall we can say that the modulus of the volume is the same as the modulus of the area perpendicular to the gradient of the flow that goes through it" Finaly in 3D, $dI = k /dV$ were $V$ stands for volume. In spherical perspective, $$ V= \int_0^r\int_0^{2\pi}\int_{-\pi/2}^{+\pi/2} p^2 \; \sin(\phi)\;dp\;d\theta\;d\phi=\frac{3k}{4\pi r^3} $$

Same as before, $3/(4\pi)$ can be merged into the constant $k$ so that $I = k / r^3$ were $k>0$ relative to Fluid environnement and some phisical proprieties. But to use it for the measure of power, you must make your observations on a area instead of a single point as the 2D method.

Same calculus apply to any kind of radiation. (Ex: electromagnetic radiation)

To conclude, 2D method of $I= k / r^2$ follows the principe that the sound is spreaded though a circle. And 3D method spead it over a sphere so $I= k / r^3$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.