# What causes volume of sound to decrease with distance?

I know about the inverse square law however I do not know why it actually decreases. Does it have to do with sound absorption, it runs out of energy so the vibrations lessen or as distance increases the area the sound hits becomes less specific etc?

The inverse square law is because we live in a 3D world. When sound is emitted from a point source, it spreads out in a sphere, covering a larger and larger areas as the sphere gets larger. This, in a way, dilutes the energy of the sound over the large area.

The area of a sphere in 3D is 4pi*R^2, so divding by this number gives something over R^2. (inverse of the radius squared)

Note, that if your source of sound is not a point source, but a line source (sony made something like this years ago, "sountina"), then it does not spread out in a sphere, but in a cylinder. In that case, the R squared law does not apply (locally), but is a 1/R relationship.

For a planar sound source, there is no drop off over distance.

(Note that a true line/planar sound source can not exist because they need to be infinitely long)

• Very interesting. I'm assuming that the sountina is thus an approximation of a line sound source. What, then, would be an approximation of a planar source? I'm trying to picture it in my mind. Mar 30 '20 at 15:21
• You can approximate whatever source shape you like with a 'phased array' of speakers (or microphones if receiving). I used to deal with LRADs (en.wikipedia.org/wiki/Long_Range_Acoustic_Device), which have many small point sources in parallel, but which can be controlled to give a wavefront that looks planar. It also uses some strange effects of air. There's a Ted talk with the creator somewhere. Mar 30 '20 at 23:10
• Thanks. Here's my mental picture! "[It reached]... a distance of half a mile from the stage without degradation." en.m.wikipedia.org/wiki/Wall_of_Sound_%28Grateful_Dead%29 Mar 31 '20 at 1:20

There is no absorption process contributing to the inverse square law. In fact it's the opposite. This is related to energy conservation.

The sound intensity (I) is related to power (P) and area (A) by I = P/A. The Power here is energy per unit time delivered to the surface. You can imagine a field of small sensors distributed over the area and taking a time average measurement over many cycles of the wave. With NO absorption or any other mechanism to steal energy the amount of energy measured at each surface over some time interval has to be equal.

P1 * T = P2 * T --> P1 = P2

This in turn implies that

I1 * A1 = I2 * A2

Now, if you consider a point source and the areas to be concentric spheres with the course at the center this gives you the famous inverse square law.

I(r) = (I0 * r0^2)/r^2

where I0 is the intensity measured at a reference position r0.

This is a classic approach to presenting the law in basic physics texts. There is a little more to the story than that.

The intensity is not the fundamental quantity describing acoustics. We have a set of equations for the pressure fluctuations, p, and the local particle displacement or velocity field, v. With a little manipulation the pressure field (in most cases) obeys a form of the wave equation in space and time, the Helmholtz equation in space, with the time derivative replaced by assuming the solution exp(i 2 pi f t). When the local sound speed is constant this has exact solutions (and in many cases where the sound speed is NOT constant). In particular, when there is a point source the solution is called the Green's function and in 3-dim looks like

G = exp(-i k R)/R

R = distance between source and point of measurement of the field.

The Intensity is proportional to the squared pressure and that also gives the 1/R^2 result. So you can derive this from more than one line of reasoning.

If you try to measure this law in a room you may be disappointed and possibly confused. It is difficult to observe this idealization except in a controlled environment in an anechoic chamber. Because...

1. Reflection off hard surfaces created reverb which adds to the local intensity. Even though the sound from each bounce follows the 1/R^2 law your microphone will not know that.

2. The atmosphere absorbs sound and the absorption depends on frequency. This will reduce the intensity by an amount usually accounted for by exp(-a R) where a is the absorption coefficient.

3. The speed of sound is not constant over long distances. This can cause wild, yet predictable, changes in propagation and even violations of the 1/R^2 result (not really a LAW).

This is observed in ordinary air as sound hits pockets of warm air it will bend (refract) back into the cold air. This bending can cause the energy that would otherwise spread out to get bunched up, i.e. louder as you get farther from the source. The effect is especially well know in underwater sound in the Pacific Ocean where there is an underwater wave guide made from high temperature near the surface of the water and high pressure near the bottom. Sound can get trapped at a depth of about 1000m to 1300m below the surface. These results are completely predictable from the wave equation and Helmholtz equation using a sound speed that is a function of position.

It has to do with the geometry of a vibration along with the source of what we perceive as sound. We'll start with the fact that sound is caused by atoms vibrating meaning it creates kinetic energy in each atom as it goes outward from the source. Now imagine the sound wave starting from a small sphere of radius 2 meters with 4 atoms vibrating on its surface (surface area of a sphere is 4*pi*r**2) all separated by 4*pi m^2 and expanding into a sphere of radius 6 with atoms still evenly distributed by 4*pi m^2 along it's surface area. The surface area of the new sphere is 144*pi m^2. Now we divide that by the distribution to get the total atoms on its surface area, 36. So lets look at our data, as the distance from the center tripled, the amount of atoms vibrating increased by a factor of 9 or in other words a square of the distance. What this tells us is that since an initial energy was needed for those 4 original atoms to vibrate, that original energy will now be needed to make 36 atoms vibrate. After all of this data we can now say that since the vibration will decrease as a square of the distance, the sound we will experience will also be observed as decreasing as a square of the distance.