Sound propagates according to the "inverse square law", that is well defined here:


So the intensity equals 1/distance^2

Like the image in the above link shows clearly, when the intensity at the distance r is 1, then the intensity at the distance 2r is 1/4

Now, my doubts started after my sound engineering teacher said that by "doubling the distance, the sound pressure become half".

that doesn't agree with the above "inverse square law". In fact, according to my teacher, the intensity at r is 1 and the intensity at r2 is 0.5.

Then, I looked a the formula for DBSpl, and that's what I found:

DBspl = 20*log(I1/I2)

Using my teacher numbers, that would be:

DBspl = 20*log(0.5/1) = 20*0.3 = 6 db

Now, I checked to some sound engineering textbooks and actually they confirm what my teacher says. In fact they say the sound decrease 6 db every time we double the distance.

But this is not the "inverse square law", isn't it?

Where is the error? I'm doing something wrong? Or is the teacher doing something wrong? Or is the textbook saying something wrong?

When I asked this same question to my teacher he couldn't help me neither, and seemed to be puzzled by the question too.

3 Answers 3


According to this article from jhsph.edu the formula for sound level is 10 *log(p2/p1)

image from the article

In that case, the halving or doubling of SPL would change by 3dB not 6dB and the inverse square law would still apply.

See the full article for more

Edit: I had mistakenly swapped the formula for sound pressure and acoustic energy in my original post.


Sound pressure becomes half at double the distance. So does sound velocity. Sound intensity is their product, so it drops with the square of the distance.

  • Velocity changes over distance?? That's an interesting concept… I wonder how it knows how far it's travelled already...
    – Tetsujin
    Sep 25, 2019 at 16:02

Well, your teacher is correct in saying that intensity falls as the inverse square of the distance. There are a lot of very well written books that have a thorough derivation (I suggest Fundamentals of Acoustics by Kinsler and Frey and Acoustics - An Introduction by Heinrich Kuttruff. Absolutely no intention to advertise here, there are a whole lot more good books out there, but these happen to have helped me a lot).

So, similarly to what Timinycricket suggested, sound pressure level is a derived metric, which depends on the intensity. The equation connecting those two is,


where z is the specific acoustic impedance. This formula can be found here. So if you replace this for the intensity into the formula of the dB you will see that you end up with with something like

enter image description here

Here, I have ignored completely the impedance because I assume it is constant (so does not change anything in respect to the "attenuation"). Now what the above result shows is that pressure does halve for every doubling of distance, but because it is a derived metric it will experience a 6 dB reduction (due to the square which is then multiplied with the whole logarithm).

Additionally, you may notice that intensity does also falls 6 dB per doubling of distance. This may be the most confusing part here. But as you can see pressure does halve every distance doubling, but the deciBels end up being the same for the aforementioned reasons.

Finally I would really suggest you go through some derivation of the said equations either through the reference I provided or one you may happen to have.

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