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enter image description here

By looking at the above picture with sound wave, I got the following question. Why is intensity at surface of the sphere the following: $$ I = \frac{P}{4πr^2} $$

While I understand that each time sound travels outward, it meets more particles. There're more particles on the surface of the sphere that has radius 2R than on the surface of the sphere that has radius R. This is because perimeter of 2R sphere is bigger, hence the surface contains more particles and energy of the initial sound is spread between more particles.

I also understand that the some part of energy is also inside the sphere above (I mean, inside, not on the surface), but when I talk, the energy gets distributed outwards, and before reaching the receiver, the energy definitely inside the sphere gets less and less as the particles stop moving. In the calculation of the intensity at surface, why is the whole area included? I would understand it, if the energy was linearly distributed in the area, but as I said, it's not.

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  • $\begingroup$ All point sources of fields in a 3D space send out spherical wave fronts. $\endgroup$
    – David White
    Commented Jun 2, 2023 at 0:50
  • $\begingroup$ I get that, but that doesn't answer my question $\endgroup$
    – Giorgi
    Commented Jun 2, 2023 at 0:53
  • $\begingroup$ "In the calculation of the intensity at surface, why is the whole area included ?" - forgive me if this is a stupid question, but are you perhaps misinterpreting "area" to mean the entire volume? $\endgroup$
    – Filip Milovanović
    Commented Jun 2, 2023 at 1:58

5 Answers 5

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I think you are envisioning this improperly. The source is continuously emitting power $P$ (Watts).

During an infinitesimal time interval $dt$, the source will emit a spherical shell of radiation with total energy $dE = P\,dt$. This shell has an energy per solid angle (steradian) of $\frac {dE}{4\pi}$ at all times as it expands outward. To find the energy "per square cm of surface area" that this shell carries at a given instant $t$, we take the current radius $r$ of that shell at time $t$, and the value is $\frac {dE}{4\pi r^2}$.

Now let's place some receiver at some distance $r$ from the source that is $A \, \rm{cm^2}$ of surface area (tangent to the sphere of energy). This means when a single shell of radiation impinges on the receiver, it receives energy:

$$ \frac {dE}{4\pi r^2}A $$

Finally, if we want to know the rate (per second) at which energy is absorbed by the receiver as it sits there, we simply divide by $dt$ again to get:

$$ \frac P{4\pi r^2}A $$

This first factor as you can see is the intensity of the source, a quantity enabling us to calculate the energy delivered to any size receiver at any given distance from the source.

(Note: to keep it simple, I did not bother with mincing radiometry terms like radiant intensity, irradiance, etc.)

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I might be misunderstanding your question, but I think you're trying to ask why it's inversely proportional to the surface area rather than to the volume since some energy is left behind in the volume in the form of heat energy, right?

That's simply because this is an approximation that assumes all of the energy remains on the wavefront, or at least that which is left behind is negligible.

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  • $\begingroup$ This helped me a lot. Though you are assuming that energy that was at wavefront of sphere of radius r-dr got spread over the wavefront of radius r. This would only be true if each collision of particles of r-dr with particles at radius r, completely give their energy which means particles at r-dr should immediatelly become with 0 KE. But this is not true. Even with zero heat loss, particles of r-dr still have some KE after the collision as they bounce off. Why do you assume that it is a model where r-dr particles give all of their KE to particles of radius r wavefront ? $\endgroup$
    – Giorgi
    Commented Jun 2, 2023 at 9:36
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    $\begingroup$ Right. It's idealized to a perfectly homogenous and perfectly elastic medium. We make these assumptions because it's accurate to a sufficient level for typical purposes and makes the math easy. If you really need to account for inelasticity or an inhomogeneous medium, then you'll ultimately have to use computational methods since it's going to result in ugly problems that can't be solved analytically. $\endgroup$
    – Logan J. Fisher
    Commented Jun 2, 2023 at 15:53
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Intensity is defined as the Power per unit area. The wave progresses equally in all directions, so at any distance $R$ you are looking at the surface area of a sphere of radius $R$.

It sounds like you might be confused with the sound dissipating over time. The diagram you give is not concerned with that; it's assuming an ideal case where the power is the same across any sphere.

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  • $\begingroup$ Yes, but remember, we are dividing the whole intensity over the whole area. Why is energy inside sphere included in the calculation ? it seems it is because we're dividing P over the whole area, but energy inside sphere is not distributed equally and while it reaches the radius of R, the energy inside sphere might be close to 0 as particles stopped moving (almost) $\endgroup$
    – Giorgi
    Commented Jun 2, 2023 at 1:10
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    $\begingroup$ @Chemistry I don't understand what you mean by "energy in the sphere". The source is producing energy at some rate. If we moved farther away from the source and looked at the energy flowing through that area we should still find the same energy (assuming no dissipation) $\endgroup$
    – BioPhysicist
    Commented Jun 2, 2023 at 1:14
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    $\begingroup$ @Chemistry I think the confusion might be that "P" is not the "energy inside the sphere". P is the rate at which energy is produced by the source $\endgroup$
    – BioPhysicist
    Commented Jun 2, 2023 at 2:04
  • $\begingroup$ I understand now the formula, but Though it is assuming that energy that was at wavefront of sphere of radius r-dr got spread over the wavefront of radius r. This would only be true if each collision of particles of r-dr with particles at radius r, completely give their energy which means particles at r-dr should immediatelly become with 0 KE. But this is not true. Even with zero heat loss, particles of r-dr still have some KE after the collision as they bounce off. Why do we assume that it is a model where r-dr particles give all of their KE to particles of radius r wavefront ? $\endgroup$
    – Giorgi
    Commented Jun 2, 2023 at 12:04
  • $\begingroup$ air particle 1 might hit the air particle 2 where air particle 1 gives part of its KE to particle 2, but bounces off and still has some KE(less than but still has it). In this way, particle 1 did not give the whole KE to particle 2. Why would not this happen in sound hitting air particles ? $\endgroup$
    – Giorgi
    Commented Jun 2, 2023 at 12:04
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I also understand that the some part of energy is also inside the sphere

For a standard compression wave in 3D space, that's not true. The spaces within the sphere destructively interfere and energy from a pulse is zero inside the (expanding) surface.

When you produce a sharp sound like a clap, the intensity you hear after the front passes goes to zero immediately. So we can model 100% of the energy as being on the surface of the sphere.

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The source may not emit sound waves isotropically, ie equally in all directions, but you can imagine the emitted wavefronts being made of lots of expanding area elements each of which follow the inverse square law.

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