I got really confused about something that I asked yesterday so I will better rephrase and explain my question. The text seems big, but wanted to write down everything so there is less space for misunderstanding. I'm not also sure why this would deserve the downvote!

Sound waves move spatially (in sphere directions - 3D). For simplicity, let's just say that no heat loss is occurring at all. So loudspeaker emits sound waves.

The notion/assumption that I have received yesterday was that if no heat loss, energy received at the surface area of $r$ radius sphere will be the same as the energy received at the surface area of $r+dr$. This means energy gets spread, but each time it spreads (dr increasing), it spreads over more particles.

My question is why the energy at the surface area of r is the same as at the surface area of r+dr ?

Let me explain why I think this is confusing to me. We know that sound waves is elastic collisions. So in order for the above to be TRUE, we must assume that while particles of r surface area hit with particles of r+dr area, they give their whole energy to them. Otherwise, if they don't, then r+dr surface area won't get the same energy spread over its particles than it was between r surface area particles. So why does the assumption hold true ? As I understand, particles of r area will collide with r+dr particles(not head-on collisions, but with some angles which will NOT cause r particles KE to become 0, so they will bounce off). This means r particles didn't give their full energy to the r+dr particles.

Where am I making a mistake ? Note that my assumption seems more correct because we know when particle 1 hits another particle 2/3/4, particle 1 doesn't stop and bounce off(we call sound wave longitudinal, so this is more logical that particle after collision comes back to equilibrum position and passes by in the opposite direction - we otherwise wouldn't get sinusoidal form of graph - This makes me believe more that energy between $r$ and $r+dr$ areas are not fully transfered (even if there was no heat loss).

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    $\begingroup$ Then, where is this energy if there is no heat loss and is not transmitted? $\endgroup$
    – nasu
    Jun 2, 2023 at 17:37

1 Answer 1


The answer to this, much like your other questions, is simply that this is an approximation for an idealized system. You're asking about complexities beyond what the model is meant to handle.

This model of sound propagation works well enough for most regular purposes, but if you want to get into the complexities of inelastic or off-center particle interactions, thermal effects, etc., then you're not going to be able to use analytical methods. You're going to have to use numerical methods (probably with computational models) instead.

  • $\begingroup$ My main struggle is that in the sound wave in 3D, do you agree with me that the energy at r+dr surface area will be much less than energy at r surface area ? If you agree with me, then I think my confusion is gone. Also, in that case, the formula we use with power divided by surface area wont be correct, but we dont care about this and idealize the system ? But in reality, if we idealize this, that means particles at r surface area stop moving right away after collision with r+dr and if they stop moving, then how do we get sinuosidal form of particle moving back and forth motion ? $\endgroup$
    – Giorgi
    Jun 2, 2023 at 17:11
  • $\begingroup$ Well, that depends on the medium. Sound dampens at different rates in different media. Next time you go swimming with someone, have them hum underwater, submerge your own head, and see how far away from them you can still hear the noise. Additionally, different media will dampen different frequencies differently. Audio physics can actually get quite involved when you get into these details. $\endgroup$ Jun 2, 2023 at 17:20
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    $\begingroup$ As for sinusoidal motion: that's a driven oscillation. That is, it's from a continuous source. If a source only releases a single pulse, the sound wave passes through the medium and there isn't any steady oscillation. $\endgroup$ Jun 2, 2023 at 17:22
  • $\begingroup$ I meant sound waves in air medium directly.how would you answer it with air medium depending on mu question in the reply above ? $\endgroup$
    – Giorgi
    Jun 2, 2023 at 17:24
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    $\begingroup$ There is energy loss due to that, due to the collisions not being perfectly elastic, and due to variations in the medium. To be perfectly clear, we're talking about the total energy across the entire wavefront. If we're just talking about a single point on the wavefront, the lower energy at r+dr is overwhelmingly simply due to the energy being more spread out. $\endgroup$ Jun 2, 2023 at 17:44

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