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Comment by dkhar | Hacker News Reader

Sound intensity actually falls off as 1/r², and here's a source:

http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.h...

If sound traveled as a sphere, then it wouldn't be able to circle the Earth. If you imagine a sphere growing from a point on the Earth's surface, most of it will wind up in space, and none of it will reach the opposite side of the Earth. The 1/r^2 falloff is probably just an approximation that holds true for short distances or smaller intensities.

In this case, the soundwave was able to follow the curvature of the Earth, which implies that it wouldn't decay as a sphere but rather a plane, which would be a 1/r falloff. I wonder why sound of massive intensity will follow Earth's curvature?

Why don't you think that the sound waves would go in all directions, and be absorbed / redirected at the edges? What you see following the curvature of the Earth is the result of that, and once you factor in the second-order wave reflection, it's still on the order of 1/r^2 isn't it?

Nah, imagine a point on the earth. Now trace rays from that point in every possible direction. All of those rays will eventually lead to space or to the ground, and none will reach the other side of the Earth. Since molecules are physically further apart the higher you go in the atmosphere, it seems unlikely that the energy would be redirected at the edges, only dispersed. That must mean the wave is literally following wherever the atmosphere is thickest rather than simply being reflected.

When I hear "energy dispersed at the edges", I think "lost" (or effectively absorbed), not "reflected back the way it came". I can't think of any obvious mechanism that would lead to significant coherent reflection of sound waves off the upper atmosphere. Similarly, I think a lot of the sound intensity would be absorbed by material on the ground rather than reflected back up. I vote for predominantly 1/r^2 behavior.

[As a physics prof, I probably ought to be doing some sort of calculation to justify that, but I don't have the time. I'm sure it's been analyzed formally somewhere already. Just off the cuff, though, the speed of sound is lower at low temperature, so refraction of a sound wave in the upper atmosphere will tend to bend it outward, away from the Earth. I'm thus imagining that loud sounds may tend to locally push the atmosphere out from the planet a little bit, and I expect that coming back to equilibrium will be a lossy process that doesn't preserve the shape of the wave.]

Sound waves can diffract around objects, which is why one can still hear someone calling even when hiding behind a tree. :

http://en.m.wikipedia.org/wiki/Difraction

Is diffraction the reason why soundwaves could travel around the whole Earth? My previous understanding of diffraction was that obstacles cause waves to propagate in different ways. But the thinning of the atmosphere isn't really an "obstacle." The molecules that soundwaves use to propagate are simply further apart from each other, meaning waves are more likely to disperse and lose energy than to keep traveling or bounce. That would imply the boom from the volcano should disperse into space and go silent rather than travel around the Earth. But since that doesn't happen, it seems like the waves follow wherever the atmosphere is thick.

I'm having trouble understanding how diffraction would cause that end result of "waves go where the atmosphere is." If waves could bounce off of the thin atmosphere near space, that would make total sense. But they can't bounce due to thin atmosphere, only disperse, so it seems like there's some other phenomenon in play.

Sound waves are points of high pressure and points of low pressure. At each point of high pressure the pressure tends to spread uniformly in all directions. Likewise at points of low pressure there is sucking from all directions. Opposite directions cancel, i.e. orthogonal to the wave direction. All in all the sum of all points of pressure creates a moving wavefront. Which naturally bends around obstacles. The earth is just a very big obstacle.

Hey, thank you for taking the time to explain this. I really appreciate it.

I'm having trouble seeing how that explanation would explain the case at hand. Your explanation is likely correct, and I'm probably just thinking about it incorrectly. Would you mind pointing out the flaw in my logic?

In this scenario, a volcano's boom was so loud that it traveled through the atmosphere, all the way around the Earth. Your explanation is perfectly reasonable for thick atmosphere. In thick atmosphere, a soundwave is a pressure differential, and since molecules are densely packed together (since the atmosphere is thick), there's no choice but for the molecules to "slosh around." The high pressure areas will spread to the low pressure areas within the thick atmosphere and create a moving wavefront, exactly like you said.

But as the soundwave travels closer to space, the atmosphere becomes thinner. There are fewer molecules for the soundwave to travel through. That means a pressure differential will have less medium through which to traverse. Since there are fewer molecules, there's more room between them to absorb a pressure differential, right? For example, the reason sound travels so well underwater is because water is extremely dense in comparison to the atmosphere, so less energy is needed to travel an equal distance underwater. Correspondingly, near space where the atmosphere is thin, more energy would be needed to traverse an equal distance. That must mean that as the wavefront approaches space, the wavefront should dissipate. Since more energy is required to travel through less atmosphere, then as the atmosphere approaches zero, the energy required for a wavefront to travel one meter should approach infinity, and that's why it seems like the wavefront should dissipate near the edge of the exosphere.

But in this case, the wavefront didn't dissipate. The volcano's boom kept on going all the way around the Earth, and it was somehow able to maintain its energy. If the soundwave travels as a sphere from its point of origin, then that sphere should have a hard time traveling all the way around Earth, shouldn't it? So it must not be travelling as a sphere, but something else.

You're saying that "something else" is diffraction. I'd like to understand that. How is it that a wavefront of such intensity can approach the exosphere where it should dissipate, yet not dissipate and instead keep traveling all the way around the Earth due to diffraction?

It is enough to think about the part of the sound wave, a ring, that travels horizontally to see that it bends with the curvature of the earth. The front of the wave will at every point "shoot" some of the energy horizontally forward, and horizontal is at every point tangential to the earth surface. What happens to the sound energy going upwards, well, energy can't disappear. And certainly not into the nothing between the air molecules in thin air. The wavefront going vertically up will eventually push some air molecules away from earth without them hitting any other molecules further out. And then gravity will pull them back. That is in fact a kind of reflection. The speed of sound changes with density, so the wavefront will not move perfectly spherically. That and the gravity induced ripples on the top of the atmosphere will distort and dissipate the energy around the earth in a somewhat complicated pattern. I think. I'm not a physicist.

This supports even more that the remaining energy is on the order of X/r^2 and attenuated with the SQUARE of the distance traveled around the globe.

The /r^2 factor is due to the fact that the surface of a sphere grows proportional to r^2. That will hold for the wavefront half way around the earth, but from there it will be reverse. The energy will converge to full strength at the point opposite from the origin. That is if there were no loss. A significant portion of the energy is converted to heat and otherwise dispersed from being a compact wavefront.

The energy is dispersed in all directions through air molecules. At the top it dissipates and at the bottom it is absorbed by the earth. The only extra boost you'd get along vectors parallel to the surface would be reflections of the energy, which other commenters have said is small ... so this term doesn't change enough to get the function out of O(r^-2) does it?

You do get a boost from the wave contracting and converging towards the opposite point from halfway around the earth. I think it is not at all obvious what the order of damping will be.

Can you back it up with some overview of the math?

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