Interestingly, the sound intensity will decay as a 1/r law, where r is the distance from the source. (I'm assuming conservation of sound energy as it travels horizontally, i.e., no loss to interactions with atmosphere and terrain).
Compare with the 1/r² laws like the strength of an electric field at a distance r from charge Q: E(r)=kQ/r².
In both cases there is a total of something (sound energy or electric field lines) and that total must be split over all possible directions. The total electric field must be split over the surface area of increasingly larger and larger spheres hence the 1/r² behaviour (1/4πr² to be precise). The sound energy is spread uniformly over a circle with circumference C=2πr, hence giving a 1/r decay, over short distances.
For longer distances, the curvature of the earth will play a role. Come to think of it, it must have been really loud somewhere diametrically opposite to Krakatoa, 17 hours after the eruption...
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.]
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.
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.
I don't think this is correct. A sound wave propagates in three dimensions the same as an electrical field (spherically) and so should also obey the inverse square law.
In this case, the atmosphere is essentially a thin (compared to the radius of the Earth) two dimensional sheet wrapping the Earth, so it's probably valid to think of the sound as propagating on a surface. There wouldn't be much loss of energy to space. Would the atmospheric interface with the ground reflect or absorb?
The divergence is probably even less that 1/r because the Earth is roughly sperical, and the sound will focus at the antipode of its source. There might even be a region, approaching the focus, where as the sound propagates its sound level is actually increasing, not decreasing!
It would be interesting to know what history has to say about Medellin, in Colombia, in 1883. Being at coordinates 6N 75E, it is roughly at the antipode of Krakatoa and could well have had an awful lot of energy dumped in it general direction a few hours after the eruption.
Would gravity affect soundwaves at all? If sound is traveling parallel to the Earth's surface, will it follow along the curvature of the Earth (implying gravity has an effect on it) or was it able to propagate around the Earth four times simply because the intensity was so great that it was able to still be heard despite the loss of energy in the "up-down" directions relative to the origin of the sound?
If I had to completely guess, I would think that gravity might have something to do with its behavior in this case, because if sound decayed at 1/r^2, it seems highly unlikely that the sound could survive four trips around the Earth. Yet we also have to explain how the sound was able to preserve itself even though sound travels "sort of spherically" (if you're up in the sky and someone shoots at you, you can hear it, so sound must be going up-down as well as left-right and every other direction) but if it was simply spherical in this case then it would decay into the upper atmosphere. If you imagine a sphere that grows from any point on the surface of the Earth, most of it will wind up in space, but since it circled the Earth, it must not have traveled exactly as sphere, which implies it didn't fall off as 1/r^2.
The most likely explanation seems to be that sound of huge intensity somehow follows the curvature of the Earth, which is very interesting. I wonder if gravity is influencing the soundwaves directly (which might actually be nonsense, since waves aren't matter, but something that travels through matter) or if gravity somehow sets up the propagation medium in a way that lets the sound curve around the earth without losing intensity as 1/r^2.
You're partially correct about Gravity, but there's more to it.
The radius of the Earth is roughly 6000 km, while the thickness of the atmosphere is about 120km. Thus, the ratio of the two heights is about 1/60. Because the ratio of atmosphere thickness to terrestrial thickness is so small, for the purposes of modeling wave propagation through the atmosphere we can approximate the atmosphere as a thin spherical shell around the Earth.
Furthermore, the Earth's surface forms a hard boundary below the shell, further confining the sound to the atmosphere, while the force of gravity also prevents the sound wave from escaping into space via the atmosphere.
As other posters have stated, this forms a 2-dimensional propagation layer which results in a linear falloff rather than an inverse square falloff.
"which might actually be nonsense, since waves aren't matter, but something that travels through matter"
I don't think that's much of a nonsense. These are waves propagating through a medium - in this case mostly air. And particles of air are affected by gravity, therefore I would very much expect them to follow Earth's curvature.
> And particles of air are affected by gravity, therefore I would very much expect them to follow Earth's curvature.
Gravity isn't the reason sound waves curve around the earth. One reason is that the curved surface is the only avenue open to them. There are a number of other, less influential reasons like different temperatures at different altitudes, but gravity isn't on the list.
If a loud noise was heard there, or nearby, it would likely have been reported...? I didn't get very far searching the web for something like that. One would probably have to go to the Bibliotecas de Bogotá and check newspapers from the day.
In both cases there is a total of something (sound energy or electric field lines) and that total must be split over all possible directions. The total electric field must be split over the surface area of increasingly larger and larger spheres hence the 1/r² behaviour (1/4πr² to be precise). The sound energy is spread uniformly over a circle with circumference C=2πr, hence giving a 1/r decay, over short distances.
For longer distances, the curvature of the earth will play a role. Come to think of it, it must have been really loud somewhere diametrically opposite to Krakatoa, 17 hours after the eruption...