4 ms·

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 ho

by ivan_ah 12y ago

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...

abduhl 12y ago
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.
- femto 12y ago
  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.
  - sillysaurus3 12y ago
    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.
    - kruczek 12y ago
      "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.
      - lutusp 12y ago
        > 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.
dkhar 12y ago
Sound intensity actually falls off as 1/r², and here's a source: http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.html#c1 http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.h...
- sillysaurus3 12y ago
  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?
  - EGreg 12y ago
    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?
    - sillysaurus3 12y ago
      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.
      - jonsen 12y ago
        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 http://en.m.wikipedia.org/wiki/Difraction
lifeformed 12y ago
> ...somewhere diametrically opposite to Krakatoa... Dang, this was your chance to say the word "antipode" in everyday conversation!
pcrh 12y ago
The antipode of Krakatoa is Bogotá, Colombia. I wonder what the news reports from there said at the time?
- junto 12y ago
  I would assume that energy follows the path of least resistance. Hence, all the energy went up and out and not down. Notably Bogota sits in a huge valley, so they are also less likely to have been affected by the surface sound wave as well.
  - pcrh 12y ago
    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. Any Colombians reading this...?

by ivan_ah 12y ago

abduhl 12y ago

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.

femto 12y ago
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.
- sillysaurus3 12y ago
  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.
  - kruczek 12y ago
    "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.
    - lutusp 12y ago
      > 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.

dkhar 12y ago

Sound intensity actually falls off as 1/r², and here's a source: http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.html#c1 http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.h...

sillysaurus3 12y ago
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?
- EGreg 12y ago
  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?
  - sillysaurus3 12y ago
    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.
    - jonsen 12y ago
      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 http://en.m.wikipedia.org/wiki/Difraction

lifeformed 12y ago

> ...somewhere diametrically opposite to Krakatoa... Dang, this was your chance to say the word "antipode" in everyday conversation!

pcrh 12y ago

The antipode of Krakatoa is Bogotá, Colombia. I wonder what the news reports from there said at the time?

junto 12y ago
I would assume that energy follows the path of least resistance. Hence, all the energy went up and out and not down. Notably Bogota sits in a huge valley, so they are also less likely to have been affected by the surface sound wave as well.
- pcrh 12y ago
  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. Any Colombians reading this...?