Question About Speed Of Movement Through Materials

[DaleSwanson]

A common thought after one learns about the limit of the speed of light is that a long rod when pushed would seem to cause the other end to move instantly, thus carrying information faster than light. The reason this isn't the case is because the movement will only travel through the rod at the speed of sound in that material, quite a bit slower than light.

I was thinking about this the other day and I realized that it must also be true that the amount of force required to move the rod at a certain speed must also be less than the amount for the full rod, at least initially. If pushing the rod with x force caused it to move at y speed you could easily work out it's mass z, and if the rod was a known density and width you'd then know the length. That could be used to send info faster than light.

My question is then how would the local end of the rod behave when receiving a single large push? It would seem as though it would have to move initially fast, and then slow down as the wave of the push moves farther and farther down the rod, and thus more and more mass is moving.

However, I can't resolve the following. Imagine a rod of known density and dimensions, this rod is a light hour long and the end points are at two space stations, with an additional space station at the midway point. At a predetermined time the local end is given a push. For the sake of simplicity let's assume the speed of sound in this rod is 0.01c, so the far end should begin to move at 100 hours, the midpoint at 50 hours. At 49:59 hours the midpoint station decides if it wants to send a 1 or 0 bit, to send a 0 it does nothing. For 1 it cuts the rod in half and moves the far half out of the way. When the wave of movement reaches the midpoint (which is now the endpoint of the half rod) the wave stops, and the local end of the rod stops slowing down. The local station observes if the rod continues slowing down or not and from this knows if the midpoint station intended to send a 1 or 0.

Because of that example I feel my thought for how the motion would travel through the rod must be wrong. Can someone please explain how the end would act?

 

[diazona]

I was thinking about this the other day and I realized that it must also be true that the amount of force required to move the rod at a certain speed must also be less than the amount for the full rod, at least initially. If pushing the rod with x force caused it to move at y speed you could easily work out it's mass z, and if the rod was a known density and width you'd then know the length. That could be used to send info faster than light.

Actually I have my doubts about that. Can you construct an explicit scenario in which determining the length of a rod in this way allows transmission of information faster than light?

However, I can't resolve the following. Imagine a rod of known density and dimensions, this rod is a light hour long and the end points are at two space stations, with an additional space station at the midway point. At a predetermined time the local end is given a push. For the sake of simplicity let's assume the speed of sound in this rod is 0.01c, so the far end should begin to move at 100 hours, the midpoint at 50 hours. At 49:59 hours the midpoint station decides if it wants to send a 1 or 0 bit, to send a 0 it does nothing. For 1 it cuts the rod in half and moves the far half out of the way. When the wave of movement reaches the midpoint (which is now the endpoint of the half rod) the wave stops, and the local end of the rod stops slowing down. The local station observes if the rod continues slowing down or not and from this knows if the midpoint station intended to send a 1 or 0.

Because of that example I feel my thought for how the motion would travel through the rod must be wrong. Can someone please explain how the end would act?

Well, the exact details probably depend on what kind of a push is given at the local end (an impulse push, a continuous constant force, a continuous force at constant velocity, etc.) but basically what you wind up with is a compression wave traveling through the rod. When that wave reaches the midpoint of the rod, it can either reflect (if there is no more rod to be pushed) or continue (if there is another half light-hour of rod). And if it reflects, it takes about 50 hours to get back to the local end of the rod. So the people at the local end have no idea whether there is more rod or not until those extra 50 hours have passed.

Given an exact formula for the wave (which depends on what kind of push is given), you could calculate the velocity of the local end of the rod over time, but I'm pretty sure you'd find that it does take that 100 hours from the initial push for any effect of a cut (or lack of one) at the middle to be felt.

P.S. Great question though ;-)

 

[DaleSwanson]

Actually I have my doubts about that. Can you construct an explicit scenario in which determining the length of a rod in this way allows transmission of information faster than light?

Well use the same setup as above, then the midpoint station simply cuts (or doesn't cut) the rod before the local station gives its push. Knowing how long the rod is at the moment you push it let's you know if the midpoint station cut it or not.

Well, the exact details probably depend on what kind of a push is given at the local end (an impulse push, a continuous constant force, a continuous force at constant velocity, etc.) but basically what you wind up with is a compression wave traveling through the rod. When that wave reaches the midpoint of the rod, it can either reflect (if there is no more rod to be pushed) or continue (if there is another half light-hour of rod). And if it reflects, it takes about 50 hours to get back to the local end of the rod. So the people at the local end have no idea whether there is more rod or not until those extra 50 hours have passed.

Given an exact formula for the wave (which depends on what kind of push is given), you could calculate the velocity of the local end of the rod over time, but I'm pretty sure you'd find that it does take that 100 hours from the initial push for any effect of a cut (or lack of one) at the middle to be felt.

P.S. Great question though ;-)

If you could please explain this a bit more. The knowledge of weather or not the rod is cut traveling back at the speed of sound is what must happen, I just don't understand the how. Specifically, how does the local end act for the first 50 hours (when the wave is traveling through the rod), and then the second 50 (after it has reached the end, and that fact is traveling back). Logic and SR tell me that there must be no change in its behavior.

I suppose this question can be simplified by removing all the cutting business. Given two otherwise identical rods, one which is 1 light hour long, and the other which is half that length (and half the mass as a result). How will they behave when given an identical push? Assume the same 0.01c speed of sound. Also a impulse push, as in a projectile hitting one end and transferring all it's momentum to the rod.

Allow me to add what I think would happen. For the first 50 hours the wave travels from the local end to the far end, during this time the speed of the local end slows down as more and more mass has begun to move, the far end remains still. At 50 hours the wave reaches the far end, and it begins moving, the wave must now travel back through the rod to the local end taking another 50 hours. During the entire 100 hours the wave takes to travel there and back the local end must continues to decelerate. I'm unsure if the far end would be at it's final speed at the 50 hour mark, or if it would take 100 hours more (although I think it would be at the final speed at 50 hours). I'm just having trouble understanding what happens during the second 50 hours when the wave is traveling back.

 

[Vanadium 50]

If pushing the rod with x force caused it to move at y speed you could easily work out it's mass z

Thinking about the rod as having a single speed is equivalent of thinking of it as infinitely rigid. Think of a rod as a spring and you'll have much clearer intuition on how it behaves.