Mowgli Posted May 25, 2005 Share Posted May 25, 2005 Shall we try a time dilation paradox, again? I have been trying in vain to locate a flaw in the thought experiment below... Imagine a train moving at a constant speed, so that its frame is an inertial frame. At the front end of the train, there is a clock with a trigger mechanism. When the train passes a pole by the track, the clock triggers and records the time reading. There are two poles by the track. Each of them has a similar clock, which triggers when the front end of the train passes it. The clocks on the poles have been carefully synchronized to avoid any ambiguity. The synchronization can be achieved, for instance, by putting the clocks next to each other at the midpoint between the poles and then moving them to the poles at the same speed, accelerating and decelerating in tandem, so that they stay synchronized. As the train passes by, we get four time readings: 1. T1 the time registered when the train passes the first pole, as read by the clock attached to the train, 2. T2 the time registered when the train passes the second pole, as read by the train clock, 3. P1 the time registered by the clock at the first pole as the train passes it, and 4. P2 the time registered by the clock at the second pole as the train passes it. Since T1 and P1 are recorded locally, in the immediate vicinity of the first pole, they should be immune to non-local effects, such as a need to synchronize clocks between their locations. The same argument of locality applies to the second pair of readings (T2 and P2) as well. Once all four time readings are recorded at the respective sites, they are compared. Since the comparison is done at a later point in time, the process of comparison cannot affect the time readings already recorded, even if it involves acceleration or deceleration. (But the necessity for acceleration can be avoided, for instance, by calling the train driver on his mobile phone asking him to read out T1 and T2, and giving him the pole clock readings P1 and P2.) For the stationary observer (staying with the poles), the time interval recorded at the train DT=T2-T1 should be smaller than his interval DP=P2-P1, because the train is in motion, and time dilation applies to its frame of reference. But for the driver moving with the train, the poles are in motion. Each pole clock is dilated, and should be running slower than the train clock. So the difference between their readings should be smaller. DP should be smaller than DT. Since T1, T2, P1 and P2 are four numbers that have already been registered and recorded, DT cannot be both greater and smaller than DP at the same time. Do you think there is a paradox here? If not, which time difference (DP or DT) do you think will be larger? - best wishes, - Mowgli Link to comment Share on other sites More sharing options...
Tom Mattson Posted May 25, 2005 Share Posted May 25, 2005 Shall we try a time dilation paradox' date=' again? I have been trying in vain to locate a flaw in the thought experiment below... [/quote'] You haven't found a flaw in it because you haven't actually worked it out. Define Event 1 to be the coincidence of the train with the first pole and Event 2 to be the coincidence of the train with the second pole. Let the poles be a distance D apart and let g stand for the Lorentz factor. First look at it from the rest frame of the poles (frame S). The train (frame S') is then an inertial frame approaching the poles with velocity v. For simplicity let t1=t1'=0. In frame S t2=D/v. In frame S' t2'=g(t2-vx2/c2)=D/(vg) Now look at it from the rest frame of the train. The poles are then an inertial frame approaching the train with velocity -v. Again let t1=t1'=0. According to S' the distance between the poles is contracted to D/g and t2'=D/(gv), just like before. Using the Lorentz transformation to get t2 yields: t2=g(t2'+vx2'/c2)=D/v, which is exactly what we got before. No paradox at all. Link to comment Share on other sites More sharing options...
Mowgli Posted May 25, 2005 Author Share Posted May 25, 2005 Thanks for the explanation, it makes sense now. Only one thing I'm still not clear about though. I thought that the time dilation was always a dilation, whether the coordinate systems were approaching or receding from each other. When Tom applies Lorentz transformation to go from t2' to t2, he gets t2 greater than t2'. Can the time dilation be a time contraction as well? Mowgli Link to comment Share on other sites More sharing options...
Jacques Posted May 25, 2005 Share Posted May 25, 2005 I thought that the time dilation was always a dilation, whether the coordinate systems were approaching or receding from each other. Me too I think that because in the gamma factor equation speed is squared Link to comment Share on other sites More sharing options...
Tom Mattson Posted May 25, 2005 Share Posted May 25, 2005 Only one thing I'm still not clear about though. I thought that the time dilation was always a dilation' date=' whether the coordinate systems were approaching or receding from each other. When Tom applies Lorentz transformation to go from t2' to t2, he gets t2 greater than t2'. Can the time dilation be a time contraction as well? [/quote'] No, according to each observer the other's clock runs slow. The thing in this case is that you are using two fixed points in frame S to define the events. Since everyone agrees that those points are closer together in S', everyone will agree that Events 1 and 2 take less time in S'. Now if you had used two fixed points that are a distance D apart in S' (instead of S), then you would have gotten the opposite result. Me too I think that because in the gamma factor equation speed is squared That's true but v is not squared in the other factor in the Lorentz transformation. Link to comment Share on other sites More sharing options...
Mowgli Posted May 26, 2005 Author Share Posted May 26, 2005 Thanks for clearing that up, Tom. Tom Mattson:No, according to each observer the other's clock runs slow. Since time dilation is the way one clock runs as seen by an observer in a different frame, is it possible to understand it as a perceptual effect? If I have a clock going away from me at a constant speed, it will appear to me to be running slower, due to the light travel time effects. Is the SR time dilation the same as this perceptual effect? Thanks again, Mowgli Link to comment Share on other sites More sharing options...
Tom Mattson Posted May 26, 2005 Share Posted May 26, 2005 Is the SR time dilation the same as this perceptual effect? No, it isn't. Time dilation has been observed with atomic clocks and with subatomic particle decays. It is so well-verified and well-understood that particle physicists actually use it on a practical level. If a charged particle is very short-lived then expeirmentalists can accelerate a beam of them to very high speeds so that they live longer in the lab frame, and hence they can be better studied. Link to comment Share on other sites More sharing options...
Mowgli Posted May 26, 2005 Author Share Posted May 26, 2005 Originally Posted by Mowgli Is the SR time dilation the same as this perceptual effect? Tom Mattson: No, it isn't. This is the part I haven't understood yet. Suppose I have a clock moving away from me at a constant speed of 0.8c. The clock is programmed to flash every second. I would expect to see the first flash at 1.8 sec (because it is emitted at 0.8c away from me), the second at 3.6 sec and so on. So I would conclude that the time is flowing at the rate of 1/1.8 at the clock. This is the perceptual time dilation. I would be wrong, of course; the first flash won't be emitted one second after the clock leaves me, but 1.67 sec after it leaves me, because of the SR time dilation at the clock. And it will be emitted at a distance of about 1.33c away from me. I will see this flash 3 sec after the clock leaves me, leading me to conclude that the time is flowing at the rate of 1/3 at the clock. So, when SR says that the real time dilation is 1/1.67, am I supposed to observe a time dilation of 1/3 and then deduce from that the time dilation must be 1/1.67? Thanks again, Mowgli. Link to comment Share on other sites More sharing options...
Tom Mattson Posted May 26, 2005 Share Posted May 26, 2005 So' date=' when SR says that the [i']real[/i] time dilation is 1/1.67, am I supposed to observe a time dilation of 1/3 and then deduce from that the time dilation must be 1/1.67? Assuming your arithmetic is correct: Yes. In an expeirment of this type you must correct for the flight time of the light from the clock face. Of course if you wanted to know at what time the flash occured, you would have to make this correction even if Galilean Relativity held. But Special Relativity predicts that even after this correction is made, there will still be a discrepancy, while Galilean Relativity predicts that there will be none. Link to comment Share on other sites More sharing options...
Mowgli Posted May 26, 2005 Author Share Posted May 26, 2005 Tom Mattson:In an expeirment of this type you must correct for the flight time of the light from the clock face. Of course if you wanted to know at what time the flash occured, you would have to make this correction even if Galilean Relativity held. But Special Relativity predicts that even after this correction is made, there will still be a discrepancy, while Galilean Relativity predicts that there will be none. Okay, thanks. For an object moving away from the observer, the light flight time effect predicts the following: The time flow appears dilated The length appears contracted The apparent speed cannot be more than the speed of light It takes an infinite energy to accelerate to an apparent recessional speed of c These look suspeciously like the consequences of the coordinate transformation in SR. That's why I was wondering if they could be the same, or at least related... Mowgli Link to comment Share on other sites More sharing options...
Mag Posted May 31, 2005 Share Posted May 31, 2005 If a charged particle is very short-lived then expeirmentalists can accelerate a beam of them to very high speeds so that they live longer in the lab frame, and hence they can be better studied. sorry to bump this thread... but im confused on that part. I thought the moving object goes slower, and time in the lab stays "the same". and if so, how could we observe it more if our time in the lab stays the same? sorry, kinda rusty on this. I havent had relativity in 2 yrs. Link to comment Share on other sites More sharing options...
swansont Posted May 31, 2005 Share Posted May 31, 2005 sorry to bump this thread... but im confused on that part. I thought the moving object goes slower' date=' and time in the lab stays "the same". and if so, how could we observe it more if our time in the lab stays the same? sorry, kinda rusty on this. I havent had relativity in 2 yrs.[/quote'] When you accelerate the particle, it is no longer in the lab frame. To an observer in lab frame, the particle is moving and its clock runs slow, so it takes longer to decay. Link to comment Share on other sites More sharing options...
Mag Posted May 31, 2005 Share Posted May 31, 2005 ohh right. Its just like the thing where if someone is in space going the speed of light for 2 years, (their POV 2 years), on earth the time is going to be longer. So to earth it looks like the people in space are going slower. correct? Link to comment Share on other sites More sharing options...
Deviation Posted May 31, 2005 Share Posted May 31, 2005 When you accelerate the particle, it is no longer in the lab frame. To an observer in lab frame, the particle is moving and its clock runs slow, so it takes longer to decay. How is the size of a frame determined ? Link to comment Share on other sites More sharing options...
swansont Posted May 31, 2005 Share Posted May 31, 2005 ohh right. Its just like the thing where if someone is in space going the speed of light for 2 years' date=' (their POV 2 years), on earth the time is going to be longer. So to earth it looks like the people in space are going slower. correct?[/quote'] Other than the fact that you can't go the speed of light, yes. Link to comment Share on other sites More sharing options...
swansont Posted May 31, 2005 Share Posted May 31, 2005 How is the size of a frame determined ? The question doesn't make any sense. There is no size of a frame of reference - it's a coordinate system related to a different one by the Lorentz transformation. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted May 31, 2005 Share Posted May 31, 2005 When you accelerate the particle, it is no longer in the lab frame[/b']. To an observer in lab frame, the particle is moving and its clock runs slow, so it takes longer to decay. No longer at rest in the lab frame? Link to comment Share on other sites More sharing options...
swansont Posted May 31, 2005 Share Posted May 31, 2005 No longer at rest[/b'] in the lab frame? Yes. I think that's redundant in the context of the discussion, though. Link to comment Share on other sites More sharing options...
Mag Posted May 31, 2005 Share Posted May 31, 2005 Other than the fact that you can't go the speed of light, yes. yeah i know, im just giving an example. anyways, thanks Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted June 1, 2005 Share Posted June 1, 2005 Yes. I think that's redundant in the context of the discussion, though. I thought I knew what you meant. (I am never "absolutely" sure with frames) I actually believe the particle can be out of that frame in some respects (or approach "out" as c is approached, again in some respects) and that your point was relevant (pun intended) but that it might have lead to the "what determines the size of a frame?" post/question. Link to comment Share on other sites More sharing options...
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