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Synchronized clocks with respect to rest frame 
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#1
Feb2612, 05:24 AM

P: 215

Hello,
Suppose, A and B is clocks at both end of train. A is at left and B is at right. Observer O is at middle of train at point M. Observer R is on platform. Train is at rest and O synchronize both clock. The clock is synchronized with respect to both observer. Now, train starts moving to right. It accelerate and after some time it runs with constant speed. Now, the clock is still synchronized with respect to O. But what about R? Is clocks synchronized with respect to R? If no, then which clock is ahead A or B? Thanks. 


#2
Feb2612, 05:44 AM

P: 3,186

 According to O, clock B in the front is now ahead on clock A in the rear. This is quickly understood with a simplified analysis from the platform: neglecting the small effect from length contraction, both clocks are about equally behind. If O sends a signal to both A and B, clock A is moving towards the signal while B is running away from it. Thus the signals will reach A before B. Consequently, A will indicate less time than B at these events which O defines as simultaneous. Harald 


#3
Feb2612, 06:22 AM

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I agree with harrylin's analysis, but want to point out that the answer depends on the details of the acceleration. This analysis assumes that in the station's frame A, B, and O all have the same acceleration profile. If they have different acceleration profiles (e.g. if the train is being pulled from the front or pushed from the back or undergoes Bornrigid acceleration) then the answer will typically be that they do not remain synchronized in the station's frame either.



#4
Feb2612, 08:16 AM

P: 98

Synchronized clocks with respect to rest frame
SR explicitly chickens out of any cases involving acceleration. As soon as the train started to accelerate, we were out of our depth. You can synchronise clocks when you're still or when you're moving, but you can't accelerate anything during the thought experiment unless you want to learn GR.



#5
Feb2612, 08:50 AM

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#6
Feb2612, 08:51 AM

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#7
Feb2612, 09:00 AM

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#8
Feb2612, 09:08 AM

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#9
Feb2612, 09:14 AM

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Yes, it was clear, I was just adding emphasis.



#10
Feb2712, 03:32 AM

P: 215

Thanks guys for your replies
Ok, so moral of the story is: Clocks synchronized in train frame at rest remain synchronized for platform frame, but not for train frame when train starts moving. (neglecting the small effect from length contraction, which will reduce more if train travels with constant speed for much more time) Clocks synchronized in train frame at motion remain synchronized for train frame, but not for platform. Ok, so here I am confused with two questions. Clocks is synchronized in train frame at rest, but now train is moving with constant speed. clocks is no longer synchronized for O, but they are for R. If O brings all clocks together, are they synchronized? If no which one is ahead? Clocks is synchronized in train frame at motion. Clocks is not synchronized for R, but they are for O. If O brings all clocks together, are they synchronized? If no which one is ahead? Thanks EDIT: please, replace "all clocks" with "both clocks" 


#11
Feb2712, 04:08 AM

P: 98

What do you mean by "bring all clocks together"? You mean the train is driving through the station? That wouldn't matter. The deal is: if you hit the gas or the brakes, the synchronisation goes wrong and you have to resync. Choose your speed, then synchronise, then stick to that speed.



#12
Feb2712, 05:05 AM

P: 3,186

For R, the easy way to solve it is to do just the same: look at the problem from O's perspective, and as the clocks are together the result must be the same from R's perspective. 


#13
Feb2712, 05:19 AM

P: 215

I restate this if it is hard to understand. Clocks is synchronized in train frame at rest, but now train is moving with constant speed. clocks is no longer synchronized for O, but they are for R. If O brings both clocks together in their frame, are they synchronized? If no which one is ahead? Clocks is synchronized in train frame at motion. Clocks is not synchronized for R, but they are for O. If O brings both clocks together in their frame, are they synchronized? If no which one is ahead? Please, tell me the answer.... synchronized or not??? 


#14
Feb2712, 05:36 AM

P: 3,186

The answer to your second question is similar, which you could have figured out for yourself without any calculation, as I explained. What is the use of hearing answers if it doesn't make you understand? Note also that for O the two clocks remain in/out of synch with each other, but will be slightly behind on clocks that didn't move. 


#15
Feb2712, 05:37 AM

P: 98

Aha! I think there's a terminology mismatch going on here. Synchronised might mean that zero on one corresponds with zero on the other, or it might mean they're going at the same speed. That's two different things.
We can all just define zero of time and space (origin) as the place and time where/when the middle of the train passes through the middle of the station. If we'd been writing down numbers before the train arrived on a different basis, we can all just note when in our old sync the origin happened and correct our notes afterwards. This is not the problem. The problem is how fast the clocks run and how long the sticks are. It's not sensible to neglect length contraction compared to time dilation or vice versa because they both happen to the same degree, namely, gamma. So let's just assume we can define the origin clearly. We can because there are no distances to worry about. We put a big red spot on the middle of the train and the middle of the platform, and when they coincide, that's everybody's origin of time and space. Sorted. Now let's get started with the real problem. We don't know how to synchronise clocks that are not at the origin. We can't carry them to the origin, sync them there and carry them elsewhere because WE DONT HAVE AN AXIOM to tell us what happens when we move clocks around. In fact, the only axiom we have that might come in useful here is the one about the constancy of c. We can use that as follows: at time t1, launch a light ray from the origin to the remote clock and back. See it arrive back at t2. Set the remote clock to (t1+t2)/2. Because of the axiom, we can even do that on the train. That's the only method we have that's guaranteed by the axioms. Now you can draw the rhombus diagram to show that the guy on the platform will think that the guy on the train screwed up. You draw the train as a world line at about 1 o'clock, and the light lines at 45 degrees. You see that the light beam reaches the front clock late because the clock is running away from the beam. So the guy on the platform thinks that readings on the front clock underestimate the actual time. Now imagine that the guy on the train wants to measure the platform. Assume that both train and platform are 100m long at rest. He'll put cameras at the front and the back of the train and tell them to go off at t=0. What happens? Options: * both cameras see the exact end of the platform * both cameras see green fields * both cameras see platform with no end in sight * one camera sees fields while the other sees platform. Figure it out for yourself. The correct answer is that both cameras see green fields and conclude that the platform is shorter than the train. Now we reverse the whole argument and see that the platform also thinks the train is too short. We now have three out of four pieces of the puzzle in place: 1) Obviously, the time axis of the train looks wonky from the point of view of the platform. That's just because the train defines his own centre of x as the centre of the train, which the platform can see is moving. 2) We've also established that the space axis of the train is wonky, that's the bad sync of the front and back clocks. 3) We see lorentz contraction resulting from the bad sync of the clocks. To complete the square, we'd like to see time dilation resulting from the disagreement about the centre of x. I'll leave that for homework ;) 


#16
Feb2712, 06:01 AM

P: 215

Can you tell me what is the reason for it? (length contraction or any other) Because, slowly bringing together is one of Einstein's conventions. 


#17
Feb2712, 06:16 AM

P: 3,186

From the point of view of O, I also gave that reason to you: both clocks are equally affected by their motion according to O, and thus they will equally slow down according to O. As a result, clocks that were out of synch stay out of synch. You can surely fill in what the result is for clocks that were in synch, especially as I gave the answer in post #16 (if you cannot, then I'm very sorry, but I won't reply anymore! ). Is that good enough for you or do you want to hear the more complex reason according to R? 


#18
Feb2712, 06:54 AM

P: 215

Suppose, clocks is synchronized by O in rest frame when train is standing still at station. So, clocks is synchronized for both observer O and R. Now, train start moving, both clocks is equally slow down for R, both remain synchronized for R, but not for O. O feels B is ahead of A, because of direction of travelling. Suppose, clocks slowly moved for some distance. now O feels B is less ahead of A. Because signal of light will take less time to reach and come back. difference between clocks decreases when clocks comes nearer and nearer. Can, you tell me what is wrong with this? 


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