## synchronized clocks with respect to rest frame

 Quote by harrylin 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.
Clocks is not moving according to O. because O and clocks are in train frame. I don't want answer for R.

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?

 Quote by mananvpanchal [..] 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?
You see a lightning stroke at a distance, and you hear the thunder 2 seconds later. Now you think that at the place that the lightning struck, the thunder happened 2 seconds after the lightning?

 Quote by harrylin You see a lightning stroke at a distance, and you hear the thunder 2 seconds later. Now you think that at the place that the lightning struck, the thunder happened 2 seconds after the lightning?
The same thing Einstein thinks: two lightning reach to him at different time means it is happened at different time.

If we bring the two clocks together to O. Is it synchronized or not?
 You have several satisfactory answers. Just not in the form you expected. But I think perhaps you already understood it in your own terms: You are right that as the train accelerates, the people on the train see a bigger and bigger discrepancy between the two clocks. Meanwhile, the guy on the platform continues to think the clocks on the train are in sync.

Recognitions:
Gold Member
I don't know which four lines of Einstein's paper you are referring to but the section I previously pointed out to you (section 4) asserts that you can determine what happens in a continuously changing situation by approximating it with a series of straight line segments. This is merely a statement about integration. That section also contains the equation:

τ = t√(1-v2/c2)

This equation computes the instantaneous Proper Time (or tick rate) on a clock moving at speed v with respect to an inertial frame with Coordinate Time t. So we can take any acceleration profile and either calculate by ordinary symbolic integration or by numerical analysis what the time on an accelerated clock will be.

 Quote by mananvpanchal The same thing Einstein thinks: two lightning reach to him at different time means it is happened at different time. [..]
No that's wrong: the thunder happens roughly at the same time as the lightning, because the lightning stroke creates the thunder sound. The time difference between the thunder and the lightning is due to the different times to reach you. Such propagation times are taken in account in theories. That is what was wrong with your analysis.

 Quote by mananvpanchal 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?
And you correct me by this

 Quote by harrylin No, that is wrong: according to the standard synchronisation convention, the clocks are now out of synch with respect to O. - Both moving clocks are now very slightly behind according to R. - 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.
So, conclusion is:
when train is stayed, both clocks is synchronized for both observer.
But, when train starts moving clocks is very slightly asynchronous with respect to R.
But, clocks is out of sync with respect to O, because of direction of motion.

So, I am just trying to understand is if train will stop the clocks becomes again synchronous for both observer?
If clocks brought near to O slowly, can O see decreasing difference of clock reading?

 Quote by mananvpanchal [..] So, conclusion is: when train is stayed, both clocks is synchronized for both observer. But, when train starts moving clocks is very slightly asynchronous with respect to R. But, clocks is out of sync with respect to O, because of direction of motion.
To be more precise: both clocks will be slightly behind (=>out of synch) as compared by R with R's clocks, and one of them may be very slightly more behind than the other one.
 So, I am just trying to understand is if train will stop the clocks becomes again synchronous for both observer?
After the train stops:
- R can compare the clocks with its own reference clocks. For R the clocks will both be slightly behind.
- O has no own valid reference clocks to compare the clocks with; however, according to O the clocks are again in synch with each other (exactly or very nearly so).
 If clocks brought near to O slowly, can O see decreasing difference of clock reading?
No.
Again: according to O, both clocks move exactly the same. How could one be affected differently from the other?

I give up. Perhaps someone else wants to try? Good luck!

 Quote by harrylin To be more precise: both clocks will be slightly behind (=>out of synch) as compared by R with R's clocks, and one of them may be very slightly more behind than the other one.
Ok, this is fine.

 Quote by harrylin After the train stops: - R can compare the clocks with its own reference clocks. For R the clocks will both be slightly behind.
Yes, that is same as twin paradox. O is less aged than R.

 Quote by harrylin No. Again: according to O, both clocks move exactly the same. How could one be affected differently from the other?
I think we both talking same. I didn't say one clock is affected differently from other.

 Quote by harrylin I give up. Perhaps someone else wants to try? Good luck!
Why? please don't do that. You have to explain me much more... just kidding.

Anyway, thanks for your replies and passion.
 I thought one of the postulates of SR is that on object in constant motion can't distingiush between that and being at rest. It would seem like that if A and B could be measured to not be in sync with O from O, then O could then conclude that it was in fact in motion. So then, only R could conclude that the clocks where not in sync but O would conclude that they are. This would be because O would measure the speed of light to be the same forwards and backwards through the train since it assumes it is at rest, and R concludes that the light traveling to the back of the train from O reaches first, since it observes the speed of light to be the same in both directions, but the velocity of the train itselfs shortens the distance it has to travel. But, does this actually puts the clocks out of sync from the frame of reference of R? All three clocks would have been seen from the frame of reference of R to all have gone the same speed that should calculate into them expereincing the same amount of time dialation. All it would meen is that a signal seen from R from O would hit A and B at different times. It would seem like each side of an object shouldn't experience different amounts of time dialation. I think it only means that a signal form O seen from R would reach A and B at two different times, but the clocks themselves would be in sync if you could check using some sort of action at a distance.

Mentor
 Quote by AdrianMay Let me give you the correct answer to your question: if the train was at rest and it starts moving, AS FAR AS WE CAN TELL UNDER THE AXIOMS OF SR, the clocks might stand on their heads and sing the Alleluliah Chorus.
This is simply untrue. See the Usenet Physics FAQ on the topic:
http://math.ucr.edu/home/baez/physic...eleration.html

 Quote by AdrianMay We have two axioms to go on and both explicity restrict themselves to inertial observers.
Also untrue. The postulates restrict themselves to inertial frames. You can have non-inertial observers and objects moving in an inertial frame, you just cannot build an inertial reference frame where they are at rest. See the FAQ linked above.

However, although neither postulate explicitly mentions non-inertial reference frames, from an inertial reference frame it is simply a mathematical transform to obtain the physics of a non-inertial reference frame. Thus, SR can deal with non-inertial reference frames as well. The two postulates do not apply directly, but the physics can nevertheless be derived from the postulates in a mathematically rigorous way.

 Quote by AdrianMay If you are asking about accelerating objects, then you are outside of the scope of the 1905 paper.
Also untrue. Einstein explicitly deals with accelerating clocks in section 4.

 Quote by AdrianMay Four lines. Hardly a sufficient treatment.
Nonsense. Exactly how many lines are required for a sufficient treatment? What if I set Einstein's treatment in a larger font with a narrower column width so that it takes the required number of lines, does the treatment suddenly become sufficient?

The sufficiency of the treatment has nothing to do with the length. If a correct result is derived or explained in a few words, then that is a credit to the treatment, not a detraction. In this case, Einstein succinctly and clearly extended the time dilation of an inertial clock to the case of an accelerating clock. It is clearly part of the 1905 paper, and trying to pretend otherwise really weakens your credibility.

 Quote by AdrianMay Acceleration and gravitation are indistinguishable under GR, at least over short intervals where tidal affects aren't observable.
This actually contradicts the point you are trying to make. The whole point of the equivalence principle is that, over a small region, GR reduces locally to SR. So the fact that you can already deal with acceleration in SR is (via the equivalence principle) what allows you to know how to deal with gravity in GR.

The Pound Rebka experiment is a classic example of this. You can analyze the Pound Rebka experiment as an experiment on an accelerating rocket far from gravity using SR. You then know immediately the result you expect in the stationary lab under gravity using GR.

 Quote by John232 I thought one of the postulates of SR is that on object in constant motion can't distingiush between that and being at rest. It would seem like that if A and B could be measured to not be in sync with O from O, then O could then conclude that it was in fact in motion. So then, only R could conclude that the clocks where not in sync but O would conclude that they are. This would be because O would measure the speed of light to be the same forwards and backwards through the train since it assumes it is at rest, and R concludes that the light traveling to the back of the train from O reaches first, since it observes the speed of light to be the same in both directions, but the velocity of the train itselfs shortens the distance it has to travel. But, does this actually puts the clocks out of sync from the frame of reference of R? All three clocks would have been seen from the frame of reference of R to all have gone the same speed that should calculate into them expereincing the same amount of time dialation. All it would meen is that a signal seen from R from O would hit A and B at different times. It would seem like each side of an object shouldn't experience different amounts of time dialation. I think it only means that a signal form O seen from R would reach A and B at two different times, but the clocks themselves would be in sync if you could check using some sort of action at a distance.
Thank you very much John232.

This is what exactly I want to describe. Somehow I could not explain myself.
One way speed of light is same for all, then clocks is synchronized with respect to O. And if R can see the clocks by sending light pulse. R can also conclude that the both clock is synchronized with each other, but the clocks is behind of his own clock due to time dilation.
 Quote by mananvpanchal 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?

Mentor
 Quote by John232 I thought one of the postulates of SR is that on object in constant motion can't distingiush between that and being at rest. It would seem like that if A and B could be measured to not be in sync with O from O, then O could then conclude that it was in fact in motion.
But O isn't in constant motion, O accelerates. That acceleration is detectable in multiple ways, including the desynchronization of A and B.

Recognitions:
Gold Member
 Quote by DaleSpam Nonsense. Exactly how many lines are required for a sufficient treatment? What if I set Einstein's treatment in a larger font with a narrower column width so that it takes the required number of lines, does the treatment suddenly become sufficient?