Synchronized clocks with respect to rest frame

[mananvpanchal]

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.

 

[harrylin]

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.

Yes, both clocks are synchronised for both reference systems according to the standard synchronisation convention.

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.

No, that is wrong: according to the standard synchronisation convention, the clocks are now out of synch 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.

- 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.

Harald

 

[Dale]

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 Born-rigid acceleration) then the answer will typically be that they do not remain synchronized in the station's frame either.

 

[AdrianMay]

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.

 

[ghwellsjr]

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.

Somebody forgot to tell Einstein that. In his 1905 paper introducing Special Relativity near the end of section 4 he describes what happens to an accelerating clock compared to an inertial clock. This was the origin of the Twin Paradox which is routinely handled by SR even though at least one of the Twins accelerates.

 

[Dale]

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.

This is not correct. SR can handle acceleration just fine. All it cannot handle is gravitation.

 

[AdrianMay]

Somebody forgot to tell Einstein that. In his 1905 paper introducing Special Relativity near the end of section 4 he describes what happens to an accelerating clock compared to an inertial clock. This was the origin of the Twin Paradox which is routinely handled by SR even though at least one of the Twins accelerates.

Four lines. Hardly a sufficient treatment.

This is not correct. SR can handle acceleration just fine. All it cannot handle is gravitation.

Acceleration and gravitation are indistinguishable under GR, at least over short intervals where tidal affects aren't observable.

 

[harrylin]

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 Born-rigid acceleration) then the answer will typically be that they do not remain synchronized in the station's frame either.

I hoped that it was clear from my analysis that the clocks will be slightly out of synch in the station's rest frame. Note also that the usual assumption of mechanics is that no plastic deformation occurs so that after some time running at constant speed, the acceleration profile doesn't matter.

 

[Dale]

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

 

[mananvpanchal]

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"

 

[AdrianMay]

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.

 

[harrylin]

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)

Not exactly. The effect from length contraction on clock time is what you will hardly measure after the train travels for a long time: it does not go away but it is a small correction on the time dilation effect because B has at any time very slightly less advanced than A. Ideally, according to the platform frame clock A in the rear remains extremely slightly more delayed than clock B in the front. However, I can imagine that if the train is pushed, B might be doing some additional swings in the process which brings them perfectly in tune again or even inverses the effect (is there by chance a math enthusiast in the room?).

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 both clocks together, are they synchronized? If no which one is ahead?

You could try to answer all those questions by calculation (a little complex) or by using the relativity principle (easy): O can assume to be "in rest", so that the situation is symmetrical. The clocks are not synchronized at the start, next they move exactly the same, thus they must stay out of synch.

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.

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, are they synchronized? If no which one is ahead? [..]

See above; I'm sure that you can now answer that yourself.

 

[mananvpanchal]

You could try to answer all those questions by calculation (a little complex) or by using the relativity principle (easy): O can assume to be "in rest", so that the situation is symmetrical. The clocks are not synchronized at the start, next they move exactly the same, thus they must stay out of synch.

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.

I cannot understand this. I don't need any calculation. I need just answer of my confusion.

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?

 

[harrylin]

I cannot understand this. I don't need any calculation. I need just answer of my confusion. [..]

I gave you the answer to your first question ("they must stay out of synch"); your question was clear!

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.

 

[AdrianMay]

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 ;-)

 

[mananvpanchal]

I gave you the answer to your first question ("they must stay out of synch"); your question was clear!

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.

Ok, so my first question's answer is : clocks is out of synch.

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.

 

[harrylin]

Ok, so my first question's answer is : clocks is out of synch.

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.

I wonder if you're right that slow clock transport is one of Einstein's conventions. Anyway, I already gave you an "Einsteinian" reason: it obeys the relativity principle. However, you seem to want a "Lorentzian" reason.
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?

 

[mananvpanchal]

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?

 

[harrylin]

[..]
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?

 

[mananvpanchal]

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?

 

[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. We have two axioms to go on and both explicity restrict themselves to inertial observers. If you are asking about accelerating objects, then you are outside of the scope of the 1905 paper. Anybody who disagrees should show me which part of that paper can answer manam's question. Four lines won't suffice.

Furthermore, there's no point talking about the train standing in the station because then they would be the same reference frame and you wouldn't need the station.

Next, clocks are not synchronised or asynchronised per se, rather, they are synchronised FROM THE POINT OF VIEW OF A PARTICULAR OBSERVER. I'd hazard a guess that if you did sync the clocks while the train was in the station, then accelerated it very gradually, then the clocks would continue to be synced from the point of view of the station, but anybody on the train would think they were getting further and further out of sync as the train accelerated. The reason for that is not about the clocks though - it's about changes to the point of view of the people on the train. But this is just a guess - I can't prove it because I don't have an axiom to go on unless I jump to GR.

If you really want to understand this, read my previous post.

 

[mananvpanchal]

Hello All,

I still don't have satisfactory answer. Please, give me answer with reason

Thanks

 

[AdrianMay]

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.

 

[ghwellsjr]

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. We have two axioms to go on and both explicity restrict themselves to inertial observers. If you are asking about accelerating objects, then you are outside of the scope of the 1905 paper. Anybody who disagrees should show me which part of that paper can answer manam's question. Four lines won't suffice.

Furthermore, there's no point talking about the train standing in the station because then they would be the same reference frame and you wouldn't need the station.

Next, clocks are not synchronised or asynchronised per se, rather, they are synchronised FROM THE POINT OF VIEW OF A PARTICULAR OBSERVER. I'd hazard a guess that if you did sync the clocks while the train was in the station, then accelerated it very gradually, then the clocks would continue to be synced from the point of view of the station, but anybody on the train would think they were getting further and further out of sync as the train accelerated. The reason for that is not about the clocks though - it's about changes to the point of view of the people on the train. But this is just a guess - I can't prove it because I don't have an axiom to go on unless I jump to GR.

If you really want to understand this, read my previous post.

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.

 

[harrylin]

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.

 

[mananvpanchal]

Ok, I asked like this

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

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?

 

[harrylin]

[..] 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!

 

[mananvpanchal]

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.

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.

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.

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.

 

[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.

 

[Dale]

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/physics/Relativity/SR/acceleration.html

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.

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.

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.

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.