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.

 

[mananvpanchal]

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.

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?

 

[Dale]

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.

 

[nitsuj]

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?

ah ha ha, reading this thread was worth that laugh.

I'm learning about "debate logic", AdrianMay you should do the same. The comment "Four lines. Hardly a sufficient treatment." I'd class as non-sequitor (lines of text has nothing to do with suffucient treatment") &/or it begs the question. I'm not too sure just learning still.

 

[Dale]

Thanks nitsuj. I like this site for learning about logical fallacies:
http://www.fallacyfiles.org/

I agree with your classification as a non-sequitor. The fallacyfiles site calls it the "red herring" fallacy.

I wouldn't classify it as begging the question because begging the question means that the conclusion (Einstein's treatment is insufficient) is contained in the premises (Einstein's treatment is 4 lines long) or that the argument assumes some point not conceded by the other side (I concede the length of the treatment). In this case, the conclusion is not contained in the premise and the premise is not controversial in the discussion, it is simply that the premise (length) is completely irrelevant to the conclusion (sufficiency).

 

[John232]

But O isn't in constant motion, O accelerates. That acceleration is detectable in multiple ways, including the desynchronization of A and B.

I don't recal ever hearing that clocks that accelerated together at the same rate would ever no longer be synchronized. Thinking back to the Einstein in the elevator thought experiment, I thought, what if he had a stepladder. Could he move up the step ladder and find that his light ray bends by a different amount? I think no at each step, but yes as he moved up the steps. Then what if he did the same experiment at Earth? Closer to the Earth the beam would bend more, since it would have a stronger gravitational pull. If he moved farther away from the Earth it would bend less being farther away from the planet.

The trains acceleration, unlike Earths gravitational pull, would act equally on the whole train if it was rigid.

 

[mananvpanchal]

Hello,

The below images show O's accelerating motion with respect to R. As speed changes desynchronization becomes bigger with respect to R.

Please, look at first frame of below image. It shows desynchronization of clocks when frame is in motion. But, it is with respect to R.

When we transform to first frame, as below image shows there is no longer desynchronization with respect to O. But, second frame is desynchronized with respect to O.

But, again we transform to second frame. there is no longer desynchronization with respect to O. But, first frame is desynchronized with respect to O.

So, desynchronization is only there when we see other frame. Our own frame cannot have desynchronization with respect to itself.

So, as I understand constant speed of frame would not create desynchronization, and changing in frame (acceleration) also would not create desynchronization.

Can you explain me how acceleration can create desynchronization with respect to O?

 

[Dale]

I don't recal ever hearing that clocks that accelerated together at the same rate would ever no longer be synchronized.

In an inertial frame this is correct. Two clocks which are initially synchronized and share the same speed profile will remain synchronized, but only in that inertial frame.

In other inertial frames or in non inertial frames they will not be synchronized. O's frame is non inertial.

 

[mathal]

|mananvpanchal first post (with quotation marks) and my comments without quotation marks.

"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."

clocks are synchronized.

"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."

At the point that the entire train is again inertial, yes the clocks, tell the same time as O and O confirms this as being true.

During acceleration A clock will appear to run behind O and B will appear to run ahead of O WRT the inertial synchronization (also B will appear larger and A will appear smaller -very minutely during acceleration).The aberration in time measurement during acceleration corrects itself when the object returns to an inertial state- there is no measurable difference in the clocks after the train return to an inertial state.

"But what about R? Is clocks synchronized with respect to R? If no, then which clock is ahead A or B?"

Clearly The clock of R will be running faster than A and B and O and continue to run at the same faster rate with the train again inertial. From the moment the train started moving WRT R, the clock of R was not synchronized with the clocks on the train.

"Thanks."

I hope this answers your question.
mathal

 

[mananvpanchal]

There is other doubt too.

Please, look at first frame of below image. It shows desynchronization of clocks when frame is in motion. But, it is with respect to R.

second image of post #36 says clocks is synchronized with respect to O, but not for R. For R clock A is ahead of clock B, because signal reaches lately to B with respect to R because of direction of motion.

Suppose, that train is coming from far left of R, it passes R and goes far right with constant velocity. So below image show how clocks becomes desync with respect to R. It also tells train coming from left and goes right not make changes in desync. A is always ahead of B with respect to R during whole journey.

But, wait if we think how R sees clocks during whole journey. Below image show that

There is matter that train coming from left and goes to right. Coming from left tells R that clock B is ahead of A, and going right tells R that clock A is ahead of B.
Look, at the point where R and O is on perpendicular line of train path way. This is the point where R thinks both clock is synchronized.

From my understanding, first image shows that "R thinks that clocks would be desynchronized like this with respect to O". Neither for R the clocks would be desynchronized like this, nor for O.

What I understand is clocks is perfectly synchronized for O and, clocks is desyhnchonized with respect to R as showed in second image.

Can anyone shade some light on this?

Thanks

 

[John232]

In an inertial frame this is correct. Two clocks which are initially synchronized and share the same speed profile will remain synchronized, but only in that inertial frame.

In other inertial frames or in non inertial frames they will not be synchronized. O's frame is non inertial.

I fail to see how that is true without an explanation. For instance, how does the relation of the distance of two objects make them observered from rest as measuring time differently when they have accelerated the same amount for the same duration? I think the proof in the pudding here would be that if you considered each object separately you would find that they both expereince the same amount of time dialation, but if you considered them together or linked to an action they would be seen to experience different amounts of time dialation. So, then why would an object require different amounts of spacetime dilation if it was seen to travel at the same speed of another object in order to measure the same speed for light? It would make it seem that the origanal answer of the amount of spacetime dilation occurred for each object would be false unless you considered your own relation to the object. Then answers that considered your relation to the object would not be interchangeable with answers that didn't consider your relation to the object, since they would give different values. Then if the object is traveling with a velocity then would the amount of spacetime dilation that occurred change as it moved to a different location? Then the whole thing falls apart, none of the results would give the same speed of light or the same amount of time dilation if it was considered to be X units away from something else, and that affect the outcome of the amount of spacetime dilation that occured.

For each velocity, there could only exist one distortion of spacetime that would conclude that there is the same value for the speed of light. For instance, two people traveling together on a train couldn't be seen to measure time differently because then they would each have different values of the speed of light. Their location as being in the front or back of the train doesn't affect how they can make that measurement. The basis of SR is that the speed of light is constant in all frames. I think clocks should only be checked if they are synchronized using an entanglement experiment, something Einstein never would have done because nothing was supposed to travel faster than the speed of light at the time.

 

[mathal]

When I posted my responce I edited myself twice- and I'm still not satisfied with what I wrote. It all boils down to the measurement problem.
If there are only 2 clocks held by A and B then while the train is inertial O will observe the same time on both clocks.
For instance if the train is dark inside and the clocks flash seconds the signals will reach O at the same moment. If O uses a flash of light to see each clock the images will show the same time.

During acceleration things are different.
Using flashing clocks the signal from A is traveling to O who is accelerating away from this flash and so will receive it later than the flash from B which is approaching him as he accelerates towards it. The impression this gives is that A clock is lagging behind B clock.
The observation when O uses flashes of light towards A and B clocks is the opposite. His flash arrives later to B then it does to A. The picture of the clock times gives the impression that A clock is running faster than B clock. (tic) The images of the clocks will arrive back to O at the same moment.
If the acceleration is constant there will be a constant lag time.

Simultaneous identical acceleration of two clocks does not break the synchronization.
mathal

 

[Dale]

I fail to see how that is true without an explanation.

Can you explain me how acceleration can create desynchronization with respect to O?

Certainly, it follows directly from the Lorentz transform. Let's analyze the scenario from the OP where at t=0 in the original frame A, B, and M all accelerate instantaneously up to v = .6c and furthermore let's use units where c=1 and where the distance from M to A and from M to B is 1 in the original frame.

So, in R's frame the worldline of A, O, and B are:
r_d=\left(t,x=\begin{cases}<br /> d &amp; \mbox{if } t \lt 0 \\<br /> 0.6 t+d &amp; \mbox{if } t \ge 0 <br /> \end{cases}<br /> ,0,0\right)
where d=-1 for A, d=0 for O, and d=1 for B.

As per the OP, the clocks are initially synchronized in R's frame such that at t=0 they all read 0. So, we can calculate the time displayed on each clock, τ, using the spacetime interval. Solving for t we get:
t=\begin{cases}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}

Substituting into the above we get:
r_d=\left(<br /> t=\begin{cases}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases},<br /> x=\begin{cases}<br /> d &amp; \mbox{if } \tau \lt 0 \\<br /> 0.75 \tau+d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> ,0,0\right)

Noting that τ does not depend on d in this frame we see immediately that the clocks remain synchronized in R's frame.

Now, boosting to the primed frame where O is at rest for τ=t>0, we obtain.
r&#039;_d=\left(<br /> t&#039;=\begin{cases}<br /> 1.25 \tau - 0.75 d &amp; \mbox{if } \tau \lt 0 \\<br /> \tau - 0.75 d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases},<br /> x&#039;=\begin{cases}<br /> 1.25 d - 0.75 \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> ,0,0\right)

Noting that τ does depend on d in this frame we see immediately that the clocks are desynchronized in O's frame. If there is some step that you do not follow then please ask for clarification. But it is quite clear the the clocks remain synchronized in the unprimed frame and are not synchronized in the primed frame.

 

[mananvpanchal]

Thanks for your reply.

Certainly, it follows directly from the Lorentz transform. Let's analyze the scenario from the OP where at t=0 in the original frame A, B, and M all accelerate instantaneously up to v = .6c and furthermore let's use units where c=1 and where the distance from M to A and from M to B is 1 in the original frame.

1. r_d=\left(t,x=\begin{cases}<br /> d &amp; \mbox{if } t \lt 0 \\<br /> 0.6 t+d &amp; \mbox{if } t \ge 0 <br /> \end{cases}<br /> ,0,0\right)

2. t=\begin{cases}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}

3. r_d=\left(<br /> t=\begin{cases}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases},<br /> x=\begin{cases}<br /> d &amp; \mbox{if } \tau \lt 0 \\<br /> 0.75 \tau+d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> ,0,0\right)

I follow this.

Now, boosting to the primed frame where O is at rest for τ=t>0, we obtain.
4. r&#039;_d=\left(<br /> t&#039;=\begin{cases}<br /> 1.25 \tau - 0.75 d &amp; \mbox{if } \tau \lt 0 \\<br /> \tau - 0.75 d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases},<br /> x&#039;=\begin{cases}<br /> 1.25 d - 0.75 \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> ,0,0\right)

But I don't follow this. I have tried to get it, but I don't succeed. That's fine, don't worry about it.

From 1st equation the calculation is done considering speed 0.6c.
So, you are trying to say is:
at t < 0, clocks is synchronized for both R and O. And at t = 0 train changes its frame and travels at 0.6c, and at t=0 the clocks becomes desynchronized for O.

Whatever you stated before you state same here with maths.
Actually, I like maths, but, equations comes from theory. It state same thing as theory states. I need to understand the theory first.

Actually I need be explained that how clocks becomes desynchronized for O, when train changes its frame.

Please, shade some light on my post #36 and #39. Please, tell me what is wrong with my understanding.

Thanks again DaleSpam.

 

[mananvpanchal]

Thanks mathal

When I posted my responce I edited myself twice- and I'm still not satisfied with what I wrote. It all boils down to the measurement problem.
If there are only 2 clocks held by A and B then while the train is inertial O will observe the same time on both clocks.
For instance if the train is dark inside and the clocks flash seconds the signals will reach O at the same moment. If O uses a flash of light to see each clock the images will show the same time.

During acceleration things are different.
Using flashing clocks the signal from A is traveling to O who is accelerating away from this flash and so will receive it later than the flash from B which is approaching him as he accelerates towards it. The impression this gives is that A clock is lagging behind B clock.
The observation when O uses flashes of light towards A and B clocks is the opposite. His flash arrives later to B then it does to A. The picture of the clock times gives the impression that A clock is running faster than B clock. (tic) The images of the clocks will arrive back to O at the same moment.
If the acceleration is constant there will be a constant lag time.

Simultaneous identical acceleration of two clocks does not break the synchronization.
mathal

That is what I think. Constant speed cannot affect synchronization. And simultaneous identical acceleration creates constant deference between clocks. As soon as the speed becomes constant the clocks again becomes synchronized.

 

[John232]

Looks to me like you rigged the equations to give different values for A, O, and B from the intitial setup of the first equation. Say for instance instead, I used different equations like the time dilation equation. Then I solved for A, O, and B. I would get the same answer for each one that would end up getting me different values than what you got...

 

[John232]

This would be because d has different values for A, O, and B. If I simply put the values of A in the time dilation equation it would have the same values for O and B, so then I would know that they should all come out to be the same since they had the same value for t and v. Distance wouldn't be an issue in my setup. I don't think it should since the distance the objects are away from each other wouldn't affect how they measured the speed of light...

 

[Dale]

Looks to me like you rigged the equations to give different values for A, O, and B from the intitial setup of the first equation. Say for instance instead, I used different equations like the time dilation equation.

There is more to relativity than just time dilation. The Lorentz transform is more general and automatically reduces to the time dilation formula whenever appropriate. I recommend against ever using the time dilation formula. It will cause mistakes, as this example would show.

 

[Dale]

But I don't follow this. I have tried to get it, but I don't succeed. That's fine, don't worry about it.

So, using the matrix form of the Lorentz transform we have:

r&#039;_d=\Lambda \cdot r_d = \left(<br /> \begin{array}{cccc}<br /> 1.25 &amp; -0.75 &amp; 0 &amp; 0 \\<br /> -0.75 &amp; 1.25 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right) \cdot \left(<br /> \begin{array}{c}<br /> t = \begin{cases}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases} \\<br /> x = \begin{cases}<br /> d &amp; \mbox{if } \tau \lt 0 \\<br /> 0.75 \tau+d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> \\<br /> 0 \\<br /> 0<br /> \end{array}<br /> \right)

r&#039;_d=\left(<br /> \begin{array}{c}<br /> t&#039;= -0.75 \begin{cases}<br /> d &amp; \mbox{if } \tau \lt 0 \\<br /> 0.75 \tau+d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> + 1.25 \begin{cases}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases} \\<br /> x&#039;=<br /> 1.25 \begin{cases}<br /> d &amp; \mbox{if } \tau \lt 0 \\<br /> 0.75 \tau+d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> - 0.75 \begin{cases}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases} \\<br /> 0 \\<br /> 0<br /> \end{array}<br /> \right)

r&#039;_d=\left(<br /> t&#039;=\begin{cases}<br /> 1.25 \tau - 0.75 d &amp; \mbox{if } \tau \lt 0 \\<br /> \tau - 0.75 d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases},<br /> x&#039;=\begin{cases}<br /> 1.25 d - 0.75 \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> ,0,0\right)

 

[Dale]

Please, shade some light on my post #36 and #39. Please, tell me what is wrong with my understanding.

If you go back to your #36 and #39, you can see that you have identified some events on A and B which are identified as being simultaneous for O.

What you have not done is indicate what the clocks at A and B read at those events. However, if you look at the length of the line from the bend, when the clocks all read 0, to the events in question you will see that they are different lengths.

Therefore A and B will not read the same at those events. They will therefore be desynchronized according to O.

 

[John232]

There is more to relativity than just time dilation. The Lorentz transform is more general and automatically reduces to the time dilation formula whenever appropriate. I recommend against ever using the time dilation formula. It will cause mistakes, as this example would show.

Thats like saying Lorentz was the real genius and Einstein was a dumb***. I went to the wiki and in the fine print under the first diagram to the right it says that it describes when one event detects another event. So then any result useing that equation would only tell what each observer see's when they detect an event. So that doesn't mean that A and B are no longer in sync it only means that the signal to R is not in sync. I wouldn't use the Lorentz Transform to ever find out any information about what clocks say to determine the amount of time dilation from one frame of reference to another, or you would always be wrong, since your just finding out what an observer detects at a certain moment.