Explore Clock Synchronisation in a Moving Train Frame

[Foppe Hoekstra]

TL;DR Summary

A clock ticking slower when it moves in respect to a substantial mass, I can understand, Hafele Keating showed it does and it is even measurable.
Still, Lorentz time dilation puzzles me.

Consider the full-circle train (radius R) of a great number (n) of wagons, coupled by clocks that can print a dot on the rail. All clocks are synchronised when the train is at rest (in regard to the rail, being the Stationary Frame SF) and then the train is accelerated to a constant speed v. After that in the centre of the circle an omnidirectional light signal is given. Every clock will print one dot on the rail and print its own reading ti (i = 1,2,3,…n) on a receipt when the signal is received.

I presume that all dots will be printed in a regular pattern, each pair of sequential dots on a distance of L = 2πR/n. But what times will be printed on the receipts? (I think it will be ti = t(i+1), as the signal comes from SF and in SF all clocks will still be synchronised.)

Now we board the train. To have the dots printed at the distance L (in SF), the printing by the front clock of any wagon cannot have been sync (in the co-moving frame of the wagon) with the printing of the rear clock of the same wagon; the front printing should be ahead vL/(c^2) relative to the rear printing, so time dilation ensures printing just on time in SF. And although every wagon should be seen as another (almost inertial) moving frame, it still means that the printing by clock 1 is ahead of the printing by clock zero, clock 2 is ahead of clock 1, et cetera, till finally clock n is ahead of clock n-1 and all together it means that the printing by clock n is ahead of the printing by clock zero. But clock n is clock zero! How does that work?

 

[phinds]

Summary: A clock ticking slower when it moves in respect to a substantial mass, I can understand, Hafele Keating showed it does and it is even measurable.

That is not correct. Clocks always tick at one second per second. What you are talking about is the APPEARANCE of ticking slower due to either gravitational time dilation or time dilation due to relative motion.

You are probably confused by differential aging, which is something that does happen but does not mean that clocks tick slower, just that they can take different paths through space-time than one another and thus have a differing number of ticks from the time they are separated to the time they come back together.

 

[PeterDonis]

what times will be printed on the receipts?

The easiest way is to analyze everything in the SF. In that frame each light signal reaches all wagons simultaneously, and the wagons are all time dilated by the same factor $\gamma$, so if $t_i$ is the time in the SF at which light signal $i$ reaches all the wagons, then each wagon will print $t_i / \gamma$ on its receipt for that signal.

Note that, as the above shows, in the SF the wagon clocks all remain synchronized; they all just slow down by the time dilation factor. That is because the SF uses the SF's simultaneity convention, not the simultaneity convention of any of the wagon comoving frames.

the co-moving frame of the wagon

Each such co-moving frame is different, and the wagon clocks are not synchronized in any of them. In fact, each wagon is only co-moving in a particular inertial frame for a single instant; the wagons are moving in a circle and always have nonzero proper acceleration. So any attempt to construct a "wagon frame" even for a single wagon in which that wagon is always at rest, requires constructing a non-inertial frame and taking particular care to do so consistently. That is a hard problem and doesn't tell you anything about the physics that isn't already told by the simple SF analysis done above.

every wagon should be seen as another (almost inertial) moving frame

For a single instant, yes. But at any given instant, all such frames are moving in different directions. So transforming between them is highly non-trivial. It certainly can't be done in the simple way you are trying to do it.

the printing by clock n is ahead of the printing by clock zero. But clock n is clock zero!

This just means you tried to construct a non-inertial frame in which clock 0/clock n are at rest, and didn't pay sufficient attention to all the details you need to pay attention to in order to do it properly. It's better just to avoid that hassle altogether and analyze things in the SF.

 

[Ibix]

All clocks are synchronised when the train is at rest (in regard to the rail, being the Stationary Frame SF) and then the train is accelerated to a constant speed v.

The devil is in the detail here. Is every wagon coupled to both the ones next to it, or is there a front and back wagon that are not coupled to each other? In the first case, the train clocks will remain synchronised in the stationary frame; in the second case they will not, and a gap will open up between the front and rear carriages.

I'm assuming you meant the first case, from what you subsequently wrote.

I presume that all dots will be printed in a regular pattern, each pair of sequential dots on a distance of L = 2πR/n. But what times will be printed on the receipts? (I think it will be ti = t(i+1), as the signal comes from SF and in SF all clocks will still be synchronised.)

If they receive the triggering pulse at time $t$ then each clock will print $t/\gamma$. They all tick slowly in the stationary frame, but there is no relative offset.

But clock n is clock zero! How does that work?

As per the last four or five times we've discussed this, you are chaining together local inertial frames. This leads to a construct that "goes wrong" somewhere around the loop.

You can use a single global inertial frame that covers the whole train and in which one carriage is instantaneously at rest. In this frame, all of the other carriages are moving at different speeds, so their tick rate varies round the loop in a consistent way.

Or you can construct a global non-inertial frame - for example using the time signal from the centre of the track as a source of simultaneity (other options are available). You cannot use Einstein synchronisation.

Or you can use Einstein synchronisation along the train and work out some way to extend it to points off the train. In any case, you need to accept that this approach cannot cover all of spacetime - there is a plane somewhere around the track where clock readings are inconsistent and this method does not give you an answer to "what time is it". That's a fundamental limitation of this approach. It's not really any more mysterious than the international date line, where "what day is it" can have two perfectly plausible answers.

 

[PeterDonis]

You can use a single global inertial frame that covers the whole train and in which one carriage is instantaneously at rest.

You could, but I think this would be adding a lot of difficulty for no real benefit.

Or you can construct a global non-inertial frame

Yes, this would be Born coordinates:

https://en.wikipedia.org/wiki/Born_coordinates

Or you can use Einstein synchronisation along the train and work out some way to extend it to points off the train.

No, you can't use Einstein synchronization along the train while it is moving in a circle.

The other option is to just do the analysis in the stationary frame (the inertial frame in which the train tracks and the light emitter at the center are at rest), in which it is easy.

 

[Ibix]

No, you can't use Einstein synchronization along the train while it is moving in a circle.

I'm not sure I agree. A very short carriage is instantaneously inertial (edit: as noted below, this should say "is covered by an inertial frame in which it is instantaneously at rest") and you can Einstein-synchronise clocks along at its ends. Nothing stops you chaining this procedure, except that when you get back to the start clock you are out of sync. But all you have to do is not apply that coordinate system on the worldsheet swept out by the radius through that point on the train. (And you need to think up some way of generalising the procedure to points off the track). That assigns one coordinate label to each point in spacetime (excluding a small region), with a smooth relationship to (for example) inertial coordinates.

That's fine, isn't it? I'm not saying it's a particularly good idea, but I think it's legitimate.

 

[PeterDonis]

A very short carriage is instantaneously inertial and you can Einstein-synchronise clocks along at its ends.

Being "instantaneously inertial" is not enough. The entire system that is being Einstein synchronized needs to be inertial with all its parts at rest relative to each other for the time it takes round-trip light signals to cross it, since that is what Einstein synchronization uses. There is no way to do this for the entire train. Of course you can do it for a sufficiently short single wagon in the train ("sufficiently short" defined by the light travel time as compared with the angular velocity of rotation), but this just amounts to using the instantaneously comoving inertial frame of one particular wagon; it's not a separate option.

 

[Dale]

And although every wagon should be seen as another (almost inertial) moving frame,

Why would you think that? Isn’t this not just false but obviously so?

How does that work?

It doesn’t. Why would you expect treating a non inertial frame as though it were inertial to work? It’s a bad assumption and that is exactly why it doesn’t work.

 

[Ibix]

Of course you can do it for a sufficiently short single wagon in the train ("sufficiently short" defined by the light travel time as compared with the angular velocity of rotation), but this just amounts to using the instantaneously comoving inertial frame of one particular wagon; it's not a separate option.

Once those clocks are synchronised they will not drift. So you can use the same process to sync the clock in the next carriage to your front clock using Einstein synchronisation. And you can continue round the train. Obviously you run into a problem when you complete the loop, which is why I said you need to exclude a point.

I don't see the problem with this for defining a foliation (of most of spacetime), except that I haven't specified how it works off the track. Are you arguing that a clock synchronisation procedure implies a foliation but a foliation does not necessarily imply a clock synchronisation convention? Or that there is something wrong with my foliation? Or am I missing something?

 

[DrGreg]

except that I haven't specified how it works off the track

I think this is where it goes wrong. If you model the train as a one-dimensional circular arc (in space) then I think you can set up a two-dimensional orthogonal coordinate system (in spacetime) using the method you describe, but you can't extend it to four dimensions.

Suppose you arrange some clocks in a circle on the floor of one of the train carriages and try to Einstein-synchronise them working your way round the circle. You'll find the same problem when you try to complete the circle -- the last clock will be out-of-sync with the first (because the carriage is rotating). And this doesn't just apply to circles, I believe it applies to any closed path that encloses a non-zero area.

 

[A.T.]

Consider the full-circle train (radius R) of a great number (n) of wagons, coupled by clocks that can print a dot on the rail. All clocks are synchronised when the train is at rest (in regard to the rail, being the Stationary Frame SF) and then the train is accelerated to a constant speed v.

Is the train forced to stay a full circle during the acceleration? How is this different from your previous scenario, that was already explained here?

https://www.physicsforums.com/threa...ontradiction-in-rt.972918/page-2#post-6188566

in the co-moving frame of the wagon

You haven't defined that frame properly, just like in your previous threads.

 

[Dale]

A very short carriage is instantaneously inertial and you can Einstein-synchronise clocks along at its ends.

An accelerometer carried on a very short carriage will read non-zero at every instant. In what sense is something with a non-zero accelerometer at a given instant instantaneously inertial?

You can construct a momentarily comoving inertial frame, but it is the frame that is inertial and not the object, not even instantaneously.

 

[PeterDonis]

Once those clocks are synchronised they will not drift. So you can use the same process to sync the clock in the next carriage to your front clock using Einstein synchronisation.

No, you can't, because the paths of light rays won't be the same relative to the wagons, since the wagons now have nonzero proper acceleration.

 

[PeterDonis]

I don't see the problem with this for defining a foliation (of most of spacetime),

Not of "most" of spacetime. Only of a small enough region. Which in practice will only be a small "world tube" surrounding the worldline of one wagon (i.e., Fermi Normal Coordinates centered on that worldline).

 

[PeterDonis]

Are you arguing that a clock synchronisation procedure implies a foliation but a foliation does not necessarily imply a clock synchronisation convention?

No. You are not just talking about general properties of clock synchronization conventions. You are talking about a very specific clock synchronization procedure: Einstein clock synchronization. That greatly restricts what can be done.

 

[PAllen]

Not of "most" of spacetime. Only of a small enough region. Which in practice will only be a small "world tube" surrounding the worldline of one wagon (i.e., Fermi Normal Coordinates centered on that worldline).

There is an alternative that I’ve never seen worked out for this case. That is, using radar coordinates based on one train car. @Dale recently posted a paper that established that in SR, radar coordinates can consistently cover the intersection of the causal future and causal past of an origin world line. Thus, for eternal uniform acceleration they have the same coverage (and foliation) as Rindler coordinates. However, for this case, the spiral world line of a single train car has all spacetime in both its causal past and causal future. Thus, radar coordinates should have global, consistent coverage. How useful they would be is another matter ...

I found this paper particularly interesting because it greatly generalized something I had worked out on my own - that radar coordinates have global coverage for any world line that is inertial in the past of some event, and also inertial in the future of some event, no matter what happens in between, or what the relative velocity is between the past inertial part and the future inertial part. Clearly, this is a very special case of the more general result of this paper.

 

[Dale]

There is an alternative that I’ve never seen worked out for this case. That is, using radar coordinates based on one train car. @Dale recently posted a paper that established that in SR, radar coordinates can consistently cover the intersection of the causal future and causal past of an origin world line. Thus, for eternal uniform acceleration they have the same coverage (and foliation) as Rindler coordinates. However, for this case, the spiral world line of a single train car has all spacetime in both its causal past and causal future. Thus, radar coordinates should have global, consistent coverage. How useful they would be is another matter ...

Here is that paper:
https://arxiv.org/abs/gr-qc/0006095
The directly relevant section is figure 3 and the associated text and equations.

 

[PAllen]

Here is that paper:
https://arxiv.org/abs/gr-qc/0006095
The directly relevant section is figure 3 and the associated text and equations.

Thanks!

 

[Ibix]

In what sense is something with a non-zero accelerometer at a given instant instantaneously inertial?

It isn't - I should have said something like "can be covered by a single inertial frame in which it is (very nearly) at rest". I've added an edit to the post you quoted.

 

[Foppe Hoekstra]

To all,
I noticed that no one denies that the printing of the front clock should be ahead of the printing of the rear clock, except that it is hard to define the proper frame for it. I think it is agreed that it is valid in the co-moving frame of a similar single wagon, when we provide that the only difference to one of the train carriages is that it moves on a straight rail. But what I understand from your comments is that a non-zero centripetal acceleration - on such an originally inertial single wagon - is the decisive game-changer.

Still, it means that (taken for granted that every clock does only one printing and that there is rotational symmetry) every single clock on the train has two different readings in the sequential frames it is acting in (as a front clock and as a rear clock), no matter how similar these sequential frames are, apart from the difference in moving direction. I just don’t see how that is possible. If two sequential wagons would both move straight on in the same direction, there would be no different readings, but bring them to even the slightest centripetal acceleration and suddenly the difference in readings is there in its full extent. The size of it is not even dependent of the degree of acceleration.

 

[Ibix]

No. You are not just talking about general properties of clock synchronization conventions. You are talking about a very specific clock synchronization procedure: Einstein clock synchronization. That greatly restricts what can be done.

I have to say that I don't see the problem with what I'm saying, except for DrGreg's point about the difficulty of extending off the track. It's not clear to me whether we're talking at cross-purposes or if there's something I'm just not understanding, so I thought I'd draw some diagrams. Below is a (2+1)d Minkowski diagram drawn in the rest frame of the circular track. The right hand half is just a detail from the left hand half.

The translucent cylinder is the worldsheet of the track - coordinate time obviously runs parallel to its axis. A single carriage is going round the track, equipped with clocks at the front and rear - the worldlines of these clocks are marked in blue. Clock ticks are marked as red crosses, and "simultaneous" ticks are joined by green lines. The definition of simultaneity in use here is that the clocks are synchronised in their instantaneously co-moving inertial frame. Thus, where they form an acute angle, a blue line and a green line form two legs of a tetrad. I don't think there's any problem with this. Both clocks have the same speed in the frame shown, so tick at the same rate and (once initialised) maintain this synchronisation without further intervention.

All I was suggesting was that you can chain this process together along the train - using the front clock as the rear clock in the next carriage, and so on. Eight carriages looks like this (again, the right is a detail of the left):

Obviously you have to have a break somewhere because, if I added enough carriages to complete the circle, the green lines wouldn't close. And you and DrGreg already explained why I can't extend this off the surface of the cylinder. But I don't see why this won't work along the length of the train. It isn't a global Einstein synchronisation, and I wasn't saying it was (or certainly wasn't intending to say so if I did). What it is is a chain of clocks, each of which is Einstein synchronised (in a slightly different frame) to each of its two neighbours.

 

[Ibix]

But what I understand from your comments is that a non-zero centripetal acceleration - on such an originally inertial single wagon - is the decisive game-changer.

No. How you choose to synchronise your clocks is a game-changer. The way you've set it up they remain synchronised in the rest frame of the track. This is not the same as the synchronisation used by the instantaneous inertial rest frame of a carriage.

So in the rest frame of the track, your printers print simultaneously. An inertial observer passing by, instantaneously at rest with respect to a carriage, will say that the printers at the ends of the carriage printed at different times and all the printers in other carriages printed at yet other different times.

 

[Ibix]

I noticed that no one denies that the printing of the front clock should be ahead of the printing of the rear clock

It's important to stress that whether the printing of one clock "should be" ahead of another is entirely your choice. The way you have set your experiment up, the clocks tick simultaneously in the rest frame of the track, and non-simultaneously in the instantaneous inertial rest frame of a carriage. But you could have set it up the other way round, so that the clocks tick simultaneously in the instantaneous inertial rest frame (that's the situation illustrated in my diagrams above), in which case the printers will not trigger simultaneously in the rest frame of the track, and the front of the pattern of dots will overlap the rear.

There is no "should be" here. There is only how you chose to set up the experiment.

 

[Dale]

I noticed that no one denies that the printing of the front clock should be ahead of the printing of the rear clock, except that it is hard to define the proper frame for it

Until you define the proper frame for it the statement is literally meaningless. A sentence with undefined words has no meaning.

You have the order wrong here. First you need to clearly define the question, only then can you find answers.

every single clock on the train has two different readings in the sequential frames it is acting in

Same issue here. What does this mean? Nobody can either affirm or deny an undefined statement.

bring them to even the slightest centripetal acceleration and suddenly the difference in readings is there in its full extent. The size of it is not even dependent of the degree of acceleration.

What difference in readings exactly? Specify mathematically what quantity represents this difference in readings. Then you can calculate if its size is independent of the degree of acceleration. Somehow, I suspect it will be difficult to reasonably define it to make this statement true.

 

[Vanadium 50]

I noticed that no one denies that

I'm sorry, but that's not how things work. The burden of proof is not on us to rebut.

How is this thread different than your two closed threads on this? At the risk of an atrocious pun, we seem to be going round and round here.

 

[pervect]

To all,
I noticed that no one denies that the printing of the front clock should be ahead of the printing of the rear clock, except that it is hard to define the proper frame for it.

I think there's an easier way to look at it. If the clocks all accelerate in a symmetrical manner in the station frame, we can say that they all print out exactly the same time.

Then it is a simple matter, if one is familiar with the relativity of simultaneity, to say that this directly implies that the clocks are NOT syncrhronized in the train frame.

I'll provide a reference to where Einstein mentioned this. Now, if you don't quite follow Einstein's argument, you would be far from the first. But at least you might have some inkling of where the rest of us are coming from, and what the issue is with your analysis.

The link is https://www.bartleby.com/173/9.html, and the reference is Einstein's book, "Relativity: the special and general theory", chapeter IX, 'The relativity of simultaneity'.

UP to now our considerations have been referred to a particular body of reference, which we have styled a “railway embankment.” We suppose a very long train traveling along the rails with the constant velocity v and in the direction indicated in Fig. 1. People traveling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. Then every event which takes place along the line also takes place at a particular point of the train. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment. As a natural consequence, however, the following question arises:

1​

Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.

So there you have it. BECAUSE the events on two neary cars are simultaneous in the station frame (when we assume symmetry), it follows, from Einstein's argument, that they can't be simultaneous in the train's frame.

The "train's frame" is only defined in the limit as the two cars are very close to each other, which has caused some discussion. But after taking said limit, there's nothing at all ambiguous about the train frame - it's moving at some velocity v along the track, and because of the effect Einstein mentioned, the fact that the clocks are synchronized in the station frame implies they cannot be synhronized in the train frame.

 

[PeterDonis]

I noticed that no one denies that the printing of the front clock should be ahead of the printing of the rear clock

"Ahead" is frame dependent. The invariant is what gets printed on each clock's receipt. The analysis in the SF that I already gave answers that question.

every single clock on the train has two different readings in the sequential frames it is acting in (as a front clock and as a rear clock)

I have no idea what you mean by this. The reading on a particular clock at a particular point in spacetime is an invariant. That particular point in spacetime will in general be labeled by different coordinates in different frames. But the invariant is the point itself and the reading of the clock at that point.

I think you would greatly benefit from forgetting all about coordinate-dependent quantities and focusing on invariants. Otherwise you will just continue to confuse yourself.

 

[PeterDonis]

A single carriage is going round the track, equipped with clocks at the front and rear - the worldlines of these clocks are marked in blue.

Yes, and what you are basically drawing here is the start of a system of Fermi Normal Coordinates centered on the chosen carriage. The "simultaneous lines" are the surfaces of constant coordinate time, and the worldlines are the "grid lines" of the time coordinate.

All I was suggesting was that you can chain this process together along the train

As I understand the OP, the train is supposed to go all the way around the circle, so it's impossible to cover the whole train this way. But I think you could cover a portion of the train this way, yes.

Also, I'm not sure that the chaining process you describe, within the ange you can extend it, will be exactly the same as Fermi Normal Coordinates centered on the chosen carriage. The simultaneity conventions might not be the same when extended beyond the initial chosen carriage; I would have to look at the detailed math.

 

[pervect]

A few more comments on the topic. If we assume that all the train cars are accelerating unfiormly (one way of doing this would be to assume they all start acclelerating with the same proper acceleration at the same time , "same time" being defined in the station frame, then in their own instantaneous frame, the nearby cars will get further apart as time goes on, in addition to the loss of Einstein synchronzation that has already been discussed.

It's somewhat similar to Bell's spaceship paradox, except to make life a little more complicated it's on a curved track rather than a straight one.

It's easiest to analyze the case of a straight track than a curved one.

In the case with a straight track, if the lead spaceship accelerates at 1 light year/year^2 (roughly one Earth gravity) , when the rear spaceship accelerates at the same rate, it does not keep a constant distance behind the lead ship in the ship frame.

The rear ship has to accelerate harder at roughly 1.1 light years/year^2, to keep a constant distance away from the lead ship in the ship frame.

This also implies that if the two ships are to stop when they have the same velocities, they accelerte for different time periods, illustrating some physical consequences to the loss of synchronization we were talking about earlier. If the lead spaceship accelerates at 1 g for one month, the tailing spaceship accelerates at 1.1g for .9 moths. So obviously, their clocks do nor remain synchronized

 

[Ibix]

Building on my earlier post, I've constructed Minkowski diagrams for a train circling a track at constant speed. Unfortunately my arithmetic must have gone wrong somewhere because the train has a small gap between front and rear. I don't think this makes much difference to anything.

Again, the diagram is drawn in the rest frame of the circular track, the worldsheet of which is the translucent cylinder. Pale blue lines show the worldlines of clocks distributed evenly along the train (and you can see the small gap because the last line doesn't quite line up with the first), with red dots marking their ticks. One set of "simultaneous" red dots are joined by a green line - simultaneous, here, means in the sense that a clock is Einstein synchronised to its neighbours in the inertial frame in which they are instantaneously at rest. I've also added a white line, which is the worldline of an observer at rest in the track frame.

Note that if these clocks are equipped with printers and print at the green-connected times, then in the track frame they print one after the other. Thus the movement of the train means that when the last-to-print clock prints, it isn't in the same place as the first-to-print clock was when it printed, and the ends of the dot pattern thus printed overlap.

The above is not the scenario outlined in the OP. In the OP, all of the printers print simultaneously in the track frame (in contrast to earlier threads by the same poster). I've illustrated that circumstance below.

Note that this diagram is identical to the above, except that I deleted the green line and added an orange line that marks simultaneity in this different sense. If the clocks print at the time indicated by the orange line then there is no overlap and the dots printed on the track are uniformly separated (except for the one small gap, in this diagram). The orange line lies in a plane of simultaneity for the trackside observer, but note that it does not pass through a line of red dots. Thus observers on the train would not regard the printers at opposite ends of a carriage as printing simultaneously, assuming they adopt their instantaneous rest frame's simultaneity convention (life gets complex beyond the end of a short carriage because we haven't truly defined a generally applicable simultaneity convention for the on-train observers).

It has to be stressed that these two diagrams are (very nearly) identical, representing identical physical situations in terms of trains on the track. The only difference is the rule we use for defining "at the same time" in the sentence "the printers print at the same time". And the rule to use is a matter of choice by the experimenter.

 

[Nugatory]

Unfortunately my arithmetic must have gone wrong somewhere because the train has a small gap between front and rear.

I presume you used a computer in an interative loop? If so, I'd bet long odds that you're just seeing accumulated rounding errors.

 

[Ibix]

I presume you used a computer in an interative loop? If so, I'd bet long odds that you're just seeing accumulated rounding errors.

Turns out to be too many tools in my workflow. I simplified a bit - still don't know quite what I did wrong, but the numbers going into the generation process were wrong. Corrected results are below:

No comments to add to my previous, except now you can see that the front of the train is coincident with the rear. Using the green "chain of local Einstein synchronisations" makes the pattern overlap due to the motion of the train while the printing happens. Using the orange "synchronised in rest frame" criterion for printing makes the pattern even, but adjacent clocks do not print simultaneously according to an observer in the carriage.

 

[Ibix]

My last post with Minkowski diagrams, as no-one seems to be responding (no disrespect to Nugatory intended).

First of all, here's a (very long) train on a (very long) straight track. It is initially stationary and then accelerates to a constant speed. The acceleration profile is chosen so that the ends of the carriages are the same distance apart in the track frame (thus they are moving apart in their own frame).

The red line represents the front carriage and the blue line the back one. This is actually the standard Bell's spaceships configuration, just with eleven carriages instead of two ships. Clocks in the carriages tick every half-year of proper time, and these events are marked. The clocks are initially synchronised and remain synchronised with one another in the track frame, although they tick slower than coordinate time clocks - there are only ten ticks in six years. Acceleration terminates at $\tau=3$, which is about $t=3.25$

Next, here is a diagram of a train accelerating so that the distance between the carriages remains constant for observers on the train. It's fairly easy to see that this requires the train observers to be a family of Rindler observers with shared horizons.

As before, clock ticks are marked with crosses. They are initially synchronised in the track frame but, due to different velocity profiles, they de-synchronise during the acceleration. Because the requirement is that the final velocity be the same along the train, the carriages stop accelerating at different times - these are marked with red crosses. Note that, in the way I've chosen to set this up, the blue clock behaves identically to the blue clock in the first diagram. You can also see that the train is undergoing length contraction.

It's also interesting to see the above in the final rest frame of the train:

You can see that the carriages stop accelerating simultaneously in this frame, but their clocks are not synchronised. You can also see that the separation of the carriages after the acceleration is, in this frame, the same as it was before in the track frame, as promised.

Finally, why am I talking about this linear track? It's because the surface of a cylinder has the same geometry as a plane, albeit with a different topology. So you can simply wrap these diagrams into cylinders with the t-axis parallel to the cylinder axis to get cylindrical diagrams like the ones I posted before. Or one can imagine slitting the cylinders parallel to their axis and spreading them out to get flat diagrams.

Because the original problem specification was that the train exactly fitted around the track, in either of the first two diagrams we can simply cut off everything to the right of $x=2$ and paste it on to the left to get a Minkowski diagram of the "unwrapped" cylinder. So at the start of the experiment the red and blue lines are coincident, and anything that passes $x=2$ moves to $x=0$. Here, then, is the version keeping constant acceleration in the track frame:

The red and blue lines are initially coincident and remain coincident (this has been rendered as black by the colour-combining process I used). You can see that the clock ticks always remain simultaneous in the track frame, as they must from the symmetry of a situation where every clock does the same thing in this frame.

What about the case where the train is free to keep its natural length? That looks like this:

Here we can see that a gap opens up as the train length contracts and the clocks de-synchronise.

In both cases, if the clocks were re-synchronised after the acceleration, their "same time" ticks would look like the red events - non-simultaneous in the track frame. Note that the string of red events goes all the way around the cylinder and a little bit more, so if these were used to time printing then the end of the pattern would overlap the beginning.

So here's the takeaway message: the clock synchronisation you get is your choice. You can re-synchronise your clocks by some procedure after the acceleration, or you can accept whatever you get from synchronising pre-acceleration. The results of either approach will differ depending on the exact physical situation (which is also your choice!)