I A clock’s double life?

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?
 
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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.
 
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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

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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.
 
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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:


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

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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.
 
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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.
 
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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.
 
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Ibix

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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

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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.

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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?


in the co-moving frame of the wagon
You haven't defined that frame properly, just like in your previous threads.
 
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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.
 
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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.
 
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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).
 
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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

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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.
 
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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:

The directly relevant section is figure 3 and the associated text and equations.
 

Ibix

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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.
 
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

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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.
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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):
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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.
 
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Ibix

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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.
 
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Ibix

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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.
 
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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.
 
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Vanadium 50

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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.
 

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