Clock synchronising by clock transport?

[Peter Strohmayer]

TL;DR Summary

To set the clocks of a reference system to zero, Einstein synchronization with reflected light beams is usually used. Is it a good idea to bring a watch to others and coordinate times?

I have a reference system A with three clocks of the same type. Two clocks are at rest in the origin of A and could be synchronized without any problems. The third clock rests at a distance in the x-direction.

Is it possible to synchronize this third clock by accelerating the second clock at rest with an acceleration of 1 over the time of 1 in the x-direction and immediately bringing it back to rest with an acceleration of -1 over the time of 1 in the x-direction at the third clock, taking into account the additional proper time elapsed due to the accelerated motion, just as I would take into account the time of propagation of the light signal in Einstein's synchronization?

 

[Peter Strohmayer]

Correction: of course not additional, but less proper time.

 

[Ibix]

How are you planning to calculate the amount by which the moving clock's proper time underestimates the lapse of coordinate time? You can, of course, measure the proper time elapsed on the transported clock, but what time "should" it read?

The answer to that depends on your choice of synchronisation convention. With Einstein's synchronisation convention you can simply repeat the experiment at slower and slower speeds and predict the result in the limit of zero speed, when there would be no time dilation. With other synchronisation conventions, moving clocks drift out of sync in a direction-dependent way even in the limit of low speed. The synchronisation you get out of this depends on which clock drift correction you apply.

 

[Ibix]

Slow clock transport with zero drift in the zero speed limit, incidentally, is used in practice for Einstein synchronising clocks where they don't have line of sight. I recall that one line of investigation in the "faster than light neutrinos" kerfuffle was whether the process had been done correctly.

 

[PeroK]

TL;DR Summary: To set the clocks of a reference system to zero, Einstein synchronization with reflected light beams is usually used. Is it a good idea to bring a watch to others and coordinate times?

I have a reference system A with three clocks of the same type. Two clocks are at rest in the origin of A and could be synchronized without any problems. The third clock rests at a distance in the x-direction.

Is it possible to synchronize this third clock by accelerating the second clock at rest with an acceleration of 1 over the time of 1 in the x-direction and immediately bringing it back to rest with an acceleration of -1 over the time of 1 in the x-direction at the third clock, taking into account the additional proper time elapsed due to the accelerated motion, just as I would take into account the time of propagation of the light signal in Einstein's synchronization?

Another idea would be to take a clock from O to X and back again, using symmetric proper acceleration profiles for both legs. When back at O, check the loss of synchronization between the clock at O and the second clock. Then, reset the second clock and take it back to X (again with the same proper acceleration profile as before) and finally adust it by half the round trip amount.

This should be equivalent to Einstein synchronisation.

 

[Sagittarius A-Star]

Is it a good idea to bring a watch to others and coordinate times?

As others mentioned, the slow clock transport method is equivalent to the Einstein synchronization method. But that does not imply, that Einstein synchronization is the only possible one. If the one-way-speed of light is defined to be anisotropic, then time dilation can no longer be ignored for small velocities.

Source:
https://en.wikipedia.org/wiki/One-way_speed_of_light#Slow_clock-transport

 

[Peter Strohmayer]

Thank you for the explanations. I am especially grateful for Perok #5's suggestion because the solution he shows does not depend on clocks being transported slowly.

 

[Sagittarius A-Star]

Thank you for the explanations. I am especially grateful for Perok #5's suggestion because the solution he shows does not depend on clocks being transported slowly.

Of course, also such a synchronization refers to the inertial frame, in which both clocks to be synchronized are at rest.

The transported clock has different (direction-dependent) time dilation factors and travel-distances with reference to other frames.

 

[Peter Strohmayer]

How are you planning to calculate the amount by which the moving clock's proper time underestimates the lapse of coordinate time?

From its point of view, the transported clock is accelerated in x' direction by 0.1 c/sec for the time of 8.6 sec and in -x' direction by 0.1 c/sec for the time of 8.6 sec.

From the point of view of the reference system F, the transported clock is at the origin at the beginning of the movement and at the end of the movement on the x-axis at the clock of the reference system F to be synchronized.

Calculate the time difference resulting from the hyperbolic movement:

c = 1

α = proper-acceleration of the transported clock

τ = proper time of the transported clock

T = coordinate time of the reference system F

T=sinh(α*τ)/α

sinh(0.1*8.6)/0.1 = 9.7 sec.

T= 19.4 sec

At the end of the movement, the transported clock shows the time 17.2 and sets the clock of the reference system F to the time 19.4 sec.

Can it work like this or have I made a mistake?

 

[Ibix]

Can it work like this or have I made a mistake?

You chose a reference system F, which means you picked a simultaneity convention (Einstein's synchronisation with respect to the origin clock, given your choice of relationship between $T$ and $\tau$). So your clocks will be synchronised according to that convention. So if what you intended was just an implementation of Einstein synchronisation without using light signal exchange, it's fine.

I haven't checked your maths in detail. At a quick glance, though, your acceleration is such that it isn't remotely slow clock transport.

 

[Peter Strohmayer]

your acceleration is such that it isn't remotely slow clock transport

The amount of acceleration should be irrelevant (#7).

 

[Vanadium 50]

You might be better served by a different approach: get hold of a copy of Taylor and Wheeler’s “Spacetime Physics” and work through it from the beginning. Do not speed through the early chapters thinking you understand that basic stuff - you don’t. The first edition is legal and free on the internet, and we can help over any hard spots if you get stuck.

This is good advice. It would be good to take it. It is not too late.

The amount of acceleration should be irrelevant (#7).

I don't see where this is supported, nor what you mean, but even if it is the case, invariant acceleration doen't mean equal invariant velocity even in Newtonian physics.,

Consider two identical systems at point A. Both are given a brief acceleration of a1. After a short time, one is given a brief acceleration a2 (a2 >> a1) and starts to go faster. As it nears point B, it is given an acceleration -a2 and slowly reaches poiny B.

At a later time, the second widget is given an acceleration a1 and almost immediately an acceleration -a1. It creeps toward B and eventually reaches it.

The first widget took longer than the second, even though they experienced accelerations of the same magnnitude. The first also had a higher average velocity than the second, even though they experienced accelerations of the same magnitude.

 

[Ibix]

The amount of acceleration should be irrelevant (#7).

You can use the process you defined with any acceleration you like, yes. It just isn't slow clock transport if you take the clock up to 0.65c. The advantage to slow transport is that, in the limit of very slow transport $T$ and $\tau$ can be made equal to arbitrarily high precision, so no clock correction is needed if you can wait long enough.

 

[PeterDonis]

I don't see where this is supported, nor what you mean

What he means is that @PeroK proposed in post #7 a method (somewhat different from the method the OP originally proposed) which does not require slow clock transport, and therefore does not require imposing any limits on the acceleration profile (although it does require the same acceleration profile to be duplicated exactly multiple times). Instead clocks are corrected after the transport is complete, by a method somewhat similar to Einstein clock synchronization, but not requiring extra light signals.

 

[Peter Strohmayer]

The (accelerated) motion of a transported clock described in #9 can be performed by a homogeneously accelerated scale in x' direction. Its material survives the acceleration unchanged. In an idealized form, the scale consists of many equally accelerated Bell spaceships, each of which transports one clock. We consider the clocks at the beginning and end of the scale. In the first part of the movement, the distance between these clocks increases ("the rope between the bell spaceships breaks"). At the end of the movement, the scale is back to its original length. Now all the clocks of the scale can be used to synchronize the clocks of the reference system F using the method described in #9.

 

[Ibix]

I think you are trying to describe a ruler with clocks spaced along its length. You are accelerating the ruler parallel to its length and decelerating it again to rest.

This does not work as you imagine. The problem is that it takes time for the "push" to propagate along the ruler, so the clocks do not start accelerating simultaneously. And the ruler will probably behave in a Born-rigid fashion, so the clocks will have different proper accelerations so that their spacings remain constant in their instantaneous rest frame, rather than remaining constant in their initial rest frame. Thus the clocks will most probably be out of synchronisation with each other when they come to rest, at least by the Einstein synchronisation convention.

You could do the process you describe by equipping each clock with a rocket with a pre-programmed burn time so that the clocks all undergo the same proper accelerations.

You still have to synchronise your clocks along the ruler somehow before you can use this process.

Why are you asking all this? There are, in practice, many ways to synchronise a pair of separated clocks, and you can use any method that works for one synchronisation convention to synchronise by any other. We just tend to use light signals in examples because the maths is easier.

 

[Peter Strohmayer]

You could do the process you describe by equipping each clock with a rocket with a pre-programmed burn time so that the clocks all undergo the same proper accelerations.

In an idealized form, the scale consists of many equally accelerated Bell spaceships, each of which transports one clock.

Why are you asking all this?

If the synchronized clocks T of a transported reference system T are accelerated in the manner described ("Bell's spaceships"), each clock T will lag behind the displays of the synchronized clocks F of the reference system F by 2.2 seconds (#9) at the end of the first transport. In this way, the clock T at the origin of the reference system T and all other clocks on the scale T can perform any number of simultaneous accelerated movements and - from the point of view of the reference system F - come to rest again at any location. At each of these locations, the clocks T will continue to lag behind the clocks F by the same amount. This will be the case in particular when the origin of the reference system T returns to the origin of the reference system F.

In other words, the comparison shown between the readings of the clocks of an inertial system F and the readings of the clocks of an accelerated system T at the points at which the clocks of the two reference systems are at rest relative to each other results in a uniformly smaller aging of the entire reference system T and of the objects at rest in T
compared to the reference system F and the objects at rest in F.

 

[Ibix]

In an idealized form, the scale consists of many equally accelerated Bell spaceships, each of which transports one clock.

Yes, but this is a different system from a ruler connecting a series of clocks, and the differences may matter.

In other words, the comparison shown between the readings of the clocks of an inertial system F and the readings of the clocks of an accelerated system T at the points at which the clocks of the two reference systems are at rest relative to each other results in a uniformly smaller aging of the entire reference system T and of the objects at rest in T
compared to the reference system F and the objects at rest in F.

This is not true in general, since there are many ways to define "an accelerated system". It's true in the specific case you are describing, but this isn't particularly relevant to anything since you'd have to generate internal stresses in your ruler to achieve it, since the spacing of the clocks is not constant in their own instantaneous rest frames. Essentially when the ruler accelerates to some speed $v$ its natural length in the original inertial rest frame shortens by a factor of $\gamma$, so to keep its length constant in that frame you have to apply forces to stretch it by a factor of $\gamma$ in its own rest frame.

In a real physical system consisting of a ruler with clocks attached along its length, the clocks will not be in synchronisation after an acceleration and deceleration phase of the kind you are describing. The whole ruler-and-clocks thing seems like adding pointless complications. If you want to Einstein synchronise a line of clocks, just move one clock along the line, accelerating and decelerating and correcting for the calculated difference between the time it shows and the coordinate time you choose as necessary.

 

[Peter Strohmayer]

the clocks will not be in synchronisation after an acceleration and deceleration phase of the kind you are describing

Once again: "In an idealized form, the scale consists of many equally accelerated Bell spaceships, each of which transports one clock." Therefore, the clocks of the accelerated system remain synchronised by definition.

 

[PeterDonis]

Once again: "In an idealized form, the scale consists of many equally accelerated Bell spaceships, each of which transports one clock." Therefore, the clocks of the accelerated system remain synchronised by definition.

They remain synchronized in the frame in which they were originally at rest. But this is not the same as the momentarily comoving inertial frame of any of the spaceships once they are moving. So the "synchronization" you are talking about here is a very peculiar one.

You need to read up on the Bell spaceship paradox.

 

[Ibix]

Once again: "In an idealized form, the scale consists of many equally accelerated Bell spaceships, each of which transports one clock." Therefore, the clocks of the accelerated system remain synchronised by definition.

And, once again, this is not the same as an experiment with clocks attached to a ruler ("a homogeneously accelerated scale...[i]ts material survives the acceleration unchanged"), to which you referred in #15. The presence or absence of a material connection between the clocks will make a difference to the behaviour (edit: unless you take active steps to deform the material as described in the first paragraph of #18). This may be important; I'm still not sure why you're adding complexity to a fairly straightforward process.

 

[Peter Strohmayer]

They remain synchronized in the frame in which they were originally at rest. But this is not the same as the momentarily comoving inertial frame of any of the spaceships once they are moving. So the "synchronization" you are talking about here is a very peculiar one.

Thank you very much for pointing out my mistake.
So let me summarize: if all clocks in system T, which have been accelerated in the described idealized way, come to rest again from the point of view of inertial system F, they (all the clocks T!) are no longer synchronized with each other. Is this correct?

 

[Ibix]

Thank you very much for pointing out my mistake.
So let me summarize: if all clocks in system T, which have been accelerated in the described idealized way, come to rest again from the point of view of inertial system F, they (all the clocks T!) are no longer synchronized with each other. Is this correct?

If they are initially at rest in an inertial frame F and synchronised in that frame, and all are accelerated with identical acceleration profiles in such a way that they return to rest in F then they will (at the end of it all) be synchronised with each other, and will all have the same offset compared to clocks that remained at rest in F.

The point is that during the acceleration process (assuming it starts simultaneously according to F) they will remain synchronised in F (albeit with a growing lag compared to F's clocks), but they will not be synchronised in their own rest frame until they return to rest in F. Not unless you deliberately construct a non-orthogonal "rest frame" for them, anyway, but you wouldn't normally call that a frame.

 

[Peter Strohmayer]

Thanks for the clarification. That's what I wanted to say in #17.

The clock transport of extended bodies reveals a special aspect of synchronization. The proper time of a clock T is defined by its proper acceleration profile. In this example, the proper acceleration profiles and thus the proper times of all clocks T are the same. This is a physical fact independent of any reference system. It has the effect that all clocks T, which are at rest again from the point of view of system F, remain synchronized despite their temporary desynchronization.

Wherever the body T comes to rest, from the point of view of the system F, alongside a body F that is always at rest there, it has aged less than the body F since the beginning of its journey.

 

[PeterDonis]

The proper time of a clock T is defined by its proper acceleration profile.

No, it isn't. It's defined by the arc length along the clock's worldline. In this particular example, equality of arc lengths happens to match up with equality of acceleration profiles (in a particular frame--see my next post), but that does not generalize to other scenarios.

 

[PeterDonis]

This is a physical fact independent of any reference system.

No, it isn't; the claim you made about synchronization for your scenario is only true in the frame in which the clocks start out and end up at rest. See post #20.

 

[Ibix]

The proper time of a clock T is defined by its proper acceleration profile.

No - you need at least one more fact to deduce the lapse of proper time from the acceleration. Typically this would be an initial or final velocity with respect to some coordinate system, but other options are possible and it would depend exactly what you measuring.

In this example, the proper acceleration profiles and thus the proper times of all clocks T are the same. This is a physical fact independent of any reference system.

That's arguable, at least. Most frames would not regard the acceleration profiles as identical because they would not regard the starts as simultaneous. You can't really construct this in a coordinate independent way because the whole process depends on synchronised clocks for the start times, which means a simultaneity convention.

Of course you can describe your synchronisation process in a coordinate independent way, but it's still a free choice of clock synchronisation methodology.

It has the effect that all clocks T, which are at rest again from the point of view of system F, remain synchronized despite their temporary desynchronization.

From the frame F the clocks on your rod were never out of sync with each other. From any other inertial frame they were never in sync. It's only from most constructions of a non-inertial frame of T that they drift out of sync and back into sync.

Wherever the body T comes to rest, from the point of view of the system F, alongside a body F that is always at rest there, it has aged less than the body F since the beginning of its journey.

This is always true; it's the Minkowski geometry equivalent of the Euclidean observation that the shortest distance between two points is a straight line.

 

[Vanadium 50]

@Peter Strohmayer, your posts tend to be full of "facts" that are simply not true. I think this has been previously brought to your attention. Posting a parade of untruths hoping to be corrected is a very inefficient way of learning (you are not the first to have tried it) and just makes everybody cross.

In previous threads, you have been pointed towards sources that will clear up your misunderstandings. But you seem not to be going in that direction. Soon people will start to wonder why.,

 

[Nugatory]

In this example, the proper acceleration profiles and thus the proper times of all clocks T are the same. This is a physical fact independent of any reference system.

It is not.
The statement "the proper acceleration profiles are the same" is equivalent to "the proper accelerations all change at the same time". As with just about any statement that includes the words "at the same time", this is frame-dependent, not a "physical fact".

 

[Peter Strohmayer]

It's [the proper time of one clock] defined by the arc length along the clock's worldline.

Isn't it the same? The curvature of the world line of a clock is determined by the proper acceleration of the clock, which is measured by an accelerometer carried on board.

No - you need at least one more fact

The acceleration profile between two events recorded by the accelerometer on the transported clock is used to calculate the amount by which the proper time τ deviates from the coordinate time T of any inertial system?

Thanks for the other clarifications @Ibix in #27.

 

[PeterDonis]

Isn't it the same?

No. Read the multiple responses you have received.

The curvature of the world line

By itself tells you nothing about synchronization with other clocks following other worldlines--which is what all of your claims are about.

 

[Peter Strohmayer]

Of course, you are right: the described synchronization with the other accelerated clocks only occurs through the (idealized) common acceleration of a ruler (a rocket) (#17, #27). This is practically the case that matters.

 

[PeterDonis]

the described synchronization with the other accelerated clocks only occurs through the (idealized) common acceleration of a ruler (a rocket) (#17, #27).

You're missing the point. The point is that "the described synchronization" is only one of an infinite number of possible synchronizations (simultaneity conventions). All of your claims are only true for that one particular synchronization.

This is practically the case that matters.

Really? Says who?

There are lots of practical cases that do not meet your specifications.

 

[Ibix]

Isn't it the same? The curvature of the world line of a clock is determined by the proper acceleration of the clock, which is measured by an accelerometer carried on board.

No, the length of a line is not the same as its curvature. You need to integrate twice to get the elapsed proper time starting with the acceleration profiles, and one of the constants of integration is the initial velocity of the object.

The acceleration profile between two events recorded by the accelerometer on the transported clock is used to calculate the amount by which the proper time τ deviates from the coordinate time T of any inertial system?

You need an initial velocity, as I keep saying. And if you want to do something with several clocks along a rigid body you need to synchronise them somehow before you start (which is where the arbitratiness comes in). But with all of that and the acceleration profiles, you can compute the instantaneous velocity in any coordinate system and hence the elapsed proper time, yes.

 

[Ibix]

Of course, you are right: the described synchronization with the other accelerated clocks only occurs through the (idealized) common acceleration of a ruler (a rocket) (#17, #27). This is practically the case that matters.

What you are describing here, with an array of clocks, is not really a synchronisation process. You need either additional communication. Or many repeats of your process, in which case you'd be better to junk all but one of the array of clocks you call T and just use one clock.