I Is it possible to have synchronized clocks in a rotating system?

Foppe Hoekstra

Summary
Consider a wagon with length L and constant speed v on the straight part of a rail. The wagon has clocks on both ends that are sync in the co-moving frame (of the wagon and the clocks). Then there is a curve in the rail with radius r. The speed of the wagon in the curve is still constant v, but it will not be an inertial system anymore. Does that desynchronize the clocks? And if so, by what amount?
Consider a wagon with length L and constant speed v on the straight part of a rail. The wagon has clocks on both ends that are sync in the co-moving frame (of the wagon and the clocks). Then there is a curve in the rail with radius r. The speed of the wagon in the curve is still constant v, but it will not be an inertial system anymore. Does that desynchronize the clocks? And if so, by what amount?

In what frame? You will probably ask first. If possible in the co-moving (rotating) frame of the wagon, or else, if possible, in the co-moving frame of the first clock.

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Dale

Mentor
To do this in the inertial frame just write down the expression for the velocity as a function of time for each clock and then integrate $\int \sqrt{1-v^2} dt$

Orodruin

Staff Emeritus
Homework Helper
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In what frame? You will probably ask first. If possible in the co-moving (rotating) frame of the wagon, or else, if possible, in the co-moving frame of the first clock.
This is not sufficiently specified to provide a unique answer.

The easiest thing to do is to analyse this in the ground frame and then transform to whatever frame you are interested in.

Ibix

What are you trying to describe here? Two pieces of straight rail joined by a curve (so just a corner in a track) or a single piece of straight rail feeding into a closed circle?

In either case, as Orodruin and Dale have said, the easiest thing to do is analyse in the rest frame of the track. If the train's speed doesn't vary then the tick rate doesn't vary in this frame, so the clock synchronisation doesn't vary. You can extend that to the inertial parts of the train's frame - its clocks must be synchronised if they were initially synchronised. However, the parts of the train's path that are non-inertial are more complex. Not because the tick rate in the inertial frame varies, but because there isn't a unique interpretation of what you mean by "the train's frame".

For example, consider a continuous chain of clocks all along the train that are initially Einstein synchronised. The train enters the turn. As long as the distance between clocks is much less than the radius of the curve then each clock remains Einstein synchronised to its neighbours. But the distance to a clock further along the train varies as the train enters the curve, so a pair of distant clocks would not be Einstein synchronised until the train comes back out of the curve and the distance returns to normal. So you could argue that the clocks are synchronised during the turn if you accept the along-the-train chain of clocks as the definition of time, or argue that they are not synchronised if you compare distant clocks.

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Foppe Hoekstra

What are you trying to describe here?
but because there isn't a unique interpretation of what you mean by "the train's frame".
The situation is a piece of straight rail feeding into a closed circle.

Consider my definition of synchronized clocks in a circle:
• We have n clocks in a closed circle( so clock number zero is clock number n)
• Every sequential pair of clocks is at equal distance
• t0 = t1, t1 = t2, ..., tn-1 = tn, tn = t0, where ti is the time read on clock i in a unique relevant frame.

This sync can be obtained in a circle at rest. But as soon as the circle has a rotating speed, there is not a unique relevant frame anymore, so I understand that it is not possible to have synchronized clocks in a rotating system.

Nevertheless you state that in the curve “each clock remains Einstein synchronised to its neighbours”, so ti-1 = ti = ti+1. Mathematically that would mean that it is possible to have synchronized clocks in a rotating system according to my definition!?

Ibix

I understand that it is not possible to have synchronized clocks in a rotating system.
It's possible. It's just that there isn't a unique way to do it and it often won't agree with obvious Einstein synchronisation methodology.
Nevertheless you state that in the curve “each clock remains Einstein synchronised to its neighbours”, so ti-1 = ti = ti+1. Mathematically that would mean that it is possible to have synchronized clocks in a rotating system according to my definition!?
I was referring to the two pieces of straight track joined by a curve in that paragraph. The problem with Einstein synchronisation around the curve only arises if you close the circle.

You are correct that you cannot Einstein synchronise a complete circle of clocks - there has to be a break somewhere (unless they are stationary with respect to the track). But you can synchronise them in other ways, for example using a master clock at the center of the circle and defining the time to be whatever you see on the clock (literally see, through a telescope) minus $r/c$. That's how GPS satellite clocks are synchronised, I believe (obviously there's only a notional central clock in this case). However, clocks synchronised this way will not be synchronised in the Einstein sense for observers comoving with a clock.

The general point is that there is no absolute method for synchronising clocks under any circumstances. It's just that for mutually at rest clocks in flat spacetime, almost any methodology leads to the same synchronisation in practice (as long as you share a notion of "at rest"). Not so for non-inertial clocks.

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vanhees71

Gold Member
Let's do a bit of math ;-)). The Minkowski pseudometric in terms of a set of observers rotating with constant angular velocity, $\omega$, is derived as follows.

Let
$$(x^{\mu})=\begin{pmatrix}\lambda \\ r \cos(\omega \lambda+\varphi) \\ r \sin(\omega \lambda+\varphi) \\z \end{pmatrix}$$
define the set of rotating observer, given in terms of components in an inertial frame of reference. Note that $\lambda$ is the coordinate time in this space. The new generalized coordinates are $\lambda$, $r$, $\varphi$, and $z$, where $(r,\varphi,z)$ are zylinder coordinates labeling the set of observers in their rest frame.

The pseudometric in terms of these coordinates is given by
$$\mathrm{d} s^2=\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}=(1-\omega^2 r^2) \mathrm{d} \lambda^2 -\mathrm{d} r^2 - r^2 \mathrm{d} \varphi^2 -\mathrm{d} z^2 - 2 \omega r^2 \mathrm{d} \lambda \mathrm{d} \varphi.$$
Of course this set of coordinates covers only part of space time, namely the cylinder with $r \omega <1$.

The proper time of the set of observers is given by
$$\mathrm{d} \tau^2=\mathrm{d} s^2|_{r,\varphi,z=\text{const}}=(1-\omega^2 r^2) \mathrm{d} \lambda^2.$$
This means that only the observers with the same $r$ have synchronized clocks. The reason is that the clocks of any two observers with different $r$ have a different time dilation relative to the time in the inertial frame and thus their clocks are not synchronized relative to each other.

Ibix

This means that only the observers with the same $r$ have synchronized clocks.
To be precise, it means that such observers have clocks that tick at the same rate. They may or may not be synchronised in the sense of having a shared notion of $t=0$.

@Foppe Hoekstra - as vanhees71 points out, clocks at different radii tick at different rates. My previous post is assuming all clocks are at the same radius and therefore ticking at the same rate.

Foppe Hoekstra

It's possible. It's just that there isn't a unique way to do it and it often won't agree with obvious Einstein synchronisation methodology.

I was referring to the two pieces of straight track joined by a curve in that paragraph. The problem with Einstein synchronisation around the curve only arises if you close the circle.

You are correct that you cannot Einstein synchronise a complete circle of clocks - there has to be a break somewhere (unless they are stationary with respect to the track). But you can synchronise them in other ways, for example using a master clock at the center of the circle and defining the time to be whatever you see on the clock (literally see, through a telescope) minus $r/c$. That's how GPS satellite clocks are synchronised, I believe (obviously there's only a notional central clock in this case). However, clocks synchronised this way will not be synchronised in the Einstein sense for observers comoving with a clock.

The general point is that there is no absolute method for synchronising clocks under any circumstances. It's just that for mutually at rest clocks in flat spacetime, almost any methodology leads to the same synchronisation in practice (as long as you share a notion of "at rest"). Not so for non-inertial clocks.
OK, I think we share some tracks now;-)

But what if we have instructed every clock to put a dot on the rail at ‘noon’ (in every clock’s own frame). Sufficiently before noon the complete train is on the circle and it precisely fills the circle, so the rear clock of the last wagon coincides with the front clock of the first wagon. What pattern would these dots have for an observer in the rest frame of the rail?

• Are all dots at the same sequential distance
• or are some sequential pair of dots at a less distance than average and other pairs at a larger distance than average
• or does the pattern show the “break somewhere” and where would that be then?

Dale

Mentor
What pattern would these dots have for an observer in the rest frame of the rail?
You can calculate that from the equation I posted and the expression for the worldline of each clock. It would be good for you to work through the math yourself.

Ibix

So I think you've Einstein synchronised clocks along the train when it's on the straight section of track, then fed it onto a circular track without changing its speed relative to the track. Edit - and once on the circle the nose of the train touches the tail.

The easiest way to understand this is to draw a spacetime diagram. If you work in the rest frame of the track you can "unwrap" the cylinder of spacetime occupied by the track into a flat 2d surface where anything that falls off one "space" end appears on the other side. Then you just draw the worldlines of your printers, a plane of simultaneity, and drop verticals to the x axis. That gives you the spacing of the dots.

Off the top of my head, you'll find that the dots are evenly spaced, but there are too many to fit in the circle so the end of the pattern overlaps the beginning - there's an area that's double-printed.

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jbriggs444

Homework Helper
Sufficiently before noon the complete train is on the circle and it precisely fills the circle, so the rear clock of the last wagon coincides with the front clock of the first wagon. What pattern would these dots have for an observer in the rest frame of the rail?
Like @Ibix, I will assume that the clocks are synchronized before the train enters the loop.

From the rest frame, the fact that the first and last wagon coincide means that the length-contracted train must be exactly one circumference in length. From the train frame (prior to entering the circle), the train's proper length will be greater than that.

According to the rest frame, the train clocks are all ticking slowly and do so throughout the scenario. The clocks at the front are incorrectly synchronized -- leading clocks lag. They remain out of synch throughout the scenario.

When the clocks strike noon and make their marks on the track, the effect will be like a ripple moving from the back of the train to the front. The string of marks will wrap all the way around the track. By the time the ripple hits the front of the train, the front wagon will have moved on. At least some portion of the track will be double (or triple or quadruple) printed.

Which matches what @Ibix expected. Just had to talk it out to be sure I agreed.

Ibix

@metastable - the example is an idealisation. You can dispose of your objections by having the track floating in zero g and an arbitrarily light train travelling on the inside of the loop. Or by making the track arbitrarily small so that the effects of the rotation of the Earth are arbitrarily small.

Mentor note: The posts that this thread is responding to have been deleted as off-topic.

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vanhees71

Gold Member
To be precise, it means that such observers have clocks that tick at the same rate. They may or may not be synchronised in the sense of having a shared notion of $t=0$.
Yes, but this I can do easily by defining
$$\tau(r)=\int_0^{\lambda} \mathrm{d} \lambda' \sqrt{1-\omega^2 r^2} = \sqrt{1-\omega^2 r^2} \lambda.$$
Then at $\lambda=0$ all observers have $\tau(r)=0$, i.e., they are synchronized at this time, $\lambda=0$. Due to the different clock rates observers with different $r$ are not synchronized at any later time (or extrapolating also any earlier time). Only those with equal $r$ will still have synchronized clocks relative to each other.

pervect

Staff Emeritus
It's impossible to pairwise Einstein-synchronize clocks on a rotating platform.emes, are even unsed.

One can start at a specific starting point, and pairwise synchronize clocks using the Einstein convention, and things go fine until one returns back to the starting point, where one winds up with two clocks that are next to each other and not synchronized.

Non-Einstein clock synchronization schemes are possible, and, due to the non-existence of Einstein clock synchronizatiton, are even used. A typical scheme will put a light source at the origin of the rotating system to define syncrhonizaton.

However, the student or layperson applying standard high school formulas to this case will find that they often do not work as expected. For instance, momentum is no longer given by p=mv, momentum and energy are not even isotropic any more.

Tensor methods work fine, but they may not be accessible to all readers.

Ibix

Only those with equal $r$ will still have synchronized clocks relative to each other.
Indeed. All I was doing was noting that you seem to be using "synchronised" to mean "ticking at the same rate", and that you did not seem to me to have specified a particular zeroing procedure.

Foppe Hoekstra

To all,
After 27 years of computer programming experience I know: before you start any calculation, first get the conditions right.

Note that before entering the circle all clocks are sync in the co-moving frame. During the entering of the circle all clocks remain to have the same amount of speed, although in different direction, over the same period of time. I presume that changing the direction of speed does not cause (additional) time dilation to single clocks (it only makes it more complicated to compare different clocks). So when the circle is closed the front clock and the rear clock (of the complete train) coincide and they are sync (*). And as they are at the same time in the same place, at the same speed in the same direction, they might as well be considered as one and the same clock in the co-moving frame.

And then time dilation in the rest frame according to t = γ(t0 + xv/c^2). The x in this equation does not so much stand for travelled distance, but is a space coordinate. In the rest frame this space coordinate is the same for both the front and the rear clock, so also in the rest frame these clocks may be considered as one and the same clock.

That means that the pattern of dots in the rest frame has to make a neatly closed figure, with no overlap of dots. Remains the question whether the dots will be regularly spaced or show a symmetric pattern of narrower and wider gaps. But as I see no possibility to determine where an irregular pattern should ‘start’, there is no other option than a regular pattern.

All together that makes the calculation of the distance d between sequential dots in the rest frame quite simple: d = 2πr/n , where r = n.αL/2π . Thus d = αL.

(* Do we have a new type of synchronisation here? “Feeder synchronisation” so to speak.)

jbriggs444

Homework Helper
So when the circle is closed the front clock and the rear clock (of the complete train) coincide and they are sync (*)
Remember that in special relativity it is not good enough to say "in synch". You have to specify a frame of reference for the synchronization. From the point of view of the rest frame, the clocks were out of synch when they were on the straight track and remain out of synch on the circular section. Even though they are co-located, the front and rear clocks have a different time reading. This is an invariant fact of the matter. Even in the train frame (to the extent that one can be constructed), this fact is evident.

Edit: The fact that the front and rear clocks are co-located yet have different readings makes it impossible to construct a "train frame" where the time coordinate is always equal to the reading of the co-located clock.

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Ibix

Note that before entering the circle all clocks are sync in the co-moving frame. During the entering of the circle all clocks remain to have the same amount of speed,
The first sentence is true, but the second is not if you are working in the initial inertial rest frame of the train. In this frame, the train is initially at rest, then changes speed a lot as it enters the circle.

If you wish to think of the train as not changing speed then you need to work in the rest frame of the track (edit: or the non-inertial frame of the train, but then you cannot use your naive analysis). In that frame, however, the clocks were not synchronised to start with (they tick at the same rate, but do not show the same time initially).

Either way, the problems come from your erroneous analysis, not the theory itself.

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Ibix

In the initial inertial rest frame of the train the clocks change speed and desynchronise because of the time dilation associated with their speed.

In the inertial rest frame of the track the clocks are not synchronised to start with.

In the non-inertial rest frame of the train, the synchronisation convention you are using includes a discontinuity in clocks' behaviour at the point where you are feeding in the trains.

jbriggs444

Homework Helper
In the non-inertial rest frame of the train,
When I was learning mathematics, we were taught not to use the definite article "the" unless there was a demonstration of existence and uniqueness available. Even referring to "the rest frame of the train" contains an error.

Ibix

When I was learning mathematics, we were taught not to use the definite article "the" unless there was a demonstration of existence and uniqueness available. Even referring to "the rest frame of the train" contains an error.
Fair point. However, I think since there's a defined synchronisation convention then the frame is defined, at least well enough for the purposes of the thought experiment. How clocks behave off the tracks isn't specified, but that's not relevant to the experiment.

jbriggs444

Homework Helper
Fair point. However, I think since there's a defined synchronisation convention then the frame is defined, at least well enough for the purposes of the thought experiment. How clocks behave off the tracks isn't specified, but that's not relevant to the experiment.
In my mind's eye, I pictured the train overlapping itself so that the front of the front car and the back of the back car coincide. The events at the front/back of the train would then have conflicting time coordinate requirements. It seems that you are contemplating a half-open "train interval" with a discontinuity but no actual conflict.

Ibix

Hm - no, you have a point. His system is valid as long as you don't include the overlap point. The requirement for a coordinate system is that it be both one-to-one and onto and cover an open interval, because you need to be able to define an infinitesimal open ball around each point covered by the system and you need an invertible relationship between the space and the coordinates.

So he's got a valid frame (subject to further definitions of clock behaviour off the track) that doesn't quite cover the complete track.

jbriggs444

Homework Helper
[...]cover an open interval
Thanks. I was trying to figure a way to phrase that quibble without becoming confrontational.

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