Rotating frames desynchronization

[PAllen]

If we start with a non rotating steel ring in flat space and start it spinning, it will length contract and the radius will shrink according to observers on the ring and according to inertial non rotating observers. If instead, we have a thin disc, and spin it up, the disc will buckle. If we have a solid cylinder and spin it up the stresses will tear it apart. All these physical effects due to length contraction induced by rotation are observer independent. That seems physical enough to me.

Back to the ring. Let us say its initial un-rotating radius is R and and the circumference is $2*\pi*R$. After being spun up in such a way that the tangential velocity of a point on the rim is v according to an inertial non rotating observer (O) at rest with the centre of the ring, then the new radius according to O is $R*\sqrt{(1-v^2)}$ and the new circumference according to O is $2*\pi*R*\sqrt{(1-v^2)}$. According to an observer (O') riding on the ring, the new radius is also $R*\sqrt{(1-v^2)}$ when measured using a tape measure, so O' agrees with O about the radius, but 0' measures the circumference of the ring using a tape measure to still be the same as when it was not spinning (2*pi*R) so O' sees the circumference of the spinning ring to larger by a factor of $1/\sqrt{(1-v^2)}$ than the circumference of the spinning ring as measured by O.

P.S. I am of course ignoring stresses due to centrifugal forces in all the above.

So much depends on your definitions. If, for example (getting very specific), you assume a rubber ring around a rigid cylinder, with lubrication, all initially at rest in an inertial frame. Then, spin up the rubber ring and let its state settle. The radius per the inertial frame cannot change by virtue of the rigid cylinder. Now the the circumference of the rubber ring measured by the inertial frame is the same as it always was; while the circumference measured by a 'rim dweller' using their own rulers laid end to end, will be increased by gamma.

 

[Mentz114]

If we have a solid cylinder and spin it up the stresses will tear it apart. All these physical effects due to length contraction induced by rotation are observer independent. That seems physical enough to me.

Does this mean that one observer will see the ring break up and another will not ?

I assume you are not saying that.

Suppose we put strain gauges around a circumference of the cylinder that measure the deformation of the material. Let them have digital readouts that are visible to anyone who looks. All observers will see the same readings - therefore the strains could no have been caused by length contraction.

 

[PAllen]

Of course there is real strain in the rubber ring around a rigid cylinder I described. My intent was that breakage would not occur due to specification of rubber - it stretches.

This is not different from Bell - everything depends on where you put the constraints. If you say the ships accelerated uniformly per an inertial frame, then the string stretches (physical strain). If you say string preserves local length along itself [no strain], then the accelerations are not uniform, and the string shrinks per the inertial frame.

There is no greater validity to one set up or another. They are just different scenarios.

 

[WannabeNewton]

For example, if the rim is rotating at a constant rate, there are two rim observers at different points on the rim, and each observer has an atomic clock that sends signals to the other, would the two rim observers find that their atomic clocks are coherent?

Yes because they both read the same kind of time (in this case proper time since the clocks are ideal). On the other hand if one clock is non-ideal and the other clock remained ideal e.g. one observer keeps the atomic clock but the other observer uses a oscillating mass-spring system that is sensitive to centrifugal forces, then clearly the clock hands will tick at different rates.

If they are coherent, then on which kinds of event pairs would the observers disagree about simultaneity (the "many"), and on which would they agree (the few), using typical simultaneity conventions? Specifically, would they agree on the simultaneity of events along the rim?

This will vary significantly depending on the choice of simultaneity or equivalently synchronization convention. Also, what exactly do you mean by "agree on the simultaneity of events along the rim"? If we have three clocks A,B, and C on the rim and define simultaneity pairwise amongst them then the substantial question is whether in the case of an event $p_A$ simultaneous with an event $p_C$ and an event $p_B$ simultaneous with $p_C$ we also have $p_A$ simultaneous with $p_B$. This is basically just a question of transitivity of the simultaneity convention. If this is not what you're talking about with regards to "agreement on simultaneity" then it would help me if you could clarify.

For example, say we have clocks laid out along the rim of the disk and we synchronize them to the central (inertial) clock i.e. we adjust the clocks until they all read the time of the central clock. Now take any two clocks on the rim and define events in their vicinities to be simultaneous if the clocks read the same time at the respective events. Then clearly the clocks will agree on simultaneity of events on the rim because all the clocks on the rim read the same time-that of the central clock; equivalently, the simultaneity surfaces of each rim clock are simply those of the central clock. In other words, transitivity is trivially satisfied.

On the other hand, transport of local Einstein synchronization around the rim of the disk yields a non-transitive relation.

Then it would be helpful to understand why coherence is sufficient to determine simultaneity in an IRF but not in a rotating frame.

Again, it isn't. Two ideal inertial clocks mutually at rest can be coherent but still not be synchronized if their zeroes are not identical. A synchronization convention trivially defines a simultaneity convention so the claim follows.

 

[WannabeNewton]

All observers will see the same readings - therefore the strains could no have been caused by length contraction.

No that's incorrect. Different observers will have different stories regarding the origin of the deformations. In the inertial frame centered on the symmetry axis, the strains come from the Lorentz contraction of the circumference upon simultaneous application of a tangential acceleration to the circumference. In the comoving frame of any given point on the circumference the deformations come from the non-simultaneous application of the tangential acceleration. This is no different at a basic level from the Bell spaceship paradox.

 

[Mentz114]

All observers will see the same readings - therefore the strains could no have been caused by length contraction.

No that's incorrect.

I presume you mean the second part.

Different observers will have different stories regarding the origin of the deformations.

And the stories all have the same ending. Is there any measurement process to measure
the bits which you mention below ?

In the inertial frame centered on the symmetry axis, the strains come from the Lorentz contraction of the circumference upon simultaneous application of a tangential acceleration to the circumference. In the comoving frame of any given point on the circumference the deformations come from the non-simultaneous application of the tangential acceleration. This is no different at a basic level from the Bell spaceship paradox.

This does not convince me of anything except that the theory is consistent. Sure, we can explain the readings, but the only thing we can measure is the physical deformation which is the same for all frames.

[edit]
Suppose we do a calculation in the two frame bases you mention and it comes down to
$S = A^{ab}C_aC_b=A'^{ab}C'_aC'_b$. Both calculations give the observed result but the components of A,C are not the same as the components of A',C'. Can we ascribe independent physical existence to these components ? I think not.

 

[yuiop]

So much depends on your definitions. If, for example (getting very specific), you assume a rubber ring around a rigid cylinder, with lubrication, all initially at rest in an inertial frame. Then, spin up the rubber ring and let its state settle. The radius per the inertial frame cannot change by virtue of the rigid cylinder. Now the the circumference of the rubber ring measured by the inertial frame is the same as it always was; while the circumference measured by a 'rim dweller' using their own rulers laid end to end, will be increased by gamma.

I agree. If we prevent the radius of the ring from length contracting, then we end up with real stresses in the ring and it will eventually break (with no change in radius or circumference according to the non rotating inertial observer (O) at rest wrt the centre of the ring).

Whether or not we allow the radius of the ring to contract, observer O always measures the circumference of the ring to be to smaller than circumference as measured by the ring rider, by a factor of gamma.

 

[PAllen]

Whether or not we allow the radius of the ring to contract, observer O always measures the circumference of the ring to be to smaller than circumference as measured by the ring rider, by a factor of gamma.

Yes, I agree and have derived several different way.

 

[yuiop]

Does this mean that one observer will see the ring break up and another will not ?

I assume you are not saying that.

Correct. If one observer sees the ring break up, then all observers see that.

Suppose we put strain gauges around a circumference of the cylinder that measure the deformation of the material. Let them have digital readouts that are visible to anyone who looks. All observers will see the same readings - therefore the strains could no have been caused by length contraction.

Consider this scenario. We have a train track in arranged in a circle and banked like a "wall of death" so that when the train is circulating, centrifugal forces keep the train firmly on the track inside of the circular wall. If the train is linked all the way round the track, then as it travels faster around the track the links will eventually break. What would you assign as the cause of this breakage, if not stresses due to length contraction?

 

[Mentz114]

Consider this scenario. We have a train track in arranged in a circle and banked like a "wall of death" so that when the train is circulating, centrifugal forces keep the train firmly on the track inside of the circular wall. If the train is linked all the way round the track, then as it travels faster around the track the links will eventually break. What would you assign as the cause of this breakage, if not stresses due to length contraction?

I'm tired so I need to think about this. One thing I can assure you, is that I will not be attributing anything breaking to length contraction. Usually a chain breaks if its end are pulled apart. The fact that two things are moving apart is frame invariant so I'll probably start there.

Later ...

 

[WannabeNewton]

I presume you mean the second part.

Yes.

And the stories all have the same ending. Is there any measurement process to measure
the bits which you mention below ?

That's irrelevant. It doesn't change the fact that it's incorrect to claim length contraction, or more precisely the resistance to length contraction and lack thereof due to non-Born rigid motion during acceleration, plays no role; it plays a role in the appropriate frame. Why would the latter be the "correct" reason for the strains and not the former? Both frames observe strains but for different reasons so again your claim about the innocuous nature of length contraction during the tangential acceleration process is off the mark because as far as the observer in the inertial frame is concerned (resistance to) length contraction is what leads to the deformations.

Just analyze the Bell spaceship paradox in the inertial frame whereupon the spaceships are simultaneously accelerated by the same proper acceleration. How would you describe the breaking of the string in this frame if not for (resistance to) length contraction of the string? If you think it's wrong to ascribe it as such then come up with an explanation for why the string breaks according to said inertial frame.

 

[Mentz114]

Why would the latter be the "correct" reason for the strains and not the former? Both frames observe strains but for different reasons so again your claim about the innocuous nature of length contraction during the tangential acceleration process is off the mark because as far as the observer in the inertial frame is concerned length contraction is what leads to the deformations.

(my bold)

The other frame does not see length contraction - so why would it be the incorrect reason ?

I've done a little calculation ( in answer to yuiop's challenge with the loco and train).

Considering a locomotive pulling a train of trucks along a flat track. The back of the train has eom

$x_b=x_0+\frac{a}{2}(t-\delta_t)^2$

and the front

$x_f=x_0+L + \frac{a}{2}t^2$

where $\delta_t=L/v_s$ and $v_s$ is the speed of sound in the train.

So the distance between the front and back is $L+\frac{a\,\delta_t\,\left( 2\,t-\delta_t\right) }{2}$

Provided $\delta_t>0$ the separation tries to increase indefinately with time.

Some of the work the loco is doing is putting tensile stress in the couplings, which eventually will break them.

If the couplings are weak enough, this will happen at sub-relativistic velocities.

If this calculation is done in a relativistic way it will give the same conclusion, but clearly no length contraction of a physical nature is required.

Now, please don't get upset with me. I'm not challenging the fact that length contraction plays a role in resolving how the correct answer is found from every frame. Only the ontological status of tensor components is questioned.

 

[WannabeNewton]

I'm not challenging the fact that length contraction plays a role in resolving how the correct answer is found from every frame. Only the ontological status of tensor components is questioned.

The only unequivocal answer that's valid in all frames is the breaking of the string or whatever consequence we're considering. The cause of the string breaking depends on the frame, simple as that. There is no substance in saying "length contraction doesn't play a role" just because it doesn't play a role in one frame-it certainly plays a role in another frame.

Now, please don't get upset with me.

I'm not getting upset by any means :) I feel like we're just miscommunicating. Perhaps this should be opened up in a new thread so as not to take this one off topic? At least if this is something you are interesting in pursuing.

 

[JVNY]

George and WBN, thanks very much for great points.

. . . what exactly do you mean by "agree on the simultaneity of events along the rim"? . . . For example, say we have clocks laid out along the rim of the disk and we synchronize them to the central (inertial) clock . . .

I mean that the observers use a convention to determine simultaneity that does not require the use of synchronized clocks. In an inertial frame, such as the Einstein train platform, observers at rest on the platform agree on whether any two lightning bolts that strike anywhere along the platform are simultaneous or not. They can do this without having synchronized their clocks, and indeed without even having coherent clocks. Each merely records the time on his clock when light from each bolt reaches him, then measures the distance to the char mark, then calculates whether the flashes struck at the same time. Each will determine whether the two flashes were simultaneous, and all observers following the same convention will reach the same answer on whether the bolts were simultaneous. George advises that you can use other methods, like the radar method, without having to look to the char marks and again without the observers even having coherent clocks. As George states in post 30,

It doesn't matter whether any clocks are synchronous or not or even if they are coherent. All that matters is whether the Coordinate Time of the events are the same. You could have the time on one clock reading 13 simultaneous with the time on another clock reading 34 and they could be simultaneous.​

Therefore George's comments support the view that two observers can agree on the simultaneity of distant events without using synchronized watches, and without even using coherent watches. At least in an IRF, and apparently also in a non-inertial frame. Condensing his post 34 (hopefully appropriately):

. . .
Yes, but their clocks don't even have to be the same type of clock. They can be running at different rates even though they are in mutual rest. . . they don't have to have coherent watches. Each observer is determining simultaneity of events according to his own clock and if they are inertially at rest with each other, then (as long as they are applying the same simultaneity convention) they will automatically establish the same set of simultaneous events. . .

Two observers that are not even inertial can use radar to determine . . . the simultaneity of remote events . . . As long as they both agree on the frame and the simultaneity convention, they will also agree on which events are simultaneous.

So, can observers on a rotating rim agree on whether two events on the rim are simultaneous without utilizing synchronized watches, just as observers on the platform can agree on the simultaneity of events that occur on the platform without using synchronized watches? Per George's post, as long as the two non-inertial observers agree on the frame and the simultaneity convention, they will agree on which events are simultaneous. Is there such a convention that rim observers can agree on, which does not require the rim observers to have synchronized watches -- without sending signals from the center to synchronize the watches; without sending signals from on the rim to synchronize the watches; without any attempt at all to synchronize the watches?

 

[WannabeNewton]

Is there such a convention that rim observers can agree on, which does not require the rim observers to have synchronized watches -- without sending signals from the center to synchronize the watches; without sending signals from on the rim to synchronize the watches; without any attempt at all to synchronize the watches?

Sure just use light signals from the central observer to define simultaneity for all the rim observers: if we have observers A and B on the rim we say events $p_A$ and $p_B$ in the vicinities of the respective observers are simultaneous if they are both simultaneous with respect to the central observer as determined by light signals and the usual radar time formula. But note that this is entirely equivalent to synchronizing the rim clocks so as to read the time of the central clock so I'm not immediately seeing the need to make a distinction.

Note that if $p_A$ and $p_B$ are simultaneous in this sense then they clearly won't be Einstein simultaneous. This is because if the associated clocks are set to read the time $t$ of the central clock, or more precisely the global time coordinate of the inertial frame of the central clock, and a light signal is emitted prograde along the rim of the disk by clock A at time $t_0$ arriving back to A at time $t_2$ after being reflected by clock B at a time $t_1$ then $t_1 - t_0 \neq t_2 - t_1$ since in the prograde direction clock B is moving away from the light signal whereas in the retrograde direction clock A is approaching the light signal. The same goes for a light signal initially emitted in the retrograde direction except now clock B is approaching the retrograde signal whereas clock A is moving away from the prograde signal.

 

[JVNY]

I agree that this is essentially synchronizing the clocks to the axis, which is not the goal. As you object earlier,

. . . Synchronizing the clocks on the equator according to a master clock at the center . . . just makes the clocks agree on simultaneity of events relative to the master clock, not on simultaneity of events relative to their own rest frames . . .

The goal is to determine whether rim observers can agree on the simultaneity of events relative to their rest frame (the rim), not relative to some other location (like the axis). We have concluded that observers at rest on the inertial platform can agree on the simultaneity of events relative to their platform without using synchronized watches, and without even using coherent watches. Can rim observers do the same? If not, why not?

 

[WannabeNewton]

Sorry I should have been more careful when I said what I did in that quote. What I meant was simultaneity amongst the clocks on the rim of the rotating disk as defined using the global time coordinate of the central clock's inertial frame will not give us Einstein simultaneity amongst the clocks on the rim. But certainly the clocks on the rim will agree on simultaneity of events on the rim if simultaneity is defined using light signals from the central clock.

 

[PAllen]

I agree that this is essentially synchronizing the clocks to the axis, which is not the goal. As you object earlier,



The goal is to determine whether rim observers can agree on the simultaneity of events relative to their rest frame (the rim), not relative to some other location (like the axis). We have concluded that observers at rest on the inertial platform can agree on the simultaneity of events relative to their platform without using synchronized watches, and without even using coherent watches. Can rim observers do the same? If not, why not?

Let me first ask your preference: Do you want to consider a ring or a disk? What I mean is, suppose rim observers want to do something analysis to simultaneity via lightning bolt strikes. Do the see the light moving directly via vacuum (ring), or do we pretend light is forced to follow effectively fiber optic paths around the rim (which is what would be needed for seeing around a disc)?

 

[WannabeNewton]

Sorry I should have been more careful when I said what I did in that quote. What I meant was simultaneity amongst the clocks on the rim of the rotating disk as defined using the global time coordinate of the central clock's inertial frame will not give us Einstein simultaneity amongst the clocks on the rim. But certainly the clocks on the rim will agree on simultaneity of events on the rim if simultaneity is defined using light signals from the central clock.

Thankfully, I found a rigorous mathematical treatment of this.

See section IV of the following paper: http://arxiv.org/pdf/gr-qc/0405139v2.pdf as well as the paragraph directly above section 3 (conclusion) of the following paper: http://arxiv.org/pdf/gr-qc/0506127.pdf

 

[pervect]

I would prefer a concrete example to clarify this conceptual mess in my head. So clocks on Earth that are at rest, when we consider them as the frames of reference, disagree on simultaneity. I get this, but what is the criteria for this. All clocks on the line of rotation have different perspectives on simultaneity? How does their perspective differ? There are many questions in my head and I doubt maths would help it since I'm not an excellent mathematician like most of you guys. Can you give me an example that is based with some clocks on earth, or something like that?

I'm not quite sure I understand what your confusion is, so it's hard to address.

Let me add a few things.

A clock that is "at rest" as in having constant lattitude and longitude is not at rest relative to a hypothetical non-rotating inertial frame based at the center of the Earth.

Einstein synchronization is based on a non-rotating and inertial (or nearly inertial) frame of reference. You were asking about "criterion" - the "criterion" for Einstein clock synchronization is is that clocks moving at a different velocity in an inertial (or nearly so) frame of reference have different concepts of simultaneity.

THe clocks "at rest" (as in having constant lattitude) all have different velocities, and hence all have different notions of "Einstein simultaneity" for points near them. Is this what you were asking about? I wasn't quite sure.

You are also asking about how we actually keep time on Earth. YOu can read about "atomic time", aka TAI time on the wikipedia. You'll note that it does NOT use Einstein synchronization, which is logical because TAI covers the whole Earth without any discontinuities, and as it's been remarked this isn't possible with Einstein clock synchronization.

All clocks on the geoid all run at the same rate, and the TAI standard defines a notion of simultaneity that is not the same as Einstein's.

THis has some consequences to the laws of mechanics, etc - I'm not going to go into detail unless this is one of the things you're interested in. Well, I will say one thing. The Einstein notion of simultaneity (which is the one we are NOT using on the Earth) is really the best/simplest one mechanics, so expect a few surprises when you write the laws of mechanics on the Earth using TAI time. Going into the details would probalby just be confusing until we get the rest of your confusions sorted, I think.

You were also asking about how we order events. This doesn't have anything to do with how we define simultaneity in relativity. Ordering of events in an observer independent manner is always done with light cones, because ordering events according to their coordinates depend on what coordinates you use (I think this is obvious?).

Briefly, I can say that "events in the past light cone are in the past" and "events in the future light cone are in the future" isn't clear enough, I suppose I could go on in more detail. But since I don't know what you're confused about and what you're interested in, it would seem to be better to wait for a question.

So you asked about a bunch of different things, and I tried to answer each "tangent". I hope that clarifies things rather than confuses you more, but I can't quite figure out what you want to know,.

 

[JVNY]

Let me first ask your preference: Do you want to consider a ring or a disk? What I mean is, suppose rim observers want to do something analysis to simultaneity via lightning bolt strikes. Do the see the light moving directly via vacuum (ring), or do we pretend light is forced to follow effectively fiber optic paths around the rim (which is what would be needed for seeing around a disc)?

I would prefer to consider light traveling around the rim (for example by traveling along a mirrored interior surface of the rim, or through fiber optic cables laid along the rim, or the like) if that is instructive. Thanks.

 

[WannabeNewton]

I would prefer to consider light traveling around the rim (for example by traveling along a mirrored interior surface of the rim, or through fiber optic cables laid along the rim, or the like) if that is instructive. Thanks.

The simplest simultaneity convention that makes use of light signals along the rim to and fro observers is Einstein simultaneity and as you know this will not give rise to a valid global time coordinate for the family of observers on the rim. On the other hand if you use the Einstein time of the inertial frame fixed to the symmetry axis as the global time coordinate for the family of observers on the rim then you will get a consistent global simultaneity convention and it will just be given by the simultaneity surfaces of the observer at the center of the disk. The observers on the disk will agree on simultaneity of events anywhere and everywhere as per this convention-it's trivially transitive because it's just the synchronous time of an inertial frame. Keep in mind this simultaneity convention only works because of axial symmetry.

 

[FactChecker]

If you had lots of clocks all around the Equator, at rest relative to the Earth's surface, and used Einstein synchronisation to sync the 2nd clock to the 1st clock, then the 3rd to the 2nd, then the 4th to the 3rd, and so on all round the Equator until you got back where you started, you would find that the last clock and the 1st clock, which are side-by-side, would be out of sync by about 200 nanoseconds.$$\frac{ \left( \frac{40 \times 10^6}{24 \times 60 \times 60} \right) \times \left( 40 \times 10^6 \right) } {\left( 3 \times 10^8 \right) ^2} \approx 2 \times 10^{-7}$$For an object rotating much faster than the Earth, the effect would be greater.

I can understand the clocks being out of sync with a clock at the center of the earth, but why out of sync with other clocks at the same distance from the axis. Aren't they all slowed an identical amount versus the clock at the center? Is this an effect related to what causes precession of Mercury's orbit?

 

[WannabeNewton]

Aren't they all slowed an identical amount versus the clock at the center?

That's not enough to guarantee global Einstein synchronization of the clocks. Just because they are slowed by the same amount doesn't mean Einstein synchronization will be transitive for the clocks on the periphery of the disk. The point is that in a rotating frame Einstein synchronization between clocks at rest in the frame will never be transitive and will therefore lead to time discontinuities.

This is due to a well known result in relativity which states that given a family of standard clocks each following an integral curve of a time-like vector field $\xi^{\mu}$, a necessary condition for the standard clocks to be Einstein synchronizable is that $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$. If you've ever seen Frobenius' theorem in differential topology then it will be evident why this is so.

However for clocks at rest in a rotating frame we have $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} \neq 0$ and so Einstein synchronization is impossible for these clocks.

This is why we compensate by synchronizing the disk clocks to the time of the central clock instead. They will no longer be standard clocks, rather they will be coordinate clocks in the rest frame of the rotating disk, but they will at least be synchronized and share common global simultaneity surfaces. This works of course because the central clock follows an integral curve of the time-like vector field $\nabla^{\mu}t$ where $t$ is the time coordinate in the rest frame of the rotating disk and hence the time read by the central clock and the inertial clocks at rest with respect to it-we see trivially that $\nabla_{[\gamma}t\nabla_{\mu}\nabla_{\nu]}t = 0$.

EDIT: See chapter 3 of Gron's notes: http://www.uio.no/studier/emner/matnat/fys/FYS4160/v06/undervisningsmateriale/kompendium.pdf

 

[pervect]

I can understand the clocks being out of sync with a clock at the center of the earth, but why out of sync with other clocks at the same distance from the axis. Aren't they all slowed an identical amount versus the clock at the center? Is this an effect related to what causes precession of Mercury's orbit?

The precession of Mercury's orbit is a GR effect - the clock synchronization issue appears within special relativity. So I don't think they are directly related.

It confuses a lot of people who hold onto the idea of "absolute time".

A close study of "Einstein's train", the issue where two lightning bolts strike the front and rear of the train simultaneously in one frame, and not simultaneously in the other, is the key to understanding the effect.

Einstein's original argument can be found at http://www.bartleby.com/173/9.html - but it doesn't give any numbers. You can get the numbers directly from the Lorentz transform though:

$t' = \gamma ( t - \beta x/c)$

where $\beta = v/c$ and $\gamma = 1 / \sqrt{1-\beta^2}$

We can break the above equation for t' into two parts or terms

term 1: $\gamma t$
term 2: $-\beta \gamma x/c$

term 1 represents "time dilation:
term 2 represents "the relativity of simultaneity" and is responsible for the issue under discussion

Given the numerical approach above, you basically apply the result from the Einstein train thought experiment around a closed loop of "trains", summing together the second term around the loop.

 

[FactChecker]

term 2 represents "the relativity of simultaneity" and is responsible for the issue under discussion

Given the numerical approach above, you basically apply the result from the Einstein train thought experiment around a closed loop of "trains", summing together the second term around the loop.

Ha! I like that.

 

[JVNY]

The simplest simultaneity convention that makes use of light signals along the rim to and fro observers is Einstein simultaneity and as you know this will not give rise to a valid global time coordinate for the family of observers on the rim. On the other hand if you use the Einstein time of the inertial frame fixed to the symmetry axis as the global time coordinate for the family of observers on the rim then you will get a consistent global simultaneity convention and it will just be given by the simultaneity surfaces of the observer at the center of the disk. The observers on the disk will agree on simultaneity of events anywhere and everywhere as per this convention-it's trivially transitive because it's just the synchronous time of an inertial frame. Keep in mind this simultaneity convention only works because of axial symmetry.

Per our follow up discussion on another thread ("Synchronizing rotating clocks"), there is another convention that does work by sending signals only around the rim, see post 70 linked here: https://www.physicsforums.com/showthread.php?t=732892&page=4. Interestingly, even though the signals only go around the rim, they likely also synchronize the clocks on the rim to an axis centered inertial coordinate system -- despite there being no signals sent from the axis or from any other point at rest with respect to the axis.