Rotating frames desynchronization

[johnny_bohnny]

I was reading a little about the behavior of non-inertial frames in relativity, since I'm interested in knowing how can we measure time on Earth and the sequence of events here. So as we know Earth is rotating and therefore the clocks on surface that are mutually at rest get desynchronized. Can somebody explain this to me better, because I don't understand it. I've always thought that frames that are mutually at rest will agree on simultaneity, but it isn't the case. How do we describe the differences and by which criteria regarding clocks that are at rest with respect to Earth's surface. What are the criteria?

It's very confusing for me to understand this so I hope somebody can simplify it. Regards, johnny

 

[A.T.]

So as we know Earth is rotating and therefore the clocks on surface that are mutually at rest get desynchronized.

At the same distance from the axis they have the same rate in a rotating frame. But this gets superimposed with the gravitational time dilation of the Earth.

I've always thought that frames that are mutually at rest will agree on simultaneity

Only inertial frames.

 

[WannabeNewton]

I've always thought that frames that are mutually at rest will agree on simultaneity, but it isn't the case.

You don't need to consider rotating frames for that. Even in linearly accelerating frames there will be clock desynchronization of spatially separated ideal clocks situated along the line of acceleration. The difference between rotating frames and linearly accelerating frames is the latter allows global Einstein synchronization in principle if you're willing to use non-ideal clocks whereas for the former there is a fundamental obstruction preventing the establishment of global Einstein synchronization even for non-ideal clocks. More generally, global Einstein synchronization cannot be achieved in any rotating frame because the rotation breaks transitivity of the synchronization. The reason for this is a famous rotating frame phenomena known as the Sagnac effect: http://en.wikipedia.org/wiki/Sagnac_effect

 

[WannabeNewton]

By the way we had a related thread some while back on clock synchronization in rotating frames that might prove helpful to you: https://www.physicsforums.com/showthread.php?t=732892

I can go into more of the mathematical details regarding the impossibility of global Einstein synchronization in rotating frames but I know not of your mathematical background so if you want the explicit calculations just say the word :)

 

[johnny_bohnny]

By the way we had a related thread some while back on clock synchronization in rotating frames that might prove helpful to you: https://www.physicsforums.com/showthread.php?t=732892

I can go into more of the mathematical details regarding the impossibility of global Einstein synchronization in rotating frames but I know not of your mathematical background so if you want the explicit calculations just say the word :)

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?

 

[johnny_bohnny]

Can somebody compare the disagreement between frames with the situation regarding the agreement of the frames mutually at rest in the inertial scenario?

 

[DrGreg]

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.

 

[WannabeNewton]

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?

For now forget about clock synchronization in relation to Earth bound orbits because this also involves gravitational effects which will necessarily complicate the matter. Let's consider something simpler.

Imagine we have a circular ring in free space rotating with some constant angular velocity about its symmetry axis relative to an inertial observer at the center of the ring. At each point on the ring we've placed an ideal clock, a concave mirror, and a radar set; we want to try and synchronize all these clocks with one another using Einstein synchronization. One way to go about this is using radar signals. Say we take some clock $A_1$ on the ring. From $A_1$ a light signal is emitted counter-clockwise towards an infinitesimally neighboring clock $A_2$ on the ring whereupon the light signal is reflected back to $A_1$-the placement of concave mirrors alongside each clock guarantees that this light signal will circulate along the ring. We then synchronize $A_1$ and $A_2$ using the Einstein synchronization formula i.e. we define the event at which the light signal reaches $A_2$ to be simultaneous with the event in the vicinity of $A_1$ that lies halfway in $A_1$-time between the round trip time of the light signal. Now a light signal is emitted by $A_2$ counter-clockwise towards an infinitesimally neighboring clock $A_3$ on the ring and we synchronize $A_2$ and $A_3$ using the same operational definition as above. We then repeat this process for each consecutive infinitesimally neighboring clock on the ring until we complete a full circuit around the entire ring and come back to $A_1$.

But what we find upon coming back to $A_1$ is that there is a gap $\Delta t_{\text{desynch}}$ between the original time $t_0$ that $A_1$ read and the time $t_f$ that it reads after performing the above synchronization full circuit around the entire ring starting from $A_1$ and ending at $A_1$. What this means is the above clock synchronization procedure fails to be transitive i.e. if $A_1$ is Einstein synchronized with $A_2$ and $A_2$ is Einstein synchronized with $A_3$ then $A_1$ will not be Einstein synchronized with $A_3$.

Compare this with what happens to ideal clocks at rest in an inertial frame. In such a case if $A_1$ is Einstein synchronized with $A_2$ and $A_2$ is Einstein synchronized with $A_3$ then $A_1$ will be Einstein synchronized with $A_3$ i.e. the procedure is transitive. Because it is transitive, all the clocks at rest in the inertial frame will agree on simultaneity and we can build a global time coordinate $t$ such that the surfaces $t = \text{const}$ correspond to global simultaneity surfaces shared by all of the clocks at rest in this inertial frame.

However in the case of the clocks at rest on the rotating ring, the synchronization procedure outline above is as already mentioned not transitive. Therefore distant clocks on the ring will not agree on simultaneity and we cannot build a global synchronous time $t$ such that the $t = \text{const}$ surfaces represent global simultaneity surfaces shared by all of the clocks on the ring.

In the case of the rotating ring in free space, the reason for the discontinuity $\Delta t_{\text{desynch}}$ that arises after attempting Einstein synchronization in full circuit around the entire ring is the already mentioned Sagnac effect. In our case this is very easy to understand. Imagine you're sitting at some point on the rotating ring and you place two clocks, one separated from you in the clockwise direction and one separated from you in the counterclockwise direction, such that both are equidistant from you. We still have concave mirrors setup at each point on the ring. You set the hands of the two clocks to the same position and temporarily lock the hands in place. Then you emit a light signal in the prograde direction and a light signal in the retrograde direction and have the clocks start ticking when the respective light signals reach them. Clearly the clock separated from you in the clockwise direction starts ticking before the clock separated from you in the counterclockwise direction starts ticking because for the former the light signal in the retrograde direction catches up to an approaching target whereas for the latter the light signal in the prograde direction catches up to a receding target. This means however that clock synchronization on the ring is path dependent and this is exactly why we get the discontinuity $\Delta t_{\text{desynch}}$ after attempting the above synchronization procedure starting and ending at the same clock on the ring. Notice that this wouldn't happen if the ring was non-rotating i.e. if the clocks at rest on the ring were also at rest in the inertial frame of the observer at the center of the ring.

Aha, DrGreg beat me to it!

 

[johnny_bohnny]

Thanks to both of you for your answers. I'll think about it more carefully and contact you in need of addition details.

 

[bcrowell]

Section 8.1 of my SR book may be helpful: http://www.lightandmatter.com/sr/

 

[A.T.]

Thanks to both of you for your answers. I'll think about it more carefully and contact you in need of addition details.

Note that there are different issues here:

- Clocks at rest in a rotating frame which have different distances to the axis, cannot be synchronized at all, because they run at different rates.

- Clocks at rest in a rotating frame which have the same distance to the axis, cannot be synchronized using a certain convention (Einstein synchronization). But they run at the same rate, so they can be synchronized by other means like sending a signal from the center.

 

[bcrowell]

- Clocks at rest in a rotating frame which have the same distance to the axis, cannot be synchronized using a certain convention (Einstein synchronization). But they run at the same rate, so they can be synchronized by other means like sending a signal from the center.

They can be synchronized with the signal, but then they're not synchronized with each other according to Einstein synchronization. The basic issue is that Einstein synchronization isn't transitive in a rotating frame.

 

[WannabeNewton]

Clocks at rest in a rotating frame which have the same distance to the axis, cannot be synchronized using a certain convention (Einstein synchronization). But they run at the same rate, so they can be synchronized by other means like sending a signal from the center.

That won't achieve what the OP wants, as Ben already pointed out. Say we have a rotating disk, ideal clocks laid out along the rim of the disk, concave mirrors laid out along the rim of the disk, and a master clock at the center of the disk. Certainly if we have the master clock send out a spherical pulse of light at some instant that travels outwards, gets reflected back when it reaches the rim of the disk, and arrives back at the master clock individually in the form of light rays then we can set each clock on the rim so that it reads the time simultaneous with the time read by the master clock halfway between the round trip time of the radar signal to and fro each clock. Because of the axial symmetry, all the clocks on the rim will be set to read the same time and hence will be synchronized according to the master clock and the global inertial frame that the master clock is at rest in.

But the key point is the clocks on the rim will only be synchronized according to the inertial frame of the master clock. In other words on each $t = \text{const}$ simultaneity surface of the master clock, all the clocks on the rim of the rotating disk will read the same time-this won't be $t$ obviously since the clocks are not at rest relative to the master clock but the clocks will all read the same time as one another on each $t$ slice regardless. But this is not the global rest frame of the clocks on the rim. If we instead go to the global rest frame of these clocks, and we can appropriately talk about the global rest frame of the clocks because they form a Born rigid time-like congruence, then there is no way to synchronize all the clocks with one another using Einstein synchronization as pointed out by Ben. This is of course because Einstein synchronization relies on the isotropy of light which doesn't hold globally in the rest frame of the clocks-it of course holds locally even in the rest frame hence why we can always Einstein synchronize neighboring clocks on the rim with one another. Anyways, so what this means is we cannot use Einstein synchronization to build a global synchronous time $t$ whose simultaneity surfaces are adapted to the rest frame of the clocks on the rim, but this is in fact what the OP was asking for. If we modify the definition of Einstein synchronization in the global rest frame of the clocks by giving up the isotropy of light inherent in this synchronization procedure, which we know doesn't exist globally in the rest frame of the clocks, then we can synchronize them.

For a more mathematical treatment see here: http://arxiv.org/pdf/gr-qc/0204063v2.pdf

 

[A.T.]

...since the clocks are not at rest relative to the master clock ...

The clock at the center is not at rest in the rest frame of the clocks at the rim?

 

[WannabeNewton]

The clock at the center is not at rest in the rest frame of the clocks at the rim?

Sure, it certainly is, but in the inertial rest frame of the master clock, which is the clock at the center, the clocks on the rim are circling around it, which is what I was referring to or at least meant to in case it came off as ambiguous. Regardless, the issue elucidated above remains.

 

[A.T.]

That won't achieve what the OP wants

Obviously not, because the OP asks about the entire surface of the Earth, which cannot be synchronized. But if we restrict ourselves to say the equator, the clocks can be synchronized by a signal from the center, and all the observers on the equator will agree about the sequence of events on the equator.

 

[WannabeNewton]

But if we restrict ourselves to say the equator, the clocks can be synchronized by a signal from the center...

This still doesn't achieve what the OP wants even if restricted to the equator and Ben and I already explained why so no need to repeat.

 

[A.T.]

This still doesn't achieve what the OP wants even if restricted to the equator

The OP is asking about clock de-synchronization and agreeing on the order of events. If we restrict ourselves to the equator and synchronize the clocks by a signal from the center, then they will not de-synchronize and all the observers on the equator can agree about the sequence of events on the equator.

 

[WannabeNewton]

If we restrict ourselves to the equator and synchronize the clocks by a signal from the center, then they will not de-synchronize and all the observers on the equator can agree about the sequence of events on the equator.

What you're describing is equivalent to the Universal Time system of synchronization. However the OP explicitly stated "I've always thought that frames that are mutually at rest will agree on simultaneity, but it isn't the case." Synchronizing the clocks on the equator according to a master clock at the center won't change this because said synchronization procedure 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 ergo why we can't define global simultaneity in the extended rest frame of the clocks mutually at rest.

 

[A.T.]

we can't define global simultaneity in the extended rest frame of the clocks mutually at rest.

What disagreement about simultaneity of events on the rim would the observers on the rim have, if their clocks were synchronized by a signal from the center? Can you give an example?

If there are clocks placed everywhere around the rim, and every event on the rim gets a t-coordinate according the local clock, then I don't see how there could be a disagreement about the order of events.

 

[WannabeNewton]

Read section 2.3 of the following paper: http://arxiv.org/pdf/gr-qc/0103076v1.pdf

 

[bcrowell]

What disagreement about simultaneity of events on the rim would the observers on the rim have, if their clocks were synchronized by a signal from the center?

It's easy to make observers agree on simultaneity and the order of events. They just have to agree on a set of coordinates. What's less trivial, and in this case impossible, is to make that notion of simultaneity agree with Einstein synchronization, which involves sending light signals back and forth.

Keep in mind that the notion of noninertial particle A being at rest relative to noninertial particle B is not as well defined as people may have been assuming in this discussion. In an inertial frame instantaneously comoving with Los Angeles, Beijing is not at rest. In the rotating frame tied to the earth, Los Angeles and Beijing are both at rest. Only when A and B both move inertially is there a natural way of deciding whether they are mutually at rest, which is to decide the question in an inertial frame tied to A (or to B).

 

[PAllen]

Read section 2.3 of the following paper: http://arxiv.org/pdf/gr-qc/0103076v1.pdf

This paper comes to the conclusion that the theory of Born rigidity is 'wrong' in the sense of not describing the actual physics of 'spinning up'. They make no mention of the expansion tensor, but their conclusions are in conflict with it.

I also wonder (since they mention the analogy of laying out a tape measure), what the author would make of my argument here:

https://www.physicsforums.com/showthread.php?p=4631865#post4631865

[edit: It appears that in the following paper, the referenced author backs off some of the conclusions I found problematic:

http://arxiv.org/abs/gr-qc/0604118

becoming more 'agnostic' about Born rigidity arguments, and modifying the disagreement on circumferential measurerment. ]

 

[Mentz114]

Read section 2.3 of the following paper: http://arxiv.org/pdf/gr-qc/0103076v1.pdf

I read it and I'm completely unconvinced that SR needs to be modified. For instance

Ehrenfest [23, 24] saw a paradox in the presumed circumferential Lorentz contraction effect
which Einstein [25] and Grøn [26] attempted to resolve by claiming that the disk circumference
tries to contract in Lorentz fashion, ...

assumes that length contraction is a physical effect causing stresses. But it is a frame dependent quantity, surely.

Tensile stress may be a well known physical phenomenon in a material body
in space, but there is certainly no such phenomenon associated with time.

That does not make sense to me.

And finally, how good a representation of the physical world is a model in which a clock
can not be synchronized with itself?

But the circumferential clocks can be synchronised, after which they will remain synchronised. It is only the Einstein synchronisation that fails.

The whole paper seems bitty and represents a solution to a problem that does not exist.

 

[WannabeNewton]

Yeah PAllen (EDIT: and Mentz..and Ben!) I agree with you in that the negative discussions in the paper regarding Born rigidity are in obvious conflict with standard calculations involving the kinematical decomposition of time-like congruences. I was intending only to reference the discussion about simultaneity relative to the comoving frames tangential to the disk circumference versus simultaneity relative to the master clock at the center.

The whole paper seems bitty and represents a solution to a problem that does not exist.

Funny enough that's rather characteristic of the entire book to which the paper belongs http://www.springer.com/physics/the...+computational+physics/book/978-1-4020-1805-3 with a few exceptions.

A.T. it seems we were talking past one another. I think we can both agree that Einstein simultaneity relative to the clocks on the rim doesn't work but synchronization relative to the master clock at the center does work. The point I was trying to make is the latter only gives a global synchronous time relative to the inertial frame of the master clock which is fine as long as one knows that it doesn't give a global Einstein time relative to the 4-velocity field of the rim clocks.

 

[bcrowell]

Read section 2.3 of the following paper: http://arxiv.org/pdf/gr-qc/0103076v1.pdf

The Klauber paper comes off like pure crank material to me. He starts off by misunderstanding what we mean in SR when we say that the speed of light is the same in all frames. All we really mean by that is that the metric is a scalar, and therefore everyone agrees when an infinitesimal displacement has ds^2=0. This is true in rotating coordinates just as much as in any other coordinates.

A better reference would be Ø. Grøn, Relativistic description of a rotating disk, Am. J. Phys. 43 869 (1975).

 

[WannabeNewton]

A better reference would be Ø. Grøn, Relativistic description of a rotating disk, Am. J. Phys. 43 869 (1975).

Thanks! But for some reason I can't seem to find it. The best I can find is the following paper: http://link.springer.com/article/10.1007/BF00708527 which for the record also discusses the master clock synchronization.

 

[JVNY]

Let's separate two uses of "synchronized," an intransitive use and a transitive use.

There are two ordinary definitions of "synchronized." "Synchronized" means to occur with exact coincidence in time or rate. Under the latter definition, clocks can be synchronized without any signaling. If two ideal clocks are at rest with respect to each other without rotation or gravity, then they will always be synchronized in their own inertial reference frame regardless of whether either sends a signal to the other. They will tick with exact coincidence in rate in their own IRF (and also in any other IRF). This is an intransitive use because no one is causing the clocks to be set in any way, such as by signals; the clocks merely tick at the same rate.

They are unlikely to tick with exact coincidence in time in any given reference frame, however, because each has to independently begin to tick at some time. Absent chance, they will only tick coincident in time if someone signals or otherwise causes them to be set to commence ticking simultaneously. With a signal they can be synchronized (caused to tick coincident in time, using "synchronize" transitively) under the first definition, at least in one IRF.

A prerequisite for concluding under either definition that the clocks are synchronized is agreement on simultaneity. If there is no agreement on simultaneity, then it is not possible to conclude that each ticks with exact coincidence in time (the first definition), and because rate depends on time it is not possible to conclude that each ticks with exact coincidence in rate.

Consider two clocks at rest with respect to each other, with no rotation or gravity, that begin to accelerate in a row to the right at the same rate in an inertial reference frame (Bell's spaceship paradox). The clocks will be synchronized in the IRF under the second definition (they will tick at the same rate in the IRF). They can also be synchronized in the IRF under the first definition (by an observer in the IRF sending signals that are simultaneous in the IRF, setting the clocks to ticking at the same time). However, neither clock will agree that the other is synchronized with it, because the two do not agree on simultaneity.

So the primary question is whether clocks at two different points on a rotating rim at the same distance from the axis agree on simultaneity. If they do not, then I do not think that it is even possible for them to tick with exact coincidence in rate except in some arbitrary reference frame, such as the frame of the ground at rest underneath the rotating rim. I believe that WannabeNewton has concluded that the two clocks cannot agree on simultaneity (they do not share the same hypersurface of simultaneity on the rim). This is from the thread that he links to in post 4. So they cannot even agree that they are synchronized in the sense of ticking at the same rate except in some reference frame like the ground. And there is no need to even think about the Sagnac effect or Einstein synchronization.

A.T. concludes that clocks at rest on a rotating rim at the same distance from the axis tick at the same rate (post 11). A.T., do you mean that the clocks agree that they tick at the same rate in their own reference frame, that is that they share a hypersurface of simultaneity? If so, then there is a significant disagreement. A.T. would conclude that the clocks are synchronized under the second definition. Whether they can be set to start ticking at the same time as determined on that hypersurface of simultaneity is a secondary question.

 

[WannabeNewton]

If there is no agreement on simultaneity, then it is not possible to conclude that each ticks with exact coincidence in time (the first definition), and because rate depends on time it is not possible to conclude that each ticks with exact coincidence in rate.

Actually no. Two ideal clocks can disagree on simultaneity but still tick at the same rate. If two clocks are at rest in an IRF but their zeroes are set differently then their hands won't have exact coincidence i.e. they won't be synchronized but the hands tick at the same rate. In other words the simultaneity surfaces of the first clock won't agree with the simultaneity surfaces of the second clock but only by an overall translation. The distances between each successive simultaneity surface relative to either clock will be the same i.e. they will tick at the same rate. We can operationally test this very easily because all it amounts to is testing the difference in periodicities of the two clocks which is independent of the zeroes of the clocks-we can for example have each clock flash a pulse of light at each tick towards the opposite clock and compare the frequency of arrivals of light pulses.

If so, then there is a significant disagreement.

There is no disagreement because there is no absolute nor even preferred notion of simultaneity and synchronization relative to non-inertial clocks in general. A.T. and I are referring to two different synchronization methods-one which works and one which doesn't. Instead of pursuing the futile attempt at having the rim clocks read a common global Einstein time orthogonal to their 4-velocity field which cannot work because the 4-velocity field is twisting, thus leading to the failure of global clock synchronization and global simultaneity that I was referring to, we can do something simpler. We can have them read the time of the central inertial clock-this is what A.T. was referring to and is basically just the Geocentric Coordinate Time synchronization used in the International Atomic Time standard. See here for a detailed description of this synchronization: http://relativity.livingreviews.org/Articles/lrr-2003-1/download/lrr-2003-1Color.pdf

 

[ghwellsjr]

Let's separate two uses of "synchronized," an intransitive use and a transitive use.

Let's don't. Private definitions invite confusion and it's totally unnecessary. We already have another word for clocks that tick at the same rate but are not synchronized: coherent.

...

A prerequisite for concluding under either definition that the clocks are synchronized is agreement on simultaneity. If there is no agreement on simultaneity, then it is not possible to conclude that each ticks with exact coincidence in time (the first definition), and because rate depends on time it is not possible to conclude that each ticks with exact coincidence in rate.

More confusion. Simultaneity has to do with events. They either are or are not simultaneous in a given reference frame. 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.

Consider two clocks at rest with respect to each other, with no rotation or gravity, that begin to accelerate in a row to the right at the same rate in an inertial reference frame (Bell's spaceship paradox). The clocks will be synchronized in the IRF under the second definition (they will tick at the same rate in the IRF). They can also be synchronized in the IRF under the first definition (by an observer in the IRF sending signals that are simultaneous in the IRF, setting the clocks to ticking at the same time). However, neither clock will agree that the other is synchronized with it, because the two do not agree on simultaneity.

More confusion. What does "sending signals that are simultaneous" mean? It's an abuse of the definition of simultaneous which applies to events. Let me remind you that an event is an instant in time (zero duration) at a point in space (no expanse).

 

[JVNY]

Thanks WBN and ghwellsjr. It does help to have the defined term coherence, and I will rephrase the post using that term.

Two observers at rest with respect to each other in an IRF with ideal but unsynchronized watches will agree on the simultaneity of events if they use the same simultaneity convention (like the convention used in the Einstein train example). Each simply notes the time on his watch at which light from the event arrives to him (like light from each lightning bolt strike), then measures the distance to the event (e.g., the distance to each char mark on the platform), then using the one way speed of light convention determines whether the two events were simultaneous.

So in this way the observers can (and do) agree that events were or were not simultaneous even though their watches are not synchronized. Two observers at rest anywhere on Einstein's platform will agree that two lightning bolts either were or were not simultaneous; each may ascribe a different clock time to the bolts, but the settings of the clocks are irrelevant to the issue of whether the bolts were simultaneous. I actually agree with what ghwellsjr stated: "You could have the time on one clock reading 13 simultaneous with the time on another clock reading 34 and they could be simultaneous."

But more fundamentally, they can only agree on the simultaneity or nonsimultaneity of the bolts because their watches are coherent -- each must take into account the difference in time at which the two light flashes reach him in order to calculate whether the bolts struck simultaneously. But, how can the two observers determine whether two watches are coherent if they disagree about the simultaneity of events? How can each agree that the event "watch one flashed 13" was simultaneous with the event "watch two flashed 34," and then also that the event "watch one flashed 14" was simultaneous with the event "watch two flashed 35," and so on?

Back to the Bell spaceship analogy, assume that there is an observer on each ship with an ideal watch. Each watch is set to zero and programmed to start ticking when struck by a flash of light. They are at rest with respect to each other in an inertial lab, then begin to accelerate to the right at the same rate. Some lab time later, and simultaneously in the lab, two lightbulbs flash, each of which is directly underneath one of the ships. Both watches begin to tick. The watches will tick coherently in the lab frame (because they are ideal and the ships have the same velocity in the lab frame). The watches will also show the same clock time simultaneously in the lab frame.

However, the observers on the ships will not agree on the simultaneity of the flashes; they will not agree that their clocks began to run at the same time; and they will not agree that their clocks are ticking coherently.

So it seems that before getting to the issue of how two observers may synchronize their clocks, one should ask whether it is possible for the clocks to be coherent for the two observers. Can they be coherent for the two observers if the observers disagree on the simultaneity of events?

 

[PAllen]

So it seems that before getting to the issue of how two observers may synchronize their clocks, one should ask whether it is possible for the clocks to be coherent for the two observers. Can they be coherent for the two observers if the observers disagree on the simultaneity of events?

Yes. A pair of rim observers fit this bill - timing exchange of signals, they may conclude that their distances are remaining constant (so no need to worry about Doppler; of course, since both rim observers feel proper acceleration, this conclusion is not strictly correct). By agreeing on clock design from fundamental physics (e.g. atomic clocks), they find that their reception rate of signals demonstrate their clocks are coherent. Yet, by any of the standard simultaneity procedures they may carry out, they disagree on simultaneity of many event pairs.

 

[JVNY]

Thanks to both. A follow up for PAllen if I might. You say that the rim observers "may" conclude that their proper distance remains constant, and that they disagree on simultaneity of "many" event pairs. Is it possible to be more concrete? 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? 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?

Note that watch coherence is sufficient to allow the platform observers to determine whether events along the platform were simultaneous under a typical simultaneity convention. Clock synchronization is unnecessary; it is irrelevant whether each platform observer ascribes the same watch time (such as "13" or "34") to the events. If the same rule applies on a rotating rim, then the rim observers would agree on the simultaneity of events along the rim in the rim frame as long as their atomic clocks are coherent. Whether it is possible to Einstein synchronize the atomic clocks would not be relevant; ascribing the same clock times is unnecessary.

But perhaps event pairs around the rim are among the many for which rim observers disagree about simultaneity. Then it would be helpful to understand why coherence is sufficient to determine simultaneity in an IRF but not in a rotating frame. I am trying to get at the simultaneity without regard to synchronization. The original post refers to agreement "on simultaneity," and AT focuses on simultaneity, and as already stated one can determine simultaneity without clock synchronization, at least in IRFs.

 

[ghwellsjr]

Thanks WBN and ghwellsjr. It does help to have the defined term coherence, and I will rephrase the post using that term.

You're welcome and I'm glad you're now using the word coherence. It really helps in understanding.

Two observers at rest with respect to each other in an IRF with ideal but unsynchronized watches will agree on the simultaneity of events if they use the same simultaneity convention (like the convention used in the Einstein train example). Each simply notes the time on his watch at which light from the event arrives to him (like light from each lightning bolt strike), then measures the distance to the event (e.g., the distance to each char mark on the platform), then using the one way speed of light convention determines whether the two events were simultaneous.

You can make this work for two observers at rest in the platform IRF because they can go back and measure the distances using rulers to the char marks on the platform from their original positions but it won't work for two train observers because they cannot go back and measure the distances in their IRF.

A way that will always work for all observers is for them to be constantly sending out radar signals (while keeping a log of the sent times) and then when they see the event, they also see the reflection of the particular radar signal that was sent and they can look in their log to see when it was sent. Then they apply the one way speed of light convention to determine the time of each event.

So in this way the observers can (and do) agree that events were or were not simultaneous even though their watches are not synchronized. Two observers at rest anywhere on Einstein's platform will agree that two lightning bolts either were or were not simultaneous; each may ascribe a different clock time to the bolts, but the settings of the clocks are irrelevant to the issue of whether the bolts were simultaneous. I actually agree with what ghwellsjr stated: "You could have the time on one clock reading 13 simultaneous with the time on another clock reading 34 and they could be simultaneous."

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.

But more fundamentally, they can only agree on the simultaneity or nonsimultaneity of the bolts because their watches are coherent -- each must take into account the difference in time at which the two light flashes reach him in order to calculate whether the bolts struck simultaneously.

No, 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.

But, how can the two observers determine whether two watches are coherent if they disagree about the simultaneity of events? How can each agree that the event "watch one flashed 13" was simultaneous with the event "watch two flashed 34," and then also that the event "watch one flashed 14" was simultaneous with the event "watch two flashed 35," and so on?

If they are mutually at rest and inertial then they can simply observe each other's clock. As long as they don't see it gaining or losing time, then they are coherent. This has nothing to do with simultaneity (although they can additionally determine any difference in the settings of their clocks).

Back to the Bell spaceship analogy, assume that there is an observer on each ship with an ideal watch. Each watch is set to zero and programmed to start ticking when struck by a flash of light. They are at rest with respect to each other in an inertial lab, then begin to accelerate to the right at the same rate. Some lab time later, and simultaneously in the lab, two lightbulbs flash, each of which is directly underneath one of the ships. Both watches begin to tick. The watches will tick coherently in the lab frame (because they are ideal and the ships have the same velocity in the lab frame). The watches will also show the same clock time simultaneously in the lab frame.

However, the observers on the ships will not agree on the simultaneity of the flashes; they will not agree that their clocks began to run at the same time; and they will not agree that their clocks are ticking coherently.

During the acceleration period, they will not see each other's clock ticking at the same rate as their own but eventually they will become coherent.

So it seems that before getting to the issue of how two observers may synchronize their clocks, one should ask whether it is possible for the clocks to be coherent for the two observers. Can they be coherent for the two observers if the observers disagree on the simultaneity of events?

Again, simultaneity of events is not dependent on coherency of clocks. It is dependent on establishing a reference frame and coordinate time for that frame.

Two observers that are not even inertial can use radar to determine not only the simultaneity of remote events but the distances to those events and establish a complete reference frame with coordinates. Then they can transform that reference frame into any other reference frame and re-establish a new set of simultaneities. As long as they both agree on the frame and the simultaneity convention, they will also agree on which events are simultaneous.

 

[yuiop]

... assumes that length contraction is a physical effect causing stresses. But it is a frame dependent quantity, surely. ...

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.

 

[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.