Synchronizing rotating clocks

[WannabeNewton]

I think the clocks themselves *will* remain synchronized permanently, in the sense that a consistent one-to-one correspondence can be set up, using the method given, between events on the two clocks' worldlines, so that we have a continuous set of pairs of events that happen "at the same time".

I agree with what you have just said regarding permanent simultaneity, but if it's not too much trouble could you address my confusion(s) in post #20, which describes why it isn't 100% evident to me why there is permanent synchronization? Thanks.

But the usual sense of "simultaneity" implies more than that. It implies being able to consistently assign time coordinates to events off the worldlines of the two clocks and the (changing) common diameter between them. If all that can be "synchronized" are events on the clocks' worldlines, and no others, I don't see why such a concept of "synchronization" is worth pursuing.

Correct me if I'm wrong but your point is in agreement with the second point I made in post #2: https://www.physicsforums.com/showthread.php?t=732892#post4630828 right?

 

[PeterDonis]

I agree with what you have just said regarding permanent simultaneity, but if it's not too much trouble could you address my confusion(s) in post #20, which describes why it isn't 100% evident to me why there is permanent synchronization?

Look at it in the global inertial frame in which the center of the ring is at rest. You should be able to convince yourself that the synchronization of the two opposite ring-riding observers at some instant will give a set of simultaneous events (along the ring diameter connecting the two observers) that all have the same coordinate time in the global inertial frame--i.e., the instantaneous simultaneity using the "opposite diameter" convention just happens to match, along the ring diameter (but *only* there--see below), with the simultaneity of the global inertial frame.

Now, if we restrict attention only to the two clocks' worldlines, having once been synchronized as above, they will remain so synchronized, because their velocities both have identical magnitudes relative to the global inertial frame, and their "clock rates" relative to that frame are determined by $v^2$ (i.e., just by the magnitude of $v$, not its direction). So they will indeed be permanently synchronized.

Of course this raises an obvious question: does this mean the "same diameter" synchronization method is actually *equivalent* to the method of just using the simultaneity of the global inertial frame? No, it isn't, because, as I've noted before, the "opposite diameter" method is different for each distinct pair of opposite observers (more precisely, different once we try to extend it off the worldlines of the observers), and only works along the (changing) diameter between them. So you can't construct any global common simultaneity by this method. But, of course, once you realize that any opposing pair of worldlines can be synchronized permanently by this method, and that the synchronization just on those worldlines is identical to that of the global inertial frame, the obvious strategy is to drop the "opposite diameter" criterion altogether, and just adopt the global inertial frame's simultaneity, since it gives the same answer for any opposite pair of ring-riding observers, plus it works globally (where "works globally" means "is transitive"--see below).

Correct me if I'm wrong but your point is in agreement with the second point I made in post #2: https://www.physicsforums.com/showthread.php?t=732892#post4630828 right?

Generally, yes. The word "synchronization" (or "simultaneity") can be used in multiple ways; I, like you, would view the transitivity property as a key requirement. But that's more a question of words than physics. Physically, I think it's clear that the "synchronization" being talked about in this case does *not* have the transitivity property. Whether "synchronization" is the right term for such a thing, or whether it still might be useful for anything, are separate questions.

 

[WannabeNewton]

Now, if we restrict attention only to the two clocks' worldlines, having once been synchronized as above, they will remain so synchronized, because their velocities both have identical magnitudes relative to the global inertial frame, and their "clock rates" relative to that frame are determined by $v^2$ (i.e., just by the magnitude of $v$, not its direction). So they will indeed be permanently synchronized.

Thank you. So this is basically the same as what I said near the end of post #20 (quoted below).

The only reason I mentioned what I called a "minor detail" is that while your explanation used the global inertial frame of the central clock, relative to which all the clocks on the rotating ring tick at the same rate, JVNY's original analysis used an MCIF at the location of clock A at some instant of clock A's time, relative to which clock B moves tangentially to the rotating ring at that instant whereas clock A is of course at rest in this MCIF.

EDIT: So at this instant clock B would be ticking slower than clock A in the MCIF because of the gamma factor at this instant that the MCIF would attribute to clock B relative to clock A, whereas in the global inertial frame of the central clock we know that both clocks A and B tick at the same rate for all time. But like I said I don't know if it's an issue at all, it's probably just me over-thinking the situation.

One might instead argue that if clocks A and B are initially synchronized as per the "comoving inertial synchronization" (so that at the instant clock A reads 1.0s, clock B is also set to read 1.0s because the event at clock B is simultaneous with the event at clock A relative to the inertial observer comoving with clock A at 1.0s) then they must remain synchronized because there is no physical reason for them to become desynchronized: the two clocks tick at the same rate because they are at the same radius from the central clock. I wouldn't immediately have any problem with this line of argument apart from the minor detail that the uniform clock rate of clocks A and B is relative to the central clock but I don't know if this is really an issue or not.

 

[PeterDonis]

JVNY's original analysis used an MCIF at the location of clock A at some instant of clock A's time, relative to which clock B moves tangentially to the rotating ring at that instant.

But this only works for an instant, and even at that instant, its surface of simultaneity only coincides with that of the global inertial frame along the diameter connecting A and B. That's why, as I said, I think it's better just to do the analysis in the global inertial frame. The MCIF analysis, IMO, ends up taking you down a dead end; the fact that, at a particular instant, the MCIF's simultaneity happens to match up with the global inertial frame's along the one diameter is, IMO, an observation that, while true as far as it goes, doesn't actually lead anywhere.

 

[WannabeNewton]

But this only works for an instant, and even at that instant, its surface of simultaneity only coincides with that of the global inertial frame along the diameter connecting A and B. That's why, as I said, I think it's better just to do the analysis in the global inertial frame. The MCIF analysis, IMO, ends up taking you down a dead end; the fact that, at a particular instant, the MCIF's simultaneity happens to match up with the global inertial frame's along the one diameter is, IMO, an observation that, while true as far as it goes, doesn't actually lead anywhere.

Sorry I was in the process of editing my post when you replied but I added an additional comment in post #38 to clarify my original concern regarding the use of the MCIF of clock A over the global inertial frame of the central clock.

Anyways, I agree with everything you've said and if I may add, the mathematical analysis is much simpler in the global inertial frame of the central clock because in this frame the clocks on the rotating ring unequivocally all run at the same rate, we can easily describe mathematically the use of light signals emitted by the central clock to synchronize the clocks on the rotating ring, and we only have to work with this one global inertial frame.

For example, let's say at a given instant $t_0$ of the central clock's time that the diametrically opposite clocks A and B on the rotating ring are located at $\phi_0$ and $-\phi_0$ respectively where $\phi_0$ is chosen such that light signals emitted by the central clock towards $\phi = \pi$ and $\phi = 0$ at $t_0$ are received by clocks A and B when they reach $\phi = \pi$ and $\phi = 0$ respectively; furthermore let's assume that clocks A and B are temporarily deactivated.

Relative to the central clock, clocks A and B will of course arrive at the respective locations $\phi = \pi$ and $\phi = 0$ simultaneously so we can use the central clock's notion of simultaneity to then activate clocks A and B when they receive the light signals emitted by the central clock at $t_0$ so that they become synchronized at this instant. Because they tick at the same rate relative to the central clock (same gamma factor relative to the central clock) they will trivially remain synchronized.

For me at least, the analysis just isn't as intuitive when using different MCIFs at different instants of clock A's time.

 

[JVNY]

This conversation is more detailed than I considered in the original post. I was thinking of three pairs: first, the pair of clocks riding on the rim; second, a pair of momentarily co-moving points in inertial motion with respect to the lab frame, each in the same direction as one of the clocks, and aligned right alongside the clock; and third, a pair of signaling devices at rest and synchronized in the lab frame, each right alongside a clock. Thus each pair is momentarily aligned along the same line (the dashed line of simultaneity in the diagram of the original post). I had assumed that the signaling devices on the ground would signal the clocks to set themselves to a specified time simultaneously in the clocks' own frame. However, the co-moving points could alternatively provide the signals, as long as they both showed the same time as they came alongside the clocks (that is, along the dashed line of simultaneity in the diagram attached to the original post). Of course this does not work because the clocks do not agree on simultaneity, but that was the thought.

After that, I did not think about the co-moving frames anymore. I just thought as follows: the clocks remain always with instantaneous perpendicular direction of motion to the diameter, so they should continue to agree on simultaneity (as if they were points moving inertially in opposite directions that ran into the dashed line of simultaneity in the diagram in the original post, and got stuck there). So they are effectively at rest with respect to each other in that diagram. Then, they are at the same distance from the hub, so neither clock should run faster than the other owing to the equivalence principle (neither being in effect higher in gravity). So they should stay synchronized in their own frame.

Put in a more practical way, if the clocks agree on simultaneity when set for the L-15 reasons, and if they agree that their simultaneity matches that of the lab underneath them for the same reasons, then they will continue to agree on simultaneity between themselves and the lab as they rotate, and the lab will also agree. They will remain synchronized to ground observers only if they remain synchronized in their own frame. And they do remain synchronized in the ground frame, as the CERN experiments of muons in a ring show. So it must be the case that they remain synchronized in their own frame. That is about as far as I thought about it, and you guys are clearly thinking through it in much more detail.

This leads to a bit more confusion for me, perhaps caused by too closely linking simultaneity and synchronization. In the CERN experiments, the muons all around the ring decayed at the same rate. Time passed for each muon at the same rate in the lab frame; decays of many muons would be simultaneous in the lab frame. However, the muons would not themselves agree on the simultaneity of their decays, or put another way the simultaneity of the ticking of their own proper clocks. Does that puzzle anyone? The muons do not agree on the simultaneity of their own time (and decay), yet they all decay simultaneously in the ground frame? If you accelerated muons in Bell's spaceship paradox the muons would decay at the same time in the ground frame, but not at the same time for the muons. That makes sense in a straight line, but it is hard to see how it works with large numbers of muons aligned all around the rim of the hypersurface of simultaneity in the Vallisneri piece.

 

[PeterDonis]

They will remain synchronized to ground observers only if they remain synchronized in their own frame.

No, they won't, because they do not remain "synchronized in their own frame" in any meaningful sense; there is no "their own frame" that works like this. That's part of the point I've been making in my posts.

In the CERN experiments, the muons all around the ring decayed at the same rate. Time passed for each muon at the same rate in the lab frame; decays of many muons would be simultaneous in the lab frame.

Yes.

However, the muons would not themselves agree on the simultaneity of their decays, or put another way the simultaneity of the ticking of their own proper clocks.

It depends on how you synchronize their clocks, and what requirements you want to put on the "simultaneity" thus obtained. If all you want is a sense of "simultaneity" that works on the muon's worldlines, but not anywhere else, you can synchronize the muon's clocks with a light signal emitted in all directions at the center of the ring; each muon's clock starts at time zero when it receives the signal. The simultaneity surfaces thus obtained will match those of the global inertial frame in which the center of the ring is at rest; and if every muon decays exactly at the average decay lifetime, all the decay events will also be in the same simultaneity surface in this sense. But those surfaces will not be "natural" simultaneity surfaces for any of the muons given their state of motion; put another way, they will not be simultaneity surfaces in any MCIF at any event on any muon's worldline.

Does that puzzle anyone? The muons do not agree on the simultaneity of their own time (and decay), yet they all decay simultaneously in the ground frame?

No, because simultaneity is relative. Or perhaps a better way to put it is: simultaneity is a convention. The muon's don't agree on simultaneity if we use the "natural" simultaneity convention of any MCIF at any event on any muon's worldline. But we can construct a simultaneity convention (though not a "natural" one in this sense for any muon) in which the muons do agree on simulaneity, as above. Neither convention is any more "physically real" than the other; one may be more useful than another for a particular purpose, but that's all.

 

[JVNY]

No, they won't, because they do not remain "synchronized in their own frame" in any meaningful sense; there is no "their own frame" that works like this. That's part of the point I've been making in my posts.

I agree with you. I was just explaining what I was thinking when I made the original post (which I now recognize is wrong). Some of the posts seem to try to understand what I was proposing, so I thought that I would attempt to make that clearer.

 

[JVNY]

. . . if every muon decays exactly at the average decay lifetime, all the decay events will also be in the same simultaneity surface in this sense. But those surfaces will not be "natural" simultaneity surfaces for any of the muons given their state of motion; put another way, they will not be simultaneity surfaces in any MCIF at any event on any muon's worldline.

Understood. And this makes total sense for me in straight line motion. For example, consider two muons accelerated toward each other in a lab (rather than into a ring). I understand how both could decay simultaneously in the lab frame even though they would not agree on simultaneity and each could observe the other to decay first. But how to extend that to a large number of muons around the rim of the undulating hypersurface of simultaneity in the Vallisneri paper really challenges my imagination.

Which is not a bad thing!

 

[WannabeNewton]

Keep in mind however that given a rotating ring and (ideal) clocks mounted on the ring, you can always synchronize two clocks on the ring that are neighboring by using Einstein synchronization locally.

More precisely, consider a thin disk in flat space-time rotating with angular velocity $\omega$ about the $z$ axis of a global inertial frame with origin at the center of the disk; the metric in the coordinates comoving with observers at rest on the disk is simply that of the rotating coordinates on the disk: $ds^{2} = -\gamma^{-2}dt^2 + 2r^2 \omega dtd\phi + r^2 d\phi^2 + dr^2$ where $\gamma^{-2} = 1 - \omega^2 r^2$. Consider an observer O at $(r,\phi)$, another observer O' at $(r+dr,\phi)$, and a third observer O'' at $(r,\phi + d\phi)$; O and O'' are riding on the same ring whereas O' is riding on the ring directly behind O. We are assuming, as is unequivocally assumed, that O, O', and O'' carry ideal clocks (c.f. section 16.4 of MTW). The clocks of both O and O'' tick at the same rate $d\tau = \gamma^{-1}(r)dt$ whereas the clock of O' ticks at the rate $d\tau' = \gamma^{-1}(r + dr) dt$.

At some event on the world line of O let O emit a light signal towards a neighboring event, have the light signal instantaneously reflected back to O at the neighboring event, and define the neighboring event to be simultaneous with the event on the world line of O whose proper time lies halfway between those of emission and reception of the light signal. Doing so we find that an event $(t,r,\phi)$ on the world line of O is simultaneous with a neighboring event $(t + dt,r + dr, \phi + d\phi)$ if and only if $dt = -\gamma ^2(r) r^2 \omega d\phi$.

It's clear that using this relation O and O'' can synchronize their clocks. O sets the zero of his clock so that $\tau = \gamma^{-1}(r)t$ whereas O'' sets the zero of his clock so that $\tau'' = \gamma^{-1}(r)t +\gamma(r) r^2 \omega d\phi$; then whenever an event p local to O is simultaneous with an event q local to O'' we have $\tau''(q) - \tau(p) = \tau''(t + dt) - \tau(t) = \gamma^{-1}(r)dt +\gamma(r) r^2 \omega d\phi =0$, where $dt = -\gamma ^2(r) r^2 \omega d\phi$.

However we will not be able to extend this synchronization globally to all clocks around the ring for obvious reasons: the tangent field to the congruence of observers riding on the ring has non-vanishing vorticity.

For O and O', the relation reduces to $dt = 0$. However O and O' can't synchronize their clocks. Even if O and O' initially synchronize their clocks at $t = 0$, so that $\tau = \gamma^{-1}(r)t$ and $\tau' = \gamma^{-1}(r + dr)t$, they will immediately become desynchronized at the next instant of coordinate time because the two clocks tick at different rates. The only possible way to synchronize them is if O' decides to use a non-ideal clock by readjusting the clock rate of his clock so that it reads $\tilde{\tau}' = \frac{\gamma^{-1}(r)}{\gamma^{-1}(r + dr)}\tau'$ but this will mess up all the formulas for 4-velocity, 4-acceleration, and other kinematical quantities so it's an undesirable alternative.

 

[ghwellsjr]

In spite of the fact that there are at least nine references in this thread to exercise L-15 in Taylor and Wheeler's Spacetime Physics (which you can view here if you don't have the book), this thread has absolutely nothing to do with that exercise. Nowhere do they use the words "synchronize" or "simultaneous" or any of their derivatives. Instead, it's about where two clocks "agree" and by that they mean where do you have to be to see both clocks always displaying the same time, assuming that they have been previously set to agree in the first place, which is not the same thing as saying that either clock agrees that the other clock is displaying its same time.

If you want to draw an analogy from the exercise to clocks on a rotating ring, then the place where clocks agree is along the axle, assuming that they have been previously set to agree in the first place. And it's not just two clocks on opposite ends of a diagonal, it's all the clocks on the rim. And the place is not just at the center of mass of the ring, it's along an infinite line passing through that point which I have referred to as the axle, for lack of a better term.

Talking about how two clocks on opposite ends of a diagonal may or may not be synchronized may be an interesting discussion but to associate it with L-15 is a mistake.

 

[JVNY]

But all with different magnitudes. The standard definition of "co-moving" in SR requires both magnitude and direction to be the same. . . The velocities are all pointed (momentarily) in the same direction, but they are of different magnitudes. So the surfaces of simultaneity of all the observers lying, momentarily, on a diameter of the ring will be "tilted" by different amounts in spacetime; they won't line up.

Yes, I see; my statement was incorrect. What I should say is this: in that instant, each clock on the radius is indistinguishable from an inertially moving clock having the same direction (perpendicular to the radius) and magnitude as it. Now, the problem arises again. Consider clocks inertially moving at different speeds in the same direction, and they happen to align in a row perpendicularly to their direction of motion at one instant. The attached diagram shows three such clocks, moving to the left along the x axis, shown in a fourth inertial frame (the lab frame). The three clocks happen to have the same x coordinate at the same time in the lab frame -- they are aligned along the dashed line. I believe that under the rationale of exercise L-15, all of the inertial frames (the three clocks and the lab frame) will agree on the simultaneity of events that occur at that moment anywhere along the line transverse to the relative motion (anywhere along the dashed line).

Do you agree?

If so, then there must be another reason why the radius clocks do not agree on simultaneity along the same line. It cannot be their different speeds. Even if you tilt the inertial clocks' lines of simultaneity, the lines will all tilt up from their location, so they will all tilt up from the same coordinate on the x axis. So they will all agree on simultaneity for events that occur at that time on a line transverse to them. Each radius clock appears superficially to be identical to its pair on the inertial row (same instantaneous velocity as its pair in the inertial row, and same position on the line perpendicular to relative motion). But the radius clocks do not agree on simultaneity of events along the radius.

 

[JVNY]

In spite of the fact that there are at least nine references in this thread to exercise L-15 . . . this thread has absolutely nothing to do with that exercise. Nowhere do they use the words "synchronize" or "simultaneous" or any of their derivatives. Instead, it's about where two clocks "agree" and by that they mean where do you have to be to see both clocks always displaying the same time, assuming that they have been previously set to agree in the first place, which is not the same thing as saying that either clock agrees that the other clock is displaying its same time.

I think that "to see both clocks . . . displaying the same time" implicitly assumes a determination of simultaneity. It implicitly says "to see both clocks displaying the same time at the same time." And the only reference frames that Taylor and Wheeler are discussing in that context are the two at issue in the exercise: the lab and the rocket. So both frames consider their clocks to display the same time at the same time.

We can skip references to L-15 if we address the question in my immediately prior post: if there are two events along the dashed line (which everywhere has the same x-axis coordinate) that occur at the same time in the lab frame, will each of the three inertial clocks agree that the events are simultaneous. I think that they will.

It is as if the flashes in Einstein's train example occur to the left and right of the train, rather than in the front and the rear. Or, put another way, if you have three trains on parallel tracks in a race, each moving inertially at a different speed as they cross the finish line but all of their fronts crossing the line simultaneously in the platform frame, won't they and everyone on the platform at the finish line agree that the race was a tie? All should agree that the trains' fronts reached the finish line at the same time, because the line is perpendicular to their motion.

 

[PeterDonis]

in that instant, each clock on the radius is indistinguishable from an inertially moving clock having the same direction (perpendicular to the radius) and magnitude as it.

But if they're indistinguishable, then they must agree on a sense of simultaneity at that instant. So this...

the radius clocks do not agree on simultaneity along the same line.

...cannot be right as you state it, because if the radius clocks are indistinguishable from momentarily comoving inertial clocks, then they must have the same definition of simultaneity at that instant. Conversely, if the radius clocks do not agree on simultaneity with momentarily comoving inertial clocks, then they are *not* indistinguishable.

Again, what's really going on here is that simultaneity is a *convention*. For inertial observers, there happens to be a unique "natural" simultaneity convention that is picked out; but for accelerated observers, there isn't. The radius clocks (which are accelerated) could adopt the same simultaneity convention as momentarily comoving inertial clocks, but since the radius clocks are accelerated, they will be comoving with *different* inertial clocks as they move, so the "tilt" of the simultaneity surfaces will change. That means that this simultaneity convention breaks down for accelerated observers for events far enough away from their worldlines (how far "far enough away" is depends on the magnitude of the acceleration); in other words, events far enough away from the radius clock's worldline will be simultaneous, by the "momentarily comoving" convention, with *multiple* events on the radius clock's worldline.

(Note, btw, that this also happens with straight linear acceleration: events far enough away from a linearly accelerated worldline will be simultaneous with multiple events on the worldline by the "momentarily comoving" convention.)

So in view of the above, I think the questions you are asking about the radius clocks' simultaneity aren't really well-defined; you are assuming that "simultaneity" is something given, but it's not; it's just a convention. Different conventions have different implications, and that's all there is to it.

 

[JVNY]

. . . I think the questions you are asking about the radius clocks' simultaneity aren't really well-defined; you are assuming that "simultaneity" is something given, but it's not; it's just a convention. Different conventions have different implications, and that's all there is to it.

But there appears to be an agreed upon definition of simultaneity for observers riding on a rotating disk (thus for any observers riding on a radius of a rotating disk). WannabeNewton refers to it, and the Vallisneri piece illustrates it. And that definition appears to say that the riders on a radius do not agree on the simultaneity of distant events even along the line of the radius. Per WBN:

the simultaneity hypersurfaces of observers riding on the rotating disk are quite a bit more complicated than the simultaneity hypersurfaces of momentarily comoving inertial observers

https://www.physicsforums.com/showthread.php?t=732892#post4632878​

Using the convention of sending light signals from the hub can cause the clocks to be set to a given time simultaneously in the lab frame, but everyone says that this does not set them to the given time simultaneously on the disk. So it does not seem right to say that there are just different conventions and leave it at that.

 

[PeterDonis]

But there appears to be an agreed upon definition of simultaneity for observers riding on a rotating disk (thus for any observers riding on a radius of a rotating disk). WannabeNewton refers to it, and the Vallisneri piece illustrates it.

Are you referring to the Marzke-Wheeler definition? I'm not sure I would call that one "agreed upon". It does have some nice properties, but it's not the only convention used in the literature. And for the rotating ring case, it has two obvious drawbacks: (1) the surfaces of simultaneity are different for each different observer on the ring, and (2) none of those surfaces match up with the simultaneity surfaces of an observer who is moving with the ring's center of mass.

And that definition appears to say that the riders on a radius do not agree on the simultaneity of distant events even along the line of the radius.

Yes, that's true of the Marzke-Wheeler convention.

Using the convention of sending light signals from the hub can cause the clocks to be set to a given time simultaneously in the lab frame, but everyone says that this does not set them to the given time simultaneously on the disk.

Not with either the "momentarily comoving" simultaneity convention or the Marzke-Wheeler convention, no. But those are not the only possibilities.

So it does not seem right to say that there are just different conventions and leave it at that.

Why not? The convention you adopt makes no difference to the results of any experiments; different conventions result in different coordinate charts being used to assign coordinates to events, but the predictions for all observables are the same.

 

[ghwellsjr]

I think that "to see both clocks . . . displaying the same time" implicitly assumes a determination of simultaneity.

No, it doesn't. Simultaneity is not something that the observers/objects/clocks in the scenario can see. We can see it on the diagram but those observers cannot see the diagram unless they send out a lot of radar pulses, make an assumption about the timing of the radar signals, make a lot of observations, collect a lot of information, do a lot of calculations and when it's all over, draw their own diagram. Then, long after the fact, they can go back and establish the simultaneity of previous events according to whatever frame they choose. (Just like we can pick any frame we choose and the simultaneity of events changes.)

Here, take a look at this diagram showing an example of a lab frame where there is a blue clock at rest and a red rocket clock traveling at 0.8c. The dots represent one-nanosecond increments of time on each of the two clocks (there are only two clocks in their example):

As you can plainly see, the blue clock at 5 nsecs is simultaneous with the red clock at 3 nsecs. Simultaneity means that two or more events have the same coordinate time, in this case, 5 nsecs. But neither clock sees both clocks at either the same time as itself or as the simultaneous time. In fact, the blue clock doesn't see the reading of 3 nsecs on the red clock until it has gotten up to 9 nsecs and the red clock doesn't see the reading of 5 nsecs on the blue clock until it has gotten up to 15 nsecs.

It implicitly says "to see both clocks displaying the same time at the same time." And the only reference frames that Taylor and Wheeler are discussing in that context are the two at issue in the exercise: the lab and the rocket. So both frames consider their clocks to display the same time at the same time.

The exercise is specifically to determine the speed that a plane surface has to travel so that any observer traveling on that plane surface will see the two clocks always displaying the same time. It has nothing to do with what observers at rest in either the lab frame or the rocket frame see of anything.

They point out that the obvious choice for this speed to be one-half of the rocket's speed is not correct. In my example, it is not one-half of 0.8c which would be 0.4c. Instead, their formula shows that the correct speed is 0.5c. Here is another diagram with an observer traveling at 0.5c in the lab frame with signals every nanosecond from both clocks arriving at the same time so that this observer always sees the clocks displaying the same time. They do not consider the rest frame of the observer and it won't matter what frame we use to illustrate this:

We can skip references to L-15 if we address the question in my immediately prior post: if there are two events along the dashed line (which everywhere has the same x-axis coordinate) that occur at the same time in the lab frame, will each of the three inertial clocks agree that the events are simultaneous. I think that they will.

It is as if the flashes in Einstein's train example occur to the left and right of the train, rather than in the front and the rear. Or, put another way, if you have three trains on parallel tracks in a race, each moving inertially at a different speed as they cross the finish line but all of their fronts crossing the line simultaneously in the platform frame, won't they and everyone on the platform at the finish line agree that the race was a tie? All should agree that the trains' fronts reached the finish line at the same time, because the line is perpendicular to their motion.

But if you are concerned with what observers actually see, then everyone on the platform will see the nearer train win. In order to determine that it was a tie, they have to do something much more complicated such as place previously synchronized clocks adjacent to each track and then separately photograph each train and clock and go back and look at the arrival times in the recordings.

 

[WannabeNewton]

But there appears to be an agreed upon definition of simultaneity for observers riding on a rotating disk (thus for any observers riding on a radius of a rotating disk). WannabeNewton refers to it, and the Vallisneri piece illustrates it

Ah but it's not an "agreed upon" definition it's simply the most natural extension to non-inertial observers of the Einstein simultaneity convention for inertial observers. The observer riding on the rotating ring sends out a light signal at a given local time, it gets reflected off of a distant event back towards the observer whereupon the observer declares the distant event to be simultaneous with the local time that occurs halfway between emission and reception of the light signal as usual. Notice how the extension of Einstein simultaneity to a non-inertial observer is sensitive to an entire segment of the observer's world line i.e. it isn't determined by just a single event on the observer's world line; furthermore because it depends on light signals it is limited to a subset of space-time because of the way light cones vary along the world line of an observer riding on the rotating ring.

But there's nothing that says this is the "canonical" or "unequivocal" definition of simultaneity to use for non-inertial observers. We can just as well use, at each event on the observer's world line, the simultaneity plane corresponding to the orthogonal projection of the observer's 4-velocity at that event but this will only work locally (i.e. nearby the world line of the observer at each event) because far away from the world line the simultaneity planes associated with different events on the observer's world line will start intersecting one another*. This is why the simultaneity planes (obtained from the orthogonal projection of the observer's 4-velocity) are better thought of as the "local simultaneity planes", "local simultaneity slices", or "local rest spaces" of the observer**. If you want a definition of simultaneity that doesn't result in overlapping simultaneity surfaces for non-inertial observers when straying far away from their world lines then you need to use, for example, the Marzke-Wheeler definition of simultaneity described above.

*http://www.vallis.org/publications/tesidott.pdf (see section 3.2)
**http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf (see p.140 and pp.163-164)

 

[JVNY]

. . . Not with either the "momentarily comoving" simultaneity convention or the Marzke-Wheeler convention, no. But those are not the only possibilities. . . The convention you adopt makes no difference to the results of any experiments; different conventions result in different coordinate charts being used to assign coordinates to events, but the predictions for all observables are the same.

Very useful, thanks. Are there any good references to read for the other conventions, and the predictions and experimental agreements?

 

[JVNY]

. . . But if you are concerned with what observers actually see, then everyone on the platform will see the nearer train win. In order to determine that it was a tie, they have to do something much more complicated such as place previously synchronized clocks adjacent to each track and then separately photograph each train and clock and go back and look at the arrival times in the recordings.

In this case, I am concerned with whether they would agree that it was a tie. Sorry for having used the word "see," and for referring to L-15. I don't think that you would disagree with the answer to the basic question: using methods like the one you describe here, all would agree that there was a tie. For all of the four (three objects and platform), the three train fronts and the finish line were in the same place on the axis of motion at the same time -- as all four define simultaneity.

Yet consider:

(a) a rotating radius just above the three trains and finish line,

(b) that happens to be parallel to the finish line and the fronts of the trains at the same time as the trains reach the finish line (same time as determined by the trains and finish line), and

(c) is rotating at a speed such that the point on the radius above each train has the same instantaneous velocity (direction and magnitude) as the train beneath it, at the same time as the trains reach the finish line (same time as determined by the trains and finish line).

Now, is there any reasonable simultaneity convention under which observers riding on the radius on the three points above the three trains would agree that they simultaneously reached the finish line? It seems not, from the earlier responses in this thread.

I am focusing on this because it will help me understand why simultaneity differs for the radius riders. Each is at the same point on the axis of motion and has the same instantaneous velocity as another observer -- the one on the train below -- for whom all events on the finish line at that instant are simultaneous.

The radius riders appear to be identical in that instant to the trains' fronts. Yet as questioned further above, can they really be identical if they disagree with the trains and platform on simultaneity? Or, is it that they are identical, and it is something that happens later (for example the fact that immediately afterward the radius riders undergo change of direction acceleration) that causes the disagreement over simultaneity? In the Dolby and Gull article, for example, one determines a hypersurface of simultaneity by sending out a signal to an event and receiving the signal back after the event reflects it. This suggests that you cannot determine simultaneity by considering events at just a single moment, and provides the start of an explanation why I am wrong to compare a radius rider with a momentarily comoving train rider to determine simultaneity.

 

[WannabeNewton]

Or, is it that they are identical, and it is something that happens later (for example the fact that immediately afterward the radius riders undergo change of direction acceleration) that causes the disagreement over simultaneity? In the Dolby and Gull article, for example, one determines a hypersurface of simultaneity by sending out a signal to an event and receiving the signal back after the event reflects it.
...
This suggests that you cannot determine simultaneity by considering events at just a single moment, and provides the start of an explanation why I am wrong to compare a radius rider with a momentarily comoving train rider to determine simultaneity.

See post #53.

Are there any good references to read for the other conventions, and the predictions and experimental agreements?

http://arxiv.org/pdf/gr-qc/0311058v4.pdf

 

[JVNY]

See post #53.

Exactly. Post 53 seems to be like the Dolby and Gull approach. Interestingly, it depends on signals going out and then returning. So in that approach simultaneity depends on the "entire segment" of the worldline (as the post states). It does not rely on only one way signals (e.g., only signals coming from the events, like lightning flashes in the Einstein lightning and train example). The other approaches might be different, though; thanks for the reference, and I will read that one for how other approaches apply in these circumstances.

 

[ghwellsjr]

. . . But if you are concerned with what observers actually see, then everyone on the platform will see the nearer train win. In order to determine that it was a tie, they have to do something much more complicated such as place previously synchronized clocks adjacent to each track and then separately photograph each train and clock and go back and look at the arrival times in the recordings.

In this case, I am concerned with whether they would agree that it was a tie. Sorry for having used the word "see," and for referring to L-15. I don't think that you would disagree with the answer to the basic question: using methods like the one you describe here, all would agree that there was a tie. For all of the four (three objects and platform), the three train fronts and the finish line were in the same place on the axis of motion at the same time -- as all four define simultaneity.

I think you still don't understand that the tie issue is not something that is intrinsic to nature. It is something that is determined by human definition (or convention). As long as a definition is consistent with itself (only gives a single answer) and can actually be implemented, then it's not that one definition is more correct or reasonable than another one. In order for the platform observers to agree that it was a tie, they first have to agree what definition of tie they are going to use and then they have to agree on the frame in which that definition is going to be applied. After that, they have to implement whatever it takes to substantiate that definition and then determine whether or not it was a tie. Even with Einstein's convention, there are other frames where the same scenario is not a tie.

Yet consider:

(a) a rotating radius just above the three trains and finish line,

(b) that happens to be parallel to the finish line and the fronts of the trains at the same time as the trains reach the finish line (same time as determined by the trains and finish line), and

(c) is rotating at a speed such that the point on the radius above each train has the same instantaneous velocity (direction and magnitude) as the train beneath it, at the same time as the trains reach the finish line (same time as determined by the trains and finish line).

Now, is there any reasonable simultaneity convention under which observers riding on the radius on the three points above the three trains would agree that they simultaneously reached the finish line? It seems not, from the earlier responses in this thread.

I am focusing on this because it will help me understand why simultaneity differs for the radius riders. Each is at the same point on the axis of motion and has the same instantaneous velocity as another observer -- the one on the train below -- for whom all events on the finish line at that instant are simultaneous.

The radius riders appear to be identical in that instant to the trains' fronts. Yet as questioned further above, can they really be identical if they disagree with the trains and platform on simultaneity? Or, is it that they are identical, and it is something that happens later (for example the fact that immediately afterward the radius riders undergo change of direction acceleration) that causes the disagreement over simultaneity? In the Dolby and Gull article, for example, one determines a hypersurface of simultaneity by sending out a signal to an event and receiving the signal back after the event reflects it. This suggests that you cannot determine simultaneity by considering events at just a single moment, and provides the start of an explanation why I am wrong to compare a radius rider with a momentarily comoving train rider to determine simultaneity.

I'm sorry, I don't understand your new scenario.

I think you need to limit your scenarios and frames to the easily defined and easily described ones before you go on to the more complicated ones. And that would be in-line one-dimensional scenarios and the Inertial Reference Frames (IRF) that you can use the Lorentz Transformation process to get from one IRF to the other. You will be able to see that Dolby and Gulls process is identical for inertial frames as the Lorentz Transformation process. Then you can venture out to non-inertial frames of accelerating observer but all still in-line. After that you might be ready to tackle 2D scenarios. But this 3D scenario is beyond my interest.

 

[WannabeNewton]

Exactly.

Do you recall our discussion about synchronization for a family of uniformly (Born-rigidly) accelerating observers? Remember that the equivalence principle equally well allows us to think of this as a system of observers at rest at different potentials in a uniform gravitational field. The 4-velocity field of this family is then given simply by $\vec{u} = \frac{1}{x}(1,0,0,0)$ where $x$ is the usual Rindler spatial coordinate. $\vec{u}$ is therefore everywhere perpendicular to the $t = \text{const}$ surfaces, where $t$ is the usual Rindler time coordinate. If we now define simultaneity for all observers in this family by $dt = 0$, then the $t = \text{const}$ surfaces constitute global simultaneity surfaces for all of these observers; what we've done here is use the fact that $\vec{u}$ is everywhere perpendicular to the $t = \text{const}$ surfaces to smoothly glue together the "local simultaneity surfaces" talked about in post #53 into a global simultaneity surface for the entire family. We can then synchronize all the clocks carried by these observers by readjusting their rates of ticking so that they all read the time coordinate $t$. Alternatively, one can define simultaneity for each observer by directly applying the Einstein simultaneity convention to all events that can receive light signals from and send light signals to each observer; this is just the Marzke-Wheeler prescription. It just so happens that for uniformly accelerating observers, this prescription gives us back the $t = \text{const}$ surfaces!

But for a family of observers riding on a uniformly rotating disk there are a plethora of problems with regards to the above. We can again consider the observers as being at rest in a certain gravitational field, wherein they have the 4-velocity field $\vec{u} = \gamma (1,0,\omega,0)$ where $\gamma = \frac{1}{\sqrt{1 - \omega^2 r^2}}$ and $\omega$ is the angular velocity of the disk (the coordinates being used are $(t,r,\phi,z)$). The main problem is that there are no level sets of the form $t = \text{const}$ that $\vec{u}$ is everywhere orthogonal to; in other words there is no way for us to smoothly glue together the "local simultaneity surfaces" of each observer riding on the disk into a global simultaneity slice for the entire family. This is the result of an important theorem from differential geometry (Frobenius' theorem) which basically states that if $\vec{\omega} := \vec{\nabla} \times \vec{u}\neq 0$, such a one-parameter family of global simultaneity slices for $\vec{u}$ cannot exist. See post #45. As a consequence, if we now resort to Marzke-Wheeler simultaneity, the observer at the center of the disk has simultaneity surfaces consisting of planes orthogonal to his world line whereas the observers elsewhere on the disk have much more complicated simultaneity surfaces that are in general non-orthogonal to their world lines.

But is there anything that says we can't use the central observer's simultaneity planes to synchronize diametrically opposite clocks placed on the rim of the rotating disk? No.

 

[JVNY]

. . . Even with Einstein's convention, there are other frames where the same scenario is not a tie.

Can you describe an inertial reference frame in which the scenario of the three trains and finish line is not a tie, using Einstein's convention? I agree that there is no point going on with the other examples if I do not understand this, and I appreciate your taking the time to help me with it.

 

[ghwellsjr]

Can you describe an inertial reference frame in which the scenario of the three trains and finish line is not a tie, using Einstein's convention? I agree that there is no point going on with the other examples if I do not understand this, and I appreciate your taking the time to help me with it.

Yes, an IRF moving along the finish line.

 

[PeterDonis]

I think you still don't understand that the tie issue is not something that is intrinsic to nature. It is something that is determined by human definition (or convention).

Exactly.

 

[JVNY]

Exactly.

Agreed. I am coming from having only studied inertial motion, and only knowing the basics of the Einstein convention. So this is an area of not knowing what I don't know, and I appreciate all your patience.

 

[WannabeNewton]

Agreed. I am coming from having only studied inertial motion, and only knowing the basics of the Einstein convention.

If I recall from previous discussions, you have access to MTW right? Read chapter 6 and work through all the problems in said chapter; it's a brilliant chapter.

 

[JVNY]

If I recall from previous discussions, you have access to MTW right? Read chapter 6 and work through all the problems in said chapter; it's a brilliant chapter.

Excellent, thanks.

 

[JVNY]

Yes, an IRF moving along the finish line.

Yes, agreed; I was not thinking broadly enough (I was just focused on frames in inertial motion along the same axis). L-15 may be the wrong citation for what I was proposing. Here is a good excerpt from Spacetime Physics, page 63, on both the line of simultaneity that I described (perpendicular to the direction of relative motion) and the IRF in the motion that you describe (in which the train finishing events are not simultaneous):

Only in the special case of two or more events that occur at the same point (or in a plane perpendicular to the line of relative motion at that point -- see Section 3.6) does simultaneity in the laboratory frame mean simultaneity in the rocket frame. When the events occur at different locations along the direction of relative motion, they cannot be simultaneous in both frames.​

Section 3.6 goes into greater detail, showing that in the plane transverse to relative motion, (a) events simultaneous in one frame are simultaneous in the other, and (b) there is no length contraction; both of these results are required by symmetry (on pages 66-67).

 

[JVNY]

. . . Next, a rim observer can locate the 180 degree point on the opposite end of the diameter as follows. First, place a two-way mirror anywhere on the rim. Simultaneously send co- and counter-rotating signals from the clock. The signals will reflect off of the mirror and return (in opposite rotations) to the clock. Record the order that the signals arrive back at the clock. Repeat the process, moving the mirror until the two signals arrive back at the clock simultaneously. The mirror is now located at the opposite end of the diameter to the clock, half the distance of a one-way trip around the rim. . .

. . . This won't work if the ring is rotating. The degree marker at which a mounted two-way mirror would reflect the counter-propagating light beams such that they arrive back at the clock (placed at the 0 degree marker) simultaneously would not read what we normally think of as 180 degrees i.e. as the degree marker diametrically opposed to the 0 degree marker. If we in fact placed the two-way mirror at what we normally think of as the 180 degree marker then the prograde beam would arrive at the clock before the retrograde beam. Using your method we would in fact place the two-way mirror at a degree marker that is shifted, from what we normally think of as the 180 degree marker, in the direction favoring the retrograde beam in order to compensate for the Sagnac time delay (which is what results in the associated phase shift). So your method would only work if the ring is non-rotating. If you really wanted to place the two-way mirror at the "normal" 180 degree mark then you would need to take the Sagnac time delay into account when recording the arrival times of the counter-propagating signals, upon reflection, as opposed to looking for simultaneous arrival times, upon reflection. . .

I have researched this further and found that the method should properly locate the 180 degree mark. The Sagnac effect depends on the area enclosed by the signal path. Where the area is zero, there is no Sagnac effect. In particular, when a signal travels outbound in one direction around the rotating rim, reflects, and then retraces its path in the opposite direction, the area is zero and the Sagnac effect is zero. Ashby and Allan describe the underlying theory in "Practical implications of relativity for a global coordinate time scale," available at: http://tf.nist.gov/timefreq/general/pdf/133.pdf.

In particular, on page 657 they describe an experiment by Besson in which the portable clock turned around and retraced its path, creating a zero area, and no effect from Earth rotation (see first attachment).


On pages 658-59 they describe an experiment taking a portable clock ("PC") from the US Naval Observatory west to the National Bureau of Standards, then back east again, retracing its path, and note that the Sagnac correction cancels out (-9.6 ns effect westward offset by +9.6 ns effect eastward) (see second attachment, from page 659).

Therefore two signals sent simultaneously in opposite directions from 0 degrees on the rotating disk that reflect off of a mirror at 180 degrees on the disk and retrace their paths back should arrive at 0 degrees simultaneously; by retracing its path, each signal covers a zero area and the Sagnac effects for each from the outbound and retrace legs perfectly offset each other.

 

[yuiop]

This won't work if the ring is rotating. The degree marker at which a mounted two-way mirror would reflect the counter-propagating light beams such that they arrive back at the clock (placed at the 0 degree marker) simultaneously would not read what we normally think of as 180 degrees i.e. as the degree marker diametrically opposed to the 0 degree marker. If we in fact placed the two-way mirror at what we normally think of as the 180 degree marker then the prograde beam would arrive at the clock before the retrograde beam.

I am afraid I have to agree with JVNY here and disagree with your conclusion. If we have a non rotating ring with a mirror fixed at the 180 degree mark, such that light signals sent out in opposite direction from the 0 degree mark, return simultaneously to the the 0 degree mark after reflection, then when the ring is spun up, the signals will return simultaneously to the 0 degree mark, whatever the angular velocity of the ring. Basically, this is due to the fact that it makes no difference whether the light follows the prograde path followed by the retrograde path, or vice versa. The round trip time is the same.

It is however true, that the retrograde beam would arrive at the double sided mirror at 180 degree mark before the prograde beam, but that is a separate issue.

Here is slightly different scenario that might clarify things. Imagine we have a double sided mirror at the 0 degree mark and a double sided source also at the 0 degree mark. If signals are simultaneously sent in opposite directions around the ring and allowed to travel all the way around and reflect off the mirrors at the 0 degree mark they will both arrive back at the 0 degree mark simultaneously, with each signal having completed 2 trips around the ring. One signal will have completed a prograde circumnavigation followed by a retrograde circumnavigation, while the other will done the same but in reverse order, but the total times will be the same.

 

[WannabeNewton]

Therefore two signals sent simultaneously in opposite directions from 0 degrees on the rotating disk that reflect off of a mirror at 180 degrees on the disk and retrace their paths back should arrive at 0 degrees simultaneously; by retracing its path, each signal covers a zero area and the Sagnac effects for each from the outbound and retrace legs perfectly offset each other.

It is however true, that the retrograde beam would arrive at the double sided mirror at 180 degree mark before the prograde beam, but that is a separate issue.

You guys are right, I was confusing the reflection events at the two-way mirror with the entire round-trip circuits.

Sorry JVNY!

 

[JVNY]

No problem, and thanks to both of you for picking this back up.

[Note: I have edited this post from last night to change the discussion of the clock initialization method]. The article also gives a method for initializing the clocks by way of signals around the rim. I used a different method:

. . . Finally, the master clock can synchronize a dependent clock located on the mirror as follows. The master clock sends a signal in the co-rotating direction that instructs the dependent clock to set itself to a specified time equal to the proper time of the master clock when it sends the signal plus half the time that a signal takes to make a complete co-rotating one-way trip (previously determined by the timed one way co-rotating trip at the beginning of this process). . .

The article, although referring to sending signals around earth, says that it is possible to synchronize clocks by a radar method. I believe that it says that the master clock signals the dependent clock to set itself at (a) the master clock's time when it sends the signal, plus (b) half the master clock's proper time that a signal takes to go out to the dependent clock and back again, plus or minus (c) the appropriate amount of the Sagnac effect for the distance from the master to dependent clock (plus or minus depending upon the signal direction). I think that you get the same result for the dependent clock at 180 degrees following the method in the quotation above. In the original of this post from last night I suggested that you needed to use instead half of the average of one complete co-rotating and one complete counter-rotating trip (since using the average of the two cancels out the Sagnac effect, as yuiop points out). But I think in hindsight that using half of the average reintroduces a Sagnac error. Using half of the master clock's proper time that a single complete circuit takes in the same direction should work as long as the signal to the dependent clock is sent in the same direction.

So it is possible to synchronize the clocks using a convention consisting only of signals sent around the rim. There is no need to send signals from the axis to the rim or otherwise have any signal travel off of the rim. The paper analyzes synchronization of clocks around the Earth (not a rim), so I am analogizing to the rim.

Interestingly, the paper states that using this convention on Earth creates an Earth centered inertial coordinate system. So by analogy one should probably conclude that using the convention on the rim creates an axis centered inertial coordinate system -- even though no signal comes from the axis, rather everything takes place on the rim.