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Synchronizing rotating clocks

  1. Jan 15, 2014 #1
    Here for your consideration is a try at synchronizing clocks on a rim rotating at a constant angular velocity. It is restricted to two clocks, each at the opposite end of a diameter of the rim.

    It is based on simultaneity as determined in SR. Taylor and Wheeler in Spacetime Physics, page 120, referring to two inertial frames (rocket and lab), state regarding simultaneity:

    At any instant there is just one plane in which both the laboratory and the rocket clocks agree . . . [T]his plane lies perpendicular to the direction of relative motion.​

    If at that instant one sets the two clocks to the same time, they will agree that the setting is simultaneous, and the clocks will be for that instant synchronized. The first diagram attached shows the world lines of two objects in inertial motion and the instant that they agree on simultaneity (at the dashed line).

    Clocks at opposite ends of a diameter have instantaneous velocity tangent to the rim, so they have instantaneous velocity in opposite directions, perpendicular to the diameter. See a diagram from wikipedia at http://en.wikipedia.org/wiki/File:Uniform_circular_motion.svg or previously posted here at post 3, https://www.physicsforums.com/showthread.php?t=730574#post4616303.

    Therefore, at any instant the two clocks agree on simultaneity of events that occur along the line perpendicular to their relative motion. That is, they agree on the simultaneity of events that occur along the diameter. Therefore one can synchronize them using synchronized clocks that are at rest underneath them (say in the lab frame in which the rim is rotating). As the rotating clocks pass over the synchronized lab clocks, the rotating clocks are set to the time shown on the lab clocks.

    Subsequently, the rim clocks travel at all times with velocity perpendicular to the diameter. So they continue to agree on simultaneity. So they remain synchronized.

    For a more detailed discussion of the relationship between simultaneity and synchronization, see: https://www.physicsforums.com/showthread.php?t=731437

    For an earlier discussion of the proposed synchronization method, see:
    https://www.physicsforums.com/showthread.php?t=730574
     

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  2. jcsd
  3. Jan 15, 2014 #2

    WannabeNewton

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    Let me make two clarifying remarks:

    (1) The proposed method of synchronization in the thread you linked is in general invalid. If we have an observer in arbitrary motion then the observer can determine, at an instant of his proper time, his simultaneity surface by using light signals and the Einstein simultaneity convention. In general this will only agree with the orthogonal complement of the observer's 4-velocity on the order of the observer's proper acceleration (in other words on the order of the path curvature of the observer's world line) and the torsion of the world line. So only locally will the simultaneity surface agree with the orthogonal complement of the observer's 4-velocity; globally they can easily differ. In SR there are only two special cases wherein the two will agree globally: that of inertial observers and that of Rindler observers. In order to see what the simultaneity surfaces look like for uniformly rotating observers (or clocks) see section 3.3 of the following paper, and page 31 in particular: http://www.vallis.org/publications/tesidott.pdf

    (2) The term "synchronization" is being used in an overly liberal fashion. The act of synchronizing clocks requires first a synchronization procedure to be defined. The synchronization procedure has to satisfy some basic properties in order to be consistent. Apart from the requirement that two initially synchronized clocks have to remain synchronized, the synchronization procedure has to also be transitive; it isn't enough to simply consider two very special comoving clocks from an entire family of comoving clocks. If we have a rotating ring and clocks placed along the ring, the synchronization has to be transitive across the entire family of clocks along the rotating ring; clearly an attempt at employing Einstein synchronization by means of light signals, or an attempt at employing slow clock transport, to the family of clocks along the rotating ring will fail to establish transitivity.
     
  4. Jan 15, 2014 #3

    DrGreg

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    To JVNY

    Yes, your method works for two clocks diametrically opposite each other. It won't work for more than two rim clocks, assuming your definition of simultaneity coincides with that of a comoving inertial observer. (For non-inertial observers, other definitions are possible.)

    To be more specific, with many clocks around the rim, your method would synchronise each clock to its diametrically opposite clock, but would fail to synchronise each clock to its neighbour along the rim. (For "comoving inertial synchronisation".)
     
  5. Jan 15, 2014 #4

    WannabeNewton

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    Correct me if I'm wrong but I don't see how. Given two observers ##O## and ##O'## diametrically opposite on the rim, the method assumes that the simultaneity surfaces for ##O## and ##O'## are both planes orthogonal to the respective 4-velocities of ##O## and ##O'##; clearly two such planes will intersect along the diameter of the disk. But these are the simultaneity surfaces of inertial observers momentarily comoving with ##O## and ##O'## respectively, they are not the simultaneity surfaces of ##O## and ##O'## themselves. The latter have more complicated surfaces of simultaneity (depicted in the paper I linked above).
     
  6. Jan 15, 2014 #5

    DrGreg

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    I think you misunderstand what I'm saying. Following JVNY's method you can sync one clock to one other clock (no more, no less), compatibly with the co-moving inertial observers' definitions of simultaneity. (The Märzke-Wheeler coordinates in the paper that you refer to aren't relevant, as we're attempting to construct a different definition of simultaneity.)

    But the rest of my post says that such a method doesn't generate a consistent sync across all clocks on the rim. Or to be more precise, not one that is everywhere locally compatible with the co-moving inertial observer's definition of simultaneity. I don't believe I am contradicting anything you've said.
     
  7. Jan 15, 2014 #6

    ghwellsjr

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    Maybe I'm totally misunderstanding the problem or the proposed solution but it seems to me that in the rest frame of the axle of the rotating rim (if there isn't one, we'll pretend like there is one), that all the clocks on the rim are traveling at the same speed and therefore will be Time Diltated by the same amount and therefore if we send a signal from the axle to all the clocks for them to reset to a given time, then they will forever stay in sync.

    I got the impression that the proposed solution claimed that clocks along the diameter would also be in sync but they will be traveling at a different speed and Time Dilated to a different extent so I don't see how that is possible.

    I know you guys know way more than I, so I'm sure I must have misinterpreted something somewhere.
     
  8. Jan 15, 2014 #7

    WannabeNewton

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    Ah ok, sorry that wasn't clear to me from the wording of the OP. I thought it was talking about the simultaneity surfaces of the diametrically opposed observers sitting on the rim as opposed to the continuous sequence of simultaneity surfaces obtained from the continuous sequence of diametrically opposed momentarily comoving inertial observers.
     
  9. Jan 15, 2014 #8

    WannabeNewton

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    If I understood correctly, the proposed method simply says the following:

    Take clock A and clock B on the rim of the disk such that A and B are diametrically opposed. Consider an event p in the vicinity of clock A and an observer O momentarily comoving with clock A at p. Then the plane perpendicular to the direction of motion of clock B relative to O at this instant is the plane cutting across the disk diameter joining A and B at this instant. All that we care about now is the fact that relative to O, clock A and clock B will unequivocally read the same time at this instant because this plane intersects the world lines of both clock A and clock B (c.f. exercise L-15 in Taylor and Wheeler, quoted by the OP in post #1). Therefore clock B is synchronized with clock A relative to O at p. This will clearly hold for all later events in the vicinity of clock A relative to the respective inertial observers momentarily comoving with clock A, so clock B is trivially synchronized with clock A for all events in the vicinity of clock A. Here I believe we are implicitly making use of the fact that clock A and clock B tick at the same rate relative to the central clock (if they ticked at different rates then simultaneity obviously does not in and of itself guarantee synchrony).

    EDIT: I agree, at face value, that this method works. I have yet to check if there are any subtleties that "defeat" said method.
     
    Last edited: Jan 15, 2014
  10. Jan 15, 2014 #9
    The suggestion is restricted to clocks on opposite ends of a diameter. It might be that those can be set to the same time and stay synchronized with each other, yet not with other clocks on the rim. For example, consider a row of clocks oriented north/south and traveling east in a lab frame, and a second row of clocks oriented east/west and traveling north in the lab frame. You can synchronize each row of clocks using simultaneous signals from the lab (because each row is aligned perpendicular to its direction of motion in the lab frame). The clocks in each row will be synchronized in their own frame and in the lab frame. But neither row will present synchronized to the other. One end of each might even coincide in space (the two rows presenting momentarily as an "L" in the lab frame). But the fact that the two ends share the same point and momentarily agree on simultaneity does not transitively cause all of the other other clocks on the two rows to agree on simultaneity.

    Also, I do not mean for the opposite clocks to travel at different speeds. They will both have the same speed and the same time dilation relative to the lab frame. However, I do mean to suggest that the clocks will agree on simultaneity when they are set by the lab signals and also thereafter. To anthropomorphize the clocks, two human observers riding the disk would agree that their wristwatches remain synchronized.

    The basic suggestion from the original post is that the same rationale that permits setting the clocks to the same time at inception (that they momentarily agree on simultaneity because they have instantaneous velocity in opposite directions, per the Taylor and Wheeler quotation) keeps them synchronized into the future. They never cease to have instantaneous velocity in opposite directions transverse to the diameter, so they never cease to agree on simultaneity, so they never cease to be synchronized.

    Here is a quotation from another thread on the relationship between simultaneity and synchronization. It may be out of context, so any error in applying it here is my own and not the author's, but it states the argument better than I can:

    To me, it seems given a definition of simultaneity, the corresponding notion of clock synchronization is a consequence: synchronized clocks must read the same time for a pair of events that are simultaneous.

    Equally reasonable to me is the reverse framework: given a defined clock synchronization, we define events to be simultaneous per that synchronization if synchronized clocks read the same time for them.

    I don't see any reason to favor one direction of definition over another. I also don't think you can plausibly define both independently - either defines the other.

    https://www.physicsforums.com/showthread.php?t=731437#post4621662
    I cannot pretend to understand fully the linked article. However, it does cause me to move now to a second step of the analysis, one that I would also like to restrict to clocks at opposite ends of a diameter. Typical uses of light for measurement utilize round trip signals -- from the emitter to a reflector then back to the emitter. The article discusses this. But the Sagnac effect occurs because of one-way signaling -- from the emitter co-rotating, or from the emitter counter-rotating. These one-way trips result in the measurement 1+v or 1-v. It might be useful instead to utilize two-way signal trips with a reflection. A zero area Sagnac interferometer eliminates the 1+v and 1-v effect, and a two-way trip with reflection might have a similar (if not the same) benefit.

    Consider a rotating rim. On an arbitrary point marked zero degrees is an emitting, receiving, reflecting, and recording master clock. This master clock sends a co-rotating signal, receives the signal after its one-way trip around the rim, and records the time for the trip. The clock then sends a counter-rotating signal, receives the signal, and records the time for that one-way trip. The clock can also send a co-rotating signal and reflect it when it arrives back at zero degrees, sending it on a return counter-rotating trip, receive it upon its arrival, and record the total time for the two-way trip. This will equal the sum of the two one-way trips described above. This is a round trip.

    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.

    Next, the clock sends a signal in either direction and records the time that the signal takes to go out, reflect off of the mirror at 180 degrees, and return. That should be simply half the time of a round trip around the rim and back again. This is the case for either direction of signaling. This is just a sanity check.

    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).

    This seems a plausible way to set the dependent clock. It does not suggest any reason why the dependent clock would fall out of synchronization with further rotation.

    Like Einstein synchronization, one should start with two-way reflected signals. This avoids the problem of the 1+v and 1-v effect. It is also independent of whether the rim's circumference is less than, greater than, or equal to 2πr. It only relies on the equality of the total time for a signal's reflected round trip travel, whether beginning co- or counter-rotating, and on first determining the time that a signal takes to complete a one-way trip around the rim so that the master clock can know that the same signal will take half that time to get to the dependent clock.

    But then, why can't one use a similar reflection method with other clocks around the rim? Say put another mirror on the rim and move it around until a signal sent to it takes one quarter of a round trip time to return; then the mirror is at 90 degrees; and so on. So WBN is right to continue to think about whether there is a problem with the suggestion.
     
  11. Jan 16, 2014 #10

    ghwellsjr

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    Why, and how, are you sending the light signal around the rim? Do you have a light tunnel? And why are you doing it that way in any case? Why not just send the signal directly to the other side? If you're worried about the mirror reflecting the signal back to the observer that sent the signal, you can always use a convex mirror that reflects the light back in all (or almost all) directions.

    Then you might consider what happens if the rim is rotating so fast that the observer has moved a significant fraction of the circumference compared to the linear distance across to the other side. You can determine the distance to the point of reflection by taking one half of the difference between the sent and received signals. Try this for other points on the rim, not just on the opposite side.

    Finally, can you explain what this means:

    I don't see how any of this is possible since in the rest frame of the lab, the lab clocks are not Time Dilated, whereas the clocks on the rim are Time Dilated and any clocks along a radius are Time Dilated to a different extent.
     
  12. Jan 16, 2014 #11

    pervect

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    I'm not sure what JVNY is trying to say, since he tends to omit what frame the clocks are synchronized in . (This is often a symptom of a belief that the frame doesn't matter, when of course according to relativity every inertial frame has a different notion of how to synchronize clocks fairly - which is the whole point.)
     
  13. Jan 16, 2014 #12
    Sorry, assume that the inside surface of the rim is mirrored. The light flash moves along the mirrored rim. This is a curved example of what West calls a "floor-mirrored Einstein-Langevin light clock," described here: http://phys.org/news102850833.html. In fact I think that West uses the mirrored rim elsewhere, but I have lost the citation.

    I use this method so that the light always travels along the circumference. This makes the method comparable to light flashes sent along a straight line in non-rotating scenarios, where there is no other "across" to send a signal. It is true that you can send a signal across rather than along the rim, and you can even do that using mirrors aligned such that there is no interference from signals sent out, reflected, and returned back (zero area Sagnac arrangements, per http://en.wikipedia.org/wiki/Common_path_interferometer). But I am trying to keep the example as close to straight line examples as possible, so that I can confront directly the problem that people claim that the Sagnac effect causes -- the problem that light signals take different amounts of time to go around the rim co-rotating versus counter-rotating. I don't think that this is a problem, as described above, as long as you use a mirror and measure two-way reflected signal paths.
     
  14. Jan 16, 2014 #13
    [sorry, the system does not copy the full post over]

    Think of a row of observers sitting on a radius, and then think of a row of Rindler observers. The radius observer at the rim end of the radius is equivalent to the Rindler observer at the rear end of the Rindler row. Their clocks run the slowest relative to their rows. The farther forward you move on the Rindler row, and the farther inward on the radius toward the hub that you move, the faster the clocks run. The Rindler observers agree that they are at rest with respect to each other on a line of simultaneity (that is, they agree that they are not moving with respect to each other, and they agree on the simultaneity of events that occur along their line of motion relative to an inertial frame).

    Might not the radius observers also agree that they are at rest with respect to each other on a line of simultaneity? They have, at all times and in every moment, instantaneous velocity in the same direction. That line is not limited to the radius, though. As described in the original post, the simultaneity also applies to objects in motion in opposite directions perpendicular to the line, so the line of simultaneity extends the whole diameter.

    Next, consider the two ends of the diameter. They have the same distance from the hub, so they should have the same time dilation in the lab frame. As described above, per problem L-15 from Spacetime Physics they agree on simultaneity momentarily. By the same rationale as that problem, they should agree on simultaneity at all other times, so shouldn't their clocks be able to be set to the same time, then stay synchronized?
     
  15. Jan 16, 2014 #14
    How one applies the term "frame" to a rotating disk is highly controversial. To put it in that term, however, the suggestion is that the two clocks at opposite ends of a diameter are in a reference frame consisting of those two objects. One can set those clocks to read the same clock time simultaneously in their own reference frame by using light signals from the ground. And the clocks will stay synchronized in the two clock reference frame.

    Consider a different example, two clocks in a row in inertial longitudinal motion relative to another inertial frame (the lab frame). You can use simultaneous signals in the lab frame to set the two clocks to the same time in their own frame (by signaling the rear clock to set itself to a time ahead of the time you signal the front clock to set itself to, by the product of the clocks' proper distance and their velocity relative to the lab). Thereafter, the clocks will be synchronized in their own frame.

    I think that the critical issue for the suggestion is in WannabeNewton's first point in post 2. It may be that the momentary position only creates a line of simultaneity for observers who stay in the same frame, not observers who change direction. Although one can use a momentarily comoving inertial frame to help analyze acceleration in some circumstances (like relative clock rates), perhaps one cannot use MCIFs in other circumstances (like simultaneity of events at a distance).

    This seems to be the common theme of the vallis.org piece linked above, the Dolby and Gull article on radar time (at http://arxiv.org/abs/gr-qc/0104077), and WannabeNewton's explanations in this thread.

    Put in crude terms, it is kind of like the question "if a tree falls in a forest and no one is around to hear it, does it make a sound?" Put in the current context, the question is:

    if you are in an inertial frame, and

    if two distant events happen in that inertial frame that are simultaneous in that frame (as confirmed by observers in the frame who are close to the events), and

    if you accelerate out of the frame after the events occur there (as confirmed by the same observers, who later receive light reflected off of your watch), but

    before you observe the two events (before any light that reflected off of the events reaches you),

    were the events simultaneous for you?​

    The simplistic view is that the events were simultaneous for you. But I think that Vallisneri, Dolby, Gull and WannabeNewton would say no, the events were not simultaneous for you. It is not enough to be in the same frame at the same frame time that the events occurred in the frame. You have to stay in the frame long enough to observe the events yourself.
     
  16. Jan 16, 2014 #15

    WannabeNewton

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    At face value I don't think that's an issue as long as you make sure to adjust the clock rates (relative to the central clock) of the clocks closer to the central clock so that they all match the clock rate of the clocks on the rim (meaning we're dealing with non-ideal clocks again if you want synchronization, just like in the thread with the clocks comoving with the accelerating rod).

    What's unconvincing to me however is the use of simultaneity to argue synchrony. For example, say we have clocks A and B diametrically opposed on the rim. At some event p in the vicinity of clock A we consider an inertial observer O momentarily comoving with clock A. The observer knows there exists an event q in the vicinity of clock B that is simultaneous with p because the plane perpendicular to the line of relative motion of B intersects the world line of B as well as the world line of A at p. Now it's clear from a diagram of the world lines of clocks A and B that this plane varies smoothly along said world lines so that's not a problem. However you then say that because q is simultaneous with p according to O, clock B can be synchronized to clock A at this instant if it already isn't. Furthermore since this simultaneity of events between clocks A and B exists everywhere along the world line of clock A, relative to observers momentarily comoving with clock A, you then say that clocks A and B are forever synchronized. Do you see why I'm bothered by this line of reasoning? If, as you argue, simultaneity ##\Rightarrow ## synchrony then why would we even have had to synchronize clocks A and B when using the momentarily comoving inertial observer O at event p that very first time? Why wouldn't simultaneity have immediately implied synchrony then? Why the need to synchronize those clocks that very first time "if they happened to not be synchronized at the simultaneous events p and q in the vicinities of clocks A and B respectively"?

    Also I'm having a hard time seeing the relevance of the lab clocks that you're using (the clocks at rest in the global inertial frame centered on the axle that the disk is spinning about). Why are we using lab clocks to synchronize? The simultaneity (and hence as per your argument synchrony) is determined relative to inertial observers/inertial clocks momentarily comoving with clock A is it not?

    As a side note, I should really read the stipulations of exercise L-15 in Taylor and Wheeler more carefully :wink:
     
  17. Jan 16, 2014 #16
    The lab clocks are located underneath the rotating clocks when the send the signal to set the rotating clocks to a specified time, so they are momentarily in the same line as the diameter. They are just a simple way of synchronizing if you believe that anything momentarily on that line, anywhere along that line, agrees on simultaneity with everyone else along the line.
     
  18. Jan 16, 2014 #17
    Yes -- after reading the Vallisneri piece, rereading Dolby and Gull from an earlier reference, and rereading the other posts in this thread, I think that I see your point, as summarized at the end of post 9 (fully retracting the suggestion from the original post). Does that final summary capture your view?
     
  19. Jan 16, 2014 #18

    WannabeNewton

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    My trouble with this has already been elucidated in post #15 so I'll just skip over this.

    Just as a remark on terminology, usually "one-way" refers to a light beam traveling from an emitter to a receiver whereas "two-way" refers to a light beam traveling from an emitter, doing whatever it needs to do, and coming back to the emitter. The Sagnac effects measures (via an interferometer) the phase shift between two counter-propagating light beams emitted from the same light source mounted on a rotating ring after the light beams complete a full circuit and come back to the light source so this would fall into the latter category.

    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.

    It doesn't but it's an inconsistent synchronization method because the synchronization is path dependent. You chose to use prograde signals for the synchronization but there's no physical reason one can't use retrograde light signals as well. Yet the two will obviously result in two different synchronization prescriptions. This is again because the world lines of the clocks mounted around the ring twist around one another. In fluid mechanics one calls this the vorticity of the flow lines of a fluid; here we are conceiving the world lines of said clocks as flow lines and the entire congruence as a space-time fluid. The point is that the non-vanishing vorticity leads to path dependence of exactly the global synchronization method you described (what makes the prograde signal any more special than the retrograde signal?).
     
  20. Jan 16, 2014 #19
    WannabeNewton, hold on -- you are responding to suggestions that I have already retracted. Can you have a look at post 14 starting with the sentence "I think that the critical issue for the suggestion is in WannabeNewton's first point in post 2." I think that I am in agreement with you on simultaneity starting there, and I try to summarize the reasons there. Have I summarized the reasons correctly? Please ignore the suggested radar method for locating the mirror -- it becomes entirely irrelevant.
     
  21. Jan 16, 2014 #20

    WannabeNewton

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    Alright let me try to get the dust to settle on the ground because I don't want to further confuse you or myself. So what I was saying in post #2 is that the simultaneity surfaces of observers riding on the ring don't agree exactly with the simultaneity surfaces of inertial observers momentarily comoving with observers riding on the ring. I'm pretty sure TSny was aware of this as well because he made sure to highlight the fact that at any given instant of time on a clock mounted on the ring, the simultaneity surface we are going to use in our "definition of synchronization" is that associated with an observer momentarily comoving with this clock at this instant. If we do this we find that there is always a unique event in the vicinity of the clock on the opposite side of the diameter simultaneous with the clock we're considering. TSny explicitly stated that this notion of simultaneity is that of the momentarily comoving inertial observer and not necessarily that of an observer riding on the ring itself (because the two differ globally).

    I don't have any problems with this; as far as simultaneity goes this is all sound to me. My only confusion arises from the act of then extrapolating from the fact that given any event in the vicinity of one clock on the ring, there is a unique event in the vicinity of the diametrically opposite clock simultaneous with this one relative to an inertial observer momentarily comoving with this clock, to then establish our "definition of synchronization". DrGreg called this definition "comoving inertial synchronisation" in post #3.

    My confusion is regarding the operational aspect of "comoving inertial synchronisation". In the original thread wherein TSny laid out this method, he said "Thus at the instant the clock at A reads 1.0 s we image an inertial frame that is co-moving with A at that instant. Then, according to this co-moving inertial frame, the diametrically opposite clock B will read 1.0 s simultaneously with clock A reading 1.0 s.". According to TSny, the clocks are already synchronized (relative to the momentarily comoving inertial observer). If so, what's there to actually synchronize?

    Alternatively, let's assume that the clocks aren't synchronized at these simultaneous events and that we then synchronize them so that they agree in their readings at these simultaneous events. We must then show that the clocks remain synchronized permanently, as per the "comoving inertial synchronization". Your argument for this is that since this simultaneity of events between clocks A and B exists everywhere along the world line of clock A, relative to observers momentarily comoving with clock A, clocks A and B are forever synchronized. But if simultaneity immediately implies synchrony, as this argument claims, then there would have been no reason for clocks A and B needing synchronization to begin with in the manner described at the start of this paragraph. But then we're left with the question at the end of the previous paragraph.

    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.

    Is the source of my confusion apparent now?

    My apologies, I totally missed your later statements.
     
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