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Clarification on synchronous clocks

  1. Oct 20, 2015 #1
    Kind of long winded, but I would appreciate a second veiw on untangling EA's thinking about this.

    § 2. On the Relativity of Lengths and Times

    We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the “time of the stationary system” at the places where they happen to be. These clocks are therefore “synchronous in the stationary system.”

    AE has previously defined the synchronization of clocks in § 1, but just to be clear at this point that I follow his reasoning, let's make sure these conclusions are consistent with his reasoning...

    - synchronous clocks to not need to display the same time indication with their hands "at the same time"; they just need to advance at the same rate, that is say they operate at the same rate when both are at rest relative to each other. I think this is clear from the method used in § 1 where it is only the differences in clock readings that are used to verify their synchrony.

    - from the above, if one desires, one may indeed set all the synchronous clocks to display the "same display indications at the same time" by taking into account c and adjusting them to do so.

    - synchronous clocks means clocks in the same inertial reference frame... all clocks at rest with respect to each other within an inertial reference frame are synchronous.

    If all that is well, it seems clear that the "clocks" placed at the A and B ends of the moving rod are not really clocks. They are indicators that use clock faces to show the time of the moving rod ends in the stationary system. They cannot be clocks in the moving system synchronous with clocks in the stationary system... why are these two rod end "clocks" needed to indicate the stationary time of the moving rod ends? Cannot observers in either the stationary or moving system read the clocks EA has imagined populate the stationary system? If the moving observers at the rod ends cannot read the stationary clocks directly, then what is the mechanism that allows the moving rod end "clock faces" to indicate the stationary clock times at the rod ends?

    In short, it looks like in § 1 synchronization of clocks was demonstrated to require that they be at inertial rest with respect to each other, but the moving rod end clocks are then somehow allowed to "synch" with the stationary clocks in order to be their proxy indicators.

    We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks. Let a ray of light depart from A at the time(*4) tA, let it be reflected at B at the time tB, and reach A again at the time t'A. Taking into consideration the principle of the constancy of the velocity of light we find that

    rB-tA=rAB/(c-v) and t'A-tB=rAB/(c+v)

    where rAB denotes the length of the moving rod — measured in the stationary system. Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous.

    EA has the two moving observers test for synchronization between the ends of the moving rod...

    - surely they are not using the time tA and subsequent times from the stationary system to do this?
    - surely they are not using the c-v and c+v in their calculation?

    It seems to me that when EA says, "Taking into consideration the principle of the constancy of the velocity of light we find that..." that this "we" is meant to be those of us in the stationary system, not the two moving observers. He even says that rAB is the rod length "measured in the stationary system", so this pair of equalities are the stationary measures, not the two observers.

    So I'm thinking that when he writes

    "We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks. Let a ray of light depart..."

    what he means is

    "We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks, to which moving system we will return in a moment. Meanwhile from the standpoint of the stationary system, let a ray of light depart..."

    Assuming all that is well, it looks like what has happened is that the stationary observer has calculated what he thinks the moving observers will find - the moving observers are doing the synch test and the stationary observer is following along from his stationary perspective.

    EA concludes that the stationary observer will find the "clocks" at the moving rod ends to be synchronized, which is to be expected since as proxy indicators of clocks in the stationary system, they must show synchrony to a stationary observer.

    He concludes that the observers moving with the rod will find the clocks are not synchronous... but these observers are at rest with respect to the rod and the "clocks" at the rod ends, so all are at rest with respect to each other. Going back to the understanding of synchrony of clocks, are not both moving rod ends subject to the same rate of time with respect to the moving observers?

    Does EA mean to suggest that "actual moving clocks" at the rod ends in the moving system would not be synchronous with clocks in the stationary system (which seems correct to me)? It would just need to be clear that the "indicator clock faces" and the actual indications of real clocks at the rod ends would differ and not be synchronous. It looks to me that if these moving rod end indicator clock faces showing the time of the stationary system are taken to be real clocks by the moving observers, then yes, the moving observers would conclude that the rod end clocks were not synchronous - because they are using measures from the stationary system!

    It seems to me that if by chance the time reading tA for both the stationary and moving systems just happened to be the same value, and then both the stationary and moving system observers tested for synchrony of real clocks at the moving rod ends, the stationary observers would find that those clocks were not in synchrony, but the moving observers would find that they were in synchrony... as would any test of two real clocks at the ends of a rigid rod with respect to which the observers were at inertial rest.

    Any learned comments on whether I am understanding EA's reasoning here is much appreciated.
    Last edited: Oct 20, 2015
  2. jcsd
  3. Oct 20, 2015 #2


    Staff: Mentor

    No. Synchronized clocks do need to display the same time indication.
  4. Oct 20, 2015 #3


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    Clocks which consistently tick at the same rate but which may not be set to show identical times are sometimes referred to as "syntonous".
  5. Oct 20, 2015 #4
    EA writes that
    "In accordance with definition the two clocks synchronize if tB-tA=t'A-tB"

    That equality looks like it is satisfied by clocks that advance at the same rate when at rest with respect to each other no matter what the actual hand positions at any moment. Any difference in time indication on their respective faces will remain constant.

    This reminds me of the proper way to use a graduated meter stick - one lays it on the two points positioning the stick randomly and taking the difference between the two readings... one does not align the zero end mark on one point and record the single reading of the other.

    EA's meaning of synchronization is important... can we just start with figuring out whether he meant that the clocks just need to share a common rate of time or if he also meant that they must also share a common time indication for any particular time within the shared common time rate?
  6. Oct 20, 2015 #5


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    Staff: Mentor

    That works when you are measuring the distance interval between two points, but not when you are assigning coordinates to points. The distance between two points is equal to the difference between the coordinates (this is the ##\Delta{x}## that you'll find in the formula for the spacetime interval ##\Delta{s}^2=\Delta{x}^2-\Delta{t}^2##) and then any constant offset in the position of the meter stick cancels out.

    You'll also notice that it only works when you're lining two points up against one meter stick; you are measuring the position of two different points in spacetime at the same time ("at the same time" in a frame in which the meter stick is at rest). The analogy with a clock is measuring the time interval between two events; you're measuring the time distance between two points in spacetime at the same place ("at the same place" in a frame in which the clock is at rest).
  7. Oct 20, 2015 #6
    Nugatory, I think I agree with all you wrote to the extent that I think I understand it...

    If so, my sense is that you would agree that any constant offset in the indicated times of the clocks cancels out during EA's convention from section one and synchronous clocks do not have to indicate zero mutual offset in a frame within which they are at rest...?

    I'm finding slightly different conceptual definitions...

    synchronize is defined as indicating the same time and happening at the same rate and exactly together:

    synchronous is less restrictively occurring at the same time, going on at the same rate and exactly together, and in Physics, Electricity - having the same frequency and zero phase difference.

    That last about zero phase difference looks like the zero offset requirement of clock indications but may be applicable in this question?

    Maybe there is some loss in the translation, but EA's convention in part one does not seem to require the clock time indications to be the same, only their rates.

    I have looked for other presentations of this to compare but all the ones I have found seem to skip these preliminary details and jump quickly into the Lorentz equations as the foundation for subsequent analysis.
  8. Oct 20, 2015 #7
    Just to be clear, EA means Elbert Ainstein?
    The clock need to "show the same time" to be Einstein-synchronized. Imagine a rod 1 light hour long, and ##t_A=##0:00. If ##t_B##=5:00 and ##t'_A##=2:00, then 5:00-0:00##\neq##2:00-5:00. This means, ##t_B## has to be 1:00.
  9. Oct 20, 2015 #8


    Staff: Mentor

    As @SlowThinker just mentioned, this is not correct. If you work it out you will see that it requires synchronized hand positions.
  10. Oct 20, 2015 #9
    Thanks SlowThinker

    That helps because it clarifies that the tB - tA = t'A - tB looks like it does test for instantaneous synchrony at B's time tB... but it also is helping me understand my troubles with this. The convention does not seem to test for inertial resting synchrony unless repeated. It looks like the time of that instantaneous test for "same time indication" is tB, but B could be moving and the clock at B could be accidentally indicating the expected time for synchrony even while running at any arbitrary rate when the reflection happens.

    What is bothering me is that EA suggests "We assume that this definition of synchronism is free from contradictions..." but the procedure to determine if two clocks are synchronized (show "the same time") is only exclusive for a particular instant (does not test for "at the same rate"), so it may be without contradiction at the instant of test, but it seems incomplete as far as establishing "the "common "time" for A and B", which includes and "same rate" and inertial rest.

    Since this is a test it is not assumed that the clock at B is yet in the same inertial frame. EA writes "We have so far defined only an “A time” and a “B time.” We have not defined a common “time” for A and B, for the latter cannot be defined at all unless we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A."

    But by the second postulate the " “time” required by light to travel from A to B equals the “time” it requires to travel from B to A even when the reflective source B is in relative motion with respect to A..

    I am not getting the feeling that EA is fully demonstrating what he intends to represent here because the convention does not seem sufficient to exclude false positives - B might be in relative motion and the reflection might just happen to occur "at the right place and time" so as to mimic implied resting clock synchrony in spite of having a different time rate.
  11. Oct 20, 2015 #10


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    Staff: Mentor

    I don't agree. The only way you can get cancellation of the offset (see SlowThinker's post above) is when you are subtracting readings from the same clock, just as your distance measurement starting away from the end of the meter stick requires that both readings be taken from the same meter stick.

    Dictionaries are good for non-technical language usage, but you're putting too much weight on definitions that weren't intended to stand up in a serious discussions of relativity. In those discussions synchronized clocks aren't offset from one another, for about the same reason that all meter sticks start at zero centimeters and end at one hundred centimeters - you'll never find us using one that starts at 19 cm and ends at 119 cm.
  12. Oct 21, 2015 #11


    Staff: Mentor

    I agree. I think that is indeed a problem with this specific formulation. I think that it is better formulated elsewhere, by other authors and even elsewhere by Einstein.

    Don't be too critical of the awkward wording of this first paper. He was breaking new ground here, and it is unreasonable to expect him to get everything perfect the first time and on his own.

    This is also the reason why learning from a good modern textbook is preferable to learning from the seminal papers.
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