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I Relativity of Time for Two Observers

  1. Jun 13, 2018 #1
    I have the following dilemma and ask for advice:

    Observer A is swings on the springs, while the observer B is stationary nearby. (See diagram: http://www.frozman.si/slike/Vzmet.jpg )

    According to the theory of relativity to observer A the time flows slower due to its speed (as in the paradox of twins), as the time flows to observer B. But this is just a half of the story.

    At each oscillation, the observers touch. When they touched, by the theory of relativity t is time of one observer, and at the other one is t'.

    Two different times in the theory of relativity can be described by two events, but they can be also only one event, depending on the circumstances of the observation. When observers observe the same event of their touch at the point of contact, two different times mean two touches, according to the theory of relativity.

    Only one touch on one period (as it really depends), in this case in accordance with the theory of relativity, occurs only if t = t '. This means that times are running synchronous.

    In the theory of relativity we can prove both: that the times of the observers runs differently and also that times are running the same speed.

    FR
     
  2. jcsd
  3. Jun 13, 2018 #2

    Doc Al

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    Why would you think that the touching of A and B is anything but a single event? Of course, the time between such events will be measured differently by A and by B. So what?
     
  4. Jun 13, 2018 #3

    russ_watters

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    This isn't correct(Relativity doesn't say that). Everyone gets to keep their own clock and record their own time for an event.
     
  5. Jun 13, 2018 #4

    Dale

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    This is incorrect. A single event in one frame is also a single event in all other frames. Different frames assign different coordinates to a given event, but they do not ever change one event into two events. At each oscillation there is a single event of them touching, and both frames agree on that fact.

    Note, sometimes in discussing the twin paradox people will incorrectly draw a coordinate system that assigns multiple coordinates to the same event, but that is flat out wrong. Even a non-inertial coordinate system, in order to be valid, must assign exactly one set of coordinates to a single event.

    You do need to take care with non-inertial observers. The coordinates for non-inertial observers like A are not simple Lorentz transforms of the coordinates for inertial observers like B. Your reasoning doesn't necessarily hold, and in fact would need to be considerably refined (by fully specifying what is meant by A's frame) even to make a firm conclusion. I can think of a non-inertial coordinate system for A where t=t' and I can also think of a non-inertial coordinate system where they are unequal.
     
  6. Jun 13, 2018 #5
    I chose such a frames where two different times, even by the theory of relativity, mean two touches of observers in one cycle. It is not general, but is valid in the chosen case. The observers touches only once. It means that different times do not describe real events.
     
  7. Jun 13, 2018 #6

    Dale

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    Such a frame does not exist. If you believe it does, then I challenge you to explicitly write the coordinate transformation between this frame and B's frame.
     
  8. Jun 13, 2018 #7
    Such frames can be found where the touch occurs at the same location and at this location are also the both observers. At this moment of touch they do not move one to another. Consequently there’s no need to make transformation between frames. The both frames are same.
     
  9. Jun 13, 2018 #8

    Grinkle

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    I suspect you are still mentally utilizing an absolute preferred frame against which all moving observers must reconcile. No such frame exists. Different amounts of time elapse between the same events for differently moving observers.
     
  10. Jun 13, 2018 #9

    Grinkle

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    Sure, the guy on the spring has zero velocity relative to the guy watching him when they touch. But they will measure different amounts of time between touches.
     
  11. Jun 13, 2018 #10

    Dale

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    Then the touch is clearly not one event in B's frame and two events in A's frame. Both frames agree that it is one event. So your previous comments are wrong.

    Again, a single touch is one event in ALL reference frames, as required by the math of relativity. Also, this frame that you are describing is not A's frame. It is called the momentarily co-moving inertial frame (MCIF) at the event of the touch. The MCIF is a very useful frame, but it is not the same as the observer's frame unless the observer is inertial, which A is not.
     
    Last edited: Jun 13, 2018
  12. Jun 13, 2018 #11

    A.T.

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    No. The proper times of some clocks can be different at the same event. That doesn't make it two events.
     
  13. Jun 13, 2018 #12
    At the touch of observers, the observers are in the same frame. So they have the same clock. The touching event of one observer can be written as (0, 0, 0, t) and the other as (0, 0, 0, t'). Because the observers are looking at the same clock, the two mentioned events can be described only by two touches. Alert me where I'm making a mistake.
     
  14. Jun 13, 2018 #13

    Doc Al

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    There's one mistake.
     
  15. Jun 13, 2018 #14

    Nugatory

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    Not so. Both of their clocks read the same at this particular moment, but all that means is that at some time in the past one of the observers set the hands of their clock so that it would come out that way. If the clocks are in motion relative to one another, this is the only moment at which they will both display the same time; at all other times they will be out of sync.
     
  16. Jun 13, 2018 #15

    Dale

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    Please read my comments above regarding the MCIF. The MCIF is not the same as A's frame since A is non-inertial.

    Also, the statement "they have the same clock" directly contradicts the statement "event of one observer can be written as (0, 0, 0, t) and the other as (0, 0, 0, t')" unless t=t' for all t, but since you haven't shown any transform that cannot be assured. Again, please produce this imagined transform.
     
    Last edited: Jun 13, 2018
  17. Jun 13, 2018 #16
    At the touching stage, the observers does not move. Therefore both are in the same frame and do not need any transformation.

    You said that observer A notice the event (0, 0, 0, t'), and the observer B (0, 0, 0, t). They notice each its own event. In the nature, both events happen together in just one event for both, not to each one. In the same frame, a single event can have only one record (0, 0, 0, ?), not two. What is this record of a single event for both observers?
     
  18. Jun 13, 2018 #17

    Doc Al

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    By your logic, if a twin takes off on some round trip and then returns to earth, his clock must match his brother's since they are no longer moving with respect to each other. Really?
     
  19. Jun 13, 2018 #18

    Grinkle

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    You are missing the point that the duration between events will be different for observers in motion with respect to each other.

    So - start the scenario with the string stretched, observer B strapped in, at rest with respect to observer A. They touch hands, and the spring is released. At the instant of release, their watches are synchronized. The next time they are at rest with respect to each other, B's watch will not have moved as much as A's watch. Still, the next time they are rest with respect to each other is a single event.

    Observer A says it took some time for B to come back, say 30 seconds for example, and observer B says for example that it took 27 seconds. They both agree that a single event happened when their hands re-touched, they just don't agree how many seconds passed from the point of spring release to the time of re-touching.
     
  20. Jun 13, 2018 #19

    Dale

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    For the third time, the MCIF is not the same as A’s frame because A is non-inertial.

    I didn’t say that, you did. I challenged you to show that by writing down the transformation between the two frames, which you did not do.

    This is true. Again, A’s frame is not the MCIF since A is not inertial.
     
    Last edited: Jun 13, 2018
  21. Jun 13, 2018 #20

    berkeman

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    @Rozman -- I see that you are on a 10-day vacation from the PF, but I think you are still able to read these posts.

    I'm of very little help on the relativity aspects of your question, but I think I can point out a misconception that you have about accelerated motion that may be contributing to your not accepting what you are being told by these very experienced physicists. When an object is oscillating back and forth, the only places where the acceleration is zero (briefly) is at the the midpoint of the oscillation. At the extremes (like when your two thought experiment people touch), those are the maximum acceleration points. That is why the frame is not inertial when they touch. I will try to find a basic tutorial about acceleration in oscillatory motion to help you understand the other replies that you are getting. :smile:

    EDIT/ADD -- Like this:

    iGqpV.gif
     
  22. Jun 13, 2018 #21

    Grinkle

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    To me, unless I am really missing something, that adds complexity rather than clarification. The observers touch after having traversed different paths through spacetime. Whether they are accelerated with respect to one another at the moment they re-unite I think is not relevant to whether they experienced different elapsed times since the previous time they touched (time zero). It might be relevant if one needs to calculate exactly how much the time difference is - I am not sure about that.
     
  23. Jun 13, 2018 #22

    Dale

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    This is true, but the OP is focused on the reference frames. His/her primary logical error is thinking that because they are momentarily co-located and at rest with respect to each other then their frames are the same.

    The OP’s reasoning from that incorrect premise is sound. Since a coordinate chart is a one to one mapping between events and coordinates, if the same chart assigns distinct coordinates then the corresponding events are also distinct.

    So it is necessary to address the non-inertial reference frame, not because it is necessary for the calculation of elapsed time, but because it gets to the heart of the OP’s incorrect premise.
     
  24. Jun 14, 2018 #23

    Grinkle

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    I see. There is an intuition trap in this example also because a human could easily step off the spring at the moment of maximum extension and join the frame of the other observer, and then both would agree how long it takes the spring to complete its next cycle (obviously).
     
  25. Jun 14, 2018 #24

    Dale

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    Even that is fraught with subtleties and depends strongly on the exact method used for constructing a non-inertial reference frame. For example, if you are using Dolby and Gull's method of constructing a non-inertial reference frame, then they would agree on the next cycle of the spring since it is nearby, but would still disagree about distant events.

    https://arxiv.org/abs/gr-qc/0104077

    This is why it was so important for the OP to actually work out the transformation between the two. It is a non-trivial task that he/she tried to trivialize.
     
  26. Jun 14, 2018 #25

    Grinkle

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    I read the paper - since I am not sure how much of it I followed, I probably didn't follow much of it, but I wouldn't expect to be able to without a lot of effort.

    That said -

    They would only disagree if they tried to pick an origin different from their current location, I hope?

    Meaning if they synch their watches and agree that where they are both standing is x/y/z = 0/0/0 and measure the time and location of any subsequent event relative to that agreed origin, they will agree on all future measurements relative to that origin as long as they stay co-located and at rest?

    Or is there some history that cannot be re-set by agreement to pick "here and now" as a mutually agreed upon origin?
     
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