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Special Relativity in a Parallel Universe

  1. Jun 29, 2005 #1
    good question

    The meaninglessness is based on the fact that remote clock comparisons can be dreamed up in many distinct ways. You mentioned one, the outcome of which is that approaching clocks are speeding up and receding clocks are slowing down. Yes, that's justifiable. There is also another distinct definition, which is in perfect agreement with experiment, where moving clocks tick at the same rate as stationary clocks.

    http://www.everythingimportant.org/relativity/
     
    Last edited: Jun 29, 2005
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  3. Jun 30, 2005 #2

    JesseM

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    Do you mean that this formulation says that if I am an inertial observer, then in my coordinate system, a moving clock will tick at the same rate as a stationary one? That would seem to be in obvious contradiction with the experimental predictions of relativity, because if a clock makes a round trip away from me and back, relativity predicts the clock will have lost time relative to mine.
     
  4. Jun 30, 2005 #3

    Aer

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    I do not know which formula you are referring to, but I would suspect that what was implied is that 1 second in my frame according to me is equivilent to 1 second in a frame moving relative to me according to itself, but each frame sees the other frames second as less than his own.
     
  5. Jun 30, 2005 #4
    Precisely, but only if no external forces are applied to the moving clock. If the clock were to accelerate out of its own inertial frame of reference, then all bets are off.

    The paper that I cited computes the effect of a round trip on moving clocks and obtains the undisputed, experimentally verifiable result.
     
  6. Jun 30, 2005 #5

    JesseM

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    So what about the case where a ship moves away from earth at constant velocity, accelerates briefly, and then returns to earth at constant velocity? Would the coordinate system in that paper say that the ship-clock was ticking at the same rate as the earth-clock during both inertial legs of the journey, and only running at a different rate during the accelerating phase?
     
  7. Jun 30, 2005 #6
    The time in the accelerating phase is considered negligible. The rates at which all inertial clocks tick are obviously equal, if you understand the paper. I haven't seen any rebuttal to the stated definition of clock time yet. It's kind of miraculous why the clocks become desynchronized on a final comparison but I don't see a mistake in the math.
     
  8. Jul 1, 2005 #7

    JesseM

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    Wait, are you saying that throughout the entire journey the two clocks are synchronized, but somehow at the instant they meet they're not in sync? That doesn't sound possible--have you actually done the math for this example?
     
  9. Jul 1, 2005 #8
    No. I'm only saying that don't see any mistakes in the paper.
     
  10. Jul 1, 2005 #9

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    Consider yourself jumping from a train going 100mph to another train going -100mph. In order for your acceleration to be instanteous, you must be smacked against the wall immediately upon jumping into the train and feel the crunch of being forced in the opposite direction. How do you think you'd hold up? Would you be able to withstand such a trip? Now consider a particle doing the exact same thing but at .9c and -.9c. Would the particle survive? Does time for a non-existing particle have any meaning to you?
     
  11. Jul 1, 2005 #10
    Aer,

    I see no reason why particles can't accelerate from .9c to -.9c in particle collisions. The math in the paper assumes point particles and also assumes, only for ease of computation, that the acceleration of a point particle is either instantaneous or only takes a negligible amount of time.

    If every atom of an astronaut were to accelerate uniformly to .5c in 1 sec., why would the astronaut feel the acceleration? We feel acceleration in a rocket when the thrusters are fired because our molecules are being forced into the seat cushions. If every point of an unstressed extended object were to accelerate uniformly so that the distances between points are always constant in every instantaneous co-moving inertial frame of reference, then no force would be felt at all.
     
  12. Jul 1, 2005 #11

    JesseM

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    Aer, it's standard in discussions of the twin paradox to assume a near-instantaneous acceleration, it's just a thought-experiment so the fact that this would crush you is not relevant.
     
  13. Jul 1, 2005 #12

    JesseM

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    But you said that the paper "computes the effect of a round trip on moving clocks and obtains the undisputed, experimentally verifiable result." So if it computes the results, does it say the clocks are synchronized throughout the trip or doesn't it? If a clock moves away from me at 0.8c for 1 year, then comes back towards me at 0.8c for another year, what would be the function tau(t) giving the reading tau of a clock onboard the ship at any time t in my coordinate system?
     
  14. Jul 1, 2005 #13

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    I do not question this. However if the particle experiences this level of high speed "particle-particle collision" (The particle necessarily is colliding with particles in the wall) Would the particle still exist in it's defined "particle" form or would it be broken apart and give off gamma rays for instance. I am not claiming to know what would happen which is why I posed the question - my response was not an answer.
     
  15. Jul 1, 2005 #14

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    Read my response to Perspicacious above.
     
  16. Jul 1, 2005 #15
    The paper asserts that your question is an outdated and tortured way of reasoning and avoids that approach altogether. The comparison you request should not be made, which is the point of this thread. Shubert's methodologies are based on a new definition of time.
     
  17. Jul 1, 2005 #16

    JesseM

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    It's still irrelevant to the thought-experiment, we're not dealing with realistic particles as imagined in quantum theory here, just idealized indestructible clocks of zero size. Anyway, by making the inbound and outbound legs of the trip arbitrarily long, you can make the period of acceleration arbitrarily small compared to the time of the entire trip, which is all that really matters--the acceleration could still be spread out over a day or something, but if the inbound and outbound legs lasted 100,000 years you can basically ignore the acceleration period when computing the difference in the twins' clocks when they reunite.
     
  18. Jul 1, 2005 #17

    JesseM

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    But earlier you said "There is also another distinct definition, which is in perfect agreement with experiment, where moving clocks tick at the same rate as stationary clocks." How can you say they "tick at the same rate" if you're not comparing the time on one clock with the time on the other clock "at the same moment" in some coordinate system?

    Also, is it really necessary to link to the paper every time you refer to the paper in a response to me? Just curious, are you trying to advertise the paper because you are actually the author of the paper yourself?
     
  19. Jul 1, 2005 #18
    If you would just read the paper you would find out. I'm not going to answer any more questions about a paper you refuse to read.
     
  20. Jul 1, 2005 #19

    Aer

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    Very well, you want to make the acceleration period infinitely small compared to the length of the trip - I have no problem with this. However, this doesn't resolve the issue that there still is an interval of time for the acceleration to occur or otherwise the dt in dv/dt would be undefined - In other words, dt cannot be thought of as 0 but it can be thought of as approaching 0. Now you may be wondering what my point is, which is as follows:
    if the trip is 100,000 years long, then according to the tripper (good word, yes - I know), right before he starts his "instantaneous acceleration" to the frame of visitee his frame should be thought of as an inertial frame and according to him, 100,000 years have passed while time for the visitee was less than this. But, when the tripper decelerates to visitee, his clock will agree that visitee has experienced 100,000 years and that he has experienced less than this. (Please note that I am not a physicists, but that this view is exactly as was described to me by a professor of physics).

    Maybe he was wrong, but then how do you explain the reciprocity required by the derivation of the lorentz transformation in special relativity. If you do not know what I am talking about, then I will elaborate further in my thread on special relativity as this is off-topic to this thread.
     
  21. Jul 1, 2005 #20

    JesseM

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    Well, I'm not going to spend hours going through a lot of tedious math in a paper that is widely deemed to be the work of a crackpot if you aren't willing to try to show there is actually something meaningful there by summarizing what it could possibly mean to say two clocks "tick at the same rate" in the absence of a coordinate system.
     
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