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Less elapsed time: Lorentz Transformation or Spacetime Interval ?

  1. Oct 17, 2009 #1

    morrobay

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    With one clock stationary at the common origin for 10 time units and the other rocket clock
    traveling out and back at v=.6c. When both clocks are compared at the origin the elapsed
    time on the traveling clock will show less time by a factor of (.8) from the Lorentz Transformation.
    The spacetime interval for the traveling clock will be S=8, sqrt( 10^2-6^2)
    Compared to the stationary clocks spacetime interval S=10.
    Note, it is understood if the rocket clocks coordinates were transformed to the inertial frame
    by: t=gamma(1.25) (t' + (v)(x')/cc and x=gamma(1.25)(x'+(v)(t') then S=8
    =sqrt(17^2-15^2).
    So the question: Is the 8 units of elapsed time in the rocket clock because the physical mechanism was moving at a slower rate (time between ticks) in the rockets frame.
    Or in the case of the spacetime interval, was the mechanism in the rockets clock ticking at the same rate (in rocket frame) as the rate of ticking in stationary frame, but the rocket clock just traveled a lesser distance in spacetime, S=8
     
    Last edited: Oct 17, 2009
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  3. Oct 17, 2009 #2

    JesseM

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    A clock is never running slow in its own inertial rest frame...but the rocket must accelerate at some point to turn around, so what do you mean when you say "the rockets frame"? If we assume the acceleration to turn around is instantaneous and that the rocket moves inertially before the turnaround and afterwards, then the rocket has two different rest frames: 1) the frame where the rocket is at rest during the journey away from the origin, and 2) the frame where the rocket is at rest during the journey back towards the origin. If you analyze things from frame 1, then the rocket's clock is ticking normally during the journey out but running slow after the turnaround when it's no longer at rest in this frame, and likewise if you analyze things from frame 2, then the rocket wasn't at rest before the turnaround so its clock was running slow then.
     
  4. Oct 17, 2009 #3

    morrobay

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    My question is not between the views of the rockets outbound and inbound reference frames
    or the relativity of simultaneity between other frames. It is based on the overall conclusion when the two clocks are compared at the common origin after the rocket clock has traveled
    outbound and inbound at v=.6c for a 10 year period from the reference frame of the earth.
    That the elapsed time on the stationary clock is 10 yrs and elapsed time on the rockets clock is 8 years.
    So is the 8 year elapsed time on the rocket because it was mechanically running slower(time between ticks) than the earth clock - from the Lorentz transformation.
    Or did the rockets clock record a proper time of 8 years because it traveled a lesser distance in spacetime, S=8, with a clock that mechanically ran at the same rate as the
    stationary clock ?
     
  5. Oct 17, 2009 #4

    Dale

    Staff: Mentor

    This is the interpretation I prefer.
     
  6. Oct 17, 2009 #5

    JesseM

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    But the Lorentz transform doesn't really say this--you can pick a frame where the rocket clock was ticking faster than the Earth clock for at least part of the journey, it's only if you look at the average rate of ticking over the entire trip that all frames agree the rocket clock was ticking slower on average.
    The distinction between these two options doesn't seem physically meaningful--they don't make any distinct predictions about any empirical results. Also, what does it even mean to ask whether a clock is "mechanically running slower" or if it "mechanically ran at the same rate" if you aren't asking this question from the perspective of a particular coordinate system? You can't compare rates of ticking without use of a coordinate system, your question is analogous to looking at two lines on a plane and asking whether they "physically have different slopes" or "physically have the same slope" without specifying a coordinate system to define "slope" in terms of dy/dx. The "slope" of a line on a 2D plane is an intrinsically coordinate-dependent notion, and I would say the same thing about the "rate of ticking" of a clock in 4D spacetime.
     
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