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Separation and relative time

  1. Apr 29, 2012 #1
    I think I understand much of SR and some GR, but I'm always stuck when this thought comes to mind.It looks like a stupid proposition and never appears in discussions, but want it cleared up.
    Consider two objects separated by some distance comparable to the speed of light. Thus each observer would actually be observing the activities of the other one's past.If they are apart by c metres, then each one would observe the activity of the other a second earlier of what's happening now in the latter's vicinity. When we lorentz transform time from that measured in one frame to another, we don't consider this difference in times.
    For example, the transformation for bodies separated by distance y should be
    t(prime)=gamma* (t-vx/c^2 + y/c)
    where x is the distance to the point as measured in the stationary observer's frame,
    y is the dstance between the point and the moving observer.

    The factor y/c is a corrective one(or so I think) added by myself to correct for difference in times due to separation.
    Last edited: Apr 29, 2012
  2. jcsd
  3. Apr 29, 2012 #2


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    The first thing you should understand is that a distance is different than a speed and so "distance comparable to the speed of light" makes no sense. You should multiply the speed by a time (e.g. one light-second) to get a distance.

    As to your question, the Lorentz transformation transforms from one moving frame to another who's (4-D) origin coincides. If the origin does not coincide (e.g. a translation in either time or space is involved), then one must use a general Poincare transformation.
  4. Apr 29, 2012 #3
    By that I meant distance comparable to 3x10^5 km. Yes, not the right way to put it.
  5. Apr 29, 2012 #4
  6. Apr 30, 2012 #5


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    What happened to gamma?
  7. Apr 30, 2012 #6
    It should actually be the clocks will differ by Δt' = -vΔx'/c2 = -vΔxγ/c2 and is often quoted as Δt' = -vL/c2, where L is the proper distance between the clocks and the gamma factor is "hidden".
  8. Apr 30, 2012 #7
    In those formulae, use opposite signs of v everywhere because the unprimed frame moves with velocity -v relative to the primed frame, and you'll end up with the difference in times being
    -vx/c^2. In the formula derived, the difference is gamma times this and then it is corrected by dividing it by gamma, because this difference is measured by the observer in the primed frame and according to him, the clocks in the other frame must run slower by a factor of 1/gamma.
    Another way to say this is that the length l (or x) is measured in the unprimed frame and according to the observer in the primed frame, this distance is lesser by a factor of 1/gamma.
    Both ways, the difference in times in the unprimed frame as observed in the primed frame is -vx/c^2.
    Last edited: Apr 30, 2012
  9. Apr 30, 2012 #8
    That is only correct in first approximation, for v<<c and at about the same height. And note that that is only true if the two frames are independently synchronized in one of the usual ways (for example with radio signals).
  10. Apr 30, 2012 #9
    Yes gravity has its effects, but I thought it's valid at all velocities(lesser than c).
  11. Apr 30, 2012 #10
  12. Apr 30, 2012 #11
    I omitted gamma for a reason. -vx/c^2 is the difference in the unprimed frame as seen from the
    primed frame. This has to be multiplied by gamma to correct for time dilation(or length contraction of x)
  13. Apr 30, 2012 #12
    Yes, it is only correct at the same height and for conventional synchronization, but it is exact for all v<c.

    Using the Lorentz transformation Δt' = γ(Δt - vΔx/c2)

    and for simultaneous times in the unprimed frame, Δt =0 so Δt' = γ(-vΔx/c2)

    This is the time interval measured in the primed frame, but we are interested in the times actually shown on the clocks in the unprimed frame and by using Δt = Δt'/γ, we get Δt = -vΔx/c2 when observed from the primed frame.
    Last edited: Apr 30, 2012
  14. Apr 30, 2012 #13
    yes, like that (and I added one missing Δ at the start) it's OK. I had misunderstood what vin meant in post #7 - should have read the link!
    Last edited: Apr 30, 2012
  15. Apr 30, 2012 #14
    Oops, I was editing my last post when you replied, so my explanation is slightly different now. Do you think this latest version makes more sense?
  16. May 1, 2012 #15
    - Your first version looks good to me and it corresponds to vin's link; I would have used the symbols L0=L' for Δx' and specified what you implied: L0 = Δx' = yΔx for Δt =0. That is, to determine the length Δx of a "moving" object according to S, we do a measurement of x1 and x2 at the same time t.

    - Your second version has Δt = Δt'/γ or Δt' = γΔt. Check when that is valid: Δt' = γ(Δt - vΔx/c2). Thus Δt = Δt'/γ is valid for Δx=0. That equation describes that a clock at rest in S appears to tick slow according to S'. But that is not what we wanted to know...
    [addition:] The Δt' in your first version is not a time interval but the difference in the readings t1' and t2' of the two clocks according to S. That is what we originally were after, as the OP indicated that x corresponds to "stationary frame".
    Of course the observations are symmetrical, and as the original comment lacked definition it's possible to say that the inverse view was meant, as indicated in post no. #7.

    Happily I took a snapshot of your first version. :smile:
    Last edited: May 1, 2012
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