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SR time dilation problem

  1. Dec 23, 2004 #1


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    Please help me resolve this problem.

    Suppose we have 3 objects travelling at constant velocity, and hence 3 inertial frames of reference. Call them object A, B and O respectively. O is at rest with respect to the observer.

    Let the relative velocity of A from the point of view of B be speed u in a "rightward" direction.

    Let the relative velocity of B from the point of view of O be speed v in a "leftward" direction.

    Let the magnitude (speed) |u| < |v|.

    A particular time interval tA is recorded by A. The equivalent interval recorded by B is tB and is given by :

    tA = tB*sqrt (1 - u^2/c^2) = tB*L1, where L1 is the Lorentz correction between B and A.

    Similarly as B records a time of tB, O records a time of tO, where :

    tB = tO*sqrt (1 - v^2/c^2) = tO*L2.

    Putting those two equations together, we get :

    tA = tO*L1*L2 ---(eq 1)

    meaning the Lorentz correction between A and O should be given by L1*L2.

    But another way of looking at it is like this :

    Let the relative velocity of A as seen by O be given by w.

    |w| = (v - u)/[1 - (uv/c^2)] by relativistic addition of the velocities.

    and the direction of w is "leftward" from the point of view of O.

    It is obvious that |w| < |v|

    tA can also be given by :

    tA = tO*sqrt (1 - w^2/c^2) = tO*L3 -- (eq 2)

    Since |w| < |v|, L3 > L2.

    But by comparing eq 1 and eq 2, we see that L3 = L1*L2

    Since L1 is necessarily less than one, L3 < L2

    So we have L3 > L2 and L3 < L2, which is a contradiction, creating a paradox.

    I got this from trying to figure out one of the accounts of the Hafale-Keating experiment, where a clock on a plane and a clock on the earth surface were compared. I found the method described strange, because the only proper time was considered to be at a hypothetical clock at the center of the earth, and everything was calculated relative to the center of the earth. I thought it should be fine to use the surface of the earth as inertial, even though there is a centripetal acceleration, there is no change in speed of the surface clock in a tangential direction. To figure out the mathematical approach I would use, I considered 3 idealised inertial reference frames and tried to compute relativistic intervals between them, expecting everything to come out OK. But I ran into the above problem that I can't resolve.

    This looks like a fairly simple problem that must have been addressed before, and I think I'm making some horrible mistake somewhere. Any ideas ? Thanks.
  2. jcsd
  3. Dec 23, 2004 #2

    Doc Al

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    One problem is that you are assuming that time intervals in two frames transform like this:
    [tex]\Delta t = \gamma(\Delta t')[/tex]
    But this is only true when [itex]\Delta x' = 0[/itex]. (The special case of a moving clock.) The general Lorentz transformation for time is:
    [tex]\Delta t = \gamma(\Delta t' + v\Delta x'/c^2)[/tex]

    (To properly relate measurements in two frames you must consider [itex]\Delta x[/itex] and [itex]\Delta t[/itex] together.)
  4. Dec 23, 2004 #3


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    Thanks for the reply, Doc. But could you please amplify further. [itex]\Delta x[/itex] has dimensions of distance but exactly what is it. Can you give me more detail or point to a page with this explanation ? Thanks so much.
  5. Dec 24, 2004 #4

    Doc Al

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    Just like two events can be separated in time, they can be separated in distance. When you made the statement
    you are talking about the time interval [itex]\Delta t'[/itex] between two events. But what about the distance ([itex]\Delta x'[/itex]) between those events? (Here I am calling the A frame the primed frame.)

    To figure out what another frame (the B frame) would measure as the time and distance between those same two events, one must apply the Lorentz transformations:
    [tex]\Delta t = \gamma(\Delta t' + v\Delta x'/c^2)[/tex]

    [tex]\Delta x = \gamma(\Delta x' + v\Delta t')[/tex]

    Where v is the speed of the A frame with respect to the B frame, and [itex]\gamma = 1/\sqrt{1- v^2/c^2}[/itex]. Note that the time measured in one frame ([itex]\Delta t[/itex]) depends on both the time and distance measured in the other frame.

    So, in general, your statement:
    is not true.

    If you wish to transform measurements made in the A frame to those made in the O frame, you are free to do so in two steps (A measurements ==> B measurements ==> O measurements) or in one (A measurements ==> O measurements) as long as you use the correct relative speeds and the correct Lorentz transformations. Either way gives the same answer.
  6. Dec 24, 2004 #5


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    ^Thanks, Doc for all your help. I worked through the algebra and I have verified that it comes out the same way.

    Happy Holidays. :)
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