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Ratio of proper clock rates?

  1. Mar 3, 2013 #1
    Some of you may be familiar with Dingle, an "opponent" of SRT. I can condense his argument (the only one I know about anyway) to a sentence or two. Solve the Lorentz Transform for delta-t-prime, then divide through by delta-t.

    delta-t-prime/delta-t = gamma - gamma*v*delta-x/delta-t

    [Dingle] "We therefore get different values for the ratio according to the (location of) events we choose... We can, by a suitable choice of events, make (delta-t-prime) greater or less than (delta-t), by any amount we like. But ... the relative rates of the clocks must be independent of the events chosen for their determination."

    SRT sure explains a lot of stuff, but I can't figure out what may be wrong with his argument.

    Any help appreciated.
     
  2. jcsd
  3. Mar 3, 2013 #2

    Nugatory

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    There's a relativity of simultaneity problem buried in the bogus argument. The ratio of delta-t to delta-t' is not the same thing as the ratio of clock rates unless the end points of the intervals across which delta-t and delta-t' are measured are simultaneous to both observers.

    (this is why the twin paradox is set up as a round trip, so that the endpoints of both twins' journeys must coincide in time and space. It's also why in the easier case of constant relative velocity it's possible for both observers to see the other clock running slow and why the question "which one is really slower?" is meaningless)
     
  4. Mar 3, 2013 #3

    Doc Al

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    His error is in assuming that Δt' for an arbitrary set of events represents the reading of a single clock. That would only be true if Δx' equals zero. In such a case, Δt = γΔt'.
     
  5. Mar 3, 2013 #4

    Dale

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    Just a general hint. 99% of the time, it is relativity of simultaneity. As Doc Al mentioned, whenever Δx≠0 you are talking about the reading on two different clocks. Those clocks are synchronized in their own reference frame, but due to the relativity of simultaneity they are not synchronized in the primed reference frame. Their desynchronization depends on position and accounts for the position-dependent portion of the effect.
     
  6. Mar 3, 2013 #5

    ghwellsjr

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    The Lorentz Transform works on the coordinates of individual events, not on the delta between pairs of events. And there are four parts to the transform, one for t, x, y, and z. (We can ignore the y and z parts in this discussion because they don't change during the transform.)
    Of course, because both t and x influence the transformed values of t' and x'.
    It's not the relative rates of clocks that we are concerned about. It's the rate of a single clock relative to the coordinate time of events. We can see this clearly if we start with a clock at rest in one frame (the unprimed frame) where its time (which is its Proper Time) is defined to be the same as the coordinate time of that frame and then transform the events of equal intervals of Proper Time to a frame moving with respect to the original frame and we plot the same events in this new frame. The events marking intervals of Proper Time on the clock in this new frame will be spaced farther apart than the coordinate time of the frame.

    Here is a diagram showing a single clock at rest in a frame with the dots marking equal intervals of time:

    attachment.php?attachmentid=56301&stc=1&d=1362281009.png

    Now here is a diagram showing the same clock in a frame moving at -0.6c with respect to the first frame:

    attachment.php?attachmentid=56302&stc=1&d=1362281009.png

    Gamma at 0.6c is 1.25. Notice how the dots are spaced at coordinate intervals of 1.25. This is easily seen by looking at the fourth dot up which coincides with the coordinate time of 5 (since 4*1.25=5).

    Remember, there's only one clock involved.
    I hope this helps. Let me know if it doesn't.
     

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  7. Mar 3, 2013 #6

    dvf

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    Thanks to its linearity, the Lorentz transformation works perfectly on coordinate differences between pairs of events.
     
    Last edited: Mar 3, 2013
  8. Mar 3, 2013 #7

    dx

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    In a given reference frame, let clock A be stationary and clock B moving with a speed v.

    Take two events:

    1. Clock A reads 0
    2. Clock A reads 1

    For these two events,

    Δx = 0
    Δt = 1

    When clock A reads 1, the reading of clock B in that frame of reference is √(1 - v2), so in this reference frame, the rate of clock B relative to A is √(1 - v2)

    The problem with the argument in the OP is that he assumes you can calculate the relative rate as Δt'/Δt, i.e. he assumes that in the given frame, when clock A ticks by an amount Δt, clock B will tick by an amount Δt'

    This is wrong. Δt' is actually 1/√(1 - v2), it is not √(1 - v2).

    (t' refers to the reference frame in which clock B is at rest)
     
    Last edited: Mar 3, 2013
  9. Mar 3, 2013 #8

    DrGreg

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    Correct.
     
  10. Mar 3, 2013 #9

    dvf

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    Worth a little texercise:
    [tex]
    \begin{align}
    \Delta x' &= x'_2 - x'_1 &\quad \Delta t' &= t'_2 - t'_1 \\
    &= \gamma \ ( x_2 - v \ t_2 ) - \gamma \ ( x_1 - v \ t_1 ) & &= \gamma \ ( t_2 - \tfrac{v}{c^2} x_2 ) - \gamma \ ( t_1 - \tfrac{v}{c^2} x_1 ) \\
    &= \gamma \ ( x_2 - v \ t_2 - x_1 + v \ t_1 ) & &= \gamma \ ( t_2 - \tfrac{v}{c^2} x_2 - t_1 + \tfrac{v}{c^2} x_1 ) \\
    &= \gamma \ ( x_2 - x_1 - v \ (t_2 - t_1) ) & &= \gamma \ ( t_2 - t_1 - \tfrac{v}{c^2} (x_2 - x_1) ) \\
    &= \gamma \ ( \Delta x - v \Delta t) & &= \gamma \ ( \Delta t - \tfrac{v}{c^2} \Delta x)
    \end{align}
    [/tex]
     
  11. Mar 3, 2013 #10
    Wow - thanks all! (And I left a c^2 out of the original post.)

    Yes, it seemed to me that Dingle was imagining an infinite string of passing clocks at one observer's location in S', rather than watching a passing clock in the un-primed system. So thus delta x-prime=0, instead of delta x=0. So he effectively was not really watching a time interval in the un-prime system. I guess that is what all of you are saying too.

    Thanks again.
     
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