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SuperDuper Problem In The Time Dilation Formula Revealed

  1. Mar 28, 2004 #1
    Let us define 'time' empirically, as that which is measured by clocks. Thus, we have an operational definition of time. We certainly have no definition of 'clock', but none is needed because clocks can be constructed.

    Stipulation 1: Let us assume that we have two clocks of identical construction, clock A, and clock B, which are at rest with respect to each other, and in an inertial reference frame.

    Assumption 1: The time dilation formula is true.

    Time dilation formula:

    [tex] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}} [/tex]

    When v=0 in the formula above, the following equation is a true statement:

    [tex] \Delta t = \Delta t^\prime [/tex]

    In the time dilation formula, let the letter v denote the relative speed of clock A to clock B. Thus, [tex] \Delta t [/tex] denotes an amount of time measured by one of the clocks, and [tex] \Delta t^\prime [/tex] denotes an amount of time measured by the other clock.

    Now, as you can see from the formula, if the relative speed of the two clocks is zero, the amount of time of one 'tick' of either clock must be identical. THIS MEANS THAT CLOCKS OF IDENTICAL CONSTRUCTION WILL HAVE A DIFFERENCE IN READINGS WHICH IS CONSTANT IN TIME. In other words, if the clocks are synchronous, then they remain in-sync forever, and if they are out of sync by an amount D (where D is the difference in clock readings) then D is constant. Let X denote the reading of clock A, and let Y denote the reading of clock B. Define D as follows:

    D = X-Y

    If the clocks are in sync, then X=Y, and hence D=0. If the clocks are out of sync then not (X=Y), and D is nonzero, but the D is constant in time, hence dD/dt=0.

    Ok, so everything so far is totally unobjectionable. Let us now suppose that clock B is subjected to a constant outside force F, for a period of time [tex] \Delta t [/tex] as measured by clock A, and then suddenly, the force applied to clock B returns to zero.

    We are under the standing assumption that the time dilation formula is true. Since clock A remains in an inertial reference frame, we can apply the time dilation formula to clock A. The mathematical conclusion is that clock B begins to tick slower than clock A, therefore the difference in clock readings D is no longer constant in time.

    Let the force be applied exactly for time delta t, according to clock A, at which point F suddenly returns to zero. Let the final speed reached by clock B relative to clock A be denoted by V. Thus, the amount of time which for which the force was applied to clock B as measured by clock B is unambiguosly given by delta t` and related to delta t by the time dilation formual as follows:


    [tex] \Delta t^\prime = (\Delta t)( \sqrt{1-V^2/c^2}) [/tex]

    Thus, delta t` is less than delta t, which means that clock B ticked slower than clock A. Let us interpret this result in terms of clock readings.

    Let us use the following initial conditions at the moment when force F was applied:

    W= reading on clock A
    R = reading on clock B

    For the sake of simplicity, let us presume that before force F was applied to clock B, that the clocks were synchronous. Thus, before the force was applied, we had W=R. For the sake of simplicity, let W=R=0.

    Now, let us define G,H as follows:

    G = reading on clock A at the moment F returns to zero
    H = reading on clock B at the moment F returns to zero

    Now, using the time dilation formula, we are forced to conclude that G > H.

    Now, the total time for which the force was applied to clock B as measured by clock A, is simply the difference in readings according to clock A, which is given by:

    G-0=G

    And the total time for which the force was applied to clock B as measured by clock B, is simply the difference in readings according to clock B, which is given by:

    H-0 = H

    Since clock A is always in an inertial reference frame, we can use the time dilation formula to conclude that G>H.

    Thus, the clocks are now out of sync by the following amount:

    G-H

    This is now the difference in readings of the two clocks. Hence, we have:

    D = G-H and (G>H)

    The relative speed of the two clocks is now constant, hence D is a non-zero constant.

    Now, let us suppose, that clock A is subject to the same constant force F that was applied to clock B, and let the force be applied in such a direction as to again bring the clocks to rest with respect to one another. Let the amount of time force F is applied to clock A as measured by clock A be equivalent to the time force F was applied to clock B as measured by clock B previously.

    During the time that force F is applied to clock A, clock B is in an inertial reference frame, and can use the time dilation formula. Thus, B will use the following formula:

    [tex] \Delta t^\prime = \frac{\Delta t}{\sqrt{1-v^2/c^2}} [/tex]

    Thus, clock A slows down, until its reading matches that of clock B, and the clocks are again in sync.

    SUPERDUPER PROBLEM IN THE TIME DILATION FORMULA REVEALED

    Suppose instead that the second time force F was applied, it was applied to clock B instead of clock A, in a direction to again bring the clocks to rest with respect to one another, and for the same amount of time clock B experienced it the first time.

    Can the clocks be brought into sync this way????????

    And if not, why does that conclusively prove that the time dilation formula is false?
     
  2. jcsd
  3. Mar 28, 2004 #2
    Your final equation with respect to clock B would determine that there is no time dilation of clock B because its velocity is zero.

    In the second part, Clock A would experience time dilation of the same magnitude as clock B experienced in the first part, but B would not experience any.

    There is no discrepancy in the time dilation formula.

    Leaving clock A stationary and applying the same force to B, twice would result in twice the time dilation.
     
  4. Mar 29, 2004 #3
    Sync clocks through timedilation? Hehe. :smile:
     
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