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Hafele and Keating Experiment

  1. Feb 6, 2004 #1
    While researching how SR effect could be applied to GPS clock after GR effect, I encountered the experiment made on 1971.

    In the section "Kinematic Time Shift Calculation", I saw an totally no sense algorithm, violating simple math.

    Ts = T0 * (1+R^2w^2/(2*c^2)), where Ts is the Earth surface time, R is the Earth's radius and w is the Earth's angular velocity of Earth's rotation.

    Ta = T0 * (1+(R^2*w^2+v^2)/(2*c^2)), where Ta is the airplane's time and v is airplane's velocity.

    Ta-Ts = T0 *( 2Rwv+V^2)/(2c^2).

    Up to here , it's totally reasonable.

    Next thing, it brought in is completely math. incorrect.

    Ta-Ts = - Ts * (2Rwv+v^2)/2c^2.

    Note the minus sign coming from nowhere. If Ta is larger than Ts from the first two equations, Ta-Ts will never be negative.

    If this math. can live in the Physics, there is definitely something wrong.
  2. jcsd
  3. Jan 15, 2007 #2


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    I think you mean Kinematic Time Shift Calculation

    Yes, this derivation is loose a litle bit.
    In the first equation, i.e. in [tex]T = \frac{T_0}{\sqrt{1 - v^2/c^2}}[/tex]
    [tex]T_0[/tex] is the time elapsed between two events occuring in the centre of the Earth, while [tex]T[/tex] stands for the time elapsed between this two events in an inertial reference frame moving with velocity v with respect the centre of the Earth.

    Really, here is nothing to do with this. We need just the opposite. If the proper time elapsed on the airplane is T (this isn't an inertial frame!), then during this, in the centre of the Earth elapses
    [tex]T_0 = \frac{T}{\sqrt{1 - v^2/c^2}}[/tex]
    time (the centre of the Earth is regarded now to move inertially)

    So, our second equation reads correctly
    [tex]T \approx T_0 \left[ 1 - \frac{v^2}{c^2}\right] [/tex].

    And so on. The sign was missed at the beginning.
  4. Jan 15, 2007 #3

    Chris Hillman

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  5. Jan 16, 2007 #4


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    Fine! Thx!
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