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The Universal Time Coordinate

  1. Mar 29, 2010 #1
    Hey

    If you have the Minkowski line equation of

    -ds^2 = c^2 d tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2

    I don't understand how you can assume from this ^ that each observer in different reference frames will experience time changing at the same rate 'dt'. I thought that tau was the relative time experienced by each observer depending on their relative velocity :S

    Am I getting confused?!

    Thanks :-)
     
  2. jcsd
  3. Mar 29, 2010 #2

    dx

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    The rate at which time is "experienced" does not depend on the observer. That would violate the principle of relativity which says that all uniformly moving observers are equivalent, and each of them can consider themselves at rest.

    If an observer B is moving away from observer A with a speed v, then A will observe B's clocks ticking at a rate √(1 - v²) relative to A's own clocks. But observer B will also see the same thing, i.e. he will observe A's clocks ticking at a rate √(1 - v²) relative to B's clock.
     
    Last edited: Mar 29, 2010
  4. Mar 29, 2010 #3

    Mentz114

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    You can manipulate that equation to give

    [tex]\left(\frac{d\tau}{dt}\right)^2= 1 - \left(\frac{dx}{c^2dt}\right)^2 - \left(\frac{dy}{c^2dt}\right)^2 - \left(\frac{dz}{c^2dt}\right)^2= \frac{1}{\gamma ^2}[/tex]

    taking square roots on both sides is easy.

    In your own frame [itex]\gamma=1[/itex] so you think your clock is in synch with t. But observing another frame which is moving wrt to you, you see a difference of a factor of [itex]\gamma[/itex] in the clock rate.
     
  5. Mar 29, 2010 #4
    Thanks guys, I think I'm clearer now.....

    So; any fundamental observer in their own frame of reference sees their clock running at the cosmological time. But then if they were to observe another observers clock which was moving relative to theirs, they would observe tau, the proper time.

    So c^2 tau^2 in the minkowski line element is expressing the length of a line in some reference frame as seen from some observer in a different frame of reference? I think I've got it? Thanks :-)
     
  6. Mar 29, 2010 #5

    dx

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    There's no such thing as the 'cosmological time', at least not in special relativity.

    If a clock is moving in some reference frame, then it will tick slower than clocks that are stationary. The proper time is an invariant; it is the same according to any observer.

    I'm not entirely sure what kind of picture you're imagining here, but just in case: consider two clocks A and B which are moving relative to each other. Let the events "B reads 0s" and "B reads 1s" be denoted E1 and E2. In a frame in which A is at rest, the time that elapses between the two events E1 and E2 is 1/√(1 - v²), but the proper time 'tau' is still 1.
     
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