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A moving clock lags behind a stationary clock

  1. Oct 22, 2006 #1
    A. Einstein states the following on page 49 in The Principle of Relativity:

    "Between the quantities x, t, and T, which refer to the position of the clock, we have evidently, x = vt and

    T = (t - v*x/c^2)/sqrt(1 - v^2/c^2).

    Therefore,

    T = t*sqrt(1 - v^2/c^2) = t - (1 - sqrt(1 - v^2/c^2))*t

    whence it follows that the time marked by the clock (viewed in the stationary system) is slow by 1 - sqrt(1 - v^2/c^2) seconds per second, or - neglecting magnitudes of fourth and higher order - by
    (1/2)*(v^2/c^2).
    From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)*(t*v^2/c^2) (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B."

    it is important to note here that "the time marked by the clock (viewed in the stationary system) is slow by 1 - sqrt(1 - v^2/c^2) seconds per second, or - neglecting magnitudes of fourth and higher order - by
    (1/2)*(v^2/c^2)." for example, if the observer moving with the clock at A measures the value of t using the clock at A (the observer at A and the clock at A are at rest relatively to each other), and the stationary observer at B measures the value of t using the stationary clock at B, the value of t obtained by the obsever at A equals the value of t obtained by the observer at B.

    but if the stationary observer at B measures the value of t using the stationary clock at B, and then the same stationary observer at B measures the value of t using the moving clock at A, then the value of t obtained by the stationary observer at B using the stationary clock at B is not equal to the value of t obtained by the same oberver using the moving clock at A. the moving clock lags behind the stationary clock "by (1/2)*(t*v^2/c^2) (up to magnitudes of fourth and higher order)." but i don't know what magnitudes of fourth and higher orde is.
     
    Last edited: Oct 22, 2006
  2. jcsd
  3. Oct 22, 2006 #2

    JesseM

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    Science Advisor

    It seems like all you're saying with this last part is that the two clocks at rest in the stationary system are synchronized, so that at any single moment in the stationary system's rest frame, according to that frame's definition of simultaneity, the clocks at A and B will both show the same time (which is almost a tautology, because these clocks are used to define the coordinate system that is 'the stationary system's rest frame').
    If t is taken to mean the time-coordinate in the stationary system's rest frame, then you can't "measure the value of t using the moving clock"--the time t of an event in the stationary system's rest frame is defined in terms of the reading of one of the clocks which is at rest and synchronized in that frame, whichever clock was next to the event as it happened. The moving clock's readings are represented using the symbol T, not t, according to Einstein's quote above.
     
  4. Oct 23, 2006 #3
    yes. it is almost a tautology, but it isn't. the point is that to say that for all of us the moving clock at A is slower than the stationary clock at B is not correct because some of us are moving along with the clock. if the color of the clock at A is red and the color of the clock at B is blue, those moving along with the red clock (let's call them the red team) would say that the blue clock is slower than the red clock. the comon A and B time would be the red (and blue) clock at rest relatively to the red and blue team respectively.

    yes. you are right. i should have used T for the time value of the event using the moving clock. thanks.
     
    Last edited: Oct 23, 2006
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