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Some Inertial Frames may br more Equivalnet than Others

  1. Apr 4, 2007 #1
    Some Inertial Frames may be more Equivalent than Others

    Silly proposition - but take the peculiar results predicted by Einstein in Part IV of the 1905 paper with a little extra added. We place two clocks at locations A and B separated by a great distance L. We identify A clock with an S' frame and B clock with an S frame. A third clock is added at point P in the S frame on a line that connects A and B. All clocks are initially synchronized and then A clock is accelerated toward B to 0.9c as measured in the S frame. A clock reaches its terminal velocity as it passes P, thereafter it continues at a uniform velocity 0.9c until it reaches B. The distance required to reach 0.9c is small compared to L.

    Einstein says the A clock will lag behind the B clock when A arrives at B. This is not a reciprocal observation - this is a permanent difference in the amount of time logged by the A clock and the B clock...the one-way time difference confirmed by all experiments with high speed particles (no turn around required).

    During the acceleration phase, A clock will lose time as measured by P due to the acceleration relative to P (It can't lose a different amount of time because it is also being monitored by B). This provides an offset, but thereafter the S and S' frames should be regarded as equivalent - yet the final age difference is based upon the time traveled from P to B (plus whatever is lost in getting up to speed). As A passes P, from P's perspective, the S' frame is simply a frame with relative velocity v and therefore both frames should be equivalent (in transit from P to B we can no longer consider the history that included the acceleration phase). So the primary contribution to the final age difference between A and B depends upon travel time of A as it moves at a uniform velocity from P to B.

    So, does the initial acceleration create asymmetry between the S and S' frames that leads to the final age difference?
     
    Last edited: Apr 4, 2007
  2. jcsd
  3. Apr 4, 2007 #2
    Hi Yogi. I have not yet studied GR and so I do not know what happens to accelerated clocks. You say that A is accelerated to 0.9c. If this acceleration continued the velocity would continue to increase, limited presumably to <c. To avoid this continued increase in speed if A is to travel no faster than 0.9c then a deceleration must occur. Have you taken this into account.

    Matheinst.
     
  4. Apr 5, 2007 #3
    The situation I intended to convey is that A reaches 0.9c at the time it is adjacent to the P clock - thereafter it coasts the rest of the trip at this velocity (0.9c). The major contribution to the difference in age at the end of the trip occurs during the coasting phase.
     
  5. Apr 5, 2007 #4

    Demystifier

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  6. Apr 5, 2007 #5
    Hi Yogi. But to coast at 0.9c it has to remove the acceleration which took it to that speed and so has at some point to decelerate.

    Matheinste.
     
  7. Apr 5, 2007 #6
    for me, the "time lag" checked between A and B will be the same that the one checked between A and the third clock since the third clock is in the same frame as B, the distance between B and the third clock should not make any difference...correct me if i am wrong????
     
  8. Apr 5, 2007 #7

    pervect

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    Haven't we been over something a lot like this before? When you say "all clocks are initially synchronized', do you mean that they've been synchronized in S or in S'? The two notions of synchronization are not equivalent.
     
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