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Special relativity simultaneity

  1. Jan 12, 2017 #1
    1. The problem statement, all variables and given/known data
    toDJKbk.png


    2. Relevant equations
    The rear clock ahead example gives vL/c^2:
    c3QBs3X.png

    3. The attempt at a solution
    I think the solution is the same, because even if there is a time dilation due to u (downward velocity) both clocks would slow down at the same rate and so the time difference would still be the same. Is that correct?
     
  2. jcsd
  3. Jan 12, 2017 #2

    Simon Bridge

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    You think the solution to the 1st one should be the same as the solution to the 2nd one?
    Can you prove it?
     
  4. Jan 12, 2017 #3
    well, all I can think for as a reason is that they are both dilated by gamma from u, but then the rate at which they tick are the same, so the difference should be the same.
     
  5. Jan 13, 2017 #4

    TSny

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    Yes, the rate of each clock is affected the same. But the difference of the clock readings at simultaneous times according to the ground reference frame is affected by the u velocity.
     
  6. Jan 13, 2017 #5

    PeroK

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    Here's a hint (or, at least, what I thought is the best way to look at it). Imagine (in the truck frame) there are two more clocks (at rest) part way up the the vertical rails. These clocks are colocated with the moving clocks at some time. In the moving frame, you could reset all four clocks to read ##0## at this point (as the moving clocks move past the stationary ones).

    Now, analyse this from the ground frame, using what you know about the time lag between the vertically at-rest clocks.
     
  7. Jan 13, 2017 #6

    Simon Bridge

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    Also: what is the relative velocity of the clock-frame wrt the ground-frame (magnitude and direction)?
    ... another way to approach this is via a space-time diagram.
     
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