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Special Relativity: Simultineity question

  1. Oct 11, 2004 #1
    A rod whose length at rest is 10 meters is moving to the right along the positive x-axis of the O-frame with velocity V = 0.70c. An observer riding on the rod (i.e. in the O'-frame) sets off two explosions simultaneously - so he says. If the rod is parallel to the x-axis how far apart are the two explosions (in meters) in the O-frame?

    I found the length the rod appears in the O-frame (about 7.1m) but am not really sure how to answer the question if I am not given how far in meters apart the explosions happen in the O'-frame. Can someone explain? thanks.
     
  2. jcsd
  3. Oct 11, 2004 #2
    This does seem to be a misprint, you should have been given the locations of the explosions in *some* frame. If the explosions were done at opposite ends of the rod, you are on the right track to use Lorentz transformations to find the new x's. You might seek clarification of the information given in the problem.
     
  4. Oct 11, 2004 #3
    Ok assuming the explosions happened at opposite ends of the rod..

    x' = (x-vt) / sqrt(1-v^2/c^2) so...

    x'2 - x'1 = ((x2-vt) - (x1-vt))/sqrt(1-v^2/c^2)

    where x2 and x1 are the points at which the explosions occur in the O-frame and x'2 and x'1 are their O'-frame counterparts so picking appropriate x2 and x1..

    x2 = 10
    x1 = 0

    x'2 - x'1 = (10-0) / sqrt(1-(.7c)^2/c^2)
    = 10/sqrt(0.51)
    = 14m approximately.

    Is this correct? Edit I have 10 submissions but I already used 4 so I am asking ahead of time so I Don't waste them unnecessarily.

    NEvermind I got it. Silly me it was 14.
     
    Last edited by a moderator: Oct 11, 2004
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