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Relativity: Breakdown in Simultaneity Question

  1. Apr 17, 2008 #1
    1. The problem statement, all variables and given/known data:

    We have a pole vaulter, and farmer, and a barn. The vaulter is traveling at 0.866c (as measured by the farmer), and carrying a pole that, in his reference frame, is 10 meters long. The barn has an open door at either side, and is 10 meters wide, in the farmer's reference frame. The farmer believes he can shut both doors simultaneously for an instant in time, with the vaulter inside, then open them again, without breaking the vaulter's pole. The vaulter believes that farmer cannot do this. Prove, that when properly analysed, the farmer and the vaulter agree on the outcome, and include both quantitative calculations and a written explanation. We are told to assume the walls (and doors) are paper thick, that the pole would break without slowing down, and that it is possible for the farmer to shut both doors simultaneously (in his reference frame) for the purposes of this question.

    2. Relevant equations:

    Obviously the one for length contraction:
    L = Lp/[tex]\gamma[/tex]
    where Lp is the proper length and gamma is:
    [tex]\gamma[/tex] = 1/[tex]\sqrt{1-(v^2/c^2)}[/tex]
    And then, part of why I am struggling with this: one to prove a breakdown in simultaneity.
    Possibly also time dilation: [tex]\Delta[/tex]t = [tex]\gamma[/tex][tex]\Delta[/tex]t'

    3. The attempt at a solution

    So far, I have, using the length contraction formula above, shown that the vaulter will measure the barn as length contracted to 5.000m(3d.p) and the farmer will measure the pole to be length contracted to 8.000m(3d.p).

    Thus the farmer can shut the vaulter in the barn for an instant in time, but in the vaulter's reference frame the two events (the first and second doors opening) will not be simultaneous.

    However, I need to mathematically prove the breakdown in simultaneity (the explanation is fine AFAIK, in terms of special relativity theory), and I don't know how to do this. Also, does time dilation affect the result? Obviously it takes place, but does it actually affect the result here?
     
    Last edited: Apr 17, 2008
  2. jcsd
  3. Apr 17, 2008 #2

    dynamicsolo

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    Homework Helper

    You won't be able to solve this by just using the dilation formulas. Simultaneity (or the lack of it) needs to be established by comparing the coordinates of specific events in each system. The point of the problem is to show that just using the formulas appears to lead to a paradox.

    Consider the following two events:

    A -- the farmer closes the lead door at position x = 10 m, t = 0 (the value for t doesn't really matter)

    B -- the farmer closes the trailing door at position x = 0 m, t = 0 (all that matters is that the time of event B is the same as for event A, in the farmer's reference frame).

    The vaulter's speed is v/c = 0.866 (along the x-axis), giving you a gamma of 2. Use the equations for transformation of coordinates to find the values of (x', t') for events A and B in the vaulter's reference frame. You should find that the distance separating the two events matches the result from the length contraction formula. You will now find, however, that the values of t' for the two events are not the same. That result is what establishes the lack of simultaneity in the vaulter's frame of reference.

    BTW, there is a problem with this statement. Since the barn and the pole have the same rest lengths and each is seen as moving at the same speed in the other reference frame, the farmer should see the pole contracted to 5 meters and the vaulter should see the barn contracted to 5 meters.
     
    Last edited: Apr 17, 2008
  4. Apr 18, 2008 #3
    Ah, I see what you're saying. I think I can take it from here.Thank you very much for your assistance.

    That would be because there is a typo in my original post. Very sorry about that. The pole, in the vaulter's frame, is 16m long, not10m. Which is why the calculation for length contraction works out at 8.0m, not 5.0m

    P.S this works out with the second door closing then opening again at (quite) approx. t'=-5... while the first is at t'=0, and both of them occur simultaneously at t=0, which proves the breakdown of simultaneity, and then also that at t'=~~-5, the the distance to the second door in the vaulters frame is around 19m, which is more than enough to avoid chopping off his pole. The first door still closes and opens just as the rear of the pole has passed under it. :) (I've only given rough values because I don't have my calculations sitting in front of me right this instant.)
     
    Last edited: Apr 18, 2008
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