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Proof problem

  1. Jan 20, 2006 #1
    Prove that a stick of proper length L has a length L' in a frame in which it moves with speed v along a line that makes an angle theta with it's length is given by

    [tex] L' = L \sqrt{\frac{1-v^2 / c^2}{1 - (v^2 / c^2)\sin^2{\theta}}}[/tex]

    My problem here is the picture I think. So the stick overall is moving with speed v, but the stick is not necessarily parallel to v, but at an angle created by the direction of v and the stick?
    Last edited: Jan 20, 2006
  2. jcsd
  3. Jan 20, 2006 #2

    Tom Mattson

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    You need to remember that your stick is only contracted in the direction of motion. So just for definiteness say that [itex]\vec{v}=v\hat{i}[/itex]. Then the stick is contracted in the x-direction but not in the y-direction.

    Once you have the two components in the frame in which the stick has velocity [itex]\vec{v}[/itex] you can find its total length.
    Last edited: Jan 20, 2006
  4. Jan 20, 2006 #3
    Got it. I went that direction earlier, but for some reason it didn't look as if the expression I derived was equivalent. BUT after some algebra..........
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