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Time dilation and length contraction problem

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Anna is on a railroad flatcar moving at 0.6c relative to Bob. Her arm is outstretched in the dirction the flatcar moves, and in her hand is a flashbulb. According to her wristwatch, the bulb goes off at 100 ns. This even according to Bob differs by 27 ns.
    a) is it earlier or later than 100 ns?
    b) how long is Anna's arm?


    2. Relevant equations

    lambdav = 1 / sqrt[(1-v2) / c2]


    3. The attempt at a solution
    a) I'm a little bit confused by time dilation. I think that since Anna is moving according to Bob's frame of reference, he sees a longer time interval for her. Therfore, his watch would read later than 100 ns. I think?


    b)lambdav = 1 / sqrt[(1-v2) / c2]
    = sqrt(1-0.62)

    =1.25

    I'm confused about where to go from here. I need the length of Anna's arm. I know that this is the length from her hand to her center of mass. So if she is moving at 0.6c relative to Bob, and he sees the light go off at 127 ns,
    d = vt
    = 0.6c(127)
    =76.2 cm

    L0 = L x lambdav
    = 76.2 x 1.25
    = 95.25 cm

    So I found the length of her arm to be 95 cm. However, I have a feeling this isn't correct. I'm not sure I used the right formula...
     
  2. jcsd
  3. Sep 26, 2011 #2
    OK, this is slightly unclear at the moment but I think the intention is that Anna ('s body) passes Bob at t = 0. This will become important!

    OK, not a bad start (although I'd call this quantity [itex]\gamma[/itex]). Any other equations that might be relevant?

    Yep, that's right.

    So far so good.

    Not quite. 0.6c is Anna's velocity; 127 ns is the time interval in Bob's reference frame between her passing him and the flash going off. So you can't multiply these numbers together directly like this. Hence why I asked whether you knew any further equations that might be useful. Later on in your post you used the (correct) formula [itex]L' = L\gamma[/itex] for the transformation of a length interval; do you know similar formulas for transforming a position or time coordinate?
     
  4. Sep 28, 2011 #3
    Okay. So does that mean that I should use the formula t = gammav x t0 ?

    Then I could plug in 100 x 10-6 for t0 since this is how long it took in anna's frame of reference?
    This gives me 1.25 x 10-5 as my t. and then can I use this t instead to find length?
    But I still get a very similar answer when I do this.
     
  5. Sep 30, 2011 #4
    This is also for a time interval. Do you understand what I mean by the difference between a coordinate and an interval? And do you recognise the formula, say,

    [itex]t' = \gamma\left(t - \dfrac{vx}{c^2}\right)[/itex] ?
     
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