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Homework Help: Relativity: disagreement between two results

  1. Sep 30, 2006 #1
    I have this problem, which I think I can solve in 2 ways, but only 1 of them seems to be correct, but I don't understand why the other isn't correct. The problem is as follows:

    "A bar with the rest length L_0 moves in its length direction with the velocity v through a lab. A particle moves along the same line but in opposite direction with the same speed v [velocity -v]. How long will it take the particle to pass the bar, [time] mearsured from the lab."

    The easy way to do this, is by saying that if L is the length of the bar seen from the lab, then the time the particle will take to pass the bar will be

    [tex] T = \frac{L}{2v} = \frac{L_0}{2v\gamma} [/tex]
    (this is also the result given in the solutions manual)

    But if we look at the situation from the bars reference frame, then the particle will be moving with a velocity

    [tex] v' = \frac{2v}{1+v^2/c^2} [/tex]

    So the time the particle will take to pass the bar seen from the bars intertial frame will be

    [tex] T' = \frac{L_0}{v'} = \frac{L_0(1+v^2/c^2)}{2v} [/tex]

    This time T' seen from the lab most be

    [tex] T = \gamma T' = \gamma(1+v^2/c^2)\frac{L_0}{2v} [/tex]

    But this last result ain't equal to the first one. Why is that so? Where did I make a mistake in my last calculations?
  2. jcsd
  3. Sep 30, 2006 #2

    Andrew Mason

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    No mistake. The observers in different inertial systems disagree on when the particle passes the bar. This is an illustration of the relative nature of simultanaeity.

  4. Sep 30, 2006 #3


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    I must say this kept me stumped for a few minutes, but the solution is simple.

    You used the time dilation formula to find T ([tex]T=\gamma T'[/tex]), but this you may not do here.
    Remember how time dilation can be derived from the Lorentz transform:
    [tex]\Delta t=\gamma(\Delta t'+\frac{v}{c^2}\Delta x')[/tex]
    For a clock stationary in S' we have [tex]\Delta x'=0[/tex], i.e. two subsequent ticks of the clock occur at the same location in S'. However, the positions of the events also play a role in determining the time interval between the events in S.

    Let's say event 1 is that the particle crosses the front of the bar and event 2 is when it crosses the back end of the bar. Then [tex]\Delta x'=-L_0[/tex] and [tex]\Delta t'=\frac{L_0}{2v}(1+v^2/c^2)[/tex]

    Plug that into the Lorentz transform above and you get the same answer.
    Last edited: Sep 30, 2006
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