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Another relativity question check =P

  1. Aug 26, 2008 #1
    1. The problem statement, all variables and given/known data

    It seems this question is a little too easy? I'm a little skeptical so could someone please do a quick check for me?
    The pion is an elementary particle that decays with a mean lifetime in its rest frame of 2.6x10^-8s. A beam of pions has a speed of 0.79c.
    a) In the frame of the pion, how far does the lab travel in one mean lifetime.
    b) what is the distance in the lab's frame

    2. Relevant equations

    distance = speed * time

    3. The attempt at a solution
    so for part a, I simply multiplied the speed and the time to determine how far the pion travels and then simply use that as the distance.

    for part b, the answer simply has to be zero because in the lab's frame, the lab doesn't move but the pion moves. I assume I can prove this through galilean transformations as well.

    These answers seem too simple for some reason and just want to make sure these answers are correct. Thanks! =D
  2. jcsd
  3. Aug 26, 2008 #2

    Doc Al

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    Staff: Mentor


    I'm sure that they mean the distance the particle moves in the lab frame. (Not how far the lab moves in the lab frame! :tongue2:)

    You'll have to use special relativity, not galilean, since the speed is a high fraction of lightspeed.
  4. Aug 26, 2008 #3
    Hehe the question didn't specify =P but in terms of the pion's distance in the frame of the lab, I used lorentz transformations to approach this problem and used the following equations:

    x' = [tex]\gamma[/tex](x - v*t)
    t = mean lifetime of the pion
    v = speed of pion
    x = location of pion itself which is zero if we used the pion's frame of reference as the original

    [tex]\gamma[/tex] = 1/[tex]\sqrt{1-(v^2/c^2)}[/tex]

    since the speed is given as .79c I imagine the c cancels out in the square root so that for [tex]\gamma[/tex] I get
    [tex]\gamma[/tex] = 1/[tex]\sqrt{1-(.79^2)}[/tex]
    with that in mind, I plugged in the other values for the lorentz transformation so that
    x' = [tex]\gamma[/tex](0m - 0.79c * 2.6X10^-8)

    I'm a little uncertain about the x but hopefully I have the general idea? Any advice would be great =P
  5. Aug 26, 2008 #4

    Doc Al

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    Staff: Mentor

    That looks fine. Here's another (equivalent) way to look at it. Consider the decaying pion as a moving clock and apply time dilation to figure out the decay time in the lab frame:

    [tex]\Delta t' = \gamma \Delta t[/tex]

    Then you can use the old "distance = speed * time", just like you did in the first part.
  6. Aug 26, 2008 #5
    ah I see, that is due to the fact both observers sees the same velocity and therefore we just need to apply the [tex]\gamma[/tex] multiplied by the time to find the t'(the term with x in it goes to zero)which will ultimately give us distance. Thanks for all the help ^_^
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