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Homework Help: Relativity Help

  1. Jan 28, 2005 #1
    A muon is an unstable elementary particle with an average lifetime of 2.20×10-6 seconds (from the moment of creation until it decays) as measured by an observer at rest with the muon. If an average muon travels a distance of 900 meters during one lifetime, according to an observer in the laboratory, what is the muon's speed (in m/s)?

    I have the Lorentz Transformations sitting right in front of me but I can't seem to make the logical jump again. If O is the observer's frame and O' is the muon's frame of reference I have the following data:

    x = ?
    t = 2.2*10^6s
    x' = 900m
    t' = ?

    I really don't remember where to go from here. I tried taking the ratio of x to t and x' to t' the Lorentz transformations but it didn't yield anything relevant. Thanks in advance.
     
    Last edited by a moderator: Jan 28, 2005
  2. jcsd
  3. Jan 28, 2005 #2
    I tried using the fact that t' = [tex]\gamma[/tex]t and x' = x/[tex]\gamma[/tex] and v = x/t to yield v = x'[tex]\gamma[/tex]/t but the answer came out imaginary. What am I doing so wrong?
     
  4. Jan 28, 2005 #3

    Doc Al

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    Staff: Mentor

    Think in terms of the time and distance intervals between the birth and death of the muon. Here's what you are given:
    [tex]\Delta t = ?[/tex]
    [tex]\Delta x = 900[/tex]m
    [tex]\Delta t' = 2.2*10^6[/tex]s
    big hint:
    [tex]\Delta x' = 0[/tex]

    Now use the LT to find [itex]\Delta t[/itex], at least in terms of v. Then realize that [itex]\Delta x/\Delta t = v[/itex]. Solve for v.
     
  5. Jan 28, 2005 #4

    Doc Al

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    Staff: Mentor

    Careful of those "facts". While it's true in this case that [itex]\Delta t = \gamma \Delta t'[/itex], [itex]\Delta x \ne \Delta x' / \gamma[/itex]. Use the full LT; don't take shortcuts (until you have more experience and know when to use them).
     
  6. Jan 28, 2005 #5
    Well I got the right answer with your advice (thanks btw) but I don't understand why [tex]\Delta[/tex]x' = 0 still. It should be obvious but I apparently forgot all of relativity.
     
    Last edited by a moderator: Jan 28, 2005
  7. Jan 28, 2005 #6

    Doc Al

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    Staff: Mentor

    Remember that the primed coordinates are observations made in the "moving" frame attached to the muon. How far does the muon move in its own frame? :wink:
     
  8. Jan 28, 2005 #7
    Argh I see now. I completely switched the perspectives. For some reason I thought the decay time was measured from an observer standing on earth or something. Thank you!
     
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