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Lorentz transform and muon half-life, Help, please!

  1. Jun 1, 2004 #1
    I am new to this matter, but I was reading a problem about how as the muon travels at .98c, that it's half life is increased by an approximate factor of 5.

    What I have trouble with is when using the Lorentz transform, why time dilation is calculated in the moving system at the same X and V so that the term VX/C^2 goes out, giving us a simple form, when it would seem that if the muon was going at .98 speed of light that it would move through space.

    Thus, how do you keep these things straight, and get these simple results? Thank you, bob
     
  2. jcsd
  3. Jun 1, 2004 #2

    HallsofIvy

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    Perhaps I do not understand the question. Yes, of course, if a muon is traveling at .98 the speed of light it is would move through space- very fast! Time dilation is NOT calculated "in the moving system" since, if it were, v would be 0 and there would be no time dilation. if v= .98 c then v2/c2= .982= .9604 so 1/√(1- v2/c2)= 5 approximately. What "simple result" are you talking about?
     
  4. Jun 1, 2004 #3
    You are transforming to a spacetime frame of reference in which the muon is at rest. This means you are comparing improper time (a physical moving clock) to proper time (a stationary clock). In that reference frame, X' remains constant.
     
  5. Jun 1, 2004 #4
    I meant the Lorentz transorm

    We have the form t' = (t-vx/c^2)/sqrt(1-(v/c)^2), yet the form simplifies to

    T' = T/(sqrt(1-(v/c)^2) so that we have dropped some terms, but now I understand that time dilation requires that velocity be 0, as you have kindly explained. Thank you!
     
  6. Jun 2, 2004 #5

    Doc Al

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    Careful with those clocks

    I'm not sure what you mean by saying that time dilation "requires that velocity be 0". It is only when a moving clock is observed that "time dilation" is evident. If the clock isn't moving (with respect to the observer) there is no time dilation (or any other special relativistic effects).

    It will be easier to understand how the "time dilation" formula is derived if you express the Lorentz transformation like this:
    Δt' = (Δt-vΔx/c^2)/sqrt(1-(v/c)^2)

    Now, if you have a clock in the unprimed frame that measures a time span of Δt, how much time will the primed system (moving with speed v) observe to have elapsed (this is Δt')? Use the LT to find out. In the unprimed system the clock just sits there, so Δx = 0. So...
    Δt' = (Δt)/sqrt(1-(v/c)^2)
    That's the so-called time-dilation formula.
     
  7. Jun 3, 2004 #6
    When two frames are in uniform relative motion, the spacetime interval in one frame is always equal to the spacetime interval in the other - and when the interval is calculated in one frame according to a clock which measures proper time in that frame, the end points of the interval are subtractive so the vx/c^2 factor disappears - one implication of this result is that is that the controversial postulate of one way light velocity is not essential to the ultimate transform of the spacetime interval from one frame to the other, that is, one can derive the interval transforms referred to in the author's original post w/o Einstein's second postulate.
     
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