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Homework Help: Lorentz Transformation Woes

  1. Sep 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Muons are created in the upper atmosphere (at a height of 3000 m) and plummet downward toward a detector at ##v=0.980c##. The mean lifetime of a muon is ##t = 2.20~\mu s##.

    Find the mean lifetime of a muon measured by an observer on the ground.
    Find the distance that the muon travels in its reference frame.

    Use the Lorentz transforms!

    2. Relevant equations

    x^{\prime} = \gamma\left(x-vt\right)\\
    t^{\prime} = \gamma\left(-\frac{v}{c^2}x + t\right)\\
    x = \gamma\left(x^{\prime}+vt^{\prime}\right)\\
    t = \gamma\left(\frac{v}{c^2}x^{\prime} + t^{\prime}\right)

    3. The attempt at a solution

    First, solve for ##\gamma##:
    \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-0.980^2}} = 5.03

    The mean lifetime of a muon from a ground based observer is easy. If we assume that the muon is in frame ##S^\prime##and the ground based observer is in frame ##S##, the time of decay as measured by the observer is:
    \Delta t = (5.03)\left(\frac{v}{c^2}(0)+(2.20~\mu s)\right)\\
    \Delta t = 11.1~\mu s
    My problem is the second part. If I use the Lorentz transform correctly:
    \Delta x = 3000~m = (5.03)\left(\Delta x^{\prime} + v(0)\right)\\
    \Delta x^{\prime} = 3000/5.03 = 596~m.
    What happens if I try to solve it the other way? Obviously the time interval is not zero:
    \Delta x^{\prime} = 596~m = (5.03)\left((3000~m) - v\Delta t\right)
    Solving for ##\Delta t##, I get:
    \Delta t = \frac{\frac{596~m}{5.03} - 3000~m}{-0.98c} = 9.80 \mu s
    Pretty cool, eh?
    What exactly does that mean? The time period between measurements for the ground observer is 9.8 microseconds. However, if you divide the distance (measured by the ground observer) by the velocity of the muon, you get:
    \frac{3000~m}{0.98c} = 10.2 \mu s
    The ground observer measures that the muon is alive for 10.2 microseconds in order to reach the ground. So... what is the physical meaning of ##\Delta t = 9.8~\mu s##? That time interval is faster than the 10 microseconds required for light to traverse 3000m.

    I am so confused. I expected that transforming from the primed to the unprimed coordinate system would give consistent answers. What did I do wrong?
    Last edited: Sep 9, 2013
  2. jcsd
  3. Sep 9, 2013 #2
    "Find the distance that the muon travels in its reference frame."

    The muon is at rest within its own frame of reference, so the distance it travels within its own frame of reference is zero. But another interesting question is "During the average time that the muon decays, how far does it travel in the frame of reference of an observer on the ground."
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