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Length contraction & muons

  1. Feb 11, 2017 #1
    1. The problem statement, all variables and given/known data
    A muon is created in the atmosphere 3 km above Earth's surface, heading downward at speed 0.98c. It survives 2.2 * 10-6s in its own frame of reference before decaying.
    Relativistitically, according to the muon, what is the distance from the point in the atmosphere where the muon is created to Earth's surface, and how much time will it take this distance to pass the muon?

    2. Relevant equations
    length contraction : L=L' /γv
    where L' is the proper length and γv is Lorentz factor : 1/sqrt(1-(v/c)2) and here is = 5.03

    3. The attempt at a solution
    What do I see in my solution is :
    The muon sees a moving 'object', specifically the region of atmosphere between the point where it is born and Earth's surface. These points are 3 km apart according to an Earth observer but according to the muon they are closer together. We should expect to divide 3 km by 5.03. It is Earth's frame in which this object is at rest, so L' =3000 m and L=L'/γv = 597m
    and the time is just t = L/v
    I misunderstand a thing! What I know: Length contraction is the reduction in length according to Albert Einstein's special relativistic theory occurs when an object is moving with great speed in relation to the measuring length.
    What if someone is standing on the earth and watching 3 km above the surface and the muon coming down, why do we say that L' = 3 km?! and I don't get this : It is Earth's frame in which this object is at rest... Why we say it is at rest in Earth's frame ?!
     
    Last edited: Feb 11, 2017
  2. jcsd
  3. Feb 11, 2017 #2

    kuruman

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    If you watch the muon come down, you are at rest relative to the Earth. The muon is moving at 0.98c. The 3 km apart is what you measure, namely the proper length. The length measured by the muon is (as you said) 597 m.
     
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