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Time it would take muon to reach the surface of the earth. Work check.

  1. Apr 17, 2013 #1
    So I'm just studying for my final on Saturday and I realized that I'm missing the answer to a question in my notes( probably missed that day). So I was that hoping someone could look over my attempt. Thanks :smile:!

    1. The problem statement, all variables and given/known data
    The muon is an unstable elementary particle with charge equal to that of the electron and mass 207 times that of the electron. In the laboratory muons have been observed to live 2.2[itex]\mu[/itex]s on average before disintegrating into an electron, a neutrino, and an anti-neutrino. Assume that cosmic muons are created 6.0 km above the earth's surface, in the upper atmosphere and that they travel with a speed of 0.998c. How much time would it take for such muons to reach the surface of the earth? Do they reach the earth?

    2. Relevant equations
    [itex]\gamma[/itex]= [itex]\frac{1}{\sqrt{1-(v/c)^{2}}}[/itex]
    [itex]\Delta[/itex]t=[itex]\gamma[/itex][itex]\Delta[/itex]t'
    d=vt


    3. The attempt at a solution
    [itex]\gamma[/itex]=[itex]\frac{1}{\sqrt{1-(0.998c/c)^{2}}}[/itex]=15.819
    [itex]\Delta[/itex]t=15.819(2.2x10-6)=3.48x10-5=34.8[itex]\mu[/itex]s
    d=vt=0.998(3x108)(3.48x10-5)=10419m=10.4 km ∴ the muon reaches the earth.

    Time it takes for muon to reach Earth:
    t= [itex]\frac{6000}{0.998(3x10^{8})}[/itex]=2.004x10-5=20.04[itex]\mu[/itex]s for the muon to reach the earth.
     
  2. jcsd
  3. Apr 17, 2013 #2

    mfb

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    The numbers are good, but "the muon reaches the earth" is problematic. Those 2.2µs are an average value - some of them decay quicker, some of them live longer. Most muons reach earth, but not all.
     
  4. Apr 17, 2013 #3

    HallsofIvy

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    Answering the first question, " How much time would it take for such muons to reach the surface of the earth?" does not require relativity because it is it is to be answered in the frame of reference of an observer on earth- and the data are given in that frame: to find the time to cover 6 km at .988 c, you have to divide the distance by the speed.

    To answer the question "Do they reach the earth?" You need to get the time in the muon's frame. If the time, in that frame, is less than 2.2μs, then the muon will reach the earth.
     
  5. Apr 17, 2013 #4

    rude man

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    As usual when dealing with speeds close to that of light, there always seems to be some ambiguity in the way problems are stated.

    I would assume that the observed lifetime should not be corrected by gamma. So their observed lifetime is 2.2 us and the distance similarly is just 6000 km. And the speed of 0.998c is similarly the speed observed from Earth. So my answer would be t = 6km * 1000m/km / 0.998*3e8 m/s = = 20 us. Which is >> 2.2 us so they never make it to Earth's surface. Which also happens to be the OP's answer, isn't that nice! But don't ask me why he/she dilated the lifetime then computed a dilated distance, only to wind up without either 'relativistic correction' to get the answer?

    Of course, if their observed lifetime is 2.2 us then their actual lifetime as measured in an inertial frame moving with the muons is only 2.2us/gamma.

    EDIT: after some research it turns out the 2.2us is not the observed lifetime in the lab but the lifetime of the muons in their own reference frame.

    So that changes things: the observed lifetime is now gamma*2.2us so they can travel 0.998c* gamma*2.2us ~ 10 km which means they will make it to earth's surface after all.
     
    Last edited: Apr 17, 2013
  6. Apr 17, 2013 #5
    I did the questions out of order. If they were in order it goes as:
    t=[itex]\frac{d}{v}[/itex]=[itex]\frac{6x10^{3}m}{0.998(3x10^{8}m/s)}[/itex]=2.004x10-5s=20.04[itex]\mu[/itex]s
    So it will take the muon 20.04[itex]\mu[/itex]s to reach the earth's surface.

    I'm not sure if this part is right, but the time in the muon's frame would be:
    [itex]\Delta[/itex]t=[itex]\gamma[/itex][itex]\Delta[/itex]t'
    where [itex]\gamma[/itex]=[itex]\frac{1}{\sqrt{1-(v/c)^{2})}}[/itex]
    so [itex]\gamma[/itex]=[itex]\frac{1}{\sqrt{1-(0.998c/c)^{2})}}[/itex]=15.82
    and, [itex]\Delta[/itex]t=15.82(2.2x10-6s)=3.487x10-5s=34.87[itex]\mu[/itex]s
    So therefore, the muon will not reach the earth's surface because it will decay first?
    I'm not very sure now and feel very confused. Can you explain to me how this should be done?

    Thank you,
    Melissa.
     
  7. Apr 17, 2013 #6
    Okay, so what I get from rude man is that the reference from the muon is actually [itex]\Delta[/itex]t'.
     
  8. Apr 17, 2013 #7

    rude man

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    That was before I found out that 2.2us was the lifetime in the muon's reference frame. In the earth frame it's gamma*2.2us.

    As I said, relativistic problems are often stated ambiguously. In this case it stated that the lifetime observed in the lab was 2.2us whereas it's really ~ 15.8 * 2.2us = 35 us. So they make it to earth's surface after all ...
     
  9. Apr 17, 2013 #8

    rude man

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    Since they live 35us as seen from earth they can cover 35us*0.998c = 10.5km but they only need to go 6km so they do make it to earth's surface.
     
  10. Apr 17, 2013 #9
    Thanks, rude man! Yeah I agree, these questions are always terribly stated.
     
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