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Question about angular muon flux

  1. Jul 16, 2013 #1
    Hi,

    The Gaisser parametrization is something like:

    [tex]\Phi_0(E,\theta)=AE^{-\gamma}\left[ \frac{1}{1+1.1E\cos(\theta)/E_{\pi}}+\frac{0.054}{1+1.1E\cos(\theta)/E_K} \right] [/tex]

    where [tex]A,\gamma,E_{\pi},E_K [/tex] are constants. My question is: Is this saying that at higher zenith angles the flux of muons is stronger? Is there a reason for this?

    Also, would this say that the intensity would be higher at an angle than at vertical for muons crossing the same opacity(=distance)?

    Any input would be appreciated.

    Thank you,

    Safinenko
     
  2. jcsd
  3. Jul 16, 2013 #2

    mfb

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    Staff: Mentor

    The flight length is shorter, so less muons decay (or get absorbed) before they reach the ground.

    Why?
     
  4. Jul 16, 2013 #3
    Hi, thank you for your input.

    From what I have always understood, [itex]\theta[/itex] in the function was the zenith angle (angle from vertical). For example, here on top of page 3 the angle is defined from the vertical. So for [itex]\theta=0[/itex], [itex] E [/itex] in the fraction is multiplied by 1; for [itex]\theta=60[/itex], [itex] E [/itex] is effectively divided by half. So this leads me to think that angular muons have higher intensity than vertical. But obviously the distance muons travel vertically is shorter than at an angle!

    Basically, this is the part that I am confused about.

    Safinenko
     
  5. Jul 16, 2013 #4

    mfb

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    The division by 2 indicates that the flight length doubled - you need twice the energy to counter this with time dilation. Therefore, E cos(θ) is a measure for the time the muon has to survive (in its own frame) to reach the ground.

    No. If you want to make this comparison, compare muons with fixed energy.
    In addition, take the solid angle into account: There is more solid angle between 79° and 80° compared to the region between 0° and 1° (as measured from the vertical axis).

    If you want to take ##\phi## into account:
    $$\Phi_0(E,\theta,\phi) =sin (\theta) \Phi_0(E,\theta)$$
     
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