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Rest frame angular distribution of meson decay into two photons

  1. Dec 3, 2012 #1
    1. The problem statement, all variables and given/known data

    consider a pion decaying into 2 photons.
    In the rest frame of the pion, the two photons must emerge back-to-back photons are equally likely to emerge in any direction.

    determine the rest frame angular distribution of the emerging photons.





    2. Relevant equations

    ΔP=f(θ)Δθ is the probability that a photon emerges with polar
    angle θ within some infinitesimal interval Δθ in the angle,

    and f(θ) is the rest frame angular distribution.

    3. The attempt at a solution

    in the pion frame, the photons emerge back to back, and since there is no preferred decay direction, the probabilities of a photon emerging in any direction are the same.

    so f(theta) is a constant?
     
    Last edited by a moderator: Dec 4, 2012
  2. jcsd
  3. Dec 4, 2012 #2

    vela

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    I think the problem is asking you to calculate ##f(\theta)## for the lab frame. The pion is moving in this frame. If you were only looking at the decay in the the pion's rest frame, then yes, ##f(\theta)## would be a constant because the decay is isotropic.
     
  4. Dec 6, 2012 #3
    where did you find this problem.
     
  5. Dec 7, 2012 #4
    I paraphrased it from a hw problem. The actual hw question asks to explain why f(θ)=sinθ is the rest frame angular distribution. Which is even less reassuring...

    but the question doesn't specify the direction in which the pion is traveling in the lab frame, and θ is the usual polar angle, relative to some choice of z-axis.
     
    Last edited: Dec 7, 2012
  6. Dec 7, 2012 #5
    Say theta is taken from the direction of the pions momentum in the lab frame (call it +z). Then the pion decays into two photons, that when boosted to the lab frame, give the same total momentum as the pion initially.

    So if the photons emerge along the z axis, could boosting to the lab frame violate cons. of momentum? I dont see how but this might give rise to the sin(theta) distribution.
     
  7. Dec 7, 2012 #6

    diazona

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    Think about this: equal probability of emerging in any direction means the probability distribution is uniform over a two-dimensional sphere. You could write this as ΔP=f(θ,φ)ΔθΔφ. What would f(θ,φ) be in that case? Remember to think the properties of spherical coordinates.
     
  8. Dec 7, 2012 #7
    oh i see. picturing a sperical surface surrounding the pion (in its own frame), the probability of a photon emerging through some infinitesimally small surface element would be constant.

    for a sphere, dA=sin(theta)d∅dθ; ΔP=sin(theta)ΔθΔ∅

    phi and theta are orthogonal, so ignoring phi, ΔP=sin(theta)Δθ
     
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