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Neutral Pion Mass from Its Decay into Two Photons

  1. Feb 11, 2017 #1
    1. The problem statement, all variables and given/known data

    A ##\pi^0## meson decays predominantly to two photons. Suppose the energies (E1, E2) and angle (##\theta##) between the emitted photons are measured. Find an expression for the ##\pi^0## mass in terms of E1, E2, and ##\theta##.

    2. Relevant equations


    3. The attempt at a solution

    In the pion rest frame, by momentum conservation, I know that the two photons are emitted with equal but opposite momenta.

    ##E'_{\pi} = m_{\pi}##
    ##E'_{\gamma} = p'_{\gamma}##

    ##E'_{\gamma 1} = p'_{\gamma 1} = E'_{\gamma 2} = p'_{\gamma 2} = \frac{E'_{\pi}}{2} = \frac{m_{\pi}}{2}##

    I know how to transform the photon energies into the lab frame, but I don't understand how to include the angular dependency.

    Thanks for any help you can give!
     
  2. jcsd
  3. Feb 11, 2017 #2

    Orodruin

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    You do not need to transform anything to the lab frame. You just need to apply conservation of 4-momentum and use invariant quantities.
     
  4. Feb 11, 2017 #3
    Photon 1 has four-momentum ##P_{1 \mu} = (E_1, p_x, p_y, p_z)##

    Photon 2 has four-momentum ##P_{2 \mu} = (E_2, -p_x, -p_y, -p_z)##

    The pion has four-momentum ##P_{\pi^0 \mu} = (m_{\pi}, 0, 0, 0)##

    I think that ##E_1 = E_2##

    ##M^2 = E_{tot}^2 - \vec{P^2}##

    ##M = 2E_1##

    I don't think that can be right, but I don't see any reason for a theta dependance...
     
  5. Feb 11, 2017 #4

    Orodruin

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    It is not right. It is not clear to me how you have accounted for 4-momentum conservation.
     
  6. Feb 11, 2017 #5

    vela

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    You're not going to get one in the CM frame, but one will appear when you Lorentz transform back to the lab frame. But that's the longer way to solve this problem. Try analyzing the problem in the lab frame.
     
  7. Feb 12, 2017 #6
    So i've done some research on using four vectors.

    ##M^2 = (P_{1 \mu} + P_{2 \mu})^2 = P^2_{1 \mu} + P^2_{2 \mu} + 2E_1E_2 - 2\vec{p_1} . \vec{p_2}##

    The squared terms are the same as dot products which are zero for a massless particle and I am left with ##M^2 = 2E_1 E_2 - 2 \vec{p_1} . \vec{p_2}##

    Since ##E' = p## for the photons I think I can write this as ##M^2 = 2 E_1 E_2 (1 - \cos \theta)##, and by employing a trig identity I can say that ##M = 2 \sqrt{E_1 E_2} \sin \frac{\theta}{2}##

    I only used (i think) properties of the four vector but I have found a theta dependence.
     
  8. Feb 12, 2017 #7

    Orodruin

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    This looks correct.
     
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