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Pion decay

  1. Feb 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Massless Particles
    A neutral pion traveling along the x axis decays into two photons, one being ejected exactly forward and the other exactly backward. The first photon has three times the energy of the second. Prove that the original pion had speed 0.5c.

    2. Relevant equations
    for m=0, E=p*c
    conservation of Energy E^2=(c*p)^2+(m*c^2)^2
    Beta = v/c

    3. The attempt at a solution

    sum(momentum photons) = sum (momentum pion)
  2. jcsd
  3. Feb 15, 2008 #2
    The photon energy is easy to compute in a frame where the pion is at rest.
    Use the formula for the relativistic doppler effect to find the frequency of both fotons in a frame that moves with a speed v. The frequency of the forward photon must be 3 times the
    frequency of the backwards photon.
  4. Feb 15, 2008 #3


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    Are you familiar with four-momentum conservation?

    You may use [tex] P_\pi = P_1 + P_2 [/tex] where 1 and 2 refer to the photons. After squaring and using P^2 = m^2 c^4 , so that P_1^2 = P_2^2 = 0, you get

    [tex] m_\pi^2 c^4 = 2 E_1 E_2 - c^2 {\vec p}_1 \cdot \vec{p_2} [/tex]

    now, using the fact that the two photons move in oppposite directions, you find that [tex] 4 E_1 E_2 = m_\pi^2 c^4 [/tex]. Using the fact that one photon has three times the energy of the other one, you then have the energy of each photon in terms of the pion mass.

    Then use [tex] E_1 + E_2 = \gamma m_\pi c^2 [/tex] to find the speed.
  5. Feb 2, 2010 #4
    I don't understand this. can you try explaining it in more detail?
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