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Particle Decay

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the decay A -> B + C (where A is not at rest). In the rest frame of A, B is emitted in a random direction (all directions have equal probability) and I need to show that in the lab frame, the energy distribution of B is uniform.

    (We assume that B has negligible mass)


    2. Relevant equations

    909c36cf59b2e9e46302e4c4fa1062ee.png

    (Let c=1)

    3. The attempt at a solution

    So I started by writing down the 4-momenta of A and B in the rest frame of A (choosing the momentum of B to be along the x axis):

    PA' = (mA,0,0,0)
    PB' = (EB',EB',0,0)

    In the lab frame:

    PA = (EA,(EA2-mA2)1/2,0,0)
    PB = (EB,EB,0,0)

    Lorentz boosting along the x-axis, I can determine the maximum and minimum energy that B can have:

    [tex]
    E_B^{\mathrm{min,max}} = \frac{E_A}{2} \bigg(1-\frac{m_C^2}{m_A^2}\bigg) \bigg(1 \pm \sqrt{1-\frac{m_A^2}{E_A^2}}\bigg)
    [/tex]

    using:

    [tex]
    \gamma = \frac{E_A}{m_A}\>,\gamma \beta = \frac{p_A}{m_A},\>E_A' = \frac{m_A^2-m_C^2}{2m_A}
    [/tex]

    I could also Lorentz boost in all other directions to get expressions for the energy. But I have no idea how to show that each of these energies in equally probable?

    Any help would be appreciated.

    Thanks!
     
  2. jcsd
  3. Aug 9, 2011 #2

    vela

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    Try calculating EA as a function of θ and show that the dE/dΩ is a constant.
     
  4. Aug 10, 2011 #3
    Ok, so using 3D spherical polars, I get [tex]E_B(\theta') = \frac{E_A}{2} \bigg(1-\frac{m_C^2}{m_A^2}\bigg) \bigg(1 + \sqrt{1-\frac{m_A^2}{E_A^2}}\>\mathrm{cos}(\theta ') \bigg)[/tex]

    (I defined it so that there is max. energy at θ'=0 and min. at θ'=π)

    Then dΩ = sin(θ')dθ'dφ' and so since EB is independent of φ', dE/dΩ is a constant?
     
    Last edited: Aug 10, 2011
  5. Aug 10, 2011 #4

    vela

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    Not quite. Nothing depends on φ', so you can integrate that out and deal only with θ', so you have dΩ' = 2π sin θ' dθ' where θ' goes from 0 to π. It's convenient to change variables to cos θ' so you have dΩ' = d(cos θ') where cos θ' runs from -1 to 1. (The minus sign goes into switching the order of the limits.) When you say that a decay is isotropic, that means that the distribution N is flat as a function of cos θ'. What you want to show is that dN/dEB is constant, using the fact that[tex]\frac{dN}{dE_B} = \frac{dN}{d(\cos \theta')} \frac{d(\cos\theta')}{dE_B}[/tex](which is simply an application of the chain rule).
     
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