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Homework Help: Pion Decay Process

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data

    A pi+ meson can undergo the following decay process:

    pi+ --> muon+ + muon neutrino

    where the rest mass of the muon neutrino can be considered as zero. Show that the energy and momentum of the muon+ in the rest frame of the pi+ are given by:

    [tex]E^{'}_{\mu}=\frac{(m^{2}_{\pi}+m^{2}_{\mu})c^{2}}{2m_{\pi}}[/tex]
    [tex]p^{'}_{\mu}=\frac{(m^{2}_{\pi}-m^{2}_{\mu})c}{2m_{\pi}}[/tex]

    2. Relevant equations

    Lorentz transformations for energy and momentum.

    3. The attempt at a solution

    In the rest frame of the pion
    [tex]p^{'}_{\mu}=-p^{'}_{\nu}[/tex]

    and i know
    [tex]E_{\nu}=p_{\nu}c[/tex]

    I have attempted the question but i can't get the mass terms in the lab frame on one side and the energy and momentum of the muon in the rest frame of the pion+ on the other side. I feel like I require more information.
     
  2. jcsd
  3. Feb 6, 2010 #2

    kuruman

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    What exactly have you attempted? Did you start by writing energy and momentum conservation equations? If not, that's what you need to do. You do not need more information.
     
  4. Feb 6, 2010 #3

    vela

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    What lab frame? The problem only involves the rest frame of the pion.
     
  5. Feb 6, 2010 #4
    This is what I have attempted:

    In the lab frame (S):
    [tex]p_{\pi}=p_{\mu}+p_{\nu}[/tex]
    [tex]E_{\pi}=E_{\mu}+E_{\nu}[/tex]

    then i tried isolating the neutrino momentum, as it doesn't appear in the answer:
    [tex]p_{\nu}=p_{\pi}-p_{\mu}[/tex]
    [tex]p^{2}_{\nu}=(p_{\pi}-p_{\mu})^{2}[/tex]
    [tex]p^{2}_{\nu}=p^{2}_{\pi}+p^{2}_{\mu}-2p_{\pi}p_{\mu}[/tex]

    The lorentz transform for:
    [tex]p^{'}_{\mu}=\gamma(p_{\mu}-\frac{vE_{\mu}}{c^{2}})[/tex]
    [tex]p^{'2}_{\mu}=\gamma^{2}(p_{\mu}-\frac{vE_{\mu}}{c^{2}})^{2}[/tex]

    I multiplied the last part out, was thinking of eliminating the muon energy term with:
    [tex]E^{2}=p^{2}c^{2}+m^{2}c^{4}[/tex]
    But, i think there is an easier way to go about it.
     
  6. Feb 6, 2010 #5

    kuruman

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    There is no easier way. Eliminating the muon energy term as you suggested is the way to go. Note that as vela suggested, what lab frame? You are to do this in the rest frame of the pion. What are the initial energy and momentum in that frame?
     
  7. Feb 6, 2010 #6
    Well, simply multiplying out the expression of the LT squared (above) gives the answer in terms of only the mass of the muon. I need another relation, to get the mass of the pion involved in the expression. That is why i have working in the lab frame...
    What I get so far:
    [tex]p^{'2}_{\mu}=(2\gamma^{2}-1)\gamma^{2}m^{2}_{\mu}v^{2}+\frac{v^{2}}{c^{2}}m^{2}_{\mu}-2\gamma^{2}v^{2}m^{2}_{\mu}[/tex]
     
  8. Feb 6, 2010 #7

    kuruman

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    Please answer my previous question
    Gamma is not involved here.
     
  9. Feb 6, 2010 #8
    In the rest frame of the pion (S'), i think
    [tex]E^{'}_{\pi}=m_{\pi}c^{2}=E^{'}_{\mu}+E^{'}_{\nu}[/tex]
    [tex]p^{'}_{\pi}=0[/tex]
     
  10. Feb 6, 2010 #9

    vela

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    You got to [itex]p^{2}_{\nu}=p^{2}_{\pi}+p^{2}_{\mu}-2p_{\pi}p_{\mu}[/itex]. What are the various terms equal to in terms of the masses and energies?
     
  11. Feb 6, 2010 #10

    kuruman

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    Correct. Now write the momentum conservation equation as

    [tex]p_{\mu}'+\frac{E'_{\nu}}{c}=0[/tex]

    Express E'μ in terms of p'μ in the energy equation and you will have a system of two equations and two unknowns, E'ν and p'μ.
     
  12. Feb 6, 2010 #11
    Thanks very much. I finally worked it out, however there's a mistake in the question.. it should be:
    [tex]p^{'}_{\mu}=-\frac{(m^{2}_{\pi}-m^{2}_{\mu})c}{2m_{\pi}}[/tex]

    The energy answer is fine.
     
  13. Feb 6, 2010 #12

    kuruman

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    What does the negative sign mean? In such problems one is looking for the magnitude of the vector rather than the vector itself, which could be positive or negative depending on how you choose your axes.
     
  14. Feb 8, 2010 #13
    yeah, that's a good point, something you don't realise when you're rushing through these problems. Since there is no directionality, the momentum takes the absolute value.
     
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