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Relativity and pi-meson decay

  1. May 25, 2008 #1
    [SOLVED] Relativity and pi-meson decay

    1. The problem statement, all variables and given/known data

    A [tex]\pi[/tex] meson of mass [tex]m_{\pi}[/tex] at rest, decays into a muon of mass [tex]m_{\mu}[/tex] and a neutrino of zero rest mass, [tex]m_{v}=0[/tex]. Find the momentum of the muon.

    2. Relevant equations

    I think this is the right equation though I'm not sure if I should use four vectors.


    3. The attempt at a solution

    Total energy before the decay is the same as the total energy after the decay, so,




    Giving a total energy of,



    So the momentum is,


    I have a horrid feeling this is all wrong as it seems too easy. Please could someone let me know if I'm wrong in my reasoning?
    Last edited: May 25, 2008
  2. jcsd
  3. May 26, 2008 #2


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    Hi Vuldoraq,

    I think there is a math error in this last line; it doesn't seem to follow from the equation you have listed in the "Relevant equations" section. (Taking the square root doesn't just cancel the squares on the right hand side.)

    Also, I think you should also use conservation of momentum in this problem.
  4. May 27, 2008 #3
    Hey Alphysicist, thanks for the pointers

    Correcting my error would give the muon energy to be,


    Which means the total energy is,


    Giving, after moving stuff around,


    Does this look better?

    How would I apply cons of momentum in this problem?
  5. May 27, 2008 #4


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    About conservation of momentum: looking at the right hand side of your final expression, you know the masses of the pion and the muon, but you do not yet know the momentum of the neutrino, so you cannot evaluate the muons momentum which is what they ask for.

    So go ahead and write down the equation for momentum conservation. What does the fact that momentum is conserved indicate about the momenta of the muon and neutrino?
  6. May 27, 2008 #5
    Does it mean they have equal and opposite momenta? So the equations then become,


    Which makes the right hand side of my equation become,


    Which after rearranging and applying the quadratic formula becomes,

    [tex]P_{\mu}=(1/2)*(m_{\pi}c \ +/- \ c\sqrt{2m_{\mu}^{2}-m_{\pi}^2})[/tex]

    Which seems to be a negative momentum? Which value gives the actual momentum? Assuming I've gone right so far.
    Last edited: May 27, 2008
  7. May 27, 2008 #6


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    They should have equal and opposite momenta. However, in terms of the energy equations you have written, I think you should substitute:

    p_{\nu} = p_{\mu}

    because I believe we want the magnitudes here. (For example, you found for the neutrino that its energy is E = p c because it's massless. Even if the momentum is negative, the energy is positive so we just use the magnitude of p.)
  8. May 28, 2008 #7
    I see what you mean, this would then give a momentum of,


    Which reduces to,


    I don't understand why it's negative though?

    Thanks again for your help! :smile:
  9. May 28, 2008 #8


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    I'm having some trouble following the work because of the changes made along the way. However, if we look back at post #3:

    This looks okay to me.

    I think there is an error here. It should have a minus sign and should be:


    Now with this, plug in the momentum relation ([itex]p_{\mu}=p_{v}[/itex]) and solve for the momentum. I think it gives the same answer but without the minus sign. Do you get the answer?
  10. May 28, 2008 #9
    Hey Alphysicist,

    I plugged in the corrected value and my answer is the same, and no longer negative,


    Thanks for your help, I would never have got there without it.

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