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Homework Help: Finding the energy and momenta of massless particles

  1. Oct 3, 2007 #1
    1. The problem statement, all variables and given/known data
    A pion spontaneously decays into a muon and an antineutrino according to pion^1- => muon^1- +antineutrino.. Current experimental evidence indicates that the mass m of the antineutrino is no greater than about 190 keV and may , in fact, be zero. Assuming that the pion decays at rest in the laboratory , compute the energies and momenta of the muon and muon anti-neutrino a) if the mass of the anti-neutrino is zero and b) if its mass is 190 keV . The mass of the pion is 139.56755 MeV/c^2 and the mass of the muon is 105.65839 MeV/c^2

    2. Relevant equations

    E=pc when (m=0)
    E^2=(pc)^2 +(mc)^2

    3. The attempt at a solution

    For part a) since the mass of the anti-neutrino is zero , I apply use the equation E=pc to find the momentum and the Energy for both the muon and anti neutrino particles. To Find E(muon), E(muon)=E(pion)-E(antineutrino) = m(pion)*c^2 - m(antineutrino)*c^2= (139.56755 MeV/c^2 )*c^2 - (105.65839 MeV/c^2)(c^2)= 39.9092 MeV => (E(muon)^2)/c^2= c^2((p(muon))^2) +(m(muon))^2, c^2 cancel out and so I'm left with (p(muon))^2 = (E(muon))^2-(m(muon))^2) => p(muon)=sqrt((E(muon))^2-(m(muon))^2) ). I don't understand how to obtain the momentum and energy of the neutrino .
    Last edited: Oct 3, 2007
  2. jcsd
  3. Oct 3, 2007 #2


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    0 = p(muon)+p(antineutrino)

    p(antineutrino) = - p(muon)

    It is important to remember that in E = pc, p is the magnitude of the momentum... so as not to be confused by the minus signs.

    assume p(muon) is towards the right and positive...

    E(pion)=E(muon)+E(antineutrino) (1)

    now using the equation: E^2 = (pc)^2 + (mc^2)^2, what is E(pion), E(muon) and E(antineutrino). You can get E(muon) and E(antineutrino) in terms of p(muon).

    plug in all 3 into equation (1), and then solve for p(muon). Remember that E(antineutrino) is not just m(antineutrino)*c^2, because the antineutrino is not at rest.
  4. Oct 3, 2007 #3
    What equation is equation(1)?
  5. Oct 3, 2007 #4


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    It's conservation of energy.
  6. Oct 4, 2007 #5
    So then , E(muon)=(p(muon)*c)
    and E^2(anti-neutrino)=(-p(muon)*c)^2 +(m(anti-neutrino)*c^2)^2=E(anti-neutrino)=sqrt((-p(muon)*c)^2 +(m(anti-neutrino)*c^2)^2). Therefore, E(pion)=E(muon)+E(neutrino)=> E(pion)= (p(muon)*c)+(sqrt((-p(muon)*c)^2+(m(anti-neutrino)*c^2)^2). Now I've run into a dilemma: How do I solved for E(pion) if all relevant terms are suppose to be in terms of p(muon)? Why wouldn't the energy of a pion be zero since it decays at rest? since I'm given the mass of the pion in the problem , would the Energy of a pion be : E(pion)=m(pion)*c^2 since the pion particle decays at rest?
    Last edited: Oct 4, 2007
  7. Oct 4, 2007 #6
    please somebody, answer my last response
  8. Oct 4, 2007 #7


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    I can't respond, because I can't make any sense out of it. I suspect other people feel the same way. E(muon)=c*p(muon) is only ok if the muon is massless. And it's not. And then it gets worse. Can you listen to what learningphysics is trying to tell you and clean that mess up?
  9. Oct 5, 2007 #8
    Okay, to find E(pion), should I assume that E(pion)=m(pion)*c^2., since the pion particle is at rest, right?

    I know , p(antineutrino)=-p(muon).

    To find E(anti-neutrino) , I used the equation E^2=(pc)^2 +(mc^2)^2. Therefore, E=sqrt((-p(muon)*c)^2 +(m(muon)*c^2)^2)

    I supposed to find E(muon) , I apply the same equation for Energy that I did for the anti neutrino particle. E^2(muon)=(p(muon)*c)^2 +(m(muon)*c^2)^2.

    On second thought, since the mass of the anti neutrino is zero in this case, I don't even need E^2(anti-neutino)=(-p(muon))*c)^2 +(m(anti-neutrino)*c^2)^2. Instead , I can say E(antineutrino)=-p(muon)*c

    Therefore, E(pion)=E(antineutrino)+E(muon)=> m(pion)*c^2 = -p(muon)*c +sqrt((p(muon)*c)^2 +(m(muon)*c^2)^2)
    Last edited: Oct 5, 2007
  10. Oct 5, 2007 #9


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    That's better. Now you didn't lose me till the last line. Let's drop the c's, ok? You can always put them back later. You have E(pi)=E(mu)+E(nu). Yes, E(pi)=m(pi). Since you figured out that the momenta are equal, you have E(mu)^2-m(mu)^2=E(nu)^2-m(nu)^2. So:


    m(pi) and m(mu) are both known. Once you put in a value of m(nu) (either 0 or 190kev) these are two equations in two unknowns E(mu) and E(nu). Just solve them.
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