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Three-particle decay and momentum

  1. Nov 15, 2016 #1
    1. The problem statement, all variables and given/known data

    The decay of a neutron into a proton, an electron, and a neutrino is an example of a three-particle decay process. Use the vector nature of momentum to show that if the neutron is initially at rest, the velocity vectors of the three must be coplanar (that is, all in the same plane). The result is not true for numbers greater than three.

    2. Relevant equations


    3. The attempt at a solution

    I knew that 0 would be equal to the net momentum of the three particles in all 3 directions, but don't really know where to go from here. I don't even know how to start this problem.
     
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  3. Nov 15, 2016 #2

    PeroK

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    To get you started, the motion of any two particles defines a plane.
     
  4. Nov 15, 2016 #3
    I think I have a tentative answer but the more I think about it the less sense it makes to me, basically what I did was take m(proton)v(proton)+m(neutrino)v(neutrino)+m(electron)v(electron)=0
    Then, I defined plane as the motion V(proton) of the proton and motion V(neutrino) of the neutrino. Next I got m(electron)v(electron)=-(m(proton)v(proton)+m(neutrino)v(neutrino)). Then converted this to momentum, thus -(P(proton)+P(neutrino))=P(electron). Thus P(electron) is opposite in direction but equal in magnitude to the addition of the other two vectors. Thus, it can be inferred that as it's opposite it's in the same plane. Is this correct or am I completely wrong. Should I write more to specify exactly what is meant?
     
  5. Nov 15, 2016 #4

    PeroK

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    Think of the momenta simply as three vectors. And think of how to define your ##x,y,z## axes.

    I'm not sure I understand your explanation.
     
  6. Nov 16, 2016 #5

    PeroK

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    An alternative approach is to use Linear Algebra. If the original particle is at rest, what can you say about the three momenta vectors?
     
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