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Conservation of momentum in relativistic disintegration

  1. Sep 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi everybody, I was studyng a photon disintegrating in one electron and one positron in the presence of a heavy atomic core; I want to find the final momentum of the atomic core



    2. Relevant equations

    Conservation of energy: Eγ=2me+Sqrt(Mn2+Pn2)
    Conservation of momentum Pγ=Pn
    Relation between the energy and momentum of a maseless particle Pγ=Eγ

    3. The attempt at a solution
    From the conservation of energy: (Eγ-2me)2=Mn2+Pn2; and since Pn=Eγ, operating after developing (Eγ-2me)2, I obtain Pn=4me-(Mn2/me). If I introduce the values in units of atomic mass, I obtain a negative value of Pn, the momentum of the atomic core.
    Can anyone please tell me what am I doing wrong?
    Thanks.
     
  2. jcsd
  3. Sep 30, 2015 #2

    haruspex

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    I get a different equation. Please post your working.
     
  4. Sep 30, 2015 #3
    This is what I have calculated:

    I start from Eγ=Pn and Eγ=2me+Sqrt[Mn^2+Pn^2]

    Eγ-2me=Sqrt[Mn^2+Pn^2]

    (Eγ-2me)^2=Mn^2+Pn^2

    (Eγ)^2+4(me^2)-4meEγ=Mn^2+Pn^2. Then, (Eγ)^2 on the left and Pn^2 on the right cancel each other.

    4me^2-4meEγ=Mn^2

    4me^2-Mn^2=4meEγ=4mePn

    Pn=(4me^2-Mn^2)/me=4me-(Mn^2)/me
     
  5. Sep 30, 2015 #4

    haruspex

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    Fine until the last step. However, correcting it will not fix your sign problem, so the error must be much earlier, in one of your initial equations. I am not a nuclear physicist, so cannot comment on those.
     
  6. Sep 30, 2015 #5

    jbriggs444

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    To be more correct, in units where c=1 and dealing with massless particles, |P| = E. If you got a negative momentum then you must have had a negative energy. If you have a negative energy then something does not add up.

    I am having a horrible time trying to decrypt the ASCII math in the original post without a cheat sheet for the variable names.
     
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