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Relativistic momentum

  1. Feb 7, 2006 #1
    I am having trouble with getting the right answer for this problem that is pretty simple and it is driving me insane.

    You start out with a pion that decays into 2 photons that split at an angle theta in opposite directions from the original pion.

    The velocity v of the pion is 2.977*10^8 m/s, with a mass m of 135 MeV.

    If E is the pions energy and E1 and E2 are the photons energy then we have:
    With E1=E1/c, E2=E2/c

    and so for momentum we have[tex] (P1-P2)Sin[\theta]=0[/tex] so we get [tex]P1=P2 so E1=E2[/tex]

    So we can write [tex]2(P1+P2)Cos[\theta]=Ppion[/tex]

    Now we can write [tex]2E/C*Cos[\theta]=\gamma*m*v[/tex]

    Which reduces to [tex]Cos[\theta]=\gamma[/tex]*m*v*c/2E but this does not give me the correct angle :( The correct angle should be 6.79 degrees for each photon but as you can see from my equation since v=2.977 I get [tex]Cos[\theta]=1/2(about)[/tex], can anyone find my problem before I go insane? Thanks
    Last edited: Feb 7, 2006
  2. jcsd
  3. Feb 8, 2006 #2

    Meir Achuz

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    E=m\gamma, not mv\gamma.
    Just write p/E=v/c=(2k\cos\theta)/(2k).
  4. Feb 8, 2006 #3
    Thanks, the book does it similar to how you solve it, but I like to be able to solve things in a way that I will remember on a test just in case I can't find the easy way. Thanks, I figured it was something stupid like that.
    Last edited: Feb 8, 2006
  5. Feb 9, 2006 #4

    Meir Achuz

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    The easy way is easier to remember. Physics is finding the easy way.
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