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Why is scalar product of momentum in electron scattering conserved?

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Reading a textbook, I come across a situation where an electron is scattered off a nucleus. The book says p.P = p'.P', where p is the momentum of the electron and P is the momentum of the nucleus.
    I don't understand how it gets the conservation of scalar product.

    It's steps are:
    four-momentum [itex] p = (E/c , \vec{p}) [/itex]
    [itex] p.p = \frac{E^{2}}{c^{2}} - \vec{p}^{2} [/itex]
    It says: there is always a rest frame to be found
    therefore, [itex]m = \sqrt{p^{2}}/c [/itex]
    and obtain: [itex] E^{2} = p^{2}c^{2} + m^{2}c^{4} [/itex]

    now conservation of four momentum: [itex] p + P = p' + P' [/itex]
    square it to make it invariant: [itex] p^{2} + P^{2} + 2pP = p'^{2} + P'^{2} + 2p'P'[/itex]

    invariant masses are unchanged. Therefore, [itex]p^{2} = p'^{2} = m_{e}c^{2}[/itex]
    and [itex]P^{2} = P'^{2} = Mc^{2}[/itex]

    Agreed so far...But now it says:
    it follows that p.P=p'.P'

    I understand all of that except that last line.
    Why does it follow?




    [Edit]
    Solved now :/
    You just plug p.p=p'.p' and P.P=P'.P' back into [itex] p^{2} + P^{2} + 2pP = p'^{2} + P'^{2} + 2p'P'[/itex]
    they cancel and you get p.P=p'.P'.
    Sorry!
     
    Last edited: Apr 4, 2013
  2. jcsd
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