1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?

    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'.
    Last edited: Apr 4, 2013
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted