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Why in scattering processes q^2 is negative?

  1. Jun 7, 2012 #1

    not sure if I'm typing the question in the right place, but I encountered this question when going through Peskin (eq. 6.44, though it's not important). If p and p' are respectively the momentum of electron before and after the scattering, then q=p'-p is the momentum of photon joining the vertex. Why is that q^2<0 ?
    Last edited: Jun 7, 2012
  2. jcsd
  3. Jun 7, 2012 #2
    I am not familiar with that textbook, but that does not make any sense.

    p, p' and q are momentum vectors. Momentum is conserved in the scattering process,
    so p = p' + q. q^2 is the (length of the photon's momentum)^2 and has to be a positive, real number.
  4. Jun 7, 2012 #3
    I guess I should've noted that p,p' and q are the four-momentums. Of course, the square of the difference of the two momenta (p'-p)^2 can be negative, but for that purpose p and p' must obey certain conditions. I don't see why the fact that the process describes a scattering puts such conditions on the four-momentum of the electrons.
  5. Jun 7, 2012 #4
    In other words, why is the virtual photon in the electron-electron scattering proccess spacelike?
  6. Jun 7, 2012 #5
    ok, eventually found the answer in an old thread of this great site. here's a long discussion https://www.physicsforums.com/showthread.php?t=372021 , particularly on the second page there's an answer why that photon should be spacelike.
  7. Jun 7, 2012 #6


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    Indeed! It's the triangle inequality.
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