Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Divergence of forward Coulomb scattering?

  1. Sep 25, 2010 #1

    I have a question about the divergence of forward Coulomb (Bhabha/Moller) scattering.

    I guess the classical analog of it is the Rutherford cross-section divergence, but that can be explained by the infinite impact parameter.

    In the QED version - the Bhabha/Moller scattering, it is the matrix element for given states that diverges - not only the cross-section, and I can't see how two plane-wave particles can have an impact parameter that could resolve this divergence.

    Also, it seems that the divergence stems from the zero-momentum divergent photon propagator here. I saw explanations that this is typical for any infinite range interaction.

    Could somebody please explain what is the solution for this divergence. Is this an unphysical one? What was the wrong assumption that caused it?

    Zero-energy propagator probably means infinite-lifetime virtual photon. Does this has anything to do with the divergence?


    Coulomb interaction between 2 charged particles is about the 1st thing we learn in high-school after the Newton laws. Doesn't this bother anyone? Am I missing something here?

    Thank you.
  2. jcsd
  3. Sep 25, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    I would love to help you out, but it seems i do not understand this phenomena very well. Could you explain to me what "Divergence" means for this exactly?
  4. Sep 25, 2010 #3
    Thanks for the response. I guess I should formulate the question better. By "divergence" I mean an infinite result. This seems unphysical for probability amplitudes.

    The phenomenon here is: one electron is scattered by another one. In the matrix element for electron-electron scattering there is a photon propagator that has 1/(p1-p2) with p1 - initial electron 4-momentum, and p2 the final electron 4-momentum. So for forward scattering (p1=p2 with no deflection), the matrix element has division by zero => infinite amplitude.

    There are cases of other infinities for very high momenta that are explained by the wrong assumption of point particles. But I think this is not applicable here.

    Does this makes the problem clearer?
  5. Sep 25, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    Would you have scattering with 0 deflection? Also, from what i've just looked up on Bhabha Scattering, it is for electron-positron scattering. Is that what you meant or can you substitute an electron for that positron?
  6. Sep 25, 2010 #5
    I think yes - a very weak scattering. Or even if we decide to call this case "no scattering" - the calculation for this case is still problematic. I wouldn't be surprised if forward scattering probability would come out close to 1, but an infinite one looks like a mathematical problem

    Bhabha is electron-positron, and electron-electron is called Moller scattering, but both have the same problem.
  7. Sep 25, 2010 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    I'm sorry, i wish i could have helped you. All i really found at all was the matrix element picture at http://en.wikipedia.org/wiki/Bhabha_scattering that has what looks like your equation. All i saw there was 1/(k1-k2)^2

    Hopefully someone else will be able to help!
  8. Sep 26, 2010 #7
    Hi again.

    Well, I'm a bit surprised I didn't get any answers.

    If the subject doesn't look that interesting - I think it is interesting. Infinite probability without explanations of the simplest electron-electron scattering looks like a serious problem in QED, which is the most accurate physical theory we have.

    If it is because nobody has an answer - this makes it even more interesting. But my guess is that I just misunderstood something here, and I'd appreciate it very much if some one could point me to the issue.

    Thank you.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook