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Electromagnetic interactions and exchange particles

  1. May 16, 2015 #1
    Protons are in very close proximity with each other in a nucleus. This means there is constant electromagnetic interaction, of which the exchange particle is a photon. What determines the wavelength of this exchange photon? How do they exist in the nucleus: constantly being emitted, or staying inside the nucleus somehow?

    Also, how can a neutral pion be an exchange particle in, say, a proton-neutron collision. Being a strong interaction, aren't gluons the gauge bosons?

    I have a hard time thinking about this kind of stuff.
  2. jcsd
  3. May 16, 2015 #2


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    They do not, most photon exchanges are virtual.

    The gluon is the gauge boson. Pions mediate a residual force between colourless states with internal colour structure.
  4. May 16, 2015 #3
    Virtual, but still there and still existent surely. What happens to the photon?

    I'm just going to research the pion question. Residual force, colourless states, colour structures....
  5. May 16, 2015 #4


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    They exist in the hadrons in the same sense that gluons do, with the difference that gluons provide a larger part of the momentum when looking at the structure functions. You cannot say that a photon "exists" in the hadron.

    You might as well ask what happens to the photon during electrostatic interaction between two charges.
  6. May 16, 2015 #5
    Would you mind if I asked that now?
  7. May 16, 2015 #6


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    Not so difficult to research... All the quark bound states, like protons, neutrons etc, are color-neutral (they don't have a color charge)= colorless states. However their constituents (quarks, gluons) are colorful particles (they have a color charge) =internal color structures for Orodruin.
    Now how does the color neutral proton interact with the color neutral proton/neutron through a strong interaction? (here comes the "pions' residual force").

    As for the internal lines of the photons, they don't exist as photons. They are virtual particles. At the end, when you want to derive their total contribution, you integrate them out (that is what someone writes in Feynman Diagrams and not the actual process).
  8. May 16, 2015 #7


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    There are good arguments to call them mathematical artifacts. That is not "existing" with the usual meaning of the word.
    It's not like there would be some number of photons you could count.
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