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How does force mediator transfer generate an attractive force?

  1. Jul 14, 2010 #1
    Hello!

    Im hoping people can answer this for me, as ive not been able to find the answer anywhere else:

    As far as I understand it (and please correct me if im wrong), the strong force is caused by the transfer of a gluon from one quark to another, electromagnetism by the transfer of photons and so.

    My problem is that I can't find anywhere why this transfer actually causes them to move closer together / attract. And with electromagnetism sometimes repel.

    Help!
     
  2. jcsd
  3. Jul 14, 2010 #2
    When a nail gets attracted to a bar magnet , Im pretty sure they don't actually exchange photons , its just a way to interpret it ,
     
  4. Jul 15, 2010 #3
    So if they dont exchange photons, then why do they move closer together? What is it that causes the attraction / repulsion?
     
  5. Jul 15, 2010 #4
    im not sure anybody knows , i hope someone can correct me ,
     
  6. Jul 16, 2010 #5
    i too want an answer.

    i want to know about gravitons (if real) are they like hands that reach out across space from the sun & hold earth in place?
     
  7. Jul 16, 2010 #6
    i thought a graviton was the excitation of the field .
     
  8. Jul 17, 2010 #7
    I was watching a video showing two guys throwing a ball between them. The ball was the photon for magnetism & they were getting closer with every throw to each other. Thats the best my understanding is. But really feel there is so much more to it.
     
  9. Jul 17, 2010 #8

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    At some point you have to simply accept that an analogy that you read in a popularization is not perfect. To actually understand this requires a lot of work - the easy path of the analogy only takes you so far.
     
  10. Jul 17, 2010 #9
    Is there any basic links online that I can read to better understand?
     
  11. Jul 17, 2010 #10
  12. Jul 21, 2010 #11
    Humanino, I read the FAQ you posted- thankyou. It helped immensely, except for I don't understand the logic behind one "step".

    "When I include all of these possibilities, it turns out that I can approximate the photon's momentum-space wave function usably well by the following: the wave function is a function proportional to the electric charge of the emitting particle (in a sense this defines what electric charge is), and it has a few big, narrow spikes in it. One spike is proportional to -i times the charge, and is to the left of the origin; the other spike is minus that and is to the right of the origin."

    Why does the momentum-space wave function look like that? Im assuming the author is glossing over why for simplification- could you (or anyone) expand please?

    Thanks
     
  13. Jul 21, 2010 #12
    What he calls the "wave-function" for the photon is just the propagator (as he mentions) which can be defined in non-relativistic scattering as the Fourier transform of the potential evaluated at the momentum transfer. There are not great depths in this sentence, it is just
    [tex]\langle p_f | {\cal H}_\text{int} | p_i\rangle = \int \text{d}x\, e^{-ip_f x}V(x)e^{ip_i x} = \int \text{d}x\, e^{iqx}V(x) = \tilde{V}(q)[/tex]
    In order to (over)simplify, he chooses the potential to be a sinusoid (whose Fourier transform is just the difference of two delta functions).

    The essential point of the discussion is that attraction or repulsion occurs only in the full amplitude including no-photon-exchange.

    I hope that helps. I may have some signs wrong w.r.t. his conventions, I certainly omitted (square root of) pi factors. If it remains obscure, please let us know.
     
  14. Jul 24, 2010 #13
    Sorry, still dont get it. What I dont understand is why the spike to the left of the origin is positive, and the spike to the right negative.

    If a photon is emitted by something my thoughts would be that its momentum wave-function would be the same in all directions, not plus in one and minus in another.
     
  15. Jul 25, 2010 #14
    It's an excellent question, right on in fact, and I wanted to discuss it more in depth, but the week end is gone and I still did not have time.

    The easiest way to see that it must be that way, Baez mentions : the potential is real, so the imaginary part of its Fourier transform must be odd.

    But it has wider implications related to causality. He briefly mentions the "trick" due to Feynman to formally write emission from left and absorption at right, while taking care of the reciprocal. This is in fact the sign difference between merely the Green function and the Feynman propagator. If you are familiar with it, it is also the [itex]-i\epsilon[/itex] prescription which produces automatically the right path around the poles.

    I hope somebody else can contribute, otherwise I'll have to come back to it later.
     
  16. Jul 27, 2010 #15
    Thanks for all the help humanino, much appreciated.
     
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