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How photon is force carrier of electromagnetic force in macroscopic level.?

  1. May 26, 2012 #1
    i have gone through that photon is force carrier of em force..but in macroscopic level as far as I studied I know em force is designated with field in macroscopic rather than a particle like interaction.so how can they justify that em force is managed by photon.If photon is more plausible is it right that what ever I read and what ever problems I ve done in undergraduate level using field like interaction is wrong :@
    I reckon maxwell's laws are also based on field concept!
    Thanks for reading out this and helping me
  2. jcsd
  3. May 26, 2012 #2


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    The best theory about elementary particles, bulk matter built by them and their interactions (except gravity) today is relativistic quantum field theory. A particle-point of view is extremely delicate and thus the attempts of popular-physics books authors to explain forces as the exchange of particles quite strange to a theoretical particle physicist.

    This becomes clear, when one asks precisely your question about the Coulomb force between macroscopic bodies. It can not be understood easily in the particle picture, let alone by the exchange of photons. It is very difficult to make sense (both theoretically and experimentally) of something like a photon position or localization since there is not even a well-defined position operator for photons to describe its location. What you can determine are reactions of a measuring device (like a photomultiplier or CCD plate) with the electromagnetic fields aka. photons. It is also not so easy to make sure that one measures really single-photon states or only very-low intensity coherent states, but that's not the topic here.

    The understanding of the Coulomb force for heavy/macroscopic charged bodies within quantum field theory amounts to the coherent resummation over infinitely many Feynman diagrams, which in the very mathematical language of quantum field theory are called photon-exchange diagrams. You sum over all the n-photon exchange diagrams with [itex]n=1,2,3,\ldots[/itex], where no photon lines cross (that's called the resummation of ladder diagrams). This resummation becomes important, when the momenta of the "exchanged photons" (these are no real photons but just mathematical descriptions of the electromagnetic field in terms of its (free) propagator) become small, because then the growing powers of electromagnetic coupling constants in the numerator of these mathematical expressions are compensated by the soft-photon-propagator denominators which becomes nearly zero in this kinematical regime. After this resummation, what comes out is nothing else than the classical Coulomb force between two heavy point particles at rest. On top of this you can evaluate quantum corrections by taking into account also other diagrams like those with crossed photon lines.

    After all this mathematical trouble one comes to the conclusion that a much better intuitive understanding is still to think in terms of fields as the "mediators" of interactions than in terms of "virtual particles" that are bouncing back and forth between the interacting bodies.

    For a very thorough explanation of the soft-photon (infrared) phenomena in QED, see

    Weinberg, Quantum Theory of Fields, Vol. I.
  4. May 26, 2012 #3


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    vanhees71, You're making the issue sound very, very complicated indeed, which it is not. The Coulomb potential e2/r arises from one-photon exchange, and that is all of QED that is required to explain it.

    Higher order diagrams are necessary only if you're interested in quantum corrections like the Lamb shift or the anomalous magnetic moment. Infrared photons are not involved either.
    Infrared divergences when they arise are canceled by other divergent diagrams of the same order, not by "growing powers of the electromagnetic coupling constant". (which is less than one anyway!)
  5. May 26, 2012 #4
    It's all part of the gauge symmetries of your lagrangian. In fact when you try to make your lagrangian (let us say of a Dirac particle/Dirac equation, like electrons) symmetrical under local gauge transformations of a phase (meaning that your Ψ'= exp[-ib X(t,r)] Ψ should not change your lagrangian) and Lorentz invariant, you will inevitably end up adding a certain field:
    Av which you know just its transformation, that is exactly the same as that of the EM field Av: A'v=Av+∂v

    Then by trying to get back your equation of motions from the lagrangian you will see that you will have to add a certain term FμvFμv which you define as Fμv=∂μAv-∂vAμ
    otherwise you would get zero 4current, or generally speaking because you want a "term of motion", which would have the derivatives of Aμ.

    Solving then again your equations of motion (for A)for your lagrangian,you will receive 2 Maxwell's Equations thanks to that F, and then the Jacobi identity will give you the rest 2!
  6. May 26, 2012 #5


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    With the single-photon exchange you don't get the approximation for, say, an electron in the hydrogen atom in the (in fact excellent) approximation that you can treat the proton simply as an external Coulomb field, in which the electron moves according to the Dirac equation with this external Coulomb field. This is in fact the starting point for calculating the higher-order corrections like the hyper-fine splitting (due to the magnetic moment of the proton) and the Lamb shift (which involves indeed radiative corrections, involving loops). From the point of view of qft that approximation is obtained by the soft-photon resummation I tried to sketch in my previous posting.

    That's what I consider the appearance of the classical-field approximation for the motion of a light particle close to a heavy particle.
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