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Photon as a guage bosons

  1. Jul 1, 2008 #1
    There is something very fundamental I am missing here. The photon is said to be chargeless. If that is the case, how does it carry the electromagnetic force? Thank you.
     
  2. jcsd
  3. Jul 1, 2008 #2
    The photon field(gauge potential [tex]A_\mu[/tex]) couples to the fermions, say, electron described by Dirac field, through the minimal coupling.
    That is, in order to have the U(1) gauge invariant, we need to replace the partial derivative with a covariant derivative, by doing this we need also introduce one so-called gauge potential. Expanding the covariant derivative, you would see the interaction term of gauge potential and Dirac fields.
    BTW, if you consider the case of scalar QED, only when the scalar field is complex, the coupling of gauge potential and scalar fields is non-vanishing.
     
  4. Jul 1, 2008 #3
    Hi, and welcome to PF :smile:

    This is either a very technical, or long question. It does not appear to me possible to render justice to it, because it is a good question, in a short simple answer.

    First, an experimentalist disclaimer : when we say that the photon has no mass, or no charge, we do not mean to say that it can not have in principle, we refer to the fact that any such quantity must be less than observable limits. Current standard model imposes zero charge and mass, experimental limits may be stringent, but there is always infinite room in between any finite limit and zero.

    Second, an unfair answer would be "why should it have a charge !?". In fact, physically we do not want the photon to have charge or mass, precisely because of the overwhelming experimental limits. Although we do not have a satisfactory final theory for this fact, charges appear quantized in unit of the electron charge (lest the quarks which do not appear free). We would not like the electron to loose a tiny bit of its charge by emitting a single photon. When we bounce around an electron radiating electromagnetic waves, we think that the wave is carrying away information about the bouncing, but only information, no charge. I am going to be provocative here : basketball players exchange a ball but the ball itself does not wear basketball shoes.

    Third, the photon field can be derived from a very important principle for the standard model of particle physics, namely the gauge principle. The gauge principle more or less says the following : the wave function, as a complex number, has an arbitrary phase. Only phase differences between two wave functions are physical. That means, we can choose the origin of the phases as we please. If we like to monkey around, we can choose this phase reference at every single point in space-time. But that affects the derivative of the wave-function ! Therefore, we need to compensate for our monkeying around the phases by introducing a modified derivative, which connects neighboring space-time points to correct the phase references. This connections is the photon field. In this case, we take the phase to simply be one real angle parameterizing the complex unit circle U(1). We could choose more elaborate phases, coming from more complicated symmetry groups than U(1) and we would end up with other interactions (and that is precisely what the gauge principle does to generate interactions from symmetry principle in the standard model). There are just two ways to wrap around the circle, clockwise and counterclockwise, and this is why there are just positive and negative electric charges. There is no more freedom than that. That means that the photon, which must generate this symmetry, can do it without carrying any non-trivial charge. There is no positive or negative photon because there is just one way to go along a circle, back and forth along only one direction. Another way to say it (I know it has gotten too technical already) is that the Lie algebra of U(1) considered as a Lie group is a trivial algebra with only one element.

    All right, I tried to address as I could cover "why the photon has no charge ?", but I have not addressed "how can it mediate the EM force ?". I'll cheat completely here
    Some Frequently Asked Questions About Virtual Particles (John Baez's Stuff)
    The bottom line is : the photon is a quantum object, and if you want to investigate force mediation in photon terms, you must go through the quantum language. The photon has no clue whether it has been emitted by a positive or a negative electric charge particle. Attraction/repulsion is decided by the interference between odd and even (including zero) number of photons exchanged.
     
  5. Jul 2, 2008 #4

    malawi_glenn

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