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Photons as the electric field. Question

  1. Jul 5, 2003 #1

    Ivan Seeking

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    Hopefully this is not a stupid question. When we talk about photon exchange as the electric field, what determines the frequency of the photons? I would tend to assume that the electric field strength goes as photon frequency, but since classically the field strength can depend entirely on the quantity of charge [assuming no magnetic field] present, which seems to mean that a stronger electric field has more photon exchanges but not stronger ones, I don't see what would detemine frequency of the photons. Is this a function of distance that manifest generally as the inverse square law?
     
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  3. Jul 5, 2003 #2

    drag

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    Re: Photons as the electric field. uestion

    Greetings !

    First of of all, it's not just the electric field if you're
    talking about photons, it's the electromagnetic field.

    As for the frequency of the photons, I believe it can
    be calculated from and described as a quality of the
    energy density of the electromagnetic field. (I'm afraid
    I can't help you with Maxwell's equations that describe this,
    my knowledge of this is strictly theoretical, for now.)

    Live long and prosper.
     
  4. Jul 5, 2003 #3

    Ivan Seeking

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    Re: Re: Photons as the electric field. uestion

    I may, but I'm very thin on this idea. I mean for a static electric field. For example, for a charged sphere. The electric field measured about the sphere can be described by a photon exchange. I understand to some extent how we get attraction and repulsion by this. Beyond these two points I get in trouble really fast.
     
  5. Jul 5, 2003 #4

    HallsofIvy

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    No, a static electric field has nothing to do with photons or light. Light consists of waves in the electromagnetic field. If you don't have both you can't have waves and so can't have light.(Strictly speaking, "electric" and "magnetic" are components of the same field. A "static electric field" is one in which the magnetic component is 0.)

    Photons are the quantization of the electromagnetic field.
     
  6. Jul 5, 2003 #5

    drag

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    Re: Re: Re: Photons as the electric field. uestion

    Greetings !
    I believe you are refering to virtual particles ?

    Well, look at it this way. The enitial mathematical discriptions
    of forces used geometrical discriptions - continous dimesnions
    that deform by the effect of the source of the force and
    thus effect other bodies with a certain "charge" relevant to
    this force. This is the way GR works - a space-time geometry
    deformed by the gravitational charge - mass.

    QM deals with things in a different manner - everything is
    quantified - cut into individual packets. This is where
    virtual particles appear - instead of discribing a geometry
    we discribe (according to QM's interpretation) individual
    packets that "transport" the electric and other forces.

    You should make a search for Feynmann Diagrams that explain
    this theorized virtual particles' interaction. They're
    easy to find online and provide a nice graphic discription
    and explanation of this process.

    Live long and prosper.
     
    Last edited: Jul 5, 2003
  7. Jul 5, 2003 #6

    Ivan Seeking

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    Re: Re: Re: Re: Photons as the electric field. uestion

    I read QED years ago but I don't remember this issue [virtual photon frequency] being addressed. I will take a look. Also, I didn't remember them as being virtual; so long as the charged body interacts with another.
     
    Last edited: Jul 5, 2003
  8. Jul 6, 2003 #7

    drag

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    Re: Re: Re: Re: Re: Photons as the electric field. uestion

    Greetings !
    I never read it so far so I can't adress it I'm afraid.
    I believe that it still relates to energy density of the
    field, but I do not know if there are precise frequencies
    for 3 of the forces or like in the case of the gravitational
    force - there is a problem of plotting (I believe it's called
    "normalising" ?) the interaction since you get any number of
    1 to infinity of virtual particles' interaction loops.

    Or something like that... On to the experts with this.:wink:

    Live long and prosper.
     
  9. Jul 6, 2003 #8

    jeff

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    Why does electromagnetic field strength grow as it's sources are approached, or in quantum theoretic terms, why are high energy photons encountered only near their sources? The answer is that the time-energy uncertainty principle allows photons to have greater energies (or equivalently, frequencies) than allowed by the relation pμpμ = -m2 = 0 satisfied classically by their 4-momenta as long as their lifetimes and hence the distances they travel are sufficiently short. Such photons are called "virtual" and their 4-momenta are said to be "off the mass shell", i.e. pμpμ ≠ 0.
     
    Last edited: Jul 6, 2003
  10. Jul 6, 2003 #9

    Ivan Seeking

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    Re: Re: Photons as the electric field. Question

    Are you saying that the inverse square law is a special case of the time-energy uncertainty relationship?
     
  11. Jul 6, 2003 #10

    jeff

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    Re: Re: Re: Photons as the electric field. Question

    No. Inverse square laws are a consequence of the structure of tree amplitudes for the exchange of bosons that result from the fact that the lagrangian densities involve two powers of the spacetime derivative. In momentum space, the relevant part of the energy E of interaction between electrically charged particles at x1 and x2 following from these amplitudes is

    E = - (1/2π)3∫d3k exp[ik⋅(x1 - x2)](ik2+m2)-1 = - (1/4πr)e-mr

    where r ≡ |x1 - x2|. Setting the mass m equal to zero for the photon in the case of the electromagnetic interaction and computing ∂E/∂r yields the familiar inverse square law of coulomb for the electromagnetic force.
     
    Last edited: Jul 7, 2003
  12. Jul 8, 2003 #11

    Ivan Seeking

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    Re: Re: Re: Re: Photons as the electric field. Question

    Then for a large charged surface we must write an equation for every combination of charged interacting pairs? Say for example if we have two large charged spheres. How does one calculate the energy for a particular photon exchange between these two spheres; thus the related frequency?
     
  13. Jul 8, 2003 #12

    jeff

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    The expression E = - (1/2π)3∫d3k exp[ik⋅(x1 - x2)](ik2+m2)-1 says that of the particles exchanged between sources a distance r apart, only those with momentum of magnitude k roughly of order 1/r or less contribute nonnegligibly to the energy. We can regard this heuristically as resulting from the cancellation of the contributions with k large compared to 1/r due to the oscillations of the phase factors exp[ik⋅(x1 - x2)]. This is consistent with the uncertainty principle since it says that higher energy particles are encountered as sources are approached. This expression also tells us that the characteristic distance of interactions mediated by field particles of mass m is just 1/m since the phase factor shows that k takes it's characteristic value m when k ≈ 1/r. Thus very light or massless particles like the photon mediate long-range interactions.
     
    Last edited: Jul 8, 2003
  14. Jul 9, 2003 #13

    Ivan Seeking

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    Jeff, thanks for all of your great answers!
     
  15. Jul 14, 2003 #14
    Feynman and some others invented the concept of virtual photons ad hoc in order to calculate a second order effect (the gyromagnetic ratio) to a high degree of precision - QED does not lead to a derivation of Coulombs inverse squared law - nor does it address the fundamental issues in a way that can be related to concepts for which we have familiar analogies - it is an alternative that provides a way to calculate the perturbation coefficients. Dirac and others proposed different explanations for the electrostatic field.
     
  16. Jul 14, 2003 #15

    jeff

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    As I've shown, QED does produce coulomb's law. If you're remark was true, QED wouldn't produce correct predictions. In fact any theory that makes correct predictions must produce the corresponding classical expressions at low energies.
     
    Last edited: Jul 14, 2003
  17. Jul 14, 2003 #16
    Jeff - where have you shown QED derives the inverse squared law - what did you use for the charge and why does it have the value it has - you would have to fudge all the properties of the virtual photons to get an inverse squared law and the correct force coefficient - and if you have to do that you have not derived anything - you simply started from the experimental result and worked backwards
     
  18. Jul 15, 2003 #17

    jeff

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    Inverse square laws are a consequence of the structure of tree amplitudes for the exchange of bosons that result from the fact that the lagrangian densities involve two powers of the spacetime derivative. In momentum space, the relevant part of the energy E of interaction between electrically charged particles at x1 and x2 following from these amplitudes is

    E = - (1/2π)3∫d3k exp[ik⋅(x1 - x2)](ik2+m2)-1 = - (1/4πr)e-mr

    where r ≡ |x1 - x2|. Setting the mass m equal to zero for the photon in the case of the electromagnetic interaction and computing ∂E/∂r yields the familiar inverse square law of coulomb for the electromagnetic force.
     
  19. Jul 16, 2003 #18
    So Jeff -- where is the physics - where did the fundamental equation come from - what have bosons to do with the electrical force between two electrons - define the parameters and how each is physically related to the problem posed -- where does your equation provide an answer to the question of why the electron has a unique charge that will repel another electron with a certain force at a certain distance - you can always gin up an inverse squared law based upon simply geometry - because the notion of a field is deemed to be reduced proportionately with the area over which is is spread ...that is not what is at issue - the point is the physics are introduced by hand

    I know the argument that the photon must be massless for the force to extend to infinity - and the notion that uncertainty allows borrowing from the medium temporarily ... but these are all hypothetical rationales ... you cannot get a qualitative value for "e" from this, or if you can - show me. As Lord Kelvin once said - "When you can talk about something with numbers, you know something about it - when you cannot, your knowledge is of a meager and unsatisfactory nature."
     
  20. Jul 16, 2003 #19

    jeff

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    As always.
     
  21. Jul 16, 2003 #20
    Jeff - if we had a complete (correct) theory we would not have to put in the electron charge by hand - the theory would predict what the charge should be in terms of other relationships e.g., when Newton combined his gravitational formulation with his laws of inertial motion - he could derive Kepler's laws - by analogy, there is a physical reason why charge has a certain value - why it cannot be something else - otherwise we are reduced to God given factors and God given constants - while there are a few occasions where the math preceded the physical model (e.g., Dirac's derivation of the gyromagnetic ratio, they are rare), In the usual case, it will we usually be more insighful to rely upon the math to test the physical model - not to generate it
     
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