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What is the Colour of Virtual photons?

  1. Aug 31, 2004 #1
    I have read some 'popular' physics and I get the impression that Electrostatic Force is caused due to transfer of virtual photons.

    Do these have a particular energy for a particular charge?
    Then they should have a particular frequency .

    Then if we place the charges in mediums with refractive index u1 and u2 and if we choose such a medium such that the photon from denser medium undergoes total internal reflection then wont force be present only in one particle? :confused:
    med 1 med 2


    Excuse my crude ASCII art

    * ** _
    * * -
    * -
    * * -
    *** par2
    Last edited: Aug 31, 2004
  2. jcsd
  3. Aug 31, 2004 #2
    Virtual photons come from QED, where the conservation of energy can be violated for a specific amount of time, thanks to the Heisenberg-uncertainty-principle.

    Yet photons do NOT carry colours, nor do they carry charge. The colour quantum-number comes from QCD, which describes the properties of quarks and they way they interact via the strong force

  4. Aug 31, 2004 #3
    Please read full post before replying

    Please read post before replying
    I am not talking about colours in strong force but each photon has some energy and in wave form it will have colour VIBGYOR,IR,UV,etc

    I mentioned the colur to bring in some flavour to the question.
    SORRY to confuse you.
    Please read full post not just the tittle before replying
    Thanks in advance
  5. Aug 31, 2004 #4

    Then please try to state your question more clearly and use English that is understandable...Personal-invented-language is not usefull...

  6. Aug 31, 2004 #5
    Besides, what the hell is this par1 par2 thing going on in your post.

    definitely the worst post ever...

  7. Aug 31, 2004 #6
    I thought that to normal people when speaking of colour (when it has been clearly indicate that it is not connected to strong force),They think of VIOLET,INDIGO,BLUE,etc.
    My Friend Marlon is surely a great physicist/Mathematician as he can do vector integrals but cant do long division.

    Par1 indicates particle 1
    and Par2 particle 2
    Last edited: Aug 31, 2004
  8. Aug 31, 2004 #7

    aahh, thank you for the "honest and sincere" recognition :secret:

    Since :wink: both energy and time are uncertain in QM one can state that the bigger the energy of a virtual foton, the smaller it's lifetime. The product of energy-interval and the time-interval equals the reduced Planck-constant. Basically, all energies are possible my friend

    Ok ????

    your Friend Marlon
    Last edited: Aug 31, 2004
  9. Aug 31, 2004 #8


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    Your question about color easy to understand, but your ASCII art isn't, especially that last thing. The answer to the question is that every possible frequency from zero to a certain maximum frequency contributes to the total probability. The maximum frequency is usually taken to be infinity in calculations, but it's more than likely that there are no photons with frequency higher than 1/(planck time).
  10. Aug 31, 2004 #9
    That's funny ! Yes Marlon, poolwin is not totally wrong when he impies we are too much into strong interaction. Indeed, I also understood color charge (as in QCD) even though I read the first post entirely.

    poolwin : I think physicists would have refered to the Energy of the photon as you do in the core of the first post, not its "color", as in the title. It is confusing to me too. Then, I am not sure I understand correctly the question. Do you mean to ask : "what is the energy spectrum (probability density) of the virtual photons involved in a static EM potential ?" and "Would the situation described, with two charges in two different otpical media, be actually affected with respect to the same situation in empty vacuum ?"
  11. Aug 31, 2004 #10


    Poolwin, this is one for YOU : now two very important physicists didn't get your very important question, yet as you say we are both your friend.

    marlon brando
  12. Aug 31, 2004 #11
    Hey Marlon !
    We are not (yet) great physicists. At least, we should keep it a secret :secret: :zip:
  13. Aug 31, 2004 #12
    OK then

    Would the situation described, with two charges in two different optical media, be actually affected with respect to the same situation in empty vacuum ?

    What about the drastic case when some of the photons undergo total internal reflection??

    Remember:Total internal reflection occurs only when light is going from denser to lighter medium not vice versa
    Thanks in advance!
    Last edited: Aug 31, 2004
  14. Aug 31, 2004 #13
    Let me give it a try, I warn you I did not make any calculation, I am just throwing thoughts :
    usualy no-one will ever try to use QFT and virtual particle-sates to describe a static situation as the one described here. This certainly leads everybody to confusion at first sight. If you insist on taking a closer look : the situation won't be different in vacuum and in the two optical media. I motivate this... opinion (I am not 100% positive this is the true answer) with the following arguments :
    1 Deflection of light is a dynamical property of the interface, it won't affect static properties
    2 The electric force is determined by the field, which can be computed by Gauss' theorem : if you consider only two charges around (the optical media are electrically neutral) nothing changes in vacuum
    3 The virtual photon exchanged between the charges are not affected by the interface : virtual particle exchange proceed only in a "pure" process : a virtual photon cannot be scattered somewhere by a mirror in the middle of the interaction. In fact, a rigourous derivation is certainly possible, but then would involve several possible process, with and without scattering at the interface, and those two class of process lead to amplitudes that must be added incoherently.

    In fact, the sole question about total internal reflection seems to me more like a contradiction disproving the idea that a static electric interaction can be affected by optical properties around.

    I did not even try to analyze dynamical properties of the interaction between the charges. I am pretty sure dynamical properties woud indeed be affected.
  15. Sep 2, 2004 #14


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    Hi Poolwin,

    First of all, let me say that, as opposed to some others (wink wink), I *did* understand what you meant by "colour", even if I am doing research in QCD and I am thinking about the colour quantum numbers all the time! And I did understand your drawing.

    First, let me warn you about the concept of "virtual photons". You will rarely hear what I am going to say (and I am sure many here will object) but it's an important point that needs to be emphasized: virtual photons are a mathematical concept, not a physical one. Sure, they make it easy to organize Feynman diagrams and to remember the rules to calculate them, but they are just that: a mathematical trick. The key point here is that one must remember that Feynman diagrams represent a perturbative expansion which is an *approximation* of the complete (and physical) calculation. As such, they must be taken with a grain of salt. When something goes wrong in a Feynman diagram (like a divergence), it's not physical, it's just a signal that the perturbative expansion is not valid and one must be more careful. The problem is *mathematical*, not *physical*. That's because each Feynman diagram is associated with a *mathematical* quantity (a single term in a series expansion), not the complete, physical, observable. For sure, they are nice and make it easy to remember the rules needed to calculate terms in the expansion. And they are nice to try to visualize the meaning of the terms of the expansion. But they are unfortunately taken too seriously in terms of their physical meaning. They are pictures used to calculate *mathematical* terms in a series expansion. Each term is not physical in itself. Only the sum of all the Feynman diagrams is observed in an actual experiment. This is why it does bother people that most Feynman diagrams are actually mathematically ill-defined, they are divergent!!
    And the virtual particles in Feynman diagrams are also mathematical concepts. The only physical things in a Feynman diagrams are the external, on-shell, particles which are observed in the lab.

    Let me now say a few words about your questions.

    First, it's true that the static electric force is often described as being due to the exchange of virtual photons. The idea is roughly (very roughly) the following: The propagator of a photon is proportional to 1/k^2, where k is the four-momentum of the photon. Recall that

    k^2 = k_0^2 - vec k^2

    where k_0 is the energy of the photon and vec k is its 3-momentum. I am using the usual units where h bar = c =1. For real (as opposed to virtual) photons, the four-momentum squared is zero, which simply comes from the fact that for real photons,

    E=cp (restoring the c)

    so that k_0 = magnitude of vec k (in natural units)

    so that k^2 =0.

    For these photons, the propagator blows up. One encounters the so-called "infrared divergences" and these can be cured by taking into account that some real photons can be emitted with arbitrarily small energies and one must combine several Feynman diagrams to get a sensible result (again, this is due to the fact that each individual Feynman diagram does not represent a physical quantity).

    On the other hand, for virtual photons, k_0 and \vec k are independent. So it's possible to go to the limit k_0 ->0 while keeping \vec k nonzero (again, this is mathematical, it does not represent a real photon that you could ever observe). Then the propagator goes like -1/vec k^2. This is in momentum space. If you Fourier transform to real space, you will find that this gives a potential that goes like 1/r, which means a force that goes like 1/r^2. Since the virtual photon connects to the charged particles with strength euqal to q (their electric charge), we get a force that goes like q_1 q_2/r^2, the usual electric force.

    Coming back to your question: virtual photons can't be observed, so we can't talk about their wavelength, strictly speaking. Hence, we can't talk about their colour in any meaningful way.

    As for your question about total internal reflection. I see what you meant. The problem is the following: in a quantum treatment of the exchange of those virtual photons (again, one must be careful about not assigning reality to those things, but I'll go along for the sake of trying to answer your question), you must sum over all possible paths, even paths that would never be taken by real photons. So even if a real photon could not go from one particle to the other because of total internal reflection, the virtual photon can follow any crazy (not necessarily straight line) motion it "wants"! So it does not have to follow the path of a real photon. Even more, it does not have the properties of a real photon and interacts the same way so I would think that it could even enter the crystal even along the path that would lead to the total reflection of a real photon.

    Hope this helps. Let me know if you have more questions.


  16. Sep 2, 2004 #15


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    Other problem... When you place QFT in a medium with optical properties, you are tacitly moving it away from vacuum calculations. QFT is very hard to do when thermodynamics enters the equation, all these interactions in principle will back react with the medium itself in highly nontrivial and complicated ways. In essence, a gigantic mess. I for one don't know how to calculate bulk problems like this, even with suitable approximations.

    The best bet in that situation is to use vanilla QM, even when dealing with relativistic particles.
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