poolwin2001 said:
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 won't force be present only in one particle?
diagram:
*********%%%%%%%
*par1*****%%par2%%%
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med 1 med 2
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.
Regards,
Pat
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
