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On one hand, in the classical electrodynamics polarization of transparent media yields in lowering the speed of light by the factor of $n=\sqrt{\epsilon_r \mu_r}$ (refractive index). On the other, in the quantum field theory the vacuum polarization does not decrease the speed of light. The thing it does is increase of strength of electromagnetic interaction. Why is it so?

My two guesses are the following:

  • In principle the vacuum polarization does decrease speed of light ($\rightarrow 0$) but we implicitly put on-shell renormalization to keep $c$ constant.
  • For some reason the process is completely different from polarization of transparent media.
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Good question. But the answer obviously can't be your first guess. QED is not the full description of the universe and photons are not the only massless particle so they can't possibly enforce the universal speed limit $c$ that all of the other particles have to obey. –  Marek Jan 6 '11 at 23:38
    
I know that $c$ is not only the speed of ''light'' but a crucial geometrical parameter. Maybe such claim implies constrains on other couplings, but still - I don't see why the first guess have to be wrong. –  Piotr Migdal Jan 7 '11 at 0:20
    
well, from the point of QFT that question doesn't make sense. Speed of propagation is not really a parameter of the theory. Only couplings (like mass and charge) are. So you can't ever renormalize speed. What you can do, however (and what happens with the photons in the media) is that particles can become effectively massive by interaction with other particles (this is the same way all the particle gain mass from Higgs interaction, by the way). And of course, massive particles propagate slower than the speed of light. –  Marek Jan 7 '11 at 0:47
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Would you like to propose an experiment to measure the speed of light in the absence of vacuum polarization.... –  dmckee Jan 7 '11 at 1:17
    
@dmckee : well, the strength of the vacuum polarization effect is momentum-dependent, so I would assume such an experiment would involve observing different speeds for radio waves and hard gamma rays. –  Jerry Schirmer Jan 3 '12 at 14:26
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3 Answers

If you accept gauge invariance to be valid then the physical propagating photon has zero mass. Special relativity requires a zero rest mass particle to move at the speed of light. Renormailzation of vacuum polarizability must be done so as to fulfill this requirement.

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The "vacuum polarization" is just like "relativistic mass".
Both concepts were introduced "to make things simpler", while they just lead to confusion.

The "vacuum polarization" is an artefact of SI units. In SI units magnetic and electric fields are measured in different units. This is very unnatural, since these fields are just components of one tensor. It is like measuring height and length in different units.

Due to this unnaturalness one have to introduce extra dimensional quantities to "convert" between magnetic, electric and other units: vacuum permittivity and permeability. And these quantities have not too much physical sence -- it is like 3.2808399 between meter and foot -- just a matter of convention.

And there is no realtion of this "vacuum polarization" to the polarization of vacuum in QFT.

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If you don`t like my answer, then can you please elaborate what is wrong with it. Do you think that I`m just wrong, or I have to be more clear in my explanations? –  Kostya Jan 8 '11 at 13:58
    
I think you missed the point of the question. I asked about effect from quantum field theory, which has nothing to do with SI units. (BTW: In QFT there is often used convention $c=\hbar=1$, very far from SI.) –  Piotr Migdal Jan 11 '11 at 22:15
    
You're misunderstanding what vacuum polarization is--it's a QED effect arising from Feynman diagrams involving internal particle loops: en.wikipedia.org/wiki/Vacuum_polarization –  Jerry Schirmer Jan 3 '12 at 14:24
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In transparent media, there is matter present. The refractive index has its origin in the tree level process of $$e^- \gamma \to e^- \to e^- \gamma$$ (with $e^-$ either a part of the atom or free as in metals). There is no need for renormalization here, it's just a standard contribution to the probability amplitude of photon propagation. There are also other processes (interaction with nuclei and whatever loop corrections you can think of) but these are only second (and more) order corrections. In any case, what happens here is that photons gain effective mass via all those interactions, which is the ultimate reason they propagate slower.

In vacuum the relevant process is a loop-level one $$\gamma \to e^+e^- \to \gamma$$. You'll obtain an UV divergence here and to take care of it introduce a renormalization of the field which is what that process corresponds to in the QED Lagrangian (the relevant term being $F^2 = (\partial A)^2$. But this is just a kinetic term, not a mass term (which would be something like $m_A^2 A^2$). You could imagine adding such a term to the Lagrangian but that is of course inconsistent with the observation. We know that photons travel at the speed of light (in the stronger sense of special relativity that is built-in in every QFT).

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And $H \gamma \rightarrow H^+ e^- \rightarrow H \gamma$ (here $H^+ e^-$ need not mean ionization, just some excitation)? –  Piotr Migdal Jan 7 '11 at 0:24
    
@Piotr: yes, that's what I meant (except it doesn't have to be $H$, but any atom of the material). The electron is to be understood as part of the atom (or molecule, more generally). But it doesn't matter much whether the electron is bound or free (like in metals), at least for this general discussion. –  Marek Jan 7 '11 at 0:51
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