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Quantum electrodynamics based upon Euler-Heisenberg or Born-Infeld Lagrangians predict photons to move according to an effective metric, which is dependent on the background electromagnetic field in a given space domain. In other words, photon trajectories are curved in presence of electromagnetic fields, meaning that an effective gravity is acting upon the photons. Charged fermions in the same space domain are obviously electromagnetically interacting; one thus wonder whether the interaction mediating virtual photons are also affected by this effective gravity. If the answer is yes, fermions will likely be affected, according to extended charge models, so that electromagnetic fields will show as another source of gravity for any kind of particles.

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I have a strong feeling of déjà vu. Is this the same question as the last one with electron replaced by photon? Couldn't you ask that at the same time? –  Marek Feb 16 '11 at 17:24
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No, it isn't the same question since the last one about fermions relied on the assumption of a positive answer for the present one. Not having obtained a satisfactory answer to the former, I get back to the first principles. Respectfully, I am looking for smart answers rather than smart comments. –  Hector Feb 16 '11 at 18:15
    
Do you think it would be nice to add a link to the "same question"? Ah, that wsn't hard: physics.stackexchange.com/questions/4815 –  Carl Brannen Feb 17 '11 at 0:26
    
@Hector: I wasn't trying to be smart with you as I don't have a first clue about the answer to your question(s). But you have to admit that looking at this and previous question side by side, it looks like copy&paste&replace. Instead, you should've mentioned (and you still can) the previous question (with link), why wasn't answer to that satisfactory, what new have you learned since then, etc. In other words, you should put some more effort into asking the question if you expect someone to help you answering it. –  Marek Feb 17 '11 at 19:18
    
In the spirit of deja vu, I’ll cut and paste my comment from the other thread: Seems to me this is taking the analogy of "effective gravity" too literally - in some equations in this situation it looks like you modified the metric to some "effective" (albeit no longer symmetric) metric, but not in all of them. As always, one has to be careful with analogies. –  user566 Feb 18 '11 at 1:55
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