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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 other words, photon trajectories are curved in presence of electromagnetic fields, meaning that an effective gravity is acting upon. If part of fermion masses is allegedly of electromagnetic origin, the question why their trajectories are not affected by this effective gravity naturally comes to mind.

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A charged fermion in the presence of a background electromagnetic field will already deviate from a geodesic trajectory! –  QGR Feb 8 '11 at 17:32
    
Effective gravity is in addition to classical electromagnetic interaction! –  Hector Feb 8 '11 at 18:26
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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 9 '11 at 7:43
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The term "effective gravity" stands for the spacetime curvature effects on photons in vacuum, due to the optical "effective metric" of Non-Linear Electrodynamics. The standard viewpoint is that neutral fermions are not affected by the optical metric. Rephrasing the main question, why should it be so, if as widely accepted, their mass is partially of electromagnetic origin? It seems the standard viewpoint cannot be conciliated with Einstein Equivalence Principle, since different non-charged particles "fall" with different accelerations. –  Hector Feb 9 '11 at 12:41
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@Hector: this is why the "effective metric" is just an analogy, it does not follow from a deep principle like the equivalence principle, it is just a tool helping rephrase certain calculations in a more familiar language. –  user566 Feb 9 '11 at 15:32
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1 Answer

In the presence of a background electromagnetic field, electromagnetic fields travel along a deformed light cone which is smaller than the "relativistic light cone". However, charged fermions can still travel faster than electromagnetic waves as long as they are still slower than the "relativistic speed of light". They emit Cherenkov radiation while doing so.

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The answer seems to be a little off-topic regarding the main question. Charged or neutral fermions can indeed travel faster than the electromagnetic waves in the presence of a background electromagnetic field, provided they are not subjected to the "effective" spacetime curvature. –  Hector Feb 9 '11 at 12:58
    
The radiative self-energy corrections due to the coupling of the charged fermion to virtual photons shows up as a correction to the anomalous magnetic moment. Given a nonuniform background magnetic field, this will contribute an additional force. –  QGR Feb 9 '11 at 14:43
    
To my knowledge, experiments have been proposed for detecting nonlinear quantum electrodynamics effects, one of them being Cherenkov radiation produced by charged fermions travelling outside the "effective" light cone. So far no results have been reported, so either these experiments were not yet performed or they did not succeed. –  Hector Feb 10 '11 at 18:34
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