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It is one of the axioms of special relativity that the photon has no rest frame; light travels at speed c when measured in any inertial frame of reference. As a corollary, it is often said that if one were in a photon's rest frame, infinite time-dilation and length-contraction would make the universe appear (from a photon's perspective) to be unchanging, with zero length in its direction of motion. In other words a measurement would not be possible in the reference frame of a photon, because there would be no time or space in which one could conduct it.

My question is: how does this accord with the fact that parallel-moving photons can interact with each other via loop diagrams[*]? Photons can therefore be used to perform measurements in the photon's rest frame. I suppose an answer may be "the act of measurement changes the photons momentum vector and therefore its co-moving frame is non-inertial" however this can be said of any measurement.

[*] Or can they? I can't find an explicit amplitude to check this (for, say, the photon-photon "box" diagram), however as far as I can tell from the Euler-Heisenberg Lagrangian nonlinear effects should be present for parallel photons.

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user1247
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In other words a measurement would not be possible in the reference frame of a photon, because there would be no time or space in which one could conduct it.

A measurement is not possible in the reference frame (rest frame) of a photon because there is no such frame. That's all there is to it. You can't make a measurement (or do anything else) in a reference frame that doesn't exist. All the stuff you may have heard about "if one were in a photon's rest frame" is just meant to appease non-physicists who have a hard time understanding why there is no such reference frame.

One of the defining conditions of an inertial reference frame is that there exists a proper orthochronous Lorentz transformation which converts that frame into any other inertial reference frame, and vice versa. But for all such transformations, the starting and ending frames are moving at some relative speed less than $c$; however, any photons that may exist will be moving at speed $c$ in both frames. There is no way to use a Lorentz transformation to produce an inertial reference frame that moves at $c$ with respect to any other inertial frame, and relatedly there is no way to use a Lorentz transformation to produce an inertial reference frame in which a photon would be at rest.

Of course, non-collinear photons do interact with each other via loop diagrams (not directly, though). You can calculate the amplitude of this interaction, which is a reference frame-independent quantity, and turn it into a cross section in any inertial frame you like. But you will never find one in which the photon is at rest.

For collinear photons, perhaps you can still calculate an interaction amplitude. I haven't tried so I don't know if it works. But if you try to compute a cross section or any physical quantity from it, you'll find that the only kinematically allowed processes are those in which the end product consists of only collinear photons with the same total energy moving in the same direction. So any such interactions are really just quantum fluctuations, not actual scatterings.

David Z
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  • Isn't a frame moving at speed c parallel to a photon in the photon's rest frame, by definition? – user1247 Mar 06 '12 at 21:11
  • But there is no inertial reference frame moving at speed c. One of the defining conditions of an inertial reference frame is that there exists a proper orthochronous Lorentz transformation which converts that frame into any other inertial reference frame, and vice versa. But for all such transformations, the starting and ending frames are moving at some relative speed less than c. There is no way to use a Lorentz transformation to produce a reference frame that moves along with a photon. – David Z Mar 06 '12 at 21:20
  • What I'm getting at in my question is the statement that there is no such frame appears to rest on the assumption that all inertial observers are timelike. Lightlike observers exemplify such an "impossible" frame. The fact that a measurement can be performed in a frame moving at speed c (by photons) directly contradicts the first sentence of your last post. – user1247 Mar 06 '12 at 21:22
  • Yes, that's another way of saying the same thing. All inertial observers follow timelike geodesics. And I would certainly hope that the fact that a measurement can be performed in a frame moving at speed c (by photons) directly contradicts my last post, because to the best of my knowledge that fact is false. – David Z Mar 06 '12 at 21:25
  • But in your response to my OP you seemed to agree that co-moving photons can interact with each other. In other words they can be used to perform a measurement on photons in their own reference frame, one that moves at speed c! – user1247 Mar 06 '12 at 21:27
  • Sure, photons can interact with each other (though I'm not sure if you can have interactions between perfectly collinear photons; any such interaction must be limited to producing virtual particles which recombine into the original photons). That does not at all imply that they have a reference frame, much less that they can be used to perform a measurement in their own reference frame. The interaction is always in the frame of an inertial observer. I've clarified my answer accordingly. – David Z Mar 06 '12 at 21:32
  • My question is precisely concerned with the case of collinear photons interacting via box diagrams and such. I see that you are now saying that such interactions have zero amplitude (is this consistent with the nonlinear classical Euler-Heisenberg terms? I thought they implied theoretically measurable effects even for collinear EM waves, but I might be wrong). If you are sure this is true, then it answers my question. Thanks! – user1247 Mar 06 '12 at 21:50
  • I'm about as sure as I can be without actually having done the calculation that interactions of the form (collinear photons -> anything else) are kinematically forbidden. (I'll do the calculation if I have time to figure it out, but I can't make any promises on that.) – David Z Mar 06 '12 at 21:55