I just got in what I thought was a silly exchange where a self-identified physicist states that the difference between "red-shifting" (in the Doppler sense) and the re-emission of light at longer wavelengths following absorption are only "semantically" different.

I'm bad at GR and I expect my (classical) understanding of the Doppler Effect cannot really account for the the red shift of light as its emitter moves away from the observer.

But maybe it can? After reading this I think it makes sense to say that as a photon red shifts it doesn't really lose energy at all; the appearance of the energy loss illustrates that energy conservation only holds within an inertial frame. This also means that an ensemble of photons maintains any phase coherences indefinitely as they propagate through space (ignoring vacuum fluctuations.)

I'm also bad at QED, so can someone explain whether the notion of a Feynmann diagram for the Doppler effect is trivial? I would figure that if there are no interactions -- the photon is simply propagating in spacetime -- then a Feynmann diagram would be featureless. Just a squiggly line on some vector? Not even a real QED problem?

When matter absorbs a photon and then emits one at a longer wavelength, this Feynmann diagram would have to look more interesting than the Doppler shift diagram, right? The absorbed and emitted photons have different identities -- different squiggles on the diagram?

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    $\begingroup$ Doppler shifting is a classical effect, not relativistic. You can see it in, e.g., Mossbauer scattering experiments. $\endgroup$ Commented Dec 17, 2013 at 13:08
  • $\begingroup$ Okay so I took off the GR tag; I realize it's not relativistic (hence appealing to my classical understanding of it) but would a photon undergoing a Doppler shift have any nodes on its Feynmann diagram? $\endgroup$
    – Ryan
    Commented Dec 17, 2013 at 19:02
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    $\begingroup$ nonsense. the classical formula for Doppler shift (radar guns) is different from special relativistic formula (astronomical redshift, z) simply because classical is only an approximation valid for v<<c. also GR redshift is yet a different ballgame but totally valid (blackholes, curved space time). these all cause a shift in the whole spectrum whereas scattering reactions occur for specific wavelengths dependent on the source element involved. $\endgroup$
    – gregsan
    Commented Dec 27, 2013 at 7:51

1 Answer 1


This is a very interesting question! First, let us be clear on the following: Red-shifts in the Doppler sense and emission and re-absorption are different. Ie they describe different physical systems and can be unambiguously distinguished from each other. Gravitational redshifts are different, but I will come back to this later.

A quick way to see that the processes are different is that the Doppler shift is dependent on the observer. Observers situated at the same point but having different velocities (relative to the source) might or might not observe a Doppler shift at all! On the other hand, the absorption and re-emission of a photon from a particle is independent of the observer and all observers will agree that these events are happening.

Regarding the QED description: As far as I know there is no Feynman diagram describing the "Doppler shift" of a photon because it doesn't really make sense to combine these two. A photon that will participate in a quantum process will have an initial energy. If it was emitted from a source far away with some relative velocity with respect to the lab frame then we can use the Doppler formula to calculate this initial energy, but this is all that is needed to do the calculation in QED!

However, there is another point that should be clarified as well. In the question you mention " "red-shifting" (in the Doppler sense)" but there is also another type of redshifting: gravitational redshifting as described by GR: A photon gets redshifted when traveling in a gravitational field. It would make sense to describe this kind of red-shifting with Feynamn diagrams because it involves the interaction of photons with gravitons, so the Feynman diagrams would be interacting ones and not just simple "squiggle lines". However, we don't really know how to describe this interaction completely since it needs a quantum theory of gravity which we do not have...

In this context maybe it makes sense thinking of the photon as being absorbed and re-emitted by a graviton but in my opinion this view is over-simplistic and definitely cannot be "proved" without a theory of quantum gravity as mentioned above.

  • $\begingroup$ Thanks for clarifying that gravitational redshifting is not the same as Doppler redshifting, and for confirming that both are disparate phenomena from absorption/emission. I had intuited that the graviton might appear on a Feynman diagram for the gravitational red shift, but also ended up deciding that, outside a QM treatment of gravity, there was no known Feynman diagram for this phenomenon. $\endgroup$
    – Ryan
    Commented Dec 31, 2013 at 2:43
  • $\begingroup$ The point I'd like to clarify now is regarding identity. I appreciate my phrasing of this is metaphysical, and I apologize for poor wording -- what WOULD be the physical proxy of 'identity'? (Information? My best guess is its phase coherence?) If a high-energy photon is absorbed and a low-energy photon is emitted, these ARE completely different photons, right? In contrast, a photon propagating through space (even if continuously interacting with gravitons) maintains a consistent identity? $\endgroup$
    – Ryan
    Commented Dec 31, 2013 at 2:51
  • $\begingroup$ Perhaps I should have read this (physics.stackexchange.com/questions/41180/…) before blathering about identity... $\endgroup$
    – Ryan
    Commented Dec 31, 2013 at 3:00

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