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?