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On another thread, users have asked for an explanation of the "red shift" of photons (the apparent loss of energy of photons due to the expansion of the universe.) All they ever got was a GENERAL RELATIVITY explanation.

So, I'll rephrase the question: if a so-called "photon" represents a discrete QUANTUM of energy exchanged between two or more atoms as a result of the the electron shell of one atom changing its energy level, then what happens to the energy of those photons that "arrive" at a lower energy level (lower "frequency") as a result of red shift?

Let me try and clarify my question based on my understanding of how "photons" work. Let say atom A releases a quantum of electromagnetic energy equal to 3 units. One atom B absorbs 1 of those energy units, and atom C absorbs 2 units. We could then say that atom A emitted two "photons." If, however, atom D absorbed all 3 energy units, then we would have to say that Atom A emitted one "photon." I think this I where people get confused--photons aren't "real" PARTICLES you can count like sheep. They're just a mathematical description of chunks of energy that get "exchanged" AS THOUGH that energy were being delivered as chunks.

So, back to my question: if the energy of EACH "photon" being received is less than it was when it was "emitted," where does that energy go? Does it become "more" photons? I know, I know, in RELATIVITY the work of bending space time robs the "photons" of some of their energy (precisely the way the bending of leaf springs in a dragster robs the axle of some of the energy directed toward the wheels.) BUT our question is: What is the QUANTUM explanation? And please, don't just point the relativistic answer we already got.

If you don't HAVE a quantum explanation, just say so.

Thanks!

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  • $\begingroup$ What, specifically, is wrong with the standard explanation? $\endgroup$
    – Jon Custer
    Commented Dec 5, 2016 at 17:41
  • $\begingroup$ Photons are as real as any other kind of particle. $\endgroup$
    – fqq
    Commented Dec 5, 2016 at 17:42
  • $\begingroup$ Jon Custer: Please read my question. I did not say there was ANYTHING "wrong" with the Relativistic explanation. I said, we have repeatedly asked for a QUANTUM MECHANICAL explanation, and all we get is Relativity. There is NOTHING "wrong" with coffee, but I ORDERED a chai latte. $\endgroup$
    – Tommy Jonq
    Commented Dec 5, 2016 at 17:48
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    $\begingroup$ A light ray isn't simply made up of photons. Attempting to describe the red shift by concentrating on the properties of individual photons is a pointless exercise. $\endgroup$
    – John Rennie
    Commented Dec 5, 2016 at 17:48
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    $\begingroup$ @TommyJonq The moderators are probably very upset with you for not understanding this site in general, for example by responding to comments as if they were answers. You probably do not want them to "moderate" the discussion as it could end in adverse consequences for you and the disrespectful way you conduct your conversation (see: "did you READ my question?" and "if YOU don't WANT to explain, then why comment at all?") etc. $\endgroup$
    – CR Drost
    Commented Dec 5, 2016 at 18:15

2 Answers 2

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Redshift is a kinematical effect, not a dynamical one. Therefore, the explanation of redshift is the same in classical mechanics and in quantum mechanics.

For example, the derivation of the red-shift formula in non-relativistic classical mechanics is valid in non-relativistic quantum mechanics. Similarly, the derivation in (special) relativistic classical mechanics is valid in (special) relativistic quantum mechanics. Finally, the derivation of the formula in curved space-times in classical mechanics is valid in quantum mechanics in curved space-times.

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Quantum mechanics is a scientific theory. As covered in my answer here, theories do not in general pin everything down but rather they create a space for modeling phenomena.

Any quantum mechanical model of redshift will say "those photons smoothly lost energy on their trajectory, therefore they must have changed frequency to satisfy the Einstein-Planck relation $E = h f.$" That is because the latter property is part of the theory of quantum mechanics. However quantum mechanics itself is not going to argue with you about why this happened.

Now, given what we know about general relativity, we can use that to inform our quantum mechanical model, saying "oh, this cannot be a case of one photon becoming two photons to conserve energy; energy is explicitly not conserved here, there is just one photon whose wavelength is changing." You seem to be very distraught by this idea, that we might try to use one theory to inform another, however it is necessary in this case because you're asking about a fundamentally gravitational phenomenon (redshift) and we do not have a theory of quantum gravity which easily can be made to model such things. So any quantum model must incorporate ideas about this phenomenon from relativity: otherwise we will simply not describe the phenomenon.

So the answer to "where does that energy go? does it become "more" photons?" is: "no, you're not understanding the phenomenon properly, the phenomenon of redshift is explicitly one where energy disappears. We can probably phrase some quantum-mechanical models where the energy doesn't disappear but probably these will ultimately give the photon an effective mass in its interaction with the other fields which absorb the photon's energy: and thus the theory will fail because the photon is massless. It's part of this phenomenon that the energy disappears. Therefore you need to model this phenomenon non-conservatively, and quantum mechanics is not going to explain it, it's going to assume it."

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  • $\begingroup$ CR Drost: "we do not have a theory of quantum gravity." First of all, thank you. That answers my question. Second of all, Jesus H Christ, it took a dozen "experts," including you, multiple pages of comments on multiple question threads to say that? That answers my question: "QM doesn't HAVE an explanation [yet, if ever.] Thank you. That's all ANYBODY needed to say. $\endgroup$
    – Tommy Jonq
    Commented Dec 5, 2016 at 18:07
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    $\begingroup$ You're welcome. Yes, it takes us multiple pages of comments in multiple question threads to articulate not just the sound-bite that's useful to you but the broader context by which we know that you're asking the wrong question. That's because you're under the narcissistic misapprehension that our answers serve you simply because you posted the question: in fact this is a public forum and my answer was written to serve anyone who has the same general question as you, not just you in particular. In that sense I'm trying to salvage some sort of redeeming value out of this. $\endgroup$
    – CR Drost
    Commented Dec 5, 2016 at 18:18
  • $\begingroup$ CR Drost: Now then, back to your QED assertion that the energy "disappears." Um, do you mean, it "disappears" so far as we are able to measure it from our observation point? That is to say, we are not going to observe a conservation of energy from where we are "standing?" Or, do you mean, it disappears from the universe all together? I was under the impression a long time ago in college that the "missing" work showed up as an increase in the observed strength of the object's gravitational effect on the intervening space time. $\endgroup$
    – Tommy Jonq
    Commented Dec 5, 2016 at 18:23
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    $\begingroup$ I don't think your questions are stupid, nor that this is a waste of my time. When I said your misapprehension was narcissistic (self-directed) I did not mean to say that you are a narcissist in general, just that the way you approached the problem was very "me-me-me"-focused leaving you ill-equipped to understand me, someone who does not care very much about you per se. As for why we don't post an authoritative answer of "I don't know", it's because (a) that's not authoritative and (b) that only helps you, not the broader community. $\endgroup$
    – CR Drost
    Commented Dec 5, 2016 at 18:31
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    $\begingroup$ @TommyJonq This has nothing to do with gravitational waves. I am considering the before and after states where the two atoms will have gained or lost a tiny bit of mass. Since they are located at different gravitational potential, there would be an equivalent tiny change in energy, compensating for the redshift lost. And what tensor? $\endgroup$
    – G. Bergeron
    Commented Dec 8, 2016 at 10:39

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