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Consider a simple model:

A single photon of light in a vacuum travels from (-1,0) to (0,0) , where it hits a reflective point at the origin at time T.

CASE 1: Suppose the velocity of light follows a continuous curve (i.e. the plot of v VS t for this photon is a continuous curve). Then, at t < T it was traveling at v=+c, and at t > T it is traveling at v=-c, so under continuity assumptions, the velocity must be 0 at some point, and so the speed must be 0 at some point, not possible.

CASE 2: Otherwise, suppose the plot of v VS t is a discrete curve (i.e. the v vs t graph is simply +c when t<T and -c when t>T). Then the photon goes from a velocity of +c to -c instantaneously at time T. Notwithstanding that its exact velocity would be incomputable at the exact time of contact (is it +c or is it -c at time T?), we know that this implies instantaneous reflection. Hence, discrete curve implies instantaneous reflection.

Thus, the contrapositive of this conclusion is that in a world with non-instantaneous reflection, we cannot have discrete curve, i.e. we must be in CASE 1. But in the real world where absorption and emission (i.e. reflection) takes time, we arrive in CASE 1, and so the mere fact that light changes directions seems to be contradictive to the fact that an individual photon of light must remain at the same speed all the time. (I understand that the energy transfer takes time and hence slows down light, but consider again how its velocity could ever go from +c to -c in the real world (i.e. non-instantaneous reflection) as that would imply it taking on a speed of 0 in between, in this continuous-curve case).

What am I misunderstanding here? It appears that it can't be the same photon that is being emitted, correct? That would actually settle the contradiction here, though it's not the accepted paradigm.

Qmechanic
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Raj R.
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    One a general note, the mental model that photons are "traveling" corpuscles is simply false. That the speed of light is a well defined quantity inside of a reflecting matter surface is also false. That's why we have concepts like "phase velocity" and "group velocity". – FlatterMann Mar 31 '23 at 06:30
  • I understand I probably have the wrong mental model. First, could you suggest any reading material that would help me visualize this better? I'm trying here but it's a bit hard to really understand the photon nature; this was my attempt which is obviously flawed :). Second, if we considered this just as a pure theoretical exercise (not practically speaking, but in theory)... then if we played the game "there is a point with a property that moves continuously and it cannot change speed. Can that same point be reflected?". That answer is (probably) obviously no then, right? – Raj R. Mar 31 '23 at 06:51
  • @BillAlsept A photon is an irreversible energy transfer between an electromagnetic field and an external system. The mental model of photons as "energy packets" is also false. It is tempting to think of the electromagnetic field as being made up of atomistic particles, but neither experiments nor the theory support that. – FlatterMann Mar 31 '23 at 07:02
  • A photon can be defined as an irreversible exchange of energy, momentum and angular momentum between the electromagnetic field and an external system. A reflecting surface is not an irreversible process, so it's a bit problematic to talk about corpuscular reflection of photons in the first place. It is much easier understood in classical wave terms using Maxwell's theory. Since such a reflection process does not involve Planck's constant in the first place, the quantum mechanical picture is completely irrelevant to begin with. – FlatterMann Mar 31 '23 at 07:06
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    @FlatterMann When you say "It is tempting to think of the EM field as being made up of atomistic particles, but [experiments don't support that]" how can you reconcile that with the pointlike detection of photons e.g. in each single hit of the double slit experiment? Not sure how to interpret your statement in a way that isn't in contradiction with such experiments. – doublefelix Mar 31 '23 at 07:17
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    @FlatterMann In your second comment, also it seems that you are saying there is no need to address reflection in a quantum mechanical picture because it can be explained classically. But, if the quantum is the more fundamental theory, it is not complete without describing all phenomena, regardless of whether something could in principle be computed without it. As such, there should be an explanation of reflection from a quantum mechanical point of view. – doublefelix Mar 31 '23 at 07:21
  • @doublefelix I never had a point-like detector. They have absolutely horrible quantum efficiency (0%). A physical detector is an extended physical system that absorbs energy from the electromagnetic field. No matter how small we make it, it still has to absorb the energy (and one Planck unit of angular momentum). The "position" of the photon is always the position of the detector. If we mount a detector in front of one of the slits of the double slit experiment, then it blocks that slit, which leaves us with one open slit and the remaining light will show the diffraction pattern of that slit. – FlatterMann Mar 31 '23 at 07:35
  • I mean that the detector gives a pointlike signal, not that the detector itself is pointlike. In the sense that if you have for example a multichannel plate, you generally get a detection in only one channel. This is the case even if the wavelength of the light is far greater than the physical separation between channels. Can this be explained without a particle model? Who knows, but I don't know of any such explanation that is backed by the math & not just story-like. Maybe you do though? – doublefelix Mar 31 '23 at 07:42
  • @doublefelix If I mount a multichannel plate in front of one of two slits, there is still only one open slit left. I generally don't understand the fuss that Feynman made about the double slit. It doesn't show us anything about quantum mechanics. Even with PMTs or a MCP all we will ever measure is the classical intensity distribution according to Maxwell. If we want to see quantum effects, then we need, at least, an experiment that is sensitive to Planck's constant OR a two-photon correlation experiment. The double slit is neither. – FlatterMann Mar 31 '23 at 07:51
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    What do we see in a single-photon run of a double slit-like experiment? For the purposes of this the slits are not relevant, I'm just talking about the detection event itself. Yes of course if we integrate many runs we can see it as an intensity distribution, but I'm looking at the physics of what happens in one run. How can this localization to only one channel of an MCP be explained without referring to particles of light? I mean maybe it can, but the burden of proof is still on you, as it was your claim that particles are not necessary. – doublefelix Mar 31 '23 at 08:12
  • Yes the burden of proof is still on someone to show why its not individual photon particles causing all light phenomena. The so called electromagnetic field is nothing more than trillions of individual photons passing through every point in the universe, every second. No one can explain the field without resorting to photons. Also individual photons can explain all light phenomena but waves cannot. You are kicking the can down the road when you speak of electromagnetic fields or waves. – Bill Alsept Mar 31 '23 at 15:10
  • I appreciate the discussion but now am even more confused than before :D any response to the original question , under the premise that light is indeed a photon “particle” ? Bill, you said “when a photon hits an electron another electron delivers its energy in a random direction” i still see the paradox of “if that direction is opposite to the original photons direction, doesn’t this violate it traveling at the speed of light ? “ . My visualization is: the original photon “stops” then “reverses directions” as a point would - Apologies if this is causing confusion , appreciate the discussion . – Raj R. Mar 31 '23 at 15:55
  • @user3000877, my comment was supposed say Photons are individual packets of energy and that's the best way to think of them. When a photon delivers its energy to an electron, another PHOTON instantly carries the energy away in some random direction. Imagine running at a constant speed carrying a baton and handing it off to another person running the opposite way at the same constate speed. – Bill Alsept Mar 31 '23 at 21:07
  • In the analogy, people are the photons and the baton is the energy. – Bill Alsept Mar 31 '23 at 21:17
  • @user3000877 You should always visualize photons as individual particles propagating from here to there. Not only does it explain every light phenomena, but also its easy to describe an actual photon particle model. No one can physically describe a models to explain a light wave without incorporating photon particles. Nor can they physically describe an electromagnetic field without photons. – Bill Alsept Mar 31 '23 at 21:33
  • @user3000877 Seems like FlatterMann is taking the "light is a wave" approach and Bill is taking the "light is a particle" approach. In my experience neither of the two perspectives alone has been sufficient to explain all phenomena. The difficulty of simultaneously thinking of particles as particles and waves is related to issues with the measurement formalism in quantum mechanics in general, and my perspective is that at the moment there is no 100% clear resolution, even >100 years after the whole debate began with Planck, Einstein & others. Not that we haven't learned since then, but – doublefelix Mar 31 '23 at 23:09
  • some things still remain unclear. I don't know the answer to your question and I think it's an interesting one. I believe that there are basic aspects of how light works at the microscopic level that are still not known today and that this question tends in that unexplained direction. That's why I upvoted and followed it. If I am wrong, and "light is only a wave" or "light is only a particle" I would be glad for Bill or FlatterMann to justify that, but neither approach is really the commonplace interpretation of today, which sees light as acting both wave- and particle-like. – doublefelix Mar 31 '23 at 23:12
  • That being said, I still think there is a pretty straightforward interpretation that resolves the paradox in your question, which is that the reflected light is absorbed and then new light is re-emitted. In between that absorption and re-emission, maybe the photon just doesn't exist. This is actually a question Richard Feynman's Dad asked him, and RF said that the photon is just gone in between the absorption and re-emission time, although I don't think that there is any experiment which really can verify this. – doublefelix Mar 31 '23 at 23:23
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    For the record, RF's dad was never happy with that answer, of just assuming the photon disappears completely from the universe in between absorption and re-emission. And I do agree that it seems a bit lacking of a deeper explanation. What that explanation might be I don't want to guess. – doublefelix Mar 31 '23 at 23:24
  • Thank you! I love hearing stories like the one you shared - are there any books with popular science stories like that? How have you heard of that one? – Raj R. Apr 01 '23 at 01:51
  • I've read a bunch of the Richard Feynman books out there (What do you care what other people think, Surely you're joking mr. feynman, a few others). I don't remember which one this was from but you might be interested in those two. If you're interested in quantum mechanics history and the dialogue and personalities of the founders who pieced together the theory, I also really recommend "The Age of Entanglement". – doublefelix Apr 01 '23 at 05:18
  • Aha! It was from a video actually! From "The pleasure of finding things out": https://www.facebook.com/watch/?v=602263401459387 . Maybe also in one of the books, who knows. – doublefelix Apr 01 '23 at 05:20
  • It's actually a full series. But it seems not to be on YouTube. I uploaded it once and it got taken down. Here is a link to all the episodes if you are interested (they are short). I have a collection of RF video/audio/books haha. https://drive.google.com/drive/folders/0Bw-X3iCjAWPoTWtvMzRYelpEX1E?resourcekey=0-wLWOMepjz106PMQVMRLBoQ&usp=sharing – doublefelix Apr 01 '23 at 05:27
  • Possible duplicates: https://physics.stackexchange.com/q/35177/2451 , https://physics.stackexchange.com/q/83105/2451 and links therein. – Qmechanic Apr 01 '23 at 10:08
  • @doublefelix The "localization" is trivial: YOU have built a segmented detector. The electromagnetic field didn't do that. All the em field can do is deposit one quantum of energy somewhere. That YOUR detector can tell "where" that happens is a function of its inhomogeneity. If all you have is a large unstructured metal plate, then it will never tell you where that irreversible energy exchange has taken place. That's why MCPs have all those tiny holes... – FlatterMann Apr 01 '23 at 10:20
  • @doublefelix There are no photons in the free field. A photon is an irreversible energy exchange and there are no irreversible processes in the free field. Einstein got this wrong in 1905 in his photoelectric effect paper and physicists have still not found a good way of talking about this, almost 120 years later. That, in my opinion, is a complete disgrace to our teaching culture. – FlatterMann Apr 01 '23 at 10:24
  • @FlatterMann In response to your first comment about the localization: Is there a way this can be derived mathematically? That in response to (e.g.) an MCP, there will be a local excitation of the EM field? Without using the Born rule. This would be convincing to me, that there may be no particle-like description necessary. But without such a derivation my skepticism would remain, justifiably, with all due respect. – doublefelix Apr 01 '23 at 16:26
  • Without the math to show that this localization phenomenon happens with only a wave-like description, I can only regard the explanation in your comment as a story & not a proof. Again I mean that only with all due respect and from a scientific POV, I have enjoyed your thoughts on things on the site in general. – doublefelix Apr 01 '23 at 16:27
  • Btw, Qmechanic has linked a duplicate post, with IMO a good answer from John Rennie. – doublefelix Apr 01 '23 at 16:30
  • @doublefelix Can the irreversible function of an MCP derived mathematically? Yes. The people who are building them are using physics to optimize their products all the time. – FlatterMann Apr 01 '23 at 16:30
  • @FlatterMann Thinking on this I am skeptical, because one could in principle set up an EM field which is symmetric with respect to the MCP's channels, and with total symmetry in the problem what would determine which MCP channel gets a signal? – doublefelix Apr 01 '23 at 16:33
  • The proof I would need to be convinced that particles are not necessary here would need to predict which channel the signal appears in just from the initial conditions of the EM field and the MCP setup. Unless you want randomness to be involved? I had in mind that you were looking at a treatment just from Maxwell's equations – doublefelix Apr 01 '23 at 16:35
  • @doublefelix The MCP itself is not symmetric. It has a symmetry breaking electric field inside. If you turn that off, it stops working. – FlatterMann Apr 01 '23 at 16:37
  • @FlatterMann And another thought, even stronger than the symmetry: Can one predict that each run with identical initial conditions will get a signal in a different MCP channel? A solution of maxwell's equations starting from the initial conditions of the MCP would at best predict the same channel to be excited every time, no? The whole thing is deterministic – doublefelix Apr 01 '23 at 16:42
  • @doublefelix Maxwell's equations are a mean field theory, so if you are making a mean field measurement with an MCP, then you will always get the same result within the statistically expected error bars. – FlatterMann Apr 01 '23 at 16:44
  • @FlatterMann I think that is changing the topic. The question we have been discussing is, "can one predict what happens in a single-detection run of an EM wave hitting an MCP, without resorting to a particle description". More than once now I have redirected back to this question in our discussion. – doublefelix Apr 01 '23 at 17:49
  • @doublefelix There are no particles in an MCP. There are only quanta of energy. We need to stop repeating Einstein's 1905 mistake over and over, again. – FlatterMann Apr 01 '23 at 18:45
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    With all due respect, you are stating your point without backing it up with proof. As such I have no reason to believe it. The question is, without using a particle description, how else can we derive mathematically that only one channel activates in the MCP, and with repeated experiments on the same initial conditions, different channels activate. This would not be possible with just Maxwell's equations, and in the QM formalism, not possible without a measurement postulate that links the final outcome to a particle. – doublefelix Apr 01 '23 at 18:57

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