• ziggurism@lemmy.world
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    1 年前

    There is no tree level photon-photon interaction. Photons scatter off electrons (or any other charged particle), not off neutral photons.

      • ziggurism@lemmy.world
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        1 年前

        The electromagnetic field does have a force carrier. It is the photon.

        The photon mediates the force between electrically charged particles. It cannot mediate any reaction between two neutral photons.

        • Ook the Librarian@lemmy.world
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          1 年前

          Ah, I see. Sorry for the snark. I was thinking more in line with the Compton effect, and thought you were talking about something like that too. (Even though it’s clear that you were explicitly not. I thought you were denying photon-virtual photon interaction because I was talking about it in a funny way.)

          I would still say it’s still philosophical whether photons experience acceleration, but I concede that photon-photon interaction is not done by virtual photon exchange.

          • ziggurism@lemmy.world
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            1 年前

            I am indeed denying the existence of photons interacting with virtual photons. I am also saying there is no tree level photon-photon interaction of on shell photons. Neither Compton scattering nor Bhabha nor pair production nor pair annihilation involves a photon-photon interaction. There is no photon-photon vertex in QED. There is no tree level Feynman diagram that you can look at and say “this is, at least philosophically, a photon changing its momentum”.

            There is a 1 loop diagram that represents photon-photon scattering. But even that doesn’t have any photon-photon vertices, instead it is mediated by electron-positron pair.

            Non-abelian gauge bosons (gluons) couple to themselves. So does gravity (gravitons). Abelian ones (photons) do not.

            Photons don’t accelerate. They are emitted or absorbed. That’s their only interaction.

            • Ook the Librarian@lemmy.world
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              1 年前

              Someone asked if diffracted light accelerated. I said no. A diffracted photon is a different photon.

              I gave some lip service to the Feynman “there is but one electron” idea, and you seemed to take that personally.

              If someone asks you if diffracted light accelerates, answer them how you want. I just thought it’d be cool to show them Feyman diagrams.

              • ziggurism@lemmy.world
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                1 年前

                Look bro. Your top level comment that I replied to was generally correct, and also very helpful. I liked it. I liked the suggestion for people to look at the Feynman diagrams. I agreed with it. I upvoted it.

                I hope I’m not giving you the impression that I’m taking a personal issue with you. I’m not. I like you and I hope we’ll still be friends when this is all over. I promise to read Discworld soon.

                The only quibble I had with what you wrote was this one sentence:

                So for a photon to change course, aka accelerate, it does it by absorbing a virtual photon and emitting another.

                Photons do not absorb virtual photons. And real on-shell photons do not interact. In Compton scattering and 1 loop photon-photon scattering, you can think of photons emitting e+e- pairs. But never do they emit or absorb other photons.

                Maybe that’s not what you meant with that sentence, and I misunderstood. If that’s the case, my bad. Maybe you didn’t need the explanation. If someone else made the same misunderstanding reading your comment that I did, then maybe my comments will help them even if you don’t need them. Or if not, if it’s just me being dumb, well c’est la vie.

                Cheers bro.

                • Ook the Librarian@lemmy.world
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                  1 年前

                  You’re right. And I’m the one being less than friendly. It’s nothing personal. It’s just something I’ve noticed about myself. It’s that I hate talking about physics on the internet.

                  I’m high on lemmy, not in my office. I read a terrible meme. So I open the comments, and see your comment. It was exactly what I was thinking. “Photons don’t accelerate.” Which I took to mean “your meme is bad and you should feel bad”. And again, I agree, it is horrible, this meme.

                  I like to shoot the shit about, say, quantum loop gravity (i’m honestly clueless about it) with people at the office, but on lemmy, academics piss me off. I don’t know why.

                  So from your reply, natural question arises: What about diffraction?

                  You went academic. I’m high. So I just steer them to a right answer while bringing up less academic (but valid (maybe)) ideas about philosophy. I did that because I hate when academics try to seriously discuss that “there is only one electron idea” and similarly unfalsifiable crap. That shit belongs on dumb internet forums with bad memes. And man did I find a bad meme. So was angling for a stupid debate about whether any particle can ever accelerate. You can’t trace them from idenitical copies. Are they the same particle after an interaction knowing that force carriers exist in the standard model? Not an actual quantum field theory debate.

                  But to give you some closure. I do see that I clearly did imply a tree-level interaction in my initial reply. It is wrong to say a photon emits anything. You were also very direct in your correction. I read it along with other comments and must have confused myself. So in all the back-peddling I was doing, I was avoiding defining “an interaction”. I was just trying to say any influence is an interaction. Not two photons touching on a diagram.

                  Also, I have a vague memory in grad school. Two people smarter than me were debating whether in a universe consisting only of 3 photons, would they be able to interact? I couldn’t focus on what was said. I was having an existantial crisis. So I had that clacking around in the back of my head. So I’m just going to stop writing now, because as I mentioned, I’m high. So I should just stop.

                  • ziggurism@lemmy.world
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                    1 年前

                    I’ll level with you. I know how to use QED to compute the cross section of a scattering reaction. But I do not remember, or perhaps never knew, what the QED theoretic description of classical wave mechanical phenomena like diffraction, reflection, refraction, and dispersion look like.

                    Well… actually of those phenomena, I think diffraction is fine. A single waveform will exhibit diffraction. It doesn’t entail any interactions. A single photon can still exhibit a diffraction pattern. It doesn’t mean that the photon has changed directions or circled around or in any way accelerated. The only reason you might think so is that you’re thinking of photons as billiard ball type classical particles, but of course they are not, they are quantum particles with spread out wavefunctions.

                    Dispersion I guess is just scattering combined with absorption re-emission (and as we discussed, even scattering is itself a form of absorption & re-emission). But as for reflection and refraction? Those are the phenomena that Entropius was pointing to elsewhere in this thread. I remember how those look in terms of solutions to Maxwell’s equations and boundary conditions, but that’s classical wave mechanics. I do not remember how to translate that into the language of QED.

                    QED is a fundamental theory, so I assume that a description exists, and of course because I know what QED looks like, so I am certain that it will still be true that in this description, photons will be absorbed & emitted by charged particles, but photons will not interact with photons. However beyond that I cannot say much. How do we describe reflection of light in a mirror as photons scattering off electrons? I don’t know exactly.

                    One thing I can say is that generally classical states are modeled in quantum mechanics as coherent states, which are eigenstates of the annihilation operator. They look something like exp(N)|0> where N is the number operator, which means that they are states with a superposition of 0 photons, 1 photons, 2 photons, etc. They don’t have a well defined number of particles. So maybe if you want a QED theoretic description of reflection, you can have it, but you won’t be able to talk about specific numbers of photons. But again, I don’t know the details of this.

                    I wonder whether this concept of classical waveforms as coherent states with a superposition of all numbers of particles will help at all with this philosophical debate about whether two photons are the same particle or not, or about whether you can have a universe with only 3 photons