Since photons are indistinguishable, it’s hard to say too much concretely, but it some sense a diffracted photon is different photon. In order for a photon to interact with say, a diffraction grating, the interaction is done with “virtual photons”.
So for a photon to change course, aka accelerate, it does it by absorbing a virtual photon and emitting another. Whether that is the “same photon” after the interaction is kinda more philosophy than physics, at least to me.
Feynman diagrams are surprisingly accessible for how much information they contain. It’s one way to think about photon (and other particle) reactions.
Are you claiming this is done without a force carrier? If you are working outside the standard model, I guess that’s fine, but I don’t want to spend time arguing with you.
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.
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.
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.
What about diffraction?
Since photons are indistinguishable, it’s hard to say too much concretely, but it some sense a diffracted photon is different photon. In order for a photon to interact with say, a diffraction grating, the interaction is done with “virtual photons”.
So for a photon to change course, aka accelerate, it does it by absorbing a virtual photon and emitting another. Whether that is the “same photon” after the interaction is kinda more philosophy than physics, at least to me.
Feynman diagrams are surprisingly accessible for how much information they contain. It’s one way to think about photon (and other particle) reactions.
There is no tree level photon-photon interaction. Photons scatter off electrons (or any other charged particle), not off neutral photons.
Are you claiming this is done without a force carrier? If you are working outside the standard model, I guess that’s fine, but I don’t want to spend time arguing with you.
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.
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.
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.
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.
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:
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.