Does a photon experience time while interacting with the weak force?

[DHO232]

Long time reader, first time posting here in physics forums.

I know that according to relativity photons have no proper time. I also know that a photon of sufficient energy can interact with a nucleus's nuclear force via the weak interaction resulting in pair production. I was looking at a Feynman diagram of the event and one axis is time. If the photon itself doesn't experience time then the time reference on the diagram must be from the perspective of the nucleus, right?
So does the photon interact with the weak field before being transformed into mass or does the interaction and creation happen instantaneously from the photon's perspective?
(I say photon's perspective even though a photon shouldn't have a perspective)

My guess is that it is the ladder, because if the photon interacted with the weak field at a separate time than the occurrence of electron and position creation then it would imply that the photon experienced time, had a proper time, and relativity had something wrong here.

I am not sure either way, I would greatly appreciate input. Thankyou.

 

[Orodruin]

If the photon itself doesn't experience time then the time reference on the diagram must be from the perspective of the nucleus, right?

It is a collision, not a decay. It has no particular time reference other than the particles colliding. You can describe this from any inertial frame.

Also, the typical leading contribution to pari production involves an exchange of an additional virtual photon attaching to the electron line.

Edit: It should also be pointed out that Feynman diagrams should not be seen as a process where particles run back and forth. It is a representation of a contribution to the expansion of the amplitude for a particular process. If you insist on that interpretation, all of the possible contributing diagrams are occurring and it is impossible to interpret a single diagram as the process (although it can represent the dominant contribution to the process) just as you cannot tell which slit a particle goes through in a double slit experiment. The amplitude is the sum of the amplitudes for all possible histories.

 

[DHO232]

It is a collision, not a decay. It has no particular time reference other than the particles colliding. You can describe this from any inertial frame.

Also, the typical leading contribution to pari production involves an exchange of an additional virtual photon attaching to the electron line.

Edit: It should also be pointed out that Feynman diagrams should not be seen as a process where particles run back and forth. It is a representation of a contribution to the expansion of the amplitude for a particular process. If you insist on that interpretation, all of the possible contributing diagrams are occurring and it is impossible to interpret a single diagram as the process (although it can represent the dominant contribution to the process) just as you cannot tell which slit a particle goes through in a double slit experiment. The amplitude is the sum of the amplitudes for all possible histories.

Aren't virtual particles just a mathematical tool used in Feynman diagrams though? I was under the impression there was no experimental evidence proving the physical existence of virtual particles. Obviously I could just be misinformed, but if there are no physical virtual particles then what are the photons colliding with?
(Unless we are assuming that photons require a collision to interact which would lead us to interoperate pair production as indirect evidence of virtual particles. But that feels a little circular to me)

 

[Orodruin]

Aren't virtual particles just a mathematical tool used in Feynman diagrams though?

I don't see how that contradicts what I said. Quite on the contrary.

 

[DHO232]

I don't see how that contradicts what I said. Quite on the contrary.

Because if they are just a mathematical tool then they aren't physical there, and without a physical particle to collide with what is the second half of the collision event?

I guess I mean: If a virtual particle is just a place holder for the unknown because we think there probably is a particle-particle interaction happening then we don't really know what's happening at all.

Obviously I don't really have any real idea myself and I could be missing something obvious but I'm missing a physical component in all of this... if it isn't a wave-field interaction then the only other option is that virtual particles are more than mathematical tools and physically exist in reality, right?

 

[Orodruin]

Because if they are just a mathematical tool then they aren't physical there, and without a physical particle to collide with what is the second half of the collision event?

I get the feeling that you are strongly over-interpreting what "physical particle" means and placing some unwarranted meaning to the word "real". Particles in quantum field theory are not some small balls or point masses that physically collide, they are particular excitations of quantum fields.

 

[DHO232]

I get the feeling that you are strongly over-interpreting what "physical particle" means and placing some unwarranted meaning to the word "real". Particles in quantum field theory are not some small balls or point masses that physically collide, they are particular excitations of quantum fields.

That is definitely what I'm doing and was thinking. I assumed a particle had to have a charge and rest mass to interact with a photon. They say you are describing it, it sounds closer to a momentary formation of energy from the nucleus into something like a finite potencial barrier. (Sorry if I worded it poorly)
If particles are just perturbing themselves into existence in time with the photon to allow for a collision then how is the charge conservation delt with? Photons requie charged particles to interact, right?

(I think virtual particles are described by a perturbation theory, not just popping into existence)