How does a photon interact with the EM field of a nucleus?

  • #1
cemtu
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How does a photon interact with the EM field of a nucleus?
The photon must be near a nucleus in order to satisfy conservation of momentum, as an electron–positron pair produced in free space cannot satisfy conservation of both energy and momentum.[4] Because of this, when pair production occurs, the atomic nucleus receives some recoil. https://en.wikipedia.org/wiki/Pair_production
How does a photon interact with EM field of a nucleus thus exchange momentum and recoil the nucleus when pair production happens?
 
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  • #2
What kind of an answer are you looking for? How does this differ from "how does a photon interact with the EM field of an electron"?
 
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  • #3
Vanadium 50 said:
What kind of an answer are you looking for? How does this differ from "how does a photon interact with the EM field of an electron"?
Exactly! How does photon interact with EM field of electron?
 
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  • #4
cemtu said:
Exactly! How does photon interact with EM field of electron?
I assume you are looking for a force between photons and electrons and maybe a virtual particle mediating this force.
I can't really help you, but I fully understand why you ask.

Scattering of a mass by another mass is due to gravity and gravitons.
Scattering of a charge by a charge is due to electric forces mediated by virtual photons.
What about scattering of photons by electrons?

Does that describe what you are wondering about?

The only answer I've found so far is that there is no such force and that scattering of photons is completely different from other types of scattering.
 
  • #5
So for the pair production cross section I see expressions here:
https://en.wikipedia.org/wiki/Gamma_ray_cross_section
said to be Maximon equation.
The problem is that Maximon equation depends on Z alone. The article states that The energy threshold for the pair production effect is k=2 (the positron and electron rest mass energy). Which is obviously false. The threshold is k=4 for electron, and necessarily bigger than 2, for the momentum and energy conservation reason, for any finite mass of nucleus.
Pair production from protons and deuterons necessarily have different energy thresholds (higher threshold for proton, lower for deuteron). For energies above the threshold for both, how does the cross-section depend on the mass of the nucleus? Other factors, like charge distribution of the nucleus?
 
  • #6
Philip Koeck said:
Does that describe what you are wondering about?
Yeah exactly!
Philip Koeck said:
The only answer I've found so far is that there is no such force and that scattering of photons is completely different from other types of scattering.
Thank you tho!
 
  • #7
snorkack said:
Other factors, like charge distribution of the nucleus?
No, not related to charge distribution, but related to proton number(Z), photon energy(E), and material's density(ρ).
 
  • #8
cemtu said:
Exactly! How does photon interact with EM field of electron?
That's easy. It's described by the interaction Lagrangian
$$L_I=eA_{\mu} \bar{\psi} \gamma^{\mu} \psi.$$
For the proton it's a bit more involved, because it's a composite object, and there are form factors involved. See, e.g.,

https://iopscience.iop.org/article/10.1088/1742-6596/299/1/012002
 
  • #9
cemtu said:
No, not related to charge distribution, but related to proton number(Z), photon energy(E), and material's density(ρ).
Material´s density (ρ) should not be directly relevant. The distance between nuclei in solids, let alone gases, is huge compared to the wavelength of photons energetic enough to form electron-positron pairs.
Or mass... How does the pair production cross-section compare between protium and deuterium?
 
  • #10
The Feynman diagram for pair production has the incident
photon emitting the pair, and then one of the charged particles absorbing a virtual photon from the atom.
 
  • #11
cemtu said:
Exactly! How does photon interact with EM field of electron?
I think what cemtu is asking is whether or not there is a force between a photon and an electron, or maybe by what mechanism momentum and energy is transferred for example in Compton scattering.
 
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  • #12
The notion of force is a non-relativistic concept. In relativistic physics we deal with local interactions and thus fields. Photons are in no way describable as localized point particles.
 
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  • #13
vanhees71 said:
The notion of force is a non-relativistic concept. In relativistic physics we deal with local interactions and thus fields. Photons are in no way describable as localized point particles.
A related question that might help to improve the mental picture:

What is the difference between a photon that's part of radiation/EM-waves and one that mediates the EM-force?

What I can see is that EM-waves can be detected with a suitable detector, but if you put a detector between 2 charges you won't see anything no matter what detector you use. Is that right?

So what is the fundamental difference? If there is a short answer to that!
 
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  • #14
Philip Koeck said:
A related question that might help to improve the mental picture:

What is the difference between a photon that's part of radiation/EM-waves and one that mediates the EM-force?

What I can see is that EM-waves can be detected with a suitable detector, but if you put a detector between 2 charges you won't see anything no matter what detector you use. Is that right?
I would rather say that you see something but you see something different.
Consider a simple dipole antenna. And suppose it is statically charged with charges at both end.
There is a strong electric field around it, which you can easily detect. It is a purely electrostatic field, and it carries no energy.
Now put the antenna oscillating. You can still measure the electrostatic field at any point and time, and now you also observe magnetic field. But when you watch the dipole from different directions, you will find that it fades in different manner in different directions. In some directions (along the dipole) it fades exponentially, in others with just the square of distance. This is the difference between freely propagating electromagnetic waves and evanescent waves.

My understanding is that real photons is free propagating electromagnetic waves quantized, and virtual ones are all other forms of electromagnetic fields viewed as quantized - evanescent waves, electrostatic and magnetostatic.
 
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  • #15
Philip Koeck said:
ha

Philip Koeck said:
What is the difference between a photon that's part of radiation/EM-waves and one that mediates the EM-force?
The photon exchanged in a Feynman diagram is virtual, an artifact in perturbation theory that does not actually exist.
 
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