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

Click For Summary
SUMMARY

The discussion centers on the interaction of photons with the electromagnetic (EM) field of atomic nuclei, particularly during pair production events. It is established that a photon must be near a nucleus to satisfy conservation of momentum, resulting in recoil of the nucleus. The conversation highlights the differences between photon interactions with electrons and nuclei, emphasizing that scattering of photons is fundamentally different from other types of scattering, such as gravitational or electric forces. Key factors influencing pair production include the proton number (Z), photon energy (E), and the mass of the nucleus.

PREREQUISITES
  • Understanding of pair production and conservation laws in physics
  • Familiarity with the concepts of virtual particles and Feynman diagrams
  • Knowledge of electromagnetic theory and photon interactions
  • Basic grasp of quantum mechanics and particle physics
NEXT STEPS
  • Research the Maximon equation and its application to pair production cross-sections
  • Explore the role of virtual photons in mediating electromagnetic forces
  • Investigate the differences between real and virtual photons in quantum field theory
  • Study the implications of charge distribution and density on photon interactions with nuclei
USEFUL FOR

Physicists, students of quantum mechanics, and anyone interested in the fundamental interactions of photons with matter, particularly in the context of particle physics and electromagnetic theory.

cemtu
Messages
99
Reaction score
7
TL;DR
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?
 
  • Like
Likes   Reactions: Philip Koeck
Physics news on Phys.org
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"?
 
  • Like
Likes   Reactions: vanhees71 and cemtu
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?
 
  • Like
Likes   Reactions: Philip Koeck
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.
 
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?
 
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!
 
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).
 
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
 
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.
 
  • Like
Likes   Reactions: cemtu
  • #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.
 
  • Like
Likes   Reactions: Philip Koeck
  • #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!
 
  • Like
Likes   Reactions: cemtu
  • #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.
 
  • Like
Likes   Reactions: vanhees71 and Philip Koeck
  • #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.
 
  • Like
Likes   Reactions: Philip Koeck and vanhees71

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K