# I How are protons, photons and quarks related electrically?

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1. Jul 12, 2016

### referframe

When protons, due to their electric charge, interact with photons are the quarks somehow also involved in this same electric interaction? After all, the quarks do have fractional electric charges.

2. Jul 12, 2016

### Staff: Mentor

For low-energetic photons (everything below ~10 MeV), the proton "looks" like a point-particle, you don't see quarks. This also means the photons won't interact with neutrons in any relevant rate although they have quarks inside.
For high-energetic photons (> 1 GeV), you basically see the quark structure only.
Everything in between is somewhere in between.

3. Jul 12, 2016

### ChrisVer

I am not quiet sure, but it is always better when saying "proton" not to put "quarks" right next to it... and vice versa...
the emission of a photon by a proton is an effective process for the quarks I'd say, but you can't keep track of what's going on [since your best accuracy can be that of the proton]....

4. Jul 12, 2016

### referframe

So, a high-energetic photon could affect the overall (center of mass) motion of a neutron?

5. Jul 13, 2016

### vanhees71

The electromagnetic interaction for quarks is as for any charged spin 1/2 particle,
$$\mathcal{L}_{\text{int}} = -\sum_{j} q_j \overline{\psi}_j \gamma_{\mu} A^{\mu} \psi_j,$$
where $q_j \in \{-1/3,2/3 \}$ is the electric charge of the various quark flavors and $A^{\mu}$ the electromagnetic field.

6. Jul 14, 2016

### referframe

Thanks. Question: Why are you summing over all types of quarks? Wouldn't you need a separate Lagrangian for each possible quark type?

7. Jul 14, 2016

### ChrisVer

that's what the sum represents... and it's for shortening the expression...
since quarks have q=1/3 or 2/3 but couple with the same form to the electromagnetic field, the sum is characterized just by the q_i's...
(the total lagrangian will be the sum of the individual lagrangians)

8. Jul 14, 2016

### Staff: Mentor

The interaction can also transform the neutron to a different particle and/or produce additional particles. At high energies that is much more likely than elastic scattering off a neutron.

9. Jul 15, 2016

### snorkack

What's the lowest excited state of neutron? Delta?

10. Jul 15, 2016

### Staff: Mentor

Δ(1232), 292 MeV heavier than the neutron, is the lightest state with the same valence quark content. Λ(1116) is lighter but then you need an additional kaon which increases the threshold energy much more.

11. Jul 15, 2016

### arivero

In this particular case I think it is safe, as part of the question is why the proton has exactly the same but opposite electric charge that the electron. We have part of the answer: thatt given the SM forces, the only possible distribution of charge among quarks and leptons is this one, because of anomaly cancellation.

12. Jul 16, 2016

### snorkack

So, the reaction n+γ->Δ0 is the lowest reaction with one particle resulting... followed by the Δ0 decays, of which prevalent are Δ0->n+π0 and Δ->p+π-
Can reactions
n+γ->n+π0
n+γ->p+π-
also take place directly, without going through Δ?

13. Jul 16, 2016

### Staff: Mentor

Δ0 has a width of 115 MeV, it does not appear as clear peak in the spectrum. For a given reaction, you can't directly say "this happened via Δ0" or "this happened directly". The reactions could also happen if there would be no Δ0.

14. Jul 16, 2016

### snorkack

So, the legal reactions for eγ would be
eγ,eγ - Compton scattering, elastic, right?
now an electron being Compton scattered is an accelerated charge, and therefore could radiate:
eγ,e2γ
From a suitable threshold energy, the legal processes would include
eγ, e+2e
The next legal process should be:
eγ, eπ0, right?
I suppose that conservation of energy and momentum sets high energy thresholds for inelastic eγ processes, compared to Nγ processes. Correct?

15. Jul 16, 2016

### Staff: Mentor

Right. More photons are always possible but the probability goes down.
You can calculate it.

16. Jul 16, 2016

### snorkack

Neutrons don´t have electric monopole charge, and also don´t have electric dipole moment (banned by CP symmetry) nor electric quadrupole moment (also somehow banned). Yet they do have magnetic dipole moments - unlike neutrinos.
Does the presence of neutron magnetic dipole allow neutrons to undergo elastic Thomson-Compton scattering off photons, while interaction of photons with neutrinos is completely impossible?

17. Jul 16, 2016

### Staff: Mentor

Maybe, I don't know. Would be some "could have happened in the universe once already" process.
It is impossible at leading order. You can always come up with more complex higher order interactions. Completely negligible of course.

18. Jul 16, 2016

### snorkack

On the limit of the photon energy being small compared to the rest energy of scattering charge, Compton scattering is Thomson scattering with the properties: cross-section independent of photon energy, being proportional to q4/m2
Is there any simplification applicable to the limit of photon energy being big compared to the rest energy of the scattering charge?

19. Jul 16, 2016

### Staff: Mentor

Well that q4 term is zero for the neutron. You need to consider smaller effects.
There is a simplification for hadrons: the cross section is about the physical size of the hadron.

20. Jul 17, 2016

### snorkack

Sure. The first effect is magnetic dipole moment.
How do low energy photons scatter off magnetic dipoles?
Hm. What´s the energy from which this applies?
In any case, the simplest charged objects to scatter photons are charged leptons. How do photons of high energy scatter off electrons, muons and tauons?
A tauon has different but similar mass to proton. Omega hyperon is even closer to tauon in mass.
How does the elastic scattering cross-section of an omega hyperon come to diverge from that of tauon at high energies, where the quark structure becomes relevant?