How are protons, photons and quarks related electrically?

[LarryS]

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.

Thanks in advance.

 

[mfb]

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.

 

[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]...

 

[LarryS]

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.

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

 

[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.

 

[LarryS]

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.

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

 

[ChrisVer]

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

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)

 

[mfb]

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

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.

 

[snorkack]

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

 

[mfb]

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.

 

[arivero]

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]...

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.

 

[snorkack]

Δ(1232), 292 MeV heavier than the neutron, is the lightest state with the same valence quark content.

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?

 

[mfb]

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.

 

[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?

 

[mfb]

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?

Right. More photons are always possible but the probability goes down.

I suppose that conservation of energy and momentum sets high energy thresholds for inelastic eγ processes, compared to Nγ processes.

You can calculate it.

 

[snorkack]

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.

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?

 

[mfb]

Maybe, I don't know. Would be some "could have happened in the universe once already" process.

while interaction of photons with neutrinos is completely impossible?

It is impossible at leading order. You can always come up with more complex higher order interactions. Completely negligible of course.

 

[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?

 

[mfb]

Well that q⁴ term is zero for the neutron. You need to consider smaller effects.

Is there any simplification applicable to the limit of photon energy being big compared to the rest energy of the scattering charge?

There is a simplification for hadrons: the cross section is about the physical size of the hadron.

 

[snorkack]

Well that q⁴ term is zero for the neutron. You need to consider smaller effects.

Sure. The first effect is magnetic dipole moment.
How do low energy photons scatter off magnetic dipoles?

There is a simplification for hadrons: the cross section is about the physical size of the hadron.

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?

 

[ChrisVer]

well you can't scatter photons on taus since taus are pretty short lived particles...but their interactions with photons shouldn't be much different to electrons' [although you'd want to keep their mass for photons/momenta exchange below 2GeV]

 

[mfb]

You can look up most of those things yourself...
Also, this is starting to get off-topic.

For high-energetic photons (GeV+), baryons "look" like a huge collection of sea quarks, within the size of the baryon. A tauon doesn't have a size.

 

[ChrisVer]

A tauon doesn't have a size.

say that to people who work on tau substructure ... [word play]
But yes, in general it becomes messy... how would you study photon elastic scattering off something that cannot be a 'target'? All that comes in my mind right now, is you get your object you want to study and bombard it with photons, measuring their angular distributions in the end... this obviously can't work for particles that are unstable.

 

[Vanadium 50]

It is impossible at leading order. You can always come up with more complex higher order interactions. Completely negligible of course.

I believe that the electric coupling to the neutrino is zero at all orders because of gauge invariance.

The magnetic dipole moment's leading term is one loop, and

[tex]
\mu_\nu = \frac{3eG_Fm_\nu}{8 \pi^2 \sqrt{2}} = \frac{3G_Fm_em_\nu}{4 \pi^2 \sqrt{2}}\mu_B = 3 \times 10^{-19} \frac{m_\nu}{1 {\rm eV}} \mu_B
[/tex]

or really, really, really small.

I leave it as an exercise to the student to explain why the neutrino mass is in the numerator.

 

[snorkack]

Note that the term q⁴/m² means that the low mass of individual up and down quarks ought to promote scattering off them, in hadrons that contain them. Correct?

 

[mfb]

A neutron doesn't consist of isolated quarks, and light quark masses are a problematic concept. Low-energetic photons don't see those isolated quarks at all (they would scatter coherently, and then only the sum of charges matters, which is zero).