# Are the gluons that make up a proton considered virtual particles?

1. Jun 30, 2011

### Spinnor

Are the gluons that make up a proton or neutron considered virtual particles?

2. Jun 30, 2011

### SpectraCat

Hmmm ... protons and neutrons are made of quarks not gluons. However, the quarks are held together by exchange of gluons, which are the fundamental particles that mediate the strong force. So, if you want to draw an analogy with the models where the electrostatic force between charged particles is mediated by the exchange of virtual photons, and call the gluons exchanged by quarks to hold them together inside protons and neutrons, then I guess that might have some validity. However, it's also worth noting that virtual particles only arise in perturbative treatments of theories that are too complex to solve exactly, and are considered to be mathematical artifacts by many physicists.

You might try asking over in the particle physics forum to see if you get a more satisfactory answer there.

3. Jun 30, 2011

### Spinnor

Thank you for that picture!

4. Jun 30, 2011

### Spinnor

Gluons carry color charge, photons carry no electric charge, does this change their virtual-ness?

5. Jun 30, 2011

### Drakkith

Staff Emeritus
No. The exchange of photons between objects that experience the electromagnetic force is referring to virtual photons.

6. Jun 30, 2011

Staff Emeritus
First, the mathematical trick that leads to a description in terms of virtual particles doesn't work for calculating proton structure. Additionally, the description of a real particle is usually in terms of a freely propagating and non-interacting wavefunction, and since gluons in a proton are neither, one could argue that they aren't exactly real either. Neither description is very good, and trying to shoehorn things into one or the other is unlikely to be enlightening.

Indeed, when you do a calculation, you have to specify a scale. This essentially forces you to decide when you treat a gluon "near" a quark as part of the color field around the quark (e.g. more like a virtual particle) and when you treat it as an almost-free particle (e.g. more like a real particle). In a world without approximations, it wouldn't matter what scale you picked (so long as you are consistent), but the sort of approximations that are necessary can introduce scale dependencies.

7. Jun 30, 2011

### Drakkith

Staff Emeritus
Ah, ok Vanadium. I didn't realize that.
Spinnor, I'd say that you only need to worry about the quarks.

8. Jul 1, 2011

### tom.stoer

All pictures based on QCD describing the nucleon (lattice QCD, structure functions) do not distinguish between "real" and "virtual" particles. These two terms are due to a perturbative picture, but the description of bound states in QCD is essentially non-perturbative.

E.g. in deep inelastic scattering the perturbative calculations are only used to probe the nucleon structure which itself is intrinsically non-perturbative.

9. Jul 1, 2011

### Islam Hassan

Off-topic but is it possible to distinguish between the mass of the quarks inside a nucleon and the mass of the field they create? From what I recall, this filed is way more massive than the quarks themselves.

IH

10. Jul 1, 2011

### tom.stoer

I guess you refer to the "current quark mass" which is observed in scattering off "asymptotically free" quarks vs. the nucleon mass.

Yes, you are right, in a certain kinematical regime the up- and down-quarks have something like a few MeV, the gluons are massless, but the nucleon itself has nearly 1GeV. In that sense the "field carries much more mass" than the "free quark", but the "free quark" does "exist" only during scattering whereas in an isolated nucleon there is nothing like a "free quark".

These two regimes cannot be described using the same terminology.

In addition when looking at a nucleon you can't say "here's the quark" and "there's the field it creates". This picture of individual objects (like electrons) and their fields (the electric field) break down in non-perturbative physics.

The QCD Hamiltonian which is awfully complicated doesn't allow for a separation of these different energy contributions; it's not something like E(Quarks) + E(Gluons).

Are you familiar with classical electrodynamics? Maxwell's equations?

11. Jul 1, 2011

### Bill_K

I'd be interested to hear what you mean by that, Tom. Of course there's an interaction term too.

12. Jul 1, 2011

### tom.stoer

Look e.g. at p 11-12 of http://eurograd.physik.uni-tuebingen.de/ARCHIV/Termine/Todtmoos/pak2010.pdf [Broken]

The problem is that fixing a physical gauge with Weyl gauge + e.g. Coulomb gauge condition in order to eliminate the longitudinal gluons produces awfully complex integral operators.

In QED you have to invert a gauge-field independent differential operator D=d² resulting in a Greens funktion ~1/k² in momentum space which is ~1/|x-y| which is the standard Coulomb potential between electric charges.

In QCD the same procedure requires to invert a gauge-field dependent differential operator D[A] resulting in a Greens funktion 1/K[A]² containing the gauge field modes in the denominator (!) which cannot be written down explicitly; therefore the "Coulomb potential" in QCD is not known explicitly but is gauge field i.e. operator-dependent. In addition instead of two pure fermionic densities in the numerator the charge density contains both quark and gluon operators. Last but not least these expressions come with nice Fadeev-Popov determinants ...

So the basic difficulty is that the static Coulomb field between two static charges in QED becomes dynamic due to coupling to gluons in QCD.

Last edited by a moderator: May 5, 2017
13. Jul 5, 2011