This is a rather artificial scenario, but it has been bugging me lately.


Due to the confinement in QCD, quarks are bound in color-neutral configurations. Any attempt to separate a quark from this bound state costs so much energy that it's enough to pair-produce new quarks, hence the quark-jets in accelerator experiments.


I'm now considering the reversed (hypothetical) scenario. Assume the you initially have two quarks (up and anti-up for instance) that are placed far away from each other. Buy far I here mean further than any other length scale in CQD. Now, let the two quarks approach each other, as in a scattering experiment.


At what distance does the two quarks start to interact, and what happens? Since the strong force is confining, the interaction should be stronger the further away the quarks are, but they cannot interact outside of their causal cones, so how does this work at really long distances?

My thoughts

I'm imagining that the "free" quarks are in a metastable state and the true ground state is the one where several pairs of quarks have pair-produced to bind with the two initial quarks. Thus the closer the two initial quarks are, the smaller the energy barrier between the metastable and the true ground state becomes. Thus at some separation $r$ there is a characteristic time-scale before pair-production occurs.

  • $\begingroup$ I think you're confusing asymptotic freedom and confinement in your background section $\endgroup$
    – innisfree
    Jun 15, 2015 at 19:29
  • 2
    $\begingroup$ secondly, what do you mean out of causal contact? you're talking about particles, not events. if you wait long enough, the world line of one will be inside the lightcone of the other $\endgroup$
    – innisfree
    Jun 15, 2015 at 19:34
  • $\begingroup$ in the reverse part of the story, where is the gluon ? $\endgroup$
    – user46925
    Jun 15, 2015 at 23:33
  • $\begingroup$ @innisfree: Sorry, I was sloppy with the notion of causal contact. I've updated the background and set-up to comply with you comments. $\endgroup$ Jun 18, 2015 at 10:36
  • $\begingroup$ @igael: The gluon should be there. This is still QCD, only with particles at very large distances. Do you now if there is something inherently unstable in just creating one isolated quark? $\endgroup$ Jun 18, 2015 at 10:38

2 Answers 2


I don't think this assumption is legit:

Assume the you initially have two quarks (up and anti-up for instance) that are placed far outside of causal contact with each other

for a real experimental setup. Indeed, what would have been the previous history of those two scorrelated (outside of each other's light cone) quarks, to be produced isolated?

If, otherwise, you assume this setup, than your question become interesting.

  • $\begingroup$ I guess this is a sort of gedanken-experiment - how would this initial configuration evolve in time? $\endgroup$
    – innisfree
    Jun 15, 2015 at 19:30
  • $\begingroup$ I'm not aware of something fundamental in CQD that prevents me from creating two quarks at a very large separation. Just like @innisfree writes this is a gedanken-experiment. $\endgroup$ Jun 18, 2015 at 10:30
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    $\begingroup$ but because of confinement, a free quark doesn't exist. So it is not possible to put 2 quarks at a very large separation. $\endgroup$
    – Paganini
    Jun 18, 2015 at 19:35

The strong force is similar to electromagnetism (the only difference is the gauge group), and like electromagnetic field lines, strong-force field lines have to originate/terminate where there is charge. Anything else would violate the strong-force version of Maxwell's equations.

In particular, if the universe contains only two quarks, and the field goes to zero at infinity, then there necessarily are field lines connecting the quarks. Of all the possible ways to do this, the lowest energy configuration is a narrow straight flux tube.

If you could somehow produce this configuration, in the next instant it would disintegrate into quark-antiquark pairs, so you would never get a chance to bring the original quarks together. Any other field configuration would have higher energy, so the result would be the same but more so.

If that decay never happened (which is absurdly unlikely but no more unlikely than producing this configuration in the first place), you would simply have a flux tube exerting a constant force of 10,000 N on each quark as usual. The force comes from the local field configuration, not from the other quark.

  • $\begingroup$ So, if i rephrase your answer, and relate it to my thoughts in the question. You are saying that the state I have created is not even meta-stable. It is unstable with no energy barrier, and the decay product is a long line of quark-anti-quark pairs that trace out the narrow straight flux-tube? $\endgroup$ Jan 21, 2019 at 9:12
  • $\begingroup$ @Mikael Fremling: I don't really understand the details. There is an energy barrier since you can make the half life of the decay arbitrarily long by increasing the mass of the lightest color-charged particle. But with SM quark masses I think it happens at the characteristic time scale of the strong force which is around $10^{-24}$ s. The result is a bunch of hadrons, something like (original antiquark + new quark 1), (new antiquark 1 + new quark 2), ..., though that may be an oversimplification. At any rate the flux tube "breaks" with new real particles at the locations of the breaks. $\endgroup$
    – benrg
    Jan 21, 2019 at 17:23
  • $\begingroup$ @Mikael Fremling: Here's a paper about what happens if the lightest color-charged particle is more massive: arXiv:1708.02243. There's still a flux tube but it can get quite long without breaking. $\endgroup$
    – benrg
    Jan 21, 2019 at 21:56

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