Why and how does a gluon become a hadron when it leaves a hadron?

Is this an analogous process to the way a quark becomes a meson, when it is "forced" out of (let's say) a proton?

  • 1
    $\begingroup$ It isn't entirely clear what level of detail you are looking for. The short answer to why is because color charged particles need to be confined, the short answer to how often involves fragmentation (but can involve any series of interactions that don't violate conservation of quantum numbers and mass-energy), and the short answer to the last question is yes. But, there are long journal articles that discuss the gory details of this messy process in great depth, and there are lots of intermediate possibilities in a potential answer. $\endgroup$
    – ohwilleke
    Oct 9, 2018 at 9:40
  • $\begingroup$ @ohwilleke Well, this is certainly a very good introduction that cleared up some of my more fundamental doubts. Thanks! $\endgroup$
    – Pregunto
    Oct 9, 2018 at 11:41

1 Answer 1


Since the coupling constant in the Quantum ChromoDynamics (QCD, i.e. the Quantum Field Theory that studies the strong interactions) is of the order of $1$, a perturbative approach as for the other interactions (Elettro-Weak) is not possible and by looking at the $\beta$-function for the QCD it is found that the shielding term (which would make the coupling constant to rise as the energy goes up) is dominated by the antishielding term, such that the coupling constant decrease as the energy goes up (Increase the energy is like going to smaller space-scales). Then, the more two quarks are apart, the more the coupling will be large and then the more they are attracted by each other.

Now, if you consider a QCD process, a gluon can be emitted and it can split itself into a pair of quark and anti-quark, which can be recombined with other quarks involved in the process, and since the strong interaction becomes stronger as the quarks go apart, then they will, eventually, always group up to become mesons and barions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.