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The strong interaction has a coupling constant of $\alpha_s(91GeV)\approx 0.1$ whereas the weak interaction has a much lower coupling constant $\alpha_w \approx 10^{-6}$. Both theories are non-abelian gauge theories, the strong interaction is based on SU(3) gauge symmetry, whereas the electroweak interaction is based on $U(1)\times SU(2)$ gauge symmetry.

What makes the strong interaction so special that it leads to confinement, whereas for the electroweak interaction it is not the case? It is certainly related with the $\beta$-function of the corresponding interaction, but why is the $\beta$-function of the electroweak interaction positive and the $\beta$-function of the strong interaction negative? Actually, I am not very familiar with use of renormalisation group arguments, so I would prefer a not too formal answer based on essentially on physics arguments.

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  • $\begingroup$ From what I can recall, Witten was able to prove confinement for a non-abelian theory but this was only the hypothesis of supersymmetry. $\endgroup$ Oct 3, 2021 at 19:59
  • $\begingroup$ @Mozibur Ullah : Interesting. Would that mean that a supersymmetric electroweak interaction could also lead to confinement in case of sufficiently high energy (to make the coupling constant large) ? $\endgroup$ Oct 3, 2021 at 20:04
  • $\begingroup$ Possibly, but the truthful answer would be - I don't know. I'll see if I can dig up the relevant paper of Witten's, it is sometime ago that I looked over it. $\endgroup$ Oct 3, 2021 at 20:11
  • $\begingroup$ @MoziburUllah: Oh, that sounds good, thank you ! $\endgroup$ Oct 3, 2021 at 20:27
  • $\begingroup$ This is the paper I was thinking of: Monopole Condensation and Confinement in N=2 Supersymmetric Yang-Mills Theory by Witten & Seiberg. The abstract states "We study the vacuum structure and dyon spectrum of N=2 supersymmetric gauge theory, with gauge group SU(2) ... the 'strongly coupled' vacuum turns out to be a weakly coupled theory of monopoles, and with a suitable perturbation confinement is described by monopole condensation." $\endgroup$ Oct 8, 2021 at 4:30

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Just to state the result for the beta-function associated to QCD
\begin{equation} \beta = -\frac{g^2}{32 \pi^2} \left(\frac{11}{3}N_c - \frac{2}{3}N_f \right) \end{equation} in which $N_c$ is the amount of colours and $N_f$ is the amount of flavours.

Essentially, both terms boil down to antiscreening and screening respectively.
A single quark can be surrounded by quark-antiquark pairs which tend to screen it from effects of the environment (much like in QED with electron-positron pairs).
However, the important piece is now the antiscreening effect due to the self-interaction of the gluons (because of the non-Abelian gauge symmetry). This tends to make the quark more susceptible to its environment.

Filling in the constants gives a negative beta-function. In other words, if we probe quarks at higher energies, the coupling constant decreases sufficiently such that we may regard them as being free (eg. a quark-gluon plasma). At small energies, the coupling constant becomes enormous and quarks are effectively bound together into hadrons - confinement.

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    $\begingroup$ Actually, the expression for $\beta$—function is positive for $N_c=3$ and $n_f$=6 like in the case of QED (also $\beta>0$). Neither I don't follow the last paragraph. Probing the quarks at higher energy should lead to a weaker interaction (that's what asymptotic freedom is about), no? $\endgroup$ Oct 3, 2021 at 12:44
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    $\begingroup$ From where did get your formula for $\beta$? Wikipedia - Beta function - Quantum chromodynamics provides a very different formula. And that formula actually gives a negative $\beta$, contrary to your formula which gives a positive $\beta$. $\endgroup$ Oct 3, 2021 at 12:59
  • $\begingroup$ Yes, you are right, my apologies. It's from a course at my university but I forgot an overall minus sign. And yes, my last paragraph is indeed wrong. I will edit it, but don't accept it as an answer because of the mistakes. $\endgroup$
    – Guliano
    Oct 3, 2021 at 13:28
  • $\begingroup$ What would be the beta function for Weingberg-Salam theory of wear interactions? $\endgroup$
    – Davius
    Jul 23, 2022 at 9:43

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