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I think the answer to my question is no, but I can't find an explicit statement about this on my books or online.
I know that gluons are the vector bosons for QCD, the $SU(3)$ gauge theory of color, and they come in $3^2-1=8$ different types, and they have interactions with themselves and with quarks, which are the only particles that aren't in the singlet representation of the $SU(3)$ color group. My question is: if there was only one quark, say the up quark, would gluons care? I wonder this because it's often said that the strong force "keeps quarks together inside nucleons" but from what I understand it doesn't necessarily mix the flavors and could easily create, say, a three-up-proton with charge $+2e$.

I read this question but it doesn't answer my doubt: I know that, since there are many flavors of quarks, the gluon can interact with all of them, my doubt is whether the gluons need so many different flavors to work, if I need to consider all of them to make a $SU(3)$ theory work.

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    $\begingroup$ If you check the QCD lagrangian you should see that Is diagonal in flavour, there is no mixing. Generally I see no problem treating the effects of gluons on a single quark, the gluon acts on the color space of internal degrees of freedom of the quark and not on the flavour one. $\endgroup$
    – Ratman
    Oct 11, 2021 at 6:26
  • $\begingroup$ @NiharKarve but Z and W bosons do require many breeds of particles, right? $\endgroup$
    – Mauro Giliberti
    Oct 11, 2021 at 6:27
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    $\begingroup$ Z and W can lead to flavour mixing, in fact their action Is reprensented on the space of weak isospin where you group together different particles (es neutrino and electron). Though should be pointed that the flavour mixing due to the Z at tree level Is not possible, you have to go to higher orders to see It $\endgroup$
    – Ratman
    Oct 11, 2021 at 6:35
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    $\begingroup$ Sure, you can definitely have a baryon with three up quarks, it's called the $\Delta^{++}$. It just happens to be unstable in our universe. $\endgroup$
    – knzhou
    Oct 11, 2021 at 6:39
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    $\begingroup$ It's the other way round: too many flavors flip the sign of the β function of QCD, so gluons don't "work" in the conventional sense: their coupling is not asymptotically free and presumably not infrared enslaving, so you lose confinement (hadronic binding). $\endgroup$
    – Cosmas Zachos
    Oct 11, 2021 at 13:45

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Flavor is entirely orthogonal to color - the gluon(s) neither "know" nor "care" about flavor, but the existence of different quarks still leads to phenomena descending from the strong force that you wouldn't get without them:

What remains of the strong force on the scale between nucleons is often called the residual strong force and can be thought of as being mediated by pions. The pions, being bound states of up- and down-quarks, evidently would not exist in a world with only a single quark.

Another phenomenon related to the strong force is that confinement depends on the number of flavors, too, see e.g. this answer.

So while on the level of the QCD Lagrangian there is nothing that would indicate at first sight that the different quark species and gluons interact in interesting ways (and indeed a theory with only a single quark species and gluons would be consistent!), there are emergent phenomena that depend on both gluons and different flavors.

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    $\begingroup$ Well, we wouldn't have charged pions. Neutral pions, which in our world are $(u\bar{u}-d\bar{d})/\sqrt{2}$, would presumably exist as $u\bar{u}$ in a $u$-only world. $\endgroup$
    – J.G.
    Oct 11, 2021 at 10:15
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    $\begingroup$ @J.G. It seems arbitrary to me to call the hypothetical $u\bar u$ meson "pion" and not e.g. "$\eta$-meson" (which also contains $u\bar u$). Whatever this $u\bar u$ does, it's different from the mesons in a world with more flavors - I think giving it the "pion" name does not really change my argument here. $\endgroup$
    – ACuriousMind ♦
    Oct 11, 2021 at 10:25
  • $\begingroup$ OK, we'd have exactly one unexcited meson species, with no real-world analogoue, whatever it should be called. $\endgroup$
    – J.G.
    Oct 11, 2021 at 10:35
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    $\begingroup$ Note that a theory with a single chiral quark is inconsistent (gauge anomaly). So if you have say, a left-handed up quark, you also need something right-handed (with appropriate quantum numbers) to cancel the anomaly. $\endgroup$
    – AccidentalFourierTransform
    Oct 11, 2021 at 16:36

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