8
$\begingroup$

The answer to this question might just be a straight-up "no, it's just a coincidence", but since coincidences are rarely a thing in physics, I thought I'd ask.

Is there a fundamental reason why (take QCD, for example) the gauge (color) group $SU(N_c)$ is the same type of group as the global (flavor) group $SU(N_f)$? (the same question applies to the weak case with isospin instead of color)

I couldn't find an answer in my textbooks, and I couldn't think of anything except that maybe we just really like this group and even if the global flavor group could be described in several different ways we choose $SU(N)$ for practicality.

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ But... the "full" flavor group of all six quarks is SU(6), terribly badly broken. Of the four lightest quarks, it's SU(4). Surely you don't expect the number of "relatively light" quarks to be deeply linked to the number of colors? $\endgroup$
    – Cosmas Zachos
    Dec 20, 2021 at 17:22
  • $\begingroup$ @CosmasZachos I'm not asking about the group index $N$, I'm inquiring about the group itself $SU(N)$ (instead of, say, $SO$ or $Sp$) $\endgroup$
    – Mauro Giliberti
    Dec 20, 2021 at 17:44
  • 1
    $\begingroup$ I believe the point for the gauge group is that $SU(N)$ is the only family of Lie groups that gives purely positive kinetic terms, which you can see by writing $\kappa_{ab}F^{mna}F_{mn}^b$ and looking at the Killing form of the different groups. I think this argument also follows for the flavour indeces when you look at the kinetic terms for the quarks, although I didn't think about this before. If that is the case, it would make the currently accepted answer incorrect. $\endgroup$
    – Joe
    Dec 20, 2021 at 22:00
  • 1
    $\begingroup$ @joe: The Killing form is (negative) definite for simple compact groups (but not for, say, $SO(1,3)$). Hence, you can (of course) have GUTs with $SO(10)$, $E_6$ etc., and for semisimple groups that are products of simple factors. $\endgroup$
    – Toffomat
    Dec 21, 2021 at 9:22
  • 2
    $\begingroup$ I see, it looks like I misunderstood this $\endgroup$
    – Joe
    Dec 21, 2021 at 9:41

3 Answers 3

Reset to default
10
$\begingroup$

It's just a coincidence.

Note that the original flavour $SU(3)$ permuting up, down and strange is broken by the differing masses, and the less badly broken $SU(2)$ is the isospin group. Both are thus approximate global symmetries. If you're fine with broken symmetries, you could add charm, bottom and top to go to $SU(6)$. However, the breaking gets progressively worse, and for the top, it would be completely useless.

The weak and strong groups $SU(3)$ and $SU(2)$, on the other hand, are exact gauged symmetries with associated gauge fields ($W$, $Z$, gluons), and so are completely different beasts.

Note also that in models beyond the standard model, gauge and flavour aspects are generally treated differently -- grand unified theories extend the gauge group to, e.g, $SU(5)$ or $SO(10)$, but the generations are just three copies of matter multiplets.

Improve this answer
$\endgroup$
1
$\begingroup$

Recall that inertial mass was shown to be equal to gravitational mass experimentally and for a long time this was thought of as a coincidence mainly because no-one could think of a better explanation. Then of course Einstein discovered his equivalence principle and the rest is history.

Likewise, this may pin to something more basic, the question is what. Perhaps string theory might have something to say here but I don't know enough string theory to say either way.

Improve this answer
$\endgroup$
0
$\begingroup$

The fundamental reason is experimental observations. The quark model was slowly built up over the years and the resonances measured in lab gave the surprising symmetries of the SU(3) group.

In general, the number of basic constituents defines the dimension n of SU(n). (As an example SU(2) for neutron and proton in nuclear physics, fitted the data). For the three quarks, SU(3). If we had found experimentally four quarks SU(4) would have been chosen. The difference between SU(n) and U(n) is mathematical, stated here:

The more general unitary matrices may have complex determinants with absolute value 1, rather than real 1 in the special case.

italics mine.

Improve this answer
$\endgroup$
2
  • 5
    $\begingroup$ I guess "The fundamental reason is experimental observations." always applies. But as the question is stated I assume OP wonders whether there is an underlying theoretical principle/argument which leads to the statement in the question. One should stress however that in the second part of your answer you refer to flavor groups. (Historically on tried to model the proton and the neutron as the same field where the different flavors refer to a specific "directions" in isospace) $\endgroup$
    – AlmostClueless
    Dec 20, 2021 at 13:26
  • $\begingroup$ @I use the nuclei example on how the SU(2) was found to describe nuclear data. $\endgroup$
    – anna v
    Dec 20, 2021 at 14:47

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.