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I was reading Scharf's Quantum Gauge Theories: A True Ghost Story when I stumbled upon the following paragraph (p. 118):

The standard example of a gauge theory with massless gauge fields is the theory of strong interactions called quantum chromodynamics (QCD) with the Lie algebra $\mathfrak g=\mathfrak{su}(3)$. The gauge fields are the $N=8$ gluon fields which interact between themselves and with the quark fields $\psi_n$. Due to [$M_{nm}=M\delta_{mn}$] the colored quarks are degenerate in mass. But the quarks also interact weakly and, therefore, have a second quantum number called flavor. The quarks with different flavor have different masses. As we will see in the following chapter, this is due to the fact that weak interactions are mediated by massive fields.

It seems to me that the author is claiming that in the absence of weak interactions, all quarks would have the same mass. I had never seen this clam before, and I find this idea confusing to say the least.

  • If the weak interactions are so weak, why are quark masses so different? Shouldn't the (relative) mass differences be of order of the weak coupling constant?

  • The fact that the limit $g_w\to0$ is discontinuous seems unphysical to me. Is it, or should I not be worried that masses suddenly become degenerate when we set $g_w=0$?

  • Also, $g_w$ runs to zero at high energies, while quark masses have different running rates. How are these two facts consistent? Shouldn't quark masses unify as $g_w$ runs to zero?

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  • $\begingroup$ And what does it say in the "following chapter " ? $\endgroup$ – my2cts May 25 '18 at 20:24
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    $\begingroup$ @my2cts the "following chapter" is 40 pages long, so he says a lot :-P but as far as I can tell, none of my questions is addressed. $\endgroup$ – AccidentalFourierTransform May 25 '18 at 20:31
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    $\begingroup$ I like the current title for its social content—it's the kind of comments that would come up in a casual chat between colleagues—but it should probably be replaced with a more descriptive one along the lines of "How does the weak interaction set the quark mass spectrum?" $\endgroup$ – dmckee --- ex-moderator kitten May 25 '18 at 20:52
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    $\begingroup$ I have to admit I missed the flavor of the pun. $\endgroup$ – dmckee --- ex-moderator kitten May 25 '18 at 20:56
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    $\begingroup$ @dmckee are you saying my sense of humour is weak? $\endgroup$ – AccidentalFourierTransform May 25 '18 at 21:00
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I don't understand why weak interaction would have anything to do with the mass difference between families unless the author is considering the Higgs as a part of what he calls "weak interactions". Without the Higgs you have no Yukawas in the SM Lagrangian and you end up with a spectrum of 6 identical massless quarks. But ignoring the Standard Model if you just give me a $SU(3)$ theory I can add as many fermions as I want with whatever masses and still get a perfectly consistent theory.

The sentence

As we will see in the following chapter, this is due to the fact that weak interactions are mediated by massive fields.

seems to confirm the fact that he is talking about the Higgs since he refers explicitly to the masses of the gauge bosons. Still it sounds weird to me, but maybe I'm being naive.

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  • $\begingroup$ Yeah, my gut tells me you're right. I'll have to do some more thinking. Cheers! $\endgroup$ – AccidentalFourierTransform May 25 '18 at 21:46
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    $\begingroup$ Well, in some sense the Higgs boson is a part of weak interaction - its three components are longitudinal polarizations of $W^\pm$ and $Z^0$ bosons. Though I still don't think it's really fits the quoted statements. $\endgroup$ – OON May 25 '18 at 22:18
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    $\begingroup$ @OON yes I know that. It is poart of the electroweak sector, but I wouldn't refer to it as "weak interaction" since its coupling with fermions is independent on the gauge couplings $g$ and $g'$ $\endgroup$ – FrodCube May 25 '18 at 22:48
  • $\begingroup$ @FrodCube one should not confuse the higgs field with the higgs boson. the weak coupling applies to interactions between particles, not fields, no? $\endgroup$ – anna v May 26 '18 at 4:51

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