Weak interactions seem the most universal after gravitation. A very few particles avoid them, only the photon and gluon, plus right leptons. The photon, however, is a part of the electroweak interactions and is a superposition of the B and W3 bosons. This leaves the gluon as one of a few particles immune to the electroweak interactions.

What is the reason for gluon not having a weak charge? Is it because the gluon is a gauge boson of a different type of interactions and different gauge bosons cannot share their charges with each other? However the Higgs boson does have the weak charge. Is this because the Higgs boson is not a gauge boson and this fact allows it to have the weak charge?

It would be great if someone could clarify the intuitive meaning of this setup and hopefully not just by formulas alone without an interpretation of their physical meaning.


As far as I can tell, all three answers are correct, just use different approaches. I am upvoting all three. The answer of @CosmasZachos is detailed, much appreciated. The answer of @annav is the closest to the question asked:

SU(3) has the gluon as the gauge boson, and it has, by construction, no weak vertices.

While to some this point may seem obvious enough to just skip and move straight to higher orders, it is worth pointing out to the rest of us. Checking Anna's answer as correct for being the closest to the actual question asked, although again, all three answers are very helpful.

To summarize, gluons do not directly interact weakly, because they are the gauge bosons of the SU(3) symmetry, which is defined independently of the SU(2) symmetry of the weak interactions. The reason for this definition is observation, but once the definition is in place, it becomes a theory prohibiting gluons from direct weak interactions. The higher order interactions (like those via virtual quarks in Anna's diagram) are not prohibited, but (as far as I understand) have not been observed.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Sep 22 '17 at 16:40

My question boils down to whether the gluon not coupling with Z is observation or theory.

The standard model of particle physics is described by a lagrangian that has the group symmetries of SU(3)XSU(2)xU(1), in order from left, strong group repersentation, weak group, electromagnetic. This is an experimental distillate, i.e. observations forced the assumtion that quarks interact independently with the three forces and their respective gauge bosons.

SU(3) has the gluon as the gauge boson, and it has, by construction, no weak vertices.

The standard model Lagrangian in a perturbative expansion represented by Feynman diagrams will give higher orders where gluon fusion can produce a Z boson.

Also gluons can end up into a Higgs

gluon fusion to higgs

gluon fusion to Higgs, where the left two vertices are with the strong coupling of top to gluon, and the right hand is the weak coupling of top to Higgs.

( which only interacts weakly) , but there are no direct vertices of Z to gluon.

The standard model is directly dependent on experimental observations.

| cite | improve this answer | |

Sigh... I'll take the bait of the comments to answer a narrow question the improbable primary question has apparently morphed to at present, possibly. So I'll only take the wild horses of the secret society to water, M Schwartz, Quantum Field Theory and the Standard Model, eqn (30.81), but who can guess at what they are keen to drink?

The Z couples to the SM combination of the conserved charges $\tau_3$ and $Y$. The quark (but not lepton) fermion bilinears in these charges also couple to gluons, so, you might think, at the level of loops, the quarks mediate effective interactions between the Z and gluons, suppressed, of course, by loop effects. After all, the uncolored Higgs is routinely produced by gluon fusion as we speak... loop effects rule.

Like all of these couplings, an interaction Zgg, absent at tree level, might be thinkable to have a small 1-loop presence, but it does not. If you call $\lambda^a$ the color matrices, the triangle anomaly $\operatorname {Tr} \tau_3 \{\lambda^a ,\lambda^b\}\propto \operatorname {Tr} \tau_3 =0$ vanishes,but, also the hypercharge one, what I called "just-so sudoku", $$ \tfrac{1}{4} \operatorname {Tr} Y \{\lambda^a,\lambda^b\}=\delta^{ab} \left (\sum_{Lq} Y_L -\sum_{Rq} Y_R\right ) = 3\Big( (1/6+1/6) -( 2/3-1/3 ) \Big )=0. $$ The factor of 3 before the parenthesis reflects the triplication of the 3 generations. Matt actually works out the sudoku in (30-85,86).

(Actually, having worked this out, and with the reassurance that it does, one may then reexpress the current coupling to the Z as a mere linear combination of the EM current, which is vector and thus not possibly anomalous, and the above trivially anomaly free $\tau_3$ one. But this is the classic stunt of writing the "just so" SM results as banal and deceptively self-evident.)

Now Higgs interactions in the wake of SSB start mooting corners of the above at higher loops involving new states entering, and of course, more gluons, as @LucJ.Bourhis suggested. Even though the basic triangle results persist to higher loops due to the Adler-Bardeen theorem, be careful when playing whak-a-mole with model builders...

My reluctance to proffer "revelation" is that I have not stumbled on such... This is a field infested with false prophets who have been proven wrong time and again by the firm voice of experimental evidence.

Edit in response to comment. Sorry, I tried to stay narrow. In general, gluons do interact with uncolored weakly interacting particles. I already admitted Zggg into the fold of perfectly fine loop induced interactions, above. The full calculation is here, pure theory. @AnnaV illustrates how gluons are relied on to produce the Higgs. My narrow point was an illustration of elaborate constraints in some interactions that the SM dictates on occasion in its idiosyncratic method in madness. If your question were "Do virtual states make all interactions communicate with each other?", then, obviously the answer is "yes".

| cite | improve this answer | |
  • $\begingroup$ While a bit dense for a non--specialist, your answer sounds like yes, there are theoretical reasons for gluons not interacting weakly. For example, what you call the "just so sudoku" of the hypercharge, in my mind still is a theoretical argument (the same one as for the right leptons, isn't it?). Thus your answer apparently is the opposite of the answer by Bob who points to observation rather than theory. Perhaps someone can clarify the difference. $\endgroup$ – safesphere Sep 22 '17 at 16:14

I think the best answer that can be currently given is that we simply don't observe effects of coupling between gluons and electroweak gauge bosons. If such effects were ever observed, we may have to rethink our current knowledge of the standard model and modify it to include some mixing of the SU$(3)$ and U$(2)$ subgroups.

From a grand unified theory standpoint, the statement is simply equivalent to the statement that the matter fields transform under representations of the SU$(3)$ and U$(2)$ subgroups of the unified gauge group without mixing between them.

Of course, there may be better reasons, but I don't know them. As far as I can tell, this question is about as hard to answer as "why does the standard model have the gauge group that it does?"

I hope this helps!

Edit: It has also occured to me that, from an effective field theory standpoint, an unobserved high-mass quard could create an apparent coupling between the gluons and the W and Z bosons. Maybe that could be interesting.

Edit: Whoops. Maybe I should have read Cosmas's wonderful answer before talking about EFT stuff. Disregard the previous edit.

| cite | improve this answer | |
  • $\begingroup$ My question boils down to whether the gluon not coupling with Z is observation or theory. Your question starts with "observation", but then you state, if this is ever observed, then we would have to rethink the theory. This is where I lose you. If there is no theory denying these interactions, why would you need to rethink it if they are found? What am I missing? Also on mixing the subgroups, quarks have both strong and weak charges. Does this not represent such a mixing already? Thanks for your help! $\endgroup$ – safesphere Sep 22 '17 at 2:28
  • $\begingroup$ Oh, I see where the confusion is coming from. By the word "theory," I mean the standard model Lagrangian. The Lagrangian itself is derived under the assumptions of observed particle charges. If we were to observe a non-zero coupling between the strong and electroweak sectors, this would be interesting and we would have to rethink how we put together such a Lagrangian. Nothing too fundamental about the underlying mathematics would change. $\endgroup$ – Bob Knighton Sep 23 '17 at 14:30
  • $\begingroup$ Yes, I got it, upvoted, and stated in the question edit that your answer was correct (among others). This is not unlike any theory. The main point of a theory is to replace an experimental set of data with an analytical expression of a pattern the data follows. For example, instead of a set of data with an object weight at different heights we just say they follow the inverse square law whether or not we understand the deep reason for this relation. In this case we don't know why the nature uses these symmetries, but they are a theory nonetheless. $\endgroup$ – safesphere Sep 23 '17 at 14:56

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.