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From a recently reignited [casual] curiosity into particle physics thanks to the Fermilab YouTube channel, I read about the g-2 experiment, followed by muons, naturally. Muons, it turns out have short lives, and they decay into an electron (and antineutrino), for example. Reading on how muons are created, I learned about the role of the cosmic rays, and intuitively I understand how 3 quarks can end up as 3 quarks and a quark–antiquark pion. [Charged] pions decay to muons, so my curiosity was satisfied, but only momentarily . . .

Until I realized this muon creation and the subsequent electron originated from quarks.

Q: How are quarks elementary when they can become leptons?

I'm either misunderstanding the meaning of a particle being elementary, or I'm missing something. (I can't get my head around that point; I also casually understand that the weak interaction is involved.)

Does it work the opposite way as well (lepton → quark)? Or, given enough time, will all elementary particles eventually decay to leptons? (Thinking out loud; not necessarily extra questions.)

I could not find the answer to my question on Wikipedia or via Google. I checked the related topics, for example, Are quarks and leptons actually fundamental particles? But it's a different question about what makes up quarks/leptons (I'm content with the fact they're as small as they get according to the Standard Model).

Thanks to @AccidentalFourierTransform for the following links (and their links):

As far as I can see those posts do not address the quark → lepton decay. All the examples given are quark → another quark (or lepton → another lepton), which I have no issue with (the muon → electron example I gave). The same for the decays of virtual/mediating particles, e.g., photons and Higgs bosons. My issue is (was) the class-changing decay of the non-virtual particles.

ymb1
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2 Answers2

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Particles are called elementary if they are not made up of other particles. However, interactions can change an elementary particle into another kind of elementary particle.

Quarks and leptons are currently believed to be elementary. (This could change if we could observe particles interacting at higher energies than, say, the LHC can achieve.) However the weak interaction can, for example, turn an up quark into a down quark, and an electron into an electron neutrino. When they change, they either emit or absorb a W or Z boson, the particles that carry the weak force.

A quark can’t directly turn into a lepton, but two quarks can indirectly produce two leptons. For example, an up quark and a down antiquark can turn into a $W^+$ boson, which can then turn into a positron and an electron antineutrino.

When an elementary particle like a muon decays into other particles, it doesn’t mean that those particles were inside the muon before it decayed. It means that the muon changed into a muon neutrino, emitting a $W^-$ boson in the process, and then that $W^-$ boson changed into an electron and an electron antineutrino.

G. Smith
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    Shouldn't it be a $W^+$ into a positron and electron neutrino, in the third paragraph? – Lucas Baldo Jul 09 '19 at 04:31
  • Thank you. I understand no sub-particles are involved (last sentence in the question). Also muon → electron I'm fine with (both leptons). So my misunderstanding was that quarks and leptons were their own "groups", but basically it's the concept of particles in general and the before/after spin/charge, as also commented, is all that matters, and they can jump groups, so to speak, right? – ymb1 Jul 09 '19 at 04:31
  • Based on your comment, I don’t think you have understood my answer. I wrote that “a quark can’t turn directly into a lepton” so I would say that quarks and leptons are separate groupings and there is no direct jumping between them. In terms of Feynman diagrams, a quark line never becomes a lepton line, or vice versa. – G. Smith Jul 09 '19 at 04:45
  • The “group jumping” always happens indirectly via a weak boson. I gave two examples of this. – G. Smith Jul 09 '19 at 04:50
  • Got it and much appreciated. I had that indirect jump in mind (from the answer); only making sure I didn't arrive at the wrong conclusion. – ymb1 Jul 09 '19 at 04:53
  • OK, sorry I misunderstood what you were thinking. – G. Smith Jul 09 '19 at 04:54
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    Would it help the OP somehow if you said that in the example you gave, no quarks (=quark + antiquark) turn into no leptons (=electron + antineutrino)? – Martin Kochanski Jul 09 '19 at 04:57
  • @ymb1 What Martin said. The total quark number of the pion is zero. We don't have quarks turning into leptons or vice versa. This "group jumping" concept isn't helpful, IMHO. FWIW, if you smash an electron into a positron with enough energy you can make pions (and other bits & pieces); you could even do it just colliding electrons, if you can supply enough energy. But in all these reactions, the various symmetries are conserved. – PM 2Ring Jul 09 '19 at 05:45
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    After reading this answer, it's not clear to me what "made up of other particles" means. What prevents me from thinking than a deuteron is an elementary particle that can "decay", or "change" into a proton and a neutron? This seems to be the crux of this answer, and it is not explained in detail. – Federico Poloni Jul 09 '19 at 13:00
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    @FedericoPoloni What should prevent you from thinking that is the experimental evidence that the deuteron has internal structure. Deuteron-electron scattering reveals that there is a proton and a neutron inside a deuteron, and quarks inside each of those. So the deuteron is observed to be a complicated composite particle. A muon, by contrast, appears to have no internal structure when we scatter electrons of it. It scatters like a single point charge. – G. Smith Jul 09 '19 at 14:22
  • @G.Smith The fact that there is no vertex where a quark can directly go into a lepton is an accident of the standard model, right? One can write down interaction vertices where there can be direct coupling between quarks and leptons consistent with all the imposed symmetries of the standard model--just that such a term would be non-renormalizable. –  Jul 09 '19 at 21:30
  • @FeynmansOutforGrumpyCat I was talking about the Standard Model. Can you explain why a hypothetical quark-lepton coupling in another model would necessarily be non-renormalizable? I’m probably missing something obvious. – G. Smith Jul 09 '19 at 22:05
  • @G.Smith Umm, I think one can make a rather general argument that there cannot be any direct fermion-fermion interaction in $d=4$. A fermionic field would have the mass-dimension $3/2$ so any term with more than two fermionic fields would be of dimension $\geq 4.5$ rendering it non-renormalizable, right? –  Jul 09 '19 at 22:15
  • @FeynmansOutforGrumpyCat That’s not what I meant. In another model could you have a triple vertex with a quark, a lepton, and a gauge boson? – G. Smith Jul 09 '19 at 23:23
  • @G.Smith Oh, I see. A triple vertex of a fermion-fermion-gauge boson would always be renormalizable so if such a term, consistent with the symmetries of the standard model, exists then it would already be there in the standard model. So, with the same symmetries as that of the standard model and the same field content, we cannot add such a vertex. But I think you are correct, you can have such a vertex if you modify the field content of the standard model or the symmetries. –  Jul 10 '19 at 06:08
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We have a plethora of data on particle interactions since last century, and laboriously have come up with a mathematical model for particle physics that works, i.e. it gives the right numerically answers for these data and, important, is successful in predicting new data, as recently happened with the Higgs boson. It is called the standard model for this reason and has an elementary particle table.

elem

These are what the successful model uses as elementary particles (together with their antiparticles), i.e. point particles carrying quantum numbers and masses that , when used to get the crossection of an interaction or the decay width or.. work beautifully and the model is continually validated.

In this model the elementary particles interact with the three interactions according to their quantum numbers, and some such interactions and the energy supplied by their masses allow them to decay to other elementary particles. The Z and W and the Higgs also decay into lower mass elementary particles.

So the answer is that it is a model that defines what elementary particles are which is working very well.

If in the future a string theory model, for example, can embed the standard model into vibrations of a string, there will be only one elementary entity, the string.

It is all in the model.

ymb1
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anna v
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    In all honesty, the OP wasn't challenging the standard model or the wisdom of the physicists who have accepted a particular set of particles as elementary based on scientific justifications (which you nicely summarize). The OP was rather trying to understand what the concept of an elementary particle is. It is natural to confuse the existence of the decay of a particle with the particle being made up of what it decays into. And I think that the question of the OP stemmed out of such a confusion. Your answer then stays orthogonal to actually addressing OP's confusion IMO. –  Jul 09 '19 at 21:40
  • @FeynmansOutforGrumpyCat Well, I have tried to say what an elementary particle is for particle physics. Having started graduate school at 1961, neutrons and protons were still considered elementary. It is the model used that defines what observations mean, imo. I have tried to state this, that what is elementary today may not be in 300 years. – anna v Jul 10 '19 at 03:12