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Let me state first that I don't think this is a duplicate of the mentioned question, though the basic thought is the same. Nevertheless, I come up with the prequark rishon theory of Harari, which accounts for all interactions, and not with a muon that consists of an electron and two neutrinos. In the first answer to the abovementioned question, it is actually confirmed that muons (or quarks and leptons in general) could actually have a substructure. To reveal the substructure of leptons you could smash them together at very high energies (because it is known they no substructure only if we probe them below a very little distance). I don't know if such experiments are done because in the framework of the standard model there is no need for them: it describes leptons (and quarks) as elementary, so why look if they are not?

A muon (${\mu}^-$) changing into an electron ($e^-$) and a photon ($\gamma$), has never been observed in Nature (or maybe it's better to say " in particle accelerators"). To explain this, physicists assumed that there are two kinds of neutrinos: one associated with the electron (${\nu}_e$) and one associated with the muon (${\nu}_{\mu}$). They assigned a "muon number" $+1$ to the muon and its associated neutrino (and a value $-1$ to their anti-particles), and an "electron number" $-1$ to the electron and its associated neutrino (and a value $+1$ to their anti-particles). These numbers had to be equal for the original particle and the particle into which it changes.

So a ${\mu}^-$ changing into a $e^-$ and a $\gamma$ (which obviously has lepton number $0$) won't occur because the muon number (and the electron number) has to be conserved (you can say it's an extension of the conservation of lepton number), which obviously isn't the case.

On the other hand, a ${\mu}^-$ can change into a $e^-$, a ${\bar{\nu}}_e$ and a ${\nu}_{\mu}$, which you can check by inspecting the (additive) muon and electron numbers for the particles.

With this rule (conservation of electron number and muon number, to which we can nowadays add a tau number) you can predict which changes can take place.

But is this really a explanation why the ${\mu}^-$ can't change into a $e^-$ and a $\gamma$?

I suppose not, so what is the conventional, deeper explanation (in the framework of the standard model) that this change can't take place?

And can't we just as well say that the $\mu^-$ is composed of three particles which each carry an electric charge (amongst other kinds of charge) of $-{\frac 1 3}$, so, because of the repulsive force of these charges, the ${\mu}^-$ can't change into $e^-$ because in a $e^-$ the three charges are closer to each other.

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    $\begingroup$ $\mu^{-} \to e^{-} \gamma$ is allowed if you introduce lepton mixing (as motivated by neutrino oscillation). The branching fraction is incredibly small, though. $\endgroup$ – dukwon Apr 17 '17 at 12:17
  • $\begingroup$ within the standard model of particle physics, there is lepton number conservation , an electron needs an anti_electron_neutrino so that the rule holds. In the SM neutrinos are massless and the process described by dukwon has zero probability. $\endgroup$ – anna v Apr 17 '17 at 13:55
  • $\begingroup$ Possible duplicate of Why are muons considered to be "elementary particles" in the Standard Model? $\endgroup$ – knzhou Apr 18 '17 at 4:37
  • $\begingroup$ 1) experiments keep on checking whether the electron is elementary or not and give limits. they do not rely on the standard model. 2) there exist models as the preon models that try to assign a complexity to what are called elementary particles. en.wikipedia.org/wiki/Preon . It just is an experimental fact that at present they are not necessary to describe the data, there is no unexplained datum that can be explained with a preon model and not a reasonable extension of the SM. So they are fringe models, not mainstream. $\endgroup$ – anna v Apr 19 '17 at 3:57
  • $\begingroup$ Well, the asymmetry of matter and anti-matter is in this (rishon) model far more simply explained. And in the standard model, this is not explained at all (in other fringe models it is though). The same for the decay of the proton. $\endgroup$ – descheleschilder Apr 19 '17 at 7:09
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There is no reason to suppose that electrons or muons are anything other than purely fundamental at the present time. There are strong upper limits on the radius of an electron, and no structure of multiplets of particles that you'd associate with such a composite structure, as led to the quark model of hadrons.

In the standard model the particles are organised into three generations. Each generation is like a pattern of particles; all three generations have the same basic pattern, but repeated with successively heavier particles. The statement that a muon is 'like the electron's heavier brother' becomes that the electron and muon are in different generations (and in the third generation, there is a tau particle that plays the same role).

In each generation, together with the charged lepton (i.e. The electron or muon or tau) there is a neutrino; the charged lepton and corresponding neutrino can change into each other with the emission or absorption of a charged W boson. This boson can then decay into a (lighter) charged lepton and neutrino. So the rules for what particles can turn into each other are set by the pattern within each generation, but the gauge bosons talk to all the generations.

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  • $\begingroup$ What would be the Feynman diagram look like associated with the change of a muon into an electron and a (real) photon? And wouldn't the standard model assign to this diagram a probability zero? $\endgroup$ – descheleschilder Apr 17 '17 at 18:45
  • $\begingroup$ The diagram drawn below is not correct within the standard model as originally conceived, because there is no vertex that connects electron and muon; this is what I was trying to tell you here. As per @dukwon 's comment, such a vertex may be introduced by the process that gives mass to the neutrino, as happens in the quark sector, but this is not presently well understood theoretically and tightly constrained experimentally. $\endgroup$ – rwold Apr 17 '17 at 23:07
  • $\begingroup$ Are you referring to muon-virtual electron-virtual Z particle vertex (that it can't exist), or to the fact that in the diagram there is no vertex that connects the real electron and the real muon? $\endgroup$ – descheleschilder Apr 17 '17 at 23:41
  • $\begingroup$ If the vertex doesn't exist it doesn't exist, for real or virtual particles. $\endgroup$ – rwold Apr 17 '17 at 23:49
  • $\begingroup$ That there are no reasons to suppose that the muon and electron have no substructure (besides the fact that it reduces the number of elementary particles to two, the most economic way from which you can build up particles) is not to say that no substructure exists. If the colliding energies in the particle accelerators is high enough these substructures may show up [as the hadron and meson (which also come in families) substructures showed up at increased collision energies]. $\endgroup$ – descheleschilder Apr 17 '17 at 23:51
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The diagram in the answer by the OP does not conserve lepton number.

Conservation of lepton number is one of the experimental facts which gave rise to accepting the standard model , SM, as the basic model of particle physics .

It is known that there exist experimental facts which are not within the SM and need an extension of it, for example CP violation in the model is not enough to explain cosmological data. Neutrino oscillations are not included in the SM because there their masses are assumed to be zero and an extension of the SM is needed.

A neutrino oscillation correct Feynman diagram, within a reasonable extension ( consistent with the rest of the data) of the standard model will allow for the reaction to happen , and is this one :

neutosc

neutrino oscillation is indicated by the cross. For the OP's channel reduce the number of vertices by eliminating the e+e- pair.

There are also other diagrams possible in Grand Unified theories, or Super symmetries that allow for the possibility of the channel under discussion .

When looking for extensions of the SM one has to embed it, i.e. follow the main structure , because the SM is an encapsulation of an enormous amount of data, and any ad hoc additions will most probably violate something that is supported by a lot of data. If , as in the proposed by the OP diagram , Z exchanges could go with lepton number violation, a lot of unusual channels would have opened at LEP, which were not seen. There would not have been a lepton number rule and there would have been a different standard model.

Note that the diagram above still depresses the probability at low energies for the decay channel, as the OP diagram, but at LEP energies a lot of violations of lepton numer would have been seen if the diagram of the OP were correct..

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  • $\begingroup$ @annav-Clear answer! So the Z- and W-particles only couple to two members of the same family? But because the neutrino isn't massless (so the neutrino can oscillate in the reaction) the reaction is actually allowed? Only the photon which exists will produce an electron anti-electron pair because of its high energy? But what is the reason a W- or Z-particle can't couple to members of different families? $\endgroup$ – descheleschilder Apr 18 '17 at 9:42
  • $\begingroup$ The Z is a gauge boson, it cannot change lepton number which is an attribute of the fermion leptons within the SM $\endgroup$ – anna v Apr 18 '17 at 9:52
  • $\begingroup$ Is the fact that it cannot change the lepton number an experimental or theoretical fact (within the framework of the unification of the e.m. and weak interaction, which according to me is a non-existent unification, but that aside). $\endgroup$ – descheleschilder Apr 18 '17 at 10:01
  • $\begingroup$ It comes from the allowed interactions in the SU(3)xSU(2)xU(1) which groups were chosen because they fit the data. Experiment forced the choice of symmetries and representations of the group. $\endgroup$ – anna v Apr 18 '17 at 10:28
  • $\begingroup$ @annav-In the Feynman diagram you show a muon changing into two electrons and a positron. But doesn't the muon change into an electron, an electron neutrino, and an anti-muon neutrino? Or can it also the change in the way you show in the diagram (in both cases the lepton number is conserved, but maybe this haven't to be the case, like in the decay of the proton the baryon number isn't conserved but can take place nevertheless according to fringe theories). $\endgroup$ – descheleschilder Apr 21 '17 at 23:53
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The Feynman diagram associated to the muon which changes in an electron and a photon would look like this:

enter image description here

In this diagram, the particles in the triangle are the virtual particles, while the lines for the incoming muon and the outgoing photon and electron represent the real particles.

Now for the change not to occur, this Feynman diagram should give a zero amplitude for the process of the muon changing into an electron and a photon. It's a straightforward but long calculation, but I can tell you that without actually computing it the amplitude is (for sure) 0, so the process won't occur.

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  • $\begingroup$ Somehow the big change of mass (which all has to be carried away by the photon and electron) gives me an uneasy feeling. $\endgroup$ – lalala Apr 17 '17 at 19:43
  • $\begingroup$ I assume you refer to the difference in mass between the muon and the electron. But in order to conserve the overall four-momentum, the four-momentum of the real electron just has to be big enough (which increases its mass), or the photon's four-momentum has to be big enough (or both) so that the real electron and the real photon can carry away the big mass of the real muon. But like I said, this process has zero probability to occur. $\endgroup$ – descheleschilder Apr 17 '17 at 20:26

protected by Qmechanic Apr 18 '17 at 5:31

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