# Why are muons considered to be “elementary particles” in the Standard Model?

According to this article, a muon decays into one electron and two neutrinos.

According to this article, elementary particles or fundamental particles are particles "whose substructure is unknown, thus it is unknown whether it is composed of other particles." I have also seen somewhere that it is a particle that cannot be reduced into other constituent particles.

While perhaps not a sure thing, seems like the decay indicates that the muon may be just a composite particle, perhaps consisting of one electron and two neutrinos?

Based on this, why does the muon fit with the above definition of an elementary or fundamental particle?

I realize there are much more complicated, historical reasons as to why it was included in the Standard Model, but this question is just related so how it fits (or doesn't fit) the stated definition above.

It seems to me that we really can only get solid evidence of elementary vs. composite when we smash the particles together and see what comes out and compare that to all the masses, energies and momentum before and after? Until we do that with muons, how can we know with much certainty?

And perhaps we'll have a better answer with a Muon collider: http://en.wikipedia.org/wiki/Muon_collider/ http://map.fnal.gov/

To that point, seems that electrons may not be fundamental after all: https://www.sciencedaily.com/releases/2016/04/160404111559.htm

• Comments are not for extended discussion; this conversation has been moved to chat. – David Z Aug 20 '16 at 4:18
• I rolled back your question; please don't add new questions or comments by editing. Edits are generally meant for clarifying the original question. – David Z Aug 20 '16 at 5:29
• The muon collider article has been moved to - en.wikipedia.org/wiki/Muon_collider – Ed999 Feb 2 at 15:34

Indeed, we can't know for sure if muons are elementary are not. In this sense, the situation is like the 1950's, where we had a zoo of mesons and hadrons, but didn't yet know they were made of quarks. We wouldn't have direct confirmation of quarks for decades to come, like how we have no indications of muon compositeness now.

Despite this, the quark model was accepted, and all attempts to make muons composite have not been. There are many scientific reasons to reject muon substructure:

• Predictive power. Your model basically says that all the charged leptons are 'really' electrons or positrons with bound neutrinos. This makes no direct predictions of new particles or processes. In contrast, the quark model predicted whole families of new particles.
• Theoretical simplicity. You need to postulate a new force to keep the neutrinos bound to the electrons, since the weak force is certainly not enough to do it. This was okay for the quark model, because we knew there was a force we didn't understand at the time (the strong nuclear force), but we have no indications for such a force. This 5th force should significantly affect the motion of neutrinos, but we haven't seen any such thing.
• Direct observation. Since muons are fairly light, whatever force is binding the electrons and neutrinos together must be fairly weak (since the binding energy $E = mc^2$ is low). This indicates that we should have been able to tear apart a muon into its constituent parts by now, or at least excite its energy levels. This hasn't been observed.
• Occam's razor. There's no need to postulate composite particles to explain decays. For example, excited atoms can decay to ground-state atoms and photons. This process can be described simply and exactly by coupling the atom to the electromagnetic field; it doesn't require a photon to be 'inside' the excited atom. Worse, there are alternative decay channels which emit two photons instead! The 'photon inside' picture isn't even self-consistent.

In short, there's no guarantee that muons aren't composite. But there are many compelling experimental and theoretical reasons suggesting so.

• "we have no indications of a fifth force now" - phys.org/news/2016-08-physicists-discovery-nature.html – John Dvorak Aug 18 '16 at 8:40
• @JanDvorak That purported force is unrelated to this question, though. It wouldn't bind neutrinos to electrons. – knzhou Aug 18 '16 at 8:42
• This is the only answer that gives the actual reasons for the choice, rather than saying that "it is true because Authority says it is true". – Martin Kochanski Aug 18 '16 at 11:20
• @JanDvorak the timing of that UCI 5th force article couldn't have been better. Now we know, it's the protophobic X boson that binds the neutrinos to the electron inside a muon! =) – Brad Cooper - Purpose Nation Aug 19 '16 at 13:09
• @knzhou thanks, appreciate you translating into plain language the high-level criteria we might use to evaluate the "fundamental-ness" of a particle and whether it meets a definition of "elementary" and/or inclusion into the Standard Model. Perhaps the Standard Model authorities would post a similar criteria overview that laypeople like myself would understand =) – Brad Cooper - Purpose Nation Aug 19 '16 at 13:58

That a particle decays into other particles is completely disjoint from it having substructure/being fundamental or composite.

Some examples: A highly energetic photon may "decay" into an electron and a positron in the presence of another object that takes the excess momentum. That doesn't mean a photon is a composite of electron and positron. A free neutron decays into a proton, an electron and an electron anti-neutrino with an average lifetime of 10 minutes, yet it is a composite state of three quarks.

Being constituted of other particles means being a bound state of these particles. Quantum field theoretic processes have no problem turning one kind of particles into other kinds of particles (subject to certain rules, of course), but this sort of process does not imply that the results actually constituted the input. In no meaningful way is a photon a bound state of electron and positron, in no meaningful way is a neutron a bound state of proton and electron, and in no meaningful way is a muon a bound state of an electron and neutrinos.

• the photon example seems to be different case, where a boson hits or interacts w/many other particles & transitions into other particles vs. a pure "decay" w/o interactions from other outside particles. In the neutron example, seems to be just another case of a composite decaying into constituents? With your "bound state," this sounds like you are introducing a new definition of "elementary" different than the one referenced. If the referenced definition of "elementary" is wrong, that is fine. But most would not think of "bound state of the particles" to define "constituent" – Brad Cooper - Purpose Nation Aug 17 '16 at 13:45
• @PurposeNation A neutron is not composed of a proton and an electron. We know it is composed of three quarks. Just as a proton is composed of three quarks. A particle decay is not the same as nuclear fission. It is not simply the case that a particle breaks up into smaller chunks. Often you can find the mass of the result is much greater than the mass of the initial particle. It's more like a decay destroys one particle and puts the energy into a more stable particle state (usually) – Jim Aug 17 '16 at 13:51
• @ACouriousMind: Also it appears the wikipedia article on free neutrons introduces another definition of a composite vs. elementary particle: "The finite size of the neutron and its magnetic moment indicate the neutron is a composite, rather than elementary, particle." Does the size of the particle and its magnetic moment define the elementary vs. composite nature? If so, seems like the first wikipedia definition really needs a lot of additional clarification? – Brad Cooper - Purpose Nation Aug 17 '16 at 13:53
• @PurposeNation We look for substructure by scattering other particle off of the target. The "scattering patterns" (so to speak) can be used to determine if there is anything inside. This is done all the time in crystallography: scatter X-rays from a solid, and infer from the scattering the arrangement of atoms in the solid. Scattering involving muons indicate no substructure. Scattering from neutrons indicate three similar internal entities (quarks), not two dissimilar entities (electron and proton). – garyp Aug 17 '16 at 14:45
• I don't think this answer is as satisfying as it could be. You're basically just repeating what we consider to be composite in the Standard Model now. This doesn't tell the OP why we concluded those particles were composite, which is essential to knowing why we haven't concluded muons are composite. – knzhou Aug 17 '16 at 19:56

The best place to seek evidence that decay doesn't equal compositness is in particle creation. Because if decay meant compositeness, then creation would require you to get the constituents together.

When you bash two nucleons together at high enough energy you get a lot of junk coming out. Some of that junk is lepton particle-antiparticle pairs, and many of them arise from interactions like $$q + \bar{q} \to l^- + l^+ \,.$$ This process is called "Drell-Yan". The leptons can be electrons, muons or tauons. Getting muons is experimentally very useful, so this process is sometimes used as a probe of the structure of the nucleon sea. (When you put protons on protons, the anti-quarks have to come from the sea as the valence content is all quark.)

If you have an electron-positron machine at high energy (i.e. the decommissioned SLC or LEP) you can also do $$e^- + e^+ \to l^- + l^+ \,,$$ with similar mathematics.

Now, at energies very much over $2m_\mu c^2$, the rate for producing electron-pairs and that for producing muon-pairs is the same, which wouldn't be the case if one were elementary and the other composite (if the muons were composite there would be a chancing of having the right bits around going into the production rate so the rate would be lower). Further the rate is in agreement with the ab initio predictions from QED for fundamental leptons. Let's take a moment to recall that QED offers the best single agreement between theory and experiment in physics (the electron's g-2).

In addition there are many other interesting predictions from QED about muons (for instance the muon g-2 which is nearly as good a theory-experiment match as that for electrons).

This is the elementary particle table of the standard model of particle physics.

Please note that it is not only the muon that decays, but also the tau and the Z and W.

They are called elementary because they are the building blocks of the Standard Model; building up all other particles and controlling interactions in the microworld where quantum mechanics is necessary to calculate and predict the behavior of particles, using the Standard Model.

Decays are not a unique indication of the existence of a substructure. The substructure is investigated in scattering experiments fitted with the standard model functions. The elementary particles in the table are called point particles because they have no substructure as a hypothesis of the model, and the model is continuously validated, i.e. it has not been falsified.As the diagram provided by @Statics shows, there is a point vertex in the weak decay of the muon ( as also of the tau and the Z and the W). All particles in the table are considered point particles when they interact. No space for constituents.

String theories aim to extend the standard model, describe elementary particles as strings, but this is a subject of research, and still there are no constituents in the string representation of a particle.

How was the compositeness of the nucleus of the atoms established? By the famous Rutherford experiment which showed deep inelastic scattering.

The same scattering experiments at higher energies showed deep inelastic scattering in the protons, establishing the existence of quarks and gluons.

In general scattering experiments established the Form factors of the targets, the "shape" in space and energy/momentum space of the targets, see figure 11 here. That the proton and neutron were not point particles was established long before the discovery of the quark content, and Feynman had proposed his parton model to model the data. Experiments showed deviations from the parton model which established the hard scattering on scattering centers by the so called high p_t data.

No such structure has appeared for the electron within the energy range of our experiments, its size point like within 10^-18 meters , and its shape spherical to great accuracy ..

The symmetries in the standard model of physics are then utilized to posit a point like structure for the elementary particles in the table. At the moment the success of the standard model in describing the experimental data does not leave space for compositeness of elementary particles. If it exists, much higher energies have to be reached in our experiments to discover it.

The muon is not a composite particle. The fact that it can decay is related to weak interaction, being possible due to th existence of $W/Z$-bosons. The muon can decay into neutrino and electon since its rest mass is larger than that of an electron. Since there is no other charged lepton with a mass lower than that of an electron, the electron can not decay into anything itself.

From all the thousands of experiments we know that a muon has pretty much the same properties as an electron and can thus be considered a lepton. Lepton are not composite particles of anything, they do not bind neutrinos and electrons together. There is no known force that would describe such bindings.

• How we do we know it is not made up of constituent particles? How do we know it is not a composite particle? – Brad Cooper - Purpose Nation Aug 17 '16 at 13:13
• @PurposeNation Perhaps that is the question you should have asked: How do we know the size of the muon is small, and that of hadrons is large (fermis)? The answer, is of course, [form factors] en.wikipedia.org/wiki/Form_factor_(quantum_field_theory) . Experimentally, you may establish limits on the size of particles through investigation of their form factors, and the size of leptons is always smaller than any scale probed so far, to date, but the size of hadrons is as big as a fermi-sized marshmallow--they are soft and squishy. A good QFT text explains that. – Cosmas Zachos Aug 17 '16 at 17:47

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