# Why are all quarks considered fundamental?

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jfmcghee
TL;DR Summary
If 5 of the quarks can eventually decay into the up (plus other stuff), why are they still considered fundamental?
Summary: If 5 of the quarks can eventually decay into the up (plus other stuff), why are they still considered fundamental?

I can't decide if I'm just misinterpreting the word fundamental in this particular usage or if there is something about the non-up quarks that makes them fundamental even though they can eventually decay into an up.

For instance: what is it about the down quark that we can't say that it is a composite particle of an up and a W-, or even an up, an electron, and an electron neutrino since that's what a W- will then decay into?

I feel like I'm glossing over something simple and obvious, I just can't find it in my notes.

Keep in mind it's finals week and I'm grading papers, so I may be brain dead right now. Don't judge me too harshly...

Homework Helper
It's simple, the being elemental is not related in any case with decays. For example, some elemental particles decay into others ##t\rightarrow W^+b## while some non-elemental particles like ##p^+## don't decay. So why ##t## is elemental and ##p^+## not? Well, because ##t## is not a bound state of other particles while ##p^+## is a bound state of two quarks up and one down. As simple as that.

vanhees71
jfmcghee
That makes sense. Plus it seems that the up and down can kind of go back and forth in some interactions, so the up wouldn't be any more fundamental than the down anyway.

Is there a specific part of the models that differentiates between the "boundedness" of the particles versus the particles just being the result of an interaction? Or is that just part of the definition?

Homework Helper
I'm not sure to understand your question, the standard model assumes that quarks are fundamental particles and tell you how they interact with, for example, the gluons (the so-called, strong interaction). With this, one should be able to deduce that there exist a bound state of three quarks with the properties of what we call a proton.

Staff Emeritus
Summary: If 5 of the quarks can eventually decay into the up (plus other stuff), why are they still considered fundamental?

I can't decide if I'm just misinterpreting the word fundamental in this particular usage or if there is something about the non-up quarks that makes them fundamental even though they can eventually decay into an up.

For instance: what is it about the down quark that we can't say that it is a composite particle of an up and a W-, or even an up, an electron, and an electron neutrino since that's what a W- will then decay into?

I feel like I'm glossing over something simple and obvious, I just can't find it in my notes.

Keep in mind it's finals week and I'm grading papers, so I may be brain dead right now. Don't judge me too harshly...

They are considered as fundamental particles (for now) because we do not know of anything "smaller" that make them up. This doesn't mean they do not or cannot change from one quark to another. That is a misapplication of the idea of "fundamental particles".

Unless you want to rewrite the Standard Model of particle physics, or want to challenge it, then that is what we have right now.

Zz.

jfmcghee
I have no intention of challenging the Standard Model, it was simply an innocent question. It's a question I could imagine my students asking and I realized I didn't have a better answer than it's because they are defined that way.

Staff Emeritus
I have no intention of challenging the Standard Model, it was simply an innocent question. It's a question I could imagine my students asking and I realized I didn't have a better answer than it's because they are defined that way.

But that is how the word "fundamental particle" is defined. At some point, there has to be a "back stop" on what things mean, and this is one of them.

Tell your students that as far as we know, leptons and quarks are not made up of anything else. THAT is the definition of "fundamental particles" in elementary particle physics. They may not buy it, but they are welcome to challenge that definition if they go into particle physics in grad school.

I am also a physics instructor, and there are more ambiguities in the concept of "energy" than there are about elementary particles in the Standard Model.

Zz.

Gold Member
2021 Award
Well, the didactical problem with HEP physics is to get the classical meaning of the word "particle" out of the students' minds (and not only the students' minds; I've to remember myself from time to time that an elementary particle is a quantum but no classical particle).

There are some quite weird things, if you forget that "elementary particle" refers to an entity that is described by quantum (field) theory, and in high-energy theory, i.e., when the interactions among these "particles" come in the realm of ## \gtrsim mc^2##, where ##m## are typical masses of the "particles" involved in a scattering experiment, it's necessarily a quantum field theory, because you deal always with the possibility to destroy some "particles" and create some new ones.

That explains why it is no contradiction that elementary particles (which are of course interacting, because otherwise we'd not notice them to begin with and wouldn't need to include them in our esteemed "standard model"!) can decay to other particles.

To understand this, you have to study relativistic QFT and the associated math (including at least some group theory and representation theory of the Poincare group). The qualitative upshot is: A "particle" in the sense of relativistic QFT are asymptotic free one-particle Fock states (I've no simpler expression for it). The idea is that if some lonesome particle comes along, where lonesome means it's far away from any other particles with which it could interact significantly, you can consider it as a free particle. In this case the math of the Poincare group tells you that you deal with something with some definite quantum numbers like mass (it's always invariant mass and nothing else we use here!) ##m \geq 0## and spin ##s \in \{0,1/2,\ldots \}##. A convenient basis for the single-particle space are the momentum-spin eigenstates (where "spin" is also a tricky concept for relativistic particles, but let's forget about this detail for the moment). If you have a particle defined in this strict sense, it will life forever as long as it doesn't react with other particles.

Then there's the case that you have some particle-like excitation, e.g., a ##Z## boson, which can e.g., be created by colliding an electron and a positron at an energy around the mass of the ##Z## boson. Now the ##Z## boson is "instable", i.e., due to the weak interaction it decays spontaneously (e.g., again to an electron-positron pair). It's thus not an asymptotic state in the strict sense, because it doesn't live forever but can spontaneously decay. Although fundamental in the sense that it is described by a quantum field in the standard model, it's not stable. Its existence is thus established by a peak when looking, e.g., at the scattering of an electron-positron pair going into any number of hadrons, around a mass of ##M_Z \simeq 90 \; \mathrm{GeV}/c^2##. See, eg. the picture on slide 15 of the following presentation:

https://www.physi.uni-heidelberg.de/~fschney/Seminar.SS09/Wagner.pdf
Everything interacting strongly is even trickier, because here the elementary "particles" (quarks and gluons) never occur as asymptotic free states, and thus the above "strict particle definition" doesn't apply either. Due to what's called "confinement" no objects carrying a non-zero color charge can be observed as asymptotic free states. The quarks and gluons thus only occur in bound states. The closest thing you can have to a "particle picture" is to use deep-inelastic electron scattering on hadrons (say on protons). At very high energies of the electrons you start to resolve the "substructure" of the proton. In the most simple picture a proton consists of three "partons" which carry the quantum numbers of the quarks in the standard-model Lagrangian (in this case two up and one down quark). However the higher and higher you make the electrons' energy in these scattering experiments the more details you resolve, and it becomes clear that the simple three-quark (constituent-quark or parton) picture is too simple. So quarks and gluons are far from being observable as "particles" in a classical sense at all.

Then there are also the neutrinos, which are also among the weirdest entities of physics. For some decades we know that neutrinos are grouped together with the charged leptons into doublets. In the standard model they are considered massless and carrying no electric and (as leptons) no color charge. They only interact through the weak interaction and thus could be considered as practically always asymptotically free and thus proper particles in the above strict sense.

However since the 1990ies it became clear that the neutrinos in fact have some mass (a very small one, but we don't know any precise values yet) and now trouble starts! The neutrinos come on one hand with definite flavor, i.e., there's a neutrino grouped with the electron (called ##\nu_e##), another one grouped with the muon (called ##\nu_{\mu}##) and finally one grouped with the tau-lepton (called ##\nu_{\tau}##). However, nothing prevents these flavor states of neutrinos to "mix" with each other, i.e., it is not forbidden by any fundamental law of nature that there are neutrino states of any superposition of the flavor eigenstates, and it turns out that the flavor eigenstates are not identical with the states of definite mass of the neutrinos.

Now the above strict definition of particles as asymptotic free states with a definite mass isn't applicable anymore either! The neutrinos are always produced in flavor eigenstates. E.g., the usual ##\beta## decay of a neutron comes (for energy reasons necessarily) with an electron. Now the lepton-flavor number is conserved in the standard model and thus since the neutron has lepton number 0 and the electron has lepton number 1, what's additionally emitted in the decay must compensate for that and thus it's definitely and anti-electron neutrino, i.e., an antiparticle with definite flavor. But then it doesn't have a definite mass! and thus it's not an asymptotic free state and cannot be interpreted as a "particle" in the above given strict sense of relativistic QFT. But since it's not a definite mass eigenstate it "oscillates", i.e., even if not interacting with anything on its path (and usually that's a pretty long "free path" because it interacts only through the weak interaction) it oscillates into an superposition of all flavor eigenstates (and also mass eigenstates, which are connected by a unitary matrix, the socalled neutrino-mixing matrix or PMNS matrix). Thus initially being an electron anti-neutrino from the decay of a neutron it may be detected at some distance as a neutrino of a different flavor. The oscillation frequency is determined by the differences of the neutrino-masses squared, and the detection can only be in terms of a definite flavor again, because the detection has to be due to some reaction with the detector material through the weak interaction, and the corresponding microscopic processes again determines uniquely the flavor of the detected (anti-)neutrino.

So the apparently obvious idea of "particle" is in fact a pretty complicated issue!

artis, protonsarecool, gleem and 3 others
jfmcghee
Thanks for that reality check.

Even though "you can't think about it classically" is my mantra to the class in that one month a year I get to talk about quantum stuff, the second I leave the classroom I make exactly the same mistakes. It's so easy to do.

I do exactly the reverse in the air. As soon as I'm with an instructor doing a flight review or refresh I suddenly forget everything: "Oh you mean that's the lever I use to make it turn?"

lpetrich
Does "fundamental" here mean "not a bound state"? That is a much more tractable question. It is easier to answer "yes" than to answer "no", because breaking up a particle immediately answers the question, while failure to break one up may because of insufficient energy to do so.

Let's look at the history so far.
• Molecules -> atoms
• Atoms -> electrons and nuclei
• Nuclei -> protons and neutrons
• Quarks have no evidence of compositeness
Dissociation relative energies:
• Molecules: ~10^(-9)
• Atoms: ~10^(-8)
• Nuclei: ~10^(-3)
• Quarks: no evidence observed
Let's estimate how much for quarks and leptons, using where the Standard Model has been confirmed for them as a lower limit. I will use 1.5 TeV, the typical parton energy in the LHC, as that lower limit.
• Charged leptons (e,mu,tau): 2.9*10^6, 1.4*10^4, 8.4*10^3
• Quarks (d,u,s,c,b,t): 3.2*10^5, 6.8*10^5,1.6*10^4, 1.2*10^3, 3.6*10^2, 8.7
• Higgs: 12
So it will be very hard for them to be composite.

ohwilleke and Astronuc
masher
They are considered as fundamental particles (for now) because we do not know of anything "smaller" that make them up. This doesn't mean they do not or cannot change from one quark to another. That is a misapplication of the idea of "fundamental particles".

Unless you want to rewrite the Standard Model of particle physics, or want to challenge it, then that is what we have right now.

Zz.
The products of a fundamental interaction does not indicate the building blocks of those particles but one result of their interaction.

Gold Member
So it will be very hard for them to be composite.
This is the bottom line point. In this context, "fundamental" simply means "not composite". Quarks (in the Standard Model of Particle Physics, at least) are fundamental because they are not composite assemblies of other particles which are themselves either fundamental or composite.

vanhees71