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3

This decay (occurring via the strong interaction) violates the charge conjugation since $J^{PC}(\pi^0) = 0^{-+}, J^{PC}(\rho^0) = 1^{--}, J^{PC}(\eta'^0) = 0^{-+}$. The charge conjugation transforms a particle in its anti-particle. In the case of the 3 particles involved in this decay, they are all their own anti-particle, and the effect of the charge ...


0

Supposedly quarks are in a well but experience minimal or no force at close range, called asymptotic freedom. If it is assumed that a quark model can be used where they are treated as "Point" particles, then they should be like three particles in a well. If QCD is believed then they are in different "color" states so can be in the same particle-in-a-well ...


-1

Is there a quark conservation law? No. In proton-antiproton annihilation (see Wikipedia) the quarks are destroyed. See this section: "...when a proton encounters an antiproton, one of its constituent valence quarks may annihilate with an antiquark, while the remaining quarks and antiquarks will undergo rearrangement into a number of mesons (mostly pions ...


2

In diagrams, you are just adding a disconnected non-interacting piece to the Feynman diagram (just a straight line for the added quark), so sure, just writing one or more additional quarks on both sides always yields an allowed reaction, although you can't be sure the quarks bind into a single state (e.g. if you added just a $u$ to your reaction instead of a ...


9

Yes, there are the quantum numbers Charm, Strangeness, Topness and Bottomness, which are conserved by strong and electromagnetic interactions, but not by weak interactions. Upness and Downness are simply the Isospin, which is also preserved for strong interactions, when the quark masses can be neglected, which is usually a very good approximation as ...


3

You're right that the reaction fails to conserve baryon number. The change in strangeness is a strike against the reaction, but not a fatal one; after all, the strange $K$ mesons decay into various mixtures of zero-strangeness mesons, charged leptons, and neutrinos. The thing to notice is that only the charged weak current, mediated by the $W$ boson, ...


1

The existence of quarks is not seriously in dispute at this point AFAICT. If you want to make something meaningful out of quarks and only quarks having fraction-of-$e$ charges, I think you pretty much have to postulate that electrons are composite. For instance, the rishon model proposes that all the "fundamental particles" of the Standard Model are ...


8

That is a good question but I think you might be a bit confused. The quark charges are quantised as they are fractional values of the electron charges, so when you refer to 2/3 and -1/3 these mean 2/3 of the electron charge and -1/3 of the electron charge respectively. As such, a Hydrogen atom with a proton in the nucleus and an electron in the shell, is ...


23

Quarks do not violate quantization of charge, it's simply that $\frac{1}{3}e$ instead of the electron charge $e$ is the smallest unit of electric charge.


1

The paper You refer to is too technical... You can imagine, two bb' quarks in Upsilon meson annihilate via gluons into u---u' (or d---d' or s---s') quark pair. Light quarks move apart from each other with velocity v=0.998c. Due to the confinement, the string of gluonic fields between them is created and it breaks via (qq') pairs creation, producing ...


1

A proton is described as a combination of three valence quarks, each with a bayrion number of 1/3 and a charge that adds up to the +1 of the proton, two up and one down. That is a primary constraint from data. Now in QCD, the theory we have developed to describe the strong interactions of quarks, it comes about that overall, in the constraining "bag" ...



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