# Is it a coincidence that quarks have exactly -1/3 or 2/3 the electron's charge? [duplicate]

Why do electron and proton have the same but opposite electric charge?

My question is different. We live in a world, where quarks can have a real fraction of the elementary charge (-1/3 or 2/3). I do understand that the experimental data fit the models and that Baryons are made up of three quarks, and that those quarks can have -1/3 or 2/3 the elementary charge. This way, the quarks can combine so that the Baryon will have an integer of the elementary charge. This way, the nucleus and the electron can be in a stable atomic state, where their electric charges cancel (attract) exactly. Any other way, the atom would not be stable.

So electron says: hey quarks, let's team up, let's make an atom.

Quarks: Hey, great idea, how much electric charge do you have? Let's call it e. OK so we will team up of three of us, so we will just take each of us -1/3 or 2/3 of your charge.

Electron says: Great!, I feel it, this way we can have a stable atom.

Quarks: Great!

I mean come on! I do understand the respect that the experimental data tells us that electrons and quarks are both elementary particles.

It seems that the quarks teamed up exactly so three of them in Baryons so they can cancel out (attract) exactly the electron's electric charge. Now I do understand that there could be Baryons made up of four quarks, and they could have then -1/4 and 3/4 charge of the electron's elementary charge. This would work too, and the neutron and proton would have the same way an integer of the electron's charge. So the atom would be stable.

We could do this with any integer number of quarks. But, come on, my question is, what if the quarks' charge could only be -1/sqrt(3) and 2/sqrt(3)? How many quarks would we then need to make up the nucleus - to make it match the electron's charge? Is it mathematically possible to have any kind of electric charge for the quark? We could simply put as many valence quarks in a Baryon to make a stable atom?

Question:

1. Is it a coincidence that quarks have -1/3 and 2/3 the electron's charge and there are three quarks in a Baryon? Could we have quarks with any kind of electric charge, like Sqrt(3)*elementary charge? Could we make a stable atom this way too? How many quarks would we put in this case into a Baryon to make atoms stable?

2. I accept and respect that currently, electrons are elementary particles. Is it impossible as of today's view that both electrons and quarks are made up of the same something smaller (strings)?

## marked as duplicate by Aaron Stevens, AccidentalFourierTransform, knzhou, Jon Custer, stafusaOct 21 '18 at 15:12

• Possible duplicate of Why do electron and proton have the same but opposite electric charge? – AccidentalFourierTransform Oct 21 '18 at 3:11
• – AccidentalFourierTransform Oct 21 '18 at 3:30
• It is unfortunate that this question is accumulating votes to close, a comprehensive answer could touch on many interesting points - why baryons have N quarks (for various gauge groups), compactness of gauge group as cause of charge quantization, and the interplay between confinement and charge; along with already mentioned topics like anomaly cancellation and grand unification. – Mitchell Porter Oct 21 '18 at 12:21
• Then Please vote to reopen. The mentioned question and answers within do not answer my question. – Árpád Szendrei Oct 21 '18 at 17:34
• Tips: 1. If you think an answer does not answer your question you should not accept it. 2. Ask only 1 subquestion per post. – Qmechanic Nov 26 '18 at 13:19

The charges of the quarks must be simple fractions of the electron charge $$e$$, because otherwise there would be a breakdown of charge conservation in quantum corrections. The fractions do not need to be $$-\frac{2}{3}$$ and $$\frac{1}{3}$$ specifically. In simple models with $$2n+1$$ quarks making up the proton (the number of quarks must be odd, so that the nucleons are still fermions with spin-$$\frac{1}{2}$$), the quarks naturally carry charges $$-\frac{n+1}{2n+1}e$$ and $$\frac{n}{2n+1}e$$. And more elaborate composite nucleon models can have constituent partons with other rational multiples of $$e$$.
However, not all fractions of $$e$$ are going to be allowed as charges of the constituent quarks. Moreover, irrational multiples of the electron charge are generally not possible. The reason is tied to the structure of the full electroweak interaction, of which electromagnetism is only one part. In relativistic quantum field theory, there are quantum corrections that involve the interactions of three separate gauge bosons (photons, $$Z^{0}$$, and $$W^{\pm}$$) with virtual fermion-antifermion pairs. The fermions involved can be quarks, electrons, muons, neutrinos, etc, and the sizes of the quantum corrections are determined (in part) by the charges of those fermion species. If the quark and electron charges are not in the correct rational ratios, certain quantum corrections will be nonzero.