0
$\begingroup$

Elementary spin 1/2 particles vary in electric charge from o(neutrinos), 1/3(anti-down-, anti-strange-, and anti-bottom-quark), 2/3(up-, charm-, and top-quark), to 1 (electrons, muons and taus). Besides them, there is the hypothetical X-boson which carries 4/3 of one unit electric charge (and which comes together with the hypothetical Y-boson, carrying a charge of 1/3).

Aren't these charges a sign that there exist truly elementary particles with charge 1/3, and charge 0, out of which the particles with a charge from 0 to 4/3 can be formed? In this context, the "elementary" particle with a charge of 1/3 isn't the anti-down-quark, but a combination of these truly elementary particles, so that they give a 1/3 charged anti-down-quark (0,0,1/3).In which case you can assign the 1/3 charged particle a charge 1, in which case quarks have charge -1(1) and 2(-2), and the electron charge-3(3).

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ Your title seems to ask (among other things) if the existence of particles with zero charge guarantee the existing of particles with zero charge. I suspect it's OK to say yes on that part... $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jan 27, 2017 at 20:55
  • 2
    $\begingroup$ Not really. The electric charge of a particle is determined by which representation it transforms under the $U(1)$ gauge group in the Standard Model. The irreps of $U(1)$ are $\rho(x)=\mathrm{exp}(inx)$ with an $n\in\mathbb Z$. This $n\in\mathbb Z$ determines the electric charge by $e=\frac n3$. $\endgroup$
    – Bass
    Commented Jan 27, 2017 at 20:56
  • 4
    $\begingroup$ Apparently you already know that there are elementary particles with charge 1/3. And your question is whether there exists an elementary particle with charge 1/3? $\endgroup$
    – Noiralef
    Commented Jan 27, 2017 at 20:56
  • 1
    $\begingroup$ Whether you call the charge of a quark "1/3" or "1" is purely a matter of convention (it's the choice of what is the "elementary charge"). I don't understand what the physics question is supposed to be here. $\endgroup$
    – ACuriousMind ♦
    Commented Jan 27, 2017 at 22:49
  • $\begingroup$ What I was wondering about is if this division of so-called elementary charges doesn't prove the existence of more elementary particles with a charge of 0 and 1/3, from which you can form all the charges from 0 to 4/3. $\endgroup$
    – Deschele Schilder
    Commented Jan 28, 2017 at 15:17

1 Answer 1

Reset to default
1
$\begingroup$

If I understand it correctly, you're asking

The electric charge of all known particles is an integer multiple of one third of the electron's charge. Does that prove the existence of an elementary particle, call it the chargino, which carries an electric charge of one third of the electron's charge, and which is the only elementary particle with a negative electric charge, and that all currently known charged particles are not elementary but composed of charginos, anti-charginos and other elementary particles? The electron would thus contain three charginos, the down quark one, and the up quark two anti-charginos.

No, it does not prove that.

However, it might be viewed as a hint for such a theory. After all, if a 19th-century physicist suddenly discovered that every atom has a core whose electric charge is an integer multiple of the hydrogen core's charge, such a physicist might conjecture the existence of the proton based on that discovery.

Note however that in modern physics there is another explanation for the fact that the electric charge is quantized. In the language of mathematical physics, the electric charge of a particle is determined by which representation it transforms under the $U(1)$ gauge group in the Standard Model. The irreps of $U(1)$ are $\rho(\phi)=\mathrm{e}^{in\phi}$ with an $n\in\mathbb Z$. This $n\in\mathbb Z$ determines the electric charge by $e=\frac n3$.

If you didn't understand that, imagine the electric charge being something like a participation in some process involving a cord which is wound several times around a lamp pole (this is just an analogy, don't take it too literally, there are not any real cords involved). If it's wound clockwise, the particle's charge is $-1/3$ (for example the down quark). If it's wound twice counter-clockwise, the particle's charge is $+2/3$ (the up quark). Three times clockwise would be $-1$, the electron. If it's not wound around the pole, the particle is electrically neutral, for example the neutrino. For composed particles, you simply sum over the constituents' number of loops.

You see, because it's forbidden to wind the cord around half of the pole, you get a quantized (integer) electric charge.

The modern explanation is in many ways better than the "chargino theory" mentioned above. The Standard Model has been confirmed by countless experiments, but AFAIK there is no single experiment that contradicts it.

Improve this answer
$\endgroup$
9
  • $\begingroup$ What I'm referring to is the Rishon Model of Haïm Harari: en.wikipedia.org/wiki/Rishon_model I understand the relationship between conservation of charge and invariance of the Lagrangian(density) under local U(1) phase (gauge) transformations, but what do you mean by $\rho(x)$? The charge density? For a point particle, this must be infinite, but renormalization comes to the rescue unless you have a way to give elementary particles a finite extension (non-stringlike). I think I know a way to do that. $\endgroup$
    – Deschele Schilder
    Commented Feb 10, 2017 at 22:29
  • $\begingroup$ Why is n restricted to 4? $\endgroup$
    – Deschele Schilder
    Commented Feb 10, 2017 at 22:35
  • $\begingroup$ @descheleschilder I'm not acquainted with the Rishon model, but it looks like some kind of "chargino theory" like I described above. If you have a question about it, please ask it in a separate question. $\rho(\phi)$ is the $U(1)$ representation the particle transforms in. All $U(1)$ representations are of the form $\mathrm{e}^{in\phi}$ where $\phi\in U(1)$ is the angle of rotation. This means you can classify all particles by the integer $n$, which is the representation the particle transforms under $U(1)$ gauge transformations. $\endgroup$
    – Bass
    Commented Feb 11, 2017 at 12:00
  • $\begingroup$ @descheleschilder Who says $n$ is "restricted to 4"? All Standard Model particles have $|n|\leq3$, but AFAIK there's no requirement for that. $\endgroup$
    – Bass
    Commented Feb 11, 2017 at 12:03
  • $\begingroup$ @Bass-Well, I was referring to the (hypothetical) X-boson (+4/3). The associated Y-boson has a charge of +1/3. The rishon model needs only two elementary particles to construct all the others from. More economic it can't be! The T-rishon has charge 1/3 and the V-rishon charge 0. And two nice properties of the model are that it says that in our Universe there are equal amounts of matter and anti-matter, and it easily explains the decay of the proton. $\endgroup$
    – Deschele Schilder
    Commented Feb 11, 2017 at 13:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.