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Is there any idea explaining why the electric charges of electron and muon are equal?

Edit:

The total charge of a particle is proportional to the integral of its own electric field flow through the sphere of a big radius surrounding the particle at rest.

The free Dirac equation describes charged fermion. It contains the mass term $m$. If $m$ tends to zero, Dirac equation tends to the pair of Weil equations that describe electrically neutral particles. Does it mean that charge somehow depends on mass? If yes, why do the electron and muon (both described by Dirac equation, but with different mass terms) have the same electric charges?

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  • $\begingroup$ Is it possible to rephrase the question(v1) in such a way that it does not overlap with this question physics.stackexchange.com/q/4238/2451 ? $\endgroup$
    – Qmechanic ♦
    Feb 1, 2012 at 16:52
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    $\begingroup$ Thanks, I didn't see that question. Give me some time to think about rephrasing/closing the question. $\endgroup$
    – Murod Abdukhakimov
    Feb 1, 2012 at 17:08
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    $\begingroup$ It's worth googling for "who ordered that" (include the quotation marks as you want to search the phrase). $\endgroup$
    – dmckee --- ex-moderator kitten
    Feb 1, 2012 at 18:17

2 Answers 2

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How do we know that there exists such a particle as the muon?

From observing its decay into an electron plus two other neutral particles, which are an antineutrino electron and a neutrino muon.

In this last sentence there are three conservation laws:

1) conservation of charge ensures that the muon has the same charge as the electron

2) lepton number conservation ensures that the number of particles with muon leptonic number and the number of particles with electron leptonic number are conserved.

These are observations, the accumulation of which together with a large number of other observations allows us to build up the standard model 0f particle physics. The Standard Model encapsulates our observations/data.

The short answer to the question is: because that is what has been observed.

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  • $\begingroup$ It is amusing that there exist people who think they are interested in physics and cannot accept an experimental fact, down voting this answer. They need convoluted theoretical arguments which in the end of course end up on the experimental fact. Data trumps theory every single time. Without data theory is science fiction. $\endgroup$
    – anna v
    Mar 24, 2012 at 4:46
  • $\begingroup$ Of course you are right, and not my downvote, I think it's an ok answer, but the question seems to be confused about the massless limit of the Dirac equation, not about particle properties. $\endgroup$
    – Ron Maimon
    Mar 24, 2012 at 7:28
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The confusion is about the Weyl limit--- a massless Weyl fermion can be charged. All the fermions in the standard model are charged Weyl fermions.

The Weyl fermion that can't be charged is the massive Weyl fermion. The reason is that the mass term in the Weyl reduction mixes up the field and its conjugate, so it isn't phase-invariant under multiplying the Weyl field by a complex phase. This type of mass, which is incompatible with charge, is more often called a Majorana mass in the literature, because it is easier to derive as the real part of the Dirac equation in a real basis.

The fact that Weyl fermions can't have mass is important--- it is the reason we see Weyl fermions in nature--- if they could be massive, they would be Planck mass massive. Instead, they are only Higgs-scale massive.

A pair of massive Weyl fermions with the same mass can together be charged, with the charge symmetry rotating one into the other. So a pair of Weyl fermions can also be massive independent of the mass, and the reason is that this is what the Dirac equation is.

The answer to the title question, about the equality of charges, at least from the theoretical standpoint is found here: What is "charge discreteness"? . Anna v. has given the experimental reason.

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