For electricity, we have one charge, for the strong force three. How many are there connected to the weak force? Three, because of the W- and Z-particles? For the weak force, there is the isospin, which plays the role as (for example) the electric charge plays for the e.m. interaction.. Then there is the relation $Y=2Q-I^3$, where Y is the weak hypercharge, Q the unit of electrical charge and $I^3$ the weak isospin along a z-axis. Weak isospin and weak hypercharge seem rather artificial to me, instead of real charges.

In the old theory of the strong force, there were three particles (like the $W^{+/-}$- and the $Z^0$-particle for the weak force) that were thought to convey the strong nuclear force: the $\pi^{+/-}$-particles and the $\pi^0$-particle. It turned out these interactions were residual.

The same can be true for the weak interaction. Maybe there is a more fundamental, very strong interaction (much stronger than the strong nuclear force), who's residual force is the weak interaction. See for example this article, page 153:

"If our conjecture is correct, the weak interactions should not be considered as one of the fundamental forces of nature. In order to prove such a conjecture, we should be able to derive the observed weak interaction phenomena from our fundamental hyper colour and colour forces. This we cannot do, at present. However, we are able to study the symmetry properties of the forces among hypercolour-singlet composite fermions."


The problem with "weak charges" is that electroweak symmetry is spontaneously broken. Before the symmetry breaking, electroweak symmetry is described by an $SU(2)_L \times U(1)_Y$ gauge group.This amounts to three charges: weak hypercharge $Y$ for $U(1)_Y$ and weak isospin (total isospin $T$ and third component $T_3$) for the $SU(2)_L$. Some examples of charge assignments:

$e_L$ (left-handed electron): $T=1/2$, $T_3=-1/2$, $Y=-1$

$\nu_L$ (left-handed neutrino): $T=1/2$, $T_3 = 1/2$, $Y=-1$

$e_R$ (right-handed electron): $T=0$, $T_3=0$, $Y=-2$

$W^1, W^2, W^3$ ($SU(2)_L$ gauge bosons): $T=1$, $T_3=-1,0,1$, $Y=0$

The Higgs mechanism breaks the $SU(2)_L \times U(1)_Y$ symmetry, and only a subgroup $U(1)_{em}$ remains unbroken, which corresponds to electromagnetism. But please notice that $U(1)_{em} \neq U(1)_Y$ (in fact, the electric charge is $Q = T_3 + \frac{Y}{2}$), and that weak interactions alone don't fit in a $SU(2)$ group.So we can't really talk about "weak charges".

  • 1
    $\begingroup$ +1 for "can't really talk about 'weak charges'". This particular line corrects the OP's misunderstanding the best. $\endgroup$ Jun 13 '16 at 14:28
  • $\begingroup$ What about cosmic strings, in which the state is that of before the breaking of symmetry, or on a smaller scale of the breaking of the colour and electro-weak? Like I comb my hair, there always has to be a singular point in my hair. Cosmic textures were shown nót to exist. $\endgroup$ Jun 13 '16 at 18:13
  • $\begingroup$ As to "can't really talk about weak charges", there's precedent to the contrary; you are correct however that the situation is relatively complicated. $\endgroup$
    – rob
    Jun 13 '16 at 23:41
  • $\begingroup$ The EW group is not explicitly broken though. So even after EWSB, the SU2xU1 original symmetry is conserved. It's not nonsensical to talk about weak charges in this sense, in my opinion. Plus, the energy scale is not specified in the question, we could well be at $T>T_{EW}$, so I dont see why it is relevant to higgs the EW group. $\endgroup$
    – xi45
    Jun 30 '16 at 13:46

You're mixing a few things up here. When you say "three" for the strong force, you're counting the number of colors of quarks, but when you guess "three" for the weak force, you're counting the number of force carriers.

These are two different things. For example, if you counted the number of gluons (the force carriers for the strong force), you'd get eight, not three.

In general, for a gauge group $G$, the number of distinct particles of some type is equal to the dimension of its representation under $G$. The force carrier is always in the adjoint representation, which has dimension equal to $G$ itself, and all matter particles we know of transform in the fundamental representation. So for the three forces, we have:

  • Electromagnetism, $G = U(1)$. One matter particle (electron), one force carrier (photon).
  • Weak force, $G = SU(2)$. Two matter particles, three force carriers ($W^+$, $W^-$, $Z$).
  • Strong force, $G = SU(3)$. Three matter particles (quarks), eight force carriers (gluons).

Then the answer to your question is two or three, depending on what you meant.

  • $\begingroup$ @kazhou What are these two charges called, and is there a connection with the six extra dimensions in string theory (1+2+3=6)? $\endgroup$ Jun 13 '16 at 5:45
  • $\begingroup$ Isn´t it possible that the weak force is a residue force, like the strong force once was ( mediated by the massive pion)? $\endgroup$ Jun 13 '16 at 7:45
  • $\begingroup$ @descheleschilder If you mean the two weak force matter particles, they are the electron and neutrino. (which also each come in 3 sub-flavours). Superstring theory has 10 dimensions (other theories differ), and the extra 6 may originate from similar Lie algebra, but is definitely not due to summing the absolute values of the above forces. $\endgroup$ Jun 13 '16 at 8:32
  • $\begingroup$ @Adam Marshall I was thinking this because in the early Kaluza-Klein theory a fifth little dimension for the electric charge was introduced (in addition to gravity), so maybe the same is done for the other five charges. $\endgroup$ Jun 13 '16 at 15:18
  • $\begingroup$ But quarks also have electric charge? $\endgroup$
    – JDługosz
    Jun 14 '16 at 3:20

I would say two, which is pleasantly consistent with the $SU(2)$ structure of the weak force. One is the coupling strength with the $Z$ boson, and one is the weak isospin which is raised and lowered by the $W^\pm$.

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    $\begingroup$ Weak force is NOT described by a $SU(2)$ group, electroweak force is described by a $SU(2)_L\times U(1)_Y$ group, but it's broken by the Higgs mechanism and no $SU(2)$ subgroup remains unbroken, only an $U(1)_{em}$ subgroup. $\endgroup$
    – Bosoneando
    Jun 13 '16 at 8:24

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