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Upon reading my book on physics, it mentions that there are only two discovered types of electric charges. I wonder if there could be a third type of elusive charge, and what type of effects could it have upon matter or similarly?

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I hope that answers don't only discuss how current theories are only consistent with two types of charges, but also how our theories would need to change in order to allow a third type of charge. –  BMS Aug 26 '14 at 21:11
@BMS : formally there is only one type of charge. The sign of charge isn't important. –  Andrew McAddams Aug 26 '14 at 21:39
Comment to the question (v1): It seems that the question should only be answered by a brief No, standard E&M has only an one-dimensional real axis of possible electric charges (which includes positive and negative electric charges and ignoring magnetic monopoles). Any attempt to justify standard E&M will likely fall short of an actual proof of uniqueness. Any discourse into non-standard E&M is likely off-topic or too broad. –  Qmechanic Aug 27 '14 at 19:56

3 Answers 3

up vote 23 down vote accepted

No, there are only positive and negative charges. Or, more carefully stated, if there is a another type of charge, then electromagnetism is not what we are currently thinking it is.1

Electromagnetism is a $\mathrm{U}(1)$-gauge theory, which relies on introducing the covariant derivative

$$ D_\mu = \partial_\mu - \frac{e}{\hbar}A_\mu$$

acting upon matter fields in representations of the $\mathrm{U}(1)$ labeled by $e$, where $A_\mu$ corresponds to the four-vector potential of electrodynamics. There is no possibility for matter fields to gain any other kind of charge here, since all representations of the circle group decompose into these one-dimensional representations of charge $e$, so charge is simply an integer $e \in \mathbb{Z}$. (The $\mathbb{Z}$ and not $\mathbb{R}$ come from the fact that $\mathrm{U}(1)$ is compact)

If there were other charges, we would need another (non-abelian, Lie) gauge group $G$ with some $\mathrm{Lie}(G)$-valued "potential" $A$ and a covariant derivative looking like

$$ D_\mu = \partial_\mu - \frac{g}{\hbar}\rho(A_\mu)$$

where now $\rho$ is some (irreducible) representation of $G$ and the $g \in \mathbb{R}$ is called the coupling constant. The charges lie within the representations and are usually thought of as the (root of the) eigenvalue of the quadratic Casimir operator in that representation.

Since $\mathrm{U}(1)$ has only one generator, its Casimir is simply that generator (squared), and we reconcile this with the above has observing that the representations of the circle group are indeed given by sending its generator to its $e$-multiple as per

$$ \rho_e : \mathrm{U}(1) \to \mathrm{GL}(\mathbb{R}) \cong \mathbb{R}, \mathrm{i} \mapsto e\mathrm{i} \text{ with } e \in \mathbb{Z} $$

Note on QCD (where the idea of "other electric charges" probably came from): The specific occurence of things like "colors" is not quite compatible with this language, as one usually identifies each dimension of a non-trivial representation with a color, but since irreducible representations have not subrepresentations, a gauge transformation will change the colors around (it won't change the quadratic Casimirs, which is why they are the proper generalization of charge, and not the colors). Nevertheless, also under this idea of charge, $\mathrm{U}(1)$ theories have only positive/negative charges, as their irreps are one-dimensional.

1 Looking at the real world, we know that electromagnetism must be a $\mathrm{U}(1)$ theory, since photon do not interact easily - they do not couple to one another on the tree-level of the quantum theory, and thus two laser beams do not significantly scatter off each other. In non-Abelian theories, the force carriers (gluons) do interact on the tree-level, and would thus deliver a wholly different force, more like the strong force, not long-range, and gluon beams would either not exist, or be very weird things. (though the details would be probably tricky for arbitrary $G$, granted, and could produced other weirdness as well)

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If $e$ is necessarily an integer, then how do the fractional quark charges come about? –  Constandinos Damalas Nov 25 '14 at 23:40
@PhotonicBoom: Choose $e = 1$ to be the quark charge, nothing here says that $e = 1$ must be the electron charge. There's also the possibility to use a different $g$ for different particles, since, comparing with the more general formula for the cov. derivative, I set $g = 1$ for the EM constant. Essentially, there's nothing fractional about the quark charge except our predisposition to take the electron charge as fundamental charge unit (which has no theoretical basis compared to the quark charge) –  ACuriousMind Nov 26 '14 at 7:10
Ah fair point. Thanks for the reply! –  Constandinos Damalas Nov 26 '14 at 8:36

In the Standard Model, electric charge $Q$ is actually part weak hypercharge $Y_W$ and part weak isospin $T_3$

$$Q = T_3 + \frac{Y_W}{2}$$

which can be either positive, zero (electrically neutral), or negative.

In this framework, that's it.

If, in fact, there is another type of electric charge (and its associated anti-charge), I believe it would be the case that the there would need to be three types of photons which would themselves be electrically charged and, thus, interact with each other.

This is in analogy with weak isospin where the three weak 'photons' ($W^+, W^0, W^-$) are isospin charged.

This would, of course change everything. But, we see only one type of photon and it is electrically neutral.

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Is there a simple way to understand how three types of "charge" require three force-mediating bosons? –  BMS Aug 26 '14 at 21:13
Out of curiosity, is the $W^0$ the same as the $Z^0$ boson? –  HDE 226868 Aug 26 '14 at 21:14
@BMS, three types of charge, e.g., red, green, blue (plus the anti-charges), requires 8 gauge bosons. Two types of charge, e.g., up, down (plus the anti-charges) requires 3 gauge bosons. –  Alfred Centauri Aug 26 '14 at 21:19
@HDE226868, the $Z^0$ is a certain mixture of the $W^0$ and $B^0$ bosons. $Z^0 = \cos (\theta_W) W^0 - \sin (\theta_W) B^0$ –  Alfred Centauri Aug 26 '14 at 21:25
@BMS, I'll work on a simple explanation for why an additional electric charge variety (and associated anti-charge) would imply three charged photons later but it's time to feed the horses. –  Alfred Centauri Aug 26 '14 at 21:27

Mathematically, electric charge current 4-vector conservation refers to the invariance of theory under U(1) transformations, so there aren't different types of electric charge (like in SU(n) theories) excepting the usual plus-minus.

Moreover, the fact of conservation of physical quantity means that corresponding operator commutes with hamiltonian which is constructed from fields operator. It's not hard to show that particle must have charge which is opposite to the antiparticle one.

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