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Is there an unit of color charge? I haven't found it, so I suppose that it doesn't exist, if this is right, why? Isn't it supposed that every measurable quantity can be expressed in terms of base quantities? What about flavour charge?

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The colour charge of quantum chromodynamics is, as far as we can tell, not experimentally measurable, because of quark confinement. More specifically, quantum chromodynamical systems are always colour neutral. Quarks do have a color charge, but they are always observed in groups of two (color + anticolor) or three (red + green + blue) for which the total colour charge is zero (i.e. "white"). If you were to observe a lone quark then you'd be able to measure its charge and (up to a messier conventions hassle than with electric charges) its "sign", i.e. colour.

What you can measure, on the other hand, is the coupling between different colour charges. This has a direct equivalent in QED: electric charge carries only artificial units, and in natural units the relevant parameter is the coupling $$\alpha=\frac{e^2}{\hbar c}.$$ Similarly, there is a QCD coupling constant, which is complicated as it depends on distance $\leftrightarrow$ enegy. Just as $e^2$, it is dimensionless in natural units, which are the only ones you want to be working with anyway. (In other units, $e^2$ has units of $\text{energy}\times\text{length}$, but I would think this does not carry over to QCD.)

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  • $\begingroup$ Now that I think about this again, I think what I was really asking is why electric charge is above the rest of charges, with its own magnitude. Whats so special about $U(1)$? Why Fermi's constant and strong couplings don't have an special charge magnitude. I guess it just because its convenient at macroscopic scale to have a dedicated unit. Btw, I'm don't think confinement is such an issue, $U(1)$ in 2D is confined, but $e$ can be still measured in Coulombs. $\endgroup$ – jinawee Nov 20 '18 at 22:48
  • $\begingroup$ @jinawee I stand by this answer - because of confinement. If you want to change the question five years after posting it, ask the new version separately. $\endgroup$ – Emilio Pisanty Nov 21 '18 at 4:33
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A coupling constant in the interaction Langangian plays the role of the charge. For example, for electrodynamics the interaction term of Langangian has the form $$\mathcal{L}_\text{int EM}=-ej^\mu A_\mu=-e\bar{\psi}\gamma^\mu\psi A_\mu$$ where $j^\mu$ is a current of electron (similarly $\bar{\psi}\gamma^\mu\psi$ is the same current calculated from the electron wave-function), $A_\mu$ is the electromagnetic field potential, and $-e$ is a numerical factor - the charge of an electron. The same way, the interaction in QCD has the form $$\mathcal{L}_\text{int C}=g\bar{q}\gamma^\mu \lambda^a q G^a_\mu$$ where the new index $a$ represents colors, and $\lambda_a$ is used to link between three colors of quarks and eight color varieties of gluons. As you see, there is a factor $g$ that plays the same role as electron charge before - it is called the coupling constant or the gauge coupling parameter. And for the electroweak interaction, there are two terms, representing interaction with $W$ and $Z$ bosons (at least, on the simplest level): $$\mathcal{L}_\text{int W}=g\bar{\psi}_L\gamma^\mu \tau^\pm \psi_L W^\pm_\mu+\dfrac{g}{\cos\theta_W}(\bar{\psi}_L\gamma^\mu \tau^3 \psi_L-q\sin^2\theta_W\,\bar{\psi}\gamma^\mu\psi)Z_\mu$$ Here the $\tau^a$ works with weak isospin and $\psi_L$ are the "left" parts of all interacting particles. $q$ is an electric charge depending on the kind of particle: 1 for electron, 0 for neutrino, fractions for quarks. And again, there are numerical factor $g$ and $g/\cos\theta_W$, which play the same role of the charge. Weak gauge charge is not called the flavour charge, but it is involved in the flavour changes the same way as color charge is involved in the color changes.

Sometimes they use another factor to represent interactions, $\alpha=g^2/4\pi$, and call it the coupling constant instead. Namely, there are three coupling constants, $\alpha_S$, $\alpha_W$, $\alpha_{EM}$.

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