We know that gluons are Lie algebra $su(3)$-valued one-form fiels $A_{\mu}$. And because of $[A_\mu,A_\nu]$ does not vanish generally for the non-Abelian case, gluons have self-interactions. Now how to understand that gluons carry color charge?

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    $\begingroup$ "The gluon fields have no $\mathrm{SU}(3)$ symmetry" I don't know what this is supposed to mean, or why it would be relevant. To "have a charge" means to transform in a non-trivial representation of the gauge group. $\endgroup$
    – ACuriousMind
    Feb 23 '16 at 20:41
  • $\begingroup$ Related: physics.stackexchange.com/q/139473/2451 $\endgroup$
    – Qmechanic
    Feb 23 '16 at 20:44

The statement that gluons have charge associated with $SU(3)$ symmetry group can be understand in the following way. Suppose you have minimal lagrangian of fermions in the fundamental representation of the local $SU(3)$ group which interact with the gluons in the adjoint representation (in the result the gluon field action is invariant under $SU(3)$ of gluons as whose which forms adjoint representation): $$ \tag 0 L = -\frac{1}{4}F_{\mu \nu}^{a}F^{\mu \nu}_{a} + \bar{\psi}_{i}(i\gamma_{\mu}D^{\mu}_{ij} - m)\psi_{j}, $$ where $$ F_{\mu \nu} = \partial_{[\mu}A_{\nu ]} - ig[A_{\mu}, A_{\nu}] \equiv t_{a}F_{\mu \nu}^{a} $$ is the gluon field strength tensor, $a$ is the generator indice, $$ D_{\mu}^{ij} \equiv \partial_{\mu} - igA_{\mu}^{a}t_{a}^{ij} $$ is the covariant derivative, $i, j$ is the indice of fundamental representations.

An action is invariant under transformations $$ \tag 1 \psi \to U\psi , \quad A_{\mu} \to \frac{i}{g}U^{\dagger}D_{\mu}U, \quad U = e^{i\alpha_{a}(x)t_{a}}, $$ so for $SU(N)$ there exist $N^{2}-1$ conserved currents $J_{\mu}^{a}$. They may be obtained by linearization of transformations $(1)$ and insertion of linearized transformation in an expression for the Noether current: $$ A_{\mu}^{a} \to A_{\mu}^{a}- f^{abc}\alpha^{b}(x)A^{c}_{\mu} = A_{\mu}^{a} + \delta A_{\mu}^{a}, \quad \psi_{i} \to \psi_{i} + i\alpha^{a}(x)t^{a}_{ij}\psi_{j} = \psi_{i} + \delta \psi_{i}, $$ $$ J_{\mu}^{a} \equiv \sum_{\varphi = \psi , \bar{\psi}, A}\frac{\partial L}{\partial (\partial^{\mu} \varphi_{n})}\frac{\partial \delta \varphi_{n}}{\partial \alpha_{a} (x)} = \bar{\psi}_{i}\gamma_{\mu}t^{a}_{ij}\psi_{j} - f_{abc}A^{\nu}_{b}F_{\mu \nu}^{c} $$ Here $f_{abc}$ are structure constants, which are defined as $$ [t_{a}, t_{b}] = if_{abc}t_{c} $$ Corresponding charges are defined as $$ Q^{a} \equiv \int J_{0}^{a}d^{3}\mathbf r $$ You see that if you disable fermions, then the charge will contain purely gluonic part. This provides the statement that gluons carry charge. It is obvious since there is self-interaction qubic and quartic on $A_{\mu}^{a}$ terms in the lagrangian, which is needed for constructing, simply speaking, gauge invariant strength. The main difference of nonabelian $SU(N)$ theory from abelian theory of electromagnetism, $U(1)$, is formally based on the fact that for $U(1)$ $f_{abc} = 0$, so there is no self-interactions in pure photon sector and hence no charge of photon.

  • $\begingroup$ Thanks for your answer. Your answer is very physical and helpful. I also recommend the more mathematical answer commented by Qmechanic. $\endgroup$
    – Wein Eld
    Feb 23 '16 at 21:32
  • $\begingroup$ @Name YYY, why did you write the linearized transformation for the connection $A^a_\mu$ as $[A_\mu,\alpha]^a$? It seems there is the $\partial_\mu \alpha^a$ term missing... $\endgroup$ Dec 20 '20 at 20:42

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