# Conservation law of colour current in Yang-Mills theories

In a Yang-Mills theory where the fermion fields transform under $$\Psi \rightarrow e^{-\theta^A t_A} \Psi$$ with $$t_A$$ generators of a Lie-algebra fulfilling $$[t_A,t_B]=f^A_{BC}t_C$$ a Noether current $$J_{\mu A}$$ of the following form can be assigned to the Dirac-equation $$(i\not\partial - m)\Psi = 0$$:

$$J^\mu_A = -\overline{\Psi} t_A \gamma^\mu \Psi.$$

For being a conserved Noether current it should fulfill $$\partial_\mu J^\mu_A=0$$.

Adding the Yang-Mills fields of the following form to the fermion fields:

$$F^{A}_{\mu\nu} =\partial_\mu A^A_\nu - \partial_\nu A^A_\mu + g f_{BC}^{\;A} A^B_\mu A^C_\nu \quad \text{with the Lagrangian}\quad {\cal L}=-\frac{1}{4}Tr(F^A_{\mu\nu}F^{A\,\mu\nu})$$

one gets the following field equations ($${\cal D}_\mu$$ being the covariant derivative)

$${\cal D}^{\mu} F^A_{\mu\nu} = -J^A_\nu$$

Curiously the Yang-Mills fields fulfill also the identity

$${\cal D}^\mu{\cal D}^\nu F^{\mu\nu} = {\cal D}^{(\mu}{\cal D}^{\nu)} F^{\mu\nu} + {\cal D}^{[\mu}{\cal D}^{\nu]} F^{\mu\nu} = 0$$

where the first term disappears since $$F_{\mu\nu}$$ is antisymmetric whereas $${\cal D}^{(\mu}{\cal D}^{\nu)}$$ is symmetric and the second term disappears because of

$$[{\cal D}_\mu,{\cal D}_{\nu}]\chi^A = g f^A_{BC} F^B_{\mu\nu} \chi^C \quad \text{for} \quad \chi^A = F^{A\,\mu\nu}.$$

because $$f^A_{BC}$$ is antisymmetric whereas $$F^B_{\mu\nu}F^{C\,\mu\nu}$$ is symmetric in the indices $$B$$ and $$C$$. But the collorary of this result is that the colour current fulfils also

$${\cal D}_\mu J_A^\mu = 0$$

How is this compatible with $${\partial}_\mu J_A^\mu = 0$$, in particular in view of $${\cal D}_\mu J^{A\,\mu}= \partial_\mu J^{A\, \mu} + g f_{BC}^A A^B_\mu J^{C\mu}$$ ? Is the connection term also zero ? Or is $${\partial}_\mu J_A^\mu = 0$$ no longer valid? If that were the case, then one might loose the conservation of the colour charge as a vanishing covariant divergence do not automatically lead to a conservation law as the famous example of the energy-momentum tensor of the GR $$T^{\mu\nu};\nu =0$$ shows.

Thank you for any help.

I think there's a bit of confusion in how you are writing the equation of motion.

TL;DR: the current usually used for the conservation of colour currents is the quark current, i.e. the one referring to the matter component of the lagrangian (as opposed to the gluon gauge field part). This current is not the same current that you'd get from Noether's theorem. And technically Noether's theorem only applies to global symmetries which is not the case for QCD.

The equation of motion for the gluon field $$F^a_{\mu\nu}$$ is: $$\tag{1} \partial^\mu F^a_{\mu\nu}(x) + f_{abc}A^\mu_bF^c_{\mu\nu}(x) = - \color{red}{j}^a_\nu(x),$$ where the lower case $$j$$ is used for the matter currents, in this case the colour currents of the quarks: $$j^a_\nu(x)= \bar\psi(x)\gamma_\nu T_a \psi(x) = \bar \psi\gamma_\nu \frac{\lambda_a}{2}\psi,$$ where $$T^a$$ are the generators of $$SU(3)$$ and $$\lambda_a$$ the Gell-Mann matrices.

Now.

In eq. 1, bring the $$f_{abc}...$$ bit on the RHS and you get: $$\tag{2} \partial^\mu F^a_{\mu\nu}(x) = - f_{abc}A^\mu_bF^c_{\mu\nu}(x) - \color{black}{j}^a_\nu(x) = \color{red}{J}^a_\nu(x).$$

Now this $$J^a_\mu = - f_{abc}A^\mu_bF^c_{\mu\nu}(x) - \color{black}{j}^a_\nu(x)$$ is:

• The current that appears in the differential form: $$\partial^\mu F^a_{\mu\nu}(x) = \color{black}{J}^a_\nu(x) \quad \leftrightarrow \quad \mathrm{d}F = J$$
• This current is the "Noether" current. Noether's (first) theorem only applies to global symmetries, whereas QCD is a local $$SU(3)$$ symmetry so Noether's formalism wouldn't strictly speaking apply as strongly.
But if you assumed a Yang-Mills lagrangian $$\mathcal{L}_{\text{YM}} = \mathcal{L}_{\text{field}} + \mathcal{L}_{\text{matter}}$$ and apply the usual formula for the Noether current $$J^\mu = \frac{\delta \mathcal{L}}{\delta(\partial_\mu \varphi_i)}\delta \varphi_i,$$ you'd get: $$J^\mu \propto \delta \mathcal{L}_{\text{YM}} \propto \delta\mathcal{L}_{\text{field}} + \delta\mathcal{L}_{\text{matter}},$$ i.e. two things, that correspond to the two bits in $$J^a_\mu$$ above the bullet points.

And, nicely, we confirm that the current associated with the matter part of the Yang-Mills Lagrangian is indeed $$j^a_\mu$$ as we had mentioned at the beginning of the answer.

Coming from Noether's theorem, this $$J^\mu$$ is also conserved according to: $$\partial_\mu J^\mu = 0.$$

So, going back to the matter current $$j^a_\nu$$. Is it "covariantly" conserved?

Luckily, we can start form eq. 1 and use the covariant derivative: $$D^{ab}_\mu = \delta^{ab}\partial_\mu + f_{abc}A^{c}_\mu$$ to re-write eq. 1 as: $$D^\mu F^a_{\mu\nu} = -j_\nu^a(x),$$ so the same thing as your third equation but with lower case $$j$$ i.e. the matter current (quarks).

And, as you have yourself showed, you do end up with: $$D_\mu j^\mu_a =0,$$ so yes, the matter current is "covariantly" conserved.

But now you may say "what if I write $$D^\mu$$ as $$\partial^\mu + \dots$$, where then $$\partial^\mu j^a_\mu =0$$ and I am left with the other bit".

The justification for $$\partial^\mu j^a_\mu =0$$ would be another Noether's theorem but only applying to the matter part of the lagrangian. So if you only consider that bit, then sure just be content with $$\partial^\mu j^a_\mu =0$$ giving you conservation of colour currents.

But if you want to bring in the covariant derivative, then you also have to consider the gauge field part of the lagrangian and then consider the $$J^\mu$$ "full" current discussed above.

And as per your GR connection at the very end, note that GR is not a Yang-Mills theory so you cannot as easily draw parallels between the two. See the end of this answer for a more quantitative discussion about this point though.