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Before equation (77.35), Srednicki's QFT book says

We define the chiral gauge current $j^{a\mu}$ [where $a$ is a color index]. Its covariant divergence (which should be zero, according to Noether's theorem) is given by $D_\mu^{ab} j^{b\mu}$ ...

But below this answer, user ACuriousMind comments

I see no reason at all why Noether's theorem would yield a covariant derivative acting on the conserved current, since the conserved currents follow from Noether's first theorem applied to the global version of the symmetry and are unrelated to the gauge theory (Noether's second theorem for gauge symmetries yields off-shell identities unrelated to conservation), and the sentence in Srednicki mystifies me.

It seems clear to me that the four-divergence that enters into the statement of charge conservation should be gauge-covariant, so Srednicki is right and we need to use the covariant derivative. But I suppose it's logically possible that despite not transforming covariantly, the expression $\partial_\mu j^{a \mu}$ could vanish on-shell in every gauge (which ACuriousMind presumably believes to be the case?). Who's right?

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  • $\begingroup$ Here's an answer by Qmechanic with a nice overview over the different identities one can obtain from Noether's first/second theorems. Also, while I think the context is relevant, I feel that maybe this question might be more useful for a generic visitor if it were phrased more like "What is the proof that the covariant derivative does/does not enter into Noether's theorem?" than its current "ACM vs. Srednicki" style ;) $\endgroup$ – ACuriousMind Jul 22 '17 at 23:35
  • $\begingroup$ @ACuriousMind I did not want to pit you two against each other gladiator-style, which is why I didn't title the question "Who's right, ACuriousMind or Srednicki?", even though that title would probably draw more views and increase the chance of someone posting a right answer :-). I was originally going to just summarize your argument, but it was so well-phrased that it seemed more direct to simply quote it. I was originally going to leave off "Who's right?", which is the only "adversarially" phrased part of the question, but I've found it's good practice to end a post with the actual question. $\endgroup$ – tparker Jul 22 '17 at 23:48
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    $\begingroup$ for a textbook exposition, see Weinberg, §15.3 $\endgroup$ – AccidentalFourierTransform Jul 23 '17 at 13:52
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We are both right!

Following AccidentalFourierTransform's suggestion in the comments, Weinberg's "Quantum Theory of Fields, Vol. II" indeed provides the relevant explanation for the discrepancy between the ordinary conservation of the Noether current in a gauge theory by applying Noether's theorem to the global version of the gauged symmetry and the covariant conservation claimed by Srednicki.

  1. Ordinary conservation of the full Noether current: Noether's theorem, taken at face value, contains no gauge covariant derivatives. The Noether current $$ J^\mu = \frac{\delta \mathcal{L}}{\delta(\partial_\mu \phi_i)}\delta \phi_i\tag{1}$$ for a global symmetry $\phi_i\mapsto \phi_i + \epsilon \delta \phi_i$ of a Lagrangian density $\mathcal{L}$ is conserved in the ordinary sense of $\partial_\mu J^\mu \approx 0$. Here the $\phi_i$ include both the gauge and matter fields.

  2. Covariant conservation of the matter part of the Noether current in a Yang-Mills theory: In a Yang-Mills theory with a Lagrangian of the form $$ \mathcal{L}_\text{YM}[A,\psi] = - \frac{1}{4} F_{\mu\nu}^a F^{a\mu\nu} + \mathcal{L}_\text{matter}(\psi,D_\mu \psi)\tag{2},$$ where $F = \mathrm{d} A + A \wedge A$ is the usual field strength tensor of a non-Abelian gauge field $A$, $\mathcal{L}_\text{matter}$ the function that yields the matter Lagrangian uncoupled to any gauge field as $\mathcal{L}(\psi,\partial_\mu \psi)$ and $\psi$ one (or several) matter fields charged under the gauge group in a representation $\rho$, the ordinary Noether current (1) under the global version of the gauge symmetry generated by the $T^a$ is more explicitly given by $$ J^{\mu a} = - C^{cab}F^{c\mu\nu}A^b_\nu - \underbrace{\frac{\delta \mathcal{L}_\text{matter}}{\delta(D_\mu \psi)}\rho(T^a)\psi}_{=:j^{\mu a}},\tag{3}$$ where the $C$ are the structure constants of the non-Abelian gauge group. Furthermore, this current is the current involved in the equation of motion $\mathrm{d}{\star}F = {\star}J$. However, one may write the equation of motion in terms of the covariant derivative of the field strength: $$ D^{ab}_{\mu} F^{b\mu\nu} = - j^{\nu a}\tag{4}$$ Contracting this equation with $D_\nu$ yields: $$ - D^{ca}_\mu j^{\nu a} = D^{ca}_\nu D^{ab}_\mu F^{b\mu\nu} = [D^{ca}_\nu ,D^{ab}_\mu] F^{b\mu\nu} = 0\tag{5},$$ where the last equality follows because $[D,D]$ is just $F$ acting in the proper representation on whatever is behind it. Since $F$ is Lie-algebra valued, it is in the adjoint representation, so we get $\rho_\text{ad}(F) F = [F,F] = 0$. This is the sought-after "covariant conservation law". I therefore conclude that Srednicki's formulae are correct, but the part where he claims this follows from Noether's theorem could perhaps have used a bit more elaboration.

Weinberg also helpfully points out that although this does not directly apply to general relativity as it is not a Yang-Mills theory, the two currents $J,j$ are directly reflected by the covariantly-conserved energy momentum tensor $T_{\mu\nu}$ and the ordinarily-conserved pseudotensor obtained by adding what Weinberg calls the nonlinear piece of $R^{\mu\nu} - \frac{1}{2}g^{\mu\nu})$ to $T_{\mu\nu}$ and what I think is usually called the Landau-Lifschitz pseudotensor. That the ordinarily-conserved object is only a pseudo-tensor is precisely the analogue of $J^{\mu a}$ not transforming in a proper linear representation, but like a gauge field (due to the appearance of $A$ in its expression).

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  • $\begingroup$ Note that the "full" current $J^\mu_a$ is also the one that equals $\partial \mathcal{L}/\partial A_{\mu a}$. $\endgroup$ – tparker Jul 24 '17 at 19:39
  • $\begingroup$ So $J^\mu$ doesn't transform covariantly, but the charge $\int d^3x\ J^{a0}$ is conserved, while $\mathcal{j}^\mu$ does transform covariantly, but the charge $\int d^3x\ \mathcal{j}^{a0}$ is not conserved? $\endgroup$ – tparker Jul 24 '17 at 21:40
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    $\begingroup$ @tparker Yup, precisely. $\endgroup$ – ACuriousMind Jul 24 '17 at 21:49
  • $\begingroup$ In your equation (5), i don't see how that's zero... In the abelian case, $F_{\mu\nu}$ is uncharged, so that the gauge covariant derivative acting on it is just the partial derivative, and then of course we have $D_\mu D_\nu F^{\mu\nu} = \partial_\mu \partial_\nu F^{\mu\nu} = 0$. But in the non-abelian case, the field strength tensor is charged, so the covariant derivative does not reduce to the usual partial derivative. In particular, since $F_{\mu\nu}$ is anti-symmetric: $D_\mu D_\nu F^{\mu\nu} = [D_\mu,D_\nu] F^{\mu \nu} = F_{\mu\nu} F^{\mu \nu}$ $\endgroup$ – Lorenz Mayer Oct 10 '18 at 12:54
  • $\begingroup$ @LorenzMayer It would appear to me that you're right. Do you have a suggestion of how to save Srednicki's statement, or is it just wrong for non-Abelian gauge theories? (It probably doesn't lead to any severe errors, because the currents of non-Abelian theories are gauge-variant anyway and hence not observable) $\endgroup$ – ACuriousMind Oct 11 '18 at 16:18

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