Derivative of a field strength tensor wrt field potential in YM gauge

Question

I'm currently following this article to cosntruct a gauge invariant energy stress tensor for pure Yang-Mills gauge:

$$ \mathcal{L} = -\frac{1}{4}F_{\mu\nu}^aF_{\mu\nu}^a, \qquad F_{\mu\nu}^a = \partial_\mu A^a_\nu - \partial_\nu A^a_\mu + g\,C^{abc}A^b_\mu A^c_\nu, $$

where $C^{abc}$ are antisymemtric structure constants. I have trouble finding the correct derivation for the result 2.18. So I'm interested in:

$$ \frac{\partial F^b_{\rho\sigma}}{\partial A_\mu^a} = \frac{\partial}{\partial A^a_\mu}\left[ \dots + g\,C^{bca}A^c_\rho A^a_\sigma \right] = -2\,C^{bca}A_\rho^c\; \delta^\sigma_\mu. $$

I can't seem to find a factor of 2 and a minus sign. I tried using some of the antisymmetric properties of $F_{\mu\nu}$ and $C^{abc}$ but without success. Any help is much appriciated!

EDIT: My best guess was to try something I used in variation of strength field tensor, namely:

$$ \begin{split} \delta F^b_{\rho\sigma} &= \dots + g\,C^{bca} \delta A_\rho^c A_\sigma^a + g\,C^{bca} A_\rho^c \delta A_\sigma^a = \dots + g\,C^{bca} \delta A_\rho^c A_\sigma^a - g\,C^{bca} A_\sigma^c \delta A_\rho^a \\ &= \dots + g\,C^{bca} \delta A_\rho^c A_\sigma^a + g\,C^{bca} A_\sigma^a \delta A_\rho^c \end{split} $$

where I get factor 2 as a result of antisymmetric properties, but I can't find the minus sign.

Answer 1

$\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$There are two mistakes in your question.

You erroneously resolved one of them your edit, by introducing the variation of $F$, which is irrelevant here.

1. The $F_{\mu\nu}^a$ is not what you claim it is in the cited paper's normalisation. In the paper the authors use $$ [T_a,T_b] = iC_{ab}^{c}\,T_c \tag{1}$$ and \begin{align} F_{\mu\nu} &= \partial_\mu A_\nu -\partial_\nu A_\mu + ig[A_\mu,A_\nu] &\implies \\[1em] F_{\mu\nu}^a T_a &= \left(\partial_\mu A^a_\nu -\partial_\nu A^a_\mu\right) T_a + ig[T_b,T_c]A_\mu^bA_\nu^c = \\ &\overset{(1)}{=}\left(\partial_\mu A^a_\nu -\partial_\nu A^a_\mu \right) T_a + ig\;iC_{bc}^a T_a\;A_\mu^bA_\nu^c &\implies \\[1em] F_{\mu\nu}^a &= \partial_\mu A^a_\nu -\partial_\nu A^a_\mu - gC_{bc}^a A_\mu^bA_\nu^c, \tag{2} \end{align} i.e. with a minus, not a plus. This is the minus you've been missing.
2. (2.18) in the paper does not claim that $$\pdv{F^a_{\rho\sigma}}{A^b_\mu}= -2g C^a_{bc} \delta^\mu_{\sigma} A_\rho^c.$$ It claims that $$ \pdv{F^a_{\rho\sigma}}{A^b_\mu} \Sigma^{\rho\sigma}_b= -2g C^a_{bc} \delta^\mu_{\sigma} A_\rho^c \; \Sigma^{\rho\sigma}_b, \tag{$\star$} $$ where $$\Sigma^{\rho\sigma}_b = \pdv{\mathcal{L}_g}{F^b_{\rho\sigma}}.\tag{3} $$ The difference is that in the latter you are contracting the $\rho$ and $\sigma$ indices, as well as the Lie index $b$ in $\pdv{F^a_{\rho\sigma}}{A^b_\mu}$ so you are basically free to rename them and use certain (anti)symmetry properties to your advantage.

For reference, using (2), you should be able to find, $$\pdv{F^a_{\rho\sigma}}{A^b_\mu} = -g C^{a}_{ac}\;\big(\delta^\mu_{\rho} A^c_{\sigma}- \delta^\mu_{\sigma} A^c_{\rho}\big).$$ Contracting with $\Sigma^b_{\rho\sigma}$ of (3) you should be able to find ($\star$), i.e. reproduce (2.18).