This post directly follows this question and is very similar in nature:
Consider the following Lagrangian:
$$\mathcal{L}=\frac14\left(\partial_\mu A_\nu-\partial_\nu A_\mu\right)\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)+\partial_\mu\phi^\ast\partial^\mu\phi$$
$$+ieA^\mu\left(\phi\partial_\mu\phi^\ast – \phi^\ast\partial_\mu\phi\right)
+e^2A_\mu A^\mu\phi^\ast\phi-m^2\phi^\ast\phi-\frac12\lambda\left(\phi^\ast\phi\right)^2\tag{1}$$
[…]The equation of motion obtained from the Lagrangian, $(1)$, for $A_\color{red}{\mu}$ is more interesting,
$$\frac{\partial\mathcal{L}}{\partial A_\nu}-\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu A_\nu}=ie\left(\phi\partial_{\color{red}{\nu}}\phi^\ast-\phi^\ast\partial_{\color{red}{\nu}}\phi\right)$$
$$+2e^2A^\nu\phi^\ast\phi-\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=0\tag{2}$$
This can be written as
$$\partial_\mu F^{\mu\nu}=ie\left[\phi\left(D^\nu\phi\right)^\ast-\phi^\ast D^\nu\phi\right]\tag{3}$$
Where $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$ and $D^\nu\phi\equiv \partial^\nu\phi + ieA^\nu\phi$ – similar to the previous question.
Before even doing any calculations I believe there are some typos present on the indices I marked in red. In particular, I think the $A_\mu$ should be $A^\nu$ as there is no equation of motion for $A_\mu$: equation $(2)$ does not even contain $A_\mu$.
I think the corrected version of $(2)$ should read
$$\frac{\partial\mathcal{L}}{\partial A_\nu}-\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu A_\nu}=ie\left(\phi\partial^\nu\phi^\ast-\phi^\ast\partial^\nu\phi\right)$$
$$+2e^2A^\nu\phi^\ast\phi-\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=0\tag{A}$$
A quick check is needed to make sure equation $(\mathrm{A})$ is actually correct, to verify this I will use it to prove eqn. $(3)$:
$$\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=\partial_\mu F^{\mu\nu}$$
$$=ie\phi\partial^\nu\phi^\ast-ie\phi^\ast\partial^\nu\phi+e^2A^\nu\phi^\ast\phi+e^2A^\nu\phi^\ast\phi$$
$$ie\phi\left(\partial^\nu\phi^\ast-ieA^\nu\phi^\ast\right)-ie\phi^\ast\left(\partial^\nu\phi+ieA^\nu\phi\right)$$
$$=ie\left[\phi\left(D^\nu\phi\right)^\ast-\phi^\ast D^\nu\phi\right]$$
which is eqn. $(3)$, so equation $(\mathrm{A})$ appears correct.
The objective is now to prove equation $(\mathrm {A})$:
To make the proof easier I first rewrite the Lagrangian, $(1)$ as
$$\mathcal{L}=-\frac14\partial_\mu A_\nu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)+\frac14\partial_\nu A_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)$$
$$+ieA_\nu\left(\phi\partial^\nu\phi^\ast – \phi^\ast\partial^\nu\phi\right)
+e^2A_\nu A^\nu\phi^\ast\phi$$
$$-m^2\phi^\ast\phi-\frac12\lambda\left(\phi^\ast\phi\right)^2+\partial_\mu\phi^\ast\partial^\mu\phi$$
where in the second line of eqn. $(1)$ I lowered and raised the respective indices on the terms involving $A^\mu \partial_\mu\phi$ and let $\mu\to \nu$ in these terms and the next term involving $e^2$.
So
$$\frac{\partial\mathcal{L}}{\partial A_\nu}=ie\left(\phi\partial^\nu\phi^\ast-\phi^\ast\partial^\nu\phi\right)+e^2A^\nu\phi^\ast\phi$$
and
$$\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu A_\nu}=-\frac14\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)$$
Putting these together gives
$$\frac{\partial\mathcal{L}}{\partial A_\nu}-\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu A_\nu}$$
$$=ie\left(\phi\partial^\nu\phi^\ast-\phi^\ast\partial^\nu\phi\right)+e^2A^\nu\phi^\ast\phi+\frac14\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=0\tag{!}$$
But this is not the same as equation $(\mathrm{A})$:
$$ie\left(\phi\partial^\nu\phi^\ast-\phi^\ast\partial^\nu\phi\right)
+2e^2A^\nu\phi^\ast\phi-\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=0\tag{A}$$
differing by a factor of $2$ and $1/4$ in the second and third terms, respectively. I'm sure equation $(\mathrm{A})$ is the correct one as I derived eqn. $(3)$ from it. It is likely I am making mistakes in $(\mathrm{!})$, but I just cannot seem to find them.
May I please have some help in proving eqn. $(\mathrm{A})$ or identifying errors in $(\mathrm{!})$?
Best Answer
The key things that lead to the correct factors in (A) are
the basic expression $A_\mu A^\mu=\eta^{\mu\rho}A_\mu A_\rho=-A_0^2+A_1^2+A_2^2+A_3^2$ in (1) from OP that you have to differentiate w.r.t. each of $A_0,A_1,A_2,A_3\,.$ This differentiation obviously yields (basic calculus) $-2A_0,+2A_1,+2A_2,+2A_3$ respectively. This can be written as $$\tag{1} \frac{\partial}{\partial A_\nu}(A_\mu A^\mu)=2\eta^{\nu\rho}A_\rho=2A^\nu $$ (c.f. (A)).
To the term $F_{\alpha\beta}F^{\alpha\beta}=(\partial_\alpha A_\beta-\partial_\beta A_\alpha)(\partial^\alpha A^\beta-\partial^\beta A^\alpha)$ we are now going to apply the same basic rules: \begin{align}\tag{2} &(\partial_\alpha A_\beta-\partial_\beta A_\alpha)(\partial^\alpha A^\beta-\partial^\beta A^\alpha)= (\partial_\alpha A_\beta-\partial_\beta A_\alpha)\,\underbrace{\eta^{\alpha\rho}\eta^{\beta\sigma} (\partial_\rho A_\sigma-\partial_\sigma A_\rho)} \end{align} From $\eta^{\mu\nu}=\operatorname{diag}(-1,1,1,1)\,,$ the underbraced term is numerically equal to \begin{align} \cases{0&$\alpha=\beta\,,$\\[2mm] -(\partial_\color{red}{0} A_\beta-\partial_\beta A_\color{red}{0})\,,&$\beta=1,2,3\,,$\\[2mm] -(\partial_\alpha A_\color{red}{0}-\partial_\color{red}{0} A_\alpha)\,,&$\alpha=1,2,3\,,$\\[2mm] (\partial_\alpha A_\beta-\partial_\beta A_\alpha)\,,&$\alpha,\beta=1,2,3,\alpha\not=\beta\,.$\\ } \end{align} This leads to \begin{align} \\[2mm] (2)&=-F_{01}^2-F_{02}^2-F_{03}^2+F_{12}^2+F_{13}^2+F_{23}^2 -F_{10}^2-F_{20}^2-F_{30}^2+F_{21}^2+F_{31}^2+F_{32}^2\\[2mm] &=-2F_{01}^2-2F_{02}^2-2F_{03}^2+2F_{12}^2+2F_{13}^2+2F_{23}^2\,. \end{align} The last line uses again that $F_{\alpha\beta}$ is anti symmetric which makes $F_{\alpha\beta}^2$ symmetric. Differentiating this w.r.t. $\partial_\mu A_\nu$ for a fixed index pair $\mu,\nu$ gives $$ \frac{\partial F_{\alpha\beta}F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)}=\cases{-4\,F_{\color{red}{0}\nu}=-4\,(\partial_\color{red}{0} A_\nu-\partial_\nu A_\color{red}{0})\,,&$\nu=1,2,3\,,$\\[2mm] -4\,F_{\mu\color{red}{0}}=-4\,(\partial_\mu A_\color{red}{0}-\partial_\color{red}{0} A_\mu)\,,&$\mu=1,2,3\,,$\\[2mm] 4\,F_{\mu\nu}=4\,(\partial_\mu A_\nu-\partial_\nu A_\mu)\,,&$\mu,\nu=1,2,3\,.$} $$ This can be written as $$ \frac{\partial F_{\alpha\beta}F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)} =4\,\eta^{\mu\rho}\eta^{\nu\sigma}\,F_{\rho\sigma}=4\,F^{\mu\nu}=4\,(\partial^\mu A^\nu-\partial^\nu A^\mu)\,. $$ This confirms again what we see in (A): the $\frac14$ cancels and a minus sign comes from $-\partial_\mu\frac{\mathcal{L}}{\partial(\partial_\mu A_\nu)}\,.$