Squaring the E&M (Maxwell) field strength tensor

Question

In Section 3.4 of Schwartz's "Quantum Field Theory and the Standard Model", the square $F_{\mu\nu}^2$ of the field strength tensor $F_{\mu\nu} = \partial_{\mu} A_{\nu} - \partial_{\nu}A_{\mu}$ is expanded. When I do this I get (after legitimate relabelling) that

$$F_{\mu\nu}^2 \equiv (\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}) = 2 \, (\partial_{\mu}A_{\nu})^2 - 2 \, \partial_{\nu}A_{\mu} \, \partial^{\mu} A^{\nu} \; .$$

Schwartz, however, appears to use

$$F_{\mu\nu}^2 \equiv = 2 \, (\partial_{\mu}A_{\nu})^2 - 2 \, (\partial_{\mu}A_{\mu})^2$$

such that in the Lorenz gauge the second term vanishes. I'm probably overlooking something elementary here, but I can't reconcile these two expansions.

PS. Apologies for the duplicate post. This entry may nevertheless help others, in view of its distinct title and the detailed answer provided by @Frobenius.

Answer 1

The answer to your question is in the following equations : ^{\begin{equation} \boxed{\:\:\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right) \boldsymbol{=}\left(\partial_{\mu}A^{\mu}\right)^2\boldsymbol{+}\overbrace{\partial_{\nu}\left(A_{\mu}\partial^{\mu}A^{\nu}\!\boldsymbol{-}\!A^{\nu}\partial^{\mu}A_{\mu}\right)}^{\textbf{4-divergence}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{a}\label{a} \end{equation}} so that for the variation of the action integral of this part of the Lagrangian density $\:\mathcal{L}$ ^{\begin{equation} \!\!\!\!\!\!\!\boxed{\:\:\delta\!\!\iiiint\limits_{\textbf{spacetime}}\!\!\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right)\mathrm d^4x \boldsymbol{=}\delta\!\!\iiiint\limits_{\textbf{spacetime}}\!\!\left(\partial_{\mu}A^{\mu}\right)^2\mathrm d^4x\boldsymbol{+}\!\overbrace{\delta\!\!\iiiint\limits_{\textbf{spacetime}}\!\!\partial_{\nu}\left(A_{\mu}\partial^{\mu}A^{\nu}\!\boldsymbol{-}\!A^{\nu}\partial^{\mu}A_{\mu}\right)\mathrm d^4x}^{0}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{b}\label{b} \end{equation}} since the 4-volume integral of this 4-divergence is equal to the $^{\prime\prime}$flux$^{\prime\prime}$ of the 4-vector function through the 4-dimensional boundary surface with its space part extending to infinity and the time part between initial and final time moments $\:t_1,t_2\:$ respectively.

This means that you could replace in the Lagrangian density $\:\mathcal{L}$ the term $\:\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right)\:$ by the term $\:\left(\partial_{\mu}A^{\mu}\right)^2\:$ without affecting the extremization of the action and so the Euler-Lagrange equations of motion. These two terms are not equal, they differ by a 4-divergence. $\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}$

Proof of equation \eqref{a}

If $\:f(\cdots,x^{\alpha},\cdots)\:$ and $\:g(\cdots,x^{\alpha},\cdots)\:$ are scalar functions and $\:x^{\alpha}\:$ is a common variable in the set of variables they depend on, then for the partial differentiation \begin{equation} \partial_{\alpha}\boldsymbol{\equiv}\dfrac{\partial}{\partial x^{\alpha} } \tag{01}\label{01} \end{equation} of their product we have \begin{equation} \partial_{\alpha}\left(fg\right)\boldsymbol{=}f(\partial_{\alpha}g)\boldsymbol{+}(\partial_{\alpha}f)g \tag{02}\label{02} \end{equation} So by a first step ^{\begin{equation} \!\!\!\!\!\left. \begin{cases} \:\:\:\partial_{\alpha} \boldsymbol{=}\partial_{\nu}\\ \:\:\:f \hphantom{_{\alpha}}\!\boldsymbol{=}A_{\mu}\\ \:\:\:g \hphantom{_{\alpha}}\!\boldsymbol{=}\partial^{\mu}A^{\nu} \end{cases}\right\} \quad\stackrel{\eqref{02}}{\boldsymbol{=\!=\!=\!=\!\Longrightarrow}}\quad \boxed{\:\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right) \boldsymbol{=}\partial_{\nu}\left(A_{\mu}\partial^{\mu}A^{\nu}\right)\boldsymbol{-}A_{\mu}\left(\partial_{\nu}\partial^{\mu}A^{\nu}\right)\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:} \tag{03}\label{03} \end{equation}} Now, if we suppose that it's permissible to interchange the partial derivatives $\:\partial_{\nu},\partial^{\mu}\:$ then for the last to the right term of equation \eqref{03} we have \begin{equation} A_{\mu}\left(\partial_{\nu}\partial^{\mu}A^{\nu}\right)\boldsymbol{=}A_{\mu}\left(\partial^{\mu}\partial_{\nu}A^{\nu}\right) \tag{04}\label{04} \end{equation} By a second step for the last to the right term of equation \eqref{04} we have ^{\begin{equation} \!\!\!\!\!\left. \begin{cases} \:\:\:\partial_{\alpha} \boldsymbol{=}\partial^{\mu}\\ \:\:\:f \hphantom{_{\alpha}}\!\boldsymbol{=}A_{\mu}\\ \:\:\:g \hphantom{_{\alpha}}\!\boldsymbol{=}\partial_{\nu}A^{\nu} \end{cases}\right\} \quad\stackrel{\eqref{02}}{\boldsymbol{=\!=\!=\!=\!\Longrightarrow}}\quad \boxed{\:A_{\mu}\left(\partial^{\mu}\partial_{\nu}A^{\nu}\right)\boldsymbol{=}\partial^{\mu}\left(A_{\mu}\partial_{\nu}A^{\nu}\right)\boldsymbol{-}\left(\partial^{\mu}A_{\mu}\right)\left(\partial_{\nu}A^{\nu}\right)\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:} \tag{05}\label{05} \end{equation}} But \begin{equation} \left(\partial^{\mu}A_{\mu}\right)\left(\partial_{\nu}A^{\nu}\right)\boldsymbol{=}\left(\partial_{\mu}A^{\mu}\right)\left(\partial_{\nu}A^{\nu}\right)\boldsymbol{=}\left(\partial_{\mu}A^{\mu}\right)^2 \tag{06}\label{06} \end{equation} while the term $\:\partial^{\mu}\left(A_{\mu}\partial_{\nu}A^{\nu}\right)\:$ by interchanging $\:\mu,\nu\:$ and inverting the position (up to down - down to up) of its four indices yields
\begin{equation} \partial^{\mu}\left(A_{\mu}\partial_{\nu}A^{\nu}\right)\boldsymbol{=}\partial^{\nu}\left(A_{\nu}\partial_{\mu}A^{\mu}\right)\boldsymbol{=}\partial_{\nu}\left(A^{\nu}\partial^{\mu}A_{\mu}\right) \tag{07}\label{07} \end{equation} Combining equations \eqref{04},\eqref{05},\eqref{06} and \eqref{07} we have \begin{equation} \boxed{\:A_{\mu}\left(\partial_{\nu}\partial^{\mu}A^{\nu}\right)\boldsymbol{=}\partial_{\nu}\left(A^{\nu}\partial^{\mu}A_{\mu}\right)\boldsymbol{-}\left(\partial_{\mu}A^{\mu}\right)^2\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:} \tag{08}\label{08} \end{equation} and inserting this expression in equation \eqref{03} we have finally \begin{equation} \boxed{\:\:\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right) \boldsymbol{=}\partial_{\nu}\left(A_{\mu}\partial^{\mu}A^{\nu}\!\boldsymbol{-}\!A^{\nu}\partial^{\mu}A_{\mu}\right)\boldsymbol{+}\left(\partial_{\mu}A^{\mu}\right)^2\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{09}\label{09} \end{equation} that is equation \eqref{a}.

$\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}$

Addendum

In a system $\:\mathrm{Oxyz}t\:$ with $\:\phi, \mathbf{A} \boldsymbol{=}\left(\mathrm{A_x},\mathrm{A_y},\mathrm{A_z}\right)\:$ the scalar and vector potentials the terms appeared in the main section of the answer are expressed as follows : ^{\begin{align} &\!\!\!\!\!\!\!\!\!\!\!\!\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right) \boldsymbol{=}\left(\dfrac{\partial\phi}{c^2\partial t}\right)^2\!\!\boldsymbol{+}\dfrac{2}{c^2}\left(\boldsymbol{\nabla}\phi\boldsymbol{\cdot}\dfrac{\partial\mathbf{A}}{\partial t}\right)\boldsymbol{+}\left(\dfrac{\partial\mathrm{A_x}}{\partial \mathrm x}\right)^2\boldsymbol{+}\left(\dfrac{\partial\mathrm{A_y}}{\partial \mathrm y}\right)^2\boldsymbol{+}\left(\dfrac{\partial\mathrm{A_z}}{\partial \mathrm z}\right)^2 \nonumber\\ &\!\!\!\!\!\!\!\!\!\!\!\! \hphantom{\boldsymbol{==}}\boldsymbol{+}2\left(\dfrac{\partial\mathrm{A_x}}{\partial \mathrm y}\dfrac{\partial\mathrm{A_y}}{\partial \mathrm x}\boldsymbol{+}\dfrac{\partial\mathrm{A_y}}{\partial \mathrm z}\dfrac{\partial\mathrm{A_z}}{\partial \mathrm y}\boldsymbol{+}\dfrac{\partial\mathrm{A_z}}{\partial \mathrm x}\dfrac{\partial\mathrm{A_x}}{\partial \mathrm z}\right) \tag{A-01}\label{A-01}\\ &\!\!\!\!\!\!\!\!\!\!\!\!\left(\partial_{\mu}A^{\mu}\right)^2 \boldsymbol{=}\left(\dfrac{\partial\phi}{c^2\partial t}\right)^2\boldsymbol{+}\dfrac{2}{c^2}\left(\dfrac{\partial\phi}{\partial t}\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{A}\right)\boldsymbol{+}\left(\dfrac{\partial\mathrm{A_x}}{\partial \mathrm x}\right)^2\boldsymbol{+}\left(\dfrac{\partial\mathrm{A_y}}{\partial \mathrm y}\right)^2\boldsymbol{+}\left(\dfrac{\partial\mathrm{A_z}}{\partial \mathrm z}\right)^2 \nonumber\\ &\!\!\!\!\!\!\!\!\!\!\!\! \hphantom{\boldsymbol{==}}\boldsymbol{+}2\left(\dfrac{\partial\mathrm{A_x}}{\partial \mathrm x}\dfrac{\partial\mathrm{A_y}}{\partial \mathrm y}\boldsymbol{+}\dfrac{\partial\mathrm{A_y}}{\partial \mathrm y}\dfrac{\partial\mathrm{A_z}}{\partial \mathrm z}\boldsymbol{+}\dfrac{\partial\mathrm{A_z}}{\partial \mathrm z}\dfrac{\partial\mathrm{A_x}}{\partial \mathrm x}\right) \tag{A-02}\label{A-02}\\ &\!\!\!\!\!\!\!\!\!\!\!\!\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right)\boldsymbol{-}\left(\partial_{\mu}A^{\mu}\right)^2 \boldsymbol{=}\dfrac{2}{c^2}\left(\boldsymbol{\nabla}\phi\boldsymbol{\cdot}\dfrac{\partial\mathbf{A}}{\partial t}\boldsymbol{-}\dfrac{\partial\phi}{\partial t}\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{A}\right)\boldsymbol{+}2\left(\dfrac{\partial\mathrm{A_x}}{\partial \mathrm y}\dfrac{\partial\mathrm{A_y}}{\partial \mathrm x}\boldsymbol{+}\dfrac{\partial\mathrm{A_y}}{\partial \mathrm z}\dfrac{\partial\mathrm{A_z}}{\partial \mathrm y}\boldsymbol{+}\dfrac{\partial\mathrm{A_z}}{\partial \mathrm x}\dfrac{\partial\mathrm{A_x}}{\partial \mathrm z}\right) \nonumber\\ & \hphantom{\boldsymbol{==}}\boldsymbol{-}2\left(\dfrac{\partial\mathrm{A_x}}{\partial \mathrm x}\dfrac{\partial\mathrm{A_y}}{\partial \mathrm y}\boldsymbol{+}\dfrac{\partial\mathrm{A_y}}{\partial \mathrm y}\dfrac{\partial\mathrm{A_z}}{\partial \mathrm z}\boldsymbol{+}\dfrac{\partial\mathrm{A_z}}{\partial \mathrm z}\dfrac{\partial\mathrm{A_x}}{\partial \mathrm x}\right) \tag{A-03}\label{A-03} \end{align}}

\begin{align} A_{\mu}\partial^{\mu}A^{\nu}\boldsymbol{-}A^{\nu}\partial^{\mu}A_{\mu} &\boldsymbol{=}\boldsymbol{H}\boldsymbol{=} \left(h^{0},h^{1},h^{2},h^{3}\right)\boldsymbol{=} \left(h^{0},\boldsymbol{h}\right) \quad \text{where} \nonumber\\ h^{0} &\boldsymbol{=}\dfrac{1}{c}\left(\mathbf{A}\boldsymbol{\cdot}\boldsymbol{\nabla}\phi\boldsymbol{-}\phi\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{A}\right) \nonumber\\ \boldsymbol{h} &\boldsymbol{=}\dfrac{1}{c^2}\left(\phi\dfrac{\partial\mathbf{A}}{\partial t}\boldsymbol{-}\mathbf{A}\dfrac{\partial\phi}{\partial t}\right)\boldsymbol{+}\left(\mathbf{A}\boldsymbol{\cdot}\boldsymbol{\nabla}\right)\mathbf{A}\boldsymbol{-}\mathbf{A}\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{A} \tag{A-04}\label{A-04} \end{align}

The expression of the rhs of equation \eqref{A-03} is the 4-divergence of the 4-vector $\:\boldsymbol{H}\:$ defined by \eqref{A-04}.

$\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}$

Link : Expanding electromagnetic field Lagrangian in terms of gauge field

$\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}$

Reference : "Field Quantization" by W.Greiner and J.Reinhardt, Springer 1996, 6.1 The Lagrangian of the Maxwell Field (Problem and Solution).