# Squaring the E&M (Maxwell) field strength tensor

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.

• the two expressions differ by total derivatives ( $\to 0$ when integrated) – tbt May 23 at 15:21
• @tbt This is the answer. It should not be a comment. – my2cts May 23 at 20:26
• Thanks for this. However Schwartz's result is given without the proviso that it is to be integrated, so (provisionally) I don't see how it justifies his result. – JayDee.UU May 23 at 20:36
• @JayDee.UU The square of the field strength tensor appears as a term in the Lagrangian density, which is then integrated to obtain the action. The equations of motion are obtained by varying the action, so although it is not explicitly seen from the Lagrange equations, two Lagrangian densities which integrate to yield the same action are completely physically equivalent to one another. – J. Murray Jun 18 at 1:00
• Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field – knzhou Jun 18 at 8:29

$$$$\boxed{\:\:\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right) \boldsymbol{=}\left(\partial_{\mu}A^{\mu}\right)^2\boldsymbol{+}\overbrace{\partial_{\nu}\left(A_{\mu}\partial^{\mu}A^{\nu}\!\boldsymbol{-}\!A^{\nu}\partial^{\mu}A_{\mu}\right)}^{\textbf{4-divergence}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{a}\label{a}$$$$ so that for the action integral of this part of the Lagrangian density $$\:\mathcal{L}$$ $$$$\!\!\!\!\!\!\!\boxed{\:\:\iiiint\limits_{\textbf{spacetime}}\!\!\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right)\mathrm d^4x \boldsymbol{=}\!\!\!\!\iiiint\limits_{\textbf{spacetime}}\!\!\left(\partial_{\mu}A^{\mu}\right)^2\mathrm d^4x\boldsymbol{+}\!\!\overbrace{\!\!\iiiint\limits_{\textbf{spacetime}}\!\!\partial_{\nu}\left(A_{\mu}\partial^{\mu}A^{\nu}\!\boldsymbol{-}\!A^{\nu}\partial^{\mu}A_{\mu}\right)\mathrm d^4x}^{\boldsymbol{=0}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{b}\label{b}$$$$ This means that you could replace in the Lagrangian density $$\:\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 $$$$\partial_{\alpha}\boldsymbol{\equiv}\dfrac{\partial}{\partial x^{\alpha} } \tag{01}\label{01}$$$$ of their product we have $$$$\partial_{\alpha}\left(fg\right)\boldsymbol{=}f(\partial_{\alpha}g)\boldsymbol{+}(\partial_{\alpha}f)g \tag{02}\label{02}$$$$ So by a first step $$$$\!\!\!\!\!\left. \begin{cases} \:\:\:\partial_{\alpha} \boldsymbol{=}\partial_{\nu}\\ \:\:\:f \hphantom{_{\alpha}}\!\boldsymbol{=}A_{\mu}\\ \:\:\:g \hphantom{_{\alpha}}\!\boldsymbol{=}\partial^{\mu}A^{\nu} \end{cases}\right\} \quad\stackrel{\eqref{02}}{\boldsymbol{=\!=\!=\!=\!\Longrightarrow}}\quad \boxed{\:\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right) \boldsymbol{=}\partial_{\nu}\left(A_{\mu}\partial^{\mu}A^{\nu}\right)\boldsymbol{-}A_{\mu}\left(\partial_{\nu}\partial^{\mu}A^{\nu}\right)\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:} \tag{03}\label{03}$$$$ Now, if we suppose that it's permissible to interchange the partial derivatives $$\:\partial_{\nu},\partial^{\mu}\:$$ then for the last to the right term of equation \eqref{03} we have $$$$A_{\mu}\left(\partial_{\nu}\partial^{\mu}A^{\nu}\right)\boldsymbol{=}A_{\mu}\left(\partial^{\mu}\partial_{\nu}A^{\nu}\right) \tag{04}\label{04}$$$$ By a second step for the last to the right term of equation \eqref{04} we have $$$$\!\!\!\!\!\left. \begin{cases} \:\:\:\partial_{\alpha} \boldsymbol{=}\partial^{\mu}\\ \:\:\:f \hphantom{_{\alpha}}\!\boldsymbol{=}A_{\mu}\\ \:\:\:g \hphantom{_{\alpha}}\!\boldsymbol{=}\partial_{\nu}A^{\nu} \end{cases}\right\} \quad\stackrel{\eqref{02}}{\boldsymbol{=\!=\!=\!=\!\Longrightarrow}}\quad \boxed{\:A_{\mu}\left(\partial^{\mu}\partial_{\nu}A^{\nu}\right)\boldsymbol{=}\partial^{\mu}\left(A_{\mu}\partial_{\nu}A^{\nu}\right)\boldsymbol{-}\left(\partial^{\mu}A_{\mu}\right)\left(\partial_{\nu}A^{\nu}\right)\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:} \tag{05}\label{05}$$$$ But $$$$\left(\partial^{\mu}A_{\mu}\right)\left(\partial_{\nu}A^{\nu}\right)\boldsymbol{=}\left(\partial_{\mu}A^{\mu}\right)\left(\partial_{\nu}A^{\nu}\right)\boldsymbol{=}\left(\partial_{\mu}A^{\mu}\right)^2 \tag{06}\label{06}$$$$ while the term $$\:\partial^{\mu}\left(A_{\mu}\partial_{\nu}A^{\nu}\right)\:$$ by interchanging $$\:\mu,\nu\:$$ and inverting the position (up to down - down to up) of its four indices yields
$$$$\partial^{\mu}\left(A_{\mu}\partial_{\nu}A^{\nu}\right)\boldsymbol{=}\partial^{\nu}\left(A_{\nu}\partial_{\mu}A^{\mu}\right)\boldsymbol{=}\partial_{\nu}\left(A^{\nu}\partial^{\mu}A_{\mu}\right) \tag{07}\label{07}$$$$ Combining equations \eqref{04},\eqref{05},\eqref{06} and \eqref{07} we have $$$$\boxed{\:A_{\mu}\left(\partial_{\nu}\partial^{\mu}A^{\nu}\right)\boldsymbol{=}\partial_{\nu}\left(A^{\nu}\partial^{\mu}A_{\mu}\right)\boldsymbol{-}\left(\partial_{\mu}A^{\mu}\right)^2\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:} \tag{08}\label{08}$$$$ and inserting this expression in equation \eqref{03} we have finally $$$$\boxed{\:\:\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right) \boldsymbol{=}\partial_{\nu}\left(A_{\mu}\partial^{\mu}A^{\nu}\!\boldsymbol{-}\!A^{\nu}\partial^{\mu}A_{\mu}\right)\boldsymbol{+}\left(\partial_{\mu}A^{\mu}\right)^2\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{09}\label{09}$$$$ that is equation \eqref{a}.

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

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