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If we start with a functional or integral action

\begin{equation} \mathscr{F}(\boldsymbol{\mathcal{A}})=\iiiint_{D} L\Biggl(x_\nu, \mathcal{A}_\mu, \frac{\partial \mathcal{A}_\mu}{\partial x_\nu}\Biggr)d^4X \end{equation} with $\nu,\mu=1,\dotsb,4$ with $X=({x}_1,{x}_2,x_3,x_4)=(x,y,z,ict)$, where $\boldsymbol{\mathcal{A}}=\Bigl(\vec{A}, \dfrac{i \varphi}{c}\Bigr)$, where $L$ is the Lagrangian density. Being

$$L'=L+\mu_{0}\square (f\boldsymbol{\mathcal{J}})$$ ($\square$ is 4-divergence operator) with $\boldsymbol{\mathcal{J}}$ the 4-vector density of current. If we define as a set of admissible functions that of the functions that assume the same value at the edge (no longer in the two extremes as for functions of a variable), why

\begin{equation} \begin{aligned} \iiiint_{D}L'\,d^{4}{X}&=\iiiint_{D}L\,d^{4}{X}+\mu_0\iiiint_{D}\square (f\boldsymbol{\mathcal{J}})d^{4}{X}=\\ &\stackrel{\color{red}{\bf ?}}{=}\mathscr{F}+\mu_{0}\int_{\partial D}(f\boldsymbol{\mathcal{J}})\cdot \hat n\, d^{4}{X}\stackrel{\color{red}{\bf ?}}{=}0, \end{aligned}\tag{?} \end{equation}

where $D$ is particular domain.

NB: If that's helpful, I'm putting in a picture of a student's notes.

enter image description here

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    $\begingroup$ so as I suspected $\square$ is not the D'Alembert operator but a divergence - a very misleading notation $\endgroup$ – fqq Jul 16 '19 at 8:31
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    $\begingroup$ The first questionable equality is correct; the second is incorrect. The student’s notes are simply wrong. The conclusion is supposed to be $\mathscr{F}’=\mathscr{F}$. $\endgroup$ – G. Smith Jul 17 '19 at 0:17
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    $\begingroup$ What is the point of the weird notation $\mathcal{X}_1$, etc. instead of the standard $x_1$ ? $\endgroup$ – G. Smith Jul 17 '19 at 0:25
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    $\begingroup$ There's no need to write four integral signs for a 4D integral... that just wastes ink and hurts the eyes. $\endgroup$ – knzhou Jul 17 '19 at 11:48
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    $\begingroup$ @G.Smith With lot of cordiality can you add a detailed answer referring to the your comment, please?. I kindly ask you for the details because you cancel the outline etc.. These are topics I did 25 years ago. Thank you very much. $\endgroup$ – Sebastiano Jul 18 '19 at 7:46

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