# Difficulty to understand a chain of equalities

$$$$\mathscr{F}(\boldsymbol{\mathcal{A}})=\iiiint_{D} L\Biggl(x_\nu, \mathcal{A}_\mu, \frac{\partial \mathcal{A}_\mu}{\partial x_\nu}\Biggr)d^4X$$$$ with $$\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{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{?}

where $$D$$ is particular domain.

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

• so as I suspected $\square$ is not the D'Alembert operator but a divergence - a very misleading notation – fqq Jul 16 '19 at 8:31
• The first questionable equality is correct; the second is incorrect. The student’s notes are simply wrong. The conclusion is supposed to be $\mathscr{F}’=\mathscr{F}$. – G. Smith Jul 17 '19 at 0:17
• What is the point of the weird notation $\mathcal{X}_1$, etc. instead of the standard $x_1$ ? – G. Smith Jul 17 '19 at 0:25
• There's no need to write four integral signs for a 4D integral... that just wastes ink and hurts the eyes. – knzhou Jul 17 '19 at 11:48
• @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. – Sebastiano Jul 18 '19 at 7:46