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In a relativistic field theory, a conserved quantity according to Noether's theorem is given by a conserved current (density) $J^\mu$, i.e. $\partial_\mu J^\mu=0$. Hence, the contradiction you suggest does not really exist. The symmetry corresponding to conservation of electric charge is indeed the global part of the $U(1)$ gauge symmetry.


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Thanks for the hint. This is what I got so far: $$\delta W=\int d^4x' \mathcal{L}'(x')-\int d^4x \mathcal{L}(x)$$ Now since $d^4x'=d^4x(1+(\delta x^\alpha)_{,\alpha})$ this equation reduces to: $$\delta W=\int d^4x \mathcal{L}'(x')-\mathcal{L}(x)+(\delta x^\alpha)_{,\alpha}\mathcal{L'}(x')$$ Whic is just: $$\delta W=\int d^4x \delta\mathcal{L}+(\delta ...


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Comments to the question (v3): OP's last formula is the standard expression $$ \delta W~=~ \int \! d^4x~ \left[ {\rm EL} \cdot\bar{\delta}u + d_{\mu} j^{\mu}\right], \tag{A} $$ for the variation of the action $W=\int\! d^4x~{\cal L}$ (= Wirkung in German). Here $$ j^{\mu} ~=~ p^{\mu} \cdot \bar{\delta}u + {\cal L} ~\delta x^{\mu}, \qquad ...



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