In the case of classical field theory, Noether's theorem ensures that for a given action $$S=\int \mathrm{d}^dx\,\mathcal{L}(\phi_\mu,\partial_\nu\phi_\mu,x^i)$$ that stays invariant under the transformation $\{ x^i\rightarrow x^i+\delta x^i \,;\, \phi_\mu\rightarrow \phi_\mu+\delta \phi_\mu\}$, namely a local symmetry, then there exist conserved quantities named currents.

If it exist a set of variables $\{\epsilon_r,X_r^i,\Phi_r^\mu\}$ (where $\epsilon_r\rightarrow 0$) such that $\delta x^i=\epsilon_r\,X_r^i$ and $\delta \phi^\mu=\epsilon_r\,\Phi_r^\mu$, then these currents are the quantities  : $$ J_r^{\,\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\nu\phi^\mu)}\Phi_r^\mu+\left[\mathcal{L}\delta_\sigma^{\,\nu}-\frac{\partial\mathcal{L}}{\partial(\partial_\nu\phi^\mu)}\partial_\sigma\phi^\mu\right]X_r^\sigma $$ These are conserved in the sens that $\partial_\nu J_r^{\,\nu}=0$, which is true on-shell, i.e. provided equations of motion are verified.

Now I'm trying to apply this result in the case of classical EM field theory. Considere the associated lagrangian  : $$ \mathcal{L}_{EM}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+A_\mu j^{\,\mu} $$ where $A_\mu=(\phi,-\textbf{A})$ is the 4-potential and $j^{\,\mu}=(\rho,\textbf{j})$ is the 4-current, describing classical charge sources. $F_{\mu\nu}$ is the EM tensor.

In that case, equations of motion are : $$\partial_\nu F^{\nu\mu}=j^{\,\mu}$$

Considering the global gauge symmetry of the classical EM field theory $A_\mu\rightarrow A_\mu+\epsilon\,\partial_\mu\chi$ where $\chi$ is an arbitrary gauge function, I'd like to apply the Noether's theorem in the case where $X_r^i=0$. Refering to this P.S.E post, then the associated conserved current reads  : $$ J^\mu=F^{\mu\nu}\partial_\nu\chi\quad\text{with}\quad\partial_\mu J^\mu=0 $$ Fine, now I have a few questions  :

• Is it actually really legit to apply Noether's theorem to global gauge symmetry? I mean, is Noether's theorem still hold even if I have a set $\{\epsilon_r,X_r^i=0,\Phi_r^\mu\}$, i.e. no transformation on coordinates?

Related : physics.stackexchange.com/questions/13870/gauge-symmetry-is-not-a-symmetry.

• In that case, what does the $J^\mu$ quantity physically means? One can compute, using equations of motion and the anti-symmetry of $F^{\mu\nu}$: $$ \partial_\mu J^\mu=\partial_\mu F^{\mu\nu}\,\partial_\nu\chi + F^{\mu\nu}\,\partial_\mu\partial_\nu\chi=j^{\,\nu}\,\partial_\nu\chi+0 $$ I would rather expect something like $$ \partial_\mu J^\mu=\partial_\mu j^{\,\mu}=0,\;\forall\chi $$ In that case I could understand that charge conservation stems directly from global gauge symmetry. There is something I'm clearly missing. Or should I trick by computing $\int\mathrm{d}x\,\partial_\mu J^\mu$ and perform an integration by part?

• I can't see anywhere clearly appearing the fact that the physics of classical EM theory is invariant under Lorentz transformation. I would expect that such invariance would be a really strong symmetry of the system.

In the case of classical field theory, Noether's theorem ensures that for a given action $$S=\int \mathrm{d}^dx\,\mathcal{L}(\phi_\mu,\partial_\nu\phi_\mu,x^i)$$ that stays invariant under the transformation $\{ x^i\rightarrow x^i+\delta x^i \,;\, \phi_\mu\rightarrow \phi_\mu+\delta \phi_\mu\}$, namely a local symmetry, then there exist conserved quantities named currents.

If it exist a set of variables $\{\epsilon_r,X_r^i,\Phi_r^\mu\}$ (where $\epsilon_r\rightarrow 0$) such that $\delta x^i=\epsilon_r\,X_r^i$ and $\delta \phi^\mu=\epsilon_r\,\Phi_r^\mu$, then these currents are the quantities: $$ J_r^{\,\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\nu\phi^\mu)}\Phi_r^\mu+\left[\mathcal{L}\delta_\sigma^{\,\nu}-\frac{\partial\mathcal{L}}{\partial(\partial_\nu\phi^\mu)}\partial_\sigma\phi^\mu\right]X_r^\sigma. $$ These are conserved in the sense that $\partial_\nu J_r^{\,\nu}=0$, which is true on-shell, i.e. provided equations of motion are verified.

Now I'm trying to apply this result in the case of classical EM field theory. Consider the associated lagrangian: $$ \mathcal{L}_{EM}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+A_\mu j^{\,\mu} $$ where $A_\mu=(\phi,-\textbf{A})$ is the 4-potential and $j^{\,\mu}=(\rho,\textbf{j})$ is the 4-current, describing classical charge sources. $F_{\mu\nu}$ is the EM tensor.

In that case, equations of motion are : $$\partial_\nu F^{\nu\mu}=j^{\,\mu}.$$

Considering the global gauge symmetry of the classical EM field theory $$A_\mu\rightarrow A_\mu+\epsilon\,\partial_\mu\chi$$ where $\chi$ is an arbitrary gauge function, I'd like to apply the Noether's theorem in the case where $X_r^i=0$. Referring to this Phys.SE post, then the associated conserved current reads: $$ J^\mu=F^{\mu\nu}\partial_\nu\chi\quad\text{with}\quad\partial_\mu J^\mu=0. $$ Fine, now I have a few questions:

• Is it actually really legit to apply Noether's theorem to global gauge symmetry? I mean, is Noether's theorem still hold even if I have a set $\{\epsilon_r,X_r^i=0,\Phi_r^\mu\}$, i.e. no transformation on coordinates?

Related: Gauge symmetry is not a symmetry?.

• In that case, what does the $J^\mu$ quantity physically means? One can compute, using equations of motion and the anti-symmetry of $F^{\mu\nu}$: $$ \partial_\mu J^\mu=\partial_\mu F^{\mu\nu}\,\partial_\nu\chi + F^{\mu\nu}\,\partial_\mu\partial_\nu\chi=j^{\,\nu}\,\partial_\nu\chi+0. $$ I would rather expect something like $$ \partial_\mu J^\mu=\partial_\mu j^{\,\mu}=0,\;\forall\chi. $$ In that case I could understand that charge conservation stems directly from global gauge symmetry. There is something I'm clearly missing. Or should I trick by computing $\int\mathrm{d}x\,\partial_\mu J^\mu$ and perform an integration by part?

• I can't see anywhere clearly appearing the fact that the physics of classical EM theory is invariant under Lorentz transformation. I would expect that such invariance would be a really strong symmetry of the system.

In the case of classical field theory, Noether's theorem ensures that for a given action $$S=\int \mathrm{d}^dx\,\mathcal{L}(\phi_\mu,\partial_\nu\phi_\mu,x^i)$$ that stays invariant under the transformation $\{ x^i\rightarrow x^i+\delta x^i \,;\, \phi_\mu\rightarrow \phi_\mu+\delta \phi_\mu\}$, namely a local symmetry, then there exist conserved quantities named currents.

If it exist a set of variables $\{\epsilon_r,X_r^i,\Phi_r^\mu\}$ (where $\epsilon_r\rightarrow 0$) such that $\delta x^i=\epsilon_r\,X_r^i$ and $\delta \phi^\mu=\epsilon_r\,\Phi_r^\mu$, then these currents are the quantities : $$ J_r^{\,\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\nu\phi^\mu)}\Phi_r^\mu+\left[\mathcal{L}\delta_\sigma^{\,\nu}-\frac{\partial\mathcal{L}}{\partial(\partial_\nu\phi^\mu)}\partial_\sigma\phi^\mu\right]X_r^\sigma $$ These are conserved in the sens that $\partial_\nu J_r^{\,\nu}=0$, which is true on-shell, i.e. provided equations of motion are verified.

Now I'm trying to apply this result in the case of classical EM field theory. Considere the associated lagrangian : $$ \mathcal{L}_{EM}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+A_\mu j^{\,\mu} $$ where $A_\mu=(\phi,-\textbf{A})$ is the 4-potential and $j^{\,\mu}=(\rho,\textbf{j})$ is the 4-current, describing classical charge sources. $F_{\mu\nu}$ is the EM tensor.

In that case, equations of motion are : $$\partial_\nu F^{\nu\mu}=j^{\,\mu}$$

Considering the global gauge symmetry of the classical EM field theory $A_\mu\rightarrow A_\mu+\epsilon\,\partial_\mu\chi$ where $\chi$ is an arbitrary gauge function, I'd like to apply the Noether's theorem in the case where $X_r^i=0$. Refering to this P.S.E post, then the associated conserved current reads : $$ J^\mu=F^{\mu\nu}\partial_\nu\chi\quad\text{with}\quad\partial_\mu J^\mu=0 $$ Fine, now I have a few questions :

• Is it actually really legit to apply Noether's theorem to global gauge symmetry? I mean, is Noether's theorem still hold even if I have a set $\{\epsilon_r,X_r^i=0,\Phi_r^\mu\}$, i.e. no transformation on coordinates?

Related : physics.stackexchange.com/questions/13870/gauge-symmetry-is-not-a-symmetry.

• In that case, what does the $J^\mu$ quantity physically means? One can compute, using equations of motion and the anti-symmetry of $F^{\mu\nu}$: $$ \partial_\mu J^\mu=\partial_\mu F^{\mu\nu}\,\partial_\nu\chi + F^{\mu\nu}\,\partial_\mu\partial_\nu\chi=j^{\,\nu}\,\partial_\nu\chi+0 $$ I would rather expect something like $$ \partial_\mu J^\mu=\partial_\mu j^{\,\mu}=0,\;\forall\chi $$ In that case I could understand that charge conservation stems directly from global gauge symmetry. There is something I'm clearly missing. Or should I trick by computing $\int\mathrm{d}x\,\partial_\mu J^\mu$ and perform an integration by part?

• I can't see anywhere clearly appearing the fact that the physics of classical EM theory is invariant under Lorentz transformation. I would expect that such invariance would be a really strong symmetry of the system.

In the case of classical field theory, Noether's theorem ensures that for a given action $$S=\int \mathrm{d}^dx\,\mathcal{L}(\phi_\mu,\partial_\nu\phi_\mu,x^i)$$ that stays invariant under the transformation $\{ x^i\rightarrow x^i+\delta x^i \,;\, \phi_\mu\rightarrow \phi_\mu+\delta \phi_\mu\}$, namely a local symmetry, then there exist conserved quantities named currents.

If it exist a set of variables $\{\epsilon_r,X_r^i,\Phi_r^\mu\}$ (where $\epsilon_r\rightarrow 0$) such that $\delta x^i=\epsilon_r\,X_r^i$ and $\delta \phi^\mu=\epsilon_r\,\Phi_r^\mu$, then these currents are the quantities : $$ J_r^{\,\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\nu\phi^\mu)}\Phi_r^\mu+\left[\mathcal{L}\delta_\sigma^{\,\nu}-\frac{\partial\mathcal{L}}{\partial(\partial_\nu\phi^\mu)}\partial_\sigma\phi^\mu\right]X_r^\sigma $$ These are conserved in the sens that $\partial_\nu J_r^{\,\nu}=0$, which is true on-shell, i.e. provided equations of motion are verified.

Now I'm trying to apply this result in the case of classical EM field theory. Considere the associated lagrangian : $$ \mathcal{L}_{EM}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+A_\mu j^{\,\mu} $$ where $A_\mu=(\phi,-\textbf{A})$ is the 4-potential and $j^{\,\mu}=(\rho,\textbf{j})$ is the 4-current, describing classical charge sources. $F_{\mu\nu}$ is the EM tensor.

In that case, equations of motion are : $$\partial_\nu F^{\nu\mu}=j^{\,\mu}$$

Considering the global gauge symmetry of the classical EM field theory $A_\mu\rightarrow A_\mu+\epsilon\,\partial_\mu\chi$ where $\chi$ is an arbitrary gauge function, I'd like to apply the Noether's theorem in the case where $X_r^i=0$. Refering to this P.S.E post, then the associated conserved current reads : $$ J^\mu=F^{\mu\nu}\partial_\nu\chi\quad\text{with}\quad\partial_\mu J^\mu=0 $$ Fine, now I have a few questions :

• Is it actually really legit to apply Noether's theorem to global gauge symmetry? I mean, is Noether's theorem still hold even if I have a set $\{\epsilon_r,X_r^i=0,\Phi_r^\mu\}$, i.e. no transformation on coordinates?

Related : physics.stackexchange.com/questions/13870/gauge-symmetry-is-not-a-symmetry.

• In that case, what does the $J^\mu$ quantity physically means? One can compute, using equations of motion and the anti-symmetry of $F^{\mu\nu}$: $$ \partial_\mu J^\mu=\partial_\mu F^{\mu\nu}\,\partial_\nu\chi + F^{\mu\nu}\,\partial_\mu\partial_\nu\chi=j^{\,\nu}\,\partial_\nu\chi+0 $$ I would rather expect something like $$ \partial_\mu J^\mu=\partial_\mu j^{\,\mu}=0,\;\forall\chi $$ In that case I could understand that charge conservation stems directly from global gauge symmetry. There is something I'm clearly missing. Or should I trick by computing $\int\mathrm{d}x\,\partial_\mu J^\mu$ and perform an integration by part?

• I can't see anywhere clearly appearing the fact that the physics of classical EM theory is invariant under Lorentz transformation. I would expect that such invariance would be a really strong symmetry of the system.