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I seem to be misunderstanding the definition of gauge or symmetry transformations. Given a lagrangian, a gauge transformation is defined to be a continuous transformation such that $$ \delta \mathcal{L} = \partial_\mu A^ \mu$$ where $A^{\mu}$, I assume, is zero at the boundaries. Now, in deriving the conserved current, we assume that the field $\phi$ is "on-shell". This means that $\phi$ is a stationary solution. So any infinitesimal variation $\phi' = \phi + \epsilon \psi$ will result in a zero change in the lagrangian. But doesn't that mean that every transformation is a symmetry transformation of the lagrangian?

However, I think a symmetry transformation doesn't need to be infinitesimal but all of the books I looked at, derive the conserved current using an infintesimal approximation to the translation. But doesn't this method of restricting it to infitesimal transformations make the method useless since any infinitesimal translation doesn't change the action?

Qmechanic
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Chandrahas
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    The transformation has to be a symmetry of the lagrangian, regardless of if the equations of motion are satisfied. The conserved current is obtained after by restricting to the scenario where the fields are 'on-shell', which classically what we expect the fields to obey. – ColourConfined Feb 04 '22 at 06:01
  • Possible duplicates: https://physics.stackexchange.com/q/438403/2451 and links therein. – Qmechanic Feb 04 '22 at 06:34

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