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In whatever text/review I happen to come across (like for example From Noether’s Theorem to Bremsstrahlung: A pedagogical introduction to Large gauge transformations and Classical soft theorems, written by Noah Miller), the explanation for the Noether's theorem(s) starts in the following way:

If you have a symmetry transformation $\delta\phi_i=\epsilon \delta_{\text{sym}}\phi_i$ where $\epsilon$ is an infinitesimal constant, then the action will change by the total derivative of some vector $J^{\mu}$.

My question is, why does the fact that a transformation is a symmetry transformation imply that the action will change by a total derivative of some vector $J^{\mu}$? I know that this is the definition of "the symmetry transformation", but what is the motivation behind this definition? Does the motivation have to do with Taylor expanding the action?

I know I have asked this earlier, but I didn't get an answer.

P.S.: I have this question because I was under the impression that a symmetry is a transformation that does not alter the classical equations of motion. But apparently, this is not the case.

Qmechanic
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schris38
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1 Answers1

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Apart from the fact that Noether's first theorem not only works for strict symmetries but also for quasisymmetries of an action $S_V$ (say, in some spacetime region $V$), one intuitive reason to only require that $S_V$ respects the symmetry modulo boundary terms is that if the model has a (horizontal) spacetime symmetry vector field $X=X^{\mu}\partial_{\mu}$, there generically will be a flux in or out of $V$ at the boundary $\partial V$ that breaks the strict symmetry.

Similarly, symmetry transformations are not required to respect boundary conditions, cf. e.g. this Phys.SE post.

Qmechanic
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