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Noether's theorem states: any differentiable symmetry of the action of a physical system has a corresponding conservation law. Is this statement invertible? I mean, if a conservation law exists, this implies there is a differentiable symmetry in the action?

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marked as duplicate by jinawee, John Rennie, JamalS, Qmechanic Jul 4 '14 at 9:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Emmy Noether proved both the theorem and its converse. Look for the book "The Noether Theorems" for a precise and discussed formulation of her statements, as well as a translation of the original paper. It seems there is a link to the pdf in the princeton math website (I don't know about copyright issues, however).

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  • $\begingroup$ An English translation of Noether's original article is available for free on the arXiv. $\endgroup$ – Qmechanic Jul 4 '14 at 9:23
  • $\begingroup$ Comment to the answer (v1) (Oversimplifying, to convey the main point in a single comment; for further details, see e.g. this Phys.SE post): Noether's theorem, in its common (as opposed to original) formulation, states essentially that a symmetry of the action leads to an on-shell conservation law. The relevant converse question is: Does an on-shell conservation law lead to a symmetry? Noether only proves that an off-shell conservation law induces a symmetry. $\endgroup$ – Qmechanic Jul 4 '14 at 12:33
  • $\begingroup$ In my understanding Olver's work is devoted to proving a one-to-one correspondence between (equivalence classes) of conservation laws and variational symmetries, thus refining Nother's theorem. In the work of Noether that to a conservation law (in general a divergence) is associated a(t least one) group of transforamtions that leaves the action invariant is already proved. If we want this to be unique up to trivial transformations we should impose the additional conditions as discussed by you in the related answer. But I'm not an expert, I may have misunderstood something ;-) $\endgroup$ – yuggib Jul 4 '14 at 13:32
  • $\begingroup$ Right, Olver does a more refined analysis of the converse Noether's theorem. $\endgroup$ – Qmechanic Jul 4 '14 at 13:57

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