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Often in textbooks Noether's theorem is stated with the assumption that the Lagrangian needs to be invariant $\delta L=0$.

However, given a lagrangian $L$, we know that the Lagrangians $\alpha L$ (where $\alpha$ is any constant) and $L + \frac{df}{dt}$ (where $f$ is any function) lead to the same equations of motion.

Can we then consider that the Lagrangian is invariant under a transformation if we find $\delta L=\alpha L$ or $\delta L=\frac{df}{dt}$ instead of $\delta L=0$?

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1 Answer 1

Here I would like to mention the notion of quasi-symmetry. In general, if the Lagrangian (resp. Lagrangian density) is only invariant up to a total time derivative (resp. space-time divergence) when performing a certain off-shell$^1$ variation, one speaks of a quasi-symmetry, see, e.g., J.V. Jose and E.J. Saletan, "Classical Dynamics: A Contemporary Approach", p. 565. Noether's first Theorem does also hold for quasi-symmetries.


$^1$ Here the word off-shell means that the Lagrangian eqs. of motion are not assumed to hold under the specific variation. If we assume the Lagrangian eqs. of motion to hold, any variation of the Lagrangian is trivially a total derivative.

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Or in other words, the quantity which is conserved for quasisysmmetries is something trivial like $0$. –  Fabian Jun 20 '11 at 12:39
For examples of non-trivial conservation laws associated with quasi-symmetries, see examples 1, 2 & 3 in the wikipedia article for Noether's theorem en.wikipedia.org/wiki/… –  Qmechanic Jul 9 '11 at 16:07
For the slightly more general notion of a quasi-symmetry of an action (as opposed to the Lagrangian density), see e.g. this Phys.SE post. –  Qmechanic Jan 20 '13 at 22:32

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