# invariance of lagrangian in Noether's theorem

Noether's theorem needs the lagrangian to be invariant.

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 $\alpha L$ or $L + \frac{df}{dt}$ instead of $L$?

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This is not a "transformation", it is two Lagrangians that give the same equation. A transformation is a change of the dynamical variables. –  Ron Maimon May 15 '12 at 16:26

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.
Or in other words, the quantity which is conserved for quasisysmmetries is something trivial like $0$. –  Fabian Jun 20 '11 at 12:39