# Invariance of Lagrangian in Noether's theorem

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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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
ok, I understand it a little better now, the on-shell variation is always a total derivative because the integrand is zero, and there are only boundary terms left. But for the sake of finding $f$ that goes into Noether currents, we are only concerned with the off-shell variations? – diffeomorphism Jun 6 '15 at 16:43