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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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Here I would like to mention the notion of quasi-symmetry. In general, if the Lagrangian (resp. Lagrangian density resp. action) is only invariant up to a total time derivative (resp. space-time divergence resp. boundary term) when performing a certain off-shell$^1$ variation, one speaks of a quasi-symmetry, see e.g. Ref. 1.

Noether's first Theorem does also hold for quasi-symmetries. For examples of non-trivial conservation laws associated with quasi-symmetries, see examples 1, 2 & 3 in the Wikipedia article for Noether's theorem.

References:

  1. J.V. Jose & E.J. Saletan, Classical Dynamics: A Contemporary Approach, 1998; p. 565.

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

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  • $\begingroup$ 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? $\endgroup$ – diffeomorphism Jun 6 '15 at 16:43
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    $\begingroup$ Noether's theorem only applies to off-shell quasi-symmetries, not to on-shell quasi-symmetries. $\endgroup$ – Qmechanic Jun 6 '15 at 17:15
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I would just like to say that the standard Noether's theorem very much applies to the case that $\delta L = \dot f$. For example, time translation is of this form. We can see this by performing the Noether procedure for a tiny time translation.

$$ q(t) \to q(t + \varepsilon) \approx q(t) + \varepsilon \dot q(t)\\ \dot q(t) \to \dot q(t) + \varepsilon \ddot q(t) $$ This sends $$ L \to L + \varepsilon \dot L $$ as promised. If we then make $\varepsilon$ into a tiny time dependent function $\varepsilon(t)$, we now have

$$ q(t) \to q(t) + \varepsilon(t) \dot q(t) \\ \dot q(t) \to \dot q(t) + \varepsilon(t) \ddot q(t) + \dot \varepsilon(t) \dot q(t). $$

After a bit of fussing around with the chain rule of multi variable calculus, we find that this sends

$$ L \to L + \varepsilon \dot L + \dot \varepsilon \dot q \frac{\partial L}{\partial \dot q} $$

We then use the fact that $\delta S = 0$ on solutions to the equations of motion, and after an integration by parts find that

$$ \frac{d}{dt} ( p \dot q - L ) = 0 $$

on solutions to the equations of motion. This is just the conservation of energy

TLDR symmetries that change $L$ by a total derivative are just incorporated into Noether's theorem without having to do anything extra. Time translations are an example of this.

However, $\delta L \propto L$ is a little more exotic. Performing the Noether procedure on the Lagrangian of a free particle $(L = m \dot q^2 / 2)$ which does have a scaling symmetry $q \to (1+\varepsilon) q$, I find that the "conservation law" (if you want to call it that) is just $m q\ddot q = 0$, which is trivially $0$ anyway on the equations of motion.

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