Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\delta L=0$?

share|cite|improve this question

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.

share|cite|improve this answer
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. – 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
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
Noether's theorem only applies to off-shell quasi-symmetries, not to on-shell quasi-symmetries. – Qmechanic Jun 6 '15 at 17:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.