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So I was reading this: Invariance of Lagrange on addition of total time derivative of a function of coordiantes and time and while the answers for the first question are good, nobody gave much attention to the second one. In fact, people only said that it can be proved without giving any proof or any.

So, if I have a Lagrangian and ADD an arbitrary function of $\dot{q}$, $q$ and $t$ in such a way that the equations of motion are the same, does this extra function MUST be a total time derivative?

EDIT Ok, I changed my question a little bit:

Question: If I have a function that obeys the Euler-Lagrange equation off-shell, this implies that my function is a time derivative? (This was used in Qmechanic's answer of this other question: Deriving the Lagrangian for a free particle, equation 14.)

Also, why people only talk about things that change the Lagrangian only by a total derivative? If this is not always the case that keeps the equation of motion the same, so why is it so important? And why in the two questions I posted about the same statement on Landau&Lifshitz's mechanics book only consider this kind of change in the Lagrangian?

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I) OP essentially asked (v1):

If two Lagrangian densities ${\cal L}$ and $\tilde{\cal L}$ have the same eqs. of motions, must they necessarily differ by a total divergence?

Answer: No, not necessarily, one e.g. can always multiply a Lagrangian density ${\cal L}$ with a constant factor $\tilde{\cal L}=\lambda {\cal L}$ different from one $\lambda\neq 1$ without altering the EL equations, but the difference $$\tilde{\cal L}-{\cal L}=(\lambda-1) {\cal L} \tag{A}$$ is not a total divergence if ${\cal L}$ is not.

II) OP essentially asked (v4):

If EL equations are trivially satisfied for all field configurations, is the Lagrangian density ${\cal L}$ necessarily a total divergence?

Answer: Yes, modulo topological obstructions in field configuration space. This follows from an algebraic Poincare lemma of the so-called bi-variational complex, see e.g. Ref. 1.

We should mention that an elementary follow-your-nose-type proof exists for Lagrangians of the form $L(q^i,\dot{q}^j,t)$ without higher-order derivatives, see e.g. Ref. 2. We stress that the proof-technique of Ref. 2 does not work in the presence of higher-order derivatives or in the case of field theory. [Also Ref. 2 seems to overlook the counterexample in eq. (A).]

III)

If two Lagrangian densities ${\cal L}$ and $\tilde{\cal L}$ have the same eqs. of motions, does there exist a constant factor $\lambda$ such that $\tilde{\cal L}-\lambda{\cal L}$ necessarily differ by a total divergence?

Answer: No, not necessarily. There are topological counterexamples.

References:

  1. G. Barnich, F. Brandt and M. Henneaux, Local BRST cohomology in gauge theories, Phys. Rep. 338 (2000) 439, arXiv:hep-th/0002245.

  2. J.V. Jose and E.J. Saletan, Classical Dynamics: A Contemporary Approach, 1998; Section 2.2.2, p. 67.

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  • $\begingroup$ I think OP is asking about classical mechanics (so $\mathcal{L}\to L$, divergence $\to$ time derivative), but good observation otherwise. $\endgroup$ – Danu Aug 21 '14 at 8:23
  • $\begingroup$ Ok, I think I have a better statement. Suppose that the Euler Lagrange equation holds off-shell for a given function of q',q and t, this implies that the function is a total time derivative? $\endgroup$ – Stephen Dedalus Aug 21 '14 at 8:30
  • $\begingroup$ Also, in the link I posted the question involves a passage from Landau, the famous one when he derives the lagrangian of free particle using the fact that a galilean boost cannot change the equation of motion. So, if after a boost my lagrangian become a scalar multiple of my old one, this of course will does not change my equation of motion. So why is this option discarded? $\endgroup$ – Stephen Dedalus Aug 21 '14 at 8:41
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    $\begingroup$ @Danu: Yeah, I couldn't resist generalizing to field theory. $\endgroup$ – Qmechanic Aug 21 '14 at 9:19
  • $\begingroup$ Wow, great answer! The first paper seems to cover a lot of things that I'm not familiar with (i took a basic course on field theory, but not on this level), however It will probably be useful in the future and to other persons that may read this thread. Saletan proof is similar to the one I was thinking, however I was just too confuse with some details that I wasn't able to finish the proof. Thank you a lot. $\endgroup$ – Stephen Dedalus Aug 21 '14 at 12:26

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