In Landau & Lifshitz's derivation of the Lagrangian of a free particle in a galilean frame of reference one finds the following argument: the equations of motion in two galilean frames must be identical; hence the respective Lagrangians must differ by the total derivative of a function of the generalized position, and time. This is essentially the converse of what the authors point to as justification, namely that adding such a term to the Lagrangian leaves the equations unchanged, and I don't really get why it holds. The only relevant answer I found on stackexchange is Qmechanics' take in Deriving the Lagrangian for a free particle, but i must admit it doesn't quite satisfy me.

Edit: I'm asking why modifications of the Lagrangian that don't change the EL equations are necessarily the addition of a total derivative (and multiplication by a scalar), like L&L claims.

  • $\begingroup$ I might be wrong, but using the Euler-Lagrange equations in this context is rather circular, no? What we should be looking to prove is that something that doesn't change the equations is a total derivative, etc., not that something that doesn't change the equations, well, doesn't change the equations. Also the answer I linked gives the final form of the Lagrangian, while Landau uses another argument after the one I mentionned. $\endgroup$
    – user115153
    Commented Apr 30, 2018 at 15:49
  • $\begingroup$ Note that my use of EL eqs. is not circular logic. It is a trick to identify terms which are total derivatives. $\endgroup$
    – Qmechanic
    Commented Apr 30, 2018 at 16:18
  • $\begingroup$ I must admit this is somewhat fuzzy, but I reckon you're not fixing Landau's method but rather yours is different entirely; he says that the Lagrangians must differ by a certain kind of function and then establishes that dL/d(v^2) is a constant, but you say that the difference must obey Euler-Lagrange and then you find a formula for it; or am I misunderstanding? $\endgroup$
    – user115153
    Commented Apr 30, 2018 at 16:31
  • 1
    $\begingroup$ possible duplicate: physics.stackexchange.com/q/368801/84967 $\endgroup$ Commented May 1, 2018 at 13:12
  • 1
    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/131925/2451 $\endgroup$
    – Qmechanic
    Commented May 1, 2018 at 15:25

1 Answer 1


L&L's logic is as follows:

  1. L&L demands$^1$ that an (infinitesimal) Galilean transformation should be a quasisymmetry (QS) of the sought-for action functional $S$.

  2. We furthermore demand that

    • the action functional is local, and
    • the position space is contractible.
  3. From this Phys.SE post, we then deduce that an (infinitesimal) Galilean transformation is in fact a QS of the Lagrangian $L$ itself. This means by definition that the change $\Delta L$ in the Lagrangian is a total time derivative, as OP wanted to show.


$^1$ This is reasonable since Newtonian mechanics has Galilean symmetry. However, there is potentially a loophole since a symmetry of EOM does not have to be a QS of the action, cf. e.g. this Phys.SE post. Then the bigger question becomes:

How much can we change the action without affecting the EOM?

That's a good question, which was essentially also asked in this Phys.SE post.


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