In Sec. 2 of Landau & Lifshitz's Mechanics, they claim that

Thus the Lagrangian is defined only to within an additive total time derivative of any function of coordinates and time.

by proving that $L$ and $$L'=L+\frac{d}{dt}(f(q,t))$$ are equivalent as far as the least action principle is concerned.

However, is this condition necessary as well as sufficient? That is, can we deduce that $$L'=L+\frac{d}{dt}(f(q,t))$$ for some $f$ from their equivalence when taking the first variation of the action?

This is important because in Sec. 4, Landau tries to find the form of $L$ by requiring $$L'=L+\frac{d}{dt}(f(q,t)).$$ If the condition is not necessary, this shouldn't work.


No, in general the Lagrangians that obtain a given set of equations of motion are not of the form $L+\partial_t f$, or even $aL+\partial_f$. I asked and ultimately answered this question about an interesting counterexample. But (from Landau's likely context) all that matters is whether a transformation behaves as $\delta L =\partial_t f$ and hence is action-preserving.


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