Equivalent Lagrangians

Given a Lagrangian $L$ for a specific system we can obtain a new lagrangian $L'$ that leads to the same Euler-Lagrange equations that $L$ if

• $L'=L+cte$

• $L'=L\cdot cte$

• $L'=L+\frac{dF}{dt}$

This is an exercise from Hand and Finch Analytical Mechanics, page 28. The third item is proven by the definition of action directly. In problem 7 of the same page it is shown that it has to do with the equivalence of inertial frames.

This is related with @olver86 page 249: two Lagrangians $L$ and $\tilde{L}$ have the same Euler-Lagrange expressions if and only if the differ by a divergence

Here $\mbox{Div}$ is the total divergence.

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es