Equivalent Lagrangians

Given a Lagrangian $L$ for a specific system we can obtain a new lagrangian $L^{\prime }$ that leads to the same Euler-Lagrange equations that $L$ if

• $L^{\prime }=L+cte$
• $L^{\prime }=L\cdot cte$
• $L^{\prime }=L+\frac{dF}{dt}$

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 $\text{Div}$ is the total divergence.

Particular case

If two Lagrangians, $L=L(x,u,u_1)$ and $\tilde{L}=\tilde{L}(u,u_1)$, give rise to the same Euler-Lagrange equations, then they must differ by a total derivative with respect to $x$. Specifically, there exists a function $f(x,u)$ such that:

This is a consequence of the fact that the Euler-Lagrange equations remain unchanged when the Lagrangian is modified by a total derivative term. Therefore, the two Lagrangians are related by what is known as a "gauge transformation" in this context.