# Show two Lagrangians are equivalent

I need to show that these two Lagrangians are equivalent:

\begin{align} L(\dot{x},\dot{y},x,y)&=\dot x^2+\dot y + x^2-y ,\\ \tilde{L}(\dot x, \dot y, x, y)&=\dot x^2+\dot y -2y^3. \end{align}

It is the case iff they differ for a total derivation like $$\frac{dF}{dt}(x,y)$$.

In this case, the difference is $$x^2+y^3$$ and I can't imagine such an $$F(x,y)$$ whose total derivative is the one above. How should I move?

I tried with the following $$F(x,y)=\frac{x^3}{3\dot x} + \frac{y^4}{4\dot y}$$, but it shouldn't have the dotted terms.

Actually, I just proved they don't give rise to the same Lagrange equations, so I can conclude they're not equivalent, right?

1. Two Lagrangians $$L_1$$ and $$L_2$$ are classical equivalent iff they give the same Euler-Lagrange (EL) equations.
2. A sufficient condition is that the difference $$L_2-L_1=\frac{dF}{dt}$$ is a total derivative, but it should be stressed that it is not a necessary condition, cf. e.g. my Phys.SE answer here.
• Thanks, but I'm a bit confused with the \text{iff} in 1), because I know that if I take as Lagrangians $L$ and $\bar{L}=\alpha L$, they give rise to same Lagrange equations, but they're not equivalent... – VoB Apr 23 at 18:08
• Right. I say they're equivalent iff they differ for a total derivation. With this definition, two equivalent Lagrangians give rise to the same Lagrange-Equation. In my exercise, I couldn't find such a function $F$, so I checked that the don't give the same Lagrange-equations, in order to conclude the're not equivalent – VoB Apr 23 at 20:53