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?


I leave it to OP and the reader to prove that OP's two Lagrangians are indeed classically inequivalent, but let me make the following general remarks:

  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.

  • $\begingroup$ 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... $\endgroup$ – VoB Apr 23 at 18:08
  • $\begingroup$ @VoB It sounds like you mean something different than Qmechanic does when you say "equivalent." According to this answer, two Lagrangians are defined to be equivalent whenever they give the same E-L equations. $\endgroup$ – probably_someone Apr 23 at 20:49
  • 1
    $\begingroup$ 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 $\endgroup$ – VoB Apr 23 at 20:53

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