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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?

Qmechanic
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1 Answers1

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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.

Qmechanic
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