I don't have the books in front of me so I only use my memory. According to my professor and if I remember well, Landau and Lifgarbagez, the Lagrangian of an isolated particle can in principle depend on [tex]\vec q[/tex], [tex]\vec \dot q[/tex] and t. Therefore one can write [tex]L(\vec q, \vec \dot q , t)[/tex]. With some non mathematical arguments, one can reach [tex]L(\vec \dot q)[/tex] because it doesn't matter where the particle is, the Lagrangian should be unchanged and it doesn't matter if we study the particle today or in 10⁹ years, the Lagrangian which describes the motion of the particle should be unchanged. Ok fine until here.(adsbygoogle = window.adsbygoogle || []).push({});

Now they go on to say that since the Lagrangian is a real number, Lmustdepends on [tex]\vec \dot q ^2[/tex]. This argument doesn't convince me. I don't think the implication is true. There are a lot of functions that takes [tex]\vec \dot q[/tex] and transform it into a real number, for instance the one that assignates [tex]\vec \dot q[/tex] to [tex]|\vec \dot q|[/tex] or [tex]\vec \dot q ^4[/tex]. I don't see why one function should be more natural than the others. To me it looks like we choose the one that allow us to later find out that [tex]L=T-V[/tex], thus to coincide with notions we already knew before.

Then they do some algebra and find out that [tex]L=k \vec \dot q ^2[/tex] where k is a constant. Then they choose [tex]k=\frac{m}{2}[/tex] to make the Lagrangian coincide with the kinetic energy of a particle in Newtonian's mechanics. But if we didn't know anything from Newtonian's mechanics, we couldn't have chose such an "k" nor such a function for the Lagrangian, right? I mean, all in all Lagrangian mechanics as we use it daily is not totally independent of Newtonian's mechanics, am I right?

I don't remember where I had read that the Lagrangian mechanics is totally independent of Newtonian's one... What do you think?

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# Lagrangian of an isolated particle, independant from Newtonian Mechanics?

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