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Re: [Phys-l] Landau on Lagrangian

> Why not :
> Ergo: L=L( | v | ) ?


No generality is lost here. Note that |v| is a function of v^2. In particular |v| = sqrt(v^2). So any real function of |v| is also a real function of v^2 as well (because a function of a function is a function).

OK, but the spirit of the question survives, unless I misunderstand you. In the "first principles" approach that's been discussed in this thread, the argument does not appear to preclude one from supposing the free classical particle Lagrangian to be (1/2)m|v|. This would seem justified (Occam's Razor?), other than the rather glaring inability to get N2, which takes us back ultimately to the only argument left: "in our universe, this apparently just doesn't seem to be the case."

Stefan Jeglinski