Re: [Phys-l] Landau on Lagrangian

[Stefan Jeglinski <jeglin@4pi.com>]

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

Indeed.


> 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