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Another (different) objection:
Quoting Landau, Stefan wrote:
. . . L must also be independent of the direction of v, and is
therefore a function only of its magnitude, i.e. v^2."
Ergo: L = L(v^2) .
Why not :
Ergo: L=L( | v | ) ?
Bob Sciamanda
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
To be sure!
I merely wanted to point out that Landau's treatment (even if
accepted as valid) does not demonstrate that the translational
kinetic energy must be proportional to the SQUARE of the speed.
The opposite is easily (but incorrectly) inferred from the
presentation.
Bob Sciamanda