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Re: [Phys-l] Landau on Lagrangian

Regarding Bob S's objection:

. . . 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 | ) ?

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).

David Bowman