How does isotropy of free space imply $L(v^2)$ for a free particle? [duplicate]

From Mechanics; Landau and Lifshitz, it's stated on page 5:

Since space is isotropic, the Lagrangian must also be indpendent of the direction of $\mathbf{v}$, and is therfore a function only of its magnitudie, i.e. of $\mathbf{v} \cdot \mathbf{v}=v^2$: $L = L(v^2)$

How is he able to exlude $L = L(| \mathbf{v}|)$?

• I don't have the book on me, but does he need to exclude this case? As in, any function of $v^2$ is equally a function of $|\mathbf{v}|$ and vice versa. In other words, are you implying that he really needs to show $L \propto v^2$ (modulo a constant) for some future result? – user10851 Mar 10 '14 at 0:27
• Related: physics.stackexchange.com/q/23098/2451 , physics.stackexchange.com/q/93902/2451 and links therein. – Qmechanic Mar 10 '14 at 7:54

for any function $f$, we have $f(|v|) = f(\sqrt{v^{2}}) = g(v^{2})$, where $g = f\circ \sqrt{}$, so we lose no generality by assuming that the function is a function of the square. Also, it generalizes more nicely to vector spaces, where $v^{2} ={\vec v}\cdot {\vec v}$ is defined, but it is not necessarily the case that the absolute value function is defined.
$\sqrt{x}$ is not an analytic function -- its first derivatives are not continuous at 0. We often require that the quantities we deal with are analytic for various physical, aesthetic or practical reasons. In this case we know that one thing we would like to do with our Lagrangian is get equations of motion from it, and it would be nice for this purpose if the kinetic energy term had a continuous first derivative.