Why does Lagrangian of free particle depend on the square of the velocity ?

Why does Lagrangian of free particle depend on the square of the velocity ? For example, $L(v^4)$ also doesn't depend on direction of $v$.

-
Your question is answered here: physics.stackexchange.com/q/23098 –  nervxxx May 5 '13 at 13:16
Also related: physics.stackexchange.com/q/93902/2451 –  Qmechanic Mar 10 '14 at 8:11

The Lagrangian should not only be independent of the direction of $\vec{v}$ but it should also change correctly under a Galilean transformation. For instance, if $K$ and $K'$ are two frames of reference with a relative velocity $\vec{V}$ then the two Lagrangians $L$ and $L'$ should differ only by a total time derivative. If $L$ is a function of fourth power of velocity then $v'^4 = (\vec{v} + \vec{V})^4 = v^4 + \{V^4 + (\dot{\vec{x}}\cdot\vec{V})^2 + 2v^2V^2 + 2(\dot{\vec{x}}\cdot\vec{V})v^2 + 2(\dot{\vec{x}}\cdot\vec{V})V^2\}$. The term in the curly brackets cannot be expressed as a total time derivative. If $L$ was $v^2$, it would have been possible.