# Why is the Lagrangian specifically a function of $v^2$?

I've been reading "L. D. Landau, E.M. Lifshitz - Mechanics (Volume 1)" and he justifies the fact that the Lagrangian is a function of $$v^2$$ with the fact that space is isotropic - that is, direction does not matter. My question is: could we choose $$L$$ to be a function of $$|v|$$ or $$v^4$$? I know that, if we choose $$v^2$$ and assume $$v_0<<1$$, (note: in the book, $$v_0$$ is a factor that relates one inertial frame of reference to another) we can nicely cancel out $$v_0^2$$. However, is this the only reason? Could we achieve the same results choosing other "v"s?

## 1 Answer

1. The speed $$v\equiv |\vec{\bf v}|\geq 0$$ is by definition the magnitude of the velocity $$\vec{\bf v}$$. Any function of speed $$v$$, or say $$v^4$$, can be easily rewritten as a function of $$v^2 \equiv |\vec{\bf v}|^2$$, or vice-versa. The latter form $$f(v^2)$$ is preferable when one tries to partial differentiate wrt. a velocity component in order to avoid square roots.

2. For why the Lagrangian for a free particle is such function $$f(v^2)$$, see this related Phys.SE post.