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

Question

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?

Qmechanic · Accepted Answer

The speed $vequiv |vec{bf v}|geq 0$ 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.
For why the Lagrangian for a free particle is such function $f(v^2)$, see this related Phys.SE post.

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

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

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