Proving that the Lagrangian of a free particle depends only on $|\boldsymbol{v}|^2$

Question

The question is NOT answered by Deriving the Lagrangian for a free particle, as the answers therein assume the quadratic dependence, which is what I am trying to prove. Additionally, while one of the deeply linked answers does attempt to argue that the Lagrangian should depend on the square of the velocity, the argument is difficult / impossible for me to follow, to the extent that my goal is to make a simpler and clearer one, with all the assumptions and steps made very explicit.

I'm going through Landau and Lifshitz's Mechanics, in particular, their derivation of the Lagrangian of a free particle, and an elaboration of their approach in Lawrie's A Unified Grand Tour of Theoretical Physics (3.1, The Action Principle in Galilean Spacetime).

I am having trouble fleshing out a proof that the Lagrangian of a free particle depends only on the magnitude of the velocity $|\boldsymbol v|$, having shown that it cannot depend explicitly on the position nor the time.

The texts do not elaborate on this point, but based on my review of other similar questions, I believe we can proceed as follows. Assuming that $\mathcal L = \mathcal L(\boldsymbol {v})$ is a function of the velocity only, the isotropy of space is implemented as a symmetry of the Lagrangian under the infinitesimal rotation $\boldsymbol v \longrightarrow \boldsymbol v + (\delta\theta)\,\boldsymbol{\hat {n}}\times\boldsymbol v $. Expanding the Lagrangian to first order: $$ \mathcal {L}(\boldsymbol v + (\delta\theta)\,\boldsymbol{\hat {n}}\times\boldsymbol v ) \approx \mathcal{L}(\boldsymbol v) + \frac{\partial \mathcal L}{\partial \boldsymbol v}\cdot \left((\delta\theta)\,\boldsymbol{\hat {n}}\times\boldsymbol v \right) $$ If the Lagrangian doesn't change, then $\frac{\partial \mathcal L}{\partial \boldsymbol v}\cdot \left(\boldsymbol{\hat {n}}\times\boldsymbol v \right) = 0$. Using properties of the triple product, $$ \frac{\partial \mathcal L}{\partial \boldsymbol v}\cdot \left(\boldsymbol{\hat {n}}\times\boldsymbol v\right) = \boldsymbol{\hat {n}}\cdot \left(\boldsymbol v \times \frac{\partial \mathcal L}{\partial \boldsymbol v}\right) = 0 $$ If this is true for all rotations, then $\boldsymbol v \times \frac{\partial \mathcal L}{\partial \boldsymbol v}= \boldsymbol 0$. In general, this requires that $ \frac{\partial \mathcal L}{\partial \boldsymbol v}$ is parallel to $\boldsymbol v$ (I think?), hence $$ \frac{\partial \mathcal L}{\partial \boldsymbol v} = f(\boldsymbol v)\,\boldsymbol v $$ where $f(\boldsymbol v)$ is some unknown scalar function.

The problem: I don't know how to finish this argument and conclude that $\mathcal {L} $ is a function of the magnitude of velocity alone. I am looking for as rigorous an argument as possible - if we need to make additional assumptions to conclude $\mathcal L = \mathcal L(|\boldsymbol {v}|^2)$, that's fine, but I would like to make them explicit.

I noticed that taking one of the components of the above relation, and rearranging things a bit gives: $$ \frac{\partial \mathcal L}{\partial v_i}\frac{1}{ f(\boldsymbol v = (v_1, v_2, v_3))}=v_1 $$

The RHS doesn't depend on $v_2$ or $v_3$, so the LHS should also not depend on them. However, I am not sure if this is helpful. I also know we can't just conclude $f$ should be a constant, as this immediately yields $\mathcal {L} = \text{const} \times \,v^2$ - something which the two texts derive using an additional argument, implementing symmetry under an infinitesimal Galilean boost.

Answer 1

More generally, this is a consequence of invariant theory. E.g.:

1. If $I=f\left(({\bf v}_i)_{i=1,\ldots,n}\right)$ is invariant under $G=SO(3)$, then $I$ is a function $I=g\left(({\bf v}_i\cdot {\bf v}_j)_{i,j=1,\ldots,n}\right)$ of the inner products.

2. If $I=f({\bf v})=g(|{\bf v}|,\theta,\phi)$ is invariant under $G=SO(3)$, then $I=h(|{\bf v}|)$ is a function of the magnitude $|{\bf v}|$ of the vector ${\bf v}$.