For freely moving particle, It's said $L$ can't depend on velocity vector, but magnitude. 

**Question 1:** I'm looking for the contra-argument. Let's say it depends on velocity vector. Then, how would Lagrangian be written in terms of velocity vector and why would it yield the wrong solution for freely moving particle? (no need for bringing potential in this).

Looking for the math proof. I know Euler-lagrange. If it depends on velocity vector, $L = \frac{1}{2}m(v_i + v_j)$ and this yields $\frac{d}{dt}(\frac{1}{2}m) = 0$. Would this be correct approach to prove what I'm asking ? I'm trying to get the idea why Landau makes it depend on $v^2$. (I'm not asking why it doesn't depend $v^4$). My main question is to rigorously show why it can't depend on vector, and if it did, what would it break for freely moving particle ? It mentions isotrophy of space, but want to see in math proof how it's wrong.

**Question 2:** From the @Qmechanic [answer](https://physics.stackexchange.com/a/23123/366606), he notes that:

$$ \vec{0}~=~\frac{\partial L}{\partial \vec{q}}
~\approx~\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \vec{v}}
~=~\frac{\mathrm d }{\mathrm dt} \left(2\vec{v}~\ell^{\prime}\right)
~=~2\vec{a}~\ell^{\prime}+4\vec{v}~(\vec{a}\cdot\vec{v}) \ell^{\prime\prime}.\tag{2}$$

$\frac{\partial L}{\partial \vec{v}} = 
2\vec v \frac{\partial L}{\partial v^2}$ <br>
$\frac{d}{dt}(2\vec v \frac{\partial L}{\partial v^2})$. 

He basically says that $\ell'$ is the same as $\frac{\partial L}{\partial v^2}$. I wonder how $\ell' = 2(\vec a \vec v)\ell''$. I mean if L only contains $v^2$, $\ell'' =0$ definitely. So what's the point of having it ? are you assuming that $L$ might be containing $v^4$ ? if so, why are you saying it's a function of $v^4$ ?  and if you still do, then the formula breaks for $v^6$(though works for $v^4$) - $\ell'$ won't be equal to $2(\vec a \vec v)\ell''$.