Derivation of law of inertia from Lagrangian method (Landau) I'm reading Landau's Book.
He tries to conclude the law of inertia from the Lagrange equations.
For that, he argues (by nice suppositions about space and time), that the lagrangian must depend only on the velocity. More specifically, only on the square of the velocity.
The point is, since the lagrange equations is:
$$\nabla_x L+(\nabla_{\dot{x}}L)' \equiv 0$$
He gets that $(\nabla_{\dot{x}}L)'\equiv 0$, which implies $\nabla_{\dot{x}}L=constant$ (in time, along the trajectory).
Now here is my problem: he concludes that $\dot{x}$ is constant. How? He doesn't know anything about $L$, besides the "symmetry" properties. For instance, $L\equiv 0 $ satisfies all of the properties required from such $L$, and we would not be able to infer that $\dot{x}$ is constant. In fact, any curve would be extremal with respect to the action.
What is his reasoning, then?
 A: You're correct. 
To find the equations of motion, we have:
\begin{align*}c_i&=\frac{\partial L(v^2)}{\partial v_i}\\
&=L'(v^2) 2 v_i
\end{align*}
so that $L'(v^2) v_i$ is constant for all of time.
Firstly, you could imagine a world in which all paths ${\bf x}(t)$ are valid mechanical paths. Then the Galilean transform of a valid mechanical path is also a valid mechanical path, and so respects Galilean relativity. (Trivially, because all paths are valid) This is what happens when $L'=0$ identically.
Another degenerate case is, for example, $L(y)=\sin(y)$. Then if $v^2=\frac{pi}{2}$, $L'(v^2)=0$ and the path is allowed to change direction at will.
Yet another degenerate case is when $L(v^2)=\sqrt{v^2}$. Then $\frac{\partial L}{\partial v_i}=\frac{v_i}{\sqrt{v^2}}$ and, for example, the one dimensional motion $x(t)=t^2$ satisfies $\frac{2t}{\sqrt{4 t^2}}$ constant (for $t>0$).
So that begs the question: what DOES that passage of Landau prove? Fortunately, though I haven't looked thoroughly at it, the next section (which proves that $L\propto v^2$) does not seem to depend on the assumption that $v={\mathrm{const}}$ for a free particle. 
A: It follows from $L$ being a function $\propto\dot{x}^2$. With this at hand, you are left with two choices:


*

*$\left(\nabla_{\dot{x}}L\right)'\sim\left(\dot{x}\right)'=0$ implies $\dot{x}=\rm const.$

*$L=0$ implies $\dot{x}=0=\rm const.$


Either way, you get that the velocity is constant in time (for this particular, free-particle case).
