I'm having trouble deriving the kinetic energy from the Lagrange equations. For reference, I'm following Landau and Lifshitz book, "Mechanics," which can be found for free at Archive. In any case, I'm comfortable with the principle of minimal action, which leads to the Lagrange equations:
$$ L^\prime_{q}(q,\dot{q},t)-\frac{d}{dt} L^\prime_{\dot{q}}(q,\dot{q},t)=0 $$
Simply, they are the first-order optimality conditions to the optimization problem $\min\limits_q \int\limits_a^b L(q,\dot{q},t) dt$. Next, by assuming time and space are homogenous, I can see how $L$ must be independent of both $t$ and $q$ and only be a function of $\dot{q}$. This leads to
$$ \frac{d}{dt} L^\prime_{\dot{q}}(q,\dot{q},t)=0. $$
Finally, by assuming space is isotropic, I can see how $L$ can not be a function of the direction of $\dot{q}$. This is all covered on page 5 of Landau and Lifshitz book that I referred to above. However, after this, they argue that $L$ must therefore be a function of $\dot{q}^2$. This is where I'm confused and have the following questions:
Why does isotropy imply that L must be a function of $\dot{q}^2$?
Why doesn't isotropy instead imply that $L$ must be a function of a scalar quantity of $\dot{q}$? A quantity like $\cos(\dot{q}^2)$ doesn't depend on direction either.
If isotropy does indeed imply that $L$ is a function of the magnitude of $\dot{q}$, then why do we use, essentially, $\|\dot{q}\|^2$? Isn't the actual magnitude of $\dot{q}$ the quantity $\|\dot{q}\|=\sqrt{\dot{q}^2}$?
Anyway, the rest of the derivation of kinetic energy sits on page 7 of the above book. Their discussion really lies on the idea that $L$ must be a function of $\dot{q}^2$, which is what I'm trying to clarify above.
Edit 1
Alright, so it looks like this question has been asked in various forms, so before it closed out it'd help me if I could clarify some points in those answers that apply to here. Yes, the short answer is that the Lagrangian needs to be invariant to the Galilean transform.
From what I can see, the Lagrangian is invariant to scalar multiplication and functions of the form $\frac{d}{dt} f(q,t)$. This last one is due to
$$ \arg\min\limits_q \int\limits_a^b L(q,\dot{q},t) + \frac{d}{dt} f(q,t)dt $$ $$ =\arg\min\limits_q \int\limits_a^b L(q,\dot{q},t)dt+f(q^{(b)},t_b)-f(q^{(a)},t_a) $$ $$ =\arg\min\limits_q \int\limits_a^b L(q,\dot{q},t) dt $$
since constants don't change the location of the minima. This is discussed around equation (2.8) in Landau. Anyway, if the Lagrangian is a scalar function in terms of $\dot{q}$, then we have that $L(q,\dot{q},t)=\phi(\dot{q})$ for some $\phi:\mathbb{R}^3\rightarrow\mathbb{R}$. Further, Galilean invariance states that the Lagrangian should be invariant to transformation $\hat{q} = q + Vt$ or $\dot{\hat{q}} = \dot{q} + V$. Combining all of this, we have that
$$ \phi(\dot{\hat{q}}) = \phi(\dot{q}+V) = \phi(\dot{q}) + \phi^\prime(\dot{q})V + o(\|V\|) $$ Hence, we need that $\phi^\prime(\dot{q})V = \frac{d}{dt} f(q,t)$ for some $\phi$ and some $f$ in order for the Lagrangian to be invariant to the Galilean transformation. Certainly, $\phi(\dot{q})=\dot{q}^2$ and $f(q,t) = 2\langle q,V\rangle$ does this. How do we guarantee that this is the only solution?
