I have a question about "independent variable" in physics, and especially variable in Lagrangian or Density Function. I read several questions about it in this forum and although I have the feeling I understand the answer I still doubt.
Introduction
Ok so I read, and maybe I understand and maybe I agree with the fact that in Lagrangian $L(q_i(t),\dot{q_i}(t),t)$ the variable $q_i(t)$ and $\dot{q_i}(t)$ are independant and are implicitly linked with $t$. When we impose the condition $\delta S = 0$ with $S$ the "Action", we link $q_i(t)$ and $\dot{q_i}(t)$ during the resolution of the least-action principle when we say $\delta (\dot{q_i}(t)) = \dot{(\delta q_i(t))}$ and later we can write what we we assume "since birth":
$$ \dfrac{\mathrm{d}q_i(t)}{\mathrm{d}t} = \dot{q_i}(t). $$
Because a priori, there is no link between all the possible $q_i(t)$ and all the possible $\dot{q_i}(t)$ before that.
Don't hesitate to correct me at this point if I said horrible things.
Vlasov Equation
So we move to the density function (in space phase) and I try to use the same argument in order to find the Vlasov equation. $N_p$ The
$$ N_p \propto \iint_{-\infty}^{+\infty}f(\vec{r},\vec{p},t) \, \mathrm{d}\vec{p} \, \mathrm{d}\vec{r} = \int_V f(\vec{r},\vec{p},t) \, \mathrm{d}V $$
With $V$ the elementary volume in the phase space. If we want conserve this number in time we can say if we considere a flux in the phase space :
$$ \dfrac{\mathrm{d}N_p}{\mathrm{d}t} = \int_V \partial{t}f \, \mathrm{d}V + \oint_S \vec{J_{\varphi}} \, \cdot \, \mathrm{d}S = 0 $$
with $\vec{J_{\varphi}} = f\vec{V_{\varphi}}$ and $\vec{V_{\varphi}} = (\dot{r_1},\dot{r_2},\dot{r_3},\dot{p_1},\dot{p_2},\dot{p_3})$ the speed vector of for each variable of the phase space. Ok so with the help of Green-Ostrogradsky we can write :
$$ \int_V \partial{t}f \, \mathrm{d}V + \oint_S \vec{J_{\varphi}} \, \cdot \, \mathrm{d}S = \int_V \partial{t}f + \nabla_{r_1,r_2,r_3,p_1,p_2,p_3} \cdot (f\vec{V_{\varphi}}) \, \mathrm{d}V $$
With vector analysis we have:
$$ \nabla_{r_1,r_2,r_3,p_1,p_2,p_3} \, \cdot \, (f\vec{V_{\varphi}}) = f \nabla_{r_1,r_2,r_3,p_1,p_2,p_3} \, \cdot \, \vec{V_{\varphi}} + \vec{V_{\varphi}} \, \cdot \, \nabla_{r_1,r_2,r_3,p_1,p_2,p_3} f $$
The difficult part for me is how we can justify properly that $p_i = \dot{r_i}$ is independent of $r_i$ and moreover $a_i = \dot{p_i}$ is independent of $p_i$.
I would say "Ok for $x_i$ and $p_i$ it's the same argument for Lagrangian part. There is a priori no link beetween theses variables moreover when the partial derivative on $x$ and $p$ impose to fix the time (no trajectories)".
For the case $p_i$ and $a_i$, in general physics give a explicite form for $a_i=F$. In classical electromagnetism:
$$ F \propto E + \dfrac{p}{m\gamma} \times B. $$
What I feel: We can say $\partial_{p_i}a_i$ is zero because of the vectorial product. there is no $p_i$ on the $i$ axis.
Do you think I'm right? Is there more arguments? Can you help me to have a more elegant and right point of view?
EDIT: Hamiltonian Point of View
In fact I read that if we use the "hamiltonian formalism" we found that the 6-current in the phase space admit 0 divergence: $\nabla \cdot \vec{V_{\varphi}} = 0$
We note:
$$ \nabla \cdot \vec{V_{\varphi}} = \dfrac{\partial \dot{q_i}}{\partial q_i} + \dfrac{\partial \dot{p_i}}{\partial p_i} $$
We assume Einstein summation on the repeated indices. With the Hamiltonian equation:
$$ \dot{q_i} = +\dfrac{\partial H}{p_i} \quad \text{and} \quad \dot{p_i} = -\dfrac{\partial H}{q_i} $$
We get:
$$ \nabla \cdot \vec{V_{\varphi}} = \dfrac{\partial H}{\partial q_i \partial p_i} - \dfrac{\partial H}{\partial p_i \partial q_i} = 0 $$
So it seems that if we assume the Hamiltonian equation (because of the least action principle) it imply $\nabla \cdot \vec{V_{\varphi}} = 0$
