Collisionless Boltzmann Equation I am looking at a derivation of the collisionless Boltzmann equation and I am unsure of how they got from one line to the next, so if someone could explain the step to me that would be much appreciated!
$$
df = \frac{∂f}{∂t}dt + \frac{∂f}{∂x_i}dx_i + \frac{∂f}{∂v_i}dv_i \tag{1}
$$
and thus,
$$
\frac{df}{dt} = \frac{∂f}{∂t} + \frac{∂f}{∂x_i}v_i - \frac{∂Φ}{∂x_i}\frac{∂f}{∂v_i} \tag{2}
$$
where Φ(x) is the gravitational potential.
 A: I think it basically boils down to the fact that the gradient of the gravitational potential energy is the relevant force for your problem. For other problems, this needn't be true. Maybe you're dealing with charged particles so you're also interested in the electric potential energy.
I'm going to rewrite things in a different notation since I think it'll make things clearer.
Your distribution function $f$ is a function of position $\vec{r}$, velocity $\vec{v}$ and time $t$. Assuming there are no collisions
$f\left(\vec{r}, \vec{v}, t\right) = f\left(\vec{r} - \vec{v}dt, \vec{v} - \frac{\vec{F}}{m}dt, t\right),$
where $\vec{F}$ is the force that accelerates your particles. Taking the derivitives
$\frac{df}{dt} = \frac{\partial f}{\partial \vec{r}} \cdot \vec{v} + \frac{\partial f}{\partial \vec{v}} \cdot \frac{\vec{F}}{m} + \frac{\partial f}{\partial t}.$
Generally speaking, $\vec{F} = \frac{\partial U}{\partial \vec{r}}$, where $U$ is the potential energy. Since you're dealing with the gravitational potential (not the gravitational potential energy), you have that $\frac{\vec{F}}{m} = \frac{\partial \Phi}{\partial \vec{r}}$.
So in your notation, $\frac{\partial \vec{v}}{\partial t}$ = acceleration = $\frac{\vec{F}}{m} = \frac{\partial \Phi}{\partial \vec{r}}$.
A: For any conservative force one can define the field, $\mathbf{A}$, associated with the force as the gradient of a scalar potential, $\phi$, or:
$$
\mathbf{A} = -\nabla \phi  \tag{0}
$$
We know from Newton's laws that a force is just the mass times acceleration and that $\mathbf{a} = d\mathbf{v}/dt$.  Thus, in your example the gravitational potential is given by:
$$
\frac{dv_{i}}{dt} = a_{i} = \frac{F_{i,grav}}{m} = -\nabla_{i} \Phi_{grav} = -\frac{d \Phi_{grav}}{dx_{i}}  \tag{1}
$$
