The 3-dimensional Vlassov -Poisson equation I am studying at university is
$$ \partial\_t f (t,x,v) + v\cdot \nabla\_x f (t,x,v) - \nabla\_x \phi (t,x) \nabla\_v f (t,x,v) =0,$$
where $$\Delta \phi = 4\pi\gamma \rho (t,x,v) \text{ and } \rho (t,x) = \int\_{\mathbb{R}^3} f(t,x,v)\ dv$$


I am trying to prove the momentum is conserved:
$$q(t) = \int\_{\mathbb{R}^6} vf(t,x,v)\ dxdv \Rightarrow q' (t)=0$$


**I need some help to prove that** This is my attempt,
I first substitute the equation of $f$ into the expression:
$$ q'(t) = \int \partial\_t [vf(t,x,v)]\ dxdv = \int v [-v\cdot \nabla\_x f (t,x,v) + \nabla\_x \phi (t,x) \nabla\_v f (t,x,v)]\ dxdv$$
Secondly, I noticed that the first term is equal to zero after an integration by parts (passing the $x$-derivative to $|v|^2$ which does not depende on $x$), therefore
$$ q'(t)= \int v\nabla\_x\phi(t,x)\nabla\_v f(t,x,v)\ dxdv $$
**What can I do next to show that $q' =0$?** Thanks in advance! 



Vlasov Poisson: linear momentum conservation
It helps to consider components of $q$, in order to keep the notations clear. From what you have we can write


$$ q\_i'(t) = \int v\_i \sum\_{j} \nabla\_{x\_j} \phi \nabla\_{v\_j} f ~\mathrm{d}x ~\mathrm{d}v $$


integrate by parts in $v$, and using that $\phi$ is independent of $v$ and $\nabla\_{v\_j} v\_i = \delta\_{ij}$ we get


$$ q\_i'(t) = - \int \nabla\_{x\_i} \phi f ~\mathrm{d}x ~\mathrm{d}v $$


perform the $v$ integral first since $\phi$ is $v$-independent


$$ q\_i'(t) = - \int \nabla\_{x\_i} \phi \rho ~\mathrm{d}x $$


by definition, and using the definition of the gravitational potential you get


$$ q\_i'(t) = - \frac{1}{4\pi\gamma} \int \nabla\_{x\_i} \phi \Delta \phi ~\mathrm{d}x $$


Integrating by parts the Laplacian you get


$$ q\_i'(t) = \frac{1}{8\pi\gamma} \int \nabla\_{x\_i} ( |\nabla\phi|^2 ) ~\mathrm{d}x = 0 $$


