Liouville's Theorem and Boltzmann equation for plasma The Boltzmann equation for a plasma can be thought of as coming from a continuity equation in the 6 dimensional phase space of the plasma with coordinates $\left\{x,y,z,v_x,v_y,v_z \right\}$. So initially you start with something like $$\frac{\partial f}{\partial t}+\nabla_6\cdot \left(f\vec{u}_6 \right)=0$$
Where the subscript 6 means we have the six coordinates in phase space given above.
If we work through this a little and assume that the force on the particles is not a function of velocity (or is the Lorentz force, despite the velocity dependance) we can obtain the collisionless Boltzmann equation which is given by 
$$\frac{\partial f}{\partial t}+\vec{v}\cdot\nabla f+\frac{\vec{F}}{m}\cdot\nabla_vf=0$$
Where $\nabla_v$ gives the derivatives with respect to the velocity coordinates and $\frac{\vec{F}}{m}=\vec{a}$(I will put in the steps if anyone asks in comments)
We can write this as a Lagrangian derivative in our 6 dimensional phase space such that $$\frac{Df}{Dt}=0$$.
As I understand it Liouville's theorem states that if we have an ensemble in phase space it evolves such that the density of particles in the phase space remains unchanged i.e. $$\frac{d\rho}{dt}=0$$
Which looks similar to above if we consider the density in phase space and the distribution function to be the same (which I think they are?). 
In general though the Boltzmann equation can have a non-zero right hand side if the plasma is collisional i.e. $$\frac{Df}{Dt}\neq0$$
So my question is what is it about collisions which stops the plasma from obeying Liouville's theorem? I know that Liouville's theorem is normally used to deal with the phase space of systems which are amenable to Hamiltonian mechanics. So can we not write down a Hamiltonian which describes a collisional plasma?
I apologise in advance if this question is either obvious/complete nonsense or totally ill-formed. I have very recently started kinetic theory and so many of the concepts are pretty new.
 A: 
Which looks similar to above if we consider the density in phase space and the distribution function to be the same (which I think they are?)

They are not the same generally. The Hamiltonian and the kinetic description (equations) are fundamentally different - kinetic description is a approximate description of Hamiltonian system and introduces irreversible evolution, something that is not present in the Hamiltonian description.
Furthermore, the space in kinetic description is 6-dimensional even for many-particle system; but for $N$ particles, Hamiltonian description in phase space uses $6N$-dimensional phase space. For $N=2,3,...$ that is a different description from description in $6$-dimensional space.
The Hamiltonian system obeys Liouville's theorem in $6N$-dimensional phase space and still may approximatelly obey the kinetic equation in 6-dimensional space.
A: When including a collision term, the phase-space volume can change. If we denote the collisional term as
$$\left(\frac{\partial f}{\partial t}\right)_{col}\equiv G-L$$
Then the gain term, $G$, describes how the fraction of particles in the cell $dxdv$ of phase space (i.e., $f\,dxdv$) increases due to the collisions of other particles in different cells. Similarly, $L$ describes the loss of the quantity $f\,dxdv$ due to a particle in $dxdv$ colliding (and hence disappearing) from that cell.
The lack of reversibility in this case is what leads to your not being able to use the Hamiltonian. The collisional Boltzmann equation is non-reversible because the processes of collisions are stochastic. I do believe that there is active research as to semi-collisional limits, but I'm not entirely familiar with this subfield.
