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.



Let us define the density of particles of species $s$ in a volume element, $d\mathbf{x} \ d\mathbf{v}$, at a fixed time, $t$, centered at $(\mathbf{x}, \mathbf{v})$ as the quantity $f_{s}(\mathbf{x},\mathbf{v},t)$. I assume this function is non-negative, contains a finite amount of matter, and it exists in the space of positive times and $\mathbb{R}^{3}$ and $\mathbb{R}_{\mathbf{v}}^{3}$, where $\mathbb{R}_{\mathbf{v}}^{3}$ is the space of all possible 3-vector velocities. Then one can see that there are two ways to interpret $f$: (1) it can be an approximation of the true phase space density of a gas (large scale compared to inter-particle separations); or (2) it can reflect our ignorance of the true positions and velocities of the particles in the system. The first interpretation is deterministic while the second is probabilistic. The latter was used implicitly by Boltzmann. Let us assume that $f_{s}(\mathbf{x},\mathbf{v},t)$ $\rightarrow$ $\langle f \rangle + \delta f$, where $\langle f \rangle$ is an ensemble average of $f_{s}$ and I have dropped the subcript out of laziness.

Liouville's Equation

I know that $\langle f \rangle$ satisfies Liouville's equation, or more appropriately, $\partial \langle f \rangle$/$\partial t = 0$. In general, the equation of motion states: $$ \begin{equation} \frac{ \partial f }{ \partial t } = f \left[ \left( \frac{ \partial }{ \partial \textbf{q} } \frac{ d\textbf{q} }{ dt } \right) + \left( \frac{ \partial }{ \partial \textbf{p} } \frac{ d\textbf{p} }{ dt } \right) \right] + \left[ \frac{ d\textbf{q} }{ dt } \cdot \frac{ \partial f }{ \partial \textbf{q} } + \frac{ d\textbf{p} }{ dt } \cdot \frac{ \partial f }{ \partial \textbf{p} } \right] \tag{1} \end{equation} $$ where I have defined the canonical phase space of $(\mathbf{q}, \mathbf{p})$. If I simplify the terms dA/dt to $\dot{A}$ and let $\boldsymbol{\Gamma} = (\mathbf{q}, \mathbf{p})$, then I find: $$ \begin{align} \frac{ \partial f }{ \partial t } & = - f \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} - \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \tag{2a} \\ & = - \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \left( \dot{\boldsymbol{\Gamma}} f \right) \tag{2b} \end{align} $$ where one can see that the last form looks like the continuity equation. If I define the total time derivative as: $$ \begin{equation} \frac{ d }{ dt } = \frac{ \partial }{ \partial t } + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \tag{3} \end{equation} $$ then I can show that the time rate of change of the distribution function is given by: $$ \begin{align} \frac{ d f }{ dt } & = \frac{ \partial f }{ \partial t } + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \tag{4a} \\ & = - \left[ f \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \right] + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \tag{4b} \\ & = - f \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} \tag{4c} \\ & \equiv - f \Lambda\left( \boldsymbol{\Gamma} \right) \tag{4d} \end{align} $$ where $\Lambda \left( \boldsymbol{\Gamma} \right)$ is called the phase space compression factor. Note that Equations 4a through 4d are different forms of Liouville's equation, which have been obtained without reference to the equations of motion and they do not require the existence of a Hamiltonian. I can rewrite Equation 4d in the following form: $$ \begin{equation} \frac{ d }{ dt } \ln \lvert f \rvert = - \Lambda\left( \boldsymbol{\Gamma} \right) \tag{5} \end{equation} $$

Relation to Hamiltonian

Most readers might not recognize Equations 4d and 5 as Liouville's equation because one usually derives it from a Hamiltonian. If the equations of motion can be generated from a Hamiltonian, then $\Lambda \left( \boldsymbol{\Gamma} \right) = 0$, even in the presence of external fields that act to drive the system away from equilibrium. Note that the existence of a Hamiltonian is a sufficient, but not necessary condition for $\Lambda \left( \boldsymbol{\Gamma} \right) = 0$. For incompressible phase space, I recover the simple form of Liouville's equation: $$ \begin{equation} \frac{ d f }{ dt } = 0 \end{equation} $$ However, Liouville's theorem can be violated by any of the following:

  • sources or sinks of particles;
  • existence of collisional, dissipative, or other forces causing $\nabla{\scriptstyle_{ \mathbf{v} }} \cdot \mathbf{F} \neq 0$;
  • boundaries which lead to particle trapping or exclusion, so that only parts of a distribution can be mapped from one point to another;
  • spatial inhomogeneities that lead to velocity filtering (e.g., $\mathbf{E} \times \mathbf{B}$-drifts that prevent particles with smaller velocities from reaching the location they would have reached had they not drifted); and
  • temporal variability at source or elsewhere which leads to non-simultaneous observation of oppositely-directed trajectories.

Source of Irreversibility

Irreversibility is somewhat of a conundrum because it arises largely due to our choice of boundary conditions, smoothing assumptions (e.g., coarse graining or mean field theory), and limits. For instance, if I assume a velocity distribution of particles can be represented by a continuous model function, the use of a continuous distribution function inserts irreversibility into the equation. One can argue that this is splitting hairs because it is obvious that irreversibility exists in nature. However, I think it is important because your question points at a deeper issue.

If I assumed perfectly elastic binary particle collisions and ignore quantum uncertainties, one could, in principle, follow the trajectories of all particles in a system forward and backward in time. There would be no irreversibility in this model, if I had strong enough computers. However, binary particle collisions are not truly elastic, so our assumption of elasticity has created a loss of information.

Another subtle point is that Boltzmann a priori defined his, now famous, H-theorem such that time would increase in the correct direction (i.e., positive time). He did not originally relate the H-theorem to entropy, that interpretation came later (I believe with Gibbs, but someone correct me if I am wrong here).

The point is that the concepts of irreversibility and entropy are coupled, but not necessarily through direct means. I am inclined to think that the irreversibility to which you refer arises from our methods of solving the math necessary for modeling dynamical statistical systems.


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  • $\begingroup$ Hi! A beginner question not related to the OP. You write $f_{s}(\mathbf{x},\mathbf{v},t)$. Isn't $\dot{ \mathbf x}(t)=\mathbf v(t)$ and therefore with $f_s$ we really mean the explicit form $f_{s}(\mathbf{x}(t),\mathbf{v}(t),t)$? $\endgroup$ – JDoeDoe Nov 22 '17 at 13:20
  • $\begingroup$ Kind of but be careful. The $\mathbf{x}$ and $\mathbf{v}$ are variables of the function $f_{s}$ but not values from the velocity moments, which can be implicitly functions of time and position. I forgot the mathematical name for them but they are not variables that track specific particles. Meaning, it's not a single $\mathbf{x}$ and $\mathbf{v}$ that follows a given particle in time. By assuming $f_{s}$, one has removed the discreteness of the system to treat it like a probability distribution that generally depends upon position and velocity. $\endgroup$ – honeste_vivere Nov 22 '17 at 16:08
  • $\begingroup$ Great! If I got you correctly, $\mathbf x$ and $\mathbf v$ are not functions of $t$. $(\mathbf x,\mathbf v, t)=(x,y,z,v_x,v_y,v_z,t)$ are just the coordinates of the 6-dimensional (7?) space, i.e. $\mathbb R^6$? $\endgroup$ – JDoeDoe Nov 22 '17 at 20:13
  • 1
    $\begingroup$ Yes, that is basically the idea. As I saw someone else mention in a comment on a question you posted, $f$ is a probability distribution with coordinates specified by $\mathbf{x}$, $\mathbf{v}$, and $t$ so they are each not functions of time (at least not in the sense you are concerned about). $\endgroup$ – honeste_vivere Nov 22 '17 at 21:29

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.


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.


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