From what I understand, Liouville's theorem is about the probability density $\rho$ of an ensemble existing in a differential volume in phase space $d\mathbf{r}d\mathbf{p}$.

So the statement for Liouville's theorem is \begin{align*} \frac{d}{dt} \rho (\mathbf{r}, \mathbf{p}, t) &= \frac{\partial \rho}{\partial t} + \sum_{i=1}^N \frac{\partial \rho}{\partial q_i}\cdot \frac{dq_i}{dt}+\frac{\partial \rho}{\partial p_i} \cdot \frac{dp_i}{dt} \quad ...(1)\\ &=0 \end{align*} which arises out of the continuity equation coupled with Hamilton's relations.

All good so far.

Now we have Boltzmann's transport equation, which is given by

$$\frac{\partial f}{\partial t} + \frac{\mathbf{p}}{m}\cdot \nabla F + \mathbf{F}\cdot \nabla _p f = \left( \frac{df}{dt} \right) _{\rm coll}$$

where $f$ "probability density function: $f(\mathbf{r}, \mathbf{p}, t)$".

If I replace the $\frac{dq}{dt}$ by $\frac{\mathbf{p}}{m}$ and $\frac{dp}{dt}$ by $\mathbf{F}$ in equation $(1)$, I recover the same equation as the Boltzmann transport equation.

My question is: What is the difference between $\rho$ and $f$? Why are there no collisions in an ensemble as seen by Liouville, but we consider collisions when thinking about Boltzmann's transport equation?


3 Answers 3


Liouville: Take a system of $N$ particles and its distribution function $f_N$ over the $6N$-dimensional phase space of the system (assuming that the particles are in a volume $V \subset \mathbb R^3$).

The $N$-particles "state" at a given time $t$ is described by a function $f_N(x_i,p_i,t)$, where $x_i\in V \subset \mathbb R^3 $, $p_i\in\mathbb R^3$. The dynamics of the phase space distribution function $f_N$ is given by the Liouville equation

\begin{equation} \frac{\partial f_N}{\partial t}=\{H,f_N \}, \end{equation}

where $H(x_i,p_i)$ is the Hamiltonian of the system and $\{ , \}$ are the Poisson brackets. The Liouville theorem is a more "geometric" statement that follows from reinterpreting the Liouville equation as a conservation (continuity) equation (the global conserved quantity being $N$, see this answer). For more details, see this answer or this question.

The $f_N$ contains full information about the system and the dynamics is reversible. The Liouville equation is, thus, "exact" (no approximation is involved).

Boltzmann: it is an equation for the one-particle distribution function $f(x,p,t)$ of a system of $N$-particles. The $f(x,p,t)$ can be obtained from $f_N$ by integrating out all the $N-1$ variables but one (marginalization over $N-1$ phase space variables). By doing this a lot of information is lost: this is why it gives rise to a thing, called entropy, that grows and measures the amount of information lost (or "growth of ignorance", i.e. ignorance about the N-body correlators) as time passes by, see the scheme in figure 1 of this paper.

You can see how the Boltzmann equation for $f$ is derived from Liouville for $f_N$ thanks to the BBGKY hierarchy, see this answer. This was done, e.g., by:

Note: for the "trivial" $N=1$ case, there is no collision integral in Boltzmann, so that $Boltzmann=Liouville$ (for $N=1$ there is no marginalization and no BBGKY hierarchy). Note also that it is the "collision" term that, at least from the "visual" point of view, is the big difference between Liouville and Boltzmann: Liouville is "exact" (there is no need to consider collisions, it already accounts for the totality of the interactions between the $N$ particles), Boltzmann is an approximation for the one-particle distribution function (so you need to include, in an approximate way, how interactions modify $f$ via "collisions", see e.g. this for a physical example). It is also worth checking the question Relation between master, Fokker-Planck, Langevin, Kramers-Moyal and Boltzmann equations.

Why is there a "collision" term in Boltzmann but not in Liouville: I quote from the original question: "Why are there no collisions in an ensemble as seen by Liouville, but we consider collisions when thinking about Boltzmann's transport equation?". This should be clear from the note above, but let me expand on this:

Liouville: $\frac{df_N}{dt}=\partial_t f_N +\sum_i ( \dot{x}_i \cdot \nabla_{x_i} f +\dot{p}_i \cdot \nabla_{p_i} f ) = 0$

Boltzmann: $\frac{df}{dt}=\partial_t f +v\cdot \nabla_x f +F \cdot \nabla_p f = C[f]$

In the Liouville case, $\dot{p}_i$ is a function of all the $x_i$, namely, it is the exact total force on the $i$-particle coming from all the other particles and, possibly, external fields. In the Boltzmann case, $F$ is just the external force acting on the particles, that's why you need the collision term $C[f]$. For a closed system of $N$ particles: $\frac{df_N}{dt}=0$ along the exact trajectories in the phase space, but for every non-closed sub-system this is not true anymore. A 1-particle subsystem is a closed subsystem only if the $N$ particles are non-interacting, otherwise $df/dt \neq 0$. In principle, the term $C[f]$ is defined as $df/dt$ for the non-closed sub-system, and it is then modelled somehow in terms of collisions. The fact that the collision integral pops out for non-closed sub-systems is explained in I.3 of "Physical Kinetics" by Lifshitz (volume 10 of the famous Landau's series).


Liouvilles equation is for a probability density in the many-particle phase space so is a function $\rho({\bf x}_1,p_1,{\bf x}_2,p_2, \ldots)$ where there are as many ${\bf x}'s$ as there are particles. In Boltzmann you just have $\rho({\bf x},{\bf p})$ which gives the expected number of particles $n$ in a small region as $n= \rho({\bf x}, {\bf p}) d^3x d^3p$. There is only one ${\bf x}$ and one ${\bf p}$ in Boltzmann.

  • 1
    $\begingroup$ So boltzmann is talking about a flux of a certain number of particles through a region in phase space given by $[\mathbf{r},\mathbf{p}]$ to $[\mathbf{r+dr}, \mathbf{p+dp}]$? $\endgroup$
    – megamence
    Commented May 16, 2021 at 0:24

The time dependent probability function $\rho(q,p,t)$ that appears in Liouville's theorem can be interpreted as an expression of uncertainty about the state of the system at any given time. That is, we are free to imagine that the system might inhabit a definite state $(q,p)$ at time $t$, and in a controlled experiment it would be utterly ridiculous if our ignorance of this influenced how the system evolved in any way.

In contrast, the Boltzmann equation describes the evolution of a physical many-body mass or number density function, $f(q,p,t)dqdpdt=<n(q,p,t)>dqdpdt$, where the average is taken over a sufficiently 'large' infinitesimal volume that $<n(q,p,t)>\approx n(q,p,t)$, modulo fluctuations of much smaller magnitude that vanish in the continuum limit. Because the function $f(q,p,t)$ describes the state of a system with multiple particles, and these particles are usually interacting, its dynamics near any given point $(q,p)$ will generally depend on the value of $f(q',p',t)$ elsewhere, and sometimes (even in systems where Galilean relativity applies) at earlier times as well.

  • $\begingroup$ Thank you for your answer @TLDR! I don't understand what you mean by "in a controlled experiment it would be utterly ridiculous if our ignorance of this influenced how the system evolved in any way." How would our ignorance of this influence an experiment? $\endgroup$
    – megamence
    Commented May 15, 2021 at 19:25
  • $\begingroup$ I also have another question: why do we need a collision term in Boltzmann transport? $\endgroup$
    – megamence
    Commented May 15, 2021 at 19:28
  • $\begingroup$ @megamence, I meant the sentence about "controlled experiments" to be (only) slightly humorous. The intended meaning is essentially that the world does what it does independently of our knowledge or ignorance of it doing so. The collision term in the Boltzmann equation arises from interactions between particles. The existence of a Markovian collision operator is actually a somewhat non-trivial property that does not hold for all physical systems (in general, the collision operator depends on the values of $f$ at all prior times.) $\endgroup$
    – TLDR
    Commented May 16, 2021 at 20:24

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