8. Liouville theorem and Boltzmann equation
8.1. Liouville theorem
Theorem 8.1 (Liouville’s theorem)
Let \(f(\mathbf{r}_i, \mathbf{v}_i, t)\) the multi-particle probability density function. Liouville’s equations reads
\[
0 = \frac{\partial f}{\partial t} + \mathbf{v}_i \cdot \nabla_{\mathbf{r},i} f + \frac{\mathbf{F}_i}{m} \cdot \nabla_{\mathbf{v},i} f \ .
\]
Let \(f(\mathbf{r}_i, \mathbf{v}_i, t)\) be the \(N\)-particle distribution of a system as a function of the \(2 d N\) coordinates and components of the velocity. Probability is normalized to 1
\[1 = \int_{\overline{\Omega}} f(\mathbf{r}_i, \mathbf{v}_i, t) d \mathbf{r}_i d \mathbf{v}_i \ .\]
The governing equations of each particle read
\[\begin{split}\begin{cases}
\dot{\mathbf{r}}_i = \mathbf{v}_i \\
\dot{\mathbf{v}}_i = \frac{\mathbf{F}_i}{m} \ .
\end{cases}\end{split}\]
A governing equation for the probability \(P_{\Omega}\) of the system to be in a state belonging to an arbitrary “material” volume \(\Omega_t\) in the state space reads
\[\begin{split}\begin{aligned}
0 = \frac{d}{dt} P_{\Omega}
& = \frac{d}{dt} \int_{\Omega_t} f = \\
& = \int_{\Omega_t} \frac{\partial f}{\partial t} + \oint_{\partial \Omega_t} \left( \hat{\mathbf{n}}_{\mathbf{r},i} \cdot \mathbf{v}_i + \hat{\mathbf{n}}_{\mathbf{v},i} \cdot \mathbf{a}_i \right) f = \\
& = \int_{\Omega_t} \left\{ \frac{\partial f}{\partial t} + \nabla_{\mathbf{r},i} \cdot \left( \mathbf{v}_i f \right) + \nabla_{\mathbf{v},i} \cdot \left( \mathbf{a}_i f \right) \right\} \ ,
\end{aligned}\end{split}\]
having used Reynolds’ transport theorem in the state space, applied the multi-dimensional version of the divergence theorem. As this balance holds for any arbitrary volume \(\Omega_t\), the differential form follows
\[
0 = \frac{\partial f}{\partial t} + \nabla_{\mathbf{r},i} \cdot \left( \mathbf{v}_i f \right) + \nabla_{\mathbf{v},i} \cdot \left( \mathbf{a}_i f \right) \ .
\]
Now, using Hamilton equations in absence of non-conservative forces, in terms of generalized coordinates and generalized momenta
\[\begin{split}\begin{cases}
\dot{\mathbf{q}}^i = \frac{\partial \mathscr{H}}{\partial \mathbf{p}_i} \\
\dot{\mathbf{p}}_i = -\frac{\partial \mathscr{H}}{\partial \mathbf{q}^i} \ ,
\end{cases}\end{split}\]
or in terms of positions and velocities
\[\begin{split}\begin{cases}
\mathbf{v}_i = \dot{\mathbf{r}}_i = \frac{1}{m} \nabla_{\mathbf{v}_i} \mathscr{H} \\
\mathbf{a}_i = \dot{\mathbf{v}}_i = -\frac{1}{m} \nabla_{\mathbf{r}_i} \mathscr{H} \ ,
\end{cases}\end{split}\]
it’s possible to find the “convective form” of the Liouville’s theorem
\[\begin{split}\begin{aligned}
0
& = \frac{\partial f}{\partial t} + \nabla_{\mathbf{r},i} \cdot \left( \mathbf{v}_i f \right) + \nabla_{\mathbf{v},i} \cdot \left( \mathbf{a}_i f \right) = \\
& = \frac{\partial f}{\partial t} + f \underbrace{\left( \nabla_{\mathbf{r},i} \cdot \mathbf{v}_i + \nabla_{\mathbf{v},i} \cdot \mathbf{a}_i \right)}_{=0} + \mathbf{v}_i \cdot \nabla_{\mathbf{r},i} f + \mathbf{a}_i \cdot \nabla_{\mathbf{v},i} f \ ,
\end{aligned}\end{split}\]
i.e.
\[\begin{split}
0 & = \frac{\partial f}{\partial t} + \mathbf{v}_i \cdot \nabla_{\mathbf{r},i} f + \mathbf{a}_i \cdot \nabla_{\mathbf{v},i} f \\
0 & = \frac{\partial f}{\partial t} + \mathbf{v}_i \cdot \nabla_{\mathbf{r},i} f + \frac{\mathbf{F}_i}{m} \cdot \nabla_{\mathbf{v},i} f \\
0 & = \frac{\partial f}{\partial t} + \mathbf{v}_i \cdot \nabla_{\mathbf{r},i} f + \frac{\mathbf{F}_i^{ext}}{m} \cdot \nabla_{\mathbf{v},i} f + \sum_{j \ne i} \frac{\mathbf{F}_{ij}}{m} \cdot \nabla_{\mathbf{v},i} f \ .
\end{split}\]
8.2. Boltzmann equation
Indistinguishable particles. State counting…
1-particle probability distribution function. Integrating all but one particle (marginalizing over all the variables expect for \(\mathbf{r}_1\), \(\mathbf{v}_1\)),
\[f_1(\mathbf{r}_1, \mathbf{v}_1, t) = \int_{\mathbf{r}_{k=2:N}} \int_{\mathbf{v}_{k=2:N}} f(\mathbf{r}_i,\mathbf{v}_i, t) d \mathbf{r}_k d \mathbf{v}_k \ ,\]
\(s\)-particle probability distribution function.
\[f_s(\mathbf{r}_{j=1:s}, \mathbf{v}_{j=1:s}, t) = \int_{\mathbf{r}_{k=s+1:N}} \int_{\mathbf{v}_{k=s+1:N}} f(\mathbf{r}_i,\mathbf{v}_i, t) d \mathbf{r}_k d \mathbf{v}_k \ ,\]
Time derivative.
\[\begin{split}\begin{aligned}
\partial_t f_s
& = \partial_t \int_{\mathbf{r}_{k=s+1:N}} \int_{\mathbf{v}_{k=s+1:N}} f(\mathbf{r}_i,\mathbf{v}_i, t) d \mathbf{r}_k d \mathbf{v}_k = \\
& = - \int_{\mathbf{r}_{k=s+1:N}} \int_{\mathbf{v}_{k=s+1:N}} \left\{ \mathbf{v}_i \cdot \nabla_{\mathbf{r}_i} f + \frac{\mathbf{F}_i^{ext}}{m} \cdot \nabla_{\mathbf{v}_i} f + \sum_{j \ne i} \frac{\mathbf{F}_{ij}}{m} \cdot \nabla_{\mathbf{v}_i} f \right\} d \mathbf{r}_k d \mathbf{v}_k = \\
\end{aligned}\end{split}\]
Using Hamilton’s equation to get a divergence, and assuming zero probability at infinity (impossible to be at infinite distance, impossible to have infinite velocity. Sounds likely),
\[\begin{split}\begin{aligned}
\int_{\mathbf{r}_k} \left\{ \mathbf{v}_i \cdot \nabla_{\mathbf{r}_i} f + \frac{\mathbf{F}_i}{m} \cdot \nabla_{\mathbf{v}_i} f \right\}
& = \int_{\mathbf{r}_k} \left\{ \mathbf{v}_i \cdot \nabla_{\mathbf{r}_i} f + \frac{\mathbf{F}_i}{m} \cdot \nabla_{\mathbf{v}_i} f + f \left( \nabla_{\mathbf{r}_i} \cdot \mathbf{v}_i + \nabla_{\mathbf{v}_i} \cdot \frac{\mathbf{F}_i}{m} \right) \right\} = \\
& = \int_{\mathbf{r}_k} \left\{ \nabla_{\mathbf{r}_i} \cdot \left( f \mathbf{v}_i \right) + \nabla_{\mathbf{v}_i} \cdot \left( f \frac{\mathbf{F}_i}{m} \right) \right\} = \\
& = \sum_{i \ne k} \int_{\mathbf{r}_k} \left\{ \nabla_{\mathbf{r}_i} \cdot \left( f \mathbf{v}_i \right) + \nabla_{\mathbf{v}_i} \cdot \left( f \frac{\mathbf{F}_i}{m} \right) \right\} = \\
& = \sum_{i \ne k} \int_{\mathbf{r}_k} \left\{ \mathbf{v}_i \cdot \nabla_{\mathbf{r}_i} f + \frac{\mathbf{F}_i}{m} \cdot \nabla_{\mathbf{v}_i} f \right\} = \\
& = \sum_{i \ne k} \int_{\mathbf{r}_k} \left\{ \mathbf{v}_i \cdot \nabla_{\mathbf{r}_i} f + \left( \frac{\mathbf{F}^{ext}_i}{m} + \sum_{j \in K} \frac{\mathbf{F}_{ij}}{m} + \sum_{j \notin K} \frac{\mathbf{F}_{ij}}{m} \right) \cdot \nabla_{\mathbf{v}_i} f \right\} = \\
& = \sum_{i \ne k} \left\{ \mathbf{v}_i \cdot \nabla_{\mathbf{r}_i} \int_{\mathbf{r}_k} f + \frac{\mathbf{F}^{ext}_i}{m} \cdot \nabla_{\mathbf{v}_i} \int_{\mathbf{r}_k} f + \sum_{j \notin K} \frac{\mathbf{F}_{ij}}{m} \cdot \nabla_{\mathbf{v}_i} \int_{\mathbf{r}_k} f \right\} + \sum_{i \ne k} \sum_{j \in K} \int_{\mathbf{r}_k} \frac{\mathbf{F}_{ij}}{m} \cdot \nabla_{\mathbf{v}_i} f = \\
& = \sum_{i \ne k} \left\{ \mathbf{v}_i \cdot \nabla_{\mathbf{r}_i} f_s + \frac{\mathbf{F}^{ext}_i}{m} \cdot \nabla_{\mathbf{v}_i} f_s + \sum_{j \notin K} \frac{\mathbf{F}_{ij}}{m} \cdot \nabla_{\mathbf{v}_i} f_s \right\} + \sum_{i \ne k} \sum_{j \in K} \int_{\mathbf{r}_k} \frac{\mathbf{F}_{ij}}{m} \cdot \nabla_{\mathbf{v}_i} f = \\
\end{aligned}\end{split}\]
\[
0 = \partial_t f_s + \sum_{i \notin K} \left\{ \mathbf{v}_i \cdot \nabla_{\mathbf{r}_i} f_s + \frac{\mathbf{F}^{ext}_i}{m} \cdot \nabla_{\mathbf{v}_i} f_s + \sum_{j \notin K} \frac{\mathbf{F}_{ij}}{m} \cdot \nabla_{\mathbf{v}_i} f_s \right\} + \sum_{i \notin K} \sum_{j \in K} \int_{\mathbf{x}_k} \frac{\mathbf{F}_{ij}}{m} \cdot \nabla_{\mathbf{v}_i} f \ .
\]
with \(K = s+1:N\).
1-particle equation. With \(K = 2:N\),
\[
0 = \partial_t f_1 + \mathbf{v}_1 \cdot \nabla_{\mathbf{r}_1} f_1 + \frac{\mathbf{F}^{ext}_1}{m} \cdot \nabla_{\mathbf{v}_1} f_1 + \sum_{j = 2:N} \int_{\mathbf{x}_k, k = 2:N} \frac{\mathbf{F}_{1j}}{m} \cdot \nabla_{\mathbf{v}_1} f \, d \mathbf{r}_k d \mathbf{v}_k \ .
\]
Force \(\mathbf{F}_{1j}\) only depends on variables of particle \(1\) and \(j\). Therefore,
\[\begin{split}\begin{aligned}
\sum_{j = 2:N} \int_{\mathbf{r}_k, k = 2:N} \frac{\mathbf{F}_{1j}}{m} \cdot \nabla_{\mathbf{v}_1} f \, d \mathbf{r}_k d \mathbf{v}_k
& = \sum_{j = 2:N} \int_{\mathbf{x}_j} \frac{\mathbf{F}_{1j}}{m} \cdot \int_{\mathbf{r}_k, k = 2:N, k \ne j} \nabla_{\mathbf{v}_1} f \, d \mathbf{r}_k d \mathbf{v}_k = \\
& = \sum_{j = 2:N} \int_{\mathbf{x}_j} \frac{\mathbf{F}_{1j}}{m} \cdot \nabla_{\mathbf{v}_1} \int_{\mathbf{x}_k, k = 2:N, k \ne j} f \, d \mathbf{r}_k d \mathbf{v}_k = \\
& = \sum_{j = 2:N} \int_{\mathbf{x}_j} \frac{\mathbf{F}_{1j}}{m} \cdot \nabla_{\mathbf{v}_1} f_2^{K/\{1, j\}} (\mathbf{r}_1, \mathbf{v}_1, \mathbf{r}_j, \mathbf{v}_j, t) d \mathbf{r}_j d \mathbf{v}_j \ .
\end{aligned}\end{split}\]
For identical particles, the summation becomes \((N-1) \int_{\mathbf{x}_2} \frac{\mathbf{F}_{12}}{m} \cdot \nabla_{\mathbf{v}_1} f_2(\mathbf{x}_1, \mathbf{x}_2,t)\)
Collision integral. Assumptions
When a collision occurs, the dynamics of a particle is not smooth anymore. Thus, is it possible to use partial derivative in time? Force is impulsive. Thus, either function in the sense of distribution and/or integration in time is required
only 2-particle collision. Good assumption for “dilute systems”.
a collision occurs when 2 particles are in the same position (i.e. the dimension of the particles is negligible), \(\mathbf{r}_2 = \mathbf{r}_1\)
\[\propto \int_{} \Gamma_{\mathbf{p}_1',\mathbf{p}_2' \rightarrow \mathbf{p}_1, \mathbf{p}_2} f(\mathbf{r}_1, \mathbf{r}_2, \mathbf{p}_1', \mathbf{p}'_2) - \Gamma_{\mathbf{p}_1 ,\mathbf{p}_2 \rightarrow \mathbf{p}_1', \mathbf{p}_2'} f(\mathbf{r}_1, \mathbf{r}_2, \mathbf{p}_1, \mathbf{p}_2)\]
with
\(\Gamma_{\mathbf{p}_1', \mathbf{p}_2' \rightarrow \mathbf{p}_1, \mathbf{p}_2}\) representing the probability of starting from \(\mathbf{p}_i'\) before the collision and ending with momenta \(\mathbf{p}_i\)
this probability is determined by the analysis of 2-particle collision dynamics
this probability is “scaled” by the probability of being in the state before the collision \(f(\mathbf{r}_1, \mathbf{r}_2 = \mathbf{r}_1, \mathbf{p}_1', \mathbf{p}_2')\)
Molecular chaos (Stosszahlansatz). Before and after the collision
\[f_2(\mathbf{v}_1, \mathbf{v}_2, t) \sim f_1(\mathbf{v}_1) \, f(\mathbf{v}_2,t)\]
8.3. H-theorem
Starting from Boltzmann equation for the 1-particle probability density function, with the molecular chaos assumption,
\[D_t f_1 = \int d \mathbf{v}_2 \int d\Omega \sigma(\Omega) |\mathbf{v}_1 - \mathbf{v}_2| \left( f'_1 f'_2 - f_1 f_2 \right)\]
Functional \(H(t)\) is defined as
\[H(t) = \int f(\mathbf{v},t) \ln f(\mathbf{v},t) d \mathbf{v}\]
Taking time derivative (integration domain is unbounded. Use divergence theorem and boundary conditions to prove that flux contributions are zero)
\[d_t H(t) = \int \partial_t \left( f \ln f \right) d \mathbf{v} = \int \partial_t f \left( 1 + \ln f \right) d \mathbf{v} \ .\]
todo include convective terms of Boltzmann equation
todo prove each step
\[\begin{split}\begin{aligned}
d_t H(t)
& = \int \partial_t f \left( 1 + \ln f \right) d \mathbf{v} = \\
& = \int \int_{\mathbf{v}_1} |\mathbf{v}_1 - \mathbf{v}_2| (f'_1 f'_2 - f_1 f_2) \left( 1 + \ln f_1 \right) \sigma_{\Omega} d \mathbf{v}_1 d \mathbf{v}_2 d\Omega = \\
& = \frac{1}{2} \int \int_{\mathbf{v}_1} |\mathbf{v}_1 - \mathbf{v}_2| (f'_1 f'_2 - f_1 f_2) \left( \ln f_1 + \ln f_2 \right) \sigma_{\Omega} d \mathbf{v}_1 d \mathbf{v}_2 d\Omega = \\
& = \frac{1}{4} \int \int_{\mathbf{v}_1} |\mathbf{v}_1 - \mathbf{v}_2| (f'_1 f'_2 - f_1 f_2) \left( \ln f_1 + \ln f_2 - \ln f'_1 - \ln f'_2 \right) \sigma_{\Omega} d \mathbf{v}_1 d \mathbf{v}_2 d\Omega = \\
& = \frac{1}{4} \int \int_{\mathbf{v}_1} |\mathbf{v}_1 - \mathbf{v}_2| (f'_1 f'_2 - f_1 f_2) \ln \frac{ f_1 f_2 }{f'_1 f'_2} \sigma_{\Omega} d \mathbf{v}_1 d \mathbf{v}_2 d\Omega = \\
& \le 0 \ ,
\end{aligned}\end{split}\]
as
\[(a-b) \ln \frac{b}{a} \le 0 \ ,\]
since, if \(a > b\) the first factor is positive and the second negative, and viceversa, and \(a = b\) implies equality.
