# Boltzmann Equation Maximum Entropy Production Principle

I am currently reading this paper on the maximum entropy production principle (don't confuse it with the maximum entropy principle). Equation 2.1 (ommiting the term with the external force) is the equation governing the change of the distribution $$f$$: $$\frac{\partial f}{\partial t}+c_i\frac{\partial f}{\partial r_i}=I(f)$$ $$c$$ is the velocity, $$r$$ the position and $$I(f)$$ is the collision integral, which can also be written as $$I(f)=-\hat{\Omega}\Phi$$, where $$f=f_0(1+\Phi(c))$$ and $$f_0$$ is the Maxwell local equilibrium function. This equation is then multiplied by $$-k\,ln(f)$$ and integrated over $$c$$ to get the following equation: $$\frac{\partial s}{\partial t}+div(j_s)=k\left[\Phi, \hat{\Omega}\Phi\right]$$ where $$s$$ is the entropy density and $$j_s$$ the entropy flux and $$\left[\Phi, \hat{\Omega}\Phi\right]\equiv\int dc\, \Phi\, \hat{\Omega}\Phi$$
I am trying to reproduce the results, but I am not quite there yet: I tried using the approximation of $$ln(1+x)\approx x$$ since I am given that $$|\Phi|<<1$$: $$\int \, dc \,ln(f)I(f)=\int dc (ln(f_0)+ln(1+\Phi))\hat{\Omega}\Phi$$ $$\approx\int dc (ln(f_0)+\Phi)\hat{\Omega}\Phi$$ Using these I get: $$\frac{\partial s}{\partial t}+div(j_s)+k\frac{\partial f}{\partial t}+kc_i\frac{\partial f}{\partial r_i}=k\left[(ln(f_0)+\Phi), \hat{\Omega}\Phi\right]$$ I was able to get the LHS and two extra terms of this equation by using the product rule, but I am not yet done, some things must cancel, but I don't see how.

It might just be a typo, but you did not write out integration over $$c$$ in the last equation. We should have $${\partial s \over \partial t}+\text{div}(j_s)+k\int{dc\bigg({\partial f \over \partial t}+\sum_{i}{c_i{\partial f \over \partial r_i}}\bigg)}=k[\ln{f_0}+\Phi,\hat{\Omega}\Phi]$$ Now we come back to your question. Since you are reading the paper of your link, you may notice there is this property about the bracket integral $$[X,\hat{\Omega}Y]=[Y,\hat{\Omega}X]$$ In addition, if some function $$X$$ is independent of the velocity $$c$$, the term $$\hat{\Omega}X$$ is $$0$$. This can be seen from the explicit form of collision integral on page 13 and 14 of the paper, just around Eq. 2.1. In physical explanation, the property is easy to understand since the collision integral depicts the effect of momentum exchange of particles before and after collisions with the function $$X$$, if $$X$$ is independent of $$c$$, there is no momentum exchange to include. This means, $$\begin{split} -\int{dc\bigg({\partial f \over \partial t}+\sum_{i}{c_i{\partial f \over \partial r_i}}\bigg)} & = -\int{dc I(f)} = \int{dc \hat{\Omega}\Phi} \\ & = [1,\hat{\Omega}\Phi] = [\Phi,\hat{\Omega}1] = 0 \end{split}$$ And we have similarly for $$[\ln{f_0},\hat{\Omega}\Phi]=[\Phi,\hat{\Omega}\ln{f_0}]=0$$. Hope this answers your question.
• Yes you are right, I did forget the integration. That's a really good answer. I understand why the extra terms which come from using the product rule vanish. However I am not so sure of the terms with the $[ln(f_0), \hat{\Omega} \Phi]=0$ because $f_0=f_0(C)$ and $C=c-u$ as can be seen in the line below equation 2.4. Or are these two $C$'s different? Sep 9, 2022 at 10:23
• I think they are the same thing. The peculiar velocity $C$ is the particle velocity $c$ minus the hydrodynamic velocity $u$. However, $C=C(r,t)$ does not depend on $c$ as supplemented after Eq. 2.3 in the paper. Sep 9, 2022 at 13:35
• To find an analogy, you can think there is a boat with a motor floating on the ocean, and you are observing it on the seashore. The current flows, so even if the motor is not turned on, it still has the velocity $c=u$ where $u$ is the hydrodynamic velocity, but when the motor starts, the boat can have a peculiar velocity $C$, and the velocity $c$, despite being $u+C$, is independent of $C$ since there is no deterministic or probabilistic relation on how $C$ should be given $c$. Sep 9, 2022 at 13:36