# Time symmetry of Boltzmann kinetic equation and H-theorem for calculating time-evolution of entropy

I know the Boltzmann kinetic equation is invariant under T transformation ($t \rightarrow -t$). Also, I could derive the entropy's time evolution from H-theorem as:

$$\frac{\partial S}{\partial t} = -k_{B} \int \int \int \ln\Bigg(\frac{f_{2} f_{1}}{f^{'}_{2}f^{'}_{1}}\Bigg) (f^{'}_{2}f^{'}_{1}-f_{2}f_{1}) g \alpha_{1} d\mathbf{e}^{'}d\mathbf{v}_{2}d\mathbf{v}_{1}$$

Where $f_{2}$ and $f_{1}$ are the probabilities before collision for two particles distributions, $f^{'}_{2}$ and $f^{'}_{1}$ are the probabilities after collision, $g$ is the relative velocity of particle 1 and 2 $g = |\mathbf{v}_{2}-\mathbf{v}_{1}|$ before collision and because the collision is elastic: $g = g^{'}$. Also, $\alpha_{1}$ is the differential cross section of the collision.

According to this equation because terms $\ln\Bigg(\frac{f_{2}f_{1}}{f^{'}_{2}f^{'}_{1}}\Bigg)$ and $(f^{'}_{2}f^{'}_{1}-f_{2}f_{1})$ have always opposite signs the integral is always negative and as a result entropy will increase always or $\frac{\partial S}{\partial t} > 0$. I know it's called Loschmidt's paradox which basically says that it is impossible to extract a time asymmetric equation from time symmetric ones. The thing which is not quite clear to me is that no matter the Boltzmann equation is time symmetric or not, this equation for entropy time evolution holds true so always we could say $\frac{\partial S}{\partial t} > 0$. So why it is not possible to prove the second law of thermodynamics by using H-theorem?

• The presence of the collision term means that the Boltzmann equation is not T invariant. The subtlety therefore appears in any attempt to derive the Boltzmann equation from T invariant microscopic equations. The derivation of the H theorem from the Boltzmann equation is straightforward and involves no subtleties – Thomas Sep 11 '18 at 22:23
• Why?! I mean collision is elastic so everything should be reversible. Otherwise you are saying that H-theorem really implies second law of thermodynamics? – GGG Sep 11 '18 at 22:25
• Just look at the equation. Left hand side $\partial f/\partial t$ is odd, right hand side $C[f]$ is even. – Thomas Sep 11 '18 at 22:37
• No $C[f]$ is not even because it is a collision rate. So it is like $\frac{\partial f}{\partial t}$. In fact, in the simplest case, collision operator could be approximated by Bhatnagar–Gross–Krook (BGK) model as: $C[f] = \frac{f-f^{eq}}{\tau}$, where $f^{eq}$ is the equilibrium distribution (i.e. Maxwell-Boltzmann equilibrium distribution) and $\tau$ is the relaxation time. So, eventually it is not even. – GGG Sep 11 '18 at 23:00
• Look, as a differential equation the Boltzmann equation clearly breaks T, because the LHS and the RHS transform differently under $t\to-t$. This means that any derivation from T invarIant classical or quantum mechanics must involve subtleties. – Thomas Sep 12 '18 at 7:15