# Time derivative of Gibbs entropy (the paradox of the constant fine-grained entropy)

In the context of classical systems, the fine-grained (or Gibbs) entropy is defined as the functional:

$$S_G(t)=-k\int_{\Gamma_t}dqdp\ \rho(p,q,t)\ln[\rho(p,q,t)]$$ (1)

I've been told (Wehrl and J. van Lith) that the Liouville's theorem ensures that this quantity is constant under an evolution governed by Hamilton's equation. I can prove this statement by using the time evolution as a change of variables in (1), then using that the Jacobian of this change of variables is 1 (one of the forms of Liouville's theorem) :

$$S_G(0)=-k\int_{\Gamma_0}dq_0dp_0\ \rho(p_0,q_0,0)\ln[\rho(p_0,q_0,0)]=-k\int_{\Gamma_t}dqdp\ {J(\frac{\partial \ q_0,p_0}{\partial \ q,p})}\rho(p,q,t)\ln[\rho(p,q,t)]=S_G(t)$$

My doubts:

1. Is the proof that I suggest correct?

2. I am trying to prove it by deriving in (1) but I am not successful:

$$\frac{d S_G(t)}{dt}=-k\int_{\Gamma_t}dqdp\ \partial_t \rho(p,q,t)(1+\ln[\rho(p,q,t)])$$

which is trivially equal to zero just in the case of equilibrium ($$\partial_t \rho(p,q,t)=0$$). Any idea on why this is not working?

(This is interesting because it implies that either the Hamiltonian evolution or the Gibbs entropy cannot describe a non-equilibrium evolution in which, according to the second law, the thermodynamic entropy should increase.)

• Hamiltonian models definitely can describe non-equilibrium, even transient from non-equilibrium to equilibrium. You can see this in numerical simulations of non-equilibrium point or hard sphere gas. These models obey the Liouville equation and manifest macroscopic irreversibility at the same time. You can't use Gibbs entropy functional constancy as evidence that thermodynamic entropy of the model is constant - they are very different things. Feb 12, 2021 at 13:57
• Nice remark @JánLalinský. I wanted to say that $\frac{dS_G}{dt}=0$ implied that either the Hamiltonian evolution OR this Gibbs entropy definition cannot describe the non-equilibrium process (I edited my original question). I did not have evidence to discern between these two possibilities, Could you attach references?. I've found authors that try to solve this problem either with different evolution schemes (via master equation) and with different definitions of the entropy functional.
– Javi
Feb 12, 2021 at 14:56
• Thermodynamic entropy isn't well understood as a function of time, but rather as a function of equilibrium macrostate. See for example Jaynes, E.T.: Gibbs vs Boltzmann Entropies, Am. J. Phys. 22, 5, 391-398, 1965 bayes.wustl.edu/etj/articles/gibbs.vs.boltzmann.pdf Feb 12, 2021 at 15:47

So I found an answer to my question.

1.- As far as I know, is indeed correct and it would work for any functional of the density $$F(\rho)$$.

2.- I just had to make use of the Liouville's theorem: $$\partial_t \rho = -\{\rho,H \}$$ and work out the expression:

$$\frac{dS_G(t)}{dt}= \int_{\Gamma_t}dpdq \ \partial_t \rho(p,q;t) \ln(e\rho(p,q;t))= -\int_{\Gamma_t}dpdq \ \{\rho,H \} \ln(e\rho)=-\int_{\Gamma_t}dpdq \ \ln(e\rho) \ \sum^{q,p} (\partial_p\rho\partial_q H-\partial_q\rho\partial_p H)$$

Integrating by parts: $$\int_{\Gamma_t}dpdq \ \ln(e\rho) \partial_p\rho\partial_q H=\int_{\Gamma_t}dq \ \rho\ln\rho \ \partial_q H |^{p\rightarrow+\infty}_{p\rightarrow-\infty}-\int_{\Gamma_t}dpdq \ \rho\ln\rho \partial_{q,p}H$$

So

$$\frac{dS_G(t)}{dt}=\int_{\Gamma_t}dp \ \rho\ln\rho \ \partial_p H |^{q\rightarrow+\infty}_{q\rightarrow-\infty}-\int_{\Gamma_t}dq \ \rho\ln\rho \ \partial_q H |^{p\rightarrow+\infty}_{p\rightarrow-\infty}$$

In order to ensure convergence: $$\lim_{|q|\rightarrow + \infty} \rho=\frac{1}{q^{1+\alpha}}$$ and $$\lim_{|p|\rightarrow + \infty}\rho=\frac{1}{p^{1+\alpha}}$$

We can assume : $$\lim_{q\rightarrow\infty}\partial_q H=0$$ and $$\partial_p H \sim p$$

Concluding $$\frac{dS_G(t)}{dt}=0$$