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

Question

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.)

Answer 1

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