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:
Is the proof that I suggest correct?
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.)