My question concerns the possibility of using the hypothesis of local thermodynamic equilibrium for calculate the entropy of a non-homogeneous equilibrium states.

For example, let's imagine the following situation: Fluid in a constant gravitational field

In the figure, $N$ particles of a fluid are inside a vessel of volume $V$ under the effect of a constant gravitational field $g$. The total energy of the particles (sum of internal energy and external potential energy) is $U_{TOT}$.

I can imagine two ways for calculating the entropy of the system.

The first one is using of the definition of entropy in statistical mechanics for the microcanonical ensemble:

S= k ln $ \Omega$

where, as it's known, $ \Omega$ is the number of microstates associated with the parameters $N$, $V$ and $ U_{TOT}$ (or the phase-space area in the classical approximation).

The second possibility is instead assuming a condition of local thermodynamic equilibrium and searching the distribution of $ \rho ( \vec {x})$ and $ T( \vec {x} )$ that maximize the total entropy:

$S= \int {s dV} = \int s (\rho,T) dV$

where $s (\rho,T) $ is the specific volume entropy for the homogeneous fluid and where the maximization is done under the constraints of total mass and total energy:

$m_{TOT}= N m_{PARTICLE}= \int {\rho dV} $

$U= \int [{u_{INT}+u_{POT}] dV}= \int {[u_{INT}(\rho,T) + \rho g ]dV}=U_{TOT}$

So, in conclusion, the question is: have the two entropies the same value?.

In the book of Phil Attard "Thermodynamics and Statistical Mechanics: Equilibrium by Entropy Maximisation", pag. 145-149, the author shows that the two entropies effectively correspond for the specific case of an ideal gas (he uses the canonical formalism, but the theme is the same I guess).

On the other hand, Arieh Ben-Naim in the Appendix-B of the article Is Entropy Associated with Time's Arrow? criticizes the use of the assumption of local thermodynamic equilibrium for fluids different from ideal gases (https://arxiv.org/ftp/arxiv/papers/1705/1705.01467.pdf).

Thus, another linked question is: does the answer to the question above depend on the kind of fluid in the vessel?

I'm an energetic engineer so, maybe, I'm not considering some fundamental concept of statistical mechanics.

Anyway this seems to me an interesting question because, if the two entropies are different, it means that the use the local thermodynamics equilibrium is questionable and doubtful non only in the context of non-equilibrium thermodynamics, but also in the more specific case of non-homogeneous equilibrium thermodynamics.

Thank you everyone for the attention and best regards.

  • $\begingroup$ No need. In my head non homogenous meant conditions that would result in changes in pressure and temperature over time to achieve equilibrium. I see now that obviously the pressure differentials in a fluid due to gravity do not change in time. I deleted my comment. $\endgroup$ – Bob D Jun 18 at 13:19
  • $\begingroup$ The term $\Omega$ depends on material. An ideal gas is particles with zero volume and no interactions; they are indistinguishable. Adding volume and shape changes the validity of this assumption, as does adding (directional) interaction energies. The stat mech definition of entropy is of any given configuration. To find equilibrium after infinite time, $S$ is maximized over all possible configurations under constraints of mass and energy. So, is the difficulty in the concept of entropy at this point in time versus at equilibrium? $\endgroup$ – Jeffrey J Weimer Jun 18 at 15:08
  • $\begingroup$ No Jeffrey. The difficulty concerns the two ways for calculating the entropy of the system at equilibrium and their eventual coincidence. The first way uses the statistical mechanics definition. The second one searches the distributions of local thermodynamic quantities that maximize the integral value of local entropy with given constraints. Do they coincide? $\endgroup$ – FraNicc Jun 19 at 7:31

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