I apologize for the length of this question that was asked here prigogine-bronsted but got no reply; anyhow, Bronsted's name is much better known among chemists than among physicists...

Prigogine's minimum entropy production principle fails in the simplest heat conduction case, because the temperature distribution that minimizes the entropy production rate formula $\sum_k J_k X_k$ contradicts Fourier's law. In fact, the resulting temperature has an exponential distribution for a homogeneous rod between two fixed temperatures, see [4]. The minimization gives the correct linear temperature dependence $T(x)=c_0+c_1T$ only if the conductivity $\kappa = \kappa(T)$ has a quadratic temperature dependence $\kappa(T) \propto T^2$, a non-physical result.

This failure is well-known and Prigogine & Glansdorff explained it away by saying that the principle should only be used for small deviations relative to steady-state and the minimizing functional is to depend on two temperature distributions simultaneously, the unknown dynamical and the stable steady state distribution in which the latter temperature distribution is the thermodynamic force (ie., the gradient $X$) while the former is used in the definition of heat transported (ie.,current $J$), see eq 2.8 in [5].

I am wondering if it were possible to save Prigogine's minimum entropy production principle for heat conduction by combining it with Bronsted's "Energetics". I know Bronsted is out of fashion but would you hear me out for a couple of minutes.

According to Bronsted we can calculate the dissipated work (he calls it "lost work") for each interaction as the drop of the extensive quantity $da_0$ from a high potential (intensive) $Y_0$ to a lower one $Y_1$: $\delta w(a)=(Y_0-Y_1)da_0 \ge 0$ He also makes the point that when entropy $dS_0$ is transferred the dissipation is $\delta w(s) = (T_0-T_1) dS_0$ where $\delta q = T_0 dS_0$ is the thermal energy, "heat", transferred into the system from $T_0$ to $T_1$.

For Bronsted the sum of these dissipated terms and then their volume integral over the system represents the total dissipation. If that happens at temperature $T_1$ (system temperature) then the entropy produced within the system is $T_1\sigma = \sum \delta w$, and in the special case when only heat transfer is important then $\sigma = \frac{1}{T_1}(T_0-T_1)dS_0$.

Now Prigogine would rewrite this as $\sigma = \frac{1}{T_1}dS_0=\left(\frac{1}{T_1}-\frac{1}{T_0}\right)\delta q$ and say that the thermodynamic force $X$ is the term $\frac{1}{T_1}-\frac{1}{T_0}$in the big parentheses as it moves "heat" $\delta q$ across the gradient the result of which is the wrong temperature distribution when minimizing entropy production.

But following Bronsted who says the thermodynamic force is really the drop (gradient) of the intensive parameter (pressure, gravitational field potential, temperature, electric/magnetic field potential, etc.) for that is what moves the extensive "charge" (volume, mass, entropy, electric charge, dipole moment, etc.,) and results in "lost work", ie., dissipation.

If we do that then instead of a linear relationship between heat and temperature gradient we should look for a linear law between the transferred entropy and temperature gradient that moves it. In other words we should take as linear constitutive relationship $\delta S = L_s \delta T$ where $L_s=L_s(T)$ is the "new" Onsager conductivity coefficient but now of entropy and not of "heat". With infinitesimal deviation (force) the entropy production is then $\delta \sigma = \frac{L_s}{T}(\delta T)^2$ and its variational integral is $$ \sigma = \int_{T_\ell}^{T_h} \frac {L_s}{T} \left(\frac{dT}{dx}\right)^2 dx \tag{1}\label{1}$$ When this $\eqref{1}$ relationship is used to minimize the entropy production formula we do get is $L_s=\kappa T$ where $\kappa$ (heat conductivity coefficient) is a constant and the minimizing temperature is a linear function of the coordinate $T(x)=T_{\ell}+(T_h-T_{\ell})x$, as was to be expected from Fourier's law, and of course we also have $q=-\kappa \frac{dT}{dx}$.

If this line of thought is correct then the problem is not with the minimum entropy production as such but rather with our choice of thermodynamic force and "charge". Instead of transferring "heat" in our calculation of entropy production we should consider only the transfer of entropy from a given temperature over a given temperature gradient, and then minimize entropy production.

Bronsted emphasizes that his way of calculating total dissipation (he calls it "loss of work") from the product of a transferred "charge" (mas, chemical species, electric charge, dipole moment, entropy, etc.,) falling down a potential drop is correct even when the potential drop $Y_0-Y_1$ or $T_0-T_1$ is not infinitesimal but finite, so may be this is the right thing to do...

I would appreciate any comment you may have as I am trying to understand both Bronsted and Prigogine.

  1. La Mer, Foss, Reiss: Some New Procedures in Thermodynamic Theory Inspired by the Recent work of J. N. Bronsted, ANNALS NEW YORK ACADEMY OF SCIENCES May 1949; https://doi.org/10.1111/j.1749-6632.1949.tb27295.x
  2. Leaf: The Principles of Thermodynamics, J. Chem. Phys. 12, 89 1944 https://doi.org/10.1063/1.1723919
  3. Bronsted: The Fundamental Principles of Energetics, https://doi.org/10.1080/14786444008520665
  4. Palffy-Muhoray: Comment on ‘‘A check of Prigogine’s theorem of minimum entropy production in a rod in a nonequilibrium stationary state’’, Am. J. Phys. 69(7), July 2001; https://doi.org/10.1119/1.1371916
  5. GLANSDORFF, PRIGOGINE: ON A GENERAL EVOLUTION CRITERION IN MACROSCOPIC, Physica 30, 351-374 PHYSICS https://doi.org/10.1016/0031-8914(64)90009-6


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