Nonequilibrium Thermodynamic Entropy is Not Shannon Entropy

Defining entropy out of equilibrium is an outstanding challenge. For certain stochastic lattice models in nonequilibrium steady states, consistent definitions of temperature and chemical potential have been verified; we define the entropy S_th via integration of thermodynamic relations, e.g., (∂S/∂E)_V,N = 1/T. In equilibrium, the thermodynamic entropy equals the Shannon entropy,
S_S = - Σ_C p(C) ln p(C) [the sum is over configurations, with p(C) the probability of configuration C]. Given this, and lacking any specific alternative, the equality S_th = S_S is widely assumed to hold out of equilibrium as well. Here we show by direct calculation of the stationary nonequilibrium probability distribution p(C) of three models - the driven lattice gas with nearest-neighbor exclusion, the Katz-Lebowitz-Spohn driven lattice gas, and the two-temperature Ising model - that S_th ≠ S_S. In the driven models, we find S_S - S_th ∝ D² for small drive D, as expected on grounds of symmetry. Since the thermodynamic entropy of these models is extensive, whereas the Shannon entropy is the (essentially unique) extensive functional of p(C), we are led to conclude that out of equilibrium, thermodynamic entropy cannot be written as a functional of the probability distribution on configuration space.