# Gibbs vs. Boltzmann entropies

I was reading Goldstein's paper Boltzmann's Approach to Statistical Mechanics and he makes the following statement:

The Second Law is concerned with the thermodynamic entropy, and this is given by Boltzmann’s entropy (1), not by the Gibbs entropy (2).

On the next page he states:

Certainly contributing to the tendency to identify the thermodynamic entropy with the Gibbs entropy is the fact that for systems in equilibrium the Gibbs entropy agrees with Boltzmann’s entropy.

However a paper by Jaynes showed that the Gibbs and Boltzmann entropy do not agree for a system in equilibrium and in fact the disagreement is proportional to the inter-particle interactions (in general $S_B \geq S_G$). It is made somewhat confusing by the fact that I'm not sure Jaynes and Goldstein are using exactly the same definition of the Boltzmann entropy. But what Goldstein does make clear is that the Boltzmann entropy is calculated from the 1-particle distribution function, as does Jaynes. This must necessarily exclude inter-particle correlations and so I think is subject to Jaynes' critique.

Does anyone know how Goldstein's statement can be justified?

• They are not using the same notion of Boltzmann entropy. Goldstein (see his equ. (1)) is using the logarithm of the volume of the macrostate: this is not based on the 1-particle distribution. Jaynes is using the notion of entropy that Boltzmann used to prove his H-theorem; this one is based on the one-particle distribution. Note that the latter definition is of course not the correct one (the one corresponding to the thermodynamical entropy). – Yvan Velenik Mar 6 '17 at 17:31
• The way Goldstein defines a "macrostate" is based on the 1-particle distribution. He writes: "[P]artition the 1-particle phase space (the q,p – space) into macroscopically small but microscopically large cells $\Delta_{\alpha}$ and specify (approximately, by providing intervals) the number $n_{\alpha}$ of particles in each cell. Each such specification determines a macrostate, the set of all phase points whose particles are distributed in the manner specified." – UtilityMaximiser Mar 6 '17 at 17:59
• You are misunderstanding him (and Boltzmann). He is partitioning the 1-particle phase space, of course, but then the macrostate is characterized by the number of particles in each cell. The latter depends on all the particles. – Yvan Velenik Mar 6 '17 at 18:07
• I don't understand you. The 1-particle distribution is calculated from $N$-particle state. This is true but besides the point. Two microstates which differ in interparticle correlations may be characterised by the same macrostate for the simple fact that 1-particle distributions cannot represent correlations. Information on correlations is necessarily "thrown out" when the 1-particle distribution is calculated. – UtilityMaximiser Mar 6 '17 at 21:27
• Think about it this way: what information on a given microstate do you need to determine the macrostate to which it belongs? You do need to know the simultaneous state of all the particles in the configuration. – Yvan Velenik Mar 7 '17 at 7:28