Gibbs, Boltzmann, and negative temperatures

This claim is not fully substantiated by Chapter XIV of Ref. 17; see pp. 171–172 and Eq. (488) on p. 174.

The crucial word here is “never.”

This statement makes more sense if we introduce the width of the energy shell Δϵ. However, the ergodic hypothesis requires only that Δϵ is so small that ω(E) is constant between E and E + Δϵ. We note that the ergodic hypothesis is not disputed by Dunkel and Hilbert in Ref. 11.

A topical example may clarify this point. Suppose that there is social unrest in the banking sector because bank employees working in branch offices (the “proletariat”) find that their wages are too low, and that those of the executives (the “bourgeoisie”) are too high. This problem could be solved by raising the average salary per employee, while decreasing the average salary (including bonuses) for executives, such that the total sum spent on salaries remains the same. That would be the “Boltzmann” solution. The “Gibbs” solution according to Ref. 11 would be to increase the maximum salary of bank employees, for example, to $1M/yr, a measure from which very few, if any, employees would benefit, whilst decreasing the maximum salary for executives accordingly (say from $10bn/yr to $1bn/yr), such that the maximum amount that could be spent on salaries remains the same. Most likely, very few executives would protest. But more importantly, this measure would not result in any substantial transfer of money (our equivalent of “heat”) from one reservoir to the other. In Marxist terminology, there is no redistribution of wealth from the bourgeoisie to the proletariat. Hence, it is not a strategy that is likely to satisfy the bank employees.

We use throughout the standard definition of Carnot efficiency. As we explain in the text, with reservoirs of opposite temperatures, heat is also extracted from the cold reservoir. If one defines the efficiency to be the ratio of the work done to the heat extracted from both reservoirs, the efficiency can never be greater than one.^7,15

Excepting that a minor improvement is required in the Kelvin-Planck formulation of the second law.^13,15

Writing dF = (N₁/N) dE_M makes this precise: –N₁/N is the analogue of p, and E_M = ϵN is the analogue of V.

With this standpoint, it is then surely illogical to discuss the Gibbs entropy in the context of heat engines.

In the spin system, the Hamiltonian is $H=ϵ∑i=1Msi$⁠, where s_i is zero or one according to whether the ith spin is in the ground state or excited state. Equation (A1) is then satisfied almost trivially since $⟨∂H/∂ϵ⟩=⟨∑i=1Nsi⟩=N1$⁠.

This demonstrates the well-known ensemble inequivalence for small systems, since in that case −∂F/∂T ≠ S_B.

Another proof proceeds by considering in addition to the Gibbs entropy $SG=kBlnTr[Θ(E−H)]$ the complementary entropy $S¯G=kBlnTr[Θ(H−E)]$⁠. By adapting Eq. (7) in Ref. 11, one can easily show that $S¯G$ also meets the consistency criterion. Then, from S_G and $S¯G$⁠, a piecewise approximation to S_B can be constructed which becomes exact in the thermodynamic limit.