Boltzmann's Entropy Hides an Ergodic Assumption That Gibbs's Does Not — Epoche B2
Review: The Assumption Hidden in Boltzmann's Entropy Two formulas for entropy sit at the heart of statistical physics. The first, carved on Boltzmann's gravestone, is $S = k\ln W$; the second, due to Gibbs [1] , is $S = -k\sum_i p_i \ln p_i$. Textbooks usually introduce the first, apply it, and mention the second only in passing: Kittel and Kroemer [2] , to take a standard one, define entropy as the logarithm of a multiplicity and build the whole subject out of that count, the probability-weighted form arriving later and briefly. The impression left is that the two are interchangeable. They are not. This review builds both from the ground up, shows exactly when they agree, and argues that Boltzmann's version carries a silent assumption — that the system explores all of its accessible microstates evenhandedly — which fails for a large and physically important class of systems. Gibbs' formula, by contrast, needs no such assumption and is the more general definition. Microstates, macrostates, and counting A microstate is a complete specification of a system: the position and momentum of every particle, or the orientation of every spin. A macrostate is what we actually control or measure — typically the total energy $E$, the volume $V$ and the particle number $N$. An enormous number of microstates are compatible with any one macrostate, and Boltzmann's insight was to identify entropy with a count of them, $$ S = k\ln W, $$ where $W$ is the number of microstates consistent with the macrostate and $k = 1.38\times10^{-23}\,\text{J K}^{-1}$ is Boltzmann's constant (the logarithm is natural throughout). Taking the logarithm turns a multiplicative count into an additive quantity, so that the entropy of two independent systems is the sum of their entropies. A concrete case makes $W$ tangible: $N$ non-interacting spin-½ moments in zero field each have two allowed orientations, so $W = 2^{N}$ and $$ S = k\ln 2^{N} = Nk\ln 2 . $$ For one mole this is $R\ln 2 \approx 5.8\,\text{J K}^{-1}\,\text{mol}^{-1}$. Boltzmann's formula is the bridge from such a microscopic count to the thermodynamic entropy defined by Clausius, $dS = dQ_\text{rev}/T$. The equal-probability postulate, and ergodicity Counting states treats every accessible microstate as carrying the same weight. That is a physical assumption, not a definition. It is the postulate of equal a priori probability : an isolated system in equilibrium is found in each of its $W$ accessible microstates with the same probability, $p_i = 1/W$. What licenses this even weighting? The usual justification is the ergodic hypothesis — that over a long enough time the system's trajectory wanders through phase space so thoroughly that the fraction of time it spends in any region equals that region's share of the accessible states. Time averages then coincide with averages over the uniform distribution, and $S = k\ln W$ follows. Ergodicity is the assumption hidden inside Boltzmann's formula. When the distribution is not uniform: Gibbs' entropy Uniform weighting is special. Consider a system in contact with a heat bath at temperature $T$, the setting of most real experiments. It is no longer isolated, its energy fluctuates, and the probability of a microstate $i$ of energy $E_i$ is the canonical (Boltzmann) distribution $$ p_i = \frac{e^{-E_i/kT}}{Z}, \qquad Z = \sum_i e^{-E_i/kT}, $$ where $Z$, the partition function, is the sum that normalises the probabilities to one. These weights are manifestly not uniform: high-energy microstates are exponentially suppressed. A bare count $k\ln W$ cannot capture this. Gibbs' definition can, because it feeds each microstate its own probability, $$ S = -k\sum_i p_i \ln p_i , $$ where the sum runs over every microstate and $p_i$ is the probability of finding the system in state $i$. The factor $-\ln p_i$ measures the "surprise" of state $i$; the entropy is $k$ times its average, so entropy is large when probability is spread thinly over many states and small when a few states dominate. Substituting the canonical weights reproduces the whole of equilibrium thermodynamics: with $U = \sum_i p_i E_i$ the mean energy and $F = -kT\ln Z$ the Helmholtz free energy, a short calculation gives $$ S = \frac{U - F}{T}, $$ which is simply the thermodynamic identity $F = U - TS$ rearranged. Gibbs' entropy delivers the correct free energy for a non-uniform distribution on which Boltzmann's counting has no direct grip. Boltzmann as the equilibrium special case The two formulas are not rivals; one contains the other. Set the distribution uniform, $p_i = 1/W$ for the $W$ accessible states and $p_i = 0$ otherwise, and Gibbs' sum collapses: $$ S = -k\sum_{i=1}^{W}\frac{1}{W}\ln\frac{1}{W} = -k\,W\cdot\frac{1}{W}\,(-\ln W) = k\ln W . $$ Boltzmann's entropy is exactly Gibbs' entropy evaluated on the uniform distribution — the distribution the equal-probability postulate hands us in equilibrium. Where that postulate holds, the two agree identically; where it fails, only Gibbs' form remains defined. Jaynes pressed the same comparison from the kinetic side [3] : the Gibbs expression reproduces the measured entropy of a gas whose molecules interact, while the entropy built from Boltzmann's single-particle distribution function agrees with it only in the dilute limit. Where the assumption fails: broken ergodicity Ergodicity is not a theorem but a property, and many systems lack it. Cool a liquid quickly past its freezing point and it may form a glass : the molecules jam before they can find the crystalline arrangement, and the system is stranded in one small pocket of configuration space, unable to surmount the tall energy barriers separating it from the rest. Spin glasses, supercooled liquids, and any ferromagnet below its Curie temperature behave the same way — the accessible phase space fractures into basins, and the trajectory samples only the basin it started in (see figure). The "all $W$ microstates" that Boltzmann's count presumes are then never visit