Equivalence and Distinction of Boltzmann and Gibbs Entropies

Two men have their names on the entropy of statistical mechanics, and if you look closely they wrote down genuinely different formulas. Boltzmann's entropy is a property of one macrostate: count the microstates $\Omega$ that realize it, take $k_{\mathrm B}\ln\Omega$ , done. Gibbs's entropy is a property of a probability distribution $\{p_i\}$ over microstates: $-k_{\mathrm B}\sum_i p_i\ln p_i$ , the same functional Shannon later found for information. One counts a set; the other averages over a cloud. They are not obviously the same object — so why does every textbook use "the entropy" as if there were only one? Because for a macroscopic system they agree to breathtaking accuracy. The number of microstates grows so violently with energy that the equilibrium macrostate crushes every competitor exponentially in $N$ ; the equilibrium ensemble is, for counting purposes, a single spike. Evaluate the Gibbs functional on that spike and you recover $k_{\mathrm B}\ln\Omega$ up to a correction of order $\ln N$ — a rounding error against an entropy of order $N$ . This article proves that agreement, on a paramagnet you can hold in your hand, and shows the gap shrinking as $\ln N/N$ . But an equivalence that holds to order $\ln N$ is a warning as much as a reassurance: the two entropies come apart exactly where $N$ is small, where forces are long-ranged, or where you ask a dynamical question. The fine-grained Gibbs entropy never changes under Hamiltonian flow — Liouville forbids it — so it cannot be the entropy that rises in the second law. And the choice between counting a thin energy shell (Boltzmann) or the whole volume below $E$ (Gibbs's "volume entropy") decides whether negative absolute temperatures are real. Knowing when Boltzmann equals Gibbs, and precisely when it does not, is knowing where statistical mechanics is safe.

⚡ Key Takeaways

Boltzmann entropy $S_{\mathrm B}=k_{\mathrm B}\ln\Omega$ scores a single macrostate by its microstate count; Gibbs entropy $S_{\mathrm G}=-k_{\mathrm B}\sum_i p_i\ln p_i$ scores a whole distribution over microstates.
On the microcanonical distribution ( $p_i=1/\Omega$ ) they are exactly equal; that identity is owned by neighbouring articles and only recalled here.
For a macroscopic system in equilibrium the two agree up to an absolute error of order $\ln N$ , i.e. a relative error of order $\ln N/N\to0$ — the entropic face of ensemble equivalence.
The gap is real and computable: on an $N$ -spin paramagnet the exact microcanonical $\ln\Omega$ and the canonical $-\sum p\ln p$ differ by $\tfrac12\ln(2\pi N pq)$ , a $\ln N$ correction demonstrated numerically here.
They genuinely diverge for small $N$ , long-range forces, non-concave $S(E)$ , and the fine-grained-vs-coarse-grained distinction that drives the second law.
Key idea: Boltzmann and Gibbs entropies are two different functionals that coincide to order $\ln N$ because the equilibrium macrostate dominates the count exponentially — so they agree for thermodynamics but part ways for small systems and for dynamics.

💡 The Core Insight

The Boltzmann entropy of the equilibrium macrostate and the Gibbs entropy of the equilibrium ensemble differ only by a term of order $\ln N$ — negligible against an entropy of order $N$ — because the microstate count is so sharply peaked that one macrostate is effectively the whole ensemble. Where that peaking fails (small $N$ , long range, dynamics), the two entropies separate.

The physical picture

Picture the whole ensemble as a cloud of probability draped over the accessible microstates. Gibbs's entropy measures the spread of that cloud — how many microstates carry appreciable weight, and how evenly. Boltzmann's entropy ignores the cloud's shape and just counts the microstates in the dominant macrostate. When the cloud is a sharp spike sitting entirely inside one macrostate, "measure the spread of the cloud" and "count the microstates under the spike" give the same number. That is the equilibrium case, and it is why the two names became interchangeable. The reason the cloud is a spike is the violence of the counting. The number of microstates with energy near $E$ grows like $e^{cN}$ ; a macrostate even slightly off equilibrium has exponentially fewer, so the canonical weight $e^{-\beta E}\,\Omega(E)$ is a product of a falling exponential and an explosively rising one, pinned to a razor-thin peak. The figure shows this for a paramagnet: the microcanonical picture (a flat plateau of $\Omega$ equally-weighted microstates) and the canonical picture (a Gaussian-sharp distribution over energy) return entropies that sit on top of each other.

Figure

Boltzmann counts the plateau, Gibbs averages the spike — same number for large N

Now the payoff — the actual size of the disagreement, plotted. On a two-level paramagnet the exact microcanonical entropy per spin and the canonical entropy per spin can both be computed in closed form, and their difference is a pure $\ln N$ effect. Divide it by $N$ and watch it die:

Figure

The gap per particle collapses as ln N / N — equivalence made quantitative

The lesson of the second figure is the whole article in one curve: the two entropies never agree exactly for finite $N$ , but their per-particle difference falls like $\ln N/N$ , so by the time $N$ is Avogadro-sized the distinction is $\sim10^{-22}$ and utterly unmeasurable. For a nanomagnet of ten spins, though, the $0.16\,k_{\mathrm B}$ per spin gap is real physics.

The core equations

Start from the two definitions, stated so their different natures are visible. The Boltzmann entropy takes a macrostate — a region of phase space of multiplicity $\Omega$ — and returns a number; it presupposes equal a priori weighting of the microstates it counts:

S_{\mathrm B}(E,V,N)=k_{\mathrm B}\ln\Omega(E,V,N) \qquad (1)

where: $S_{\mathrm B}$ — Boltzmann (surface) entropy; $\Omega$ — the number of microstates on the energy shell $[E,E+\delta E]$ , owned and derived in Boltzmann Entropy; $k_{\mathrm B}=1.380649\times10^{-23}\ \mathrm{J/K}$ (exact, 2019 SI). This is a function of the macrostate alone — no probability distribution appears. The Gibbs entropy takes a distribution and returns a number; it is defined for any ensemble, in or out of equilibrium:

S_{\mathrm G}=-k_{\mathrm B}\sum_i p_i\ln p_i \qquad (2)

where: $S_{\mathrm G}$ — Gibbs entropy; $p_i$ — probability of microstate $i$ ; the sum runs over all microstates, derived and owned in Gibbs Entropy. Read it as an average of the "surprise" $-\ln p_i$ : broad distributions score high, sharp ones score low. On the microcanonical distribution the two collapse into one. Setting $p_i=1/\Omega$ on the $\Omega$ accessible microstates and zero elsewhere:

-k_{\mathrm B}\sum_{i=1}^{\Omega}\tfrac1\Omega\ln\tfrac1\Omega = k_{\mathrm B}\ln\Omega = S_{\mathrm B} \qquad (3)

where: the collapse is exact, not asymptotic — it is the identity that lets us speak of a single entropy for an isolated system. This equality is proved in Gibbs Entropy and Equal A Priori Probabilities; here it is the anchor from which the interesting inequivalence is measured. The substantive comparison is between the canonical Gibbs entropy and the microcanonical Boltzmann entropy — two different ensembles for the same physical system. The bridge is the standard thermodynamic identity, obtained by Laplace-evaluating the partition function:

S_{\mathrm G}^{\mathrm{can}} = \frac{\langle E\rangle - F}{T},\qquad F=-k_{\mathrm B}T\ln Z,\qquad Z=\sum_i e^{-\beta E_i} \qquad (4)

where: $F$ — Helmholtz free energy; $Z$ — canonical partition function; $\beta=1/k_{\mathrm B}T$ . The Laplace/saddle-point evaluation of $Z=\int\!\mathrm dE\,\Omega(E)e^{-\beta E}$ is dominated by the energy $E^*$ where $\beta=\partial_E\ln\Omega$ , tying the canonical entropy back to $\ln\Omega(E^*)$ . Carrying the saddle-point evaluation to the next order produces the central result of this article — the size and sign of the discrepancy:

S_{\mathrm G}^{\mathrm{can}} = S_{\mathrm B}(E^*) + \tfrac12 k_{\mathrm B}\ln\!\big(2\pi k_{\mathrm B}T^2 C_V\big) + \cdots \qquad (5)

where: $C_V$ — heat capacity; the correction is of order $\ln N$ (since $C_V\propto N$ ), so $S_{\mathrm G}^{\mathrm{can}}-S_{\mathrm B}=O(\ln N)$ and the relative gap is $O(\ln N/N)$ . This is equivalence of ensembles wearing its entropy face: the fluctuation term is a $\ln N$ afterthought on an $O(N)$ entropy. Finally the volume entropy, Gibbs's alternative to the surface entropy, counts everything below $E$ rather than a shell at $E$ :

S_{\Sigma}=k_{\mathrm B}\ln\Phi(E),\qquad \Phi(E)=\int_{H\le E}\frac{\mathrm d^{3N}q\,\mathrm d^{3N}p}{N!\,h^{3N}} \qquad (6)

where: $\Phi$ — the phase-space volume enclosed by the energy surface; $S_\Sigma$ agrees with $S_{\mathrm B}$ to $O(\ln N)$ for normal systems but differs in sign of $\partial S/\partial E$ for systems with a bounded spectrum — the root of the negative-temperature debate.

Derivation ABoltzmann equals Gibbs to order ln N — the paramagnet made exact

This is the article's owned structural result: for the two-level paramagnet, both entropies are known in closed form, and their difference is exactly a $\ln N$ term — no hand-waving, no unspecified "corrections." We compute the microcanonical (Boltzmann) entropy from the exact binomial multiplicity, the canonical (Gibbs) entropy from the exact partition function, match them at equal mean energy, and read off the gap. The point is to turn "equivalence up to $O(\ln N)$ " from a slogan into a formula you can evaluate at $N=10$ and $N=10^6$ and watch converge.

Step 1. Count the macrostate. With $N$ non-interacting two-level spins ( $\varepsilon=0$ or $\varepsilon$ ), a macrostate of $n$ excited spins has multiplicity $\Omega(n)=\binom{N}{n}$ , so the Boltzmann entropy per spin is $s_{\mathrm B}=\tfrac1N\ln\binom{N}{n}$ . Why: the macrostate fixes only the number excited; the microstates are the ways to choose which spins are up, a binomial coefficient — this is the counting owned by Counting and Combinatorics of Microstates.

Step 2. Apply Stirling to leading order only, $\ln\binom{N}{n}=N\!\left[-x\ln x-(1-x)\ln(1-x)\right]+O(\ln N)$ with $x=n/N$ . The bracket is the binary entropy $H(x)$ ; the discarded piece is $-\tfrac12\ln\!\big(2\pi Nx(1-x)\big)$ . Why: the extensive part is the physics; the $\tfrac12\ln N$ remainder is precisely the term we are hunting — we keep track of it rather than dropping it (that tracking is owned by The Stirling Approximation).

Step 3. Now the canonical route. The single-spin partition function is $z=1+e^{-\beta\varepsilon}$ , so $Z=z^N$ and the Gibbs entropy is $S_{\mathrm G}^{\mathrm{can}}=k_{\mathrm B}N\big[\ln z+\beta\varepsilon\,\bar n\big]$ with $\bar n=e^{-\beta\varepsilon}/z$ the mean excitation fraction. Written in $x=\bar n$ , $s_{\mathrm G}=H(x)$ exactly — the leading binary-entropy term, with no $\ln N$ piece. Why: the canonical ensemble sums over all energies with smooth weights, so its entropy is a clean per-spin function; the factorial's finite-size tail simply never enters.

Step 4. Match at equal mean energy ( $n=Nx$ in both) and subtract. The difference is entirely the term Stirling left behind:

S_{\mathrm B}-S_{\mathrm G}^{\mathrm{can}} = -\tfrac12 k_{\mathrm B}\ln\!\big(2\pi N x(1-x)\big) + O(1/N)

Why: the extensive $N\,H(x)$ pieces cancel identically, leaving a pure $-\tfrac12\ln N$ correction. Its relative size is $\ln N/(2NH)\to0$ : equivalence. Its absolute size is a fraction of $k_{\mathrm B}$ per system — real for a nanomagnet, invisible for a mole. Note the sign: the surface (Boltzmann) entropy sits below the canonical Gibbs entropy by this half-log, because the shell is a slice of the fluctuating canonical distribution. *Verified: exact $\ln\binom{N}{n}$ vs $N H(n/N)$ and the $\tfrac12\ln(2\pi Nx(1-x))$ gap computed for $N=10,10^2,10^3,10^6$ in verify_aq-sm-boltzmann-gibbs-entropy-comparison.py; per-particle gap falls as $\ln N/N$ . Cf. Pathria §3.7, Kardar §4.*

Derivation BWhy the fine-grained Gibbs entropy cannot be the second-law entropy

The most important distinction between the two entropies is dynamical, and it is a theorem, not a subtlety: under exact Hamiltonian evolution the Gibbs entropy of the full phase-space density is a constant of the motion. It literally cannot increase. Any entropy that rises toward equilibrium — the thermodynamic one — must therefore be a coarse-grained object, not the fine-grained Gibbs functional. This derivation establishes the constancy and names what has to give.

Step 1. Write the Gibbs entropy of a continuous density, $S_{\mathrm G}[\rho]=-k_{\mathrm B}\!\int\rho\ln\rho\,\mathrm d\Gamma$ , and evolve $\rho$ by the Liouville equation $\partial_t\rho=\{H,\rho\}$ . Why: the ensemble is now a smooth density on phase space, transported by the exact flow — the setting owned by Liouville's Theorem and Liouville Equation.

Step 2. The flow is measure-preserving: $\mathrm d\Gamma(t)=\mathrm d\Gamma(0)$ because the Jacobian of Hamiltonian flow is $1$ (Liouville). Along a trajectory $\rho$ is constant, $\rho(\Gamma(t),t)=\rho(\Gamma(0),0)$ . Why: this is exactly the incompressibility proved in Liouville's Theorem; probability rides the flow like an incompressible fluid.

Step 3. Therefore $S_{\mathrm G}[\rho(t)]=-k_{\mathrm B}\!\int\rho\ln\rho\,\mathrm d\Gamma$ is time-independent: change variables to the initial coordinates, unit Jacobian, and the integral is frozen. $\mathrm dS_{\mathrm G}/\mathrm dt=0$ . Why: any functional of $\rho$ that integrates over phase space is conserved when $\rho$ is merely relabelled by a volume-preserving map — the values of $\rho$ are shuffled around, never changed.

Step 4. So the second law's rising entropy is not $S_{\mathrm G}[\rho]$ . It is the coarse-grained entropy $\tilde S=-k_{\mathrm B}\!\int\bar\rho\ln\bar\rho\,\mathrm d\Gamma$ , where $\bar\rho$ is $\rho$ averaged over finite cells. By Jensen's inequality $\tilde S\ge S_{\mathrm G}$ , and as mixing filaments $\rho$ below cell resolution, $\bar\rho\to$ uniform and $\tilde S$ climbs to $S_{\mathrm B}$ . Why: Gibbs's own stirred-ink image — the ink (fine-grained $\rho$ ) is conserved, but the ink blurred to the eye (coarse-grained $\bar\rho$ ) fills the glass. The increase is in the resolution we throw away, and the full story is owned by Statistical Origin of Irreversibility. Verified: $\int\rho\ln\rho$ invariance under a unit-Jacobian (Liouville) map and the Jensen inequality $\tilde S\ge S_{\mathrm G}$ checked on a baker's-map coarse-graining in the verify script. Cf. Gibbs 1902 Ch. XII; Jaynes 1965.

Limiting cases

Case	Result	Meaning
Microcanonical distribution	$S_{\mathrm G}=S_{\mathrm B}=k_{\mathrm B}\ln\Omega$ exactly	on equal-weight microstates the Gibbs functional collapses to the Boltzmann count — no gap at all
Macroscopic equilibrium, $N\to\infty$	$S_{\mathrm B}-S_{\mathrm G}^{\mathrm{can}}=O(\ln N)$ , relative $O(\ln N/N)\to0$	ensemble equivalence: the two entropies agree per particle in the thermodynamic limit
Small $N$ (nanomagnet, cluster)	gap $\sim\tfrac12\ln(2\pi N)$ per system, non-negligible	the $\ln N$ correction is real physics; ensembles are measurably inequivalent
Bounded spectrum, $S(E)$ non-monotone	$S_\Sigma$ (volume) and $S_\Omega$ (surface) disagree on $\mathrm{sign}(\partial S/\partial E)$	the Boltzmann-vs-Gibbs temperature controversy: whether $T<0$ is admissible
Long-range forces / first-order transition	$S(E)$ non-concave; ensembles inequivalent even as $N\to\infty$	negative microcanonical specific heat; canonical ensemble cannot see it

⚠️ “The Gibbs entropy increases as the system equilibrates”

The fine-grained Gibbs entropy $-k_{\mathrm B}\!\int\rho\ln\rho\,\mathrm d\Gamma$ is exactly conserved by Hamiltonian dynamics — Liouville's theorem makes $\mathrm dS_{\mathrm G}/\mathrm dt=0$ an identity. It does not, and cannot, rise as a gas relaxes. What increases is the coarse-grained Gibbs entropy (or, in a reduced one-particle description, Boltzmann's $H$ -functional). Conflating the two is the single most common error in this subject: it makes the second law look like it contradicts mechanics, when in fact the rising entropy lives entirely in the information you discard by coarse-graining. Fine-grained: frozen. Coarse-grained: climbs.

Worked examples

1 · The actual size of the gap for a ten-spin nanomagnet (real number)

Take $N=10$ non-interacting spins at the maximum-entropy point $x=\bar n/N=1/2$ . Compute the exact microcanonical entropy per spin from $\binom{10}{5}$ , the canonical entropy per spin from the binary entropy, and the gap between them. Then repeat at $N=10^6$ to see the collapse.

Step 1. Exact microcanonical count at $n=N/2=5$ : $\binom{10}{5}=252$ , so $S_{\mathrm B}=k_{\mathrm B}\ln 252$ , i.e. $\ln 252=5.5294$ and $s_{\mathrm B}=S_{\mathrm B}/Nk_{\mathrm B}=5.5294/10=\boxed{0.5529}$ per spin.

Step 2. Canonical entropy per spin at $x=1/2$ is the binary entropy $H(1/2)=-\tfrac12\ln\tfrac12-\tfrac12\ln\tfrac12=\ln 2=0.6931$ , so $s_{\mathrm G}=\boxed{0.6931}$ per spin.

Step 3. The gap per spin is $s_{\mathrm G}-s_{\mathrm B}=0.6931-0.5529=0.1402$ . Compare the prediction $\tfrac{1}{2N}\ln(2\pi N x(1-x))=\tfrac{1}{20}\ln(2\pi\cdot10\cdot\tfrac14)=\tfrac{1}{20}\ln(15.71)=\tfrac{1}{20}(2.754)=0.1377$ — agreement to a percent, the small residue being the $O(1/N)$ Stirling tail. At ten spins the two entropies differ by about $0.14\,k_{\mathrm B}$ per spin — a fifth of $\ln 2$ .

Step 4. Now $N=10^6$ : the predicted per-spin gap is $\tfrac{1}{2\times10^6}\ln(2\pi\cdot10^6\cdot\tfrac14)=\tfrac{1}{2\times10^6}(14.27)=7.1\times10^{-6}$ . The distinction that was a fifth of $\ln 2$ at ten spins is seven parts per million at a million — and $\sim10^{-17}$ at a mole. Equivalence is not approximate for macroscopic matter; it is exact for every purpose.

2 · Surface versus volume entropy for an ideal gas (exam level)

For a monatomic ideal gas, $\Phi(E)\propto E^{3N/2}$ (volume below $E$ ) and $\omega(E)=\partial_E\Phi\propto E^{3N/2-1}$ (surface). Show the surface entropy $S_\Omega=k_{\mathrm B}\ln(\omega\,\delta E)$ and the volume entropy $S_\Sigma=k_{\mathrm B}\ln\Phi$ differ only at order $\ln N$ , and that both give the same temperature to leading order.

Step 1. Volume entropy: $S_\Sigma=k_{\mathrm B}\ln\Phi=k_{\mathrm B}\big[\tfrac{3N}{2}\ln E+\text{const}\big]$ . Its temperature: $\tfrac{1}{T_\Sigma}=\partial_E S_\Sigma=k_{\mathrm B}\tfrac{3N}{2}\tfrac1E$ , so $E=\tfrac32 Nk_{\mathrm B}T_\Sigma$ — the equipartition result exactly.

Step 2. Surface entropy: $S_\Omega=k_{\mathrm B}\ln(\omega\,\delta E)=k_{\mathrm B}\big[(\tfrac{3N}{2}-1)\ln E+\text{const}\big]$ . The two logs of $E$ differ by one unit of $\tfrac{3N}{2}$ — i.e. by $\ln E$ , an $O(1)$ piece against the $O(N)$ entropy.

Step 3. Difference: $S_\Sigma-S_\Omega=k_{\mathrm B}\ln E + \text{const}=O(\ln N)$ once $E=\tfrac32 Nk_{\mathrm B}T$ is inserted (so $\ln E=\ln N+\text{const}$ ). The two definitions of "the entropy of the ideal gas" differ by a $\ln N$ term — exactly the size found for the paramagnet, and negligible per particle.

Step 4. Temperatures: $\tfrac{1}{T_\Omega}=k_{\mathrm B}(\tfrac{3N}{2}-1)\tfrac1E$ versus $\tfrac{1}{T_\Sigma}=k_{\mathrm B}\tfrac{3N}{2}\tfrac1E$ ; the fractional difference is $1/(3N/2)=O(1/N)$ . For a normal (monotone- $\Omega$ ) system the choice is immaterial. It becomes decisive only for a bounded spectrum, where $\omega(E)$ turns over and $\partial_E S_\Omega$ can go negative while $\partial_E S_\Sigma$ stays positive — the crux of the negative-temperature dispute.

Edwin Thompson Jaynes

1922–1998 · Stanford and Washington University in St. Louis

Jaynes reframed the whole Boltzmann–Gibbs question as one about information. In two 1957 papers he showed that the Gibbs entropy $-k_{\mathrm B}\sum p_i\ln p_i$ is not a physical assumption at all but the least-biased distribution consistent with known constraints — maximize it subject to fixed $\langle E\rangle$ and out drops the canonical ensemble; fix nothing but normalization and you get the microcanonical uniform distribution, recovering Boltzmann. From this angle the equivalence of the two entropies is the statement that, for a macroscopic system, the constraints pin the distribution so tightly that "count the microstates" and "maximize the information entropy" give the same answer. Jaynes was equally sharp on the distinctions: he stressed that the fine-grained Gibbs entropy is conserved by Liouville flow and that the second law's increase is a coarse-graining effect — the stirred-ink picture made rigorous. His maximum-entropy principle is now a standard tool far beyond physics, in image reconstruction, spectral analysis, and machine learning.

⏳ History

1865

— Clausius names entropy and states the second law in its classic form.

1877

— Boltzmann identifies entropy with the logarithm of a microstate count.

1902

— Gibbs's Elementary Principles in Statistical Mechanics gives the ensemble entropy

-k_{\mathrm B}\sum p\ln p

and the coarse-graining (stirred-ink) resolution of its constancy.

1948

— Shannon derives the same

-\sum p\ln p

functional as the unique measure of information.

1957

— Jaynes recasts statistical mechanics as maximum-entropy inference, unifying the Boltzmann and Gibbs entropies.

1965

— Jaynes's "Gibbs vs Boltzmann Entropies" clarifies which entropy rises in the second law.

Where this lives today

The comparison is not a settled historical footnote — it is an active research front wherever systems are small or forces are long-ranged. Experiments on cold atoms in optical lattices realize bounded-spectrum systems whose apparent temperature depends entirely on whether you use the surface (Boltzmann) or volume (Gibbs) entropy; a 2013 report of negative absolute temperature for motional degrees of freedom, and the vigorous 2014–2016 debate that followed, turned precisely on which $\ln\Phi$ -versus- $\ln\omega$ definition is thermodynamically consistent. The stakes are practical: the two definitions predict different signs of heat flow for such systems. At the nanoscale the $\ln N$ gap this article computes is directly relevant. Single-molecule and single-electron thermodynamics operate at $N$ of tens, where per-particle corrections of $0.1\,k_{\mathrm B}$ are measurable and where ensemble inequivalence — negative microcanonical specific heat in small clusters and in self-gravitating models — is observed rather than hypothesized. Knowing that Boltzmann and Gibbs agree only to order $\ln N$ is what keeps small-system thermodynamics honest.

Check yourself

Going further

Texts. Pathria & Beale, Statistical Mechanics, 3rd ed., §1.4, §3.7 (fluctuations and ensemble equivalence) · Kardar, Statistical Physics of Particles, §4.4–4.6 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 3, 5 · Gibbs, Elementary Principles in Statistical Mechanics (1902), Ch. XI–XIV · Jaynes, "Information Theory and Statistical Mechanics" (Phys. Rev. 106, 620; 1957) and "Gibbs vs Boltzmann Entropies" (Am. J. Phys. 33, 391; 1965).

Provenance. All results are standard. A verifier should confirm: the exact identity $-\sum\frac1\Omega\ln\frac1\Omega=\ln\Omega$ ; the paramagnet gap $\ln\binom{N}{n}-NH(n/N)=-\tfrac12\ln(2\pi Nx(1-x))+O(1/N)$ computed for $N=10,10^2,10^3,10^6$ (verify script); the worked numbers ( $\ln 252=5.5294$ , $s_{\mathrm B}=0.5529$ , $\ln 2=0.6931$ , gap $0.140$ vs prediction $0.138$ at $N=10$ ; $7.1\times10^{-6}$ at $N=10^6$ ); the ideal-gas $\Phi\propto E^{3N/2}$ , $\omega\propto E^{3N/2-1}$ and the $O(\ln N)$ surface–volume gap; the Liouville-invariance of $\int\rho\ln\rho$ . The only fundamental constant is $k_{\mathrm B}=1.380649\times10^{-23}\ \mathrm{J/K}$ (exact, 2019 SI).

Prerequisites. Boltzmann Entropy · Gibbs Entropy · Statistical Ensembles · Microstates and Macrostates · Liouville's Theorem · Fundamental Postulate of Equal A Priori Probabilities. Leads to. Ergodic Theory and Mixing · Accessible Phase Space and Energy Shells · The Ergodic Hypothesis.

If they agree so well, why keep both entropies at all? Isn't one redundant?

Because they answer different questions and only one generalizes. The Boltzmann entropy is defined for a single macrostate of an isolated system; it says nothing about a system in a heat bath, out of equilibrium, or driven. The Gibbs entropy is defined for any probability distribution, so it is the one that survives into the canonical ensemble, non-equilibrium steady states, and information theory. For an isolated macroscopic system in equilibrium they coincide, so you use whichever is easier — usually Boltzmann's count. Everywhere else, Gibbs's functional is the general object and Boltzmann's is the special case.

The article says the gap is "order ln N." Where does the ln N actually come from?

From the finite-size tail of Stirling's approximation. The extensive part of $\ln\binom{N}{n}$ is $N$ times the binary entropy — that piece agrees exactly with the canonical result. But Stirling has a next term, $-\tfrac12\ln(2\pi N x(1-x))$ , coming from $\ln N!\approx N\ln N-N+\tfrac12\ln(2\pi N)$ . That half-log is the whole gap. It reflects the difference between a thin energy shell (Boltzmann) and the full fluctuating distribution over energy (canonical Gibbs): the shell is one slice of a Gaussian of width $\propto\sqrt N$ , and slicing costs you a $\tfrac12\ln N$ .

Is the sign of S_B − S_G physically meaningful, or just bookkeeping?

It is meaningful but tiny. The surface (Boltzmann) entropy sits below the canonical Gibbs entropy by $\tfrac12\ln(2\pi k_{\mathrm B}T^2C_V)$ , because the microcanonical shell corresponds to a single energy while the canonical ensemble spreads over a band of width $\sqrt{k_{\mathrm B}T^2C_V}$ — extra spread means extra entropy. For a macroscopic system this is a $\ln N$ correction to an $N$ entropy and never affects thermodynamics. For a small system it shifts computed free energies and heat capacities at the percent level, which is exactly why nanoscale thermodynamics has to specify which ensemble it means.

How can two "entropies" give different temperatures? Isn't temperature unique?

For normal matter it is — the surface and volume entropies give temperatures differing only at $O(1/N)$ . The definitions part company only for systems with a bounded energy spectrum (spins, some cold-atom lattices), where the density of states $\omega(E)$ rises, peaks, and falls. Past the peak $\partial_E\ln\omega<0$ , so the surface (Boltzmann) temperature goes negative, while the volume entropy $\ln\Phi$ keeps rising and its temperature stays positive. Which is "the" temperature is genuinely debated; the practical test is which one makes the second law and heat flow consistent, and that argument is still live.

Does coarse-graining make the increase of entropy subjective — dependent on how I choose the cells?

The amount of increase depends mildly on the cell size, but the fact of increase does not, and this is not a defect. Any partition into finite cells small compared to macroscopic scales gives a coarse-grained entropy that starts at the fine-grained value and rises to $k_{\mathrm B}\ln\Omega$ as mixing threads $\rho$ below the cell resolution. The endpoint — the equilibrium Boltzmann entropy — is cell-independent; only the transient depends on resolution. Physically, "coarse-graining" is just the statement that real instruments have finite resolution, so the entropy that rises is the one an actual observer can access.

Why does ensemble inequivalence survive the thermodynamic limit for long-range systems but not short-range ones?

Equivalence rests on the entropy density $s(e)=\lim \tfrac1N S(Ne)$ being concave. For short-range forces energy is additive and $s(e)$ is concave, so the microcanonical and canonical descriptions share the same thermodynamics as $N\to\infty$ . Long-range forces (gravity, unscreened Coulomb, mean-field magnets) make energy non-additive; $s(e)$ can develop a convex intruder, and there the microcanonical ensemble shows negative specific heat that the canonical ensemble — which only ever sees the concave hull — cannot reproduce. The two ensembles then disagree about the phase diagram itself, no matter how large $N$ is.

Where does the Shannon entropy fit — is it a third entropy?

No — it is the Gibbs entropy wearing an information-theory hat. Shannon's $-\sum p_i\ln p_i$ (up to the choice of $k_{\mathrm B}$ versus $\log_2$ and a unit) is identical to the Gibbs functional. That is why erasing information has a thermodynamic cost (Landauer's $k_{\mathrm B}T\ln 2$ ) and why Jaynes could rebuild statistical mechanics as inference. So the landscape is really two entropies, not three: Boltzmann's count of a macrostate, and the Gibbs/Shannon functional of a distribution — the same object whether you call its argument a probability or an information weight.