Mind the Gap:
Boltzmannian versus Gibbsian Equilibrium

Charlotte Werndl
Department of Philosophy (KGW), University of Salzburg

Franziskanergasse 1, 5020 Salzburg Austria and
Department of Philosophy, Logic and Scientific Method

London School of Economics, Houghton Street,
WC2A 2AE London; charlotte.werndl@sbg.ac.at

Roman Frigg
Department of Philosophy, Logic and Scientific Method,

London School of Economics, Houghton Street,
WC2A 2AE London; r.p.frigg@lse.ac.uk

Abstract

There are two main theoretical frameworks in statistical mechanics, one associated with

Boltzmann and the other with Gibbs. Despite their well-known differences, there is a

prevailing view that equilibrium values calculated in both frameworks coincide. We show

that this is wrong. There are important cases where the Boltzmannian and Gibbsian

equilibrium concepts yield different outcomes. Furthermore, the conditions under which

equilibria exists are different for Gibbsian and Boltzmannian statistical mechanics. There

are, however, special circumstances under which it is true that the equilibrium values

coincide. We prove a new theorem providing sufficient conditions for this to be the case.



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1 Introduction

There are two main theoretical frameworks in statistical mechanics (SM), one associated with

Boltzmann and the other with Gibbs. One of the crucial differences is the characterisation of

equilibrium. While in the Boltzmannian framework equilibrium corresponds to the largest

macro-region, the Gibbsian framework associates equilibrium with the stationary probability

distribution of maximum entropy. The Boltzmannian picture is usually accepted as correct

when it comes to answering foundational questions, in particular in connection with the

approach to equilibrium. At the same time the Gibbsian framework is dismissed as ‘thoroughly

misguided’ (Goldstein 2001, 39) and stands accused of introducing theoretical instruments that

we ‘neither have nor [...] need’ (Lebowitz 1993, 38). Yet, the Gibbsian framework is the

undisputed workhorse of the practitioner.

This would be not be particularly worrisome if the two formalisms were equivalent or

inter-translatable as, for instance, the Schrödinger and the Heisenberg picture in quantum

mechanics. But they are not. Not only do they disagree on how equilibrium is conceptualised;

they do not even study the same entities. While the Boltzmannian framework investigates

individual systems, the object of study in the Gibbs approach are ensembles, and there is no

obvious way to translate results from one framework into the other. This creates an awkward

situation: how can we think that the Boltzmannian framework gives the right foundational

account of SM and at the same time rely on the Gibbsian framework for calculations if the two

accounts are in fact at odds with each other?

A common response is to play down the severity of the problem with the argument that the

two frameworks lead to the same results. One can, so the argument goes, use the Gibbs

formalism as effective tool for the calculation of equilibrium values of a wide spectrum of

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physical quantities, because these coincide with the values that would come out of the

Boltzmann machinery if one was able to make the calculations. However, the reasoning behind

this assertion remains unclear. Either it is simply asserted as an obvious truism (Davey 2016

and Wallace quoted in Werndl and Frigg (forthcomingc)), or arguments are given only for

special cases (Lavis 2005 and Malament and Zabell 1980). So the claim that Gibbsian and

Boltzmannian equilibrium values generally coincide has the status of an article of faith. Faith

need not be wrong, but it can be. The project for this paper is to investigate whether the

purported equivalence of results really holds.

Our answer is negative. If understood as a general proposition, the claim is provably wrong.

After briefly introducing the Boltzmannian and the Gibbsian notions of equilibrium (Section 2)

we present two examples in which Boltzmannian and Gibbsian equilibrium values come apart:

the baker’s gas (Section 3) and the Ising model (Section 4). The differences between the two

frameworks have gone unnoticed partly because Boltzmannian and Gibbsian calculations are

seldom carried out on the same systems. We make a first step towards rectifying this situation

by discussing our examples from both theoretical vantage points, which makes visible how the

two frameworks differ. Things get worse still. Not only do equilibrium values fail to coincide;

equilibria don’t necessarily exist under the same conditions. There are systems that have a

Boltzmannian equilibrium but fail to have a Gibbsian equilibrium and vice versa (Section 5).

This raises the question whether there is even a grain of truth in common wisdom, and the good

news is that there is. We prove a new theorem establishing that under certain special

circumstances the results of the two formalisms indeed coincide (Section 6). We describe these

circumstances in some detail and give examples. We end by offering some conclusions

(Section 7).

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2 Equilibrium in Gibbs and Boltzmann

In this section we briefly introduce the Boltzmannian and the Gibbsian notions of equilibrium

(for details see Uffink, 2007; Frigg, 2008). Throughout the paper we consider systems with a

phase space X. Let µX denote the probability measure on X. Further, let Tt(x) denote the state

of the system after t time units that starts in x. The dynamics Tt is usually deterministic.

Sometimes, for instance in the Ising model, dynamics are considered that are stochastic

processes {Zt}. For ease of presentation we state general definitions and results for the

deterministic case; but all definitions have stochastic equivalents and the results equally also

hold in the stochastic case (see Werndl and Frigg (forthcominga) for a discussion of stochastic

systems).

In Gibbsian SM the object of study is an ensemble, an infinite collection of independent

systems that are all governed by the same equations but are in different states. The ensemble is

described by a probability density ρ(x, t), x ∈ X, reflecting the probability of finding the state of

a system chosen at random from the ensemble in a certain part of X at time t. Physical

observables are associated with real valued functions f : X × t →R. The phase average of

such a function is:

f̄ (t) =
∫

X
f (x, t)ρ(x, t)d x. (1)

According to the standard understanding of the formalism, phase averages are observed in

experiments. The Gibbs entropy of a distribution is

S G[ρ] = −k
∫

X
ρ(x, t) log[ρ(x, t)]d x, (2)

where k is Boltzmann’s constant. A distribution ρ(x, t) is stationary if it does not depend on

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time: ρ(x, t) = ρ(x) for all t. In Gibbsian SM equilibrium is the property of an ensemble. The

ensemble is in equilibrium if the distribution is stationary and has maximum entropy given the

constraints imposed on the system. The most common equilibrium distributions are the

microcanonical, canonical and grand-canonical distributions (Uffink 2007).

It is customary to begin a presentation of Boltzmannian SM with the combinatorial

argument. However, it is now recognised that combinatorial considerations do not provide a

good general definition of Boltzmannian equilibrium (Uffink 2007) and we therefore work with

our own alternative definition (Werndl and Frigg 2015a, 2015b).1 Consider the same dynamical

system as above and assume that the measure µX is stationary. At the macro level the system is

characterised by a set of macro-variables {v1, ..., vl} (l ∈N). They are measurable functions

vi : X →R, associating a value with each x ∈ X. We use capital letters Vi to denote the values

of vi. A particular set of values {V1, ..., Vl} defines a macro-state MV1,...,Vl . The region of phase

space corresponding to the macro-state MV1,...,Vl , its macro-region, is denoted by ZMV1,...,Vl .

The equilibrium macro-state is defined as the state in which a system spends most of its

time. Let LFR be the fraction of time a system spends in region R ∈ X in the long run:

LFR(x) = lim
t→∞

1
t

∫ t
0

1A(Tτ(x))dτ, (3)

where 1A(x) is the characteristic function of R: 1A(x) = 1 for x ∈ R and 0 otherwise. The notion

of ‘most’ can be read in two different ways, leading to two different notions of equilibrium.

The first introduces a lower bound of 1/2 for the fraction of time spent in equilibrium, leading

to the notion of an α-ε-equilibrium:

1Combinatorial considerations are of course still useful to construct macro-states, and we

rely on them below.

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Let α be a real number in the interval ( 12, 1], and let ε be a very small positive real

number.2 If there is a macro-state MV∗1,...,V∗l satisfying the following condition, then

it is the α-ε-equilibrium state of S : There exists a set Y ⊆ X such that

µX (Y ) ≥ 1 −ε, and all initial states x ∈ Y satisfy LFXMV∗1 ,...,V∗l
(x) ≥ α.

The second reading takes ‘most’ to refer to the fact that the system spends more time in

equilibrium than in any other state (this can be less that 50% of its time). This provides the

γ-ε-equilibrium:

Let γ be a real number in (0, 1] and let ε be a small positive real number (ε < γ). If

there is a macro-state MV∗1,...,V∗l satisfying the following condition, it is the γ-ε

equilibrium state of S : There exists a set Y ⊆ X such that µX (Y ) ≥ 1 −ε and for all

initial conditions x ∈ Y : LFZMV∗1 ,...,V∗l
(x) ≥ LFZM(x) + γ for all macro-states

M , MV∗1,...,V∗l .

It is obvious that on both notions of equilibrium the value associated with the equilibrium

macro-state is the observed value in equilibrium. It can be proven that equilibrium states thus

defined are the largest states in the system in the following sense: their measure is larger than

α(1 −ε) > 1/2 for an α-ε-equilibrium and their measure is γ−ε larger than the measure of any

other macro-region for a γ-ε-equilibrium.3

2We assume that (1 −α)ε > 1/2.
3Proofs for the deterministic case are given in Werndl and Frigg (2015b) and for the

stochastic case in Werndl and Frigg (forthcominga). The problem of the existence of such

equilibrium states is discussed in Werndl and Frigg (forthcomingb).

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3 Example 1: The Baker’s Gas

The baker’s gas consists of N identical particles that evolve independently according to the

baker’s transformation (Lavis 2005). Its micro-states are of the form x = (b1, c1, . . . , bN, cN ),

where bi ∈ [0, 1] is the momentum and ci ∈ [0, 1] is the position coordinate of the i-th particle.

The system’s phase space therefore is X = [0, 1]2N , which is endowed with the uniform

probability measure µX (the 2N-dimensional Lebesgue measure). Time is discrete and the

evolution to the next time step is given by applying the baker’s transformation to each

coordinate: x = (. . . bi, ci . . .) evolves into Λ(x) = (. . .θ(bi, ci) . . .), where

θ(bi, ci) = 2bi,
ci
2

if 0 ≤ bi ≤
1
2

and 2bi − 1,
ci + 1

2
otherwise. (4)

Let us begin with the Boltzmannian treatment of the baker’s gas. Here µX is the stationary

measure. We now use combinatorial considerations to construct the system’s macro-state.

Consider a partition of the unit square (the phase space for one particle) into cells of equal size

δω whose dividing lines run parallel to the position and momentum axes. This results in a finite

partition Ωbg := {ω
bg
1 , ...,ω

bg
k }, k ∈N. The coarse-grained micro-state of a particle is the cell in

which a particle’s state lies. An arrangement is given by a specification of the coarse-grained

micro-states of all the particles. A distribution is a specification of how many particles’ states

lie in a given cell. Consider the distribution Dbg = (N1, N2, . . . , Nk), where Ni is the number of

particles in cell ωi. The number G(Dbg) of arrangements that lead to the same distribution Dbg

is G(Dbg) = N! / N1!N2! . . . , Nk!.

One can now define a partition on X by grouping together in one cell all points that have the

same distribution. It is easy to see that the cell Xu corresponding to the uniform distribution (i.e

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where Ni = N/k)4 is larger than any other cell. We now introduce a macro-variable V as

follows: V (x) = 0 for x ∈ Xu and for all other cells of the partition V (x) takes values between

106 and 108 so that no two cells have the same value. The baker’s gas is ergodic (Lavis 2005),

and hence spends more time in Xu than in any other cell. Therefore the long run fraction of

time for which the value of V is 0 is larger than the long run fraction for any other value. Thus

the macro-state defined by V = 0 is a γ-0-equilibrium, and V = 0 is the Boltzmannian

equilibrium value.5

Let us now turn to the Gibbsian treatment of the baker’s gas. The stationary maximum

entropy distribution is the uniform distribution ρ(x) = 1. The phase space average V̄ for the

macro-variable V will be greater than (1-µX (Xu)) × 106. Lavis (2005) has shown that

µX (Xu) ≈ 0.47 (for large N), and hence 0.53 × 106 = 530000 is a lower bound for V̄ .

So we find Boltzmannian and Gibbsian equilibrium values that are very different! The

macro-variable of this system is admittedly contrived and so one might argue that the problem

does not arise in practice. The grain of truth in this remark is that much depends on the choice

of the macro-variable and for sufficiently restrictive classes of macro-variables the problem can

indeed be avoided (see Section 6 for a characterisation of such classes). However, there are real

physical systems that do not fall into these classes. Thus the relevant contrast is not between

‘mathematical contrivance’ and ‘sensible physics’. The example we discuss in the next section

is a case in point.

4We assume that N = k × r for some r ∈N.
5It is not an α-ε-equilibrium because the equilibrium macroregion takes up less than half of

the phase space for large N (Lavis 2005).

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4 Example 2: The Ising Model

The two-dimensional Ising model is an important system in SM. We here consider a version of

the model with nearest neighbour interactions in the absence of an external field. Despite being

only two-dimensional, this model provides a realistic description of crystals such as K2NF4 and

RB2MnF4, which have strong horizontal and weak vertical interactions (Baxter 1982).

Consider a regular two-dimensional lattice with N grid points. The lattice is assumed to lie

on a two dimensional torus so that every grid point has exactly four nearest neighbours

(allowing us to neglect border effects). At every grid point there is a spin pointing either up

(σ = 1) or down (σ = −1). The system’s micro-state is given by

σ = {σ1, . . . ,σN}, (5)

and its Hamiltonian is:

H(σ) = −J
∑
nn

σiσ j. (6)

The sum is over all nearest-neighbour pairs and the constant J ≥ 0 is the energy associated with

the nearest-neighbour interaction (Baxter 1982).

We treat the model stochastically and hence we first introduce probabilities. The probability

distribution has the form of a Gibbsian distribution, but it is important to emphasise that at this

point this is nothing more than a formal definition that is neither Gibbsian nor Boltzmannian.

We begin with the partition function:

Z =
∑
σ

e−βH(σ). (7)

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The sum is taken over all possible configurations σ of the model and β = 1/kT is a constant,

where T is the temperature and k is Boltzmann’s constant. The probability of finding the

system in a certain configuration σ is given by the canonical distribution:

P(σ) =
e−βH(σ)

Z
. (8)

For large values of β (low temperature), the probabilities of the lower energy states are

dominant. For small values of β (high temperature) the probability distribution is flattened out

and all configurations are more or less equally likely (Baxter 1982; Cipra 1987).

The Boltzmannian treatment is as follows. The probability P(σ) is the stationary measure

which defines the dynamics of the stochastic dynamical system of Section 2. It can be shown

that this dynamics is in fact an irreducible Markov chain, the stochastic equivalent to ergodicity

(Berger 2001). As the relevant macro-variable we choose the internal energy:

E(σ) =
∑
n,n

−Jσiσ j, (9)

where the sum is over all nearest-neighbour interactions. Let σ̄ be the micro-state for which

σi = 1 for all i, and let σ̂ be the micro-state for which σ j = −1 for all j. A := {σ̄, σ̂} is then the

macro-state for which the internal energy has value E = −2JN (because there are 2N

nearest-neighbour pairs for a quadratic lattice with N sites for N ≥ 9). As noted above, the

higher β the larger the probabilities of the lower energy states. Since the dynamics is

irreducible, for large enough β the system spends most of its time in A. Hence A is a

Boltzmannian γ-0-equilibrium and the value of the internal energy in equilibrium is −2JN.

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Let us now turn to the Gibbsian treatment. Here P(σ) is the stationary measure of

maximum entropy. As above the value of E is lowest for σ̄ and σ̂, namely −2JN. Since all

other micro-states have higher energy, the Gibbsian phase average of E over all micro-states is

higher than −2JN (Baxter 1982; Cipra 1987; Sekular 2015).

It follows that the equilibrium value of the internal energy in a Boltzmann equilibrium

(−2JN) is lower than in Gibbsian equilibrium. This difference arises because the

macro-variable takes the lowest value in the largest macro-region and a higher value in all other

macroregions.

The differences between the equilibrium values is not negligible. By way of illustration

consider the case of a two-dimensional lattice with four grid-points: σ = (σ1,σ2,σ3,σ4) where

(σ1, σ2) constitute the first row and (σ3, σ4) constitute the second row. Let β = 1/2 and J = 1.

There are four nearest-neighbour pairs. The Boltzmannian equilibrium macro-state is

{(1, 1, 1, 1), (−1,−1,−1,−1)}, and the equilibrium value is −4. The value obtained by phase

space averaging can be obtained as follows. There are two states which have energy −4 and

whose probability is e4/2/Z ≈ 7.389/Z: (1, 1, 1, 1), (−1,−1,−1,−1). There are twelve states

which have energy 0 and whose probability is e0/2/Z = 1/Z: (−1, 1, 1, 1), (1,−1, 1, 1),

(1, 1,−1, 1), (1, 1, 1,−1), (−1,−1,−1, 1), (−1,−1, 1,−1), (−1, 1,−1,−1), (1,−1,−1,−1),

(1, 1,−1,−1), (−1,−1, 1, 1), (1,−1, 1,−1), (−1, 1,−1, 1). Finally, there there are two states

which have energy 4 and whose probability is e−4/2/Z ≈ 0.1353/Z: (−1, 1,−1, 1), (1,−1, 1,−1).

Hence the phase average is:

−4 × 2 × 7.389 + 0 × 12 × 1 + 4 × 2 × 0.1353
2 × 7.389 + 12 × 1 + 2 × 0.1353

= −2.1454 (10)

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So the Gibbsian equilibrium value (-2.1454) is almost half of the Boltzmannian equilibrium

value (−4).

Nothing depends on the choice N = 4. It remains the case for large N that Gibbsian phase

averages will be different from the Boltzmannian equilibrium values. Indeed this situation

remains unchanged even in the limit N →∞, where one also finds that the Boltzmannian and

Gibbsian values differ.6

5 Existence Under Different Conditions

We have shown that Gibbsian and Boltzmannin equilibrium values fail to coincide. And things

get worse: there are systems that have a Boltzmannian equilibrium but fail to have a Gibbsian

equilibrium and vice versa.

5.1 Gibbs does not imply Boltzmann

Consider the baker’s gas (Section 3) with an even number of particles with one macro-variable

V indicating whether there are more particles on the left side of the container than on the right

side: V assumes value 1 if there are more particles on the left and it assumes value 0 if there are

more particles on the right. The corresponding macro states are M1 and M0. Both

macro-regions have the same measure, namely 1/2. The dynamics is ergodic and therefore the

system spends half of the time in M0 and half of the time in M1. For this reason the system has

no Boltzmannian equilibrium (neither of the α-ε nor of the γ-ε kind). However, there exists a

6More specifically, one finds that for arbitrary large N the macro-value closest to the value

obtained by Gibbs phase space averaging will always be different to the macro-value

representing the Boltzmannian equilibrium.

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Gibbs equilibrium: the uniform measure µB is the stationary distribution of maximum entropy.

The Ising Model (Section 4) provides another example. Consider again the above example

with four sites and assume β = 0.7 Suppose that there are two macro-states: (i) one where all

the molecules point upwards, all the molecules point downwards or there are an equal number

of molecules that point upwards and downwards (eight micro-states in total), and (ii) where this

is not the case (again eight micro-states in total). The usual dynamics considered for the Ising

models are irreducible Markov chains. Recall that irreducibility is the stochastic equivalent to

ergodicity, and hence, as in the previous example, there is no Boltzmannian equilibrium

because both macro-regions have the same measure 1/2. However, a Gibbsian equilibrium

exists. The probability P is the canonical distribution, which is the stationary distribution of

maximum entropy.

5.2 Boltzmann does not imply Gibbs

Boltzmannian equilibrium is defined for systems with a stationary measure µX , and so there

always exists at least one stationary distribution. Nevertheless, a stationary distribution of

maximum entropy need not exist, as the following example shows.

Consider a system with phase space [0, 2] × [0, 1]. The dynamics Tt of the system consists

of two ‘copies’ of the baker’s gas in the following sense: the restriction of Tt to [0, 1] × [0, 1] is

the standard baker’s transformation (see Section 3) and the restriction of Tt to (1, 2] × [0, 1] is

the baker’s transformation with the position coordinate shifted by one unit to the right. The

7This corresponds to an infinite temperature, which is often considered as an approximation

to very high temperatures (e.g. Baxter 1982). We work with β = 0 for reasons of simplicity;

similar examples could be given for low but non-zero values of β.

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measure of the system is 1/3 × 1[0,1]×[0,1] + 2/3 × 1(1,2]×[0,1], where 1A is the characteristic

function of A. Since the baker’s system is ergodic, clearly, the phase space consists of two

components ([0, 1] × [0, 1] and (1, 2] × [0, 1]), restricted to each of which the dynamics is

ergodic. Suppose that the macro-variable takes the value 0 if the position coordinate is in

[1/4, 7/4], 100 if the position coordinate is in [0, 1/4), −100 if the position coordinate is in

(7/4, 2]. Clearly, there exists a Boltzmannian equilibrium. The value 0 corresponds to an

0.75-0-equilibrium: since the system is ergodic on each component and the 0 macro-state takes

up three-quarters on each component, three quarters of the time the system takes the value 0.

Yet suppose that the class of distributions of interest are all densities over [0, 2] × [0, 1] of

the form α1[0,1]×[0,1] + β1(1,2]×[0,1] where α,β ≥ 0, α + β = 1 and the uniform distribution (the

case α = β = 1/2) is excluded. Then there is no Gibbsian equilibrium. It is clear that all

distributions are stationary (because µB of the baker’s gas is stationary), but by construction,

there is no distribution of maximum entropy: the closer α and β are to 1/2, the higher the

entropy (2). Yet there is no maximum since the uniform distribution (α = β = 1/2) is not

among the class of distributions under consideration.

6 When Boltzmann and Gibbs Agree

This section focuses on special cases where the Boltzmannian and Gibbsian calcuations agree.

We first consider the main case discussed in the literature. Then we present a new theorem

specifying a set of conditions under which Boltzmannian and Gibbsian values coincide. We

show that many standard examples satisfy these conditions and that they fail in the cases

discussed in previous sections.

Suppose that the relevant observable is such that it takes the value of the phase average

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nearly everywhere on phase space. Furthermore assume that a Boltzmannian equilibrium

exists. In such a situation it will be the case that the value of the observable in the

Boltzmannian equilibrium macro-state is equal to the phase average. The question is under

what circumstances something like this is the case. One such situation is described by

Khinchin (1949). He argued that phase functions have to satisfy strong symmetry requirements

and therefore ought to have small dispersion for systems with a large number of constituents

(we refer to this as the ‘Khinchin condition’). This, or something like it, is often taken to be an

explanation of why calculations in both framworks coincide (Davey 2015; Malament and

Zabell 1980; Wallace quoted in Werndl and Frigg forthcomingc).

This is correct, but the question is how far it goes. The point to note is that the conditions

not only rule out artificial examples, but also realistic physical models. Examples of systems

with macro-variables that do not satisfy the Khinchin condition include the Ising model with

the internal energy or the magnetisation as macro-variable, the six-vertex model with the

internal energy or the polarization as macro-variable (cf. Baxter 1982), and the KAC-ring with

the standard macrostate structure of the number of black and white balls.

So we need other conditions next to the Khinchin condition. Let us look at a situation in

which a system has both a Boltzmann and a Gibbs equilibrium. In this case the following

theorem provides sufficient conditions for the equilibrium values of both equilibria to coincide:

Equilibrium Equivalence Theorem (EET). Suppose that the system (X, Tt,µX ) is

composed of N ≥ 1 constituents. That is, the state x ∈ X is given by the N

coordinates x = (x1, . . . , xN ); X = X1 × X2 . . .× XN , where Xi = Xoc for all

1 ≤ i ≤ N (Xoc is the one-constituent space). Let µX be the product measure

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µX1 ×µX2 . . .×µXN , where µXi = µXoc is the measure on Xoc. Suppose that an

observable κ is defined on the one-particle space Xoc and takes the values κ1, . . . ,κk

with equal probability 1/k, k ≤ N.8 Suppose that the macro-variable K is the sum

of the one-component observable, i.e. K(x) =
∑N

i=1 κ(xi). Then the value

corresponding to the largest macro-region as well as the value obtained by phase

space averaging is Nk (κ1 + κ2 + . . .κk).

The proof is stated in full in the Appendix; an intuitive sketch is as follows. Because the

observable on the one-constituent space takes the values {κ1, . . . ,κk} with equal probability,

combinatorial considerations show that the Boltzmannian equilibrium macro-region, i.e. the

macro-region of largest size, is the one where there are N/k constituents taking the value κ1,

N/k constituents taking the value κ2, . . . , N/k constituents taking the value κk. That is, the

Boltzmannian equilibrium value is (κ1 + . . . + κk)N/k. The Gibbsian equilibrium value obtained

by phase averaging is also (κ1 + . . . + κk)N/k because one simply takes the average over all

sequences of the {κ1, . . . ,κk} and the κi are equally probable.

The proof does not make any assumptions about the dynamics of the system; in particular it

doesn’t assume that the system is ergodic. The crucial assumptions of the theorem are (i) that

the macro-variable is a sum of the observable on the constituent space and (ii) that the

macro-variable on the constituent space corresponds to a partition with cells of equal

probability. This theorem is important because it applies to many examples in SM. Consider,

for instance, the baker’s gas discussed in Section 3. The gas involves a partition of the unit

square (the phase space for one particle) into cells {ωbg1 , ...,ω
bg
k }, k ∈N, of equal size δω.

Suppose that a particle in ωbgi takes value κi for all i = 1, ..., k. Now the macro-variables often

8 N is assumed to be a multiple of k, i.e. N = k × s for some s ∈N.

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considered for the baker’s gas are of the form K(x) =
∑N

i=1 κ(xi) (e.g. Lavis 2005).
9 Then the

assumptions of Theorem 1 are satisfied. Thus the Boltzmannian equilibrium value is the same

as the value obtained by Gibbsian phase space averaging, namely (κ1 + . . . + κk)N/k.

The KAC-ring with the standard macro-state structure given by the magnetisation and the

ideal gas with the standard macro-states afford further examples of systems satisfying the

conditions of the theorem (these examples are discussed in Werndl and Frigg 2015a). In the

Boltzmannian framework one often considers macro-variables of the type as assumed in the

theorem (Frigg 2008, 110). In these cases the Gibbsian and Boltzmannian equilibrium

calculations lead to the same results.

The examples discussed earlier in the paper, showing that equilibrium calculations in

Boltzmannian and Gibbsian SM do not lead to the same results, violate the relevant

assumptions. In the baker’s system with the macro-variables as discussed in Section 3 and in

the Ising model with the internal energy as an observable significant chunks of the phase space

are taken up by non-equilibrium states, which results in the Khinchin condition not being

satisfied. These two examples do not satisfy the assumptions of EET either because the

macro-variables of both the baker’s gas and the Ising model are not the sum of a

one-component macro-variable whose outcomes have equal probability. What does the work in

these two examples is that significant parts of the phase space are taken up by non-equilibrium

states and that the Boltzmannian equilibrium state has the lowest value of the macro-variable.

This results in the phase average being different from the lowest value of the macro-variable

(the Boltzmannian equilibrium value).

These examples show that Gibbsian and Boltzmannian calculations can but need not

9Note that this macro-variable is very different from the one in Section 3.

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provide the same results.10 An important task of the foundations of SM is to classify under

which conditions the two frameworks lead to the same results and under which conditions they

do not. The Khinchin condition and EET provide partial answers to this question. The answers

are partial because both offer only sufficient but not necessary conditions for Gibbsian and

Boltzmannian results being the same.

7 Conclusion

There is widespread belief that Gibbsian and Boltzmannian SM provide the same equilibrium

values. We argued that this is false. There are important cases where the Boltzmannian and

Gibbsian equilibrium calculations will lead to different results. Furthermore, the conditions

under which an equilibrium exists are different for Gibbsian and Boltzmannian SM. It is,

however, true that the equilibrium values coincide under special circumstances. We proved a

new theorem giving conditions under which this is the case.

This raises the question of what happens in cases where the two do not coincide. Is the

Gibbsian or the Boltzmannian equilibrium the correct one? For the Ising model the correct

empirical conclusions follow from the Boltzmannian and not the Gibbsian calculations, and we

suspect that this will be the same for other cases. Which framework provides the correct values

and under which circumstances is a question for future research.

10This discrepancy is not an artefact of the use of our own definition of the Boltzmannian

equilibrium. The same results would follow if one adopted the standard definition of the

equilibrium as the largest macro-region.

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8 Appendix: Proof of Theorem 1

First, we determine the value of the largest macro-region. Recall that N is a multiple of k and

that the observable on the one-constituent space takes the values {κ1, . . . ,κk} with equal

probability. Hence for the macro-region of largest size there are N/k particles taking the value

κ1, N/k particles taking the value κ2, . . . , N/k particles taking the value κk. Therefore, the value

of the largest macro-region is (κ1 + . . . + κk)N/k.

Let us now determine the phase average. The proof is by mathematical induction. We will

show that the sum over all sequences of length N whose elements are in {κ1, . . . ,κk} is

kN−1 N(κ1 + . . . + κk). Because each sequence has equal probability 1/kN , the desired result

follows that the phase average is

1
kN

kN−1 N(κ1 + . . . + κk) =
N
k

(κ1 + . . . + κk). (11)

N = 1: Because k ≤ N, k = 1, and the the sum over all sequences of length N is

κ1 = 11−11κ1 = κ1.

N → N + 1: We need to determine the sum over all sequences of length N + 1. One obtains

sequences of length N + 1 by adding to sequences of length N one element at the end of the

sequences. One can add k possible elements at the end and so a contribution to the sum over all

sequences of length N + 1 is k times the sum of the sequences of length N, i.e.

k × kN−1 N(κ1 + . . . + κk). (12)

What is still missing is the contribution from the element at the end. There are kN sequences of

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length N to which a κi can be added, and hence this contribution is

kN (κ1 + . . . + κk). (13)

Adding these two contributions leads to:

k × kN−1 N(κ1 + . . . + κk) + k
N (κ1 + . . . + κk) = k

N (N + 1)(κ1 + . . . + κk). (14)

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