How Do Macrostates Come About?

Abstract

This paper is a further consideration of Hemmo and Shenker’s ideas about the proper conceptual characterization of macrostates in statistical mechanics. We provide two formulations of how macrostates come about as elements of certain partitions of the system’s phase space imposed on by the interaction between the system and an observer, and we show that these two formulations are mathematically equivalent. We also reflect on conceptual issues regarding the relationship of macrostates to distinguishability, thermodynamic regularity, observer dependence, and the general phenomenon of measurement.

Appendix

Proof of Proposition 1

We prove Proposition 1 via proving four lemmas.

Lemma 1

P_O ^A∼ and P_S ^A∼ are partitions of A_O and A_S, respectively, that is $P_{O}^{A\sim }\in \mathbf{P}^{A_{O}}$ and $P_{S}^{A\sim }\in \mathbf{P}^{A_{S}}$.

Proof

We show that the sets of P _O ^A ∼ are disjoint and add up to A _O . (For P _S ^A ∼ the proof is similar.)

Suppose that sets of P _O ^A ∼ are not disjoint, that is there exists a b ∈ (B _O ∩ B _O ^′) with B and B′ being different elements in P ^A ∼. Then there exist an x ∈ B and an x′ ∈ B′ such that x _O  = x _O ^′ = b. But then x ∼ x′. Contradiction.

Suppose that sets of P _O ^A ∼ are not adding up to A _O , that is there exists a b ∈ A _O such that b ∉ B _O for any B _O  ∈ P _O ^A ∼. Then $\mathcal{A}^{o}(b) \cap B =\emptyset$ for any B ∈ P ^A ∼, that is P ^A ∼ is not a partition of A. Contradiction. ■

Lemma 2

The partition P ^A∼ can be “reconstructed” from its projections in the sense that $\mathcal{A}^{o}[P_{O}^{A\sim }] = \mathcal{A}^{s}[P_{S}^{A\sim }] = P^{A\sim }$ .

Proof

We show that $\mathcal{A}^{o}[P_{O}^{A\sim }] = P^{A\sim }$. (For $\mathcal{A}^{s}[P_{S}^{A\sim }] = P^{A\sim }$ the proof is similar.)

Suppose to the contrary that $\mathcal{A}^{o}[P_{O}^{A\sim }]\neq P^{A\sim }$. This means that there exist an x ∈ A and a B ∈ P ^A ∼ such that either (i) $x \in \mathcal{A}^{o}[B_{O}]$ and x ∉ B, or (ii) $x\notin \mathcal{A}^{o}[B_{O}]$ and x ∈ B.

As for case (i), since $x \in \mathcal{A}^{o}[B_{O}]\setminus B$ and therefore x _O  ∈ B _O , there exists an x′ ∈ B such that x _O ^′ = x _O . But then x ∼ x′ and hence x ∈ B. Contradiction.

Case (ii) can be excluded since for any region B in A, $B \subseteq \mathcal{A}^{o}[B_{O}]$. ■

The following Lemma is a straightforward corollary of Lemma 2.

Lemma 3

$\mathcal{A}^{os}[P_{O}^{A\sim }] = P_{S}^{A\sim }$ and $\mathcal{A}^{so}[P_{S}^{A\sim }] = P_{O}^{A\sim }$ .

Proof

$\mathcal{A}^{os}[P_{O}^{A\sim }] =\big (\mathcal{A}^{o}[P_{O}^{A\sim }]\big)_{S} =\big (P^{A\sim }\big)_{S} = P_{S}^{A\sim }$ and similarly for $\mathcal{A}^{so}[P_{S}^{A\sim }]$. ■

Obviously, Lemma 3 shows that macrostates according to Definition 3 are macrostates also according to Definition 1. Our next Lemma demonstrates that the converse is also true.

Lemma 4

Suppose that $\{P^{A_{O}},P^{A_{S}}\}$ are partitions of A _O and A _S , respectively, such that $\mathcal{A}^{os}[P^{A_{O}}] = P^{A_{S}}$ and $\mathcal{A}^{so}[P^{A_{S}}] = P^{A_{O}}$ . Then there exists a P ^A∼ ∈ P ^A∼ such that $P^{A_{O}} = P_{O}^{A\sim }$ and $P^{A_{S}} = P_{S}^{A\sim }$ .

Proof

(i) First we show that $\mathcal{A}^{o}[P^{A_{O}}]$ is a partition of A, that is $\mathcal{A}^{o}[P^{A_{O}}] \in \mathbf{P}^{A}$.

Suppose that sets of $\mathcal{A}^{o}[P^{A_{O}}]$ are not disjoint, that is for a given $B_{O},B_{O}^{{\prime}}\in P^{A_{O}}$ there exists an x such that $x \in (\mathcal{A}^{o}[B_{O}] \cap \mathcal{A}^{o}[B_{O}^{{\prime}}])$. But then $x_{O} \in \big (B_{O} \cap B_{O}^{{\prime}}\big)$, that is $P^{A_{O}}$ is not a partition of A _O . Contradiction.

Suppose that sets of $\mathcal{A}^{o}[P^{A_{O}}]$ are not adding up to A, that is there exists an x in A such that x ∉ B for any $B \in \mathcal{A}^{o}[P^{A_{O}}]$. Then x _O ∉ B _O for any $B_{O} \in P^{A_{O}}$, that is $P^{A_{O}}$ is again not a partition of A _O . Contradiction.

Hence, $\mathcal{A}^{o}[P^{A_{O}}]$ is a partition of A. Similarly, $\mathcal{A}^{s}[P^{A_{S}}]$ is a partition of A.

(ii) Next we show that $\mathcal{A}^{o}[P^{A_{O}}] = \mathcal{A}^{s}[P^{A_{S}}]$.

Let $B_{O} \in P^{A_{O}}$ and let B _S be its corresponding element $B_{S} = \mathcal{A}^{os}[B_{O}]$. Suppose that $\exists x \in \mathcal{A}^{s}[B_{S}]$ such that $x\not\in \mathcal{A}^{o}[B_{O}]$. Then x _O  ∉ B _O and hence x _O  ∈ B _O ^′ for some B _O ^′ ≠ B _O , $B_{O}^{{\prime}}\in P^{A_{O}}$. But since $\mathcal{A}^{os}[B_{O}^{{\prime}}] = B_{S}^{{\prime}}$ for some B _S ^′ ≠ B _S , $B_{S}^{{\prime}}\in P^{A_{S}}$, we would have x _S  ∈ B _S ^′ and thus $x \in \mathcal{A}^{s}[B_{S}^{{\prime}}]$ where $\mathcal{A}^{s}[B_{S}^{{\prime}}]\neq \mathcal{A}^{s}[B_{S}]$, a contradiction. The argument is similar for the case when $\exists x \in \mathcal{A}^{o}[B_{O}]$ such that $x\not\in \mathcal{A}^{s}[B_{S}]$. Hence for all $B_{O} \in P^{A_{O}}$ and for their corresponding $B_{S} \in \mathcal{A}^{os}[B_{O}]$ we have $\mathcal{A}^{o}[B_{O}] = \mathcal{A}^{s}[B_{S}]$, and thus $\mathcal{A}^{o}[P^{A_{O}}] = \mathcal{A}^{s}[P^{A_{S}}]$.

(iii) Finally, we show that $\mathcal{A}^{o}[P^{A_{O}}] = \mathcal{A}^{s}[P^{A_{S}}] \in \mathbf{P}^{A\sim }$.

Suppose that $\mathcal{A}^{o}[P^{A_{O}}]\not\in \mathbf{P}^{A\sim }$, that is there exists an x ∈ B and an x′ ∈ B′ such that x ∼ x′ and B and B′ are distinct elements of $\mathcal{A}^{o}[P^{A_{O}}]$. Let {x _n } _n = 1 ^N be the sequence connecting x and x′ and suppose that (x ₂) _S  = x _S (the argument is similar when (x ₂) _O  = x _O ). Now, x ₂ cannot be in a $B_{2} \in \mathcal{A}^{o}[P^{A_{O}}]$ such that B ₂ ≠ B, otherwise (x ₂) _S were in B _S ∩ (B ₂) _S and consequently B _S and (B ₂) _S were not disjoint. So x ₂ ∈ B. By induction, we obtain that x ₃, … x _N  = x′ are all in B. Hence B and B′ are not distinct. Contradiction. ■

By this we also complete the proof of Proposition 1. ■