Lenker T D . Caratheodory's concept of temperature[J]. Synthese, 1979

This is a discussion in the article:

let $S_i$ be the state space of the thermodynamic system $\sum i = 1, 2, 3$, and assume there exists a set of sufficiently smooth relations $F_{ij}: S_i×S_j→R$ such that the systems $∑ i$ and $∑ j$ are in thermal equilibrium while in states $s_i$ and $s_j$ respectively if and only if $F_{ij}(s_i, s_j)=0$ for $1≤i$, $j≤3$ and $i≠j$. These relations satisfy the conditions:

(1.1) $F_{ij}(s_i, si_j)=0$ if and only if $F_{ji}(s_j, s_i)= 0$.

(1.2) If $F_{ij}(s_i, s_j)=0$ and $F_{jk}(s_j, s_k) = 0$, then $F_{ik}(s_i, s_k)=0$.

Does the existence of these relations imply that there are continuous functions

$t_i$:$s_i→R, i = 1, 2, 3$ such that $t_1(s_1)=t_2(s_2)=t_3(s_3)$ exactly when

$F_{12}(S_1, s_2) = F_{13}(s_l, s_3) = F_{23}(S_2, s_3) = 0$?

The following example due to Whaples [8] shows that without more restrictive assumptions, such functions need not exist.

EXAMPLE: Let $s_1 = s_2 = s_3 = R$ and let $F_{ij}(s_i, s_j) = sin [π(s_i - s_j)]$ for $i, j = 1, 2, 3$.

Assume there exist continuous functions

$t_i$: $Si→R$, $i = 1, 2, 3$ such that $t_1(s_1)=t_2(s_2)=t_3(s_3)$ if and only if

$F_{12}(s_1, s_2)= F_{13}(s_1,s_3) = F_{23}(s_2, s_3)=0$.

Thus $t_1(s_1)=t_2(s_2)=t_3(s_3)$ if and only if

$s_1 ≡s_2 mod 1 ≡ s_3 mod 1$.

Consider the triples $(0, 2, 3)$ and $(1, 2, 3)$ belonging to $s_1 × s_2 × s_3$. This implies $t_1(0) = t_1(1)$ and $t_1(s) ≠ t_1(0)$ for $s ∈ (0, 1)$. Thus ti is not continuous.

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I have some questions:

①why choose $F_{ij}: S_i×S_j→R$ as a smooth relations,is multiplication has some special meaning?

②why $t_1(S_1)= t_2(s_2)= t_3(s_3)$ if and only if $s_1 ≡s_2 mod 1 ≡ s_3 mod 1$ but not $s_1 ≡s_2 ≡ s_3$

③why the triples $(0, 2, 3)$ and $(1, 2, 3)$ belonging to $s_1 × s_2 × s_3$,What is the role of these triples?

  • $\begingroup$ We do have MathJax here that makes your equations look better. You can see the notation page in help center for details, if you aren't familiar with it. $\endgroup$ – Kyle Kanos Apr 3 at 17:00
  • $\begingroup$ Thank you, your advice help me a lot. $\endgroup$ – 地山谦 Apr 7 at 9:56

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