Accessibility Foundation — Examples

For $a,b>0$, apply [F6] to $\log a,\log b$, then use [F7] and monotonicity of $\log$, to obtain $\log ((1-t)a+tb)\ge (1-t)\log a+t\log b$ for $0\le t\le 1$. Apply this inequality to the energy ratios $u\mathrm{/}{u}_{\star ,i}$ and to the work ratios $v\mathrm{/}{v}_{\star ,i}$; multiplying by $c_i,r_i>0$ and adding $b_i$ proves concavity of $s_i$. This verifies the convex-combination accessibility condition. Since $\log$ is strictly increasing, raising $u$ at fixed $v$ gives a strict successor, proving S1.

The boundary equation $s_i(u^{\mathrm{\prime }},v^{\mathrm{\prime }})=s_i(u_x,v_x)$ rearranges to $u^{\mathrm{\prime }}=g(v^{\mathrm{\prime }})=K(v^{\mathrm{\prime }}{)}^{-\alpha }$, where $K=u_x{v}_x^{\alpha }>0$ and $\alpha =r_i\mathrm{/}{c}_i>0$. By [F8], $g^{\mathrm{\prime }}=-\alpha K(v^{\mathrm{\prime }}{)}^{-\alpha -1}$ and $g^{\mathrm{\prime }\mathrm{\prime }}=\alpha (\alpha +1)K(v^{\mathrm{\prime }}{)}^{-\alpha -2}>0$. Thus $g^{\mathrm{\prime }}$ is increasing by [F9]. For $v^{\mathrm{\prime }}\mathrm{\ne }{v}_x$, the mean value theorem on the interval between them and the monotonicity of $g^{\mathrm{\prime }}$ give $g(v^{\mathrm{\prime }})\ge g(v_x)+g^{\mathrm{\prime }}(v_x)(v^{\mathrm{\prime }}-v_x)$: when $v^{\mathrm{\prime }}<v_x$, both the derivative comparison and the sign of $v^{\mathrm{\prime }}-v_x$ reverse. Hence the forward sector $u^{\mathrm{\prime }}\ge g(v^{\mathrm{\prime }})$ lies above the tangent support $u^{\mathrm{\prime }}-u_x+(r_i{u}_x\mathrm{/}(c_i{v}_x))(v^{\mathrm{\prime }}-v_x)=0$. A vertical line cannot support this sector because its work coordinate varies on both sides at an interior boundary point. If a nonvertical line with slope parameter $q$ supports there, its inequality on graph points gives $g(v^{\mathrm{\prime }})-g(v_x)+q(v^{\mathrm{\prime }}-v_x)\ge 0$ for both signs of $v^{\mathrm{\prime }}-v_x$; dividing by that difference and taking limits forces $q=-g^{\mathrm{\prime }}(v_x)$. This proves uniqueness and gives $P_i=r_i{u}_x\mathrm{/}(c_i{v}_x)$. On any closed rectangle with $u,\tilde{u}\le u_1$ and $v,\tilde{v}\ge v_0>0$, $\mathrm{\mid }u\mathrm{/}v-\tilde{u}\mathrm{/}\tilde{v}\mathrm{\mid }\le \mathrm{\mid }u-\tilde{u}\mathrm{\mid }\mathrm{/}{v}_0+u_1\mathrm{\mid }v-\tilde{v}\mathrm{\mid }\mathrm{/}{v}_0^2$, so the slope is locally Lipschitz. The graph over $v^{\mathrm{\prime }}>0$ is arcwise connected, proving S3.

Put $C=c_i+c_j$. For any split $u_i+u_j=u$, [F6] with weights $c_i\mathrm{/}C,c_j\mathrm{/}C$, followed by [F7], gives $\frac{c_i}{C}\log \frac{u_i}{c_i}+\frac{c_j}{C}\log \frac{u_j}{c_j}\le \log \frac{u}{C},$ with equality exactly when $u_i\mathrm{/}{c}_i=u_j\mathrm{/}{c}_j$. Since $u_i+u_j=u$, that equality forces the unique split $u_i=c_{i}u\mathrm{/}C,u_j=c_{j}u\mathrm{/}C$. Multiply by $C$ and add the constants $c_i\log (c_i\mathrm{/}{u}_{\star ,i})+c_j\log (c_j\mathrm{/}{u}_{\star ,j})+b_i+b_j+r_i\log (v_i\mathrm{/}{v}_{\star ,i})+r_j\log (v_j\mathrm{/}{v}_{\star ,j})$; the resulting inequality is exactly the pair entropy bounded by its value at that split. By the declared total-entropy comparison, this proves T1 and shows that the maximizing split is reversible with the join, establishing the join-splitting clause of T2. For the equal-state scaled split, tagged extensivity gives $s_{ti}(tu,tv)=t{s}_i(u,v)$ and $s_{(1-t)i}((1-t)u,(1-t)v)=(1-t)s_i(u,v)$, so the pair is reversible with $x$ by the example order. The maximizing-split criterion also applies to the two scaled types: their energy coefficients are $t{c}_i,(1-t)c_i$, and their energy-to-coefficient ratios both equal $u\mathrm{/}{c}_i$. Their join is therefore reversible with the pair, which is precisely the scaled-join clause of T2. The binary join entropy is $C\log u+r_i\log v_i+r_j\log v_j+C_{ij}$ for a constant $C_{ij}$; its energy-to-coefficient ratio is $u\mathrm{/}C=u_i\mathrm{/}{c}_i=u_j\mathrm{/}{c}_j$ at the maximizing split.

By the declared total-entropy comparison, every finite product has a real-valued order. Reflexivity and transitivity follow from the real order; adding the same spectator adds the same number to both sides; positive scaling multiplies the sums by a positive coefficient; and reversible splitting preserves the extensive sum. If $A+{ϵ}_n{Z}_0\prec B+{ϵ}_n{Z}_1$, then $s(A)-s(B)\le {ϵ}_n(s(Z_1)-s(Z_0))$. Taking $n\to \mathrm{\infty }$ yields $s(A)\le s(B)$ and hence $A\prec B$. This verifies the six adopted order conditions.

For a single system the entropy has the displayed logarithmic form. If a nested join has energy coefficient $C$ and is joined with a system of coefficient $c>0$, the maximum calculation of step 1.3 gives the unique maximizing split at equal energy-to-coefficient ratios; substitution adds $c$ to the energy coefficient and adds the new work term. Induction over the finite joins therefore gives coefficient $C={\sum }_k{c}_k$ and entropy $C\log u+{\sum }_k{r}_k\log v_k+C_0$ for the positive work coordinates $v=(v_1,\dots ,v_m)$. Its boundary is $u=g(v)=K{\prod }_k{v}_k^{-{\alpha }_k}$ with ${\alpha }_k=r_k\mathrm{/}C>0$. Direct differentiation gives, for any vector $h$, $D^{2}g(v)[h,h]=g(v)[{\left({\sum }_k{\alpha }_k\frac{h_k}{v_k}\right)}^2+{\sum }_k{\alpha }_k{\left(\frac{h_k}{v_k}\right)}^2]\ge 0.$ Thus along each segment in the positive work orthant, the derivative of $g$ is nondecreasing by [F9], and the mean value theorem gives $g(v^{\mathrm{\prime }})\ge g(v)+\mathrm{\nabla }g(v)\cdot (v^{\mathrm{\prime }}-v)$. The forward sector $u\ge g(v)$ lies above this tangent support; uniqueness follows by applying the support inequality to graph points in each direction $h$ and $-h$, which forces the support functional to equal $\mathrm{\nabla }g(v)$. Its finite slope components are $P_k=r_{k}u\mathrm{/}(C{v}_k)$; on rectangles bounded away from all coordinate axes, the quotient estimate from step 1.2 applies to each component and proves local Lipschitz regularity. The positive orthant and its graph are arcwise connected. Increasing $u$ gives S1, and the coordinatewise concavity argument of step 1.1 gives the convex-combination accessibility condition. Hence each join satisfies the same simple-system geometry.

Applying the two-weight maximum calculation of steps 1.3 and 2.2 to any two primitive or recursively joined types $A,B$ shows that $x=(u_A,v_A)$ and $y=(u_B,v_B)$ are thermally equivalent exactly when $u_A\mathrm{/}{C}_A=u_B\mathrm{/}{C}_B$, where $C_A,C_B>0$ are their energy coefficients. Equality of ratios is transitive, proving T3. For any such simple type with coefficient $C$ and at least one work coordinate $v_k$, fix a state $x=(u,v)$ and choose $0<v_k^{-}<v_k<v_k^{+}$, keeping $u$ and all other coordinates fixed. The entropy formula of step 2.2 strictly increases with $v_k$, so $x_{-}\prec \prec x\prec \prec x_{+}$. Both outer states have the same ratio $u\mathrm{/}C$ and hence are thermally equivalent. This proves T4 for primitive and recursively joined simple types.

Let $A,B$ be any primitive or recursively joined simple types, with energy coefficients $C_A,C_B>0$. Given $x=(u_A,v_A)$ and any allowed work-coordinate vector $v_B$, choose $u_B=C_B{u}_A\mathrm{/}{C}_A$. The two-weight maximum calculation in steps 1.3 and 2.2 makes $x$ thermally equivalent to $y=(u_B,v_B)$, proving T5. For any equilibrated pair $x,y$, the same calculation gives $u_A\mathrm{/}{C}_A=u_B\mathrm{/}{C}_B$. Their joined state has energy coefficient $C_A+C_B$ and energy $u_A+u_B$, so its ratio equals the common ratio; the two-weight maximum criterion therefore makes that joined state thermally equivalent to $x$, and likewise to $y$. This proves C1 for every admitted join.

Let $\pi (U_1,U_2)=(U_2,U_1)$. Then $A_{\pi u}=\pi A_u$ and $B_{\pi u}=\pi B_u$, since the mean is unchanged; linearity of $\pi$ also gives $\mathrm{conv}(A_{\pi u}\cup B_{\pi u})=\pi \mathrm{conv}(A_u\cup B_u)$. Thus the declared relation is independent of the representative of $[u]$; target representatives are already quantified over, and reflexivity follows from $u\in A_u$. Consider the distinct classes $[(1,4)]$ and $[(2,2)]$. From $u=(1,4)$, every point of $A_u$ has coordinate sum at least $5$, and every point of $B_u$ has both coordinates at least $\bar{U}=5\mathrm{/}2$, hence sum at least $5$. Finite convex combinations preserve this sum bound. Every representative of $[(2,2)]$ has sum $4$, so $[(1,4)]\prec ̸[(2,2)]$.

In the one-system space, strict accessibility is exactly strict increase of $U$, while thermal equilibrium is equality of normalized energies. If $U_0\prec \prec U\prec \prec U_1$, then $U_0<U<U_1$, so $U_0\mathrm{\ne }{U}_1$ and the two cannot be thermally equivalent. Thus T4 fails.

From $u=(2,2)$, both $A_u$ and $B_u$ consist of points whose two coordinates are at least $2$; finite convex combinations preserve both inequalities. Every representative of $[(1,4)]$ has a coordinate equal to $1$. Therefore $[(2,2)]\prec ̸[(1,4)]$. Together with step 1.1, this proves incomparability of distinct product states and failure of the comparison hypothesis.

By [F4], $U$ is dimensionless, so the single-system function $S(U)=\log U$ is well-defined; it is strictly increasing and represents $U\le U^{\mathrm{\prime }}$. If its additive sum represented product accessibility, the states $[(1,4)]$ and $[(2,2)]$ would have equal summed entropy $\log 1+\log 4=2\log 2=\log 4$, forcing accessibility both ways. Step 1.1 and step 2.1 show neither direction holds. Therefore this single-system entropy does not represent the product relation. Since the example fails T4 by step 1.2, it does not satisfy the hypotheses of a product-comparison result that includes T4.