Phases And Limits — Examples

The finite sum gives $Z=1+e^{-\beta ϵ}>1$, so $F$ is finite for every $\beta >0$. Differentiating with [F2] yields

whose terms are continuous on $\beta >0$.

Each derivative of $e^{-\beta ϵ}$ is $(-ϵ{)}^m{e}^{-\beta ϵ}$ by induction on $m$, so $Z$ is $C^{\mathrm{\infty }}$. Since $Z>0$, repeated application of the logarithm derivative formula and quotient/product rules makes every derivative of $\log Z$ continuous on $\beta >0$. The same induction gives smoothness of ${\beta }^{-1}$ there. Thus $F=-{\beta }^{-1}\log Z$ is $C^{\mathrm{\infty }}$ on $\beta >0$. To establish the stronger analytic conclusion, fix ${\beta }_0>0$ and $Z_0=Z({\beta }_0)>0$. The exponential series and [F5] make $Z$ real analytic. On a sufficiently small neighborhood of ${\beta }_0$, $h(\beta ):=(Z(\beta )-Z_0)\mathrm{/}{Z}_0$ has $\mathrm{\mid }h\mathrm{\mid }<1$, and $\log Z(\beta )=\log Z_0+\log (1+h(\beta ))$ by the logarithm law. The convergent log series and local analytic composition make this real analytic. The reciprocal ${\beta }^{-1}$ is analytic on $\beta >0$ by the nonzero-quotient clause, so $F$ is real analytic there by [F5]. Since $T\mapsto (k_{B}T{)}^{-1}$ is likewise an analytic nonzero quotient on $T>0$, the same holds in the temperature variable. Thus no nonanalytic free-energy point occurs at positive finite temperature; this says nothing about a sequence limit.

At fixed $t$, differentiating the rational equation gives $\frac{\mathrm{\partial }p}{\mathrm{\partial }v}=-\frac{24t}{(3v-1{)}^2}+\frac{6}{v^3}\mathrm{.}$ At $t=0.9$ and $v=1$, this derivative is $-24(0.9)\mathrm{/}4+6=0.6>0$.

The second derivative is ${\mathrm{\partial }}_v^{2}p=144t\mathrm{/}(3v-1{)}^3-18\mathrm{/}{v}^4$. At a simultaneous zero, the first derivative gives $t=(3v-1{)}^2\mathrm{/}(4v^3)$ and the second gives $t=(3v-1{)}^3\mathrm{/}(8v^4)$. Equality and $v>1\mathrm{/}3$ imply $2v=3v-1$, hence $v=1$, $t=1$, and $p=1$. Conversely these values make both derivatives zero. Thus $T_c$ is the model critical temperature in this stated sense, and $t=0.9<1$ is below it.

The derivative is continuous near $(t,v)=(0.9,1)$ because $3v-1\mathrm{\ne }0$ there. Hence it remains positive on some open neighborhood within $v>1\mathrm{/}3$. Since $P=P_{c}p$ and $V=V_{c}v$ with positive scales, $(\mathrm{\partial }P\mathrm{/}\mathrm{\partial }V{)}_T=(P_c\mathrm{/}{V}_c)(\mathrm{\partial }p\mathrm{/}\mathrm{\partial }v{)}_t>0$ there as well. This verifies a positive-slope branch of this phenomenological model only.

Let ${\lambda }_{\beta }:=n_{\beta }\mathrm{/}N$ and ${\lambda }_{\alpha }:=n_{\alpha }\mathrm{/}N$. Positivity and amount additivity give ${\lambda }_{\alpha }+{\lambda }_{\beta }=1$, with both fractions in $(0,1)$. The adopted $V_i=n_i{v}_i$ relation and volume additivity give

All terms have units ${\mathrm{m}}^3\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$.

Here $\bar{v}=2.5\times {10}^{-3}{\mathrm{m}}^3\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$. Solving the affine relation in step 1.1 yields

The fractions are positive and sum to one, so $n_{\alpha }=n_{\beta }=1\mathrm{m}\mathrm{o}\mathrm{l}$ satisfies the given total amount and volume.

Species $A$ is active because $n_A({\xi }_{\ast })=0$. By [F1], a tangent extent $h$ is feasible exactly when $(\nu h{)}_A=-h\ge 0$, or $h\le 0$. The derivative data give ${\nu }^{\mathsf{T}}\mu =-{\mu }_A+{\mu }_B=-1\mathrm{J}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$, so $D\mathrm{\Phi }(0)[h]=({\nu }^{\mathsf{T}}\mu )h\ge 0$ for every feasible tangent extent. The units are joules because $h$ is in moles.

Directly, $\mathrm{\Phi }(\xi )-\mathrm{\Phi }(0)=-(1\mathrm{J}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1})\xi \ge 0$ for every feasible $\xi \in [-1,0]$ mol, so the boundary point is a global minimum on this interval. The tangent-cone inequality is satisfied even though the interior equality ${\nu }^{\mathsf{T}}\mu =0$ fails.

Since $0<s_0-a\le s(x)<s_0$ on $K$, every feasible $s$ is nonnegative and bounded. For $n\ge 2$, put $A_n:=\{x\in K:\mathrm{\mid }x\mathrm{\mid }\ge 1\mathrm{/}n\}$. Their union is $K$; since $P(K)=1$, countable additivity implies $P(A_n)>0$ for some $n$. On $A_n$, $s\le s_0-a\mathrm{/}{n}^2$, and on its complement $s\le s_0$. For any nonnegative simple function $\phi \le s$, the measure integral definition [F1] therefore gives

Indeed, refine the finitely many level sets of $\phi$ by $A_n$ and its complement; each simple-function coefficient is bounded by the corresponding pointwise bound on that piece, so the simple-integral formula [F1] gives the inequality. The strict upper bound is independent of $\phi$, hence its supremum also satisfies ${\int }_{K}sdP<s_0$.

For each $0<ϵ<1$, the measure $P_{ϵ}=\frac{1}{2}{\delta }_{-ϵ}+\frac{1}{2}{\delta }_{ϵ}$ is a Borel probability measure on $K$, has mean zero, and satisfies ${\int }_{K}sd{P}_{ϵ}=s_0-a{ϵ}^2$. Letting $ϵ\downarrow 0$ shows that the supremum is $s_0$. Step 1.1 shows that no feasible measure attains it. Thus the false statement fails even with a bounded positive continuous concave extension, because the admissible generator set excludes its maximizing barycenter.

Write $w_0=P(\{0\})$ and $w_1=P(\{1\})$. By [F2], $w_0+w_1=1$ and the barycenter equation is $w_1=\frac{1}{2}$. Hence the only feasible measure is $P_{\ast }=\frac{1}{2}{\delta }_0+\frac{1}{2}{\delta }_1$.

Its entropy per mole is the finite simple-function integral

Since it is the only feasible mixture, it attains the maximum. Its two support points equal $d+1$ for $d=1$, as in [F1].

For each branch choose a reference amount $n_i>0$ and a full open neighborhood ${\mathcal{O}}_i$ of $(T_0,p_0,n_i)$ in $(0,\mathrm{\infty })\times \mathbb{R}\times (0,\mathrm{\infty })$. Define ${\chi }_i(T,p,n)=(U_i(T,p,n),V_i(T,p,n),n)$ by the Gibbs assignments. At the center, differentiation gives $S_i\mathrm{/}{n}_i=s_i$, $V_i\mathrm{/}{n}_i=v_i$, and ${\mu }_i=g_i=g_0$. Also $U_{i,T}=n_{i}c$, $V_{i,p}=-n_i\kappa$, and $V_{i,T}=0$ there. The full Jacobian of ${\chi }_i$ is block triangular with last diagonal entry $1$, so its determinant is $-n_i^{2}c\kappa \mathrm{\ne }0$. By [L1], shrink ${\mathcal{O}}_i$ to an open inverse patch retaining positive temperature, volume and amount. Its numerical state patch is $O_{c_i}={\mathrm{\Phi }}_{c_i}^{-1}({\chi }_i({\mathcal{O}}_i))$, using the SI unit map in [F1]'s phase-branch definition; define its entropy by the displayed Gibbs assignment through this inverse. Thus $\mathrm{\Delta }s=3\mathrm{J}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $\mathrm{\Delta }v=3.0\times {10}^{-3}{\mathrm{m}}^3\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}\mathrm{\ne }0$.

For entropy $\mathsf{s}$ near $n_i{s}_i$, positive amount $n$ near $n_i$, and volume $V$ near $n_i{v}_i$, solve the natural assignments explicitly by $T(\mathsf{s},V,n)=T_0+(T_0\mathrm{/}c)(\mathsf{s}\mathrm{/}n-s_i)$ and $p(\mathsf{s},V,n)=p_0+(v_i-V\mathrm{/}n)\mathrm{/}\kappa$. Set ${\mathcal{U}}_i(\mathsf{s},V,n)=n{g}_i(T,p)+T\mathsf{s}-pV$. These maps are $C^2$ by [L2], and substitution gives $S_i=\mathsf{s}$, $V_i=V$, and $U_i={\mathcal{U}}_i$. On a sufficiently small neighborhood $T,n,V$ are positive. Scaling $(\mathsf{s},V,n)$ by $\lambda$ leaves both ratios unchanged, so $T,p$ are unchanged and ${\mathcal{U}}_i(\lambda \mathsf{s},\lambda V,\lambda n)=\lambda {\mathcal{U}}_i(\mathsf{s},V,n)$ whenever both states remain in this local patch. This explicitly supplies the required fixed-tag $C^2$ homogeneous energy representation.

The Gibbs assignments and [L3] give $d{U}_i=Td{S}_i-pd{V}_i+g_{i}dn$ on the inverse patch of step 1.1. Consequently its entropy partials are $S_{i,U}=1\mathrm{/}T$, $S_{i,V}=p\mathrm{/}T$, and $S_{i,n}=-g_i\mathrm{/}T$. The inverse coordinates are $C^1$ by [L1]; the reciprocal is $C^1$ and the polynomial $g_i$ is smooth by [L2]. Hence these coefficients are $C^1$, giving $C^2$ entropy with $S_{i,U}>0$ and the intended derivative-defined intensives. Conversely the explicit formulas in step 2.1 invert the energy-to-Gibbs coordinates, so the natural-variable chart is valid. The branch maps are therefore legitimate local phase and Gibbs charts with the regularity required by [F1].

Apply the slope clause of [F1] with [A1] and the verified charts:

For the latent heat, [F2] on the specified forward and reverse complete-compound records gives zero total entropy change. All components except the phases and reservoir have zero entropy change, and the latter contributes $-Q\mathrm{/}{T}_0$, so conversion of $n>0$ moles gives $0=n\mathrm{\Delta }s-Q\mathrm{/}{T}_0$. Thus ${\mathrm{\ell }}_{\alpha \to \beta }:=Q\mathrm{/}n=T_0\mathrm{\Delta }s=300(3)=900\mathrm{J}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$, also the latent-heat formula of [F1]. Both values are exact constructed-model predictions under the stated preparation assumptions.

At $(T_0,p_0)$, differentiation gives $S_i\mathrm{/}{n}_i=s_i$, $V_i\mathrm{/}{n}_i=v_0$, and ${\mu }_i=g_i=g_0$. Also $U_{i,T}=n_{i}c$, $V_{i,p}=-n_i\kappa$, and $V_{i,T}=0$, so the derivative determinant for $(T,p)\mapsto (U_i,V_i)$ at fixed positive amount is $-n_i^{2}c\kappa \mathrm{\ne }0$. Thus [L1] supplies valid local state charts. The branch equality function is

whose zero set is $T=T_0$ with every nearby chart pressure permitted.

Here $\mathrm{\Delta }v=v_{\beta }-v_{\alpha }=0$, so the zero set is not the graph of a unique pressure as a function of temperature on an open interval about $T_0$, and the quotient $\mathrm{\Delta }s\mathrm{/}\mathrm{\Delta }v$ is undefined. The nonzero-jump hypothesis of [F1] fails, so this model does not contradict that theorem.

From [A2], at $(T_0,p_0)$, $S_i=n_i{s}_0$, $V_i=n_i{v}_i$, $U_{i,T}=n_{i}c$, $V_{i,p}=-n_i\kappa$, and $V_{i,T}=0$. Thus the $(T,p)$ derivative matrix of $(U_i,V_i)$ at fixed amount has determinant $-n_i^{2}c\kappa \mathrm{\ne }0$. The full $(T,p,n)\mapsto (U_i,V_i,n)$ Jacobian has this same determinant, since its last row is $(0,0,1)$. Thus [L1] supplies the claimed open full-state inverse patch about $(T_0,p_0,n_{\ast })$.

On that patch the Gibbs assignments and the chain rule give $d{G}_i=-S_{i}dT+V_{i}dp+g_{i}dn$ and hence $d{U}_i=Td{S}_i-pd{V}_i+g_{i}dn$. Pulling these identities through the $C^1$ inverse, $d{S}_i=T^{-1}d{U}_i+(p\mathrm{/}T)d{V}_i-(g_i\mathrm{/}T)dn$. Each coefficient is $C^1$ by [L2], since the inverse coordinates and $g_i$ are $C^1$ and $T>0$. Thus the constructed entropy is $C^2$, with $S_{i,U}=1\mathrm{/}T>0$, $S_{i,V}=p\mathrm{/}T$, and $S_{i,n}=-g_i\mathrm{/}T$, so its derivative-defined intensive quantities reproduce the Gibbs assignments.

The only coexistence equation is $F(T,p):=g_2(T,p)-g_1(T,p)=(v_2-v_1)(p-p_0)=0$. It vanishes at the center and has ${\mathrm{\partial }}_{p}F=v_2-v_1\mathrm{\ne }0$, so the $1\times 1$ minor is invertible.

Here the ambient dimension is $d=2+P(C-1)=2$ and the equation count is $q=C(P-1)=1$. Applying [F1] gives local equation-set dimension $d-q=1$, exactly $C-P+2=1$. This verifies the regular equation-set rank calculation only.