Classical configuration, phase and reference spaces — Examples

For $r$ particles in position volume $W$, separate the $r$ position integrals and use [F1] in all $dr$ momentum coordinates. For one coordinate put $a=\beta \mathrm{/}(2m)$; integrating $\frac{d}{dp}(p{e}^{-a{p}^2})=e^{-a{p}^2}-2a{p}^2{e}^{-a{p}^2}$ over $\mathbb{R}$ gives $⟨p^2⟩=m\mathrm{/}\beta$, as the boundary term vanishes by Gaussian decay. Thus the mean kinetic energy is $dr\mathrm{/}(2\beta )$. The position law factorizes into $r$ uniform position laws. With ${\lambda }_T=a_0\sqrt{\beta \mathrm{/}(2\pi m)}$, the partition function is $Z_r(W)=\frac{W^r}{r!a_0^{dr}}{\left(\frac{2\pi m}{\beta }\right)}^{dr\mathrm{/}2}=\frac{(W\mathrm{/}{\lambda }_T^d{)}^r}{r!}\mathrm{.}$ For equal total particle count $2N$ before and after, the $\beta U$ terms in [F2] cancel, so the entropy change is $k_B$ times the logarithm of the partition-function ratio.

Initially the ratio is $Z_{2N}(2V)\mathrm{/}{Z}_N(V{)}^2$ $=4^N(N!{)}^2\mathrm{/}(2N)!$; for two fixed distinguishable species it is $[Z_N(2V)\mathrm{/}{Z}_N(V){]}^2=4^N$. The first ratio exceeds one because $(2N)!\mathrm{/}(N!{)}^2=\left(\genfrac{}{}{0px}{}{2N}{N}\right)<{\sum }_{j=0}^{2N}\left(\genfrac{}{}{0px}{}{2N}{j}\right)=4^N$. Under the final positional law, the $2N$ particles independently fall in either half with probability $1\mathrm{/}2$, so the central binomial probability is $\left(\genfrac{}{}{0px}{}{2N}{N}\right)\mathrm{/}{4}^N$, the reciprocal of the identical-particle ratio.

For each integer $n\ge 1$, monotonicity of $\log x$ on unit intervals gives $n\log n-n+1\le \log (n!)\le n\log n-n+1+\log n\mathrm{.}$ Indeed the lower bound integrates $\log x\le \log j$ on $[j-1,j]$ for $j=2,\dots ,n$, and the upper bound integrates $\log x\ge \log j$ on $[j,j+1]$ for $j=1,\dots ,n-1$, then adds $\log n$. Substitution for $n=N$ and $n=2N$ cancels the terms linear in $N$ in $N\log 4+2\log (N!)-\log ((2N)!)$, leaving a quantity bounded in absolute value by a constant multiple of $\log (N+1)$. Therefore $\mathrm{\Delta }{S}_{\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}\mathrm{/}N\to 0$ as $N\to \mathrm{\infty }$.

In the kinetic shell theorem set $U=0$, $C_m=(2m{)}^{n\mathrm{/}2}$ and $\alpha =n\mathrm{/}2$. The configuration integral of $1$ is $V^N$, so its exact density formula gives the displayed $D(E)$. For each $j$ the conditional component mean is $E\mathrm{/}n$, and on the shell ${\sum }_j{p}_j^2\mathrm{/}(2m)=E$ pointwise; therefore $K=E$ and its variance is zero.

Integrating $D(s)=C{s}^{\alpha -1}$ from $0$ to $E$ gives cumulative phase volume $W(E)={\int }_0^{E}C{s}^{\alpha -1}ds=\frac{C}{\alpha }{E}^{\alpha },$ where $C=V^N(2m{)}^{n\mathrm{/}2}\alpha B_n\mathrm{/}(N!a_0^n)$. Thus $\log (E_{0}D(E))=\log (E_{0}C)+(\alpha -1)\log E$ and $\log (W(E)\mathrm{/}{W}_0)=\log (C\mathrm{/}(\alpha W_0))+\alpha \log E$. Differentiation gives the two displayed derivatives. Since their difference is $k_B\mathrm{/}E>0$, they are distinct at every finite $E>0$.

Separate the position and momentum integrals and apply [F1] in each of the $dN$ momentum coordinates. This gives $Z_N=\frac{V^N}{N!a_0^{dN}}{\left(\frac{2\pi m}{\beta }\right)}^{dN\mathrm{/}2}=\frac{(V\mathrm{/}{\lambda }_T^d{)}^N}{N!}\mathrm{.}$ The position density is uniform on the region and each momentum factor is a centered Gaussian of variance $m\mathrm{/}\beta$.

For $\beta$ in a compact interval $[{\beta }_0,{\beta }_1]\subset (0,\mathrm{\infty })$, the factors $H_N^r{e}^{-\beta H_N}$ for $r\le 2$ are bounded by a constant times $e^{-{\beta }_0{H}_N\mathrm{/}2}$, since $x^r{e}^{-{\beta }_{0}x\mathrm{/}2}$ is bounded for $x\ge 0$; the latter has a finite Gaussian integral by [F1]. Thus [A3] verifies the canonical response theorem's envelope, which gives $U=-{\mathrm{\partial }}_{\beta }\log Z_N=dN\mathrm{/}(2\beta )$, $\mathrm{Var}(H_N)={\mathrm{\partial }}_{\beta }^2\log Z_N=dN\mathrm{/}(2{\beta }^2)$, and the chain rule gives $C_V=dN{k}_B\mathrm{/}2$.

The grand sum has sector weight $e^{\beta \mu j}{Z}_j=a^j\mathrm{/}j!$, so its exponential series gives $\mathrm{\Xi }=e^a$ and $P(N=j)=e^{-a}{a}^j\mathrm{/}j!$. Summing the shifted series gives $\mathbb{E}[N]=a$ and $\mathbb{E}[N(N-1)]=a^2$, hence $\mathrm{Var}(N)=a$.

On a compact $(\beta ,\mu )$ set, $a$ is bounded by some $A<\mathrm{\infty }$. For each fixed $r$, the ratio of successive terms in ${\sum }_{j\ge 0}(1+j^r)A^j\mathrm{/}j!$ tends to zero, uniformly once $j$ is large; therefore the series converges uniformly. Combined with the Gaussian bound in step 2.1, this verifies the grand theorem's energy-number envelope. Applying its response identity gives ${\mathrm{\partial }}_{\mu }⟨N⟩=\beta \mathrm{Var}(N)=\beta a$. Within each number sector, dividing the sector Boltzmann density by its sector probability cancels the grand factor, leaving precisely the canonical conditional phase law.

Integrating the momentum factors gives [F1]; integrating the unit shape coordinates gives one, while the configuration scaling contributes $V^N$. Hence ${\mathrm{\Delta }}_N=\frac{1}{N!V_0{a}_0^n}{\left(\frac{2\pi m}{\beta }\right)}^{n\mathrm{/}2}{\int }_0^{\mathrm{\infty }}{V}^N{e}^{-sV}dV,s=\beta P_{\mathrm{e}\mathrm{x}\mathrm{t}}>0.$ Let $I_j(s)={\int }_0^{\mathrm{\infty }}{V}^j{e}^{-sV}dV$. Direct integration gives $I_0=1\mathrm{/}s$; integration by parts, with $V^j{e}^{-sV}\to 0$ at infinity and vanishing lower boundary for $j\ge 1$, gives $I_j=j{I}_{j-1}\mathrm{/}s$. Thus $I_j=j!\mathrm{/}{s}^{j+1}$, and $I_N=N!\mathrm{/}{s}^{N+1}$ cancels the $N!$ in the reference; the displayed finite positive normalizer follows.

The normalized volume density is $V^N{e}^{-sV}\mathrm{/}{I}_N(s)$. Applying the same integral formula to $I_{N+1}$ and $I_{N+2}$ gives $⟨V⟩=(N+1)\mathrm{/}s$ and $⟨V^2⟩=(N+1)(N+2)\mathrm{/}{s}^2$. Subtracting the square of the mean gives $\mathrm{Var}(V)=(N+1)\mathrm{/}{s}^2$. Since $s=\beta P_{\mathrm{e}\mathrm{x}\mathrm{t}}$, these are the claimed volume moments.

To check [F2], first use fixed sufficient statistics $(K,V)$ with coefficients $(-\beta ,-\beta P_{\mathrm{e}\mathrm{x}\mathrm{t}})$. On a compact positive $(\beta ,P_{\mathrm{e}\mathrm{x}\mathrm{t}})$ set, choose lower bounds ${\beta }_0,P_0>0$. For $K,V\ge 0$, absorbing the quadratic polynomial into half of each exponential gives $e^{-\beta (K+P_{\mathrm{e}\mathrm{x}\mathrm{t}}V)}(1+K^2+V^2)\le C{e}^{-{\beta }_{0}K\mathrm{/}2-{\beta }_0{P}_{0}V\mathrm{/}2}\mathrm{.}$ This is integrable against the full reference ${\nu }_N$: the momentum integral is Gaussian by [F1] and the volume integral is $I_N({\beta }_0{P}_0\mathrm{/}2)<\mathrm{\infty }$ by step 1.1. At any fixed pressure $P_{\ast }>0$, the statistics $(B_{\ast },V)=(K+P_{\ast }V,V)$ satisfy $u{B}_{\ast }+vV=uK+(u{P}_{\ast }+v)V$. In a small compact neighborhood of $(u,v)=(-\beta ,0)$ both coefficients are bounded strictly below zero; $1+B_{\ast }^2+V^2\le C_{\ast }(1+K^2+V^2)$ gives the same Gaussian-volume majorant. Thus the exact local exponential envelopes for [F2] hold, including the kinetic-energy factor.

The momentum factors are independent of $V$. For $a=\beta \mathrm{/}(2m)>0$, the boundary term $p{e}^{-a{p}^2}$ vanishes at both infinities, and integration of $(p{e}^{-a{p}^2}{)}^{\mathrm{\prime }}=e^{-a{p}^2}-2a{p}^2{e}^{-a{p}^2}$ gives $\int p^2{e}^{-a{p}^2}dp=(2a{)}^{-1}\int e^{-a{p}^2}dp$. Dividing by the positive Gaussian normalizer [F1] and summing the $n$ components gives $⟨K\mid V⟩=n\mathrm{/}(2\beta )$. Step 2.1 gives $P_{\mathrm{e}\mathrm{x}\mathrm{t}}⟨V⟩=(N+1)\mathrm{/}\beta$; adding the two finite means proves $⟨K+P_{\mathrm{e}\mathrm{x}\mathrm{t}}V⟩=(n\mathrm{/}2+N+1)\mathrm{/}\beta$. The envelope check in step 2.2 also licenses the response identities [F2] for this preparation.