Finite gravitational phase and entropy reference

Use SI dimensions for the numerical model parameters: Newton constant $G$ has units ${\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, each mass $m_i$ has units $\mathrm{k}\mathrm{g}$, each position $q_i$ has units $\mathrm{m}$, each momentum $p_i$ has units $\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-1}$, energy $H$ has units $\mathrm{J}$, Boltzmann constant $k_B$ has units $\mathrm{J}{\mathrm{K}}^{-1}$, and inverse energy $\beta$ has units ${\mathrm{J}}^{-1}$. Fix an action reference $h_{\ast }>0$ in $\mathrm{J}\mathrm{s}$; it is a normalization convention and is not identified with Planck's constant.

For an integer $N\ge 1$, any Lebesgue measurable position region $B\subseteq {\mathbb{R}}^3$, and equal particle mass $m>0$, define the ordered phase space ${\mathrm{\Gamma }}_N=B^N\times {\mathbb{R}}^{3N}$ with coordinates $(q_1,\dots ,q_N,p_1,\dots ,p_N)$. This phase space and measure are allowed even when $B$ is unbounded or has infinite volume; finite canonical preparations separately require a positive finite partition integral. The dimensionless reference measure for identical classical particles is $d{\nu }_N=\frac{d^{3N}q{d}^{3N}p}{N!h_{\ast }^{3N}}\mathrm{.}$ For a measurable real Hamiltonian $H:{\mathrm{\Gamma }}_N\to \mathbb{R}$ and $\beta >0$ with $0<Z_N(\beta ):={\int }_{{\mathrm{\Gamma }}_N}{e}^{-\beta H}d{\nu }_N<\mathrm{\infty }$, define the canonical density $p_{\beta }=Z_N(\beta {)}^{-1}{e}^{-\beta H}$ relative to ${\nu }_N$. A probability density relative to ${\nu }_N$ is a measurable function $\rho :{\mathrm{\Gamma }}_N\to [0,\mathrm{\infty })$ with ${\int }_{{\mathrm{\Gamma }}_N}\rho d{\nu }_N=1$. Set $0\log 0:=0$. For such a density with ${\int }_{{\mathrm{\Gamma }}_N}\mathrm{\mid }\rho \log \rho \mathrm{\mid }d{\nu }_N<\mathrm{\infty }$, define its reference-relative entropy by $S(\rho )=-k_B{\int }_{{\mathrm{\Gamma }}_N}\rho \log \rho d{\nu }_N\mathrm{.}$

When $H$ is fixed and the Gibbs-volume function $\mathrm{\Omega }(E):={\nu }_N\{H\le E\}$ satisfies $0<\mathrm{\Omega }(E)<\mathrm{\infty }$, define the Gibbs-volume entropy by $S_G(E)=k_B\log \mathrm{\Omega }(E)$. Where $\omega (E):={\mathrm{\Omega }}^{\mathrm{\prime }}(E)>0$, define the distinct surface-entropy convention, for a fixed energy width $\mathrm{\Delta }E>0$ measured in the same units as $E$, by $S_B(E)=k_B\log (\omega (E)\mathrm{\Delta }E)$. Since ${\nu }_N$ and $\mathrm{\Omega }$ are dimensionless, $\omega$ has units inverse to energy and $\omega \mathrm{\Delta }E$ is dimensionless. For either $\alpha \in \{G,B\}$, where $S_{\alpha }$ is differentiable and $S_{\alpha }^{\mathrm{\prime }}(E)>0$, define $T_{\alpha }(E):=[S_{\alpha }^{\mathrm{\prime }}(E){]}^{-1}$; where also $T_{\alpha }^{\mathrm{\prime }}(E)\mathrm{\ne }0$, define $C_{\alpha }(E):=dE\mathrm{/}d{T}_{\alpha }=1\mathrm{/}{T}_{\alpha }^{\mathrm{\prime }}(E)$. These are finite-system conventions, and their temperatures and capacities need not agree.

The measure is dimensionless because $d^{3N}q{d}^{3N}p$ and $h_{\ast }^{3N}$ have the same units. The definition supplies no physical assertion that the chosen Hamiltonian describes nature and no mechanism of dynamical equilibration.