Classical ideal-gas partition function

For $N$ indistinguishable noninteracting classical particles, $Z_N=z_1^N/N!$, where $z_1$ is the normalized one-particle phase-space integral.

For a structureless particle with Hamiltonian $p^2/(2m)+ar$, $a>0$, the classical phase space integral gives the displayed one-particle partition function. Its spatial integral is $4\pi {\int }_0^{\infty }{r}^2{e}^{-ar/(k_{B}T)}dr=8\pi (k_{B}T/a{)}^3$, while the momentum integral gives ${\lambda }_T^{-3}$. For $N$ indistinguishable dilute particles, $Z_N=Z_1^N/N!$, $F\simeq -N{k}_{B}T[\ln (Z_1/N)+1]$, $U=(9/2)N{k}_{B}T$ and $S\simeq N{k}_B[\ln (Z_1/N)+11/2]$. These thermodynamic formulas use leading Stirling accuracy; the classical phase-space integral additionally requires temperatures above the discrete trap level scale.

For a dilute classical trapped gas, $\mu \simeq k_{B}T\ln (N/Z_1)$ with energy zero at the potential minimum. Thus $z=e^{\mu /(k_{B}T)}=N/Z_1$. Quantum occupation effects begin when $z$ is no longer small, giving the displayed order-of-magnitude particle threshold. Equivalently, the central density $n(0)=N/[8\pi (k_{B}T/a{)}^3]$ satisfies $n(0){\lambda }_T^3\sim 1$. This is a degeneracy crossover criterion rather than an exact Bose condensation threshold. In the continuum approximation, spinless bosons have excited-state saturation $N_c=\zeta (9/2)Z_1$, from the trap density of states proportional to $E^{7/2}$. A quantum ground energy should be subtracted consistently when discreteness matters.

At fixed atom number, the entropy of the classical ideal gas in a spherical linear trap depends on temperature and strength through $Z_1\propto T^{9/2}{a}^{-3}$. Reversible adiabatic change therefore holds this product constant, giving the displayed relation. Reducing trap strength lowers the temperature while increasing cloud size. Its peak classical phase-space occupation $N/Z_1$ remains constant, so this cooling alone does not make the gas more quantum degenerate. Heat exchange, particle loss or irreversible entropy production would change this relation.

For the potential $U=r^{2n}/V^{2n/3}$, rescaling $r=V^{1/3}(kT{)}^{1/(2n)}u$ in the classical one-particle partition function gives the formula above. The parameter-conjugate force is $P=NkT/V$, the internal energy is $(3/2+3/(2n))NkT$, and $C_P-C_V=Nk$. The kinetic term contributes $3NkT/2$, while the potential term contributes $3NkT/(2n)$. For indistinguishable particles the $N!$ factor changes free energy and entropy but not these force and heat-capacity formulas.

The three-dimensional translational momentum integral is $1/{\lambda }^3$ per unit volume, where

Equating thermal de Broglie wavelength $h/\sqrt{2\pi m{k}_{B}T}$ to mean particle spacing $n^{-1/3}$ gives the displayed quantum-degeneracy scale. It is an onset estimate rather than a universal phase-transition temperature. Different typical-momentum conventions change its order-one factor; density, mass, spin degeneracy and interactions determine the detailed crossover or condensation criterion.

The translational concentration scale for a nonrelativistic particle in three dimensions, per internal state. It is the inverse cube of the thermal de Broglie wavelength. A classical component with number density $n$ and external one-particle energy $U$ has chemical potential $\mu =k_{B}T\ln (n/n_Q)+U$. An additional internal degeneracy must be included explicitly; it is not hidden in this definition. The dilute classical condition is $n/n_Q\ll 1$ for each occupied internal component.

For energy $pc+U(x)$,If $U(r)=r^{2n}/V^{2n/3}$, thenConsequently $pV=N{k}_{B}T$, while the mean energy and variance are $NA{k}_{B}T$ and $NA(k_{B}T{)}^2$ with $A=3+3/(2n)$.

The radial density is proportional to $r^2{e}^{-r^{2n}/(V^{2n/3}{k}_{B}T)}$, whose mode is

In the classical dilute regime, a monatomic ideal gas has entropy