Classical ideal gas in a spherical linear trap ($Z_1=8\pi (k_{B}T/a{)}^3/{\lambda }_T^3$)

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.