Bose condensation temperature from a fermionic density scale

Compare uniform three-dimensional ideal gases at the same number density n, with fermion mass $m_F$, spin degeneracy $g_F$, boson mass $m_B$ and a single boson spin component. The Fermi temperature has $k_B{T}_F={\hslash }^2(6{\pi }^{2}n/g_F{)}^{2/3}/(2m_F)$, while Bose saturation gives $k_B{T}_B=2\pi {\hslash }^2[n/\zeta (3/2){]}^{2/3}/m_B$. Eliminating density yieldsThe degeneracy is essential; using a spinless Fermi sea changes the numerical estimate. This is an ideal-gas comparison and does not remove interaction effects in a dense liquid.