= Bose condensation temperature from a fermionic density scale
{c}
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 $k_BT_F=\hbar^2(6\pi^2n/g_F)^{2/3}/(2m_F)$, while Bose saturation gives $k_BT_B=2\pi\hbar^2[n/\zeta(3/2)]^{2/3}/m_B$. Eliminating <density> yields
$$
\frac{T_B}{T_F}=\frac{4\pi}{(6\pi^2/g_F)^{2/3}\zeta(3/2)^{2/3}}\frac{m_F}{m_B}.
$$
The 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.
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