= Solution
For the unpolarized gas write $x=n/2$. The positive direct <Hartree energy> before background cancellation and the <kinetic energy> are
$$
\frac{E_H}{V}=\frac\gamma2(2x)^2=2\gamma x^2,
\qquad \frac TV=2Ax^{5/3}.
$$
Their ratio is therefore
$$
\boxed{r_s=\frac{E_H}{T}=\frac\gamma A\left(\frac n2\right)^{1/3}.}
$$
This compares the separate direct and kinetic terms. The net Hartree-plus-background contribution after the stipulated cancellation is zero, so it would be incorrect to interpret $r_s$ as that residual contribution divided by $T$. The exchange-energy magnitude in the unpolarized state is $E_H/2$, not $E_H$.
Back to article page
