Ideal Fermi continuum and zero-temperature endpoint

Under the adopted preparation [A1] and the exact periodic continuum state [F1], for fixed $\beta >0$, the integrand is continuous in $(k,\mu )$ and strictly increases at every $k$ with $\mu$. The Gaussian tail bound from [F2] is uniform when $\mu$ ranges over a compact interval, so the integral is continuous in $\mu$. As $\mu \to -\mathrm{\infty }$, the pointwise bound $f_F\le e^{\beta \mu }{e}^{-\beta \alpha \mathrm{\mid }k{\mathrm{\mid }}^2}$ makes $\rho \to 0$. As $\mu \to +\mathrm{\infty }$, on the ball $\alpha \mathrm{\mid }k{\mathrm{\mid }}^2\le \mu$ the occupation is at least $1\mathrm{/}2$; the ball volume tends infinity, so $\rho \to \mathrm{\infty }$. Strict monotonicity and the intermediate value theorem give a unique chemical potential for every positive $\rho$.

Fix $\mu >0$ and let $k_{\mu }=\sqrt{\mu \mathrm{/}\alpha }$. Off any annulus $\mathrm{\mid }\mathrm{\mid }k\mathrm{\mid }-k_{\mu }\mathrm{\mid }<\delta$, the Fermi occupation converges uniformly on bounded sets to the indicator of $\mathrm{\mid }k\mathrm{\mid }<k_{\mu }$, since the exponent has a positive gap from zero there. The annulus volume tends zero with $\delta$ and the occupation is bounded by one, so its number and bounded-energy contributions vanish as $\delta \downarrow 0$. Fix ${\beta }_0>0$ in J${}^{-1}$. For $\mathrm{\mid }k\mathrm{\mid }$ sufficiently large, $\alpha \mathrm{\mid }k{\mathrm{\mid }}^2\ge 2\mu$ and, for every $\beta \ge {\beta }_0$, $f_F\le e^{-\beta \alpha \mathrm{\mid }k{\mathrm{\mid }}^2\mathrm{/}2}\le e^{-{\beta }_0\alpha \mathrm{\mid }k{\mathrm{\mid }}^2\mathrm{/}2}$. The latter Gaussian and its energy-weighted counterpart $\alpha \mathrm{\mid }k{\mathrm{\mid }}^2{e}^{-{\beta }_0\alpha \mathrm{\mid }k{\mathrm{\mid }}^2\mathrm{/}2}$ are integrable and bound the respective number and energy tails uniformly for $\beta \ge {\beta }_0$. Splitting into the bounded region and this tail proves convergence of both integrals to their filled-ball integrals.

Spherical integration of the limiting indicator gives ${\rho }_0(\mu )=g(2\pi {)}^{-3}(4\pi k_{\mu }^3\mathrm{/}3)=g{k}_{\mu }^3\mathrm{/}(6{\pi }^2)$ and $u_0(\mu )=g(2\pi {)}^{-3}4\pi \alpha k_{\mu }^5\mathrm{/}5=g\alpha k_{\mu }^5\mathrm{/}(10{\pi }^2)=(3\mathrm{/}5){\rho }_0(\mu )\mu$. Solving ${\rho }_0(E_F)=\rho$ gives the displayed $k_F,E_F$. For any $0<\delta <E_F$, the zero-temperature limits at $E_F-\delta$ and $E_F+\delta$ strictly bracket $\rho$; for sufficiently large $\beta$ the finite-temperature densities at these endpoints still bracket $\rho$, so step 1.1 traps $\mu (\rho ,\beta )$ between them. The energy density is increasing in $\mu$ pointwise; therefore $u_{\beta }(E_F-\delta )\le u_{\beta }(\mu (\rho ,\beta ))\le u_{\beta }(E_F+\delta )$. Step 1.2 gives the limits of both bounds, and continuity of $u_0(\mu )$ as $\delta \downarrow 0$ proves $u\to (3\mathrm{/}5)\rho E_F$.

Let $L(r)=\log (1+e^{-\beta (\alpha r^2-\mu )})$. Then $L^{\mathrm{\prime }}(r)=-2\beta \alpha r{f}_F(r)$. The boundary term $r^{3}L(r)$ vanishes at $r=0$ and at infinity, the latter because $L(r)\le e^{-\beta (\alpha r^2-\mu )}$. Integration by parts on $[0,R]$ followed by $R\to \mathrm{\infty }$ gives ${\int }_0^{\mathrm{\infty }}{r}^{2}L(r)dr=(2\beta \alpha \mathrm{/}3){\int }_0^{\mathrm{\infty }}{r}^4{f}_F(r)dr$. Substitution into the radial definitions of $p$ and $u$ gives $p=2u\mathrm{/}3$. Passing to the limit in step 2.1 yields $p\to 2\rho E_F\mathrm{/}5$. These are ideal-model predictions; no material pressure is inferred.