= Solution
To derive the <nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture>, take fully ionized <hydrogen> and <helium> with <hydrogen mass fraction> $X$, the <electron> <number density> is
$$
n_e=\frac{X\rho}{m_p}+2\frac{(1-X)\rho}{4m_p}=\frac{(1+X)\rho}{2m_p}.
$$
Here <helium> nuclei have <mass> $4m_p$ in the approximation used in the question. Filling the two spin states of each <momentum> mode up to the <Fermi momentum> $p_0$ gives
$$
n_e=\int_0^{p_0}\frac{8\pi p^2}{h^3}dp=\frac{8\pi p_0^3}{3h^3},\qquad p_0=h\left(\frac{3n_e}{8\pi}\right)^{1/3}.
$$
The isotropic <momentum> flux gives <pressure> $\frac13\int pv\,dn$. For nonrelativistic <electrons> $v=p/m_e$, so the <electron degeneracy pressure> is
$$
\begin{aligned}
P_e&=\frac13\int_0^{p_0}\frac{p^2}{m_e}\frac{8\pi p^2}{h^3}dp=\frac{8\pi p_0^5}{15m_eh^3}\\
&=\frac{h^2}{5m_e}\left(\frac3{8\pi}\right)^{2/3}n_e^{5/3}.
\end{aligned}
$$
Substitute the composition-dependent $n_e$ and simplify powers of two:
$$
\boxed{P_e=K\rho^{5/3},\qquad K=\left(\frac3{2\pi}\right)^{2/3}\frac{h^2(1+X)^{5/3}}{40m_em_p^{5/3}}.}
$$
Comparing $5/3=1+1/n$ gives \b[the <polytropic index> $n=3/2$]. The result assumes complete <electron> degeneracy, $p_0\ll m_ec$, and negligible nuclear <mass> corrections.
Now adopt the stated approximate partially degenerate <equation of state>, keeping composition and hence $K,\mu$ fixed during contraction. Set $q=m/M$ and define the mass-averaged <temperature> $\overline T_M=\int_0^1T\,dq$. The <stellar virial theorem> from part (a) gives
$$
3K\int_0^1\rho^{2/3}dq+\frac{3\mathcal R}{\mu}\overline T_M=-\frac{\Omega}{M}.
$$
For an index-$3/2$ <stellar polytrope>, the permitted gravitational-energy formula is $\Omega=-6GM^2/(7R)$. Also, using $\bar\rho=3M/(4\pi R^3)$,
$$
\rho^{2/3}=\frac{M^{2/3}}{R^2}\left(\frac{3\rho}{4\pi\bar\rho}\right)^{2/3}.
$$
Thus, with $A=3K\int_0^1[3\rho/(4\pi\bar\rho)]^{2/3}dq$,
$$
\boxed{\overline T_M=\frac{2\mu GM}{7\mathcal RR}-\frac{\mu A M^{2/3}}{3\mathcal RR^2}.}
$$
The normalized <density> $\rho/\bar\rho$ is a fixed function of $q$ for this homologous polytropic structure. Therefore its integral and $K$ fix $A$, which is independent of $M$ and $R$ for a common composition. The mean here is mass-weighted, rather than a volume average.
For the <mass-averaged temperature maximum in a partially degenerate polytrope>, put $\alpha=2\mu GM/(7\mathcal R)$ and $\delta=\mu A M^{2/3}/(3\mathcal R)$, both positive. As a function of $s=1/R$ the <temperature> is the concave quadratic
$$
\overline T_M=\alpha s-\delta s^2=\frac{\alpha^2}{4\delta}-\delta\left(s-\frac\alpha{2\delta}\right)^2.
$$
It reaches its maximum when $R=2\delta/\alpha=7A M^{-1/3}/(3G)$, with
$$
\boxed{\overline T_{M,\max}=\frac{\alpha^2}{4\delta}=\frac{3\mu G^2M^{4/3}}{49\mathcal RA}.}
$$
At large <radius>, contraction increases the <temperature> as in a thermally supported <star>. Once <electron> degeneracy provides a sufficiently large fraction of the support, further contraction can lower the thermal <temperature>. The formal zero-temperature <radius> is $R_{\mathrm{deg}}=\delta/\alpha=R_{\max}/2$, with $R_{\mathrm{deg}}\propto M^{-1/3}$. The continuation to negative temperatures would be unphysical; the adopted sequence only describes its positive-temperature part.
The <contraction-limited central temperature> has the same <mass> scaling. Indeed, if the total polytropic <pressure> is $P=K_{\mathrm{tot}}\rho^{5/3}$, the approximate <equation of state> gives $T=(\mu/\mathcal R)(K_{\mathrm{tot}}-K)\rho^{2/3}$. The fixed <density> profile therefore fixes $T_c/\overline T_M$. A sufficiently small <mass> never reaches the central <temperature> required for sustained <hydrogen burning>, producing a <hydrogen-burning minimum mass>. Such an object becomes a cooling, degeneracy-supported <brown dwarf> instead of settling onto the <main sequence>. More massive contracting objects can ignite <hydrogen> before reaching this degeneracy-limited maximum. This conclusion uses the nonrelativistic, fixed-composition model; it does not require extrapolation into the relativistic regime.
Back to article page
