= Solution
For the <Schwarzschild criterion>, displace a small gas parcel upwards. Assume its composition is unchanged, it remains in <pressure> balance with its surroundings, and it moves quickly enough that heat exchange is negligible. The parcel obeys the <adiabatic equation of state> $P\rho^{-\gamma}=\mathrm{constant}$. Combining this with the <ideal gas> equation gives $T\propto P^{(\gamma-1)/\gamma}$, so its <adiabatic temperature gradient> is $\nabla_{\mathrm{ad}}=(\gamma-1)/\gamma=2/5$.
Let the surrounding <temperature gradient> be $\nabla=d\log T/d\log P$. For an upward displacement $\delta\log P<0$, the parcel's <temperature> relative to its new surroundings is
$$
\delta\log T_{\mathrm{parcel}}-\delta\log T_{\mathrm{environment}}=(\nabla_{\mathrm{ad}}-\nabla)\delta\log P.
$$
If $\nabla<\nabla_{\mathrm{ad}}$, this is negative: the parcel is cooler and, at equal <pressure> and composition, denser than its surroundings. It therefore sinks back; a downward displacement likewise gives a restoring <buoyancy> force. If the inequality is reversed, the displaced parcel continues to rise. The stability condition is thus
$$
\boxed{\nabla=\frac PT\frac{dT}{dP}<\frac25.}
$$
Equality is the neutral boundary. There is no composition-gradient term because the ideal-gas parcel and the background have the same <mean molecular weight>.
In the thin upper <stellar atmosphere>, take a plane-parallel approximation with constant $g$, optical depth increasing inward, and zero overlying <pressure> $P(0)=0$. <Hydrostatic equilibrium in optical depth> gives $dP/d\tau=g/\kappa$. The <ideal gas> equation converts the stipulated <opacity> into
$$
\kappa=\frac{\kappa_0\mu P}{\mathcal R}T^{4\beta},\qquad \frac{dP^2}{d\tau}=\frac{2\mathcal Rg}{\kappa_0\mu}T^{-4\beta}.
$$
Define $u=1+3\tau/2$, so the given <grey atmosphere> relation is $T^4=T_e^4u/2$. Integrating from the surface gives
$$
\begin{aligned}
P^2&=\frac{2^{\beta+1}\mathcal Rg}{\kappa_0\mu T_e^{4\beta}}\int_0^\tau(1+3t/2)^{-\beta}dt\\
&=\frac{2^{\beta+2}\mathcal Rg}{3\kappa_0(\beta-1)\mu T_e^{4\beta}}\left(1-u^{1-\beta}\right).
\end{aligned}
$$
This proves the required <pressure> law. In particular $P^2\propto gT_e^{-4\beta}$ times a function only of $u$ and $\beta$.
For the <grey-atmosphere convection onset with density-linear opacity>, differentiate the two profiles to obtain the radiative <temperature gradient>:
$$
\frac{d\log T}{d\tau}=\frac3{8u},\qquad \frac{d\log P}{d\tau}=\frac{3(\beta-1)u^{-\beta}}{4(1-u^{1-\beta})},\qquad \nabla_{\mathrm{rad}}=\frac{u^{\beta-1}-1}{2(\beta-1)}.
$$
It increases monotonically from zero, since $\beta>1$, and first reaches the <adiabatic temperature gradient> at $u_b^{\beta-1}=(4\beta+1)/5$. Therefore
$$
\boxed{\tau_b=\frac23\left[\left(\frac{4\beta+1}{5}\right)^{1/(\beta-1)}-1\right].}
$$
Below this point the radiative stratification is unstable and <convection> must transport some of the outward energy flux.
At the outer <convection> boundary, $u_b$ depends only on $\beta$, so $T_b=T_e(u_b/2)^{1/4}\propto T_e$ and $P_b\propto g^{1/2}T_e^{-2\beta}$ for fixed $\mu,\kappa_0,\beta$. The matching constant in $P=KT^{5/2}$ is $K=P_b/T_b^{5/2}$. Consequently
$$
\boxed{K\propto g^{1/2}T_e^{-2\beta-5/2}.}
$$
This <temperature> form of the polytropic relation follows from an ideal monatomic adiabat; its $K$ is not the density-polytrope coefficient in Question 3.
For the <deep radiative boundary beneath a power-law-opacity convective envelope>, negligible envelope <mass> and absent local energy generation give $m\simeq M$ and $L_r\simeq L$. Divide the radiative <temperature> equation by the <pressure> equation to get the <stellar radiative temperature gradient>
$$
\nabla_{\mathrm{rad}}=\frac{3\kappa L P}{16\pi acGM T^4}.
$$
This is the gradient that would be needed if radiation alone carried $L$, and it is the quantity to compare with $2/5$ even while the actual region is convective. The deeper <Kramers' opacity law> gives $\kappa=\kappa_1\mu PT^{-9/2}/\mathcal R$. Substituting $P=KT^{5/2}$ yields
$$
\nabla_{\mathrm{rad}}=\frac{3\kappa_1\mu K^2L}{16\pi ac\mathcal RGM T^{7/2}}.
$$
As $T$ rises inward this decreases, so the transition back to a stable radiative interior is at
$$
\boxed{\frac{3\kappa_1\mu K^2L}{16\pi ac\mathcal RGM T_{\mathrm{in}}^{7/2}}=\frac25.}
$$
Thus $T_{\mathrm{in}}^{7/2}\propto K^2L/M$. Using the outer matching and the <Stefan–Boltzmann law>,
$$
K^2\frac LM\propto gT_e^{-4\beta-5}\frac LM,\qquad g\frac LM=\frac{GM}{R^2}\frac{4\pi R^2\sigma T_e^4}{M}=4\pi G\sigma T_e^4.
$$
The <mass> and <radius> cancel. Hence, at fixed <opacity> coefficients and composition,
$$
\boxed{T_{\mathrm{in}}\propto T_e^{-(8\beta+2)/7}.}
$$
This uses the same adiabat through the <convection> zone and distinguishes its inner transition <temperature> from its outer onset <temperature>.
Back to article page
