= Solution
Consider a small outward displacement $\delta r$ of a fluid element. It adjusts to the ambient <pressure> while retaining its <specific entropy> and <stellar composition>, so an <adiabatic process> gives $d\log\rho_{\rm parcel}=d\log P/\gamma$. Its excess <mass density> over the new surroundings is, to first order,
$$
\rho_{\rm parcel}-\rho_{\rm ambient}
=\left[\frac\rho{\gamma P}\frac{dP}{dr}-\frac{d\rho}{dr}\right]\delta r.
$$
Restoring <buoyancy> requires this bracket to be positive. With $\gamma=5/3$, the <Schwarzschild criterion> is therefore
$$
\boxed{\frac{dP}{dr}>\frac{5P}{3\rho}\frac{d\rho}{dr}.}
$$
Equality is neutral stability. For uniform <mean molecular weight>, the <ideal gas> law gives $d\log\rho=d\log P-d\log T$. Since <hydrostatic equilibrium> <pressure> decreases outward, the inequality is equivalent to $\nabla=d\log T/d\log P<\nabla_{\rm ad}=2/5$. Dividing the <radiative diffusion in a star> and <hydrostatic equilibrium> <stellar structure equations> gives
$$
\nabla_{\rm rad}=\frac{3\kappa L_rP}{16\pi acGmT^4},\qquad
\boxed{\frac{3\kappa L_rP}{16\pi acGmT^4}<\frac25.}
$$
This is the <stellar radiative temperature gradient> test for <stellar convective stability>.
In the thin upper <stellar atmosphere> take $m\simeq M$, $L_r\simeq L$ and use <gas pressure> dominance. The prescribed <opacity> becomes $\kappa=\kappa_0\mu PT^{12}/\mathcal R$. Dividing the two structure equations then yields
$$
P\frac{dP}{dT}=\frac{16\pi acGM\mathcal R}{3\kappa_0L\mu}T^{-9}.
$$
At zero <optical depth>, the atmospheric boundary has $T_s^4=T_e^4/2$ and negligible <gas pressure>. Integrating from that boundary, with $T_s^{-8}=4T_e^{-8}$, gives
$$
\boxed{P^2=\frac{4\pi acGM\mathcal R}{3\kappa_0L\mu T_e^8}
\left(4-\frac{T_e^8}{T^8}\right).}
$$
To locate the onset, put $y=T_e^8/T^8$. <Logarithmic derivative> gives $d\log P/d\log T=4y/(4-y)$, hence
$$
\nabla_{\rm rad}=\frac{4-y}{4y}=\frac{T^8}{T_e^8}-\frac14.
$$
This starts at zero at the outer boundary and increases inward. It first reaches $2/5$ at
$$
\boxed{T_b=(13/20)^{1/8}T_e.}
$$
The corresponding <optical depth> is $\tau_b=\tfrac43(\sqrt{13/20}-\tfrac12)>0$, so the onset lies inside the <stellar atmosphere>. This is <grey-atmosphere convection onset with thirteenth-power opacity>.
Let the interior of the <fully convective star> have $P=K_TT^{5/2}$, with $K_T$ spatially constant. At its boundary $T_b/T_e$ is fixed, so the atmospheric <pressure> formula gives
$$
K_T^2=\frac{P_b^2}{T_b^5}\propto\frac M{LT_e^{13}}.
$$
The <Stefan–Boltzmann law>, $L=4\pi R^2\sigma T_e^4$ with $\sigma=ac/4$, consequently implies
$$
K_T^2\propto MR^{13/2}L^{-17/4}.
$$
The <ideal gas> equation transforms the interior relation into
$$
P=K_\rho\rho^{5/3},\qquad K_\rho=(\mathcal R/\mu)^{5/3}K_T^{-2/3}.
$$
For completeness, the mass-radius scaling follows by taking $\rho=\rho_c\theta^{3/2}$ and $r=\alpha\xi$ with $\alpha^2=5K_\rho\rho_c^{-1/3}/(8\pi G)$. <Hydrostatic equilibrium> and <mass conservation> reduce to the <Lane-Emden equation> $(\xi^2\theta')'=-\xi^2\theta^{3/2}$, with $\theta(0)=1$, $\theta'(0)=0$. Its fixed <dimensionless> profile gives $R\propto\alpha$ and $M\propto\alpha^3\rho_c$. Eliminating $\rho_c$ gives $R\propto K_\rho G^{-1}M^{-1/3}$, or $K_\rho\propto GM^{1/3}R$. Thus $K_T^2\propto M^{-1}R^{-3}$ at fixed <stellar composition>. Equating this with the atmospheric expression gives $L^{17/4}\propto M^2R^{19/2}$, and hence
$$
\boxed{L\propto M^{8/17}R^{38/17}.}
$$
This <Hayashi relation with thirteenth-power opacity> concerns the interior of a <fully convective star> beneath a thin <stellar atmosphere>. The coefficient $K_T$ may vary between stars; spatially constant <specific entropy> does not mean equal <specific entropy> across the whole sequence.
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