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In the vertical equations, $\mu$ denotes dynamic viscosity, while ${\mu }_m$ is the mean molecular weight in units of the proton mass $m_p$. The symbol $k$ denotes the Boltzmann constant. For the intended hydrostatic equilibrium calculation, put $P=K{\rho }^{\gamma }$ with $K>0$, $\gamma >1$, and set $\mathcal{R}=k/({\mu }_m{m}_p)$. Dividing the hydrostatic equation by $\rho$ and integrating from the symmetric midplane givesThe specific enthalpy vanishes at the free surface. Hence the polytropic vertical structure in stellar gravity isOn $|z|<H$, the ideal gas relation gives $T=T_m(1-z^2/H^2)$ and $P=P_m(1-z^2/H^2{)}^{\gamma /(\gamma -1)}$; the vacuum extension sets $\rho =P=0$ outside. The formula needs $\gamma >1$ for a finite positive semi-thickness and assumes the harmonic vertical gravity remains valid through this height.

There is a genuine inconsistency if all four printed structure equations are imposed literally with $F=0$ and finite positive opacity: radiative diffusion then requires $dT/dz=0$, whereas the polytropic temperature above has a nonzero gradient away from the midplane. The intended result is valid for the hydrostatic polytropic equation of state with the radiative closure replaced or neglected, for example an adiabatic opaque limit. It is not an exact simultaneous solution of the stated finite-opacity radiative system. With that radiative system and a nonvacuum ideal gas, $\gamma =1$ instead gives an isothermal solutionwhich has no finite free surface. For $\gamma \ne 1$, constant $T$ and constant $K$ would force constant density, contradicting hydrostatic balance on a nontrivial vertical interval. This is the zero-radiative-flux obstruction for a polytropic ideal gas.