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Put $\Delta =R-r\ll R$. The near-surface mass density and mean free path from a and b areA simple local escape estimate sets the remaining distance equal to the mean free path, $\ell \simeq \Delta$. It givesFor the Sun's parameters this is $\boxed{\Delta R_{\mathrm{esc},\mathrm{local}}\simeq 1.94\mathrm{km},\Delta R_{\mathrm{esc},\mathrm{local}}/R\simeq 2.79\times 10^{-6}}$.

A more precise operational definition uses the outward radial optical depth. In this varying mass density,The chance of an outward radial photon escaping without another scattering is $e^{-\tau }$. Thus the conventional $\tau =1$ last-scattering depth givesThere is no sharp escape surface: $\tau \ll 1$ gives nearly free propagation, and oblique rays have different optical depths. The local $\ell =\Delta$ estimate actually has $\tau \simeq 1/2$ in this model, not $1$; the factor $\sqrt{2}$ is a convention-dependent order-one correction, not a discrepancy in the Thomson cross-section. Neither scattering criterion establishes a thermalization surface.

For a spherical star in Newtonian hydrostatic equilibrium, the total supporting pressure satisfieswhere zero external pressure is the idealized vacuum boundary condition. Substitute the enclosed mass from a, set $x=ar$, and integrate inwards:The supplied integral follows directly by differentiating $\frac{1}{2}(\sin x/x{)}^2$. It yieldsThus $P_c=\pi G{M}^2/(8R^4)$, about $4.42\times 10^{14}\mathrm{Pa}$ for the Sun's parameters. This is precisely the equation of state of a polytrope of index one.

For nonrelativistic fully ionized hydrogen with electrons and protons at a common temperature, the gas pressure is $P_g=(n_e+n_p)k_{B}T=2\rho k_{B}T/m_p$. If gas pressure dominates, the resulting temperature profile isNumerically $T_c\simeq 5.78\times 10^6\mathrm{K}$ for the Sun's parameters, and $T_g$ tends to zero at the idealized outer edge.

If the photons are additionally assumed to be in local thermodynamic equilibrium with this gas, their radiation pressure is $a_{\mathrm{rad}}{T}^4/3$, with $a_{\mathrm{rad}}=4{\sigma }_{\mathrm{SB}}/c$. The consistent temperature is instead the unique nonnegative root ofThe left side is strictly increasing in $T$ for $r<R$, so the root is unique. Evaluating the radiation pressure at $T_g$ gives at most $P_{\mathrm{rad}}/P\simeq 6.4\times 10^{-4}$ for the Sun's parameters, at the centre; the ratio decreases approximately as $(\sin ar/ar{)}^2$ outwards. The gas-only profile is consequently an excellent approximation within this imposed stellar model.

Hydrostatic equilibrium by itself fixes total pressure, not how it is divided between gas and radiation. Thomson scattering alone does not enforce a common gas–radiation temperature or a Planck photon distribution. Without either neglecting radiation pressure or specifying the radiation field and its thermodynamic closure, a unique gas temperature cannot be inferred from the given mass density alone.