 Thomson escape layer of an index-one polytrope

For a polytrope of index one with $\rho (r)={\rho }_c\sin (\pi r/R)/(\pi r/R)$ and fully ionized hydrogen, Thomson scattering gives the central mean free path $L=m_p/({\sigma }_T{\rho }_c)$. At small depth $\Delta =R-r$, $\ell \simeq LR/\Delta$ and the radial optical depth is $\tau \simeq {\Delta }^2/(2LR)$. Thus $\ell =\Delta$ gives $\Delta =\sqrt{LR}$ whereas $\tau =1$ gives $\Delta =\sqrt{2LR}$. Neither condition by itself guarantees thermalization.