= Thomson escape layer of an index-one polytrope
{c}
{title2=$\Delta R\sim\sqrt{LR}$}
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.
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