= Solution
For a slowly rotating central star, the <accretion-disk boundary layer> radiates a <luminosity> $L_{\rm BL}\sim GM_*\dot M/(2r_*)$. A layer of thickness comparable to the <disk scale height> has emitting area $A_{\rm BL}\sim r_*H$ times a geometric constant. Applying the <Stefan–Boltzmann law> to this <blackbody> area gives
$$
\sigma T_{\rm BL}^4\sim\frac{GM_*\dot M}{r_*^2H}.
$$
Comparing with $\sigma T_{\rm in}^4=3GM_*\dot M/(8\pi r_*^3)$ gives \b[the boundary-layer <temperature> scaling]
$$
\boxed{T_{\rm BL}\sim\left(\frac{r_*}{H}\right)^{1/4}T_{\rm in}.}
$$
Only the scaling is fixed: for a two-faced annulus of area $4\pi r_*H$, for example, $T_{\rm BL}^4/T_{\rm in}^4=r_*/(3H)$. A surface belt gives a different order-one coefficient.
For a <thin disk>, $H/r_*\ll1$, so a comparable <luminosity> emerges from a smaller emitting area at a higher <effective temperature>. \b[The boundary-layer emission is harder than the disk emission], with its <Planck law> peak shifted to higher <frequency>. It is also closer to a single-<temperature> component than the broad <multitemperature blackbody disk> spectrum, under the adopted uniform-<temperature> approximation. Rapid stellar rotation weakens the heating and this conclusion's <temperature> contrast.
Back to article page
