 Solution

For a slowly rotating central star, the accretion-disk boundary layer radiates a luminosity $L_{\mathrm{BL}}\sim G{M}_{\ast }\dot{M}/(2r_{\ast })$. A layer of thickness comparable to the disk scale height has emitting area $A_{\mathrm{BL}}\sim r_{\ast }H$ times a geometric constant. Applying the Stefan–Boltzmann law to this blackbody area givesComparing with $\sigma T_{\mathrm{in}}^4=3G{M}_{\ast }\dot{M}/(8\pi r_{\ast }^3)$ gives the boundary-layer temperature scalingOnly the scaling is fixed: for a two-faced annulus of area $4\pi r_{\ast }H$, for example, $T_{\mathrm{BL}}^4/T_{\mathrm{in}}^4=r_{\ast }/(3H)$. A surface belt gives a different order-one coefficient.

For a thin disk, $H/r_{\ast }\ll 1$, so a comparable luminosity emerges from a smaller emitting area at a higher effective temperature. 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.