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An accretion-disk boundary layer connects nearly Keplerian rotation to a stellar surface rotating at a different angular speed. For a slowly rotating star, dissipation of the incoming orbital kinetic energy gives $L_{\mathrm{BL}}\simeq GM\dot{M}/(2r_{\ast })$, comparable to the standard thin-disk luminosity. If the emitting area scales as $r_{\ast }H$, its blackbody effective temperature scales as $T_{\mathrm{BL}}\sim (r_{\ast }/H{)}^{1/4}{T}_{\mathrm{in}}$. The coefficient depends on geometry, and stellar rotation reduces the available energy.

Integrating the standard thin-disk dissipation flux over annular area, with both faces already included in $\mathcal{H}$, givesThus the disk luminosity for $r_{\mathrm{out}}\gg r_{\ast }$ isThe Newtonian gravitational potential decreases by approximately $G{M}_{\ast }/r_{\ast }$ per unit mass from a distant outer edge to the surface, giving a potential-energy release rate $G{M}_{\ast }\dot{M}/r_{\ast }$. The standard thin-disk luminosity is half of this.

The missing half remains as kinetic energy of nearly circular orbital motion: at the inner edge $v_K^2=G{M}_{\ast }/r_{\ast }$, so the specific orbital kinetic energy is $G{M}_{\ast }/(2r_{\ast })$. Equivalently, circular motion has total specific mechanical energy $-G{M}_{\ast }/(2r_{\ast })$. Matter joining a slowly rotating star must shed this orbital motion in an accretion-disk boundary layer, producing approximately another $G{M}_{\ast }\dot{M}/(2r_{\ast })$ of luminosity. A rotating star can retain some energy in spin, so equal disk and accretion-disk boundary layer luminosities assume slow stellar rotation. For a central black hole, there is no material surface, and energy can instead be carried inward.

The stated standard thin-disk dissipation flux is summed over the two faces, so the total luminosity isThis equals the Newtonian standard thin-disk luminosity and the orbital binding energy delivered per unit time at the inner edge. It is half the magnitude $GM\dot{m}/R_{\mathrm{ISCO}}$ of the potential-energy decrease because the other half appears as orbital kinetic energy. In the zero-torque model that remaining mechanical energy passes through the inner edge rather than being dissipated at larger radii. The associated Newtonian radiative efficiency of black-hole accretion is $GM/(2R_{\mathrm{ISCO}}{c}^2)$.