 Standard thin-disk dissipation flux

Consider a steady Keplerian disk around a central mass $M$, with positive inward accretion rate $\dot{M}$ and zero viscous torque in an accretion disk at $R_{\mathrm{in}}$. Angular momentum conservation gives the outward internal torqueDifferential rotation dissipates energy at the rate $\mathcal{T}(-d\Omega /dR)dR$ in an annulus. Dividing by its projected area $2\pi RdR$ and using $\Omega =\sqrt{GM/R^3}$ gives the flux summed over both faces:The energy flux through one face is $F_{\mathrm{diss}}/2$. In a locally radiating, optically thick disk, one therefore sets $\sigma T_{\mathrm{eff}}^4=F_{\mathrm{diss}}/2$, rather than equating a single-face blackbody flux to the two-face expression.^[ref]

The bracket enforces the zero-torque inner boundary. Far outside it, $F_{\mathrm{diss}}\propto R^{-3}$. Internal torque transports energy between annuli as well as transporting angular momentum, so local dissipation is not simply the local loss of orbital binding energy. A nonzero inner torque, strong advective transport or a mass outflow changes this model.