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.

A Keplerian accretion disk with a zero-torque inner boundary condition and local blackbody cooling has $T_{\mathrm{eff}}^4=3GM\dot{M}(1-\sqrt{r_{\mathrm{in}}/r})/(8\pi \sigma r^3)$. The Stefan–Boltzmann law uses the flux through one face; the standard thin-disk dissipation flux includes both faces. The scale $T_{\mathrm{in}}=[3GM\dot{M}/(8\pi \sigma r_{\mathrm{in}}^3){]}^{1/4}$ is a normalization, not the actual temperature at the stress-free inner edge.

For the effective-temperature profile of a zero-torque disk, differentiation of $y^{-3}(1-y^{-1/2})$ with $y=r/r_{\mathrm{in}}$ gives $y_{\max }=49/36$. Thus $T_{\max }=(6/7{)}^{3/2}{7}^{-1/4}{T}_{\mathrm{in}}\simeq 0.488T_{\mathrm{in}}$. This distinguishes a genuine maximum effective temperature from the normalization used in a power-law approximation.

For the steady zero-inner-torque disk, the standard thin-disk dissipation flux already includes both faces. Integrate it over projected annuli, rather than multiplying by a second factor of two:This is the decrease in orbital energy per unit time as matter arrives from arbitrarily large radius on a circular orbit. Half the decrease in gravitational potential energy remains as orbital kinetic energy at the inner edge. Accretion onto a slowly rotating stellar surface can release additional energy in a boundary layer; a black-hole inner boundary requires the appropriate relativistic treatment.^[ref]

For a finite outer radius $R_{\mathrm{out}}$, integration of this same specified flux gives insteadIt tends to zero as $R_{\mathrm{out}}\downarrow R_{\mathrm{in}}$ and to $L_{\mathrm{thin}}$ as $R_{\mathrm{out}}\to \infty$. This finite-annulus result inherits the same steady torque profile; it does not impose an additional zero-torque outer boundary.