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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.

The standard thin-disk dissipation flux, summed over both faces, follows by substituting the Keplerian accretion disk profile into the viscous heating rate:The Stefan–Boltzmann law gives $\mathcal{C}=2\sigma T^4$ because there are two emitting faces. Local radiative equilibrium consequently gives the effective-temperature profile of a zero-torque diskHere $\sigma$ is the Stefan-Boltzmann constant. The scale $T_{\mathrm{in}}$ is not the temperature exactly at the inner boundary: the zero-torque inner boundary condition makes that formal temperature zero. Maximizing $T^4$ shows that the maximum effective temperature of a zero-torque disk occurs at $r=49r_{\ast }/36$, with $T_{\max }\simeq 0.488T_{\mathrm{in}}$.

At equal central mass and accretion rate, characteristic effective temperatures scale as $r_{\ast }^{-3/4}$. Thus, comparing corresponding values of $r/r_{\ast }$,The neutron-star disk is about 180 times hotter. The Planck law and Wien displacement law move its characteristic emission to about 180 times higher frequency, or 180 times shorter wavelength. A white dwarf disk commonly emits in optical and ultraviolet bands, while the hotter neutron star disk can emit in X-rays. Absolute bands require an actual accretion rate; the relative shift follows directly from the stated scaling.

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 usual Pressure-opacity zones of a Shakura--Sunyaev thin disk, in order of increasing cylindrical radius, are an inner region dominated by radiation pressure and electron-scattering opacity, a middle region dominated by gas pressure and electron-scattering opacity, and an outer region dominated by gas pressure and the Kramers' opacity law, often approximated by free-free opacity. The boundaries depend on mass, mass accretion rate, and the alpha disk viscosity parameter. At sufficiently large radius, accretion-disk self-gravity or changes in ionization can invalidate this three-zone model.

For the inner region writeHere $F$ is the standard thin-disk dissipation flux through one face, and $R_{\ast }$ is the innermost stable circular orbit treated as a zero-torque inner boundary. The given factor $1/2$ in the dissipation rate is appropriate to one face. For Keplerian rotation, it is $(9/8)\nu \Sigma {\Omega }_K^2$.

Let $H$ denote the disk scale height or vertical half-thickness, not the full thickness. In vertical hydrostatic equilibrium, the leading vertical acceleration is ${\Omega }_K^{2}z$, so a one-zone estimate givesIn the inner zone, $P_c\simeq P_{\mathrm{rad},\mathrm{c}}=a_{\mathrm{rad}}{T}_c^4/3$. For an optically thick medium with constant electron-scattering opacity, radiative diffusion gives the emergent one-face fluxEquivalently, combine $F=-c({\kappa }_{\mathrm{es}}\rho {)}^{-1}d{P}_{\mathrm{rad}}/dz$ with vertical pressure balance. Equating the radiative flux to the dissipated flux yields the radiation-supported inner-disk height:Order-unity vertical-profile factors depend on the precise height convention. The central mass cancels at fixed physical mass accretion rate, because both vertical gravity and local dissipation are proportional to $G{M}_{\mathrm{BH}}/R^3$.

If ${\dot{m}}_{\mathrm{Edd}}=L_{\mathrm{Edd},\mathrm{es}}/({\eta }_0{c}^2)$ and $r_g=G{M}_{\mathrm{BH}}/c^2$, this can also be writtenThe radiation-supported height rises from the formal zero at $R_{\ast }$ toward the approximately constant value $H_0=3{\kappa }_{\mathrm{es}}\dot{m}/(8\pi c)$. That zero is not a reliable description of the physical plunging region: the radiation-dominated approximation and the Newtonian zero-torque formula fail very close to the inner edge.

Beyond the inner zone, the height grows slowly with radius rather than staying on that plateau forever. Far from the inner boundary, the standard gas-pressure scalings give $H\propto R^{21/20}$ in the electron-scattering zone and $H\propto R^{9/8}$ in the Kramers zone. These follow from $H\propto T_c^{1/2}{R}^{3/2}$ with $T_c\propto R^{-9/10}$ and $R^{-3/4}$, respectively. A schematic profile is

vertical half-thickness H
  ^
  |                                               / outer gas/Kramers
  |                                          ____/
  |                                     ____/ middle gas/scattering
  |              ______________________/
  |          ___/ inner radiation/scattering plateau
  |       __/
  |     _/
  +----|----------------------------------------------> radius R
      R_*              zone boundaries are schematic

The height formula also tests the thin disk assumption. For a nonrotating hole with $R_{\ast }=6r_g$, the largest $H/R$ in this inner approximation occurs at $R=(9/4)R_{\ast }$. At $\dot{m}={\dot{m}}_{\mathrm{Edd}}$ and ${\eta }_0=0.1$, it is about $0.37$, so a disc this close to the conventional Eddington accretion rate is only marginally thin. Slim accretion disk effects and advected heat can then matter.

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 $GM/(2R_{\mathrm{ISCO}}{c}^2)$.

Integrating the standard thin-disk dissipation flux givesHalf of the Newtonian potential-energy decrease remains as orbital kinetic energy until the inner edge.