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For the Rayleigh-Jeans spectrum of a finite blackbody disk, the Rayleigh-Jeans law replaces $(e^{h\nu /(kT)}-1{)}^{-1}$ by $kT/(h\nu )$. Here $h$ is the Planck constant and $k$ the Boltzmann constant. The multitemperature blackbody disk therefore hasFor example, setting $R=r_{\mathrm{out}}/r_{\ast }$ makes its frequency-independent coefficient proportional to the finite dimensionless integralAlthough the exact effective temperature vanishes at the inner edge, the very narrow cold rim where the Rayleigh-Jeans law fails makes a negligible contribution in this limit. More formally, after dividing the integrand by ${\nu }^2$, the inequality $e^x-1\ge x$ bounds it by $(k/h)rT$, so dominated convergence justifies the result even at that edge.

At large radius, $T\propto r^{-3/4}$, and the contribution per logarithmic interval is $r^{2}T\propto r^{5/4}$. The outer disk dominates the low-frequency emission because its much greater area outweighs its lower effective temperature.

For intermediate frequencies use the allowed power-law approximation to the effective temperature and introduceThensoThe lower limit is much smaller than one and the upper limit much larger than one. Extending them to zero and infinity leaves a constant: near zero the integrand behaves as $x^{2/3}$, and at infinity it decays exponentially. Hence the intermediate spectrum isThis is the one-third spectrum of a multitemperature disk. Much of the emission comes from radii where $kT(r)$ is of order $h\nu$, moving inward as the frequency rises. With the exact inner-edge profile, a broad intermediate interval also requires frequency well below $k{T}_{\max }/h$; the supplied approximation captures its slope away from the hottest annuli.

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.

An optically thick annulus radiates as a blackbody from both faces, soFor a steady Keplerian accretion disk with a zero-torque inner boundary condition,ConsequentlyandAway from the inner edge, $T\propto R^{-3/4}$.

Ignoring inclination and distance factors, the multitemperature blackbody disk spectrum isSet $x=h\bar{\nu }/(k_{B}T)$. Since $T\propto R^{-3/4}$, $RdR\propto {\bar{\nu }}^{-8/3}{x}^{5/3}dx$, whereas the Planck function contributes ${\bar{\nu }}^3/(e^x-1)$. In the stated intermediate-frequency range the radial endpoints become $0$ and $\infty$, leaving a frequency-independent convergent integral. Therefore

For a finite multitemperature blackbody disk, the low-frequency Rayleigh-Jeans law gives $F_{\nu }\propto {\nu }^2\int rT(r)dr$. If $T\propto r^{-3/4}$ away from the inner edge, outer annuli dominate because the weight per logarithmic radial interval grows as $r^{5/4}$. A measure-zero zero-temperature inner edge does not invalidate the leading law when $rT(r)$ is integrable.