 Solution

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.