For $r_{\rm in}\ll r\ll r_{\rm out}$, part (i) gives $\bar\nu\Sigma\propto r^{-1/2}$. Since $\Omega^2\propto r^{-3}$, the dissipative flux obeys
$$
\sigma T^4\propto\bar\nu\Sigma\Omega^2\propto r^{-7/2},
\qquad
\boxed{T(r)\simeq T_0(r/r_{\rm in})^{-7/8}}.
$$
In the spectrum, introduce the dimensionless variable
$$
x=\frac{\epsilon\omega}{T(r)}
=\frac{\epsilon\omega}{T_0}
\left(\frac r{r_{\rm in}}\right)^{7/8}.
$$
Then $r\,dr$ is proportional to $\omega^{-16/7}x^{9/7}dx$, so
$$
F_\omega\propto\omega^{3-16/7}
\int\frac{x^{9/7}}{e^x-1}\,dx.
$$
At intermediate frequencies the inner limit is much smaller than one and the outer limit much larger than one, allowing them to be replaced by zero and infinity. Thus
$$
\boxed{F_\omega\propto\omega^{5/7}
\int_0^\infty\frac{x^{9/7}}{e^x-1}\,dx
\propto\omega^{5/7}}.
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 321 / 1 / b / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-321/1/b/iv/solution