An optically thick annulus radiates as a <blackbody> from both faces, so
$$
2\sigma_{\rm SB}T_{\rm BB}^4=F_{\rm diss}.
$$
For a steady <Keplerian accretion disk> with a <zero-torque inner boundary condition>,
$$
\nu\Sigma=\frac{\dot M}{3\pi}
\left[1-\left(\frac{R_{\rm in}}R\right)^{1/2}\right],
\qquad
R^2\left(\frac{d\Omega}{dR}\right)^2=\frac94\Omega_K^2.
$$
Consequently
$$
F_{\rm diss}=\frac{3GM\dot M}{4\pi R^3}
\left[1-\left(\frac{R_{\rm in}}R\right)^{1/2}\right]
$$
and
$$
\boxed{T_{\rm BB}(R)=
\left\{\frac{3GM\dot M}{8\pi\sigma_{\rm SB}R^3}
\left[1-\left(\frac{R_{\rm in}}R\right)^{1/2}\right]\right\}^{1/4}.}
$$
Away from the inner edge, $T\propto R^{-3/4}$.

Ignoring inclination and distance factors, the <multitemperature blackbody disk> spectrum is
$$
S_{\bar\nu}\propto\int_{R_{\rm in}}^{R_{\rm out}}
2\pi R B_{\bar\nu}[T(R)]\,dR.
$$
Set $x=h\bar\nu/(k_BT)$. Since $T\propto R^{-3/4}$, $R\,dR\propto\bar\nu^{-8/3}x^{5/3}dx$, 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
$$
\boxed{S_{\bar\nu}\propto\bar\nu^{1/3}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 347 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-347/2/a/solution