For a dusty shell, direct ultraviolet absorption, infrared trapping, and incomplete ultraviolet absorption give the radiation force
$$
F_{\rm rad}=\frac Lc\left(1+\tau_{\rm IR}-e^{-\tau_{\rm UV}}\right).
$$
Balancing it against gravity gives the general <critical luminosity of a dusty shell>
$$
\boxed{L_{\rm crit}=\frac{GcM(<r)M_s(r)}
{r^2[1+\tau_{\rm IR}-e^{-\tau_{\rm UV}}]}.}
$$

For a <singular isothermal sphere>,
$$
M(<r)=\frac{2\sigma^2r}{G},
\qquad
M_s=f_{\rm gas}M(<r)=\frac{2f_{\rm gas}\sigma^2r}{G},
$$
so
$$
F_{\rm grav}=\frac{GM M_s}{r^2}=\frac{4f_{\rm gas}\sigma^4}{G}
$$
and
$$
\tau_i=\frac{\kappa_iM_s}{4\pi r^2}
=\frac{\kappa_i f_{\rm gas}\sigma^2}{2\pi Gr}.
$$
In the infrared-thick limit $F_{\rm rad}\simeq\tau_{\rm IR}L/c$; in the single-scattering limit it is $L/c$; and in the ultraviolet-thin limit it is $\tau_{\rm UV}L/c$. Therefore
$$
\boxed{L_{\rm crit,IR}=\frac{8\pi c\sigma^2r}{\kappa_{\rm IR}},
\qquad
L_{\rm crit,s.s.}=\frac{4cf_{\rm gas}\sigma^4}{G},
\qquad
L_{\rm crit,UV}=\frac{8\pi c\sigma^2r}{\kappa_{\rm UV}}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 347 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-347/3/b/solution