OWAO 2023, Теория — Задача 6. Space Vodka

• The methanol maser is larger, about $2 \cdot 10^4$ times larger than the water maser.

An astrophysical maser is a naturally occurring source of stimulated spectral line emission. Brightness temperature is the temperature at which a black body would have to be in order to duplicate the observed intensity in a spectral line; it is not the real thermodynamic temperature of some body!

Distance $d$ to the radio source is inversely proportional to its parallax $\varpi$: $d \propto \varpi^{-1}$ [2 pt].

Flux density $S_{\nu} = B_{\nu}\Omega$ [2 pt], where $\Omega$ is the solid angle corresponding to the source, $\Omega \propto R^2d^{-2}$ [2 pt]. $R$ is the spatial size of the source. $B_{\nu}$ is spectral radiance which depends on the brightness temperature $T$ and emission wavelength $\lambda$. As brightness temperature corresponds to some black body and energy of photons in radio is significantly lower than thermal energy of the matter ($hc/\lambda \ll k_BT$), $B_{\nu}$ can be expressed via the Rayleigh–Jeans law$^{5}$ [2 pt]:

Therefore, the size of the source

The size ratio is

So, the methanol maser is much bigger than the water one [2 pt].

$^{5}$ The flux density is measured in janskys, essential non-SI units excessively used in radio astronomy; $1\ \mathrm{Jy} = 10^{-26}\ (\mathrm{W/m^2})/\mathrm{Hz}$. That is why in this problem, the radiation laws should be in terms of frequencies. If one uses $B_{\lambda}$ instead of $B_{\nu}$, the result changes several times, just as blackbody emission peaks at wavelength $\lambda_{\max}$, or at frequency $\nu_{\max} \neq c/\lambda_{\max}$.