= Solution
<Interstellar dust> both hides and reveals galaxies. Absorption and scattering remove short-wavelength photons from a line of sight, causing <interstellar extinction>, reddening, anisotropic scattering, and polarization. These effects obscure embedded <star formation>, bias luminosities, colours, stellar masses, and star-formation rates, and can make an edge-on or dusty galaxy appear older and fainter. Dust also converts absorbed ultraviolet and optical power into far-infrared and submillimetre <thermal radiation>. That reradiation exposes otherwise hidden star formation, constrains dust temperature and mass, and helps map cold molecular material; polarization traces magnetic-field orientation. Grain surfaces also catalyse molecular-hydrogen formation, while photoelectric emission from grains heats interstellar gas.
One spherical grain has emitting area $4\pi a^2$. Since isotropic <blackbody> intensity $B_\nu$ gives surface flux $\pi B_\nu$, its spectral luminosity is
$$
L_{\nu,g}=4\pi^2a^2Q_\nu B_\nu(T).
$$
For $N=M_d/m_d$ optically thin grains at distance $D$, the <inverse-square law> gives
$$
F_\nu=\frac{NL_{\nu,g}}{4\pi D^2}
=\frac{M_d}{m_d}\frac{\pi a^2Q_\nu B_\nu}{D^2}.
$$
Defining the grain <mass absorption coefficient>
$$
\kappa_\nu=\frac{\pi a^2Q_\nu}{m_d},
$$
the <optically thin dust-mass estimator> is
$$
\boxed{M_d=\frac{F_\nu D^2}{\kappa_\nu B_\nu(T)}
=\frac{m_dF_\nu D^2}{\pi a^2Q_\nu B_\nu(T)}}.
$$
For a constant source function and no incident background, the <radiative transfer equation> integrates to
$$
\boxed{I_\nu(\tau_\nu)=B_\nu(T)(1-e^{-\tau_\nu})}.
$$
If the cloud subtends <solid angle> $\Omega$, uniform intensity gives
$$
F_\nu=\Omega B_\nu(T)(1-e^{-\tau_\nu}).
$$
The physical projected area is $A=\Omega D^2$. Its grain column density is $N/A=M_d/(m_d\Omega D^2)$, so the <dust optical depth> is
$$
\tau_\nu=\frac{M_d\pi a^2Q_\nu}{m_d\Omega D^2}
=\frac{\kappa_\nu M_d}{\Omega D^2}.
$$
Eliminating $\tau_\nu$ gives the finite-optical-depth result
$$
\boxed{M_d=-\frac{\Omega D^2}{\kappa_\nu}
\log\left(1-\frac{F_\nu}{\Omega B_\nu(T)}\right)}.
$$
It requires $F_\nu<\Omega B_\nu$, as demanded by the <blackbody> brightness limit. When $\tau_\nu\ll1$, $1-e^{-\tau_\nu}=\tau_\nu+O(\tau_\nu^2)$, so $F_\nu\simeq\Omega B_\nu\tau_\nu=\kappa_\nu M_dB_\nu/D^2$ and the result reduces to the optically thin estimator.
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