As I understand it astronomers estimate the mass of stars by studying the orbits of their companions (other stars or planets). This helps them to derive relationships between the luminosities and masses of stars. This allows them to infer the mass of stars of known luminosity.

But how do astronomers estimate the mass of cosmic dust in particular clouds or in entire galaxies?

  • $\begingroup$ Are you just interested in dust? Most of the mass in the ISM is not in the form of dust. A typical gas-to-dust ratio would be 100. $\endgroup$ – Rob Jeffries Mar 5 '15 at 11:07
  • $\begingroup$ @Rob. I thought I would try & keep it simple by addressing dust first :) I have come across many mentions of the 1:100 ratio. Also while gas estimation methods were quite easy to find on the web, methods for dust were not so obvious. $\endgroup$ – steveOw Mar 5 '15 at 17:52

Dust absorbs stellar light (primarily in the ultraviolet), and is heated up. Subsequently it cools by emitting infrared, "thermal" radiation. Assuming a dust composition and grain size distribution, the amount of emitted IR light per unit dust mass can be calculated as a function of temperature. Observing the object at several different IR wavelengths, a Planck curve can be fitted to the data points, yielding the dust temperature. The more UV light incident on the dust, the higher the temperature.

The result is somewhat sensitive to the assumptions, and thus the uncertainties are sometimes quite large. The more IR data points obtained, the better. If only one IR point is available, the temperature cannot be calculated. Then there's a degeneracy between incident UV light and the amount of dust, and the mass can only be estimated to within some orders of magnitude (I think).

If lines from various atomic or molecular transitions are seen as well, the composition can be better constrained. The size distribution can be determined from fitting the theoretical spectrum of a given distribution to observed dust spectra. This information is often not available in a given high-redshift galaxy, so here we can be forced to assume that the dust is similar in nature to "local" dust, i.e. in the Milky Way and our nearest neighbors.

If you're interested in the relevant equations, they can be found many places, e.g. here.

Another way to estimate the dust mass is to measure the metallicity of the gas with which the dust is is mixed, either from emission lines or absorption lines if a background source is available. The dust mass is then found from an assumed dust-to-metals ratio, which is pretty well-established in the local Universe, and to some extend also at higher redshifts.

  • $\begingroup$ Very helpful, thanks. Does the detected IR flux intensity depend on the amount of stellar light incident on the dust and thickness*density of the intervening dust cloud (causing absorbtion & scattering)? $\endgroup$ – steveOw Mar 3 '15 at 23:49
  • $\begingroup$ Yes, it does, and my answer was somewhat incomplete. If you only have one IR data point, you have to make severe assumptions and will probably only constrain the dust mass within several orders of magnitude. But for two, and preferably many, IR point, you can fit a Planck curve to your data, and thus get the temperature of the dust. With the temperature and the flux, you can calculate the total mass (modulo dust composition and size distribution, but those are of lesser importance). But in general I think it's fair to say that dust, especially extragalactic dust, it pretty obscure… $\endgroup$ – pela Mar 4 '15 at 10:07
  • $\begingroup$ @steveOw, I edited my answer to include a discussion about your relevant remark. Concerning the thickness of the dust cloud, I think this is less relevant, as the temperature tells how much UV light is absorbed, no matter how the dust is distributed spatially. But of course for a very thick dust layer and small UV fluxes, you'll have a temperature gradient from the side closest to the stars to the other side, and this will also complicate the fitting. $\endgroup$ – pela Mar 4 '15 at 10:22
  • $\begingroup$ Thanks. Presumably if a high-density, ~ planar, dust cloud is observed edge-on and has great extent along the direction to the observer, then the proximal (to observer) parts of the cloud will obscure the IR flux coming from the distal parts. Thus the (observed-flux)/mass ratio could decrease as mass increases? $\endgroup$ – steveOw Mar 4 '15 at 17:48
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    $\begingroup$ OK, thanks. Great example, here is some more about it. $\endgroup$ – steveOw Mar 5 '15 at 1:56

Even if the accepted answer is well posed and clear, I would mention another method for dust measurements. X-ray spectra can also help to infer the amount of dust, analyzing the absorption at low energy of the observed spectra.

Indeed, the neutral hydrogen column density $N_H$ and the extinction $A(V)$, which is caused by the dust, is found in our Galaxy to be (Guver & Ozel, 2009):

$N_H = 2.2\times 10^{21} A(V)\,$mag

From the extinction one can get the dust optical depth (see here), and from that the volumetric dust density.

A good example of how to obtain $N_H$ is given in the following plot (S. Ikeda et al. 2009, ApJ 692 608): enter image description here

As it can be seen, the spectra is differently modified by different amounts of neutral hydrogen column density $N_H$. This, of course, can be related to different aboundances, to best-fit the spectra and find the metals amounts.

Also, this method is tightly related to the iron K$\alpha$ line at $6.4\,$keV, which is another measurement of the metals amount (Fabian et al. 2000): enter image description here

To distinguish among different ionization states, high-resolution spectroscopy is very common.

  • $\begingroup$ It is not dust that is absorbing the X-rays. $N_H$ is the column density of neutral hydrogen atoms. To get from gas-mass to dust mass requires an assumption about the gas-to-dust ratio. The question asks about measuring the dust mass. $\endgroup$ – Rob Jeffries Mar 5 '15 at 11:05
  • $\begingroup$ @RobJeffries, thanks for pointing this out. What I meant is that, $N_H$ and extinction are related, therefore one helps to infer the other. I will edit my answer. $\endgroup$ – Py-ser Mar 5 '15 at 13:41

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