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First a couple of disclaimers:

  • My title explains the idea of my question, but I will pose it slightly differently to make it less subjective.
  • This ends up being in the style of a homework exercise (and I've tagged it as such), but it is in fact motivated by a research-level problem.

With that out of the way, perhaps I should state my question precisely (this is the tl;dr):

If I choose a line of sight on the sky at random, what galaxy stellar mass divides the set of all galaxies into two populations such that I have an equal chance of having my sight line intersect a galaxy from each population?

  • This is a bit better posed than in the question title, just because trying to draw a line between dwarfs and giants is a bit arbitrary. Instead, I'm asking where do you draw the line such that hitting a galaxy on either side of the line is equally probable.
  • The intention is that not only the number density of galaxies matters, but also their size, which can be defined in any reasonable way (consistently for all galaxies). There are a lot more small galaxies out there, but they are also smaller than big galaxies, so which one am I more likely to hit?
  • The answer doesn't necessarily have to be in stellar mass, any reasonably equivalent quantity (other masses, luminosities, etc.) is good enough.
  • Obviously the most likely galaxy you'll hit is the Milky Way, and there will be other biases from near field structure. I'm more interested in at least moderately distant objects, so I'm asking this question from the point of view of an observer placed randomly in the cosmos - so the argument should be based on number densities averaged over large volumes, etc.
  • I am interested in accounting for redshift evolution of relevant quantities. Galaxies at early times are smaller on average. Angular diameter distance starts to matter at higher redshift (and there's a lot of volume out there!).
  • Using scaling relations to go back and forth between various masses, sizes, etc. is fine.

Finally, a bit of background on why I'm interested in the question. Over coffee, a colleague and I got talking about DLA systems. These occur when a quasar (bright point source) has a galaxy in the foreground, so that one measures an absorption spectrum of the galaxies gas content. The galaxy doing the absorption is usually otherwise undetectable, because it is (1) intrinsically faint and (2) has a honking bright quasar right on top of it in the image. Quasars are more or less randomly scattered across the sky, so the question is, am I more likely to be measuring the absorption spectrum of a large or a small galaxy, on average? So an even more interesting answer would use the cross section of galaxies that has a high enough gas column density to produce a DLA system instead of some other measure of size. If anyone manages that you can expect an upvote + accept + generous bounty from me :)

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    $\begingroup$ Try submitting this to Randall Munroe here. He answers bizarre questions and is good at digging for information. $\endgroup$ – Pyraminx Jan 16 '15 at 0:45
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    $\begingroup$ Excellent question! I wish I had the time to dig for the answer, I think it is possible and could be a fun problem to solve. But time consuming. $\endgroup$ – Thriveth Jan 17 '15 at 20:49
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    $\begingroup$ Ont thing is unclear, though. Quasars are more or less randomly distributed across the sky, but not in redshifts - they are way more abundant at redshifts 2-4 than at higher or lower redshifts, IIRC. So do you mean random sight line, or sight line to random quasar? $\endgroup$ – Thriveth Jan 17 '15 at 20:55
  • $\begingroup$ @Thriveth am happy to start with random line, but adding in the quasar distribution would be interesting too! $\endgroup$ – Kyle Oman Jan 18 '15 at 16:48
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The typical galaxy is small.

The argumentation (which is not my own thinking) goes as follows:

The distribution of the galaxies' luminosities is given by the luminosity function $\phi(L)$. Assuming a relation between the luminosities and sizes $R \propto L^{0.4}$ (Holmberg 1975), this can be translated to a size distribution $\phi(R)$. The typical size of an absorber is then found by maximizing $R^2\phi(R)$ (squared, since area $\propto$ radius$^2$). Since $\phi$ is also a function of redshift, you have your redshift dependency.

Mo et al. (2010) have a discussion in Sec. 16.5.4 about absorber sizes in the Lyman $\alpha$ forest (showing that typical LAF absorbers are much larger than individual galaxies). An online version can be found here.

Fynbo et al. (1999) discuss this in relation to damped Lyman α absorbers.

Note that the model is rather simple, and that there are some caveats. For instance, the Holmberg relation is determined locally, not at high redshift. But there are indications that this holds at higher redshift as well (I can probably dig out references for this if you want). Also, the faint end slope $\alpha$ of the LF is poorly determined at high z, giving rise to uncertainties. Your result is quite sensitive to $\alpha$ (for $\alpha \le -2$, the maximum diverges). Finally, galaxies are not perfect spheres, $\phi$ may not be given by a Schechter function, etc., etc.

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In the simplest case, assuming homogeneous distributions, the likelihood of finding a dwarf galaxy (D) or a giant galaxy(G) will be in direct proportion to their relative abundance. For example, if there are 10 times more Ds than Gs, then the likelihoods are 10/11 for Ds, and 1/11 for Gs, in any random direction. If the distribution is not homogenous, then you have to find sectors that are "close" to being homogenous, find the relative abundance of Ds and Gs in the sector and calculate as above. This will apply only in the direction of the selected sector. Repeat the above for other sectors of interest.

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  • $\begingroup$ This is kind of the point of my second last bullet point - the redshift evolution of all the relevant quantities. $\endgroup$ – Kyle Oman Jan 23 '15 at 19:43
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    $\begingroup$ Why won't the chance be based on relative size in addition to relative abundance? $\endgroup$ – Math chiller Feb 24 '15 at 1:22

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