Let's assume that I have observations of $N$ galaxies in a projected area $A$, within a redshift range $\Delta z = z_{max} - z_{min}$. What is the correct way of computing their number volume density? I guess the volume has to be derived multiplying $A$ by the differential Luminosity Distance $D_{L}$ computed at $z_{max}$ and $z_{min}$, i.e., $D_{L}(z_{max}) - D_{L}(z_{min})$, but I am not sure. Also, what is the more appropriate way of evaluating the "significance" of any over-density. For instance, assuming that the observed galaxies are Lyman-$\alpha$ emitters, does it exist any computation of the "standard" number density of Lyman-$\alpha$ emitters, to compare with?


1 Answer 1


Luminosity function

Your approach is in principle correct, but would give the number density $n$ of Lyman $\alpha$ emitters (LAEs) in a "luminosity volume". Usually, the comoving volume is used, since you can then more easily compare densities at different redshifts $z$. If $n$ changes with $z$, you then know that it's not just because of the expansion of the Universe, but due to some astrophysical process.

Moreover, since $n$ is a strong function of galaxy luminosity $L$, you would do this calculation in bins of luminosity. You then get the so-called luminosity function (LF), $n(L,z)$. LFs are often fitted with the completely phenomenological Schechter function, though there is evidence that the bright end of the LF is less steep than this form would suggest.

Filter shape

Note though that, observationally, your approach is only approximate: If your redshift range $\Delta z$ comes from the width of your (presumably) narrowband filter, then this method is only exact in the case of a perfect "square" filter, i.e. one that transmits 100% in some range, and 0% outside this range. In reality, filters are described by some non-square transmission function (could be a Gaussian, but really it has to be measured, as it's not a simple functional form).

Hence, galaxies with redshifts falling in the wings of your filter has to be brighter in order to be detected, than galaxies falling in the center. In other words, your LF will be less complete the fainter the galaxies, and your inferred number density will be a function of brightness.

To account for this, you must in principle fold in the exact shape of the filter. In practice, it is not uncommon to neglect this effect though, and just use an "effective" width of the filter, typically the FWHM.


To compute an overdensity requires you to assume a "reference" LF. There isn't one standard LF, as this will depend on whose observations you prefer. But a good choice could be e.g. Wisotski et al. (2018) parametrizes the LAE LF based on deep MUSE exposure of the Hubble (Ultra) Deep Field.

  • $\begingroup$ Thanks @Pela, very useful! In my case, there is not the "problem" of Filter Shape, because I have spectroscopic redshifts for galaxies in the field. But I still have problems with the calculation of comoving volumes, trying to reproduce numbers in papers. For instance in Umehata+15 (arxiv.org/pdf/1510.08861.pdf; beginning of page 4), they estimate a number density of (in this case) SMGs of 1e-3 Mpc-3, starting from 8 SMGs at z=3.08-3.1 in a 1.5'x3' field. But I find 9e-3, where for the "comoving distance" between z=3.08-3.1, I used home.fnal.gov/~gnedin/cc/ (dist. between two z). $\endgroup$
    – mark polo
    Jan 3, 2021 at 11:06
  • $\begingroup$ @markpolo Their estimate of 1e-3 is the total number (N), not number density (n), and it is the number they get when they assume the Simpson+ 14 number density (4e-6/Mpc³) and multiply by their own volume (~300 Mpc³). That is, this is the "expected" number density. Note that they here only consider logL>12.5, so their "own" number of SMGs is 3, not 8. And N=3 is roughly three orders of magnitude larger than N=1e-3. $\endgroup$
    – pela
    Jan 5, 2021 at 15:47
  • $\begingroup$ Thank you very much @pela for the explanation! I can finally reproduce their numbers :-) But following this way, I still find inconsistencies with arxiv.org/pdf/1910.01324.pdf (Section S8), where they find SMGs volume densities of 2e-3 Mpc-3, starting from 5 SMGs in a 2'x3' area at z=3.08 - 3.1. I calculate a volume of ~410 Mpc3, so n would be n = 5 / 410 Mpc3 ~ 1.2e-2 Mpc-3, quite bigger than their result. $\endgroup$
    – mark polo
    Jan 7, 2021 at 10:54
  • $\begingroup$ @markpolo Hmm, yeah I agree that seems wrong. I can't seem to find which cosmology they use, but that shouldn't change the result by more than ~10%, and I get similar values to yours, i.e. a factor ~6 from theirs. The published version contains the same value. $\endgroup$
    – pela
    Jan 7, 2021 at 15:25
  • $\begingroup$ Thanks @pela ! Cool, it looks like we just found an error in a Science paper :-) $\endgroup$
    – mark polo
    Jan 7, 2021 at 15:36

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