From the article "Large-Scale Galaxy Bias", I try to deduce the equation that my teacher told me which links 2 quantities:

  1. the global number density of galaxies

  2. the local number density of galaxies

  3. the contrast of Dark matter density

The relation that I would like to find (the relation given by my teacher) is very simple:

$$N_{1} = n_{1} b_{1}\,\delta_{\text{DM}} \tag{1}$$

$N_{1}$ is the local number density of galaxies in Universe,
$n_{1}$ is the global number density,
$b_{1}$ is the bias (cosmological bias of galaxies) and
$\delta_{\text{DM}}$ the contrast in dark matter density.

When I say "local", I mean in the volume of scale that I consider (in a cluster of galaxies, for example, doesn't it ?)

for the moment, I can't find this equation.

Into the article above, they define the bias by doing the relation $(1.1)$ (equation reference on the article):

$$\delta_{g}(\vec{x}) = \dfrac{n_{g(\vec{x})}}{\overline{n_{g}}}-1 = b_{1}\,\delta_{\text{DM}}(\vec{x}) = b_{1}\left(\dfrac{\rho_{m}(\vec{x})}{\overline{\rho_{m}}}-1\right) \tag{2}$$

with $b_{1}$ the bias.

As you can see, in this article, authors are reasoning with the contrast of density number of galaxies ($\delta_{g}(\vec{x}))$ and the contrast of matter density of Dark matter ($\delta_{\text{DM}}(\vec{x})$).

I tried to modify this equation $(2)$ to get $(1)$ but I am stuck by the following difference: on one side, one takes number densities and on the other one, they take contrasts of density (with contrast density number and Dark matter contrast).

Multiplying the both by the volume $V$ is not enough since there is the value "-1" in the definition of contrast: I don't know if I have to write:

$$\text{Global Number of galaxies} = \overline{n_{g}}\quad V$$


$$\text{Local Number of galaxies} = \overline{n_{g}}\quad V$$


I think that I have to use the following relations: $N_{g}\equiv N_{1}$ and $\overline{n_{g}}=n_{1}$ in the relation of my teacher but I am not sure.

Anyone could help me to find the equation (1) from the equation (2) of an article cited?


If I take the relation eq$(2)$, I can write:

$$n_{g(\vec{x})} = \overline{n_{g}}\,b_{1}\,\delta_{\text{DM}}+\overline{n_{g}} \tag{3}$$

As you can see, $(3)$ is not equal to the equation $(1)$ that I would like to get (since a second term $\overline{n_{g}}$)

With the notations of the equation$(1)$, in order to be coherent, I think that I have to assimilate $N_{1}$ to $n_{g}(\vec{x})$ (local density) and $n_{1}$ to $\overline{n_{g}}$ (global or mean density).

How can I circumvent this issue about the presence of this second term into eq$(3)$ compared to eq$(1)$?

I am near from the equality between both, this is frustrating. Maybe it is a problem of the convention about the factor $b_1$?

UPDATE 2: I put here the demonstration of the expression inferred from the equation $(1)$ suggested by my teacher. Given his relation may be wrong, what it follows could be proved surely with another way, I mean with valid other definition joining the density of galaxies and the density contrast of Dark Matter with the concept of "bias" in cosmology.

1) Starting relations:

Actually, suppose we have 2 samples of galaxies clusters. So, assuming we have:

$$\delta_{g}(\vec{r},z) = b(z)\,\delta_{\text{DM}}(\vec{r},z)\quad(4)$$ where $b(z)$ is the bias depending on redshift (for the moment I only consider this dependance to make it simple).

Given 2 points correlation function $\xi(\vec{r})$ is the inverse Fourier transform of matter power spectrum, we have:

$$\xi_{g}(\vec{r},z) = b(z)^{2} \xi_{\text{DM}}(\vec{r},z)\quad(5)$$

Now, let's express the mean of the squared sum of galaxies density from 2 different samples with $N=N_{1}+N_{2}$:

\begin{align} <N^2> = <(N_{1}+N_{2})^{2}> & = <(N_{1}+N_{2})(N_{1}+N_{2})> \notag \\ & = <N_{1}^{2}> + <N_{2}^{2}> + 2\,<N_{1}\,N_{2}> \notag \\ & = (n_{1}\,b_{1}\,\delta_{\text{DM}})^2 + (n_{2}\,b_{2}\,\delta_{\text{DM}})^2 + 2\,(n_{1}\,n_{2}\,b_{1}\,b_{2}\,\delta_{\text{DM}}^2) \notag \\ & = n^{2}\,\delta_{g,1+2}^{2}\quad(6) \end{align}

with $n=n_{1}+n_{2}$ the global density of the 2 samples.

So, using $(4)$ and $(5)$, one can write:

$$\xi_{12}=b(z)^{2}\,\xi_{\text{DM}} = \dfrac{\delta_{g,1+2}^{2}}{\delta_{\text{DM}}^{2}}\,\xi_{\text{DM}}\quad(7)$$

2) Computing the bias representing the "cross-correlation" between the 2 samples (I don't know if "cross-correlation" is the right expression):

using $(4)$ involving $\delta_{g,1+2}$, we would write that, with the "merging" of the 2 samples, we have:

\begin{align} \xi_{12} & = \dfrac{1}{n^2}\,\big(n_{1}^2\,b_{1}^2 + n_{2}\,b_{2} + 2\,n_{1}\,n_{2}\,b_{1}\,b_{2}\big)\,\delta_{\text{DM}}^{2}\,\dfrac{1}{\delta_{\text{DM}}^{2}}\,\xi_{\text{DM}} \notag \\ & = \dfrac{1}{n^2}\,\big(n_{1}^2\,b_{1}^2 + n_{2}^{2}\,b_{2}^{2} + 2\,n_{1}\,n_{2}\,b_{1}\,b_{2}\big)\,\xi_{\text{DM}} \notag \\ & = \dfrac{1}{n^2}\,\big(n_{1}\,b_{1} + n_{2}\,b_{2}\big)^{2}\,\xi_{\text{DM}}\quad(8) \end{align}

So, we could conclude with $(5)$ that the bias representing the 2 samples for a given redshift is expressed as:

\begin{equation} b(z) = \dfrac{1}{n}\,\big(n_{1}(z)\,b_{1}(z) + n_{2}(z)\,b_{2}(z)\big) = \dfrac{1}{n_{1}+n_{2}}\,\big(n_{1}\,b_{1} + n_{2}\,b_{2}\big)\quad(9) \end{equation}

This "merging" bias is actually the weighted mean of bias $b_{1}$ and $b_{2}$ with weights equal to the density of galaxies of each sample.

To conclude, Steps 1) and 2) in the demonstration seems to be good except the fact that, like @Javier said, $\delta$ could be negative when I write : $N_{1} = n_{1} b_{1}\,\delta_{\text{DM}}$.

So maybe my teacher implies we manipulate implicit absolute values for $\delta$: I don't know. That's why I have difficulties with the relation $(1)$ compared to relation $(2)$ coming from a cited article at the beginning of my post.

What do you think about this ? Are there others ways of demonstration with the relation $\delta_{g}(\vec{r},z) = b(z)\,\delta_{\text{DM}}(\vec{r},z)$.

ps: I hope that I have been clear in this little demonstration. Don't hesitate to ask me for further information if needed.

UPDATE 3: @Javier : could you tell me please a method to find the equation $(9)$ (weighted average) showing a "representative" bias of 2 samples and mostly involving the density contrast of galaxies and the density contrats of Dark Matter. I would really appreciate since my demonstration is not correct but the result is reasonable as you said. Regards

  • 1
    $\begingroup$ Im pretty sure dark matter is, you know, a free parameter. It shouldn't be used like that. Bizarre assignment. $\endgroup$ – user236221 Jul 6 at 2:21
  • $\begingroup$ @user236221 So you think that the equation (1) given by my teacher is wrong and that right equation is equation (2) ? however, I am near from the equality between both, this is frustrating . Maybe it is a problem of convention about the factor "$b_{1}$" ? Thanks $\endgroup$ – youpilat13 Jul 6 at 3:07
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    $\begingroup$ I don't think equation (1) can be right, because $\delta$ can be negative while $N$ can't. I would expect either a relation between $\delta$'s or between $n$'s, but not one that mixes both (unless I'm misunderstanding what $N$ is). $\endgroup$ – Javier Jul 26 at 18:39
  • $\begingroup$ @Javier Thanks, maybe my teacher assumes an implicit absolute value for density contrast $\delta$. I am going to give more informations about this relation that he gave to me : indeed, this allows me to prove a weighted average for 2 different samples of galaxies for global bias. I try to add an update as soon as possible. Regards $\endgroup$ – youpilat13 Jul 26 at 18:52
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    $\begingroup$ I don't follow step (6), particularly when you state $\langle N_1^2 \rangle = (n_1 b_1 \delta_\text{DM})^2$. This is the crucial step, where you set $N$ proportional to $\delta$, which doesn't make sense to me (not even if you take the absolute value because $\delta$ can still be zero). I'm also iffy about taking two samples of galaxies and adding the densities: this is a theoretical relation, who cares about samples? Overall, I find the whole thing a bit confusing. Teachers can make mistakes, and sometimes they don't say what they mean. $\endgroup$ – Javier Aug 1 at 23:37

The only answer I can give is that the relation

$$N_{1} = n_{1} b_{1}\,\delta_{\text{DM}} \tag{1}$$

cannot be correct, assuming that "$N_1$ is the local density of galaxies" means that $N_1$ represents some kind of amount of galaxies per unit volume. The reason is simple: $\delta_\text{DM}$ is the DM density contrast: it can be positive, negative or zero, depending on how the local density relates to the average. But a number density can never be negative, and it can only be zero in empty places, which don't necessarily correspond to $\delta_\text{DM} = 0$.

However, this is easily fixed: the correct relation is

$$N = n (b\, \delta_\text{DM} + 1). \tag{2}$$

We get this from the definition of bias, which says that $b\, \delta_\text{DM} = \delta_g = n_g / \bar{n}_g -1$. What you call $N$ is $n_g$ (the local density), and what you call $n$ is $\bar{n}_g$, the global average. That is, using your names we have

$$b\, \delta_\text{DM} = N/n -1,$$

which is clearly equivalent to $(2)$.

  • $\begingroup$ thanks ! I will ask to my teacher this error. I suppose we can however find with this definition the same conclusion about the equation of weighted mean for 2 samples : could you confirm that ? Regards $\endgroup$ – youpilat13 Aug 2 at 3:30
  • $\begingroup$ @youpilat13 To be honest, I'm not quite sure, because I don't completely understand what the two samples represent, and where the average is being taken. The derivation cannot be right, since it uses the wrong equation $(1)$, but the result for the bias seems reasonable. $\endgroup$ – Javier Aug 2 at 13:11

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