# Formula for the bias of galaxies

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:

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

where
$$N_{1}$$ 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:

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

or

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

???

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?

UPDATE 1:

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_{1}+N_{2})^{2}> & = <(N_{1}+N_{2})(N_{1}+N_{2})> \notag \\ & = + + 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:

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

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

• Im pretty sure dark matter is, you know, a free parameter. It shouldn't be used like that. Bizarre assignment. – user236221 Jul 6 at 2:21
• @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 – youpilat13 Jul 6 at 3:07
• 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). – Javier Jul 26 at 18:39
• @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 – youpilat13 Jul 26 at 18:52
• 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. – 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 $$(2)$$.

• 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 – youpilat13 Aug 2 at 3:30
• @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. – Javier Aug 2 at 13:11