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I need to apply a correction on $\sigma_{8}$ between linear and non-linear regime to keep it fixed (I make change the values of cosmological parameters at each iteration). I have to compute $\sigma_{8}$ from the $P_{k}$ and found the following relation (I put also the text for clarify the context) :

Part of this Klein Onderzoek is aimed at finding an estimate of the cosmological parameter $\sigma_{8}$ from peculiar verlocity data only. $\sigma_{8}$ is defined as the r.m.s. density variation when smoothed with a tophat-filter of radius of $8 \mathrm{h}^{-1} \mathrm{Mpc} > .[9]$ The definition of $\sigma_{8}$ in formula-form is given by:

$$ \sigma_{8}^{2}=\frac{1}{2 \pi^{2}} \int W_{s}^{2} k^{2} P(k) d k $$

where $W_{s}$ is tophat filter function in Fourier space:

$$ W_{s}=\frac{3 j_{1}\left(k R_{8}\right)}{k R_{8}} $$

where $j_{1}$ is the first-order spherical Bessel function. The parameter $\sigma_{8}$ is mainly sensitive to the power spectrum in a certain range of $k$ -values. For large $k,$ the filter function will become negligible and the integral will go to zero. For small $k,$ the factor $k^{2}$ in combination with the power spectrum factor $k^{-3}$ will make sure that the integral is negligible. In other words, $\sigma_{8}$ is mostly determined by the power spectrum within the approximate range $0.1 \leq k \leq 2 .$ since $\sigma_{8}$ is only sensitive to a certain range of $k,$ any difference in the values of the Hubble uncertaintenty, the baryonic matter density and the total matter density will influence the found estimate.

Question 1) What numerical value have I got to take for $R_{8}$ in my code : for the instant, I put $R_{8}= 8.0/0.67$ : is this correct ?

Question 2) The other issue is, for each correction on $A_{s}$, that I find with this expression a value roughly around : $\sigma_{8} = 0.8411 ........$ instead of standard (fiducial) value $\sigma_{8} = 0.8155 ........$ : there is a 4 percent of difference between both values : is the expression above right ?

Could anyone tell me a good way to compute $\sigma_{8}$ from $P_{k}$ generated by CAMB-1.0.12?

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  • $\begingroup$ please, there is a little typo error in the description of my bounty : this is $\sigma_{8}^{2}$ and not $s\igma_{8}^{2}$. Could a moderator correct this error ? Regards $\endgroup$
    – user87745
    Commented Jan 1, 2020 at 9:41
  • $\begingroup$ May I ask you what $n$ value you put for the primordial power spectrum: $$P=Ak^n$$ $\endgroup$
    – Annie
    Commented Jul 26, 2022 at 15:38

1 Answer 1

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Firstly, the variance of density perturbations can be calculated using the matter power spectrum $P(k)$. To do so, we coarse-grain our density field with a suitable window function $W_s(k;R)$, where $R$ represents the size of our window. (see Galaxy Formation and Evolution by H. Mo, F. Bosch, S. White) Thus:

$$‎\delta_s(\textbf{x};R) \equiv \int \delta(\textbf{x}^{\prime})W_s(|\textbf{x}^{\prime}-\textbf{x}|;R) d^3\textbf{x}^{\prime}$$

Note that there are several options for the window function. An easy and indeed natural option would be a Top-hat in real space, which gives Bessel's function in Fourier space: $$‎W(x;R) = \begin{array}{lr}‎ ‎ \frac{3}{4\pi R^3} & \mbox{($x \le R$)}\\‎ ‎ 0 & \mbox{($x > R$)} \nonumber‎ ‎\end{array}‎$$

Then, $$‎S \equiv \sigma^2(R) = \langle\delta_s^2(\textbf{x};R)\rangle‎ ‎= \int‎\Delta^2(k)\ ‎W_s^2 (k;R) \ d(\ln k)$$ where $\Delta^2(k) \equiv k^3P(k)/2\pi^2$ is the dimensionless power spectrum. In addition, the power spectrum changes with time (or redshift). So, with the help of a transfer function (see Eisenstein & Hu for an example), one can build $P(k,z)$ as well. Subsequently, one can calculate $\sigma^2(R,z)$.

Secondly, $\sigma_8^2$ is the variance for a window size of $8 h^{-1}Mpc$ (i.e. $\sigma^2(R=8 h^{-1})$) at $z=0$. This size is about the size when structure formation's regime, changed from linear to non-linear. Well, it appears that in standard texts people use a real-space Top-Hat filter to calculate this, but one shall use any other appropriate option.

As an answer to your first question, it depends on the normalization you have used for physical parameters in your code, but if you are working with CAMB, putting $R_8$ as $8 h^{-1}$ is OK.

About the second question, I think one peculiarity is the numerical integration of the interpolated $P(k)$ (noting that CAMB returns an interpolated object) from zero to infinity (so feel free to sample as many points as you wish, for a precise result). Of course, the integrand is neglectable in low and high $k$ but taking these ranges into account as well might play a role. Plus, I have recently experienced a "bug" in extrapolating the non-linear $P(k)$ for $k>100$ in CAMB. It seems that CAMB uses a power-law function to extrapolate the tail of $P(k)$ which is reasonable but doesn't work properly. It is worthwhile to mention that comparing the obtained $P(k)$ with observations might give a clue as well.

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  • $\begingroup$ Thanks for your quick answer. So, taking, in $W_{s}$ impying Bessel function, $R=8.0/0.67 = 11.94\,$Mpc is correct ? Secondly, you say : "$8 h^{-1}$Mpc (i.e $\sigma^2(R=8)$ at $z=0$) : which $h$ value do you take to say that ? Classicaly, isn't rather $h=H_{0}/100\,\neq\,1$, I mean more generally $h(z)=H(z)/100\,\neq\,1$. Do you agree ? Or maybe you talk about $h=h_{0.67}=H_{0}/67$ ? Let me know please what you want to mean , Regards $\endgroup$
    – user87745
    Commented Jan 1, 2020 at 19:55
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    $\begingroup$ @youpilat13 Thanks for your note. The value of little h depends on the cosmological model and is defined as $H(z=0) = H_0 = 100 h\ km\ s^{-1} \ Mpc^{-1}$. Assuming that the cosmology is $\Lambda$CDM, according to [arxiv.org/abs/1807.06209 ] Planck 2018, it has a value of around 0.67 (please check the reference for exact report). Thus, (for the standard model) $R = 8.0/0.67 Mpc$ is the size of our desired window. $\endgroup$
    – Nim
    Commented Jan 2, 2020 at 13:52
  • $\begingroup$ ok for $h$ definition. And what about my second remark : you say : "8$h^{-1}\,$Mpc (i.e $\sigma^{2}(𝑅=8)$ at $z=0$) : we should rather say $\sigma^{2}(𝑅=11.94)$ at $z=0$), shoudn't we ? Sorry to insist, but I don't understans why at $z=0$, we define $\sigma^{2}(𝑅=8)$ instead of $\sigma^{2}(𝑅=11.94)$ . Maybe $h^{-1}$ is implicitly removed and we only keep the $8\,Mpc$ . Thanks for your patience. $\endgroup$
    – user87745
    Commented Jan 2, 2020 at 15:24
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    $\begingroup$ @youpilat13 You are totally right. I have edited my answer. $\sigma_8^2$ at $z = 0$ for $\Lambda CDM$ cosmology, is equal to $\sigma^2(R = 11.94)$. $\endgroup$
    – Nim
    Commented Jan 3, 2020 at 13:17
  • $\begingroup$ Thanks for your correction ! $\endgroup$
    – user87745
    Commented Jan 3, 2020 at 13:29