Relation between Matter Power spectrum and Angular power spectrum

  • #1
263
4
Summary:
I would like to go deeper in the relationship between Matter power spectrum and Angular power spectrum.
From a previous post about the Relationship between the angular and 3D power spectra , I have got a demonstration making the link between the Angular power spectrum ##C_{\ell}## and the 3D Matter power spectrum ##P(k)## :

1) For example, I have the following demonstration,
##
C_{\ell}\left(z, z^{\prime}\right)=\int_{0}^{\infty} d k k^{2} j_{\ell}(k z) j_{\ell}\left(k z^{\prime}\right) P(k)
##
where ##j_{\ell}## are the spherical Bessel functions.

Given

## \tag{1}
C_{\ell}\left(z, z^{\prime}\right)=\int_{0}^{\infty} d k k^{2} j_{\ell}(k z) j_{\ell}\left(k z^{\prime}\right) P(k)
##
Question: how to invert the integral to find the function ##P(k)##?
==>

The closure relation for spherical Bessel function:

## \tag{2}
\int_0^\infty x^2 j_n(xu) j_n(xv) dx = \frac{\pi}{2u^2} \delta(u-v).
##

Multipy Eq.(1) with ##z^2 j_\ell(qz)## and integral over ##z##:

\begin{align}
\int_0^\infty z^2 j_\ell(qz) C_{\ell}\left(z, z^{\prime}\right) dz =&\int_{0}^{\infty} d k k^{2} \left\{ \int^0_\infty z^2 dz j_\ell(qz) j_{\ell}(k z)\right\} j_{\ell}\left(k z^{\prime}\right) P(k) \\
=&\int_{0}^{\infty} d k k^{2} \left\{\frac{\pi}{2q^2} \delta(q-k)\right\} j_{\ell}\left(k z^{\prime}\right) P(k) \\
=& q^{2} \frac{\pi}{2q^2} j_{\ell}\left(q z^{\prime}\right) P(q) \tag{3}.
\end{align}

Once again multiply Eq.(3) with ##z'^2 j_\ell(q'z')## and integral over ##z'##

\begin{align}
\int_0^\infty z'^2 dz' j_\ell(q'z') \int_0^\infty z^2 j_\ell(qz) C_{\ell}\left(z, z^{\prime}\right) dz
=& \frac{\pi}{2} \left\{\int_0^\infty z'^2 dz' j_\ell(q'z') j_{\ell}(q z') \right\} P(q).\\
=& \frac{\pi}{2} \left\{ \frac{\pi}{2q'^2} \delta(q-q') \right\} P(q) \tag{4}.\\
\end{align}

To move the ##\delta## function in the right-hand-side, we multiply Eq. (4) (note that only ##q=q'## has contribution) with ##q'^2## and integral over ##q'##:

\begin{align}
\int_0^\infty dq' q'^2\int_0^\infty z'^2 dz' j_\ell(q'z') \int_0^\infty z^2 j_\ell(q'z) C_{\ell}\left(z, z'\right) dz
=& \frac{\pi^2}{4} \int_0^\infty dq' \delta(q-q') P(q).\\
=& \frac{\pi^2}{4} P(q) \tag{5}.
\end{align}

The left-hand-side of Eq.(5);

\begin{align}
\int_0^\infty dq' & q'^2\int_0^\infty z'^2 dz' j_\ell(q'z') \int_0^\infty z^2 j_\ell(q'z) C_{\ell}\left(z, z'\right) dz \\
= & \int_0^\infty z'^2 dz' \int_0^\infty z^2 dz \left\{ \int_0^\infty dq' q'^2 j_\ell(q'z') j_\ell(q'z) \right\} C_{\ell}(z, z') \\
= & \int_0^\infty z'^2 dz' \int_0^\infty z^2 dz \left\{ \frac{\pi}{2z^2} \delta(z-z') \right\} C_{\ell}(z, z') \\
= & \frac{\pi}{2} \int_0^\infty z^2 dz C_{\ell}(z, z). \tag{6}
\end{align}


Combine Eq.(5) and Eq.(6)

##
P(q) = \frac{2}{\pi} \int_0^\infty z^2 dz C_{\ell}(z, z).
##

2) I am surprized that ##C_{\ell}## has no dependence in π‘˜ scale ? only angular dependent and redshift dependent ? since only redshift 𝑧 appears in this expression ?

in cosmology, the angular power spectrum depends on multipole noted 𝑙 (Legendre transformation) which is related to angular quantities (πœƒ and πœ™). But the matter power spectrum is dependent of π‘˜ wave number (with Fourier transform).

I think I am wrong by saying that, in definition of 𝐢ℓ, one writes 𝐢ℓ(𝑧,𝑧′) where 𝑧 and 𝑧′ could be understood like redshift.

But here, we talk about the ##C_{\ell}## of matter fluctuations and not temperature fluctuations, do you agree ?

What do 𝑧 and 𝑧′ represent from your point of view in the expression 𝐢ℓ(𝑧,𝑧′) ?

Where is my misunderstanding ?

Thanks in advance for your help and don't hesitate to ask me for further informations if I have not been clear enough.
 
Last edited:

Answers and Replies

  • #2
kimbyd
Science Advisor
Gold Member
1,238
698
The angular power spectrum is a projection of the 3D power spectrum onto the surface of a sphere. Essentially, each ##C_\ell## is a sum over many different wavelengths that contribute to it depending upon their orientations relative to the sphere.

I believe it's a function of redshift because the power spectrum evolves over time.
 
  • #3
263
4
When you say " Angular power spectrum" and "projection of the 3D power spectrum", you talk about the "Matter Angular power spectrum", that is to say, about the fluctuations of matter and not the fluctuations of temperature like in the usual CMB Angular power spectrum ?

Best regards
 

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