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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 $P(k)$:

Maybe this is due to the fact that we talk about the $C_{\ell}$ of matter fluctuations and not temperature fluctuations (like in CMB angular power spectrum), could anyone confirm this ambiguity ?

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

Given $C_{\ell}\left(z, z^{\prime}\right)=\int_{0}^{\infty} d k k^{2} j_{\ell}(k z) j_{\ell}\left(k z^{\prime}\right) P(k)$

Question: how to invert the integral to find the function $P(k)$ ?

==> The closure relation for spherical Bessel function: $\int_{0}^{\infty} x^{2} j_{n}(x u) j_{n}(x v) d x=\frac{\pi}{2 u^{2}} \delta(u-v)$

Multipy $(1)$ with $z^{2} j_{\ell}(q z)$ and integral over $z$ :

$$ \begin{aligned} \int_{0}^{\infty} z^{2} j_{\ell}(q z) C_{\ell}\left(z, z^{\prime}\right) d z &=\int_{0}^{\infty} d k k^{2}\left\{\int_{\infty}^{0} z^{2} d z j_{\ell}(q z) j_{\ell}(k z)\right\} j_{\ell}\left(k z^{\prime}\right) P(k) \\ &=\int_{0}^{\infty} d k k^{2}\left\{\frac{\pi}{2 q^{2}} \delta(q-k)\right\} j_{\ell}\left(k z^{\prime}\right) P(k) \\ &=q^{2} \frac{\pi}{2 q^{2}} j_{\ell}\left(q z^{\prime}\right) P(q)\quad(3) \end{aligned} $$ Once again multiply $(3)$ with $z^{\prime 2} j_{\ell}\left(q^{\prime} z^{\prime}\right)$ and integral over $z^{\prime}$ $$ \begin{aligned} \int_{0}^{\infty} z^{\prime 2} d z^{\prime} j_{\ell}\left(q^{\prime} z^{\prime}\right) \int_{0}^{\infty} z^{2} j_{\ell}(q z) C_{\ell}\left(z, z^{\prime}\right) d z &=\frac{\pi}{2}\left\{\int_{0}^{\infty} z^{\prime 2} d z^{\prime} j_{\ell}\left(q^{\prime} z^{\prime}\right) j_{\ell}\left(q z^{\prime}\right)\right\} P(q) \\ &=\frac{\pi}{2}\left\{\frac{\pi}{2 q^{\prime 2}} \delta\left(q-q^{\prime}\right)\right\} P(q) \quad(4) \end{aligned} $$ To move the $\delta$ function in the right-hand-side, we multiply (4) (note that only $q=q^{\prime}$ has contribution) with $q^{\prime 2}$ and integral over $q^{\prime}:$ $$ \begin{aligned} \int_{0}^{\infty} d q^{\prime} q^{\prime 2} \int_{0}^{\infty} z^{\prime 2} d z^{\prime} j_{\ell}\left(q^{\prime} z^{\prime}\right) \int_{0}^{\infty} z^{2} j_{\ell}\left(q^{\prime} z\right) C_{\ell}\left(z, z^{\prime}\right) d z &=\frac{\pi^{2}}{4} \int_{0}^{\infty} d q^{\prime} \delta\left(q-q^{\prime}\right) P(q) . \\ &=\frac{\pi^{2}}{4} P(q)\quad(5) \end{aligned} $$

The left-hand-side of Eq.(5); $$ \begin{aligned} &\int_{0}^{\infty} d q^{\prime} q^{\prime 2} \int_{0}^{\infty} z^{\prime 2} d z^{\prime} j_{\ell}\left(q^{\prime} z^{\prime}\right) \int_{0}^{\infty} z^{2} j_{\ell}\left(q^{\prime} z\right) C_{\ell}\left(z, z^{\prime}\right) d z \\ &=\int_{0}^{\infty} z^{\prime 2} d z^{\prime} \int_{0}^{\infty} z^{2} d z\left\{\int_{0}^{\infty} d q^{\prime} q^{\prime 2} j_{\ell}\left(q^{\prime} z^{\prime}\right) j_{\ell}\left(q^{\prime} z\right)\right\} C_{\ell}\left(z, z^{\prime}\right) \\ &=\int_{0}^{\infty} z^{\prime 2} d z^{\prime} \int_{0}^{\infty} z^{2} d z\left\{\frac{\pi}{2 z^{2}} \delta\left(z-z^{\prime}\right)\right\} C_{\ell}\left(z, z^{\prime}\right) \\ &=\frac{\pi}{2} \int_{0}^{\infty} z^{2} d z C_{\ell}(z, z) . \end{aligned} \quad(6) $$

Combine (5) and (6) :

$P(q)=\frac{2}{\pi} \int_{0}^{\infty} z^{2} d z C_{\ell}(z, z)$

  1. I am surprized that $C_{\ell}$ has no dependence in $k$ 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 $(\theta$ and $\phi)$. But the matter power spectrum is dependent of $k$ wave number (with Fourier transform).

I think I am wrong by saying that, in definition of $C \ell$, one writes $C \ell\left(z, z^{\prime}\right)$ where $z$ and $z$' could be understood like redshift.

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

What do $z$ and $z^{\prime}$ represent from your point of view in the expression $C \ell\left(z, z^{\prime}\right) ?$

Where is my misunderstanding?

UPDATE : in the following expression above :

$$P(q)=\frac{2}{\pi} \int_{0}^{\infty} z^{2} d z C_{\ell}(z, z)$$

What does variable $z$ represent in the factor $C_{\ell}(z, z)$ : $C_{\ell}$ should depend only on one variable, that is to say, the multipole $\ell$, shoudn't it ? Moreover, variable $q$ into $P(q)$ represents the $k$ scale, doesn't it ? since it doesn't appear in the integral (so there would a relation between $z$ and $q$ ?) :

$$P(q)=\frac{2}{\pi} \int_{0}^{\infty} z^{2} d z C_{\ell}(z, z)$$

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

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  • $\begingroup$ Link to the Astronomy SE question. $\endgroup$ – User123 Jun 20 at 21:18
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    $\begingroup$ Care to link your previous question you refer to? $\endgroup$ – planetmaker Jul 2 at 6:12
  • $\begingroup$ You said: "I am surprized that $C_l$ has no dependence in $k$ scale ?" Probably the answer is that according to your Eq. (1) you have integrated out $k$. $\endgroup$ – verdelite Jul 8 at 13:35
  • $\begingroup$ @verdelite . Thanks for your remark. If you look at my UPDATE, I have an issue of understanding : in the expression $P(q)=\frac{2}{\pi} \int_{0}^{\infty} z^{2} d z C_{\ell}(z, z)$, What does variable 𝑧 represent in the factor $C_\ell(z,z)$ : $C_\ell$ should depend only on one variable, that is to say, the multipole $\ell$, shoudn't it ? Best regards $\endgroup$ – youpilat13 Jul 8 at 19:28
  • $\begingroup$ @verdelite Could have you see my last remark above about the quantity $C_\ell(z,z)$ ?, maybe you are not connected ... $\endgroup$ – youpilat13 Jul 10 at 15:49

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