# Relationship between 3 quantities: Density Matter power spectrum, Density Matter angular power spectrum and Temperature angular power spectrum

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.

• – User123 Jun 20 at 21:18
• Care to link your previous question you refer to? – planetmaker Jul 2 at 6:12
• 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 $k$. – verdelite Jul 8 at 13:35
• @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 – youpilat13 Jul 8 at 19:28
• @verdelite Could have you see my last remark above about the quantity $C_\ell(z,z)$ ?, maybe you are not connected ... – youpilat13 Jul 10 at 15:49