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I am looking for an explanation on the angular power spectrum. I found this extract that is interesting but not fully understood for me (I will cite the step that I didn't understand)

"what is done is we take the map of the CMB sky, and perform a spherical harmonic transform on it. A spherical harmonic transform is basically the same general concept as a Fourier transform, but the spherical harmonics are mutually orthogonal on the surface of a sphere, like so: $$ f(\theta, \phi)=\sum_{l m} a_{l m} Y(\theta, \phi)_{l}^{m} $$ Here the "I" index denotes the number of oscillations, and the "m" index is a way of encoding the direction of oscillation on the sphere, and varies from "-l" to "I". For example, $I=0$ is zero oscillation: this is the monopole that sets the overall scale. $I=1$ is a dipole: one full oscillation over the sphere, and there are three possible directions $(x, y$, and $z)$. Go to higher and higher I values, and you get more (and therefore smaller) oscillations and more possible directions for those oscillations. The way they are typically written, the spherical harmonic functions $$ Y(\theta, \phi)_{l}^{m} $$ are complex functions, and the coefficients are therefore complex. This is a minor issue, though. In order to build the power spectrum, we average over directions. This is done as follows: $$ C_{l}=\frac{1}{2 l+1} \sum_{m=-l}^{l} a_{l m} a_{l m}^{*} $$

Finally, in building that plot, you may notice that the vertical axis is not $$ C_{l} l $$ but is instead $$ C_{l} l(l+1) / 2 \pi $$ It turns out if we had a power spectrum that was uniform in a logarithmic interval in I, then multiplying said function by $$ l(l+1) / 2 \pi $$ would give us a constant. Thus this multiplication allows us to interpret the function more easily, because inflation predicts that the primordial power spectrum, the one initially generated by inflation, would be nearly a constant in this space.

If inflation is true, then, all of the features you see in a power spectrum written as above that deviate from a constant stem from the dynamics of the universe between inflation and the emission of the big bang (plus some very slight modification between us and the CMB). For example, the long damping tail at high I stems from the fact that the surface of emission of the CMB is not instantaneous: the phase transition from a plasma to a gas happened over time, and the resultant blurring of the signal damps the small-scale fluctuations. There's also the ratio between the even and odd peaks of the power spectrum. This comes about because of the differences in the physics between normal matter and dark matter: dark matter just falls into potential wells, while normal matter bounces. The failure of dark matter to bounce causes a reduction of the evennumbered peaks relative to normal matter."

1) First problem :

What I have difficulties to grasp is the signification of "vertical axis" is not : $$ C_{l} l $$ but is instead $$ C_{l} l(l+1) / 2 \pi $$

What does "vertical axis" mean in this context ? How to proove that ?

2) Second problem :

And after they say : "It turns out if we had a power spectrum that was uniform in a logarithmic interval in I, then multiplying said function by $$l(l+1) / 2 \pi$$ would give us a constant. Thus this multiplication allows us to interpret the function more easily, because inflation predicts that the primordial power spectrum, the one initially generated by inflation, would be nearly a constant in this space."

I don't understand which is the trick to make the product by $$l(l+1) / 2 \pi$$ being constant. It is very disturbing since the angular power spectrum is not constant for example with the angular spectrum of CMB.

3) Just a precision :

in one link of my previous post ( previous link ) Why does the writter say "How to write the 3D power spectrum, $P_{k}$, as an integral of the angular power spectrum, $C_{\ell}$ ?" whereas from another author, it is not the direct relation beween angular power spectrum and matter power spectrum.

By, the way, could anyone write the complete formula that links these 2 quantities ($C_{\ell}$ and $P_k$) ?

I would be grateful since I am a little confused between angualr correlation function and matter power spectrum and spherical Bessel functions.

If someone could give me clarifications on these 2 points, I would be grateful.

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I think I can sort of answer your first/second question. It's a bit hard to guess what your background is, but I hope you have seen or derived somewhere that the $a_{lm}$ coefficients can be written as $$\oint \Theta(\hat{x}) Y_{lm}^*(\hat{x}) d\hat{x}$$ where $\Theta$ is the temperature fluctuation (as seen in the CMB) and $Y_{lm}^*$ is (the complex conjugate of) a spherical harmonic. Note that $\vec{x} = (\theta, \phi)$, $\hat{x}$ is the unit vector parallel to $\vec{x}$ and $\hat{k}$ is the unit vector parallel to $\vec{k}$. $k$ is the length of $\vec{k}$. In cosmology there is an effect called the Ordinary Sachs-Wolfe effect, which has to do with redshifting of blueshifting due to matter overdensities. In this theory, you find an expression for $\Theta$ which is $$\Theta = \frac{1}{3}\Phi = \sum_k \frac{1}{3}\Phi_k e^{i\vec{k}\cdot \vec{x}}$$ where $\Phi$ is the Bardeen potential. You can plug this expression into the equation for the $a_{lm}$ and then you can use the expansion of a plane wave, which goes like this: $$e^{i\vec{k}\cdot \vec{x}} = 4\pi\sum_{lm} i^l j_l(kx) Y_{lm}(\hat{x})Y_{lm}^*(\hat{k})$$ So then you get $$a_{lm} = \frac{4\pi}{3}\sum_k \Phi_\vec{k} \sum_{lm} i^l Y_{l'm'}^*(\hat{k}) j_{l'}(kx)\oint Y_{lm}^*(\hat{x}) Y_{l'm'}(\hat{x}) d\Omega$$ Due to orthonormality of the spherical harmonics this becomes $$a_{lm} = \frac{4\pi}{3}i^l \sum_k \Phi_\vec{k} j_l(kx) Y_{lm}^*(\hat{k})$$

You may have also come across the expression for the $C_l$: $$ C_l = \frac{1}{2l + 1}\sum_m E[|a_{lm}|^2]$$ which due to the above is proportional to $$C_l \propto \frac{1}{2l + 1}\sum_m \sum_{kk'} E[\Phi_\vec{k} \Phi{\vec{k}'}] j_l(k'x) j_l(kx) Y_{lm}^*(\hat{k}) Y_{lm}(\hat{k}')$$ The E[ ] is the covariance matrix. If $\Phi_\vec{k}$ is a random variable, then $E[\Phi_\vec{k} \Phi_\vec{k'}] \propto k^{n - 4}\delta_{kk'}$ I'm not sure how to show this quickly. I'm afraid you need quite a few derivations to get to this point. It follows in part from the assumption of a power spectrum of the form $P(k) \propto k^n$. Using this you find $$C_l \propto \sum_k k^{n - 4}j_l(kx)^2 \rightarrow \int_0^\infty k^{n - 2} j_l(kx)^2 dk$$ if we approximate the sum by an integral. For a Harrison-Zel'dovich spectrum, which is a special kind of powerspectrum, defined as $P(k) \propto k$ (so n=1), this becomes $$ C_l \propto \int_0^\infty j_l(kx)^2 \frac{dk}{k} = \frac{1}{2l(l + 1)}$$ so in the case of a Harrison-Zel'dovich spectrum, the quantitiy $l(l + 1)C_l/2\pi$ is constant for the Ordinary Zachs-Wolfe effect. They put $l(l + 1)C_l/2\pi$ on the y-axis to make spotting this effect easy, I suppose.

I'd be very interested to see a detailed answer to your last question.

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  • $\begingroup$ Thanks a lot for your help, it is kind from your part to give a detailled answer. I am waiting also like you for a potential answer of my last question. Best regards $\endgroup$ – youpilat13 Apr 20 at 13:11
  • $\begingroup$ 1) Could you replace please all the $x$ variable by $\theta$ for more visibility and clarity : sorry I don't want to be boring but this is for convenience 2) the $k$ is the wavenumber, we are allright ? Thanks in advance. Best regards $\endgroup$ – youpilat13 Apr 22 at 8:15
  • $\begingroup$ @youpilat13, yes k is the wavenumer. It's a vector, actually, and so is x, so each time you see them multiplied, I'm referring to a dot product where $\vec{x} = (\theta, \phi)$ $\endgroup$ – theWrongAlice Apr 22 at 11:14
  • $\begingroup$ Just a last precision : when you write $j_l(kx)^2$, how can I understand it ? : is it equal to write $j_l(k\theta)^2$ or $j_l(k\phi)^2$ ? Regards $\endgroup$ – youpilat13 Apr 22 at 16:04
  • $\begingroup$ @youpilat13 sorry, may have confused myself a little with my own notation. I checked my cosmology notes and added all the proper symbols now. I hope this clears up the confusion somewhat. $\endgroup$ – theWrongAlice Apr 22 at 17:34

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