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What is the relation between multipole moment $l$ and angular scale $\theta$ of the Cosmic Microwave Background? Somewhere on the web I found that $\theta\propto\frac{180^{\circ}}{l}$ but I need exact relation.

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    $\begingroup$ $l$ and $\theta$ are related simply from the definition of the spherical harmonics $Y_{lm}$, independently of whether you are studying the CMB. Is this what you are looking for, or are you looking for something CMB-specific? $\endgroup$ – user10851 Feb 16 '13 at 18:14
  • $\begingroup$ @ChrisWhite I believe this is what I am looking for. So how do I relate $l$ to $\theta$ using spherical harmonics? $\endgroup$ – Luke Feb 16 '13 at 18:24
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You may not get a relation so exact as you'd like, and here's why:

First, you have to understand spherical harmonics. These are complex-valued functions $Y_{lm}$ (sometimes written $Y_l^m$) defined on a sphere. You can write them as functions of a polar coordinate $\theta$ (like latitude, or more precisely some form of colatitude, which makes one of the poles $0$) and $\phi$ (essentially longitude). Spherical harmonics are used to decompose functions defined on spheres into components, just as sines and cosines decompose functions defined on the real line (or intervals thereof) in Fourier analysis. Once you have such a decomposition, the coefficients (or rather the squares of their magnitudes) tell you how much power is in each component - lots on large scales (small $l$) and little on small scales (large $l$) or whatever.

Now you are basically asking, "What is the typical angular size of a feature with a given $l$?" The thing is, there are two indices, $l$ and $m$, where $m$ runs from $-l$ to $l$. $l$ sets the number of nodal lines, so intuitively higher $l$-values divide the sphere into more (and thus smaller) regions, but $m$ determines how these are arranged. Imagine cutting a sphere into $2l$ pieces of equal area. These pieces can be long and skinny or they can be more well-rounded, and so it is not clear how you would say what the 1D angular size $\theta$ is. You can't pick just one way of cutting things either - you need all allowed $m$ values to accurately decompose any function, and each one specifies a different way of cutting the sphere up.

Now you can ask what the average solid angle is corresponding to a feature with a given $l$. This is because there will be $2l$ internodal regions in $Y_{lm}$, regardless of $m$. In this case, the average solid angle of such a feature is $$ \Omega = \frac{4\pi}{2l} = \frac{2\pi}{l} = \frac{2(180^\circ)^2}{\pi l}. $$

However, we often desire regular old-fashioned 1D angles $\theta$. It would not be terribly wrong (or terribly right) to take the square root of this: $$ \theta = \sqrt{\Omega} = 180^\circ \sqrt{\frac{2}{\pi l}}. $$ Alternatively, you can examine the cut that makes $2l$ equal strips (much like you would get by cutting Earth's surface along $2l$ equally spaced lines of longitude), in which case the widest part (at the "equator") of these rather thin strips (thinner than the segments you'd get with a different cutting method) is $$ \theta = \frac{360^\circ}{2l} = \frac{180^\circ}{l}. $$

The methods don't quite agree because the situation is not perfectly well-defined. In practice, a perfectly circular feature on the sky, which clearly only has one angular size $\theta$ associated with it, will contribute to multiple spherical harmonics of different $l$'s and $m$'s.

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    $\begingroup$ Might be worth mentioning that in CMB analysis they typically use a different decomposition (in terms of Lagrange polynomials $P_\ell(\cos\theta)$) with coefficients $C_\ell=\sum_{m=-\ell}^\ell a^*_{\ell,m}a_{\ell,m}$. It's intimately related to the spherical harmonic decomposition, but depending on the paper this may or may not be clear. $\endgroup$ – Kyle Oman Mar 19 '14 at 19:38
  • $\begingroup$ Aaaaand just realized this is over a year old. But suddenly in vogue with the BICEP2 announcement. $\endgroup$ – Kyle Oman Mar 19 '14 at 19:39
  • $\begingroup$ The Plank Study gives an angular scale of 0.0104148. Using your formula, that gives the first multipole of 301 $l$. Yet the graph of the Power Spectrum in the Plank report shows the first multipole at roughly 220. Would you please show the math of how you go from $0.60^\circ$ degrees to 220 $l$? $\endgroup$ – Quarkly Dec 24 '16 at 22:54
  • $\begingroup$ @MikeDoonsebury: When you see "user#####" with no link to a profile, that signifies that the user has left this particular Stack Exchange and cannot respond to the comment (because he's not been alerted to it because he has no way to be alerted). $\endgroup$ – Kyle Kanos Dec 28 '16 at 16:02

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