# Power Spectrum of CMB angle-l

1. Feb 1, 2015

### ChrisVer

http://pdg.lbl.gov/2013/reviews/rpp2013-rev-cosmic-microwave-background.pdf

Here one reads in sec. 26.2.4 that:

However it states that a single $Y_{lm}$ corresponds to angular variations of $\theta \sim \pi /l$.

I am not getting these statements. Also I find it difficult to understand, since most of the times, the power spectrum (eg. Fig. 26.1 in the above reference ) shows a figure of the amplitude versus on the lower axis the multipoles $l$ and on the upper axis the angles $\theta$. If there is no one-to-one correspondence between $\theta \text{-} l$ how does these figures make sense?

2. Feb 1, 2015

### Chalnoth

It's an approximate correspondence. There can't be a one-to-one correspondence because the fact that it's on the surface of a sphere prevents that.

3. Feb 2, 2015

### bapowell

The $C_\ell$'s receive contributions from perturbations across a range of scales. On large angular scales where the Sachs-Wolfe effect dominates, each $C_\ell$ is dominated by contributions from perturbations subtending an angle of $\theta = \pi/\ell$. The approximate correspondence follows.

4. Feb 2, 2015

### Chalnoth

One way of looking at it is to examine the spherical harmonics where $\ell = |m|$. For these spherical harmonics, all of the variation of the signal is in the $\phi$ direction (angle of longitude): $e^{im\phi}$. The waves of the same $\ell$ but different $m$ represent different "directions" of an oscillation of approximately the same wavelength. The peak-to-trough distance of a wave $e^{i\ell\phi}$ is $\pi/\ell$. I believe this is considered a full wavelength in context because it is the amplitude squared that represents the real field: $Y_\ell^mY_\ell^{m*}$, and squaring the amplitude halves the wavelength.