CMB: L to Angle Conversion Simplified

[nicksauce]

In discussion of the CMB it is often claimed that a spherical harmonic l roughly corresponds to $l = \frac{\pi}{\theta}$. Does anyone know a simple way to show this?

 

[Chalnoth]

In discussion of the CMB it is often claimed that a spherical harmonic l roughly corresponds to $l = \frac{\pi}{\theta}$. Does anyone know a simple way to show this?

First of all, if we pick any of the various $Y^l_m$'s, we know that the size of the variations for any $m$ for a given $l$ is the same. So we can pick one particular $Y^l_m$ that has a particularly simple functional form, $m = \pm l$:

$$Y^l_{\pm l} \left(\theta, \phi\right) \propto e^{\pm il\phi}$$

So here we have a situation where all of the variation is in the $\phi$ direction, with the typical with of a peak being $\theta = \frac{\pi}{\l}$.