The BICEP experiment's recent announcement included the preprint of their paper,

BICEP2 I: Detection of $B$-mode polarization at degree angular scales. BICEP2 Collaboration. To be submitted. BICEP-Keck preprint, arXiv:1403.3985.

Gravitational lensing of the CMB’s light by large scale structure at relatively late times produces small deflections of the primordial pattern, converting a small portion of E-mode power into B-modes. The lensing B-mode spectrum is similar to a smoothed version of the E-mode spectrum but a factor 100 lower in power, and hence also rises toward sub-degree scales and peaks around $\ell$ = 1000.

I think the $\ell$ is this:

For example $\ell=10$ corresponds to roughly 10 degrees on the sky, $\ell=100$ corresponds to roughly 1 degree on the sky. (From CMB introduction, by Wayne Hu.)

But how does that apply here? When BICEP looks for something with an $\ell$ around 80, does that mean a "multipole moment" which spans 80 degrees across the sky?

  • $\begingroup$ The local peak from the primordial gravitational waves is normally expected around $\ell=90$. The spherical harmonic $Y_{90,90}$, for example, has the $J_z$ aligned "maximally vertically, so it is spinning maximally vertically among the $Y_{90,m}$ harmonics, and on this one, you see that the angular dependence contains $\exp(90 i\phi)$ which contains 90 maxima around the circle. So the "wavelength" of the component spans 360/90=4 degrees on the sky. The resolution has to be a bit better to actually "see" the shape of these waves. $\endgroup$ Mar 19, 2014 at 6:05

2 Answers 2


It's the same $\ell$ that indexes the spherical harmonics $Y_{\ell m}$ (or $Y_\ell^m$ if you prefer). We can decompose functions defined on the sphere (like anything defined on the sky) into a countably infinite sum of appropriately weighted spherical harmonics. $\ell$ counts the number of nodes, while different values of $m$, $0 \leq \lvert m \rvert \leq \ell$, give different arrangements of those nodes.

Higher values of $\ell$ correspond to components that have more nodes and fluctuations. The angular scale of variations corresponding to a given $\ell$ scale like $1/\ell$. For more information, you might want to look at an answer I wrote to Relation between multipole moment and angular scale of CMB.

One thing cosmologists do is plot correlations between different quantities as a function of $\ell$. You can imagine decomposing two functions \begin{align} f(\theta, \phi) & = \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell a_{\ell m} Y_{\ell m}(\theta, \phi) \\ g(\theta, \phi) & = \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell b_{\ell m} Y_{\ell m}(\theta, \phi), \end{align} where the $a$'s and $b$'s are complex numbers. Then you might plot quantities like $$ Q_\ell = \sum_{m=-\ell}^\ell a_{\ell m}^* b_{\ell m} $$ over a run of $\ell$ for which you have good data, comparing theory to observation. $Q_{80}$, for example, will be built from information about ${\sim}16^\circ$ scales.

BICEP doesn't look at the whole sky, by the way, so they can't even measure the low-$\ell$ components of anything. What they focus on is the high-$\ell$ stuff that might be harder to get with a space-based mission designed to scan the whole sky at lower resolution. The assumption is that the high-$\ell$ signal you get in one part of the sky is representative of the high-$\ell$ signal everywhere. (If this weren't the case we'd live in a very weird universe indeed.)

  • $\begingroup$ There are about 41,000 "square degrees" over the full sky. If the 2 degree region is a circle with a 2 degree diameter, then that would be about 3.14 square degrees. I can't figure out any meaningful connection to 80 here. Maybe 80^2? $\endgroup$ Mar 19, 2014 at 18:44
  • $\begingroup$ Oops - I was using formulas from my other answer, one of which was wrong, the other of which was misleading. $\endgroup$
    – user10851
    Mar 19, 2014 at 19:04

Here are some BICEP2 details to augment Chris White's answer:

BICEP2 accomplishes the task of measuring angular variations in polarization by converting those angular variations to a time domain signal. It does so by scanning its telescope across the sky at a constant rate.

Specifically, the telescope scans at a fixed rate of $2.8^\circ $/ second in azimuth (angle along the horizon), at constant declination. Because the telescope is aiming high in the sky (at its south pole location, average elevation = - average declination = $57.5^\circ$), the actual sky scan rate is approximately $2.8 * \cos(57.5^\circ) = 1.5^\circ /s$.

Therefore, a feature with angular size $\Delta \theta$ appears in the instrument data stream as a signal with time duration $\Delta t = \Delta \theta / 1.5$.

Since the $l$th multipole has $l$ nodes in $180^\circ$, the angular size of a "period" of this multipole (comprising two nodes) is approximately $180^\circ/(l/2) = (360/l)^\circ$, appearing in the data stream as a signal with period $T = (360/1.5)/l = 240/l$, or a frequency $$f=1/T=(l/240) \, \, Hz$$

Thus, the target multipole range $l=20-240$ appear in the data as signal frequencies of approximately $0.083-1 \, Hz$, or time periods ranging from 12 to 1 seconds.

The telescope performs many scans, at varying declinations, with the data being combined to form the final polarization map. Each individual scan takes data over $56.4^\circ$ in azimuth, or approximately $30^\circ$ in the sky. An individual scan therefore takes $56.4/2.8= 20$ seconds, less than 2 periods worth of signal at $l=20$. Thus the scan size limits low-$l$ data collection.

References are the BICEP2 results (especially section III A) and experiment (section 12.2) papers.


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