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 ExB in BICEP2
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Anze Slosar

Joined: 24 Sep 2004
Posts: 205
Affiliation: Brookhaven National Laboratory

Posted: April 20 2014

I am having arguments with some colleagues of mine about the following:

To me, the fact that ExB and TxB are (mostly) consistent with zero is a good argument in favour of BICEP2 seeing primordial fluctuations rather than foregrounds. One would expect that a generic foreground would push roughly the same amount of power into all cross-correlations. However, some people insist that ExB=0 by construction because of this paragraph:

 Paper, Secition 8: wrote: Once differential ellipticity has been corrected we notice that an excess of TxB and ExB power remains at > 200 versus the ΛCDM expectation. The spectral form of this power is consistent with an overall rotation of the polarization angle of the experiment. While the detector-to-detector relative angles have been measured to differ from the design values by < 0.2◦ we currently do not have an accurate external measurement of the overall polarization angle. We therefore apply a rotation of ∼ 1◦ to the final Q/U maps to minimize the T B and EB power (Keating et al. 2013; Kaufman et al. 2013). We empha- size that this has a negligible effect on the BB bandpowers at < 200.

Ok, they have one degree of freedom and they can minimize some power. Say a "loop" produces a uniform polarization contribution, that wold be a k=0 mode and you could get rid of that that way. But it is one mode and you definitely cannot kill all correlations with just one d.o.f.. So, is ExB=TxB=0 a good argument in favour of "It cannot all be foregrounds."?
Antony Lewis

Joined: 23 Sep 2004
Posts: 1352
Affiliation: University of Sussex

 Posted: April 22 2014 I think I would agree that if they've just changed a constant rotation angle by a degree, that will not significantly change any foreground argument at low L - as they say it doesn't have a big affect at L~100 anyway where most of their interesting signal is. However I think having TB and EB consistent with zero is a pretty weak check - T must be dominated by primordial CMB temperature so the sensitivity to a small foreground contamination in T must be small compared to cosmic variance from the primary signal and their B noise (might be more interesting if they correlated their B with high- and low-frequency T from Planck in the same patch of sky with CMB projected out -did they try that?). Likewise even for E: If we assume as an extreme case that observed E is Eo = E + f, and observed Bo = f, so $\langle E_o B_o\rangle= \langle ff\rangle= C_f$, the number of sigmas at which you could expect to see the EB in the null hypothesis of zero B is something like $\sigma \sim \sqrt{\frac{n C_f^2}{C_E N}}$ where n is the number of modes observed and N the (lensing+) noise power. If O(f)~O(E)/6 is the case you are trying to check against (where all the excess B is foregrounds), then Cf ~ CE / 36, so for N ~ CE / 100 you'd need at least 50 modes to be able to tell at σ > 2. I don't think they say exactly how many modes they have, but for L~100 with 2% of the sky, it's not much bigger than 50 (and certainly not per L bin), and this was a very extreme back-of-the-envelope toy case (if I did it right).
Anze Slosar

Joined: 24 Sep 2004
Posts: 205
Affiliation: Brookhaven National Laboratory

 Posted: April 23 2014 I see, I didn't appreciate just how much more power there is in EE. BB at ell=100 is something like 0.015+−0.003, while EB point there is around 0.01+−0.01. Averaged over all bins, you might actually get a better competitive error, but then there is this 2sigma point at l=50. Thanks.
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