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Posted: April 20 2014
by Anze Slosar
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&#9702;
we currently do not have an accurate external measurement of
the overall polarization angle. We therefore apply a rotation
of &#8764; 1&#9702; 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."?

Re: ExB in BICEP2

Posted: April 22 2014
by Antony Lewis
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 [tex]E_o = E + f[/tex], and observed [tex]B_o=f[/tex], so [tex]\langle E_o B_o\rangle= \langle ff\rangle= C_f[/tex], 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 [tex]n[/tex] is the number of modes observed and [tex]N[/tex] 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 [tex]C_f[/tex] ~ [tex]C_E/36[/tex], so for [tex]N[/tex] ~ [tex]C_E/100[/tex] you'd need at least 50 modes to be able to tell at [tex]\sigma > 2[/tex]. 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).


Posted: April 23 2014
by Anze Slosar
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.