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## Re: Question about inflation theory

 Subject: Re: Question about inflation theory Craig Markwardt 26 Mar 2006 05:23:52 -0600 sci.astro
 ```"Thomas Smid" <[email protected]/* */> writes: > Craig Markwardt wrote: > > Unfortunately, your claims are still incorrect. > > > > "Thomas Smid" <[email protected]/* */> writes: > > > > > Craig Markwardt wrote: > > > > > > > Quoting the web page: > > > > : However if two identical detectors measure a homogeneous background > > > > : noise (which is the case for the CMB radiation), the difference > > > > : between the count rates will not be zero but on average be equal to > > > > : the square root of the count rates (the statistical variance). > > > > > > > > This statement is incorrect for several reasons: > > > > 1. The data are not counts, and the uncertainties are not Poissonian, > > > > therefore the statistical variance will not be "the square root of > > > > the count rates". > > > > ref. N. Jarosik, et al., 2003, ApJS, 148, 29 (statistics) > > > > ref. C.L. Bennett, et al., 2003, ApJS, 148, 97 (foregrounds) > > > > > > Yes, it is probably technically not correct to use the term 'variance' > > > here. so I changed this to 'standard deviation' on my webpage. But the > > > standard deviation should, for a large number of data points, be the > > > same for a Poisson or Normal (Gaussian) distribution, and thus the > > > difference of two (on average) identical noise signals should be > > > proportional to sqrt(noise_signal). > > > > This is of course incorrect. If one takes the difference of two > > random variables with equal expectation values (mu) and equal standard > > deviations (sigma), the result is a random variable with an > > expectation value of *zero*, and a standard deviation of > > sqrt(2*sigma). > > First of all, I am sure you mean sqrt(2)*sigma here. True. > The point is that in this case the standard deviation *is* the observed > variable: > if telescope A measures a temperature T with standard deviation dT and > telescope B measures the same, then the most probable absolute value > for the difference signal is sqrt(2)*dT (assuming the measurements are > uncorrelated, which they should be as the telescopes are pointing into > different directions) Your claims continue to be erroneous. Let's take them one at a time. "standard deviation *is* the observed variable" -- False. If you had done some research on WMAP, you would know that the observable is the time-dependent flux difference between the "A" and "B" telescopes as the spacecraft spins. This is an inescapable consequence of the design of the observatory. [ E.g. http://map.gsfc.nasa.gov/m_mm/ob_techradiometers.html ] "the most probable absolute value for the difference signal is sqrt(2)*dT" -- False. It's a simple computer experiment to show that the difference you describe is not sqrt(2)*dT. HOWEVER, Since WMAP's receivers and analog electronics do not measure the absolute value, but rather the *signed* difference between the A and B fluxes, your whole line of reasoning is irrelevant. If T1 and T2 are two random variables with the same expectation (mu) and same standard deviation (sigma), then the signed difference, T1-T2, has an expected value of *zero* and standard deviation of sqrt(2)*sigma. Thus, the whole basis of your claim is erroneous, and the conclusions you draw are irrelevant. > > Also, it is not even clear what a "noise signal" is. Typically noise > > and signal are two different quantities, with the experimenter > > striving to design an experiment that minimizes the former and > > maximizes the latter. > > If the data have a statistical characteristic according to a Gaussian > distribution (which is generally assumed for the CMB) then I think it > is justified to use the term 'noise' for this (whatever its physical > relevance may be). Every signal that is absolutely isotropic on all > scales should really appear as 'noise' if one takes the difference of > the two telescope readings. However, since you are discussing detections of Jupiter, the measurement is clearly not noise. Jupiter has a systematically strong signal. Your discussions of gaussian distributions are irrelevant in that case. And to be clear, one WMAP feed detects the strong signal of Jupiter (plus the CMB); while the other feed detects only the CMB. The measurements displayed by Page et al are the difference signals, averaged over many observations and scan patterns. (see Page et al 2003) > > The plots you refer to in Page et al (2003) are certainly not noise, > > since they are observations of a very strong signal (Jupiter). Which > > is the whole point: to observe a bright source in order to > > characterize the optical beam profile. > > These plots are derived from the telecopes individually. That's why > they can not be applied to the differential signal of the CMB. False. See above. > > > .. So if noise_signal=exp(-theta^2) > > > gives the angular beam profile for the individual telescopes, then the > > > angular beam profile for the differential signal should be proportional > > > to sqrt[exp(-theta^2)] = exp(-0.5*theta^2). > > > > The premise of this statement is also incorrect. In fact, the beam > > profiles you refer to in Page et al (2003) are the measurement of > > strong *signal* (Jupiter) compared to the weak sky background noise. > > Again, the beam profiles (shown in Fig.4 on my page > http://www.physicsmyths.org.uk/wmap.htm ) are for an individual > telescope and are not appropriate for the differential signal of the > CMB. Since there is no way to derive the signal from an individual telescope on WMAP, you claim is erroneous. ... deletions ... > > > ... However, the WMAP data > > > analysis assumes also the differential beam profile to be given by > > > exp(-theta^2). This assumption should lead to an apparent residual > > > signal peaking near about theta=0.3 deg as shown on my page > > > http://www.physicsmyths.org.uk/wmap.htm > > > > This claim is again false. Since the observations of Jupiter are > > taken from the WMAP differencing assemblies, they are in fact > > *already* "differential," and thus accounted for. > > No, the Jupiter observations are taken from the telescopes individually > (how would you want to make a differential observation of Jupiter > anyway?; there is only one Jupiter in the sky, and the telecopes are > pointing into completely different directions.) It's not a question of "wanting" to make a differential measurement of Jupiter. There is no other way of observing any object with WMAP. > > And further, the authors surely do *not* assume that the beam profile > > is given by "exp(-theta^2)", as Page et al (2003) make a point of > > discussing. > > Essentially, the beam profiles are assumed as Gaussian (see Eq.(12) in > that paper), although with some distortions obviously. The exact beam > profile is however irrelevant for my argument which also would hold for > a rectangular profile for instance. The point why I used a Gaussian > here was merely in order to be more realistic when looking at the > effect of subtracting the differential from the individual beam profile > (Fig.5 on my page http://www.physicsmyths.org.uk/wmap.htm ). It's strange that you refer to equation 12 of Page et al (2003), which shows a series expansion. True, it has a gaussian as a part of it, but it also has a series of Hermite polynomials, which are not gaussians. But to summarize, since it is inherent in the design of WMAP to take differences between the two receiver feeds, the beam profiles are indeed actually differential beam profiles. Also, you are making a fundamental error regarding the way two variables are subtracted (signed difference vs. absolute value). Finally, you are imputing gaussian noise statistics on a measurement which is inherently non-gaussian (i.e. measurements of the strong Jupiter signal). When the premises of your arguments are erroneous, then the conclusions you draw are irrelevant. It may be worth it for you to read some more about how WMAP and statistics actually work before making such erroneous and unsubstantiated claims. CM ```