[Frost] Question

Wed Aug 17 21:19:19 EDT 2016

Dear Michael and Barry,

Let me comment on the things you brought up in the meantime.

(1) Oversimplification.  I completely agree with Michael’s remarks that final expressions for the estimators of the polarization observables will be more complicated if more realistic assumptions are made.  That, however, was not the question.  The question was whether it is necessary to include the azimuthal angle in the determination of the Q-factors (in my opinion it is not) and whether or not Q-factors can be included in a moments methods (in my opinion they can).  As mentioned earlier, I crafted my example using the same assumptions of a perfect detector, polarized cross section with one observable, and constant background as Michael did in his example. The goal was only to illustrate the technique.

(2) cos^2(alpha) term.  Here, I don’t see the problem.  Exactly because of the relation cos^2(alpha) = 1/2 + cos(2*alpha)/2 my expression and the expression Michael is deriving (see, p. 4, https://userweb.jlab.org/~dugger/frost/12-6-2012.pdf) are algebraically equivalent.  Where Michael has, e.g.,  2 / (Xm0 + Xm2) = 2 / (sum 1 + sum cos(2*alpha)) I have 1 / (sum cos^2(alpha)), which is the same.  Now I made my simplified expression under the assumption of A = 1 and I did not make an attempt to cancel acceptance effects by taking differences.  That certainly can be done.  The cos^2(alpha) term is not the issue.

All the best,
Steffen

> On Aug 17, 2016, at 7:59 PM, Michael Dugger <dugger at jlab.org> wrote:
> 
> 
> Hi,
> 
> Another way to look at it:
> 
> When you weight your expression by
> 
> cos^2(alpha)
> 
> you have to remember that cos^2(alpha) = 1/2 + cos(2*alpha)/2
> 
> For the case where the efficiency is perfect (A=1) and the cross section is
> 
> sigma_{so}(1 + P cos(alpha)) ,
> 
> then weighting by cos^2(\alpha) is the same as weighting by a factor of 1/2.
> 
> However, if the efficiency is not flat, then weighting by cos^2(alpha) will introduce cos(2*alpha) terms associated with the detector efficiency.
> 
> Having a cos(2*alpha) terms associated with the detector efficiency is not a big problem, but you do have to account for it in the final expression when the efficiency is not equal to a constant.
> 
> Take care,
> Michael
> 
> On Wed, 17 Aug 2016, Barry Ritchie wrote:
> 
>> Steffen, just a note: I should think that another possible worry arises in the approach you’ve outlined by the use of non-orthogonal polynomials. For any expansion of the yield in terms of powers of cos α , one should use the (orthogonal) Legendre polynomials, Pn (cos2α), I’d think, for the same reason one uses the (orthogonal)  cos mα terms in the Fourier moment method. Otherwise, since cos2α is not orthogonal to cos α, the cos2α yield term will automatically be picking up pieces both of the zeroth-order P0 (cos α)and the cos2α part of the fourth-order P4 (cos α) term. For a perfect detector (meaning perfect efficiency and 4π acceptance), this is not crucial, but I would worry that it manifestly would be problematic for the real CLAS detector.  ---BGR
>> 
>> Professor Barry G. Ritchie
>> Department of Physics
>> Arizona State University
>> Tempe, AZ 85287-1504
>> 
>> Phone: (480) 965-4707
>> Fax: (480) 965-7954
>> 
>> From: Frost [mailto:frost-bounces at jlab.org] On Behalf Of Steffen Strauch
>> Sent: Wednesday, August 17, 2016 3:17 PM
>> To: Michael Dugger <dugger at jlab.org>
>> Cc: frost at jlab.org
>> Subject: Re: [Frost] Question
>> 
>> Dear Michael,
>> 
>> Yes, this is what I originally suggested for the experimentally determined sums over all events.
>> 
>> In the meantime, I think I need to revise the expressions and the numerator sum should read: cos(\alpha) and the denominator sum should read Q cos^2(\alpha).  This and the earlier expressions give the same results for constant background or if the observable does not change over the kinematic bin; like in our simplified examples.  Practically, the differences will be small.  However, when there are variations of the observable or of the background over a large bin (like in the double-pion case), the latter expressions give in my opinion results for the observable which are more meaningful.
>> 
>> I will update my slides and we can discuss this more during tomorrow’s meeting.
>> 
>> Steffen
>> 
>> 
>> On Aug 17, 2016, at 1:21 PM, Michael Dugger <dugger at jlab.org<mailto:dugger at jlab.org>> wrote:
>> 
>> 
>> Steffen,
>> 
>> I just want to make sure I understand.
>> 
>> You weight each numerator event by
>> 
>> Q cos(\alpha)
>> 
>> and each denominator event by
>> 
>> [Q cos(\alpha)]^2 .
>> 
>> Is this correct?
>> 
>> Take care,
>> Michael
>> 
>> On Wed, 17 Aug 2016, Steffen Strauch wrote:
>> 
>> 
>> Dear Michael,
>> 
>> It is important that the weights in the moments in the denominator contain an additional factor of Q compared to the numerator.
>> 
>> I don’t think there is a typo here. If you compare on page 4 the two sets of moment ratios, you see that the first set already contains Q in the integral.  Weighing the cos(alpha) moment with an additional factor of Q, gives you the Q^2 in the integral of the numerator in the second set.  The denominator of the first set does not contain any Q in the integral.  So, an additional factor of Q^2 in the weight of the moment gives you the Q^2 in the integral of the denominator of the second set.
>> 
>> For the correct expression that the extracted value is the weighted average of P and not Q*P.
>> 
>> Thanks,
>> Steffen
>> 
>> 
>> 
>> 
>> On Aug 17, 2016, at 12:06 PM, Michael Dugger <dugger at jlab.org<mailto:dugger at jlab.org>> wrote:
>> 
>> 
>> Steffen,
>> 
>> On last line of slide 4 and slide 5 of your pdf:
>> 
>> https://www.jlab.org/Hall-B/secure/g9/g9_strauch/mtg/FROST_meeting_2016_08_18.pdf<https://urldefense.proofpoint.com/v2/url?u=https-3A__www.jlab.org_Hall-2DB_secure_g9_g9-5Fstrauch_mtg_FROST-5Fmeeting-5F2016-5F08-5F18.pdf&d=CwMFaQ&c=AGbYxfJbXK67KfXyGqyv2Ejiz41FqQuZFk4A-1IxfAU&r=NC99X3Muut85jp1nyEEaKzrqGMedseDv3USQMbrzzMU&m=uKTOtlnnx_ZYFBxdgkWlssxnYsmNeTBkCDiI5B_qXYI&s=rNmvyekg4GCUDmS6Ot0NLHq7ce474-Bn80L5NOU7gko&e=>
>> 
>> you show on the left hand side
>> 
>> Y_{Q cos\alpha}/Y_{Q^2 cos^2 \alpha},
>> 
>> where there is a factor of Q in the numerator and a factor of Q^2 in the denominator.
>> 
>> However, in the middle of the line you have Q^2 in both the numerator and denominator. Is there a typo?
>> 
>> If you have Q^2 in both the numerator and denominator, the expression will not work.
>> 
>> Take care,
>> Michael
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