[Frost] Question

Wed Aug 17 23:28:49 EDT 2016

Steffen,

I actually think we are on the same page.

The use of the simple example has been helpful in this discussion. I think
we can agree on some points:

* The manner that Q-factors can be used in determining asymmetries is
dependent on how the distance to nearest neighbors is defined.

* Final expressions will have to take into account issues that are not
included in the simple example.

* There are many ways to analyze asymmetry data!

Take care,
Michael

> Dear Michael and Barry,
>
> Let me comment on the things you brought up in the meantime.
>
> (1) Oversimplification.  I completely agree with Michaelâ€™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 donâ€™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 youâ€™ve outlined by the use of non-orthogonal
>>> polynomials. For any expansion of the yield in terms of powers of cos
>>> Î± , one should use the (orthogonal) Legendre polynomials, Pn (cos2Î±),
>>> Iâ€™d think, for the same reason one uses the (orthogonal)  cos mÎ±
>>> terms in the Fourier moment method. Otherwise, since cos2Î± is not
>>> orthogonal to cos Î±, the cos2Î± yield term will automatically be
>>> picking up pieces both of the zeroth-order P0 (cos Î±)and the cos2Î±
>>> part of the fourth-order P4 (cos Î±) term. For a perfect detector
>>> (meaning perfect efficiency and 4Ï€ 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
>>> tomorrowâ€™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 donâ€™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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>
>