[Frost] Question

[Michael Dugger]

Steffen,

I think that when we present expressions for the observables we should be 
careful to describe the observable and the reaction signature.

For the reaction

gamma p -> p pi+ pi-

once the missing mass is shown to be zero, the background is just coming 
from the bound nucleons.

For the reaction

gamma p -> p omega

with omega -> pi+ pi- (pi0)

there is background under the omega that does not have to be from the 
bound nucleons.

In addition, for the case where the observable is the target asymmetry 
(or any observable that requires a target polarization), it is safe to 
assume that the cross section of the bound nucleons has no azimuthal 
dependence.

However, for the case where the observable does not require target 
polarization, it is not save to assume that the cross section of the bound 
nucleons is free of azimuthal dependence.

The reason I bring this up is that I have been talking about the beam 
asymmetries for the omega reaction and I think that you have been talking 
about the two pion reaction for observables that require target 
polarization.

Take care,
Michael


On Wed, 17 Aug 2016, Steffen Strauch wrote:

> 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> 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> 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
>>>>
>>>> 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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>>>
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