[Frost] Question

[Michael Dugger]

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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>
>