here are the results of the test vitaliy requeste. it's not exactly the
same test because i just used the 'show_ep_asym.C' script, which just
fits to a model: A_meas = Pb*Pt*A_calc
but this should give an idea of the value of the fit wrt spin angle.
maybe chris or adrian have gotten to the point where they can say
something about fitting to background?
these numbers were all calculated using the 'Helastic_L' and
'Helastic_R' cut, which can be found in 'cuts.C' (includes a zcut and a
charge cut and a coplanarity cut). the target was in single state, so
these were beam asymmetry measurements, and the runs i used were from
3787-4744 (not all runs in that range, only the good, outbending runs,
as listed in the run list) - about 23500 C.
left right
spin angle: Pb*Pt chi^2
Pb*Pt chi^2
---------------------------------------------------------------------------------------------------------
50 0.20
1.47 0.18 2.22
45 0.18 1.47
0.17 2.26
40 0.16
1.47 0.16 2.29
35 0.15
1.47 0.15 2.28
30 0.14
1.46 0.14 2.28
vitaliy ziskin wrote:
> Jason,
> Can you do the same study for the data that we have on e-p elastic.
>
> Cheer, Vitaliy
>
> vitaliy ziskin wrote:
>
>> Per Genya's request I fitted vector asymmetry in each sector to the
>> data varying target angle. My fitting model is
>> A_{measured}(Q^2) = P0 \frac{A^{V}_{ed}}{1+P1*Q^4}. Here, P0 is a
>> Pe*Pz and P1 quantifies a contribution from background (asumming
>> background is constant). Here are my results:
>>
>>
>> right: left:
>> \theta = 45 \Chi^2 =
>> 10.8/4 \Chi^2 = 5.48/4
>> P0 = 0.545 \pm
>> 0.048 P0 = 0.494 \pm 0.03
>> P1 = 2.629 \pm
>> 1.59 P1 = 3.953 \pm 1.3
>>
>> \theta = 40 \Chi^2 =
>> 11.36/4 \Chi^2 = 5.46/4
>> P0 = 0.431 \pm
>> 0.04 P0 = 0.4634 \pm 0.03
>> P1 = 2.96 \pm 1.7
>> P1 = 3.787 \pm 1.3
>>
>> \theta = 35 \Chi^2 =
>> 11.83/4 \Chi^2 = 5.453/4
>> P0 = 0.45 \pm
>> 0.038 P0 = 0.44 \pm 0.03
>> P1 = 3.2 \pm 1.78
>> P1 = 3.64 \pm 1.3
>>
>> \theta = 30 \Chi^2 =
>> 12.2/4 \Chi^2 = 5.48/4
>> P0 = 0.42 \pm 0.036
>> P0 = 0.422 \pm 0.025
>> P1 = 3.5 \pm 1.86
>> P1 = 3.6 \pm 1.2
>>
>> \theta = 25 \Chi^2 =
>> 12.6/4 \Chi^2 = 5.48/4
>> P0 = 0.399 \pm 0.035
>> P0 = 0.4075 \pm 0.024
>> P1 = 3.68 \pm 1.92
>> P1 = 3.395 \pm 1.3
>>
>> Draw your own conclusions. Vitaliy
>
>
>
-- -------------------------------------------------------------------- jason seely 26.650.b massachusetts institute of technology 77 massachusetts avenue cambridge, ma 02139-4307email: seely@mit.edu phone: 617.253.4798/6734 html: web.mit.edu/seely/www --------------------------------------------------------------------
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