Re: [BLAST_SHIFTS] vector asymmetry with changing target angle

From: vitaliy ziskin (vziskin@mit.edu)
Date: Thu Mar 18 2004 - 11:06:29 EST


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



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