Re: [BLAST_ANAWARE] tensor asymmetry

From: Hauke Kolster (hauke@lns.mit.edu)
Date: Sat Mar 20 2004 - 23:22:12 EST


Vitaly,
There are more systematic studies you can do by calculating
the other two asymmetries A^d_V and A^ed_T in QE. These are the
vector asymmetry for the unpolarized beam and the tensor
asymmetry for the polarized beam. Both are expected to be
close to zero for the phi=0 scattered particles and should
vanish in the scattering plane. All four asymmetries, vector
and tensor with polarized and unpolarized beam are calculated
from the same six spin states. If there is a normalization
problem as you indicated by the 'inefficient MFT theory' it
should show up as a non zero asymmetry A^ed_T or A^d_V. For a
calculation of these asymmetries see Aarons summary or my draft
about the three state mode on the ABS web page:
http://blast.lns.mit.edu/targets/abs_web/documents/BlastDTarget.ps
When I tried to use my copy of your show_deep_asym.C macro this
week it coulnd't read charge files so I have not been able to
produce results from the latest data myself. The week before
everything looked reasonably close to zero but I was missing
statistics then. It would be interesting to see all four
asymmetries with your cuts.

Hauke

On Thursday, March 18, 2004, at 08:08 PM, vitaliy ziskin wrote:

> People, more food for thought
>
> I find it an increadible coincidence that tensor asymmetry on the
> quasielastic has the same pathology as elastic only the sectors are
> reversed. After hearing that Genya sees the asymmetry difference
> reverse itself with the target reversal I found out from him that for
> a normal target orientation quasielastic and elastic tensor
> asymmetries have opposite sign. That is, electron-left tensor
> asymmetry is positive in elastic and negative in quasielastic and vice
> versa. This now seems to make sence. It seems that this difference
> follows the sign of the asymmetry in each sector (verified by Genya
> today for reversed target). On the hunch I decided to look at tensor
> asymmetry in a quasielastic channel but in the region of extrimely low
> missing momentum where the tensor asymmetry is expected to be zero (or
> very small). And in fact it is very small < 1% in all Q^2 bins below
> 0.4 GeV^2/c^2. So, this to some degree eliminates a "false asymmetry
> between tensor plus and tensor minus due to MFT efficiency" theory.
> However, I still think that left-right differences in tensor asymmetry
> in elastic and quasielestic channels are NOT unrelated. It could
> still be the ABS problem, but now, less likely.
>
> Cheers, Vitaliy
>



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