[b1_ana] Follow up on Re: Fwd: Re: A_PV with Deuteron Target

Elena Long ellie at jlab.org
Fri May 24 10:14:11 EDT 2013


Good morning,

I've put together a note about these other asymmetries which can be 
found on the wiki under Tech Notes, and also at this address: 
http://nuclear.unh.edu/~elong/analysis_files/2013-05-24/b1_extra_asyms.pdf 
<http://nuclear.unh.edu/%7Eelong/analysis_files/2013-05-24/b1_extra_asyms.pdf>

Please let me know if there are any corrections that need to be made.

Thank you,
Ellie

Elena Long, Ph.D.
Post Doctoral Research Associate
University of New Hampshire
elena.long at unh.edu
ellie at jlab.org
http://nuclear.unh.edu/~elong
(603) 862-1962

On 05/23/2013 03:18 PM, O. A. Rondon wrote:
> Hi Ellie,
>
> The additional asymmetry is eq. (28) of Arenhoevel, A_T^{ed}. It
> involves P_b*P_zz, not P_z, see eq. (23). In his notation
>
> A_V^d    = doesn't appear in Jaffe or HERMES
> A_T^d    = A_zz,
> A_V^{ed} = A_1, (see below about A2)
> A_T^{ed} = doesn't appear in Jaffe or HERMES
>
> I've also added A2, since we are measuring A_parallel = D*(A1 +
> \eta*A2), where \eta = \epsilon*sqrt(Q^2)/(E - E'*\epsilon). \epsilon is
> the usual longitudinal virtual photon polarization.
>
> A1 and A2 are exactly what Arenhoevel has in eq. (27), in terms of F'_T
> and F'_{LT}. We know
> A1 = (sigma_T^(1/2) - sigma_T^(3/2))/sigma_T and
> A2 = sigma_{LT}/sigma_T.
>
> F'_{LT} is not the same as the F_{LT} of eq. (25).
>
> Note that HERMES eq. (1) uses \simeq because, although they measured
> A_para like we propose, they neglected A2. The \eta factor must be
> favorable at 27 GeV.
>
> Also, the fact that A1 and A2 are just other names for F'_T, F'_{LT}
> shows that it's perfectly all right to measure A_zz with longitudinal
> field, although I don't know which of the F's in eq. (26) could be
> interpreted as the A_zz = 2/3 b1/F1.
>
> Npol
> = sigma'_U*[1 + (D*(A1 + \eta*A2)*P_z + A_T^{ed}*P_zz)*P_b + A_v^d*P_z +
> A_zz*P_zz]
>
> and
> Nu   = sigma_U*(1 + APV*P_b),
>
> Note that in Arenhoevel, there is no APV, or any beam-only asymmetry.
>
> Cheers,
>
> Oscar
>
>
>
>
>
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