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<div class="moz-cite-prefix">Good morning,<br>
<br>
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:
<meta http-equiv="content-type" content="text/html;
charset=ISO-8859-1">
<a
href="http://nuclear.unh.edu/%7Eelong/analysis_files/2013-05-24/b1_extra_asyms.pdf">http://nuclear.unh.edu/~elong/analysis_files/2013-05-24/b1_extra_asyms.pdf</a><br>
<br>
Please let me know if there are any corrections that need to be
made.<br>
<br>
Thank you,<br>
Ellie<br>
<pre class="moz-signature" cols="72">Elena Long, Ph.D.
Post Doctoral Research Associate
University of New Hampshire
<a class="moz-txt-link-abbreviated" href="mailto:elena.long@unh.edu">elena.long@unh.edu</a>
<a class="moz-txt-link-abbreviated" href="mailto:ellie@jlab.org">ellie@jlab.org</a>
<a class="moz-txt-link-freetext" href="http://nuclear.unh.edu/~elong">http://nuclear.unh.edu/~elong</a>
(603) 862-1962</pre>
On 05/23/2013 03:18 PM, O. A. Rondon wrote:<br>
</div>
<blockquote cite="mid:519E6B84.9040806@virginia.edu" type="cite">
<pre wrap="">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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</pre>
</blockquote>
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