%Format:latex
\documentclass[12pt,epsfig]{article}
%\input phyzzx
\def\today{\ifcase\month\or
  January\or February\or March\or April\or May\or June\or
  July\or August\or September\or October\or November\or December\fi
  \space\number\day, \number\year}
\def\al{$A_1$}
\def\at{$A_2$}
\def\qsq{$Q^2$}
\def\deg{^\circ}
\def\bul{$\bullet$}
\def\apt{$A^p_2$}
\def\gt{$g_2$}
\def\gtp{$g^p_2\ $}
\def\et{{\it et al.}}
\def\gl{$g_1$}
\def\gql{g_1(x,Q^2)}
\def\gqt{g_2(x,Q^2)}
\def\gww{$g_2^{WW}$}
\def\glp{$g^p_1\ $}
\def\gld{$g^d_1\ $}
\def\gln{$g^n_1\ $}
\def\apar{$A_{\parallel}$}
\def\aper{$A_\perp$}
\def\alp{$A_1^p$}
\def\atp{$A_2^p$}
\def\rss{{\it RSS}}

\begin{document}
\begin{flushright}
RSS T.N. \# 2005-01\\ 
\today\\
\end{flushright}

\begin{center}
%\title{Spin Asymmetries \al\ and \at\ and\\ Measured Asymmetries \apar\ and
%\aper}
{\sl DRAFT}\\
{\bf The $^{15}$N correction to the measured asymmetries}\\
\vskip20pt
Mark Jones, TNJAF and Oscar A. Rondon, INPP - UVA
\vskip20pt
\end{center}

\section{\bf Introduction}

The presence of $^{15}$N in the ammonia used in the polarized targets introduces
an unwanted asymmetry because the nitrogen is partly polarized relative to the 
hydrogen or deuterium. To correct the proton or deuteron asymmetry we need to 
estimate or measure the nitrogen contribution to the measured asymmetry. The 
corrections for each target type are discussed below

\section{\bf Proton}
For the proton measurements on ordinary ammonia NH$_3$, the $^{15}$N 
contribution can be determined from our measured asymmetry in the region below 
the $e-p$ elastic peak where no proton asymmetry is expected because scattering 
on free protons is forbidden.

We call $A^M(W)$ the measured counts asymmetry $\varepsilon$ corrected for 
the beam $P_b$ and (proton) target $P_1$ polarizations
\begin{equation}
A^M(W) = \frac {\varepsilon(W)}{P_b P_1}
\label{eq:am}
\end{equation}
$A^M(W)$ for protons is tabulated for $W<$ 1.073 GeV in the output files of the 
analysis. $A^M(W<W_{el})$ depends only on the nitrogen contribution, where 
$W_{el}$ is the lowest value of invariant mass for elastic scattering. $W_{el}$ 
= 0.85 GeV is a value that works for both parallel and perpendicular 
asymmetries.

The expected number of counts from all the target components for each beam 
helicity $L(W)$ or $R(W)$ for every bin in $W$ can be written as
\begin{equation}
L(R) = \Phi (N_{15}\sigma_{15}^{L(R)}+N_1\sigma_1^{L(R)}+\sum N_A\sigma_A)
\end{equation}
where $\Phi$ is a flux factor, $N_A$ are the numbers of scattering nuclei of 
mass $A$, $\sigma_A^{L(R)}(W)$
represent polarized $e-nucleus$ cross sections (elastic or
inelastic for H, quasielastic or inelastic for N). 
All unpolarized nuclei are lumped in the sum. The polarized
cross sections can be written in terms of nucleon polarized and 
unpolarized cross sections %for $\sigma_A(W)$  and asymmetries $A_A(W)$ 
\begin{eqnarray}
\nonumber \sigma_A^{L(R)}(W) 
= &(Z-Z^{pol})\sigma_{p(A)}+(N-N^{pol})\sigma_{n(A)}
+Z^{pol}\sigma_{p(A)}^{L(R)}+N^{pol}\sigma_{n(A)}^{L(R)}\\
\end{eqnarray}
where $Z$, $N$ refer to the number of unpolarized protons and neutrons in the 
nucleus and $Z^{pol}$, $N^{pol}$ represent the polarized ones; $\sigma_{p,n(A)}$ 
are $e-nucleon$ unpolarized cross sections for nuclear species $A$. The 
polarized nucleon cross sections $\sigma_{p,n(A)}^{L(R)}$ can be expressed in 
terms of unpolarized nucleon cross sections $\sigma_{p,n(A)}$ and nucleon 
asymmetries $A_{p,n(A)}(W)$
\begin{eqnarray}
\sigma_{p,n(A)}^{L(R)}(W) = \sigma_{p,n(A)}(1\pm P_bP_AA_{p,n(A)}(W))
\end{eqnarray}
where $P_b$ is the beam polarization and $P_A$ is the polarization of nucleus 
$A$.

For $^{15}$N which has only one polarizable proton, we can simplify this 
notation using $\sigma_{15}^{L(R)} =\sigma_{15}\pm \sigma_{p(15)}P_bP_{15} 
A_{15}$, where  $A_{15}\equiv A_{p(15)}$.$\sigma_{15}$ is just the
unpolarized $e- ^{15}$N cross section. The rates for each helicity are
\begin{equation}
L(R) = \Phi (N_{15}(\sigma_{15}\pm \sigma_{p(15)}P_bP_{15}A_{15})+N_1\sigma_1(1\pm 
P_bP_1A_1)+\sum N_A\sigma_A).
\end{equation}
%where $P_1, P_{15}$ are the hydrogen and nitrogen polarizations. 

The difference over the sum of counts for each helicity is the counts asymmetry 
$\varepsilon(W)$
\begin{equation}
\varepsilon = \frac {L-R}{L+R} = 
\frac{P_b(N_{15}\sigma_{p(15)}A_{15}P_{15}+N_1\sigma_1A_1P_1)} 
{N_{15}\sigma_{15}+N_1\sigma_1+\sum N_A\sigma_A}
\end{equation}

The customary approach is to factor out the product $N_1\sigma_1$ and write 
$\varepsilon$ in terms of the proton dilution factor $f_1$
\begin{eqnarray}
\nonumber \varepsilon =
f_1P_b(\frac{N_{15}\sigma_{p(15)}}{N_1\sigma_1}A_{15}P_{15}+A_1P_1)\\
f_1 = \frac{N_1\sigma_1}{N_{15}\sigma_{15}+N_1\sigma_1+\sum N_A\sigma_A}
\end{eqnarray}
However, this form becomes undefined for $\sigma_1(W<W_{el})=0$. The alternative
solution is not to factor out $N_1\sigma_1$ but define a nitrogen dilution 
factor $f_{15}$, as used by C. Harris~\cite{chris}
\begin{eqnarray}
 \varepsilon(W) =
P_b(f_{15}(W)P_{15} A_{15}(W)+f_1(W)P_1 A_1(W));
\label{eq:chris}\\
%\end{eqnarray}
%with
%\begin{eqnarray}
\nonumber f_i(W) = 
\frac{N_i\sigma_i(W)}{N_{15}\sigma_{15}(W)+N_1\sigma_1(W)+\sum N_A\sigma_A(W)} 
\end{eqnarray}
where $i$= 1 or $p(15)$. 

With ~(\ref{eq:chris}) it becomes possible to correct the 
measured asymmetry for the nitrogen contribution, using the average nitrogen 
asymmetry
$\langle A_{15}(W)\rangle$ measured for $W<W_{el}$, assuming that the asymmetry 
is independent of $W$:
\begin{eqnarray}
\nonumber \varepsilon(W<W_{el}) &= P_b P_{15} f_{15}(W<W_{el}) 
A_{15}(W<W_{el})\\ &= P_b P_1 A^M(W<W_{el})
\end{eqnarray}
in terms of the measured asymmetry defined in~(\ref{eq:am}). The nitrogen 
asymmetry is then
\begin{equation}
\langle A_{15}(W<W_{el})\rangle = \frac{P_1}{P_{15}}\langle 
\frac{A^M(W<W_{el})}{f_{15}(W<W_{el})}\rangle 
\label{eq:nitro}
\end{equation}
and the proton asymmetry for $W\ge W_{el}$ can be obtained from
\begin{eqnarray}
\varepsilon(W) =
P_b(f_{15}(W)P_{15} \langle A_{15}\rangle +f_1(W)P_1A_1(W))
\end{eqnarray}
or
\begin{eqnarray}
\nonumber
A_1(W) =&\displaystyle{\frac{1}{f_1(W)}\Bigl(\frac{\varepsilon(W)} {P_b P_1} - 
f_{15}(W)\frac{P_{15}}{P_1}\langle A_{15}\rangle\Bigr)}\\
=& \displaystyle{\frac{1}{f_1(W)}\Bigl(A^M(W) - 
f_{15}(W)\langle\frac{A^M(W<W_{el})}{f_{15}(W<W_{el})}\rangle\Bigr)}.
\end{eqnarray}

$A_{15}$ can also be estimated from models of $^{15}$N~\cite{oara}. From the 
angular momentum decomposition of the $p_{1/2}$ level that is populated by the 
unpaired proton in the single particle shell model, one expects $A_{15}^{model}(W)= 
P_{p(15)}A_1(W) = -A_1(W)/3$, where $A_1$ is the proton asymmetry and the 
factor $-1/3$ is the effective polarization 
of the unpaired proton in nitrogen $P_{p(15)}$. In order to
solve for $A_1$ using $A_{15}^{model}$ one needs to go back to~(\ref{eq:chris}),

\begin{eqnarray}
\nonumber \varepsilon(W) = &
P_b(-\displaystyle{\frac{1}{3}}f_{15}(W)P_{15} A_{1}(W)+f_1(W)P_1A_1(W))\\
\nonumber = & P_b P_1 f_1(W)
(\displaystyle{-\frac{1}{3}\frac{f_{15}(W)}{f_1(W)}\frac{P_{15}}{P_1}} 
+1)A_1(W)\\ = & P_b P_1 f_1(W) C_N A_1(W)
\end{eqnarray}
and the customary nitrogen correction $C_N$ for DIS is recovered
\begin{eqnarray}
\nonumber C_N = 1 -\frac{1}{3}\frac{f_{15}(W)}{f_1(W)}\frac{P_{15}}{P_1}\\
=1 -\frac{1}{3}\frac{1}{3}\frac{P_{15}}{P_1}g_{EMC}(W)
\label{eq:cn}
\end{eqnarray}
where the last form comes from 
\begin{equation}
\frac{f_{15}(W)}{f_1(W)} = 
\frac{N_{15}}{N_1}\frac{\sigma_{p(15)}}{\sigma_1}=\frac{1}{3} g_{EMC}(W)
\end{equation}
where the ratio of the DIS $e-p$ cross sections on nitrogen and on hydrogen is 
approximated by the EMC effect parameterization $g_{EMC}(W)$ (more correctly 
$g_{EMC}(x)$).

It is important to 
keep track of the opposite relative signs of $P_{15}$ and $P_1$. $P_1$ and 
$P_{15}$ are related by fits to experimental data, for example
\begin{equation}
|P_{15}|\% = -\Bigl( 0.0312\% + 5.831\times 10^{-2}|P_1| + 8.935\times 
10^{-4}|P_1|^2 + 8.685\times 10^{6}|P_1|^3 \Bigr)
\end{equation}
which is based on E143 plus PSI measurements of positive and negative 
enhancements. $P_1$ needs to be in \%.

Applying this correction to the  asymmetries in the resonances, where 
\begin{equation}
A^M(W) = \frac {\varepsilon(W)}{f_1(W) P_t P_b}
\end{equation}
the inelastic proton asymmetry is
\begin{equation}
A_1(W) = \frac {A^M(W)}{C_N}.
\end{equation}
This form actually is valid at any $W\ge W_{el}$, with 
\begin{eqnarray}
C_N = 1 
+\frac{A_{15}(W)}{A_1(W)}\frac{f_{15}(W)}{f_1(W)}\frac{P_{15}}{P_1}.
\label{eq:seventeen}
\end{eqnarray}

If the expected value of $A_1$ is known from other measurements, such as the 
elastic asymmetry $A_1= A^{el}(e-p)$, the absolute size of the nitrogen 
asymmetry can be estimated from $A^{model}_{15}$ to compare with the measured 
quantity
\begin{eqnarray}
A^M(W) =  \frac {\varepsilon(W)}{P_b P_1} = f_1(W)A_1(W) - \frac{1}{3} 
f_{15}(W)\frac{P_{15}}{P_1}A_1(W) 
\end{eqnarray}
For example, for $W<W_{el}$ where only nitrogen contributes 
($f_1=0$),
$f_{15}\simeq 0.09$, $P_{15}/P_1\sim-0.165$, $A_1=0.21$ for parallel data, and 
$A^M(W<W_{el}) \simeq 0.001 $. For perpendicular data, $A_1=-0.103$ and 
$A^M(W<W_{el}) \simeq -.0004 $

The predicted values is not inconsistent with the measured values $A^M(W<W_{el}) \simeq 
0.003\pm0.0024$ for parallel data.  $A^M(W<W_{el}) \simeq 0.006\pm0.003$ for 
perpendicular data is somewhat less consistent, however, a  significantly more precise 
measurment is needed to make a conclusive statement.


\section{Deuteron}

For ND$_3$, we start with the counts asymmetry, including the contributions of 
$^{15}N$, $^{14}N$ and unsubstituted protons, all of which polarize together 
with the deuterium
\begin{eqnarray}
\nonumber \varepsilon=\displaystyle{\frac {L-R}{L+R}} =& 
\displaystyle{\frac{1}{N_{15}\sigma_{15}+N_2\sigma_2+\sum N_A\sigma_A}}
P_b\Bigl(N_2P_2\sigma_2A_2+N_{15}P_{15}\sigma_p^{15}A_p^{15}&\\
&+      N_{14}P_{14}(\sigma_p^{14}A_p^{14}+\sigma_n^{14}A_n^{14})+
N_1P_1\sigma_1A_1\Bigr),&
\end{eqnarray}
where the notation has been changed slightly, $\sigma_p^A\equiv\sigma_{p(A)}$, 
etc., to fit in the margins. Collecting the common terms, we have
\begin{eqnarray}
\nonumber \varepsilon  =&
\displaystyle{
f_2P_b\Bigl(P_2A_2+\frac{N_{15}\sigma_p^{15}}{N_2\sigma_2}P_{15}A_p^{15}}&\\
&\displaystyle{+\frac{N_{14}}{N_2\sigma_2}P_{14}(\sigma_p^{14}A_p^{14}
+\sigma_n^{14}A_n^{14})
+\frac{N_{1}\sigma_{1}}{N_2\sigma_2}P_{1}A_{1}\Bigr)};&\\
\nonumber f_2 =&
\displaystyle{\frac{N_2\sigma_2}{N_{15}\sigma_{15}+N_2\sigma_2+\sum 
N_A\sigma_A}}.
\end{eqnarray}
We need to write $A_{p,n}^{14,15}$ in terms of proton  $A_1$ and neutron $A_n$ 
asymmetries. For $A_p^{15}$ we use the SPS model discussed earlier for NH$_3$:
$A_{15}(W) = -A_{1}(W)/3$. For $A_{p,n}^{14}$ we have contributions from the 
proton and the neutron, both of which can also be described with the SPS
model too. However, there are no data on $A_n$ at RSS kinematics, 
so  we make use of the relation between the deuteron asymmetry and 
the proton and neutron asymmetries
$\sigma_{2}A_{2} =\gamma_2(\sigma_1^{(2)}A_1 + \sigma_n^{(2)}A_n)$ to solve for 
$A_n$
\begin{eqnarray}
\nonumber \sigma_p^{14}A_p^{14}+\sigma_n^{14}A_n^{14}
=& \displaystyle{-\frac{1}{3}\Bigl(\sigma_p^{14}A_1\bigl(1 - 
\frac{\sigma_n^{14}}{\sigma_n^{(2)}}\frac{\sigma_1^{(2)}}{\sigma_p^{14}}\bigr)+
 \frac{\sigma_2A_2}{\gamma_2}\frac{\sigma_n^{14}}{\sigma_n^{(2)}}\Bigr)},
\end{eqnarray}

The factor $\gamma_2= 0.924$ represents the effective polarization of 
the nucleons in the deuterons (not exactly unity due to the 
deuteron D-state, where the nucleons are aligned antiparallel with respect to 
the  deuteron spin~\cite{oara}).

The ratio 
$(\sigma_n^{14}\sigma_1^{(2)})/(\sigma_n^{(2)}\sigma_p^{(14)})$ is almost exactly 
one, so we neglect the term involving $A_1$. Substituting and collecting terms

\begin{eqnarray}
\nonumber \varepsilon =&
\displaystyle{f_2P_b(P_2A_2
-\frac{1}{3}\frac{N_{15}\sigma_p^{15}}{N_2\sigma_2}P _{15}A_{1}
-\frac{1}{3}\frac{N_{14}}{N_2}\frac{1}{\gamma_2}
\frac{\sigma_n^{14}}{\sigma_n^{(2)}}P_{14}A_{2}
+\frac{N_{1}\sigma_{1}}{N_2\sigma_2}P_{1}A_{1})}\\
\nonumber =&
\displaystyle{f_2P_bP_2\Bigl((1 
-\frac{1}{3}\frac{N_{14}}{N_2}\frac{1}{\gamma_2}\frac{P_{14}}{P_2}
\frac{\sigma_n^{14}}{\sigma_n^{(2)}})A_2
-(\frac{1}{3}\frac{N_{15}\sigma_p^{15}}{N_2\sigma_2}\frac{P _{15}}{P_2}
-\frac{N_{1}\sigma_{1}}{N_2\sigma_2}\frac{P_1}{P_2})A_{1}\Bigr)}
\\
\label{eq:twosix}
\end{eqnarray}

Before proceeding to the full solution of eq.(~\ref{eq:twosix}) for $A_2$, we 
can estimate the magnitude of the corrections involved, by looking at the 
numerical values of the different factors as they apply to $RSS$.
\vskip20pt
\begin{center}
\begin{tabular}{l c c c}
%\centering
\multicolumn{4}{c}{\centering{\bf Table 1. Numeric values of 
factors.}}\\ \hline \hline
Labels & Values & Ratio & Alternative ratio* \\
%&  moment [$\mu_N$] & component & component ($\times1/2$) \\
\hline
$N_1/N_2$ & 1\%/99\% & 0.01  & \\
$N_{14}/N_2$& 2\%/(3*99\%) & 0.0067  & \\
$N_{15}/N_2$& 98\%/(3*99\%)  &.330  & \\
$P_1/P_2$& & 4.4  &  2  \\
$P_{14}/P_2$& & 0.48  & 0.33    \\
$P_{15}/P_2$& & -0.50  & -0.4  \\
\hline
\hline
\end{tabular}
\end{center}
*Polarization ratio from EST; alternative ratio from E143 technical run.
\vskip20pt
The main correction comes from the $^{15}$N contribution. Neglecting the others 
one gets

\begin{eqnarray}
\nonumber \frac{\varepsilon}{f_2P_bP_2} =&
\displaystyle{A_2
-\frac{1}{3}\frac{N_{15}\sigma_p^{15}}{N_2\sigma_2}\frac{P _{15}}{P_2}
A_{1} = A_2 - \frac{1}{3}0.33\times(-0.5) 
\frac{\sigma_p^{15}}{\sigma_2}}A_1\\
\nonumber\frac{\varepsilon(W)}{f_2(W)P_bP_2} =& \displaystyle{A_2(W)\Bigl(1 + 0.055 
\frac{\sigma_p^{15}(W)}{\sigma_2(W)}\frac{A_1(W)}{A_2(W)}\Bigr)}\\
\simeq& A_2\bigl(1.055\bigr)
%\label{eq:twosix} 
\end{eqnarray}
taking the proton to deuteron resonances cross section ratio $\sim 0.5$, and the
$RSS$ measured ratio of raw proton to deuteron asymmetries $\sim 2$. So the 
$^{15}$N correction represents a $\sim 6\%$ relative reduction to the measured 
deuteron asymmetry.

Collecting the coefficients of $A_1$ and $A_2$ into
\begin{eqnarray}
\nonumber C_1(W) = 
\frac{1}{3}\frac{N_{15}\sigma_p^{15}(W)}{N_2\sigma_2(W)}\frac{P
_{15}}{P_2} -\frac{N_{1}\sigma_{1}(W)}{N_2\sigma_2(W)}\frac{P_1}{P_2}\\
C_2(W) = 1 
-\frac{1}{3}\frac{N_{14}}{N_2}\frac{1}{\gamma_2}\frac{P_{14}}{P_2}
\frac{\sigma_n^{14}(W)}{\sigma_n^{(2)}(W)}
\end{eqnarray}
the deuteron asymmetry corrected for the contributions of other polarized nuclei
is
\begin{eqnarray}
A_2(W) = 
\frac{1}{C_2(W)}\Bigl(\frac{\varepsilon(W)}{f_2(W)P_bP_2}+C_1(W)A_1(W)\Bigl)
\end{eqnarray}

Ignoring for the time being the $W$ dependence of $C_{1,2}$, we can estimate the
relative size of these corrections:
\begin{eqnarray}
\nonumber C_1\sim 
\frac{1}{3}\times 0.330\times 0.5\times(-0.5) 
- 0.01\times 0.5\times 4.4 = -0.050\\ 
C_2 = 1 
-\frac{1}{3}\times 0.007\frac{1}{0.924}0.48\times 1 = 1 - 0.001
\end{eqnarray}
from which we conclude that the most important correction is $C_1$; $C_2$ is 
entirely negligible.

\bibliographystyle{phaip}

\begin{thebibliography}{9}

\bibitem{chris}
C. Harris, Thesis, unpublished.
\bibitem{oara}
O. Rondon, Phys. Rev. C {\bf 60}, 035201 (1999).

 \end{thebibliography}


 \end{document}

\+\hskip15mm where\hskip6mm&$\eta_{15} = {\hbox{\rm \# of\/ \rm$^{14}$N}
         \over{\hbox{\rm \# of\/ \rm$^{14}$N + \rm \# of \rm$^{15}$N }}} $\cr
\small
\+&${\hbox{$P_1$}\over\hbox{$P_2$}}={\hbox{polarization of the hydrogen}\over
                                  \hbox{polarization of the deuteron}}$ \cr
\small
\+&${\hbox{$P_{15}$}\over\hbox{$P_2$}}$&$={\hbox{polarization of }\hbox{$^{15}$N}\o
   ver
                                  \hbox{polarization of the deuteron}}=$\cr
\small
\+&&$=-{\hbox{polarization of }\hbox{$^{14}$N}\over
                                  \hbox{polarization of the deuterons}} < 0$ \cr
\small\small
Note the minus sign in the last expression.  The polarization of $^{14}$N is par
   allel
to the one of the deuterium, the polarization of $^{15}$N, however, is however
antiparallel to the deuterium polarization.
\small
$g_{_{\rm EMC}}(x)$ is again taken at atomic number~15.
 
\huge


we use a similar expression as eq.(~\ref{eq:chris}) for the dilution 
factor 
\begin{equation}
f_2(W)={N_2\sigma_2(W)\over N_2\sigma_2(W)+N_{15}\sigma_{15}(W)+\sum 
N_A\sigma_A(W)} \end{equation}

\begin{eqnarray}
\nonumber \sigma_{14}(W)A_{14}(W) =& -\frac{1}{3}\bigl(\sigma_{1}(W)A_{1}(W) + 
\sigma_{n}(W)A_{n}(W)\bigr)\\
 =& \displaystyle{-\frac{1}{3}(\sigma_1(W)A_1(W)\bigl(1 - 
\frac{\sigma_n^{14N}}{\sigma_n^{2H}}\bigr)+
 \frac{\sigma_2(W)A_2(W)}{\gamma_2}\frac{\sigma_n^{14N}}{\sigma_n^{2H}}\bigr)},
\end{eqnarray}





keeping in mind that the dilution factor is a ratio of radiated rates.

We obtain with similar definitions as above:
\begin{eqnarray}
A_2(W)=\frac{1}{C_N(W)}\frac{\varepsilon(W)}{f_2(W)P_bP_2}%\\
%A_2(W)&=\displaystyle{{1\over1-\eta_1+C_n/\gamma} \left[{\varepsilon(W)\over
%f_2(W)P_bP_2}+(C_n-C_p)A_1(W)\right] + R_2(W)}%\\
%A_2(W)&=\displaystyle{{1\over1-\eta_1+C_n/\gamma} \left[{\varepsilon(W)\over
%f_2(W)P_bP_2}+{F_2^p\over F_2^d}(C_n-C_p)(A_1(W)-R_1)\right] + R_2(W)}%\\
%  A_n&={1\over \gamma(1-\eta_1)+C_n}{F_2^d\over F_2^n}
%    \left[{\varepsilon\over fP_bP_2}
%    + R_2 - \left(\gamma(1-\eta_1)+C_p\right){F_2^p\over F_2^d}A_1\right]
\end{eqnarray}
$\varepsilon$ is the deuterium count asymmetry, $A_1$ is taken from the NH$_3$
measurements, $R_2$ is the deuterium radiative correction.
%$F_2^p$ and $F_2^d$ are the structure functions radiated to the spectrometers.

The factor $\gamma\approx0.924$ comes from the the deuteron D-state, where the 
nucleons are aligned antiparallel with respect to the normal deuteron
polarization~\cite{oara}.

The correction factors $\eta_1$, $C_n$ and $C_p$ are necessary to correct
for the polarization of $^{15}$N and of the unsubstituted $^{14}$N and 
hydrogen in the target. 
\begin{eqnarray}
\nonumber \eta_1 &= \displaystyle{\frac {\rm unsubstituted~ 
protons}{\rm deuterons + unsubstituted~ protons}}=0.01\\
\end{eqnarray}
$C_p$ comes from the asymmetry of the unpaired 
proton in $^{15}$N, just as in the case of the proton (see the second term in 
eq.(~\ref{eq:twelve}) or (~\ref{eq:seventeen})). 
\begin{equation}
C_p = \frac{A_{15}(W)}{A_2(W)}\frac{f_{15}(W)}{f_2(W)}\frac{P_{15}}{P_2}.
\end{equation}
However, in addition to the 
proton in nitrogen, there are unsubstituted free protons (i.e. ordinary ammonia) 
in ND$_3$ that polarize and contribute to the asymmetry; $C_p$ combines 
both corrections.
\begin{eqnarray}
\nonumber 
C_p&=\eta_1{P_1\over P_2}+(2\eta_{15}-1){P_{15}\over P_2}
                                 {g_{_{\rm EMC}}(x)\over9}\\
  C_n-C_p&={P_{15}\over P_2}{g_{_{\rm EMC}}(x)\over9}(1-\eta_{15})-
                                        \eta_1{P_1\over P_2}\cr
\end{eqnarray}


where$\eta_{15} = {\hbox{\rm \# of\/ \rm$^{14}$N}
         \over{\hbox{\rm \# of\/ \rm$^{14}$N + \rm \# of \rm$^{15}$N }}}$

${\hbox{$P_1$}\over\hbox{$P_2$}}={\hbox{polarization of the hydrogen}\over
                                  \hbox{polarization of the deuteron}}$ 

${\hbox{$P_{15}$}\over\hbox{$P_2$}}$ $={\hbox{polarization of }\hbox{$^{15}$N}\over       
\hbox{polarization of the deuteron}}=$
$=-{\hbox{polarization of }\hbox{$^{14}$N}\over
                                  \hbox{polarization of the deuterons}} < 0$

\begin{eqnarray}
C_n&=\displaystyle{\eta_{15}\frac{P_{15}}{P_2}\frac {f_{15}(W)}{f_2(W)}}\\
\end{eqnarray}

Note the minus sign in the last expression.  The polarization of $^{14}$N is par
   allel
to the one of the deuterium, the polarization of $^{15}$N, however, is however
antiparallel to the deuterium polarization.
\small
$g_{_{\rm EMC}}(x)$ is again taken at atomic number~15.
 
%\huge
{\bf Estimating the magnitude of the corrections for $\bf ^{15}$ND$\bf _3$:}
%small
The contamination of $^{15}$N with $^{14}$N is about~2\%
($\eta_{15}\approx0.02$), and the contamination of deuterium with hydrogen
is about~1.5\% ($\eta_1\approx 0.015$). $P_1/P_2 \approx 1.5$,
$P_{15}/P_2 \approx -0.4$, $g_{_{\rm EMC}}(x)\approx1$, $F_2^p/F_2^d\approx0.5$.
%\tiny
This leads to $C_p \approx 0.07$, $C_n \approx -0.0009$, to
$$ {1\over1-\eta_1+C_n/\gamma} \approx 1.02 \qquad
   {F_2^p/F_2^d}(C_n-C_p) \approx -0.03$$
and finally to
$$A_2\approx 1.02 \left[{\varepsilon\over fP_bP_2}-0.03\, (A_1-R_1)\right]+R_2.$$

\subsection{$^{15}$N model extension}
The model can be readily extended to include mixing of the 
$p_{3/2}$ state (only the $M=\pm 1/2$ substates are allowed, to preserve the 
total nuclear
 $^{15}$N spin $I = (1/2)^-$). The $M=\pm 1/2$ angular momentum 
substates are (notation C-G$|m_l,m_s;M\rangle$)
\begin{eqnarray}
\nonumber\displaystyle
\sqrt{1\over3} | 1, -1/2; ~1/2 \rangle &+& \sqrt{2\over3} | 0, ~1/2; ~~1/2 
\rangle\\
\sqrt{1\over3} | -1, 1/2; -1/2 \rangle &+& \sqrt{2\over3} | 0, -1/2; -1/2 
\rangle \end{eqnarray}
 It is easily seen
 that 2/3 of the time the proton and N have parallel spins ($m_s=+M$,) and 1/3 
anti parallel, for a net 1/3
 of the time having parallel spins in the $M = \pm 1/2$ substates. 

The net normalized number of aligned protons in $^{15}$N 
can be written in terms of the contributions of the two states $p_{1/2}$ and 
$p_{3/2}$ 
\begin{eqnarray}
\frac{N_{\parallel}}{N} = 
-\frac{1}{3}P_{p(1/2)}+\frac{1}{2}\frac{1}{3}P_{p(3/2)} 
\end{eqnarray}
where $P_i$  represents the probability of the proton being in each state 
($\sum P_i = 1$,) and the extra 1/2 reflects the restriction to $M=\pm1/2$ 
substates.

Using~(\ref{eq:nitro})
%\begin{eqnarray}
%A_{15} = \frac{A^M(W<W_{el})} {f_{15}(W<W_{el})}\frac{P_{1}}{P_{15}}
%\end{eqnarray}
and substituting the measured values, one gets $A_{15} =-0.026$ for parallel 
data and $A_{15}= 0.086$ for perpendicular data. Since the substate 
probabilities $P_i$ add to unity, there is only one unknown, 
\begin{eqnarray}
\frac{N_{\parallel}}{N} = 
-\frac{1}{3}+\frac{1}{2}P_{p(3/2)} = \frac{A_{15}}{A_1}
\end{eqnarray}
where $A_1$ is the elastic asymmetry for free protons. $P_{3/2} = 0.42$ for 
parallel data but it is negative (-0.91) for perpendicular, indicating that the 
ground state of nitrogen may be more complicated than the model.
