# CLAS12 Tritium SIDIS Proposal
## For Free Proton and Neutron:
Starting from the DIS and SIDIS Cross Sections:
$$
\begin{align}
\frac{d\sigma}{dxdQ^2} &=\frac{2\pi \alpha^2}{(x_B s)^2}\frac{1-y+\frac{y^2}{2}}{y^2}\sum_{q}e^2_qf^q_1(x)\\
\frac{d\sigma^h}{dxdQ^2dz}&=\frac{2\pi \alpha^2}{(x_B s)^2}\frac{1-y+\frac{y^2}{2}}{y^2}\sum_{q}e^2_q[f^q_1(x)\otimes D_q^{h}(z)]
\end{align}
$$
or,
$$
\begin{align}
\frac{d\sigma}{dxdQ^2} &=\frac{4\pi \alpha^2s}{Q^4}(1-y+\frac{y^2}{2})\sum_{q}e^2_qf^q_1(x)\\
\frac{d\sigma^h}{dxdQ^2dz}&=\frac{4\pi \alpha^2s}{Q^4}(1-y+\frac{y^2}{2})\sum_{q}e^2_q[f^q_1(x)\otimes D_q^{h}(z)]
\end{align}
$$
Assuming we match the bins of different targets with different hadrons, normalize the exprimental dependent quantities, e.g. beam charge and target luminosity, then the normalized yield, $N_p^{\pi^+}$ or so, is proportiional to the sum term.

For proton:
$$
\begin{align}
N_p &\propto e_u^2 \cdot [u(x) + \bar{u}(x)] +e_d^2 \cdot [d(x) + \bar{d}(x)] +e_s^2 \cdot [s(x) + \bar{s}(x)] \\
N_p^{\pi^\pm} &\propto e_u^2 \cdot u(x) \cdot D_u^{\pi^\pm}(z) + e_u^2 \cdot \bar{u}(x) \cdot D_\bar{d}^{\pi^{\pm}}(z)
\\ &+ e_d^2 \cdot d(x) \cdot D_d^{\pi^\pm}(z) + e_d^2 \cdot \bar{d}(x) \cdot D_\bar{d}^{\pi^\pm}(z)
\\ &+ e_s^2 \cdot s(x) \cdot D_s^{\pi^\pm}(z) + e_s^2 \cdot \bar{s}(x) \cdot D_\bar{s}^{\pi^\pm}(z)
\end{align}
$$

For neutron, assuming charge asymmetry, simply do $u_n(x)\rightarrow d_p(x)=d(x)$, $d_n(x)\rightarrow u_p(x)=u(x)$, $\bar{u}_n(x)\rightarrow\bar{d}_p(x)=\bar{d}(x)$, $$
\begin{align}
N_n &\propto e_u^2 \cdot [d(x) + \bar{d}(x)] +e_d^2 \cdot [u(x) + \bar{u}(x)] +e_s^2 \cdot [s(x) + \bar{s}(x)] \\
N_n^{\pi^\pm} &\propto e_u^2 \cdot d(x) \cdot D_u^{\pi^\pm}(z) + e_u^2 \cdot \bar{d}(x) \cdot D_\bar{u}^{\pi^{\pm}}(z)
\\ &+ e_d^2 \cdot u(x) \cdot D_d^{\pi^\pm}(z) + e_d^2 \cdot \bar{u}(x) \cdot D_\bar{d}^{\pi^\pm}(z)
\\ &+ e_s^2 \cdot s(x) \cdot D_s^{\pi^\pm}(z) + e_s^2 \cdot \bar{s}(x) \cdot D_\bar{s}^{\pi^\pm}(z)
\end{align}
$$
Here, $e_u = 2/3$, $e_d=-1/3$, and $e_s=-1/3$

People define:
$$
D^{fav} = D_u^{\pi^+}=D_\bar{d}^{\pi^+}= D_d^{\pi^-}=D_\bar{u}^{\pi^-}, \\ D^{unfav} = D_d^{\pi^+}=D_\bar{u}^{\pi^+}= D_u^{\pi^-}=D_\bar{d}^{\pi^-},\\D_s = D_\bar{s}
$$
Hence,
$$
\begin{align}
N_p^{\pi^+} &\propto e_u^2 u(x) D^{fav}(z)+ e_u^2 \bar{u}(x) D^{unf}(z) + e_d^2 d(x) D^{unf}(z)+ e_d^2 \bar{d}(x) D^{fav}(z) + e_s^2 s(x) D^{s}(z)+ e_s^2 \bar{s}(x) D^{s}(z) \\
N_p^{\pi^-} &\propto e_u^2 u(x) D^{unf}(z)+ e_u^2 \bar{u}(x) D^{fav}(z) + e_d^2 d(x) D^{fav}(z)+ e_d^2 \bar{d}(x) D^{unf}(z) + e_s^2 s(x) D^{s}(z)+ e_s^2 \bar{s}(x) D^{s}(z) \\
N_n^{\pi^+} &\propto e_u^2 d(x) D^{fav}(z)+ e_u^2 \bar{d}(x) D^{unf}(z) + e_d^2 u(x) D^{unf}(z)+ e_d^2 \bar{u}(x) D^{fav}(z) + e_s^2 s(x) D^{s}(z)+ e_s^2 \bar{s}(x) D^{s}(z) \\
N_n^{\pi^-} &\propto e_u^2 d(x) D^{unf}(z)+ e_u^2 \bar{d}(x) D^{fav}(z) + e_d^2 u(x) D^{fav}(z)+ e_d^2 \bar{u}(x) D^{unf}(z) + e_s^2 s(x) D^{s}(z)+ e_s^2 \bar{s}(x) D^{s}(z)
\end{align}
$$

When doing the charge sum of $\pi^{\pm}$ yields:
$$
\begin{align}
N_p^{\pi^+} + N_p^{\pi^-} &\propto [e^2_u\cdot (u(x)+ \bar{u}(x)) + e^2_d\cdot (d(x)+ \bar{d}(x))]\cdot[D^{fav} + D^{unf}]+2e^2_s\cdot[s(x)+\bar{s}(x)]D^s(z) \\
N_n^{\pi^+} + N_n^{\pi^-} &\propto [e^2_u\cdot (d(x)+ \bar{d}(x)) + e^2_d\cdot (u(x)+\bar{u}(x))]\cdot[D^{fav}+ D^{unf}]+2e^2_s\cdot[s(x)+\bar{s}(x)]D^s(z)
\end{align}
$$
At x>0.1, the strangeness contribution can be ignored, and the ratio will only contain the PDF part:
$$
R_{p/n}^{\pi^{+}+\pi^{-}} = \frac{N^{\pi^+}_p + N_p^{\pi^{-}}}{N^{\pi^+}_n + N_n^{\pi^{-}}} =\frac{4 [u(x)+ \bar{u}(x)] + [d(x)+ \bar{d}(x)]}{4[d(x)+ \bar{d}(x)] + [u(x)+\bar{u}(x)]}
$$

The multiplicity is defined as:
$$
M^{\pi}_{p,n}(x,z) = \frac{N_{p,n}^{\pi^+} + N_{p,n}^{\pi^-}}{N_{p,n}}
\propto (D^{fav} + D^{unf})
$$

When doing the charge difference, the strange quark terms are gone w/ assumption, so we have:
$$
\begin{align}
N_p^{\pi^+} - N_p^{\pi^-} &\propto [e^2_u\cdot (u(x)- \bar{u}(x)) - e^2_d\cdot (d(x)- \bar{d}(x))]\cdot[D^{fav} - D^{unf}]\\
N_n^{\pi^+} - N_n^{\pi^-} &\propto [e^2_u\cdot (d(x)- \bar{d}(x))- e^2_d\cdot (u(x)-\bar{u}(x))]\cdot[D^{fav}- D^{unf}]
\end{align}
$$

Now do the ratio of charge asymmetries which are our observables for free protons and neutrons:
$$\begin{align}
R^{\pi^{+}-\pi^{-}}_{p/n} & =\frac{N^{\pi^+}_p - N_p^{\pi^{-}}}{N^{\pi^+}_n - N_n^{\pi^{-}}} =\frac{4 [u(x) - \bar{u}(x)]- [ d(x) - \bar{d}(x)] }{4 [d(x) - \bar{d}(x)]-[u(x) - \bar{u}(x)]}
\end{align}
$$
Then, we have the super ratio:
$$ \begin{align}
\frac{d_v(x)}{u_v(x)} &= \frac{d(x)-\bar{d}(x)}{u(x)-\bar{u}(x)} =\frac{e_u^2 + e_d^2\cdot R^{\pi^{+-}}_{pn} }{e_u^2\cdot R^{\pi^{+}-\pi^{-}}_{pn} + e_d^2} = \frac{4+R^{\pi^{+}-\pi^{-}}_{pn}}{1+4R^{\pi^{+}-\pi^{-}}_{pn}}
\end{align}
$$
So this ratio is sensitive to pure valance-like quark flavors.

The last quantity is the charge ratio:
$$
r^{\pi^{+}/\pi^{-}}_{p}=\frac{N_p^{\pi^{+}}}{N_p^{\pi^{-}}}=\frac{4u(x)D^{fav}+4\bar{u}(x)D^{unf}+d(x)D^{unf}+\bar{d}(x)D^{fav}}{4u(x)D^{unf}+4\bar{u}(x)D^{fav}+d(x)D^{fav}+\bar{d}(x)D^{unf}} \\
r^{\pi^{+}/\pi^{-}}_{n}=\frac{N_n^{\pi^{+}}}{N_n^{\pi^{-}}}=\frac{4d(x)D^{fav}+4\bar{d}(x)D^{unf}+u(x)D^{unf}+\bar{u}(x)D^{fav}}{4d(x)D^{unf}+4\bar{d}(x)D^{fav}+u(x)D^{fav}+\bar{u}(x)D^{unf}}
$$
And their super-ratio:

##
## Kaon Detection
For Kaon, the sea-quaks ($K^+=u\bar{s}$, $K^-=\bar{u}s$) start to play role. For a safe assumption, we can exchange the FFs between $d,\bar{d}$ and $s,\bar{s}$:
$$
\begin{align}
N_p^{K^+} &\propto e_u^2 u(x) D^{fav}(z)+ e_u^2 \bar{u}(x) D^{unf}(z) + e_d^2 d(x) D^{s}(z)+ e_d^2 \bar{d}(x) D^{s}(z) + e_s^2 s(x) D^{unf}(z)+ e_s^2 \bar{s}(x) D^{fav}(z) \\
N_p^{K^-} &\propto e_u^2 u(x) D^{unf}(z)+ e_u^2 \bar{u}(x) D^{fav}(z) + e_d^2 d(x) D^{s}(z)+ e_d^2 \bar{d}(x) D^{s}(z) + e_s^2 s(x) D^{fav}(z)+ e_s^2 \bar{s}(x) D^{unf}(z) \\
N_n^{K^+} &\propto e_u^2 d(x) D^{fav}(z)+ e_u^2 \bar{d}(x) D^{unf}(z) + e_d^2 u(x) D^{s}(z)+ e_d^2 \bar{u}(x) D^{s}(z) + e_s^2 s(x) D^{unf}(z)+ e_s^2 \bar{s}(x) D^{fav}(z) \\
N_n^{K^-} &\propto e_u^2 d(x) D^{unf}(z)+ e_u^2 \bar{d}(x) D^{fav}(z) + e_d^2 u(x) D^{s}(z)+ e_d^2 \bar{u}(x) D^{s}(z) + e_s^2 s(x) D^{fav}(z)+ e_s^2 \bar{s}(x) D^{unf}(z)
\end{align}
$$

When doing the charge sum and difference, we have:
$$
\begin{align}
N_p^{K^+} + N_p^{K^-} &\propto [e^2_u\cdot (u(x)+ \bar{u}(x)) + e^2_s\cdot (s(x)+ \bar{s}(x))]\cdot[D^{fav} + D^{unf}]+2e^2_d\cdot[d(x)+\bar{d}(x)]D^s(z) \\
N_n^{K^+} + N_n^{K^-} &\propto [e^2_u\cdot (d(x)+ \bar{d}(x)) + e^2_s\cdot (s(x)+\bar{s}(x))]\cdot[D^{fav}+ D^{unf}]+2e^2_d\cdot[u(x)+\bar{u}(x)]D^s(z)
\end{align}
$$

\begin{align}
N_p^{K^+} - N_p^{K^-} &\propto [e^2_u\cdot [u(x)- \bar{u}(x)] - e^2_s\cdot [s(x)- \bar{s}(x)]]\cdot[D^{fav} - D^{unf}]\\
N_n^{K^+} - N_n^{K^-} &\propto [e^2_u\cdot [d(x)- \bar{d}(x)] - e^2_s\cdot [s(x)- \bar{s}(x)]]\cdot[D^{fav} - D^{unf}]
\end{align}

Now do the ratio of charge asymmetries which are our observables for free protons and neutrons:
\begin{align}
R^{K^{+}-K^{-}}_{pn} & =\frac{N^{K^+}_p - N_p^{K^{-}}}{N^{K^+}_n - N_n^{K^{-}}} =\frac{4u_v(x)-[ s(x) - \bar{s}(x)] }{4d_v(x)- [s(x) - \bar{s}(x)]} = \frac{1-\frac{e^2_s}{e^2_u}R^{S}_{u_v}}{R^{d_v}_{u_v}-\frac{e^2_s}{e^2_u}R^{S}_{u_v}}
\end{align}
where $\frac{e^2_s}{e^2_u}=\frac{1}{4}$, $R_{u_v}^{d_v}=\frac{d-\bar{d}}{u-\bar{u}}$ which can be measured from the $\pi^{\pm}$ data, and $R_{u_v}^{S}=\frac{s-\bar{s}}{u-\bar{u}}$ which is sensitive to the strange-quarks. If $s(x)=\bar{s}(x)$, we should have $R_{u_v}^{S}=0$.

##
## For $^{3}H$ and $^{3}He$:
$^{3}H$ and $^{3}He$ are both A=3 systems, and we can safely use the charge symmetry similar to proton and neutron's, i.e.
$$U^{^{3}He} = U^{^{3}H} = U^{A=3}\simeq 2u(x)+d(x), S^{^{3}He} = S^{^{3}H} = S^{A=3}\simeq 3s(x)$$,
and so on.
$$\begin{align}
R^{\pi^{+}-\pi^{-}}_{A}\mid_{A=3} & =\frac{N^{\pi^+}_{^{3}He} - N_{^{3}He}^{\pi^{-}}}{N^{\pi^+}_{^{3}H} - N_{^{3}H}^{\pi^{-}}} = \frac{4U_v^{A}(x) - D_v^{A}(x)}{4D_v^{A}(x)-U^{A}_v(x)}\mid_{A=3}
\end{align}
$$
Here I have assumed the nuclear Fragmentation funtions, $D^{fav}_A$ and $D^{unfav}_A$, also depdends only on the nuclear number ($A$), so they can still be cancelled in the ratio. Hence, the valance-like d/u ratio:
$$ \begin{align}
\frac{D_v^{A}}{U_v^{A}}\mid_{A=3} =\frac{D^{A}(x)-\bar{D}^{A}(x)}{U^{A}(x)-\bar{U}^{A}(x)} =\frac{4+R^{\pi^{+}-\pi^{-}}_{A}}{1+4R^{\pi^{+}-\pi^{-}}_{A}}\mid_{A=3}
\end{align}
$$
Based on the formular above, we can use the flavor-tagged SIDIS with $^{3}H$ and $^{3}He$ to study the u and d quarks in A=3 nulcear system.

We also can perform the favor tagging of the fragmentation function using these two isotopes:

So we can measure the Fragmentation function in A=3:
$$
M^{\pi}_A = \frac{N_A^{\pi^+} + N_A^{\pi^-}}{N_{A}}
\propto (D_A^{fav} + D_A^{unf}), \\
\frac{M^{\pi}_{^{3}He}}{M^{\pi}_{^{3}H}} = \frac{D_{^{3}He}^{fav} + D_{^{3}He}^{unf}}{D_{^{3}H}^{fav} + D_{^{3}H}^{unf}}
$$
In $^{3}H$ and $^{3}He$:

\begin{align}
R^{^{K^{+}-K^{-}}}_{A} \mid_{A=3}& =\frac{N^{K^+}_{^{3}He} - N_{^{3}He}^{K^{-}}}{N^{K^+}_{^{3}H} - N_{^{3}H}^{K^{-}}} =\frac{4U_v^A(x)-[S^A(x) - \bar{S}^A(x)]}{4D_v^A(x) - [S^A(x) - \bar{S}^A(x)]} \\&= \frac{1-\frac{e^2_s}{e^2_u}R^{S}_{u_v}}{R^{d_v}_{u_v}-\frac{e^2_s}{e^2_u}R^{S}_{u_v}}\mid_{A=3}.
\end{align}

##
## EMC Effect in light nuclei
For DIS and SIDIS yield in a nucleus A:
\begin{align}
\frac{d\sigma_A}{dxdQ^2} &=\frac{4\pi \alpha^2s}{Q^4}(1-y+\frac{y^2}{2})\sum_{q}e^2_qf^{q,A}_1(x)\\
\frac{d\sigma^h_A}{dxdQ^2dz}&=\frac{4\pi \alpha^2s}{Q^4}(1-y+\frac{y^2}{2})\sum_{q}e^2_q[f^{q,A}_{1}(x)\otimes D_{q,A}^{h}(z)]
\end{align}

In a naive model, the SIDIS cross sections of a nucleus is simply the sum of protons and neutrons within the nuclei. e.g., $\sigma(A) = Z\cdot\sigma(p)+N\cdot\sigma(n)$. However, the PDFs in the nucleus should be modified. To distinguish them from the free PDFs, the nuclear PDFs are labeled as, $u_A$, $d_A$, $s_A$, etc. The normalized yields for $\pi^{\pm}$ productions for a proton and a neutron in the nucleus are:
$$
\begin{align}
N_p^{\pi^+} &\propto e_u^2 u_A (x) D^{fav}_A (z)+ e_u^2 \bar{u}_A (x) D^{unf}_A (z) + e_d^2 d_A (x) D^{unf}_A (z)+ e_d^2 \bar{d}_A (x) D^{fav}_A (z) + e_s^2 s_A (x) D^{s}_A (z)+ e_s^2 \bar{s}_A (x) D^{s}_A (z) \\
N_p^{\pi^-} &\propto e_u^2 u_A (x) D^{unf}_A (z)+ e_u^2 \bar{u}_A (x) D^{fav}_A (z) + e_d^2 d_A (x) D^{fav}_A (z)+ e_d^2 \bar{d}_A (x) D^{unf}_A (z) + e_s^2 s_A (x) D^{s}_A (z)+ e_s^2 \bar{s}_A (x) D^{s}_A (z) \\
N_n^{\pi^+} &\propto e_u^2 d_A (x) D^{fav}_A (z)+ e_u^2 \bar{d}_A (x) D^{unf}_A (z) + e_d^2 u_A (x) D^{unf}_A (z)+ e_d^2 \bar{u}_A (x) D^{fav}_A (z) + e_s^2 s_A (x) D^{s}_A (z)+ e_s^2 \bar{s}_A (x) D^{s}_A (z) \\
N_n^{\pi^-} &\propto e_u^2 d_A (x) D^{unf}_A (z)+ e_u^2 \bar{d}_A (x) D^{fav}_A (z) + e_d^2 u_A (x) D^{fav}_A (z)+ e_d^2 \bar{u}_A (x) D^{unf}_A (z) + e_s^2 s_A (x) D^{s}_A (z)+ e_s^2 \bar{s}_A (x) D^{s}_A (z)
\end{align}
$$

The total nomarlized yied for $\pi^{\pm}$ is:
$$
N_A^{\pi^{\pm}} = Z\cdot N_p^{\pi^\pm} + N\cdot N_n^{\pi^\pm}
$$
which gives:
$$
\begin{align}
N_A^{\pi^{+}} \propto & e_u^2D^{fav}(Zu_A+Nd_A)+e^2_uD^{unfav}(Z\bar{u}_A+N\bar{d}_A) \\ +&e_d^2D^{unfav}(Zd_A+Nu_A)+e^2_dD^{fav}(Z\bar{d}_A+N\bar{u}_A) \\ +&e_s^2D^s_A\cdot A (s_A+\bar{s}_A) \\
N_A^{\pi^{-}} \propto & e_u^2D^{unfav}(Zu_A+Nd_A)+e^2_uD^{fav}(Z\bar{u}_A+N\bar{d}_A) \\ +&e_d^2D^{fav}(Zd_A+Nu_A)+e^2_dD^{unfav}(Z\bar{d}_A+N\bar{u}_A) \\ +&e_s^2D^s_A\cdot A (s_A+\bar{s}_A)
\end{align}
$$

The charge differences are:
$$
\begin{align}
N_A^{\pi^+} + N_A^{\pi^-} \propto&
[(u_A+\bar{u}_A)(Ze^2_u + Ne^2_d)+(d_A+\bar{d}_A)(Ze^2_d + Ne^2_u)] \cdot (D_A^{fav}+D_A^{unfav}) \\
N_A^{\pi^+} - N_A^{\pi^-} \propto&
[(u_A-\bar{u}_A)(Ze^2_u - Ne^2_d)-(d_A-\bar{d}_A)(Ze^2_d - Ne^2_u)] \cdot (D_A^{fav}-D_A^{unfav}) \\
\end{align}
$$

Assuming we measure two nuclei, A and Deutron (D), their ratio of charge difference is then given as:
\begin{align}
R_{A/D}^{\pi^{+}+\pi^{-}} (x,z, Q^2)=& \frac{N_A^{\pi^+} + N_A^{\pi^-} }{N_D^{\pi^+} + N_D^{\pi^-} }
= \frac{1}{5}\frac{(4Z+ N)(u_A+\bar{u}_A)+(Z + 4N) (d_A+\bar{d}_A)}{(u_D+\bar{u}_D)+(d_D+\bar{d}_D) } \cdot \frac{D_A^{fav}+D_A^{unfav}}{D_D^{fav}+D_D^{unfav}}
\end{align}

$$
\begin{align}
R_{A/D}^{\pi^{+}-\pi^{-}} (x,z, Q^2)=& \frac{N_A^{\pi^+} - N_A^{\pi^-} }{N_D^{\pi^+} - N_D^{\pi^-} }
= \frac{1}{3}\frac{(4Z - N)u_{v}^A-(Z - 4N)d_v^A}{u_v^D + d_v^D } \cdot \frac{D_A^{fav}-D_A^{unfav}}{D_D^{fav}-D_D^{unfav}}
\end{align}
$$
where
$$
A^{+}_{A/D}(x, Q^2)= \frac{1}{5}\frac{(4Z+ N)(u_A+\bar{u}_A)+(Z + 4N) (d_A+\bar{d}_A)}{(u_D+\bar{u}_D)+(d_D+\bar{d}_D) } $$
and
$$B^{+}_{A/D}(z, Q^2)=\frac{D_A^{fav}+D_A^{unfav}}{D_D^{fav}+D_D^{unfav}}
$$
$$ A^{-}_{A/D}(x, Q^2)= \frac{1}{3}\frac{(4Z - N)\cdot u_{v}^A-(Z - 4N)\cdot d_v^A}{u_v^D + d_v^D }$$,
and
$$ B^{-}_{A/D}(z, Q^2)= \frac{D_A^{fav}-D_A^{unfav}}{D_D^{fav}-D_D^{unfav}}
$$
Generalize into any meson productions:
$$
\begin{align}
R_{A/D}^{h^{+}+h^{-}} (x,z, Q^2)= \frac{N_A^{h^+} + N_A^{h^-} }{N_D^{h^+} + N_D^{h^-} } = A^{h+}(x, p_T)\otimes B^{h+}(z, p_T) \\
R_{A/D}^{h^{+}-h^{-}} (x,z, Q^2)= \frac{N_A^{h^+} - N_A^{h^-} }{N_D^{h^+} - N_D^{h^-} } = A^{h-}(x, p_T)\otimes B^{h-}(z, p_T)
\end{align}
$$
And Combine w/ the multplicity ratio:
$$ B^{h+}_{A/D}(z, Q^2)=\frac{M^{h}_A(z, Q^2)}{M^{h}_D(z, Q^2)}=\frac{(N_{A}^{h{+}}+N_{A}^{h^{-}})/N_{A}}{(N_{D}^{h^{+}}+N_{D}^{h^{-}})/N_{D}}
$$

In A=3 system:

$$A^{+}_{H3/D}(x, Q^2)= \frac{1}{5}\frac{6(u_{H3}+\bar{u}_{H3})+9(d_{H3}+\bar{d}_{H3})}{(u_D+\bar{u}_D)+(d_D+\bar{d}_D) } $$

$$A^{+}_{He3/D}(x, Q^2)= \frac{1}{5}\frac{9(u_{He3}+\bar{u}_{He3})+6(d_{He3}+\bar{d}_{He3})}{(u_D+\bar{u}_D)+(d_D+\bar{d}_D) } $$

$$ A^{-}_{H3/D}(x, Q^2)= \frac{1}{3}\frac{2 u_{v,H3}+7 d_{v,H3}}{u_v^D + d_v^D }$$

$$ A^{-}_{He3/D}(x, Q^2)= \frac{1}{3}\frac{7 u_{v,He3}+2 d_{v,He3}}{u_v^D + d_v^D }$$

## Medium Effect of Strange-Quark in light nuclei
Just like how we construct the pion-production in SIDIS above, we have the Kaon-Production experimental observables:

$$
\begin{align}
N_A^{K^{+}} \propto & e_u^2D^{fav}(Zu_A+Ns_A)+e^2_uD^{unfav}(Z\bar{u}_A+N\bar{s}_A) \\ +&e_s ^2D^{unfav}(Zs_A+Nu_A)+e^2_s D^{fav}(Z\bar{s}_A+N\bar{u}_A) \\ +&e_d^2D^s_A\cdot A (d_A+\bar{d}_A) \\
N_A^{K^{-}} \propto & e_u^2D^{unfav}(Zu_A+Ns_A)+e^2_uD^{fav}(Z\bar{u}_A+N\bar{s}_A) \\ +&e_s^2D^{fav}(Zs_A+Nu_A)+e^2_s D^{unfav}(Z\bar{s}_A+N\bar{u}_A) \\ +&e_d^2D^s_A\cdot A (d_A+\bar{d}_A)
\end{align}
$$