\chapter{Chargino production}

\section{Lagrangian representation}

In this chapter I consider the production cross section
for the lightest charginos by the process
\bel{charginoProdProc}
e^+e^- \rightarrow \chi^+\chi^- ,
\ee

where I denote by $\chi$ the lightest chargino $\chi_1$. Also I will refer
to $\chi^+$ as the chargino particle and to $\chi^-$ as
the corresponding antiparticle.

\P
At the tree level, the process is represented by the following
interaction terms from the MSSM Lagrangian,

\bel{eToChi}
L_{int}
  = L[e, \gamma] + L[e, Z]
  + L[\chi, \gamma] + L[\chi, Z]
  + L[e, \tilde{\nu_{e1}}, \chi]
.
\ee

Here $L[e, \gamma]$ and $L[e, Z]$  represent the usual coupling
of electrons with photons and $Z^0$ bosons and the remaining terms are given
by~\rf{charginoCoupling} and~\rf{ChiLSNu}.

\P
To simplify the cross-section calculations I found it useful to introduce
the following notations,

\bem
L[e, \gamma] & = & -\overline{\psi}_e e\gu{\rho} \psi_e A_\rho
\nel
&\equiv&
\overline{\psi}_e \gu{\rho}(a_\gamma - b_\gamma \gd5 ) \psi_e D^\gamma_\rho,
\nel
L[e, Z] &=& - \frac{g}{4\cos\theta_W}
  \overline{\psi}_e \gu{\rho} [(1 - 4\sin^2\theta_W) - \gd5] \psi_e Z_\rho
\nel
& \equiv &
\overline{\psi}_e \gu{\rho}(a_Z - b_Z \gd5 ) \psi_e D^Z_\rho,

\ee

with
\bel{chargeGabIs}
a_\gamma = -e \equiv - \abs{e}, \quad b_\gamma = 0,
\ee

\beml{chargeZabIs}
a_Z  =  -\frac{g}{4\cos\theta_W}(1 - 4\sin^2\theta_W),
\quad
b_Z  =  -\frac{g}{4\cos\theta_W},
\ee

and
\be
D^\gamma_\rho =  A_\rho, \quad D^Z_\rho =  Z_\rho
.
\ee

According to~\rf{charginoCoupling} I can write
in the case of the  coupling of the lightest chargino $\chi_1$
with the gauge bosons
\beml{charge-L-with-chi-and-gamma}
L[\chi, \gamma] &=&
\overline{\psi}_{\chi_1} e\gu{\rho} \psi_{\chi_1} A_\rho
\nel
& \equiv &
\overline{\psi}_{\chi}\gu{\rho}(A_\gamma - B_\gamma \gu5)\psi_{\chi}
   D^\gamma_\rho,
\nel
L[\chi,Z] &=&
    \frac{g}{4 \cos \theta_W}
    \overline{\psi}_{\chi_1} \gu{\rho}
\nel && \times
    \Bigl\lbrace
          [2\cos (2\theta_W)
            + \abs{Y_{11}}^2 + \abs{X_{11}}^2]

          + \gu5 [\abs{Y_{11}}^2 - \abs{X_{11}}^2]\Bigr\rbrace

    \psi_{\chi_1} Z_\rho
\nel
& \equiv &
  \overline{\psi}_{\chi}\gu{\rho}(A_Z - B_Z \gu5)\psi_{\chi}
      D^Z_\rho,
\ee

where

\bel{chargeGABIs}
A_\gamma  = e, \quad B_\gamma =   0,
\ee

and
\beml{chargeZABIs}
A_Z &=& \frac{g}{4 \cos \theta_W}
             [2\cos (2\theta_W) + \abs{Y_{11}}^2 + \abs{X_{11}}^2],
\nel
B_Z &=& - \frac{g}{4 \cos \theta_W} [\abs{Y_{11}}^2 - \abs{X_{11}}^2].
\ee

The interaction term between electrons and positrons, charginos and
scalar electron neutrinos is given by~\rf{lagranChiLSNu}
(I write in the following $\tilde\nu^L_e$ simply as $\tilde\nu$)

\bem
L[e, \tilde{\nu}_{e1}, \chi] & = &
  \frac{g}{2}
    [-\psi^T_{\chi_1}C (\Vc^\chi_{e1L} + \Ac^\chi_{e1L}\gd{5}) \psi_{e}
                    \tilde{\nu}^\hc
       + \overline{\psi}_e (\Vc^{\chi\hc}_{e1L} - \Ac^{\chi\hc}_{e1L} \gd{5})
                 C^{-1}\overline{\psi}_{\chi_1}^T \tilde{\nu}]
\nel
& \equiv &
        -\psi^T_{\chi}C (F + G\gd{5}) \psi_e
                    \cc{\tilde{\nu}}
       + \overline{\psi}_e (F^\hc - G^\hc \gd{5})
                 C^{-1}\overline{\psi}_{\chi}^T \tilde{\nu}.
\ee

Here
\be
F = \frac{g}{2}\Vc^\chi_{e1L},
\quad
G  =  \frac{g}{2}\Ac^\chi_{e1L}
.
\ee

Under the approximation
\be
\abs{Y_{i2}\frac{m_e}{\sqrt{2} m_W \cos\beta}}
\le \frac{m_e}{\sqrt{2} m_W \cos\beta} \approx 0
\ee

($\abs{Y_{i2}} \le 1$ as an element of the unitary matrix $Y$),
the coefficients $F$ and $G$ become
\bel{chargeFandG1}
F \approx -{g \over 2} X_{11}^\hc,
\quad
G  \approx {g \over 2} X_{11}^\hc
\ee

\P
The expressions for
$X_{11}^\hc$, $\abs{X_{11}}^2$ and $\abs{Y_{11}}^2$ are given by~\rf{lagXandY},
\bel{chargeFandG2}
X_{11}^\hc = \cos\theta_X,
\quad
\abs{X_{11}}^2 = \cos^2\theta_X,
\quad
\abs{Y_{11}}^2 = \cos^2\theta_Y
,
\ee

\P
With this notations $L_{int}$ becomes

\bel{charge-useful-form-for-L}
L_{int} = \sum_{k = \gamma, Z} (L_e^k + L_\chi^k)
            + L_{\tilde\nu} + L_{\tilde\nu}^\hc,
\ee

where I set
\beml{charge-useful-form-for-L-2}
L_e^k & = &
\overline{\psi}_{e}\gu{\rho}(a_k - b_k \gu5)\psi_{e}D^k_\rho
,
\nel
L_\chi^k & = &
\overline{\psi}_{\chi}\gu{\rho}(A_k - B_k \gu5)\psi_{\chi}D^k_\rho
,
\nel
L_{\tilde\nu} & = &  \overline{\psi}_e (F^\hc - G^\hc \gd{5})
                 C^{-1}\overline{\psi}_{\chi}^T \tilde{\nu}
,
\nel
L_{\tilde\nu}^\hc & = & -\psi^T_{\chi}C (F + G\gd{5}) \psi_e \tilde{\nu}^\hc
.
\ee

\section{Production Amplitude}

According to~\cite{SMTextBook} the tree-level transition amplitude
for the process \rf{charginoProdProc}
is given by the second-order $S$-matrix element,

\bem
\lefteqn{
S_2(e^+e^- \rightarrow \chi^+\chi^-) =
\frac{(-i)^2}{2!} \int d^4\!x\, d^4\!x'
}
\nel &&{} \times
\bra{\chi^+(\ppa, \rpa), \chi^-(\ppb, \rpb)}
  \Torder\lbrace
    \Norder[L_{int}(x)] \Norder[L_{int}(x')]\rbrace
\ket{e^+(\pa, \ra), e^-(\pb, \rb)}
.
\ee

Here $\pa$, $\ra$, $\pb$, $\rb$ denote the initial momentum and polarization
state for the positron and electron fields,
$\ppa$, $\rpa$, $\ppb$, $\rpb$ are used
for the final states of antichargino and chargino.
The functional $\Torder$ gives
the time ordered product of its arguments
and $\Norder[L_{int}]$ stands for the normal-ordered form
of the interaction Lagrangian.

\P

I must note here that there is an ambiguity in the definition of
a multi-fermion state.
For example, for any two fermions $f_1$, $f_2$
one can take either
\bel{charge-multi-fermion-state-is}
\ket{f_1, f_2}
    = (2\pi)^{3\over 2} c^\hc(f_1) (2\pi)^{3\over 2} c^\hc(f_2)\ket{0}
    = (2\pi)^3 c^\hc(f_1) c^\hc(f_2)\ket{0}

\ee

or
\be
\ket{f_1, f_2}
    = (2\pi)^{3\over 2} c^\hc(f_2) (2\pi)^{3\over 2} c^\hc(f_1)\ket{0}
    = -(2\pi)^3 c^\hc(f_1)c^\hc(f_2)\ket{0}.
\ee
where the factor $(2\pi)^{3\over 2}$ reflects the unit volume normalization of
the states.

\P
This causes an uncertainty in the sign of the S matrix for processes
with more than one fermion in the initial or final state
although measurable quantities do not have this ambiguity.
Nevertheless, from the computational point of view the sign should be fixed
somehow. Thus I choose the first possibility~\rf{charge-multi-fermion-state-is} and

\bem
\bra{f_1, f_2} & \equiv & \left(\ket{f_1, f_2}\right)^\hc
    = \left((2\pi)^3 c^\hc(f_1)c^\hc(f_2)\ket{0}\right)^\hc
\nel & = &
    (2\pi)^3 \bra{0}c(f_2)c(f_1)
\ee

\P
I may write now for the initial and final states of
the process~\rf{charginoProdProc},
\bem
\bra{\chi^+(\ppa, \rpa), \chi^-(\ppb, \rpb)}
&=&
(2\pi)^3 \bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa),
\nel
\ket{e^+(\pa, \ra), e^-(\pb, \rb)}
&= &
(2\pi)^3 d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0},
\ee

where $c$ and $d$ are fermion and antifermion annihilation operators
and $c^\hc$ and $d^\hc$ are creation operators.

\P
The explicit form~\rf{charge-useful-form-for-L} of $L_{int}$
and the commutational and anticommutational properties of different fields
it depends on give

\bem
\lefteqn{
S_2 = \frac{(-i)^2}{2!} \int d^4\!x\, d^4\!x'
    \bra{\chi^+(\ppa, \rpa), \chi^-(\ppb, \rpb)}
}
\nel
&& {} \times
\Torder\Bigl\lbrace
  \sum_{k = \gamma, Z} \bigl(
    \Norder[L_e^k(x)]\Norder[L_\chi^k(x)(x')]
    +  \Norder[L_\chi^k(x)(x)]\Norder[L_e^k(x')] \bigr)
\nel && \qquad \qquad {}
+ \Norder[L_{\tilde\nu}(x)]\Norder[L_{\tilde\nu}^\hc(x')]
+ \Norder[L_{\tilde\nu}^\hc(x)]\Norder[L_{\tilde\nu}(x')]
 \Bigr\rbrace
\nel && \times
\ket{e^+(\pa, \ra), e^-(\pb, \rb)}

\nel
& = & -\int d^4\!x\, d^4\!x'
\bra{\chi^+(\ppa, \rpa), \chi^-(\ppb, \rpb)}
\nel && {} \times
  \Torder\Bigl\{
  \sum_{k = \gamma, Z} \Norder[L_e^k(x)]\Norder[L_\chi^k(x)(x')]
    + \Norder[L_{\tilde\nu}^\hc(x)]\Norder[L_{\tilde\nu}(x')]
 \Bigr\}
\nel && \times
\ket{e^+(\pa, \ra), e^-(\pb, \rb)}
,
\ee

where I have taken into account that
the terms like $\Norder[L(x)]$ here contains an even number (two)
of fermion fields so I can freely move them to any position
in the $\Torder$-product. Thus if I simply rename $x \leftrightarrow x'$
and change the integration order, I find, for example,
\bem
\int\! d^4\!x\, d^4\!x'
  \Torder\Bigl\lbrace
     \Norder[L_\chi^k(x)(x)]\Norder[L_e^k(x')]
\Bigr\rbrace
& = &
\int\! d^4\!x\, d^4\!x'
  \Torder\Bigl\lbrace
     \Norder[L_e^k(x')]\Norder[L_\chi^k(x)(x)]
\Bigr\rbrace
\nel
& = &
\int\! d^4\!x'\, d^4\!x
  \Torder\Bigl\lbrace
     \Norder[L_e^k(x)]\Norder[L_\chi^k(x')(x)]
 \Bigr\rbrace
\nel
& = &
\int\! d^4\!x\, d^4\!x'
  \Torder\Bigl\lbrace
     \Norder[L_e^k(x)]\Norder[L_\chi^k(x')(x)]
 \Bigr\rbrace
,
\ee

and in the same way
\bem
\int d^4\!x\, d^4\!x'
    \Norder[L_{\tilde\nu}(x)]\Norder[L_{\tilde\nu}^\hc(x')]
& = &
\int d^4\!x\, d^4\!x'
    \Norder[L_{\tilde\nu}^\hc(x)]\Norder[L_{\tilde\nu}(x')]
.
\ee

At this stage
I can represent $S_2$ according to~\rf{charge-useful-form-for-L-2}
and~\rf{charge-multi-fermion-state-is} as,

\bel{charge-S2-as-two-term}
S_2 = \sum_{k = \gamma, Z} S_k + S_{\tilde\nu},
\ee

with
\beml{charge-S-D-k-is}
S_k & = & -(2\pi)^6 \int d^4\!x\, d^4\!x'
\bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel && \times
    \Torder\Bigl\lbrace
        \Norder[\overline\psi_{\chi}
                \gu{\rho}(A_k - B_k\gd{5})
                \psi_{\chi}
                D_\rho]_{x}
        \Norder[\overline\psi_e
                \gu{\sigma}(a_k - b_k\gd{5})
                \psi_e
                D_\sigma]_{x'}
    \Bigr\rbrace
\nel && \times
    d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0}
,
\ee

and
\beml{charge-S-nu-is}
S_{\tilde\nu} & = & (2\pi)^6 \int d^4\!x\, d^4\!x'
\bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel && \times
    \Torder\Bigl\lbrace
        \Norder[\psi^T_{\chi}C(F + G\gd{5})
                \psi_{e}\cc{\tilde\nu}]_x
        \Norder[\overline{\psi}_e (F^\hc - G^\hc\gd{5})
                C^{-1}\overline{\psi}_{\chi}^T{\tilde\nu}]_{x'}
    \Bigr\rbrace
\nel && \times
    d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0}
.
\ee

\P
I follow the usual approach and calculate $S_2$ by the means of
Wick's theorem. According to~\rf{gammaFermionField} I have,
\bem
\lefteqn{
\psi(x) \; = \; \posPart\psi(x) + \negPart\psi(x)
}
\nel & = &
\sum_{r = +, -}\int d^3p \left(\frac{m}{(2\pi)^3 E_p}\right)^{\frac{1}{2}}
    [ u_r(p)c_r(p) e^{-ipx} + v_r(p)d_r^\hc(p) e^{ipx} ] ,
\nel
\lefteqn{
\overline{\psi}(x) \; = \;
\overline{\posPart\psi}(x) + \overline{\negPart\psi}(x)
}
\nel & = &
\sum_{r = +, -}\int d^3p \left(\frac{m}{(2\pi)^3 E_p}\right)^{\frac{1}{2}}
    [ \ub_r(p)c_r^\hc(p) e^{ipx} + \vb_r(p)d_r(p) e^{-ipx} ]
    ,
\ee

where $u_r(p) = u_r(p/m)$ and $v_r(p) = v_r(p/m)$.

\P
Thus I obtain for first term $S_k$ in \rf{charge-S2-as-two-term}
(I suppress here the index $k$) :
\beml{charge-ref11}
S_k & = & -(2\pi)^6\int d^4\!x\, d^4\!x'
\bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel &&{} \qquad\times
\Torder\Bigl\lbrace
            \Norder[\overline\psi_{\chi}
                    \gu{\rho}(A - B\gd{5})
                    \psi_{\chi}
                    D_\rho]_{x}
            \Norder[\overline\psi_e
                    \gu{\sigma}(a - b\gd{5})
                    \psi_e
                    D_\sigma]_{x'}
        \Bigr\rbrace
\nel &&{} \qquad\times
    d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0}
\nel & = &
-(2\pi)^6 \int d^4\!x\, d^4\!x'
    \bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel &&{} \qquad\times
        \Torder\Bigl\lbrace
            [\overline{\posPart\psi}_{\chi}
                    \gu{\rho}(A - B\gd{5})
                    \negPart\psi_{\chi}
                    D_\rho]_{x}
            [\overline{\negPart\psi}_e
                    \gu{\sigma}(a - b\gd{5})
                    \posPart\psi_e
                    D_\sigma]_{x'}
        \Bigr\rbrace
\nel &&{} \qquad\times
    d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0}
\nel
& = &
-(2\pi)^6 \int d^4\!x\, d^4\!x'
    \bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel &&{} \qquad\times
        \Norder\Bigl\lbrace
            [\overline{\posPart\psi}_{\chi}
                    \gu{\rho}(A - B\gd{5})
                    \negPart\psi_{\chi}]_{x}
            [\overline{\negPart\psi}_e
                    \gu{\sigma}(a - b\gd{5})
                    \posPart\psi_e]_{x'}
        \Bigr\rbrace
            iD_{\rho\sigma}(x' - x)
\nel &&{} \qquad\times
    d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0},
\ee

where $D_{\rho\sigma}(x' - x)$ stands for $\gamma$ or $Z^0$
propagator. Normal ordering of the fermion operators in~\rf{charge-ref11}
gives

\beml{charge-ref11-a}
S_k
& = &
-(2\pi)^6 \int d^4\!x\, d^4\!x'
    \bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel &&{} \qquad\times
        [\overline{\negPart\psi}_e
                \gu{\sigma}(a - b\gd{5})
                \posPart\psi_e]_{x'}
        [\overline{\posPart\psi}_{\chi}
                \gu{\rho}(A - B\gd{5})
                \negPart\psi_{\chi}]_{x}
        iD_{\rho\sigma}(x' - x)
\nel &&{} \qquad\times
    d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0} .
\ee

From the anticommutational relations between fermion operators
and~\rf{gammaFermionField} I obtain for the scalar product in~\rf{charge-ref13}:
\beml{charge-ref12}
\lefteqn{
\bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
[\overline{\negPart\psi}_e\dots \posPart\psi_e]_{x'} \}
[\overline{\posPart\psi}_{\chi} ... \negPart\psi_{\chi}]_{x}
d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0}
} \;
\nel & = &
-
\left(\frac{m_e^2}{\Ea\Eb}\right)^{\frac{1}{2}}
\bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
[\overline{\posPart\psi}_{\chi} ... \negPart\psi_{\chi}]_{x} \ket{0}
\nel && \qquad \times
[\vb_{\ra}(\pa)e^{-i\pa x'} \dots u_{\rb}(\pb)e^{-i\pb x'}]

\nel & = &
-
\left(\frac{m_e^2 m_\chi^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\dirProd{0}{0}
\nel && \quad \times
[\ub_{\rpb}(\ppb)e^{i\ppb x} \dots v_{\rpa}(\ppa)e^{i\ppa x}]
[\vb_{\ra}(\pa)e^{-i\pa x'} \dots u_{\rb}(\pb)e^{-i\pb x'}]
,
\ee

where $\Ea$ and $\Eb$ are the energies of the incoming particles and
$\Epa$ and $\Epb$ are the energies of the outgoing ones. In the
center-of-mass frame one will have $\Ea = \Eb$ and $\Epa = \Epb$.

\P
Thus from the equations~\rf{charge-ref12} I find

\beml{charge-S2-boson-tmp1}
S_k
& = &
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\int d^4\!x\, d^4\!x'
    \dirProd{0}{0}
\nel & & \qquad \times \,
    [\ub_{\rpb}(\ppb)e^{i\ppb x}
        \gu{\rho}(A - B\gd{5})
         v_{\rpa}(\ppa)e^{i\ppa x}]
\nel & & \qquad \times  \,
    [\vb_{\ra}(\pa)e^{-i\pa x'}
        \gu{\sigma}(a - b\gd{5})
        u_{\rb}(\pb)e^{-i\pb x'}]
    iD_{\rho\sigma}(x' - x)
.
\ee

To calculate the integral in~\rf{charge-S2-boson-tmp1}
I use the Fourier transformation to write
\be
iD_{\rho\sigma}(x' - x) = (2\pi)^{-4}
\int  d^4\!k\, e^{-ik(x' - x)} iD_{\rho\sigma}(k)
.
\ee

It gives for~\rf{charge-S2-boson-tmp1}
\beml{charge-boson-S-is}
S_k & = &
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
(2\pi)^{-4}
\int d^4\!x\, d^4\!x' d^4\!k\,
    e^{ix(\ppa + \ppb)}e^{-ix'(\pa + \pb)}e^{-ik(x' - x)}
    iD_{\rho\sigma}(k)
\nel & & \times\,
    [\ub_{\rpb}(\ppb)\gu{\rho}(A - B\gd{5})v_{\rpa}(\ppa)]
    [\vb_{\ra}(\pa)\gu{\sigma}(a - b\gd{5})u_{\rb}(\pb)]
\nel
& = &
(2\pi)^{4}
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\int d^4\!k\,
    \delta^4(k + \ppa + \ppb)\delta^4(k + \pa + \pb)
    iD_{\rho\sigma}(k)
\nel & & \times\,
    [\ub_{\rpb}(\ppb)\gu{\rho}(A - B\gd{5})v_{\rpa}(\ppa)]
    [\vb_{\ra}(\pa)\gu{\sigma}(a - b\gd{5})u_{\rb}(\pb)]
\nel
& = &
(2\pi)^{4}
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\delta^4(\ppa + \ppb - \pa - \pb)
\left. iD_{\rho\sigma}(k) \right|_{k = - \pa - \pb =  - \ppa - \ppb}
\nel & & \times\,
    [\ub_{\rpb}(\ppb)\gu{\rho}(A - B\gd{5})v_{\rpa}(\ppa)]
    [\vb_{\ra}(\pa)\gu{\sigma}(a - b\gd{5})u_{\rb}(\pb)]
    .
\ee

\P
Handling of the $S_{\tilde\nu}$ term
in~\rf{charge-S2-as-two-term} is slightly
different due to the presence of the charge conjugation operation.

\beml{charge-ref13}
S_{\tilde\nu} & = & (2\pi)^6 \int d^4\!x\, d^4\!x'
\bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel&&\qquad {} \times
       \Torder\Bigl\lbrace
            \Norder[\psi^T_{\chi}C(F + G\gd{5})
                    \psi_{e}{\tilde\nu}^\hc]_x
            \Norder[\overline{\psi}_e (\cc{F} - \cc{G}\gd{5})
                    C^{-1}\overline{\psi}_{\chi}^T\tilde{\nu}]_{x'}
        \Bigr\rbrace
\nel&&\qquad {} \times
    d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0}
\nel
& = &
(2\pi)^6\int d^4\!x\, d^4\!x'
    \bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel&&\qquad {} \times
        \Torder\Bigl\lbrace
            [(\negPart{\psi})^T_{\chi}C(F + G\gd{5})
                    \posPart{\psi}_{e}\cc{\tilde{\nu}}]_x
            [\overline{\negPart{\psi}}_e (\cc{F} - \cc{G}\gd{5})
                    C^{-1}(\overline{\posPart{\psi}}_{\chi})^T\tilde{\nu}]_{x'}
        \Bigr\rbrace
\nel & & \qquad \times \,
    d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0}
\nel
& = &
(2\pi)^6 \int d^4\!x\, d^4\!x'
    \bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel&&\qquad {} \times
        \Norder\Bigl\lbrace
            [(\negPart{\psi})^T_{\chi}C(F + G\gd{5})
                    \posPart{\psi}_{e}]_x
            [\overline{\negPart{\psi}}_e (\cc{F} - \cc{G}\gd{5})
                    C^{-1}(\overline{\posPart{\psi}}_{\chi})^T]_{x'}
        \Bigr\rbrace
\nel&&\qquad {} \times
        i\Delta_{\tilde{\nu}}(x' - x)
    d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0},
\ee

where $\Delta_{\tilde\nu}$ is the sneutrino propagator that has the usual
form of the propagator for a scalar particle. If I take into account
that $\psi_\chi$ anticommutes with $\psi_e$
I will find from~\rf{gammaFermionField} in a way similar to~\rf{charge-ref12}
that
\bem
\lefteqn{
\bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\Norder\Bigl\lbrace
  [(\negPart{\psi})^T_{\chi} \dots \posPart{\psi}_{e}]_x
  [\overline{\negPart{\psi}}_e \dots (\overline{\posPart{\psi}}_{\chi})^T]_{x'}
\Bigr\rbrace
d^\hc_e(\pa,\ra)c^\hc_e(\pb,\rb)\ket{0}
}
\nel
& = &
-
\left(\frac{m_e}{\Ea}\right)^{\frac{1}{2}}
\bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel && \qquad{}\times
\Norder\Bigl\lbrace
  [(\negPart{\psi})^T_{\chi} \dots \posPart{\psi}_{e}]_x
  e^{-i\pa x'}[\vb_{\ra}(\pa) \dots (\overline{\posPart{\psi}}_{\chi})^T]_{x'}
\Bigr\rbrace
c^\hc_e(\pb,\rb)\ket{0}
\nel
& = &
\left(\frac{m_e^2}{\Ea\Eb}\right)^{\frac{1}{2}}
\bra{0}c_{\chi}(\ppb,\rpb)d_{\chi}(\ppa, \rpa)
\nel && \qquad{}\times
\Norder\Bigl\lbrace
  [(\negPart{\psi})^T_{\chi} \dots u_{\rb}(\pb)]_xe^{-i\pb x}
  e^{-i\pa x'}[\vb_{\ra}(\pa) \dots (\overline{\posPart{\psi}}_{\chi})^T]_{x'}
\Bigr\rbrace
\ket{0}
\nel
& = &
\left(\frac{m_e^2 m_\chi^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\dirProd{0}{0}
\nel && \qquad{}\times
e^{i\ppa x}[v^T(\ppa, \rpa) \dots u_{\rb}(\pb)]e^{-i\pb x}
  e^{-i\pa x'}[\vb_{\ra}(\pa) \dots \ub^T(\ppb, \rpb)]e^{i\ppb x'}
.
\ee

Thus~\rf{charge-ref13} gives

\bem
S_{\tilde\nu} & = &
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\int d^4\!x\, d^4\!x'
    \dirProd{0}{0}  i\Delta_{\tilde{\nu}}(x' - x)
\nel & & {}\times
    e^{i\ppa x}e^{-i\pb x}
    [{v^T(\ppa, \rpa)}C(F + G\gd{5})u_{\rb}(\pb)]
\nel & & {}\times
    e^{-i\pa x'}e^{i\ppb x'}
    [\vb_{\ra}(\pa)(\cc{F} - \cc{G}\gd{5})
                    C^{-1}\ub^T(\ppb, \rpb)]
.
\ee

The Fourier transformation of $i\Delta_{\tilde{\nu}}(x' - x)$ leads to
\bem
S_{\tilde\nu}
& = &
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
(2\pi)^{-4}
\int d^4\!x\, d^4\!x' d^4\!k\,
    e^{ix(\ppa - \pb)}e^{ix'(\ppb - \pa)}e^{-ik(x' - x)}
    i\Delta_{\tilde{\nu}}(k)
\nel & & \qquad \times\,
    [{v^T(\ppa, \rpa)}C(F + G\gd{5})u_{\rb}(\pb)]
    [\vb_{\ra}(\pa)(\cc{F} - \cc{G}\gd{5})
        C^{-1}\ub^T(\ppb, \rpb)]
\nel
& = &
(2\pi)^{4}
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\int d^4\!k\,
    \delta^4(\ppa - \pb + k)\delta^4(\ppb - \pa - k)
    i\Delta_{\tilde{\nu}}(k)
\nel & & \qquad \times\,
    [{v^T(\ppa, \rpa)}C(F + G\gd{5})u_{\rb}(\pb)]
    [\vb_{\ra}(\pa)(\cc{F} - \cc{G}\gd{5})
        C^{-1}\ub^T(\ppb, \rpb)]
\nel
& = &
(2\pi)^{4}
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\delta^4(\ppa + \ppb - \pa - \pb)
\left. i\Delta_{\tilde{\nu}}(k) \right|_{k = \pb - \ppa =  \ppb - \pa}
\nel & & \qquad \times\,
    [{v^T(\ppa, \rpa)}C(F + G\gd{5})u_{\rb}(\pb)]
    [\vb_{\ra}(\pa)(\cc{F} - \cc{G}\gd{5})
        C^{-1}\ub^T(\ppb, \rpb)]
.
\ee

Now, if I use~\rf{gammaSpinorRelation}, I will obtain
$v^T(p, s) C = - \ub$, $C^{-1}\ub^T = v$,
\bem
S_{\tilde\nu} & = &
-(2\pi)^{4}
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\delta(\ppa + \ppb - \pa - \pb)
\left. i\Delta_{\tilde{\nu}}(k) \right|_{k = \pb - \ppa =  \ppb - \pa}
\nel & & \quad \times\,
    [{\ub_{\rpa}(\ppa)}(F + G\gd{5})u_{\rb}(\pb)]
    [\vb_{\ra}(\pa)(\cc{F} - \cc{G}\gd{5})v_{\rpb}(\ppb)]
.
\ee

So the whole amplitude can be represented in the standard form,
\beml{charginoProductionAmpl}
\lefteqn{
S_2(e^+e^- \rightarrow \chi^+\chi^-) \; = \;
\sum_{k = \gamma, Z} S_k + S_{\tilde\nu}
}
\nel
& = &
i(2\pi)^{4}
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\delta(\ppa + \ppb - \pa - \pb)
\nel & & {}\times
\Bigl\lbrace

\sum_{k = \gamma, Z}
D^k_{\rho\sigma}(\pa + \pb)
\nel & & {}\qquad\qquad \times
    [\ub_{\rpb}(\ppb)\gu{\rho}(A_k - B_k\gd{5})v_{\rpa}(\ppa)]
    [\vb_{\ra}(\pa)\gu{\sigma}(a_k - b_k\gd{5})u_{\rb}(\pb)]

\nel & & \quad {}

-
\Delta_{\tilde{\nu}}(\pb - \ppa)
    [\ub_{\rpa}(\ppa)(F + G\gd{5})u_{\rb}(\pb)]
    [\vb_{\ra}(\pa)(\cc{F} - \cc{G}\gd{5})v_{\rpb}(\ppb)]
\Bigr\rbrace

\nel
& \equiv & -i(2\pi)^{4}
\left(\frac{m_e^2m_{\chi}^2}{\Ea\Eb\Epa\Epb}\right)^{\frac{1}{2}}
\delta(\ppa + \ppb - \pa - \pb)M_{e\chi},
\ee

where I set
\beml{charge-M-is}
M_{e\chi} & = & \sum_{k = \gamma, Z} M_k + M_{\tilde\nu},
\ee

and
\beml{charge-M-channels-is}
M_k & = &
-D^k_{\rho\sigma}(\pa + \pb)
\nel&&{} \quad\times
    [\ub_{\rpb}(\ppb)\gu{\rho}(A_k - B_k\gd{5})v_{\rpa}(\ppa)]
    [\vb_{\ra}(\pa)\gu{\sigma}(a_k - b_k\gd{5})u_{\rb}(\pb)]
    ,
\nel
M_{\tilde\nu}
&= &\Delta_{\tilde{\nu}}(\pb - \ppa)
    [\ub_{\rpa}(\ppa)(F + G\gd{5})u_{\rb}(\pb)]
    [\vb_{\ra}(\pa)(\cc{F} - \cc{G}\gd{5})v_{\rpb}(\ppb)]
    .
\nel
\ee

In the unitary gauge the photon and $Z$ propagators have the
form~\cite{SMTextBook}
\bel{photonPropogator}
D^\gamma_{\rho\sigma}(k)
= - {g_{\rho\sigma}\over k^2}
= (-g_{\rho\sigma} + \omega_\gamma k_\rho k_\sigma ) \Delta_\gamma(k^2),
\ee
\be
\omega_\gamma = 0,
\qquad \Delta_\gamma(k^2) = {1 \over k^2},
\ee

\be
D^Z_{\rho\sigma}(k) =
{(-g_{\rho\sigma} + k_\rho k_\sigma/\mB^2)\over
k^2 - \mB^2 + i\mB\Gamma_Z}
= (-g_{\rho\sigma} + \omega_Z k_\rho k_\sigma ) \Delta_Z(k^2),
\ee

\bel{ZPropogator}
\omega_Z = {1 \over \mB^2},
\qquad \Delta_Z(k^2) = {1 \over k^2 - \mB^2 + i\mB\Gamma_Z},
\ee

where $\mB$ is the mass of $Z$ and $\Gamma_\B$ is its decay rate,
it reflects the instability of $Z$.

I also include the decay term to the sneutrino propagator,
\bel{chargeSneutrinoDelta}
\Delta_{\tilde{\nu}}(k) =
\Delta_{\tilde{\nu}}(k^2) =
{1 \over k^2 - m_{\tilde{\nu}}^2 + im_{\tilde{\nu}}\Gamma_{\tilde{\nu}}}
.
\ee

The delta function in \rf{charginoProductionAmpl} implies
the usual conservation law $\pa + \pb = \ppa + \ppb$.

\P
Now with the help of \rf{gammaPOnSpinors}, $\pc u = m u$ and $\pc v = -m v$,
I find from $k = \pa + \pb$ applied to the electron current
\bem
\lefteqn{
k_\sigma [\vb_{\ra}(\pa)\gu{\sigma}(a_k - b_k\gd{5})u_{\rb}(\pb)]
\; = \; [\vb_{\ra}(\pa)\slsh{k}(a_k - b_k\gd{5})u_{\rb}(\pb)]
}
\nel & = &
[\vb_{\ra}(\pa)\pca(a_k - b_k\gd{5})u_{\rb}(\pb)]
+
[\vb_{\ra}(\pa)\pcb(a_k - b_k\gd{5})u_{\rb}(\pb)]
\nel & = &
[\vb_{\ra}(\pa)\pca(a_k - b_k\gd{5})u_{\rb}(\pb)]
+
[\vb_{\ra}(\pa)(a_k + b_k\gd{5})\pcb u_{\rb}(\pb)]
\nel & = &
-[\vb_{\ra}(\pa)\ma(a_k - b_k\gd{5})u_{\rb}(\pb)]
+
[\vb_{\ra}(\pa)(a_k + b_k\gd{5}) \mb u_{\rb}(\pb)]
\nel & = &
2m_eb_k[\vb_{\ra}(\pa)\gd{5}u_{\rb}(\pb)]
.
\ee

In the same way from $k = \ppa + \ppb$ I have
\bem
\lefteqn{
k_\rho [\ub_{\rpb}(\ppb)\gu{\rho}(A_k - B_k\gd{5})v_{\rpa}(\ppa)]
}
\nel & = &
[\ub_{\rpb}(\ppb)\mpb(A_k - B_k\gd{5})v_{\rpa}(\ppa)]
-
[\ub_{\rpb}(\ppb)(A_k + B_k\gd{5})\mpa v_{\rpa}(\ppa)]
\nel & = &
-2m_\chi B_k[\ub_{\rpb}(\ppb)\gd{5}v_{\rpa}(\ppa)]
.
\ee

Thus

\beml{chargeMBoson}
M_k & = &
\Delta_k(s)\bigl\{
    [\ub_{\rpb}(\ppb)\gu{\rho}(A_k - B_k\gd{5})v_{\rpa}(\ppa)]
    [\vb_{\ra}(\pa)\gd{\rho}(a_k - b_k\gd{5})u_{\rb}(\pb)]
\nel &&
{} + 4m_e m_\chi \omega_k b_k B_k
  [\ub_{\rpb}(\ppb)\gd{5}v_{\rpa}(\ppa)][\vb_{\ra}(\pa)\gd{5}u_{\rb}(\pb)]
  \bigr\}
  ,
\ee

and
\beml{chargeMTriple}
M_{\tilde\nu}  &=& \Delta_{\tilde{\nu}}(t)
    [\ub_{\rpa}(\ppa)(F + G\gd{5})u_{\rb}(\pb)]
    [\vb_{\ra}(\pa)(\cc{F} - \cc{G}\gd{5})v_{\rpb}(\ppb)]
    ,
\ee

with
\be
s = (\pa + \pb)^2 = (\ppa + \ppb)^2,
\qquad
t = (\ppb - \pa)^2 = (\pb - \ppa)^2
.
\ee

It is expected that the lightest chargino mass would not exceed
the order of some TeV. So I may use the approximation
\be
m_e m_\chi \omega_Z = {m_e m_\chi \over \mB^2} \ll 1
.
\ee

At the same time I have from \rf{gammaProj} that the spinors $u$, $v$
do not depend on masses. Thus under the approximation
\be
{m_e m_\chi \over \mB^2} \approx 0
\ee
I may write,

\bel{chargeMBosonApprox}
M_k \approx
\Delta_k(s)
    [\ub_{\rpb}(\ppb)\gu{\rho}(A_k - B_k\gd{5})v_{\rpa}(\ppa)]
    [\vb_{\ra}(\pa)\gd{\rho}(a_k - b_k\gd{5})u_{\rb}(\pb)]
    .
\ee

\section{Cross section}

From the usual form of \rf{charginoProductionAmpl}
according to appendix \rf{CrossSectionApp}
I may express the center-of-mass differential and
total cross sections for the process \rf{charginoProdProc}
through $X_{e\chi}$ that reads
\beml{chargeFullX}
X_{e\chi}
&=&  {1 \over 4}\sum_\ra\sum_\rb\sum_\rpa\sum_\rpb
   M_{e\chi}^\hc M_{e\chi}
= \frac{1}{4}\sum_{\ra, \rb, \rpa, \rpb}
\abs{\sum_{k = \gamma, Z} M_k + M_{\tilde\nu}}^2
\nel
&=&
\sum_{k, k' = \gamma, Z} X_{kk'} + X_{\tilde\nu\tilde\nu}
 + 2\sum_{k = \gamma, Z} X_{k\tilde\nu}
.
\ee

Here the contribution from boson ($\gamma$ and $Z^0$) reaction
channels is given by
\beml{charge-boson-X-definition}
X_{kk'} & = &
\frac{1}{4}\sum_{\ra, \rb, \rpa, \rpb} \Re(M_k M_{k'}^\hc).
\ee

and similarly the $\tilde\nu$ channel is given by
\beml{charge-sneutrino-X-definition}
X_{\tilde\nu\tilde\nu} & = &
\frac{1}{4}\sum_{\ra, \rb, \rpa, \rpb} M_{\tilde\nu} M_{\tilde\nu}^{\hc}
.
\ee

The interference term between boson and sneutrino channels has the form
\beml{charge-interference-X-definition}
X_{k\tilde\nu} & = &
\frac{1}{4}\sum_{\ra, \rb, \rpa, \rpb} \Re(M_k M_{\tilde\nu}^{\hc}) .
\ee

\P

To represent the calculations of the terms in \rf{chargeFullX}
in a compact form I suppress the index $k$ in
the expressions~\rf{chargeMTriple} and \rf{chargeMBosonApprox},

\bel{charge-M-k-form}
M_k = \Delta_k(s)
    [\ub_{\rpb}(\ppb)\gu{\rho}(A - B\gd{5})v_{\rpa}(\ppa)]
    [\vb_{\ra}(\pa)\gd{\rho}(a - b\gd{5})u_{\rb}(\pb)].
\ee

So $M_{k'}^\hc$ is given by
\bem
M_{k'}^\hc  &=& \Delta_{k'}^\hc (s)
    [\ub_{\rpb}(\ppb)\gu{\rho}(A' - B'\gd{5})v_{\rpa}(\ppa)]^\hc
    [\vb_{\ra}(\pa)\gd{\rho}(a' - b'\gd{5})u_{\rb}(\pb)]^\hc
\nel & = &
\Delta_{k'}^\hc (s)
    [\vb_{\rpa}(\ppa)(A' + B'\gd{5})\gu{\sigma}u_{\rpb}(\ppb)]
    [\ub_{\rb}(\pb)(a' + b'\gd{5})\gd{\sigma}v_{\ra}(\pa)],
\ee

In the same way I find $M_{\tilde\nu}^{\hc}$

\beml{chargeMShortExp}
M_{\tilde\nu}^{\hc}
&=& \Delta_{\tilde{\nu}}^\hc(t)
    [\ub_{\rpa}(\ppa)(F + G\gd{5})u_{\rb}(\pb)]^\hc
    [\vb_{\ra}(\pa)(F^\hc - G^\hc\gd{5})v_{\rpb}(\ppb)]^\hc
\nel & = &
 \Delta_{\tilde{\nu}}^\hc(t)
    [\ub_{\rb}(\pb)(F^\hc - G^\hc\gd{5})u_{\rpa}(\ppa)]
    [\vb_{\rpb}(\ppb)(F + G\gd{5})v_{\ra}(\pa)]
,
\ee

where I used \rf{gammaBProperties} to calculate the hermitian conjugated
expression.

\subsection{Contribution from boson channels}

By using the spin polarization sum formulae  \rf{gammaSpinPolarization},
\beml{SpinSums}
\sum_r {\v_k}(p, r){\vb_l}(p, r) & = &
 \frac{1}{2m}(\slsh{p} - m)_{kl}
    = \frac{1}{2}(\slsh{q} - 1)_{kl},
\nel
\sum_r {\u_k}(p, r){\ub_l}(p, r) & = &
\frac{1}{2m}(\slsh{p} + m)_{kl}
   = \frac{1}{2}(\slsh{q} + 1)_{kl},
\quad q \equiv p/m,
\ee

where $k$ and $l$ denote spinor indices , I have

\bel{charge-X-boson-1}
X_{kk'} =
\frac{1}{4}\Re\left[\sum_{\ra, \rb, \rpa, \rpb}
M_k M_{k'}^\hc \right]
\approx
\frac{1}{4}\Re[\Delta_k(s)\Delta_{k'}^\hc(s)
L^{\rho\sigma} N_{\rho\sigma}],
\ee

with ``electron'' ($L^{\rho\sigma}$) and ``chargino''
($N_{\rho\sigma}$) tensors given by
\bem
L^{\rho\sigma} & = & \sum_{\rpa, \rpb}
    [\ub_{\rpb}(\ppb)\gu{\rho}(A - B\gd{5})v_{\rpa}(\ppa)]
    [\vb_{\rpa}(\ppa)(A' + B'\gd{5})\gu{\sigma}u_{\rpb}(\ppb)]
\nel
& = &
{1 \over 4}
    \Tr\Bigl[(\qcpb + 1)\gu\rho(A - B\gd5)
             (\qcpa - 1)(A' + B'\gd5)\gu\sigma\Bigr] ,
\ee

and
\bem
N_{\rho\sigma} & = & \sum_{\ra, \rb}
    [\vb_{\ra}(\pa)\gd{\rho}(a - b\gd{5})u_{\rb}(\pb)]
    [\ub_{\rb}(\pb)(a' + b'\gd{5})\gd{\sigma}v_{\ra}(\pa)]
\nel
& = &
{1 \over 4}
\Tr\Bigl[(\qca - 1)\gu\rho(a - b\gd5)(\qcb + 1)(a' + b'\gd5)\gu\sigma\Bigr]
.
\ee

The chargino tensor $N_{\rho\sigma}$
can be expressed through $L^{\rho\sigma}(A, A', B, B', \qpa, \qpb)$ via
\bel{NthroughL}
N_{\rho\sigma}
   = L_{\rho\sigma}(A\rightarrow a, A'\rightarrow a',
                   B \rightarrow b, B' \rightarrow b',
                   \qpa \rightarrow -\qb, \qpb \rightarrow -\qa)
.
\ee
Thus it is necessary to calculate only one quantity.

\P

Applying properties of $\gamma$-matrices one finds for $L^{\rho\sigma}$
\bem
L^{\rho\sigma} & = &
\frac{1}{4}
    \Tr\Bigl[
        \qcpb\gu\rho(A - B\gd5)\qcpa(A' + B'\gd5)\gu\sigma
        - \gu\rho(A - B\gd5)(A' + B'\gd5)\gu\sigma\Bigr]
\nel & = &
\frac{1}{4}
    \Tr\Bigl[
        \qcpb\gu\rho(A - B\gd5)(A' - B'\gd5)\qcpa\gu\sigma
        - \gu\rho(A - B\gd5)(A' + B'\gd5)\gu\sigma\Bigr]
\nel & = &
\frac{1}{4}
    \Tr\Bigl[
        \qcpb\gu\rho\lbrace
            AA' + BB' - \gd5(AB' + BA')
        \rbrace \qcpa\gu\sigma
\nel & & \qquad{}
        - \gu\rho\lbrace
            AA' - BB' + \gd5(AB' - BA')
        \rbrace \gu\sigma\Bigr]
\nel & = &
\frac{1}{4}
    \Tr\Bigl[
        \qcpb\gu\rho\lbrace
            AA' + BB' - \gd5(AB' + BA')
        \rbrace \qcpa\gu\sigma
        - \gu\rho(AA' - BB')\gu\sigma\Bigr]
\nel & = &
\frac{1}{4}\Bigl[
    (AA' + BB')\Tr(\qcpb\gu\rho\qcpa\gu\sigma)
    -(AB' + BA')\Tr(\qcpb\gu\rho\qcpa\gu\sigma\gd5)
\nel & & \qquad {}
    -(AA' - BB')\Tr(\gu\rho\gu\sigma)
    \Bigr]
    .
\ee

I calculate the traces with the help of \rf{gammaTraces},

\bem
L^{\rho\sigma} & = &
    (AA' + BB')
        [\qpb^\rho \qpa^\sigma + \qpb^\sigma \qpa^\rho
        -
        (\qpb\cdot\qpa)g^{\rho\sigma}]
\nel & & {} +
    i(AB' + BA')
        \varepsilon^{\mu\rho\nu\sigma}\qpb_\mu \qpa_\nu
    -(AA' - BB')g^{\rho\sigma}
\nel
 & = & L_S^{\rho\sigma} + L_A^{\rho\sigma},
\ee

where
\bem
L_S^{\rho\sigma}  & = &
    (AA' + BB')[2\qpb^{(\rho} \qpa^{\sigma)}
     - (\qpb\cdot\qpa)g^{\rho\sigma}]
    -(AA' - BB')g^{\rho\sigma}
,
\nel

L_A^{\rho\sigma}  & = &
    i(AB' + BA')
        \varepsilon^{\mu\rho\nu\sigma}\qpb_\mu \qpa_\nu
\ee

In the last expression I explicitly write the symmetric part
$L_S^{\rho\sigma}=L_S^{(\sigma\rho)}$ and the asymmetric one
$L_A^{\rho\sigma}=L_A^{[\sigma\rho]}$.

\P
From~\rf{NthroughL} I have for the chargino tensor $N_{\rho\sigma}$,
\bem
N_{\rho\sigma} & = & {N_S}_{\rho\sigma} + {N_A}_{\rho\sigma},
\nel
{N_S}_{\rho\sigma} & =  &
    (aa' + bb')[2\qa_{(\rho} \qb_{\sigma)}
     - (\qb\cdot\qa)g_{\rho\sigma}]
    -(aa' - bb')g_{\rho\sigma}
,
\nel
{N_A}_{\rho\sigma} & =  &
    i(ab' + ba')
        \varepsilon_{\eta\rho\theta\sigma}\qa^\eta \qb^\theta
        .
\ee

Two products will be required. I first consider the symmetric parts:

\bem
L_S^{\rho\sigma} {N_S}_{\rho\sigma}
& =  &
\{
    (AA' + BB')[2\qpb^{(\rho} \qpa^{\sigma)}
     - (\qpb\cdot\qpa)g^{\rho\sigma}]
    -(AA' - BB')g^{\rho\sigma}
\}
\nel && {} \times
\{
    (aa' + bb')[2\qa_{(\rho} \qb_{\sigma)}
     - (\qb\cdot\qa)g_{\rho\sigma}]
    -(aa' - bb')g_{\rho\sigma}
\}

\nel
& = &
(aa' + bb')(AA' + BB')
\nel && \qquad {} \times
[2\qpb^{(\rho} \qpa^{\sigma)} - (\qpb\cdot\qpa)g^{\rho\sigma}]
[2\qa_{\rho} \qb_{\sigma} - (\qb\cdot\qa)g_{\rho\sigma}]
\nel
&& {}
- (aa' - bb')(AA' + BB')[2(\qpb \cdot \qpa) - 4(\qpb\cdot\qpa)]
\nel && {}
- (aa' + bb')(AA' - BB')[2(\qb \cdot \qa) - 4(\qb\cdot\qa)]
\nel
&& {}
+ 4(aa' - bb')(AA' - BB')
\nel
& = &
(aa' + bb')(AA' + BB')
[2(\qpb \cdot \qa) (\qpa \cdot \qb) + 2(\qpb \cdot \qb) (\qpa \cdot \qa)
\nel
&& \qquad {}
- 2 (\qb\cdot\qa) (\qpb \cdot \qpa) - 2 (\qb\cdot\qa) (\qpb \cdot \qpa)
 + 4 (\qb\cdot\qa) (\qpb \cdot \qpa)]
\nel
&& {}
+ 2(aa' - bb')(AA' + BB')(\qpb\cdot\qpa)
\nel
&& {}
+ 2(aa' + bb')(AA' - BB')(\qb\cdot\qa)
+ 4(aa' - bb')(AA' - BB')
\nel
& = &
2(aa' + bb')(AA' + BB')
[(\qpb \cdot \qa) (\qpa \cdot \qb) + (\qpb \cdot \qb) (\qpa \cdot \qa)]
\nel
&& {}
+ 2(aa' - bb')(AA' + BB')(\qpb\cdot\qpa)
\nel
&& {}
+ 2(aa' + bb')(AA' - BB')(\qb\cdot\qa)
+ 4(aa' - bb')(AA' - BB')
.
\ee

To calculate $L_A^{\rho\sigma} {N_A}_{\rho\sigma}$ I use the identity
\be
\varepsilon^{\mu\rho\nu\sigma} \varepsilon_{\eta\rho\theta\sigma}
= -2
(\delta^\mu_\eta \delta^\nu_\theta - \delta^\mu_\theta \delta^\nu_\eta)
.
\ee

Thus
\bem
L_A^{\rho\sigma} {N_A}_{\rho\sigma}
& = &
-(ab' + ba')(AB' + BA')
        \varepsilon^{\mu\rho\nu\sigma}\qpb_\mu \qpa_\nu
        \varepsilon_{\eta\rho\theta\sigma}\qa^\eta \qb^\theta
\nel
& = &
2(ab' + ba')(AB' + BA')
(\delta^\mu_\eta \delta^\nu_\theta - \delta^\mu_\theta \delta^\nu_\eta)
\qpb_\mu \qpa_\nu
\qa^\eta \qb^\theta

\nel
& = &
2(ab' + ba')(AB' + BA')
[(\qpb \cdot \qa) (\qpa \cdot \qb) - (\qpb \cdot \qb) (\qpa \cdot \qa)]
.
\ee

Now I use~\rf{contractionWithSymAnti} to simplify the product
$L^{\rho\sigma} N_{\rho\sigma}$,
\beml{chargeLMIs}
\lefteqn{
L^{\rho\sigma} N_{\rho\sigma}
=
(L_S^{\rho\sigma} + L_A^{\rho\sigma})({N_S}_{\rho\sigma} + {N_A}_{\rho\sigma})
= L_S^{\rho\sigma} {N_S}_{\rho\sigma} + L_A^{\rho\sigma} {N_A}_{\rho\sigma}
}
\nel
& = &
2(aa' + bb')(AA' + BB')
[(\qpb \cdot \qa) (\qpa \cdot \qb) + (\qpb \cdot \qb) (\qpa \cdot \qa)]
\nel
&& {}
+ 2(ab' + ba')(AB' + BA')
[(\qpb \cdot \qa) (\qpa \cdot \qb) - (\qpb \cdot \qb) (\qpa \cdot \qa)]
\nel
&& {}
+ 2(aa' - bb')(AA' + BB')(\qpb\cdot\qpa)
\nel
&& {}
+ 2(aa' + bb')(AA' - BB')(\qb\cdot\qa)
\nel
&& {}
+ 4(aa' - bb')(AA' - BB')
.
\ee

\P
Under the approximations $m_\chi \gg m_e$, $s \gg 4m_e$
I may use the formula~\rf{cs-q-scalar-products}
where $m = m_e$, $m' = m_\chi$,
to calculate the kinematic quantities in \rf{chargeLMIs},
\bem
(\qpb \cdot \qa)(\qpa \cdot \qb) + (\qpb \cdot \qb) (\qpa \cdot \qa)
&\approx& {1 \over 4m_e^2 m_\chi^2}[2(m_\chi^2 - t)^2 + s(s + 2t - 2m_\chi^2)],
\nel
(\qpb \cdot \qa)(\qpa \cdot \qb) - (\qpb \cdot \qb) (\qpa \cdot \qa)
&\approx& {s \over 4 m_e^2 m_\chi^2}(s + 2t - 2m_\chi^2),
\nel
(\qb\cdot\qa) &\approx& {s \over 2 m_e^2}
,
\ee

\be
t = (\ppa - \pa)^2 = (\pb - \ppb)^2
.
\ee

I also may neglect $2(aa' - bb')(AA' + BB')(\qpb\cdot\qpa)$ and
$4(aa' - bb')(AA' - BB')$ under
the same approximation because they do not contain $m_e^{-2}$. Thus
\bem
\lefteqn{
L^{\rho\sigma} N_{\rho\sigma}
}
\nel
& \approx  &
{1 \over 2m_e^2 m_\chi^2} \bigl\{
(aa' + bb')(AA' + BB')[2(m_\chi^2 - t)^2 + s(s + 2t - 2m_\chi^2)]
\nel && \qquad {}
+ (ab' + ba')(AB' + BA')s(s + 2t - 2m_\chi^2) \bigr\}
\nel
&& {}
+ {s \over m_e^2}(aa' + bb')(AA' - BB')

\nel
& = &
{(m_\chi^2 - t)^2 \over m_e^2 m_\chi^2} (aa' + bb')(AA' + BB')
\nel && {}
+
{s(s + 2t - 2m_\chi^2) \over 2m_e^2 m_\chi^2}
[(ab' + ba')(AB' + BA') + (aa' + bb')(AA' + BB')]
\nel
&& {}
+ {s \over m_e^2}(aa' + bb')(AA' - BB')
,
\ee

which is a real quantity so $X_{kk'}$ of the equation~\rf{charge-X-boson-1}
is given by

\beml{charge-X1}
\lefteqn{
X_{kk'} \approx \Re[\Delta_k(s)\Delta_{k'}^\hc(s)]
}
\nel
& \times & \Bigl\{
{(m_\chi^2 - t)^2 \over 4m_e^2 m_\chi^2}
(a_{k}a_{k'} + b_{k}b_{k'})(A_{k}A_{k'} + B_{k}B_{k'})
\nel
&& {}
+
{s(s + 2t - 2m_\chi^2) \over 8m_e^2 m_\chi^2}
\nel
&& \quad{}\times
[(a_{k}b_{k'} + b_{k}a_{k'})(A_{k}B_{k'} + B_{k}A_{k'})
    + (a_{k}a_{k'} + b_{k}b_{k'})(A_{k'}A_{k'} + B_{k}B_{k'})]
\nel
&& {}
+ {s \over 4m_e^2}(a_{k}a_{k'} + b_{k}b_{k'})(A_{k}A_{k'} - B_{k}B_{k'})
\Bigr\}
.
\ee

\subsection{Contribution from the scalar neutrino channel}

The calculation of $X_{\tilde\nu\tilde\nu}$
(see~\rf{charge-sneutrino-X-definition})
follows the same pattern as for $X_{kk'}$,

\be
X_{\tilde\nu\tilde\nu} =
\frac{1}{4}\Re\left[\sum_{\ra, \rb, \rpa, \rpb}
M_{\tilde\nu} M_{\tilde\nu}^{\hc}\right]
=
\frac{1}{4}\abs{\Delta_{\tilde\nu}(t)}^2\Re[L N],
\ee

where $L$ and $N$ come from~\rf{chargeMTriple} and~\rf{chargeMShortExp},
\bem
L & = & \sum_{\rpa, \rb}
    [\ub_{\rpa}(\ppa)(F + G\gd{5})u_{\rb}(\pb)]
    [\ub_{\rb}(\pb)(F^\hc - G^\hc\gd{5})u_{\rpa}(\ppa)]
\nel
& = &
\frac{1}{4} \Tr
\Bigl[(\qcpa + 1)(F + G\gd{5})
      (\qcb + 1)(F^\hc - G^\hc\gd{5})\Bigr]
,
\ee

and
\beml{charge-N-tmp23}
N & = &  \sum_{\ra, \rpb}
    [\vb_{\ra}(\pa)(F^\hc - G^\hc\gd{5})v_{\rpb}(\ppb)]
    [\vb_{\rpb}(\ppb)(F + G\gd{5})v_{\ra}(\pa)]
\nel
& = &
\frac{1}{4}\Tr
\Bigl[(\qca - 1)(F^\hc - G^\hc\gd{5})
      (\qcpb - 1)(F + G\gd{5})\Bigr]
\nel & = &
L(\qpa \rightarrow -\qpb, \qb \rightarrow -\qa).
\ee

Now I simplify $L$:
\bem
L
& = &
\frac{1}{4} \Tr
\Bigl[\qcpa(F + G\gd{5})\qcb(\cc{F} - \cc{G}\gd{5})
      + (F + G\gd{5})(\cc{F} - \cc{G}\gd{5})
\Bigr]

\nel & = &
\frac{1}{4} \Tr
\Bigl[\qcpa\qcb(F - G\gd{5})(\cc{F} - \cc{G}\gd{5})
      + \abs{F}^2 - \abs{G}^2
\Bigr]

\nel & = &
\frac{1}{4} \Tr
\Bigl[\qcpa\qcb(\abs{F}^2 + \abs{G}^2)
      + \abs{F}^2 - \abs{G}^2
\Bigr]

\nel & = &
(\abs{F}^2 + \abs{G}^2)(\qpa \cdot \qb) + \abs{F}^2 - \abs{G}^2.
\ee

The relation~\rf{charge-N-tmp23} gives for $N$
\bem
N
& = &
(\abs{F}^2 + \abs{G}^2)(\qpb \cdot \qa) + \abs{F}^2 - \abs{G}^2
\; = \; L
.
\ee

Thus I have
\bem
X_{\tilde\nu\tilde\nu} & = &

\frac{1}{4}\abs{\Delta_{\tilde\nu}(t)}^2\Re\left\{
[(\abs{F}^2 + \abs{G}^2)(\qpa \cdot \qb) + \abs{F}^2 - \abs{G}^2]^2
\right\}

\nel
& = &
\frac{1}{4}\abs{\Delta_{\tilde\nu}(t)}^2
[(\abs{F}^2 + \abs{G}^2)(\qpa \cdot \qb) + \abs{F}^2 - \abs{G}^2]^2

\nel & \approx &
\frac{1}{4}\abs{\Delta_{\tilde\nu}(t)}^2
(\abs{F}^2 + \abs{G}^2)^2(\qpa \cdot \qb)^2
,
\ee

or

\bel{charge-X2}
X_{\tilde\nu\tilde\nu} \approx
\abs{\Delta_{\tilde\nu}(t)}^2{(s + t - m_\chi^2)^2 \over 16 m_e^2 m_\chi^2}
(\abs{F}^2 + \abs{G}^2)^2
,
\ee

where I used \rf{csKinApproximation} in the last step.

\subsection{Interference contribution.}

To find $X_{k\tilde\nu}$ I write for it according to~\rf{chargeMShortExp},
\be
X_{k\tilde\nu} \approx
\frac{1}{4}\Re\left[\sum_{\ra, \rb, \rpa, \rpb}
M_k M_{\tilde\nu}^{\hc}\right]
=
\frac{1}{4}\Re[\Delta_k(s)\Delta_{\tilde\nu}^\hc(t) K]
,
\ee

with $K$ given by
\bem
K
& = & \sum_{\ra, \rb, \rpa, \rpb}
    [\ub_{\rpb}(\ppb)\gu{\rho}(A - B\gd{5})v_{\rpa}(\ppa)]
    [\vb_{\ra}(\pa)\gd{\rho}(a - b\gd{5})u_{\rb}(\pb)]
\nel && \qquad
{} \times
    [\ub_{\rb}(\pb)(F^\hc - G^\hc\gd{5})u_{\rpa}(\ppa)]
    [\vb_{\rpb}(\ppb)(F + G\gd{5})v_{\ra}(\pa)]
.
\ee

I transform $K$ to the form necessary to calculate the spin
polarization sums with the
help of~\rf{gammaCProperties} and~\rf{gammaSpinorRelation},

\beml{charge-before-spin-sums-1}
[\ub_{\rpb}(\ppb)\gu{\rho}(A - B\gd{5})v_{\rpa}(\ppa)]
& = &
[\ub_{\rpb}(\ppb)\gu{\rho}(A - B\gd{5})v_{\rpa}(\ppa)]^T
\nel & = &
[v_{\rpa}^T(\ppa)(A - B\gd{5}^T)\gu{\rho}^T\ub_{\rpb}^T(\ppb)]
\nel & = &
[v_{\rpa}^T(\ppa)CC^{-1}(A - B\gd{5}^T)\gu{\rho}^TCC^{-1}\ub_{\rpb}^T(\ppb)]
\nel & = &
-[\ub_{\rpa}(\ppa)(A - B C^{-1}\gd{5}^T C)C^{-1}\gu{\rho}^TC v_{\rpb}(\ppb)]
\nel & = &
[\ub_{\rpa}(\ppa)(A - B \gd{5})\gu{\rho} v_{\rpb}(\ppb)]
.
\ee

Now I can represent $K$ as a trace
\bem
K
& = & \sum_{\ra, \rb, \rpa, \rpb}
    [\ub_{\rpa}(\ppa)(A - B \gd{5})\gu{\rho} v_{\rpb}(\ppb)]
    [\vb_{\ra}(\pa)\gd{\rho}(a - b\gd{5})u_{\rb}(\pb)]
\nel && \qquad
{} \times
    [\ub_{\rb}(\pb)(F^\hc - G^\hc\gd{5})u_{\rpa}(\ppa)]
    [\vb_{\rpb}(\ppb)(F + G\gd{5})v_{\ra}(\pa)]
\nel
& = & \sum_{\ra, \rb, \rpa, \rpb}
    [\ub_{\rpa}(\ppa)(A - B \gd{5})\gu{\rho} v_{\rpb}(\ppb)]
    [\vb_{\rpb}(\ppb)(F + G\gd{5})v_{\ra}(\pa)]
\nel && \qquad
{} \times
    [\vb_{\ra}(\pa)\gd{\rho}(a - b\gd{5})u_{\rb}(\pb)]
    [\ub_{\rb}(\pb)(F^\hc - G^\hc\gd{5})u_{\rpa}(\ppa)]
\nel
& = & {1 \over 16} \Tr \Bigl[
    (\qcpa + 1)(A - B \gd{5})\gu{\rho} (\qcpb - 1)(F + G\gd{5})
\nel && \qquad
{} \times
    (\qca - 1) \gd{\rho}(a - b\gd{5}) (\qcb + 1)(F^\hc - G^\hc\gd{5})
    \Bigr]
.
\ee

\P

To simplify $K$ I use $m_\chi \gg m_e$, thus I can omit terms without
$m_e^{-2}$ that come from $\qca = \pca/m_e$, $\qcb = \pcb/m_e$. I also
keep only terms with an even number of Dirac matrices,
\bem
K
& \approx &
{1 \over 16} \Tr [
    \qcpa(A - B \gd{5})\gu{\rho}\qcpb(F + G\gd{5})
    \qca \gd{\rho}(a - b\gd{5}) \qcb(F^\hc - G^\hc\gd{5})
    ]
\nel && {}
-
{1 \over 16} \Tr [
    (A - B \gd{5})\gu{\rho}(F + G\gd{5})
    \qca \gd{\rho}(a - b\gd{5}) \qcb(F^\hc - G^\hc\gd{5})
    ]
\nel& = &
{1 \over 16} (K_1 - K_2),
\ee

where $K_1$ and $K_2$ are given by
\bem
K_1 & = &
\Tr [\qcpa(A - B \gd{5})\gu{\rho}\qcpb(F + G\gd{5})
    \qca \gd{\rho}(a - b\gd{5}) \qcb(F^\hc - G^\hc\gd{5})]
\nel & = &
\Tr [\qcpa\gu{\rho}\qcpb(A - B \gd{5})(F + G\gd{5})
    \qca \gd{\rho}\qcb (a + b\gd{5}) (F^\hc - G^\hc\gd{5})]
\nel & = &
\Tr [\qcpa\gu{\rho}\qcpb\qca \gd{\rho}\qcb
     (A + B \gd{5})(F - G\gd{5})
     (a + b\gd{5}) (F^\hc - G^\hc\gd{5})]
,
\ee

\bem
K_2 & = &
\Tr [(A - B \gd{5})\gu{\rho}(F + G\gd{5})
    \qca \gd{\rho}(a - b\gd{5}) \qcb(F^\hc - G^\hc\gd{5})]
\nel & = &
\Tr [\gu{\rho}(A + B \gd{5})(F + G\gd{5})
    \qca \gd{\rho}\qcb(a + b\gd{5}) (F^\hc - G^\hc\gd{5})]
\nel & = &
\Tr [\gu{\rho}\qca \gd{\rho}\qcb(A - B \gd{5})(F - G\gd{5})
    (a + b\gd{5}) (F^\hc - G^\hc\gd{5})]
.
\ee

To find the sandwich-like products $\gu{\rho} \dots \gd{\rho}$ I use
the basic properties \rf{gammaAlgebra} of the Dirac algebra,

\be
\gu{\rho}\qca \gd{\rho} = -\gu{\rho}\gd{\rho}\qca + 2\gu{\rho} \qa_\rho
= -4\qca + 2\qca = - 2\qca,
\ee

and
\be
\gu{\rho}\qcpb\qca \gd{\rho}
=
-\gu{\rho}\qcpb\gd{\rho}\qca + 2 \gu{\rho}\qcpb \qa_\rho
= 2\qcpb \qca + 2 \qca \qcpb = 2\{\qcpb, \qca\} = 4 (\qpb \cdot \qa)
.
\ee

It gives for $K_1$ and $K_2$

\bem
K_1 & = & 4 (\qpb \cdot \qa) \Tr [\qcpa\qcb
     (A + B \gd{5})(F - G\gd{5})
     (a + b\gd{5}) (F^\hc - G^\hc\gd{5})]
\nel & = &
4 (\qpb \cdot \qa) \Tr [\qcpa\qcb
\nel && \qquad {} \times
     \{AF - BG +\gd5(BF - AG)\}
     \{aF^\hc - bG^\hc +\gd5(bF^\hc - aG^\hc)\}]
\nel & = &
4 (\qpb \cdot \qa) \Tr [\qcpa\qcb
     \{(AF - BG)(aF^\hc - bG^\hc) + (BF - AG)(bF^\hc - aG^\hc) \} ]

\nel & = &
16 (\qpb \cdot \qa)(\qpa \cdot \qb)
[(AF - BG)(aF^\hc - bG^\hc) + (BF - AG)(bF^\hc - aG^\hc)]
\nel & = &
16 (\qpb \cdot \qa)(\qpa \cdot \qb)
[(aA + bB)\abs{F}^2  + (aA + bB)\abs{G}^2
\nel & & \quad {}
- GF^\hc(bA + aB) - FG^\hc(bA + aB)]
\nel & = &
16 (\qpb \cdot \qa)(\qpa \cdot \qb)
[(aA + bB)(\abs{F}^2  + \abs{G}^2) - 2\Re(FG^\hc)(bA + aB)]
,
\ee

and
\bem
K_2 & = &
-2 \Tr [\qca\qcb(A - B \gd{5})(F - G\gd{5})
    (a + b\gd{5}) (F^\hc - G^\hc\gd{5})]
\nel & = &
-2 \Tr [\qca\qcb
     \{AF + BG - \gd5(BF + AG)\}
     \{aF^\hc - bG^\hc +\gd5(bF^\hc - aG^\hc)\}]
\nel & = &
-2 \Tr [\qca\qcb
     \{(AF + BG)(aF^\hc - bG^\hc) + (BF + AG)(aG^\hc - bF^\hc)\}]
\nel & = &
-8(\qb \cdot \qa)
[(AF + BG)(aF^\hc - bG^\hc) + (BF + AG)(aG^\hc - bF^\hc)]
\nel & = &
-8(\qb \cdot \qa)
[(aA -bB)\abs{F}^2 + (aA - bB)\abs{G}^2
\nel & & \quad {}
+ GF^\hc(aB - bA) + FG^\hc(aB - bA)]
\nel & = &
-8(\qb \cdot \qa)
[(aA - bB)(\abs{F}^2  + \abs{G}^2) + 2\Re(FG^\hc)(aB - bA)]
,
\ee

where I used~\rf{gammaTraces} and~\rf{gammaZeroTraces}
together with $\gd5^2 = 1$. Thus

\bem
K & \approx &
{1 \over 16} (K_1 - K_2)
\nel
& = &
(\qpb \cdot \qa)(\qpa \cdot \qb)
[(aA + bB)(\abs{F}^2  + \abs{G}^2) - 2\Re(FG^\hc)(bA + aB)]
\nel && {} +
{1 \over 2} (\qb \cdot \qa)
[(aA - bB)(\abs{F}^2 + \abs{G}^2) - 2\Re(FG^\hc)(bA - aB)]
\nel & \approx &
{(s + t - m_\chi^2)^2 \over 4 m_e^2 m_\chi^2}
[(aA + bB)(\abs{F}^2  + \abs{G}^2) - 2\Re(FG^\hc)(bA + aB)]
\nel && {} +
{s \over 4m_e^2}
[(aA - bB)(\abs{F}^2 + \abs{G}^2) - 2\Re(FG^\hc)(bA - aB)]
\ee

from~\rf{csKinApproximation}.

\P
Now I have for the interference contribution $X_{k\tilde\nu}$

\beml{charge-X3}
X_{k\tilde\nu} & \approx &
\frac{1}{4}\Re[\Delta_k(s)\Delta_{\tilde\nu}^\hc(t) K]
\nel & \approx &
{(s + t - m_\chi^2)^2 \over 16 m_e^2 m_\chi^2}
\Re[\Delta_k(s) \Delta_{\tilde{\nu}}^\hc(t)]
\nel && \qquad {} \times
[(a_k A_k + b_kB_k)(\abs{F}^2  + \abs{G}^2) - 2\Re(FG^\hc)(b_k A_k + a_k B_k)]
\nel && {} +
{s \over 16 m_e^2}
\Re[\Delta_k(s) \Delta_{\tilde{\nu}}^\hc(t)]
\nel && \qquad {} \times
[(a_k A_k - b_k B_k)(\abs{F}^2 + \abs{G}^2) - 2\Re(FG^\hc)(b_k A_k - a_k B_k)]
.
\ee

\subsection{Differential cross section}

According to~\rf{chargeFullX} and~\rf{csDifApproximation}
the differential cross section for the chargino production process
is given by

\be
\left({d\sigma \over d\Omega }\right)_{\rm c.m.}
\approx {m_e^2 m_\chi^2  \over 4\pi^2 s} \sqrt{1 - 4\rho}X_{e\chi}(s, t),
\ee

where
\be
\rho = {m_\chi^2 \over s},
\ee

and
\bel{charge-differential-as-a-sum}
X_{e\chi}(s, t)
= X_{\gamma\gamma} + X_{ZZ} + 2X_{\gamma Z} + X_{\tilde\nu\tilde\nu}
+ 2X_{\gamma\tilde\nu} + 2X_{Z\tilde\nu}
.
\ee

The first three terms in~\rf{charge-differential-as-a-sum} are given
by~\rf{charge-X1},

\beml{charge-d-gamma-gamma}
X_{\gamma\gamma} & \approx & { \Delta_\gamma^2(s) \over 4m_e^2 m_\chi^2 }
\Bigl\{
(m_\chi^2 - t)^2 a_\gamma^2 A_\gamma^2
+ {1 \over 2} s(s + 2t - 2m_\chi^2) a_\gamma^2 A_\gamma^2
+ s m_\chi^2 a_\gamma^2 A_\gamma^2 \Bigr\}
\nel
& = &
{e^4 \over 4s^2m_e^2 m_\chi^2 }
[(m_\chi^2 - t)^2  + s m_\chi^2
+ {1 \over 2} s(s + 2t - 2m_\chi^2)
],
\ee

where I use~\rf{chargeGabIs}, \rf{chargeGABIs} and~\rf{photonPropogator},
\be
-a_\gamma = A_\gamma = e, \quad b_\gamma = B_\gamma = 0,
\quad \Delta_\gamma(s) = {1 \over s}.
\ee

The explicit form of $X_{ZZ}$ is
\bem
X_{ZZ} & \approx & {\abs{\Delta_Z(s)}^2 \over  4m_e^2 m_\chi^2 }
\Bigl\{
(m_\chi^2 - t)^2 (a_Z^2 + b_Z^2)(A_Z^2 + B_Z^2)
\nel
&& {}
+
{1 \over 2} s(s + 2t - 2m_\chi^2)
[(2a_Z b_Z)(2A_Z B_Z) + (a_Z^2 + b_Z^2)(A_Z^2 + B_Z^2)]
\nel
&& {}
+ sm_\chi^2(a_Z^2 + b_Z^2)(A_Z^2 - B_Z^2)
\Bigr\}.
\ee

I find for the coefficients in the last expressions

\bem
a_Z^2 + b_Z^2 & = &
{g^2 \over 16\cos^2\theta_W}[(1 - 4\sin^2\theta_W)^2 + 1]
\nel &=& {g^2 \over 8\cos^2\theta_W}(1 - 4\sin^2\theta_W + 8\sin^4\theta_W)
\quad (\mbox{from \rf{chargeZabIs}})
,
\ee

and
\be
a_Z b_Z =
{g^2 \over 16\cos^2\theta_W}(1 - 4\sin^2\theta_W)
.
\ee

The definitions~\rf{chargeZABIs} lead to
\bem
A_Z^2 \pm B_Z^2 &= &
{g^2 \over 16\cos^2\theta_W}
\{
[2\cos (2\theta_W) + \abs{Y_{11}}^2 + \abs{X_{11}}^2]^2
\pm [\abs{Y_{11}}^2 - \abs{X_{11}}^2]^2
\}
,
\nel
A_Z B_Z &= &
-{g^2 \over 16\cos^2\theta_W}
[2\cos (2\theta_W) + \abs{Y_{11}}^2 + \abs{X_{11}}^2]
[\abs{Y_{11}}^2 - \abs{X_{11}}^2]
.
\ee

Thus
\beml{charge-d-Z-Z}
X_{ZZ} & \approx & {a_Z^2 + b_Z^2 \over 4m_e^2 m_\chi^2}\abs{\Delta_Z(s)}^2
[
(m_\chi^2 - t)^2 (A_Z^2 + B_Z^2) + s m_\chi^2 (A_Z^2 - B_Z^2)
]
\nel &&{} +
{s(s + 2t - 2m_\chi^2) \over 8m_e^2 m_\chi^2} \abs{\Delta_Z(s)}^2
[4 a_Z b_Z A_Z B_Z + (a_Z^2 + b_Z^2)(A_Z^2 + B_Z^2)] .
\nel
\ee

\P
The interference between the $\gamma$ and $Z^0$ channels reads,
according to the equation~\rf{charge-X1}
\beml{charge-d-gamma-Z}
\lefteqn{
X_{\gamma Z}
}
\nel & \approx & {\Delta_\gamma(s)\Re(\Delta_Z^\hc(s)) \over 4m_e^2 m_\chi^2}
\Bigl\{
(m_\chi^2 - t)^2(a_\gamma a_Z)(A_\gamma A_Z)
\nel
&& {}
+
{1\over 2} s(s + 2t - 2m_\chi^2)
[(a_\gamma b_Z)(A_\gamma B_Z) + (a_\gamma a_Z)(A_\gamma A_Z)]
\nel
&& {}
+ s m_\chi^2(a_\gamma a_Z)(A_\gamma A_Z)
\Bigr\}
\nel
& = &
-{e^2 a_Z A_Z \over 4sm_e^2 m_\chi^2} \Re(\Delta_Z(s))
\nel&& {} \quad \times
\Biggl[
(m_\chi^2 - t)^2
+
{1 \over 2}s(s + 2t - 2m_\chi^2)
\left( {b_Z B_Z\over a_Z A_Z} + 1\right)
+ sm_\chi^2
\Biggr]
.
%\nel
\ee

\P
The expression for $X_{\tilde\nu\tilde\nu}$ is given by~\rf{charge-X2}.
If I substitute there the definitions~\rf{chargeFandG1} and~\rf{chargeFandG2},
then
\be
\abs{F}^2 + \abs{G}^2 \approx {g^2 \over 2}\cos^2\theta_X,
\quad
2\Re(FG^\hc) \approx {g^2 \over 2}\cos^2\theta_X,
\ee

where according to~\rf{lagrangian-theta-X-and-Y}
and~\rf{lagrangian-eta-X-and-Y}
\be
\cos^2\theta_X = {1 \over 2 + 2 \eta_X^2 - 2\sqrt{1 + \eta_X^2}},
\quad
\eta_X =
   \frac{m_{\tilde\Lambda}^2 - \abs{\mu}^2 + 2m_W^2\cos(2\beta)}
        {2\sqrt{2}m_W\abs{m_{\tilde\Lambda}\sin\beta + \mu^* \cos\beta}}
        .
\ee

Thus
\beml{charge-d-nu-nu}
X_{\tilde\nu\tilde\nu} &\approx &
\abs{\Delta_{\tilde\nu}(t)}^2{(s + t - m_\chi^2)^2 \over 16 m_e^2 m_\chi^2}
(\abs{F}^2 + \abs{G}^2)^2
\nel
& = &
g^4 \cos^4\theta_X
\abs{\Delta_{\tilde\nu}(t)}^2{(s + t - m_\chi^2)^2 \over 64 m_e^2 m_\chi^2}
.
\ee

Now I use~\rf{charge-X3} to write the photon--$\tilde\nu$
interference term $X_{\gamma\tilde\nu}$

\beml{charge-d-gamma-nu}
X_{\gamma\tilde\nu} & \approx &
{(s + t - m_\chi^2)^2 \over 16 m_e^2 m_\chi^2}
\Delta_\gamma(s) \Re[\Delta_{\tilde{\nu}}^\hc(t)]
\nel && \qquad {} \times
[(a_\gamma A_\gamma)(\abs{F}^2  + \abs{G}^2)]
\nel && {} +
{s \over 16 m_e^2}
\Delta_\gamma(s)\Re[\Delta_{\tilde{\nu}}^\hc(t)]
[(a_\gamma A_\gamma)(\abs{F}^2 + \abs{G}^2)]

\nel & = &
-{e^2 (\abs{F}^2 + \abs{G}^2) \over 16s m_e^2 m_\chi^2}
\Re[\Delta_{\tilde{\nu}}(t)]
\left[(s + t - m_\chi^2)^2 + s m_\chi^2\right]

\nel & = &
-{e^2 g^2 \cos^2\theta_X \over 32 s m_e^2 m_\chi^2}
\Re[\Delta_{\tilde{\nu}}(t)]
\left[(s + t - m_\chi^2)^2 + s^2\rho \right]
.
\ee

In a similar way $X_{Z\tilde\nu}$ becomes
\beml{charge-d-gamma-nu1}
X_{Z\tilde\nu} & \approx &
{(s + t - m_\chi^2)^2 \over 16 m_e^2 m_\chi^2}
\Re[\Delta_Z(s) \Delta_{\tilde{\nu}}^\hc(t)]
\nel && \qquad {} \times
[(a_Z A_Z + b_Z B_Z)(\abs{F}^2  + \abs{G}^2) - 2\Re(FG^\hc)(b_Z A_Z + a_Z B_Z)]
\nel && {} +
{s \over 16 m_e^2}
\Re[\Delta_Z(s) \Delta_{\tilde{\nu}}^\hc(t)]
\nel && \qquad {} \times
[(a_Z A_Z - b_Z B_Z)(\abs{F}^2 + \abs{G}^2) - 2\Re(FG^\hc)(b_Z A_Z - a_Z B_Z)]
\nel
& = &
{1 \over 16 m_e^2 m_\chi^2 } \Re[\Delta_Z(s) \Delta_{\tilde{\nu}}^\hc(t)]
\nel && {}\times
{g^2 \cos^2\theta_X \over 2}
\Bigl[
(s + t - m_\chi^2)^2(a_Z A_Z + b_Z B_Z - b_Z A_Z - a_Z B_Z)
\nel && {}\qquad
+
sm_\chi^2(a_Z A_Z - b_Z B_Z - b_Z A_Z + a_Z B_Z)
\Bigr]
\nel
& = &
{a_Z - b_Z \over 16 m_e^2 m_\chi^2 } \Re[\Delta_Z(s) \Delta_{\tilde{\nu}}^\hc(t)]
\nel && {}\times
{g^2 \cos^2\theta_X \over 2}
\Bigl[
(s + t - m_\chi^2)^2(A_Z - B_Z)
+
sm_\chi^2(A_Z + B_Z)
\Bigr]
.
\ee

From~\rf{chargeZabIs} and~\rf{chargeZABIs} I have
\be
a_Z - b_Z  =  {g \sin^2 \theta_W \over \cos \theta_W},
\ee

and
\bem
A_Z + B_Z &= &
{g \over 4\cos\theta_W}
\{[2\cos (2\theta_W) + \abs{Y_{11}}^2 + \abs{X_{11}}^2]
- [\abs{Y_{11}}^2 - \abs{X_{11}}^2]\}
\nel
& = &
{g \over 2\cos\theta_W}[\cos(2\theta_W) + \abs{X_{11}}^2]
=
{g \over 2\cos\theta_W}[\cos(2\theta_W) + \cos^2\theta_X]
,
\nel
A_Z - B_Z & = &
{g \over 2\cos\theta_W}[\cos(2\theta_W) + \abs{Y_{11}}^2]
=
{g \over 2\cos\theta_W}[\cos(2\theta_W) + \cos^2\theta_Y],
\ee

Thus
\beml{charge-d-Z-nu}
X_{Z\tilde\nu} & \approx &
{g^4 \tan^2 \theta_W \cos^2\theta_X \over 64 m_e^2 m_\chi^2}
 \Re[\Delta_Z(s) \Delta_{\tilde{\nu}}^\hc(t)]
\nel && {} \times
\Bigl[
(s + t - m_\chi^2)^2(\cos(2\theta_W) + \cos^2\theta_Y)
+
sm_\chi^2(\cos(2\theta_W) + \cos^2\theta_X)
\Bigr]
\nel
.
\ee

\subsection{Total cross section}

I have for the total cross-section from~\rf{csTotalApproximation},
\bel{chargeCross}
\sigma^T_{e\chi}(s)
= {m_e^2 m_\chi^2 \over \pi s^2} \int_{t_-}^{t_+}
X_{e\chi}(s, t) \, dt
= \sigma^T_{\gamma\gamma}
+ \sigma^T_{ZZ}
+ 2\sigma^T_{\gamma Z}
+ \sigma^T_{\tilde{\nu}\tilde{\nu}}
+ 2\sigma^T_{\gamma \tilde{\nu}}
+ 2\sigma^T_{Z \tilde{\nu}}
,
\ee

with the limits of integration
\bel{charge-t-plus-minus-is}
t_{\pm} \approx {s \over 2} (2\rho - 1 \pm \sqrt{1 - 4\rho})
.
\ee

I use \rf{csInt1}--\rf{csInt65} to calculate the integrals
in~\rf{chargeCross} together with the
expressions~\rf{chargeSneutrinoDelta} and~\rf{ZPropogator} for
$\Delta_Z(s)$ and $\Delta_{\tilde{\nu}}(t)$,
\be
\Delta_Z(s)
= {1 \over s - \mB^2 + i\mB\Gamma_Z}
= {s - \mB^2 - i\mB\Gamma_Z \over (s - \mB^2)^2 + \mB^2\Gamma_Z^2},
\ee

\bel{charge-x-is}
\Delta_{\tilde{\nu}}(t) = {1 \over t - x} ,
\quad x =  m_{\tilde{\nu}}^2 - im_{\tilde{\nu}}\Gamma_{\tilde{\nu}}
.
\ee

I obtain for the photon-channel contribution from~\rf{charge-d-gamma-gamma}

\bem
\sigma^T_{\gamma\gamma} & = &
{m_e^2 m_\chi^2 \over \pi s^2} \int_{t_-}^{t_+} X_{\gamma\gamma} dt
\nel
&\approx&
{e^4 \over 4\pi s^4 }
\int_{t_-}^{t_+}[(m_\chi^2 - t)^2  + s m_\chi^2
+ {1 \over 2} s(s + 2t - 2m_\chi^2)]dt
\nel&=&
 {e^4 \over 4\pi s^4 }
[{s^3 \over 3} (1 - \rho)\sqrt{1 - 4\rho} + s^2 \rho(t_+ - t_-) + 0]
\nel&=&
{e^4 \over 4\pi s^4 }
[{s^3 \over 3} (1 - \rho)\sqrt{1 - 4\rho} + s^3 \rho\sqrt{1 - 4\rho}]
,
\ee

where the integration was carried out with the help
of~\rf{csInt2} and~\rf{csInt1}. So I find

\bel{charge-total-gamma-gamma}
\sigma^T_{\gamma\gamma}(s) =
{e^4 \over 4\pi s} \sqrt{1 - 4\rho}{1 + 2\rho \over 3}
.
\ee

Similarly for the $Z^0$-channel contribution, the equation~\rf{charge-d-Z-Z}
yields
\bem
\sigma^T_{ZZ} & = &
{m_e^2 m_\chi^2 \over \pi s^2} \int_{t_-}^{t_+} X_{ZZ} dt
\nel
& \approx & {a_Z^2 + b_Z^2 \over 4\pi s^2}\abs{\Delta_Z(s)}^2
\int_{t_-}^{t_+}[
(m_\chi^2 - t)^2 (A_Z^2 + B_Z^2) + s m_\chi^2 (A_Z^2 - B_Z^2)
]dt + 0
,
\ee

where again the term in $X_{ZZ}$ proportional to $(s + 2t - 2m_\chi^2)$
does not contribute to the integral. Thus,
\bem
\sigma^T_{ZZ} & = &
{a_Z^2 + b_Z^2 \over 4\pi s^2}\abs{\Delta_Z(s)}^2
\biggl[{s^3 \over 3} (1 - \rho)\sqrt{1 - 4\rho}(A_Z^2 + B_Z^2)
\nel&& {} \qquad
  + s^3\rho \sqrt{1 - 4\rho}(A_Z^2 - B_Z^2)
\biggr]
\nel
& = &
{a_Z^2 + b_Z^2 \over 4\pi} s\abs{\Delta_Z(s)}^2
\left[{1 \over 3} (1 - \rho)(A_Z^2 + B_Z^2)
  + \rho (A_Z^2 - B_Z^2)
\right]\sqrt{1 - 4\rho},
\ee

so
\bel{charge-total-Z-Z}
\sigma^T_{ZZ}(s) =
{a_Z^2 + b_Z^2 \over 4\pi} s\abs{\Delta_Z(s)}^2\sqrt{1 - 4\rho}
\left[A_Z^2{1 + 2\rho \over 3} + B_Z^2{1 - 4\rho \over 3}\right].
\ee

The calculation of $\sigma^T_{\gamma Z}$ is also straightforward.
I obtain from~\rf{charge-d-gamma-Z}
\bem
\sigma^T_{\gamma Z}(s) & = &
{m_e^2 m_\chi^2 \over \pi s^2} \int_{t_-}^{t_+} X_{Z\gamma} dt
\nel
&\approx&
-{e^2 a_Z A_Z \over 4\pi s^3} \Re(\Delta_Z(s))
\int_{t_-}^{t_+}[(m_\chi^2 - t)^2 + sm_\chi^2]dt
\nel&=&
-{e^2 a_Z A_Z \over 4\pi s^3} \Re(\Delta_Z(s))
\left[{s^3 \over 3} (1 - \rho)\sqrt{1 - 4\rho}
   + s^3 \rho\sqrt{1 - 4\rho}\right],
\ee

or
\bel{charge-total-gamma-Z}
\sigma^T_{\gamma Z}(s)
=
-{e^2 a_Z A_Z \over 4\pi} \Re(\Delta_Z(s))
\sqrt{1 - 4\rho}{1 + 2\rho \over 3}
.
\ee

\P
Calculations of the terms in~\rf{chargeCross} that involve scalar neutrinos
is more complicated due to the integrations over $\Delta_{\tilde\nu}(t)$-like
terms. By the equation~\rf{charge-d-nu-nu}

\bem
\sigma^T_{\tilde{\nu}\tilde{\nu}} & = &
{m_e^2 m_\chi^2 \over \pi s^2} \int_{t_-}^{t_+} X_{\tilde\nu\tilde\nu} dt
\nel
&\approx &

{g^4 \cos^4\theta_X\over 64 \pi s^2}
\int_{t_-}^{t_+} \abs{\Delta_{\tilde\nu}(t)}^2(s + t - m_\chi^2)^2 dt
\nel
& = &
{g^4 \cos^4\theta_X\over 64 \pi s^2}
\int_{t_-}^{t_+}
{1 \over \abs{t - x}^2}[t - s(\rho - 1)]^2
dt.
\ee

According to~\rf{csInt65} and from the definition~\rf{charge-x-is} of $x$ I get
\beml{charge-total-nu-nu}
\sigma^T_{\tilde{\nu}\tilde{\nu}} & \approx  &
{g^4 \cos^4\theta_X\over 64 \pi s^2}
\Biggl\{
{[\Re(x) + s(1 - \rho)]^2 - \Im(x)^2 \over \Im(x)}
\phi(x)
\nel&&
\qquad{}
+ 2[\Re(x) + s(1 - \rho)]\ln\abs{t_+ - x \over t_- - x}
+ s\sqrt{1 - 4\rho}
\Biggr\}
\nel & \equiv  &
{g^4 \cos^4\theta_X\over 64 \pi s^2}
\Biggl\{
-{[m_{\tilde{\nu}}^2 + s(1 - \rho)]^2
    - m_{\tilde{\nu}}^2\Gamma_{\tilde{\nu}}^2
  \over m_{\tilde{\nu}}\Gamma_{\tilde{\nu}}}
\phi(x)
\nel&&
\qquad{}
+ 2[m_{\tilde{\nu}}^2 + s(1 - \rho)]\ln\abs{t_+ - x \over t_- - x}
+ s\sqrt{1 - 4\rho}
\Biggr\}
,
\ee

where $\phi(x)$ is defined in~\rf{csImLog}. In this case it becomes
\bem
\phi(x) & = & \arctg{s\sqrt{1 - 4\rho}\Im(x)
         \over m_\chi^4 + \abs{x}^2 - s(2\rho - 1)\Re(x)}
\nel
& \equiv &
-
\arctg{s\sqrt{1 - 4\rho}\, m_{\tilde{\nu}}\Gamma_{\tilde{\nu}}
         \over s^2 \rho^2
            + m_{\tilde{\nu}}^2[m_{\tilde{\nu}}^2 + \Gamma_{\tilde{\nu}}^2
             + s(1 - 2\rho)]}
.
\ee

\P
Finally I consider the interference between the bosons and sneutrino channels.
To find $\sigma^T_{\gamma\tilde{\nu}}$ from~\rf{charge-d-gamma-nu}
I apply~\rf{csInt3} and~\rf{csInt4}
in the case when $z = 1$, so
\bem
\sigma^T_{\gamma\tilde{\nu}}
& = &
{m_e^2 m_\chi^2 \over \pi s^2} \int_{t_-}^{t_+} X_{\gamma\tilde\nu} dt
\nel
& \approx &
-{e^2 g^2 \cos^2\theta_X \over 32\pi s^3 }
\int_{t_-}^{t_+}
\Re[\Delta_{\tilde{\nu}}(t)]
\left[(s + t - m_\chi^2)^2 + s m_\chi^2\right]
dt
\nel
& = &
-{e^2 g^2 \cos^2\theta_X \over 32\pi s^3 }
\int_{t_-}^{t_+}
\Re\left({ 1 \over t - x}\right)
\left\{[t - s(\rho - 1)]^2 + s m_\chi^2\right\}
dt
\nel
& = &
-
{e^2 g^2 \cos^2\theta_X \over 32\pi s^3 }
\Biggl\{
\Re[(x - s(\rho - 1))^2] \ln\abs{t_+ - x \over t_- - x}
\nel
&&\qquad{}
+ \Im[((x - s(\rho - 1))^2] \phi(x)
\nel
&&\qquad{}
+ s\sqrt{1 - 4\rho}[\Re(x) - 2s(\rho - 1) - s(1 - 2\rho)/2]
+ s^2\rho\ln\abs{t_+ - x \over t_- - x}
\Biggr\}
,
\ee

Thus
from~\rf{csReImForSub} I have,

\beml{charge-total-gamma-nu}
\sigma^T_{\gamma\tilde{\nu}} & \approx  &
-{e^2 g^2 \cos^2\theta_X \over 32\pi s^3 }
\Biggl\{
[(\Re(x) + s(1 - \rho))^2 - \Im(x)^2 + s^2\rho]
\ln\abs{t_+ - x \over t_- - x}
\nel
&&\qquad{}
- 2\Im(x)[\Re(x) + s(1 - \rho)] \phi(x)

+ s\sqrt{1 - 4\rho}[\Re(x) - s\rho + {3 \over 2}s]
\Biggr\}
\nel
& \equiv &
-{e^2 g^2 \cos^2\theta_X \over 32\pi s^3 }
\Biggl\{
[(m_{\tilde\nu}^2 + s(1 - \rho))^2
- m_{\tilde\nu}^2\Gamma_{\tilde\nu}^2 + s^2\rho]
\ln\abs{t_+ - x \over t_- - x}
\nel
&&\qquad{}
+ 2m_{\tilde\nu}\Gamma_{\tilde\nu}[m_{\tilde\nu}^2 + s(1 - \rho)] \phi(x)

+ s\sqrt{1 - 4\rho}[m_{\tilde\nu}^2 - s\rho + {3 \over 2}s]
\Biggr\}
.
\ee

To calculate $\sigma^T_{Z\tilde{\nu}}$ I rewrite it from~\rf{charge-d-Z-nu} as
\beml{charge-total-Z-nu}
\sigma^T_{Z\tilde{\nu}} & = &
{m_e^2 m_\chi^2 \over \pi s^2} \int_{t_-}^{t_+} X_{Z\tilde\nu} dt
\nel
& \approx &
{g^4 \tan^2 \theta_W \cos^2\theta_X \over 64 \pi s^2}
\int_{t_-}^{t_+} \Re\left[\Delta_Z^\hc(s) { 1 \over t - x}\right]
\nel &&{}\times
\Bigl\{
[t - s(\rho - 1)]^2[\cos(2\theta_W) + \cos^2\theta_Y]
+
s^2\rho[\cos(2\theta_W) + \cos^2\theta_X]
\Bigr\}
dt
\nel
& = &
{g^4 \tan^2 \theta_W \cos^2\theta_X \over 64 \pi}
\{f_1[\cos(2\theta_W) + \cos^2\theta_Y]
    + f_2[\cos(2\theta_W) + \cos^2\theta_X]\}
.
\nel
\ee

The formulas~\rf{csInt3} and~\rf{csInt4} give for $f_1$ and $f_2$,

\bem
f_1 & = & {1 \over s^2}\int_{t_-}^{t_+}
 \Re\left[\Delta_Z^\hc(s){ 1 \over t - x}\right][t - s(\rho - 1)]^2 dt
\nel
& = &
\Re\{z[x/s + 1 - \rho]^2\} \ln\abs{t_+ - x \over t_- - x}
-\Im\{z[x/s + 1 - \rho]\} \phi(x)
\nel
&&{}
+ {1 \over s}
  \sqrt{1 - 4\rho}\{\Re(zx) - \Re(z)[2s(\rho - 1) + s(1 - 2\rho)/2]\}
\nel
& = &
\Re\{z[x/s + 1 - \rho]^2\} \ln\abs{t_+ - x \over t_- - x}
-\Im\{z[x/s + 1 - \rho]^2\} \phi(x)
\nel
&&{}
+ \sqrt{1 - 4\rho}[\Re(zx/s) + ({3\over 2} - \rho)\Re(z)]
,
\nel
f_2 & = & \rho \int_{t_-}^{t_+}
 \Re\left[\Delta_Z^\hc(s){ 1 \over t - x}\right] dt
=
\rho\left[\Re(z)\ln\abs{t_+ - x \over t_- - x}
- \Im(z) \phi(x)\right],
\ee

where
\be
z = \Delta_Z^\hc(s) = {1 \over s - \mB^2 - i\mB\Gamma_Z}
.
\ee


%!TEX root = ../notes.tex
\section{February 4, 2022}
\subsection{Fast Powering \emph{continued}}
\begin{example}
    \recall we wanted to compute $3^{37}\mod 100$
    \begin{align*}
        3^1    & \equiv 3\pmod{100} \\
        3^2    & \equiv 9           \\
        3^4    & \equiv 81          \\
        3^8    & \equiv 61          \\
        3^{16} & \equiv 21          \\
        3^{32} & \equiv 41
    \end{align*}
    so we have
    \[37=1+4+32 \qquad 3^{37} = 3^1\cdot 3^4\cdot 3^{32} \equiv 3\cdot 81\cdot 41\equiv 63\]
\end{example}
How might we do this as an algorithm? We want to keep track of a few things, such as $g$ (the current power), $p$ (the multiple we are building), $a$ (the remaining powers). This is akin to \emph{deconstructing the power in binary and composing our product}.

\begin{algorithm}[Fast Powering Algorithm]
    ~\lstinputlisting[numbers=none,language=Python]{code/pow_mod.py}
\end{algorithm}

\begin{example}
    $37=100101_2$, so we peel off last digits and multiply $g$ into $p$.

    Thinking about iterations, we have
    \[\begin{array}{llll}
            g  & p          & a  & a_2                \\ \hline
            3  & 1          & 37 & 10010\underline{1} \\
            9  & 3          & 18 & 10010\underline{0} \\
            81 & 3          & 9  & 100\underline{1}   \\
            61 & 43         & 4  & 10\underline{0}    \\
            21 & 43         & 2  & 1\underline{0}     \\
            41 & 43         & 1  & \underline{1}      \\
               & \boxed{63} & 0  & \underline{0}
        \end{array}\]
\end{example}

This algorithm takes approximately $\log_2(a)$ time to run, since it does as many steps for each digit in the binary representation of $a$.

\subsection{Fun Integers}
\recall An integer $p$ is \ul{prime} if $p\geq 2$ and
\[a\mid p\Rightarrow a = \pm 1, \pm p\]

\begin{proposition}
    Let $p$ be prime. Then $p\mid ab\Rightarrow p\mid a$ or $p\mid b$.
\end{proposition}
\begin{example}
    $p$ is not prime, this doesn't work. $p = 6$. $p\mid 4\cdot 9 = 36$
    but $6\nmid 4$ and $6\nmid 9$.
\end{example}

\begin{proof}
    Let $g = \gcd(p, a)$. $g$ is either $1$ or $p$.

    If $g = p$, then we have that $p = g\mid a$.

    If $p = 1$, we can write this as
    \begin{align*}
        1 = g & = p\cdot u + a\cdot v   \\
        b     & = p\cdot ub + ab\cdot v
    \end{align*}
    since $p$ is a multiple of $p$ and $ab$ is a multiple of $p$, we have that $p\mid b$.
\end{proof}

\begin{theorem}[Fundamental Theorem of Arithmetic]
    Any integer $a\geq 1$ can be factored into product of primes
    \[a = p_1^{e_1}\cdots p_n^{e_n}\]
    and this product of primes is \emph{unique} up to rearrangement.\footnote{This is to say, $\ZZ$ is a UFD! }
\end{theorem}
\begin{example}
    Instead of thinking about integers, we think about $\ZZ[\sqrt{-5}]$, like
    \[\ZZ[\sqrt{-5}] = \{a + b\sqrt{-5}\mid a, b\in \ZZ\}\]
    Consider
    \[6 = (1 + \sqrt{-5})(1-\sqrt{-5}) = 2\cdot 3\]
    and each of $(1 + \sqrt{-5})$, $(1-\sqrt{-5})$, $2$, $3$ have no divisors besides themselves and $\pm 1$ (units).
\end{example}
\begin{proof}
    We begin by working out an example:
    \begin{example}
        Let's factor $60$, we can write this as \[60=6\cdot 10 = (2\cdot 3)\cdot (2\cdot 5) = 2^2\cdot 3\cdot 5.\]
    \end{example}
    What if we had different answers
    \[p_1p_2\cdots p_t = a = q_1q_2\cdots q_s\]
    We have that
    \begin{align*}
        p_1\mid p_1\cdots p_t & = q_1\cdots q_s      \\
                              & = q_1(q_2\cdots q_s)
    \end{align*}
    So we have that $p_1\mid q_1$ or $p_1\mid q_2\cdots q_s$, and we go on. So $p_1$ has to divide \emph{one} of $q_i$. But both are primes, so they are equal $p_1 = q_i$. We rearrange so $q_i$ is $q_1$. We strip off $p_1$ and $q_1$ and we have
    \[p_2\cdots p_t = q_2\cdots q_s\]
    we continue until we have no factors left\footnote{We could also have taken a well-ordering approach to this statement, taking $a$ to be the least such non-uniquely factorizable number and showing that by peeling off $p_1$ and $q_1$, we get a smaller such $a$, which is a contradiction. }
\end{proof}

\begin{definition}[Order]\label{defn:order-of-prime-factor}
    We define the \ul{order}
    \[\ord_p(a) = \text{the power of $p$ in the factorization of $a$}\]
    such that we have
    \[a = \prod_p p^{\ord_p(a)}\]
    \emph{(This makes sense since $\ord_p(a)$ is finite for finitely many $p$.)}
\end{definition}

\begin{theorem}[Fermat's Little Theorem]\label{theorem:flt}
    Let $p$ be prime, $a\in \ZZ/p\ZZ$,
    \[a^{p-1}\equiv \begin{cases}
            0 & \text{if $a\equiv 0$} \\
            1 & \text{otherwise}
        \end{cases}\]
\end{theorem}
In abstract algebra, this directly follows from Lagrange's Theorem for $\ZZ/p\ZZ$, we give another argument.

\begin{proof}
    If $a\equiv 0$, this is sufficiently clear.

    Let $a\not\equiv 0$. We look at the numbers
    \[a, 2a, 3a, \dots, (p-1)a\]
    We consider 2 questions:
    \begin{enumerate}[i.]
        \item Are any of these divisible by $p$?

              No! $p\nmid a$ and $p\nmid i$ so $p\nmid ia$ for $1\leq i < p$.
        \item Are any of these equal? i.e. $ia\equiv ja\mod p$.

              No again! $a$ has an inverse mod $p$.
    \end{enumerate}
    So we have that this list is a permutation of $\{1, 2, \dots, p-1\}$, that is,
    \[\{1, 2, \dots, p-1\} = \{a, 2a, \dots, (p-1)a\}\mod p\]
    we multiply these sets together\footnote{This is truly a pro-gamer move},
    \begin{align*}
        1\cdot 2\cdot 3\cdots (p-1) & \equiv a\cdot 2a\cdots (p-1)a \mod p \\
                                    & \equiv (1\cdot 2\cdots p-1)a^{p-1}
        1\cdot 2\cdot 3\cdot (p-1)(a^{p-1}-1)\equiv 0\mod p                \\\implies a^{p-1}&\equiv 1\mod p.
    \end{align*}
    Which is as desired.
\end{proof}


%----------------------------------------------------------------------
% COMPUTATIONAL PHYSICS - PROJECT 5
% GRAVITATIONAL LENSING BY POINT MASSES
%----------------------------------------------------------------------

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\title[Gravitational Lensing]{Gravitational Lensing by Point Masses}
\author{Michael Papasimeon}
\date{29 October 1998}
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% BEGIN DOCUMENT
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\begin{document}
\maketitle

\begin{frame}{Gravitational Lens in Abbell 2218}
    \begin{center}
        \includegraphics[width=0.7\textwidth]{images/hubble.eps}
    \end{center}
\end{frame}

\begin{frame}{The Lens Equation}
    The angular deflection $\alpha$ of a gravitational lense is given by
    \begin{equation}
        \alpha = \frac{4GM}{c^2 \xi}
    \end{equation}
    where
    \begin{itemize}
        \item $M = $ mass of the deflecting object
        \item $G = $ gravitational constant
        \item $c = $ speed of light
        \item $\xi = $ impact radius of the incoming photon
        \item $\alpha = $ deflection angle
    \end{itemize}
\end{frame}

\begin{frame}{Angular Deflection of a Photon}
    \begin{center}
        \includegraphics[width=0.7\textwidth]{images/gldiag.eps}
    \end{center}
\end{frame}

\begin{frame}{Deflecting Mass}
    The deflecting mass may be viewed as an optical thin lens, made up
    of a two dimensional mass distribution.
    \begin{equation}
        M = \int_{R^2} \Sigma(\vec{\xi '})
            \frac{\vec{\xi} - \vec{\xi '} }{|\vec{\xi} - \vec{\xi '}|^2} d^2 \xi '
    \end{equation}
    where
    \begin{itemize}
        \item $\Sigma = $ surface density
        \item $R^2 = $ surface area
    \end{itemize}
    If the deflecting mass, is perfectly symmetric, the deflection angle becomes
    \begin{equation}
        \vec{\alpha} = \frac{4GM(<\xi)\vec{\xi}}{c^2|\vec{\xi}|^2}
    \end{equation}
\end{frame}

\begin{frame}{The Lens Equation}
    The lens equation is
    \begin{equation}
        \psi D_{os} + \alpha D_{ds} = \beta D_{os}
    \end{equation}
    It is often the case in many gravitational lensing problems that
    the images form do not depend on the distances between source, observer
    and deflector directly. Rather a specific ratio of these distances
    is the quantity which needs to be considered. This is known as the
    effective distance $D$ and is defined as:
    \begin{equation}
    \label{eq:effd}
        D = \frac{D_{od} D_{ds} }{D_{os}}
    \end{equation}
\end{frame}

\begin{frame}{Setup for the Gravitational Lens Equation}
    \begin{center}
        \includegraphics[angle=-90,width=0.8\textwidth]{images/setup.eps}
    \end{center}
\end{frame}

\begin{frame}{Einstein Ring}
    If a distant star is in perfect alignment with a point mass gravitational
    lens and an observer, the light from the distant star is lensed
    perfectly symmetrically forming a ring image of the star known as an
    Einstein ring. The radius of the Einstein ring is given by
    \begin{equation}
        \theta_{E} = \sqrt{\frac{4GM}{c^2} \frac{D_{ds}}{D_{od} D_{os}}}
    \end{equation}
    If however there isn't a perfect alignment, for a point mass two
    images are produced for each point on the source plane.
    The angular positions at which these images is given by:
    \begin{equation}
        \theta_{\pm} = \frac{\xi}{D_{od}}
                 = \frac{\beta}{2} \pm \sqrt{\beta^2 + 4\theta^{2}_{E}}
    \end{equation}
    where $\beta$ is the angular position of the source as shown in figure 3.
    The magnification of each of the images is given by
    \begin{equation}
        \mu_{\pm} = \frac{1}{4}
                \left[
                    \frac{y}{\sqrt{y^2 + 4}}
                    +
                    \frac{\sqrt{y^2 + 4}}{y}
                    \pm 2
                \right]
    \end{equation}
    where the source and image angles have been scaled such that
    $y = \psi / \alpha_0$ and $x = \alpha / \alpha_0$.
\end{frame}

\begin{frame}{Reformulating the Lens Equation}
Using this notation the lens equation can be rewritten as the scaled
lens equation given by
\begin{equation}
    \vec{y} = \vec{x} - \vec{\alpha}(\vec{x}).
\end{equation}
The vector notation used represents the coordinate in a cartesian plane.
Therefore
\begin{itemize}
    \item $\vec{y} = (y_1, y_2)$ is a coordinate in the source plane and
    \item $\vec{x} = (x_1, x_2)$ is a coordinate in the deflecting plane.
\end{itemize}
\end{frame}

\begin{frame}{Complex Notation}
This can also be written using complex number notation. A position in
deflecting plane can be denoted as $z = x_1 + ix_2$, and a position in the
source plane as $z_s = y_1 + iy_2$.

The magnification is given by:
\begin{equation}
    \mu(\vec{x}) = \frac{1}{det A(\vec{x})}
\end{equation}
where $det A(\vec{x})$ is the Jacobian determinant of the Hessian matrix
given by
\begin{equation}
    A(\vec{x}) = \frac{\partial \vec{y}}{\partial \vec{x}}
\end{equation}
\end{frame}

\begin{frame}{Critical Curves and Caustics}
\begin{itemize}
    \item Critical curves are the set of all points in the deflection plane where $A(\vec{x}) = 0$.
    \item The corresponding curves in the source plane (obtained from the lens equation) are known as caustics.    \item Critical curves are where the gravitational lens infinitely magnifies the light passing through
          that point.
    \item Source plane caustics are the pre-image of the critical curves.
    \item Any light emitted near a caustic will be greatly magnified as it passes through the lens.
\end{itemize}
\end{frame}

\begin{frame}{The Chang-Refsdal Lens}
The Chang-Refsdal lens model describes gravitational lensing using a modification
of the point mass model. This model says that when a source crosses a fold caustic
the lensing is due to a point mass but with an additional external shear applied.
The corresponding lens equation for the Chang-Refsdal model is:
\begin{equation}
    \vec{y} = \left[
                \begin{array}{cc}
                    1 + \gamma & 0   \\
                    0          & 1 - \gamma
                \end{array}
              \right] \vec{x} - \frac{\vec{x}}{|\vec{x}|^2}
\end{equation}
This can also be written in the complex notation as
$z_s = z + \gamma \bar{z} - \frac{\epsilon}{\bar{z}}$, where
\begin{itemize}
    \item $\gamma$ is a constant determining the amount shear.
    \item $\epsilon = (\frac{\kappa_s}{1 - \kappa_c})$, where $\kappa_s$ is
          the density of compact objects such as stars and $\kappa_c$ is the
          surface density of continuously distributed matter.
\end{itemize}
\end{frame}

%\end{multicols}

\begin{frame}{The Effects of Distance on Gravitational Lensing}
The Dyer-Roeder equation is given by:
\begin{equation}
    (z+1)(\Omega z + 1) \frac{d^2D}{dz^2} +
    \left( \frac{7}{2}\Omega z + \frac{1}{2}\Omega + 3 \right) \frac{dD}{dz} +
    \frac{3}{2} \tilde{\alpha} \Omega D = 0
\end{equation}
This relates the angular diameter distance of a lensing system with the
redshift $z$ of the source object.
\begin{itemize}
    \item $\Omega$ is the ratio of the mean mass density to the critical
          density of the universe and
    \item $\tilde{\alpha}$ is the clumpiness parameter, which determines
          the amount of matter between the source and the observer.
\end{itemize}
\end{frame}

\begin{frame}{Initial Conditions for Dyer-Roeder Equation}
The initial conditions of the Dyer-Roeder equation are:
\begin{equation}
    D_{ii} = 0
\end{equation}
\begin{equation}
    \left[ \frac{dD_{ij}}{dz} \right]_{z_j = z_i} =
         \frac{\mathrm{sgn} (z_j - z_i)}{(z_i + 1)^2 \sqrt{\Omega z_i + 1} }
\end{equation}

The Dyer-Roeder equation needs to be solved for three different cases.
\begin{enumerate}
    \item  $\Omega = 0$ $(D_{I})$
    \item  $\Omega = 1$, $\tilde{\alpha} = 1$ $(D_{II})$
    \item  $\Omega = 1$, $\tilde{\alpha} = 0$ $(D_{III})$
\end{enumerate}
\end{frame}

\begin{frame}{Solving the Dyer-Roeder Equation}
This leads to three different equations which need to be solved.
\begin{equation}
    (z+1)\frac{d^2D}{dz^2} + 3 \frac{dD}{dz} = 0
\end{equation}

\begin{equation}
    (z+1)^2 \frac{d^2D}{dz^2} + \frac{7}{2}(z+1)\frac{dD}{dz} + \frac{3}{2}D = 0
\end{equation}

\begin{equation}
    (z+1)^2 \frac{d^2D}{dz^2} + \frac{7}{2}(z+1)\frac{dD}{dz} = 0
\end{equation}
\end{frame}

\begin{frame}{Computational Solution to the Dyer-Roeder Equation}
\begin{itemize}
\item A FORTRAN program was written
to numerically solve these equations.
\item The algorithm used to solve these
equation numerically was the \emph{Runge-Kutta-Nystr\"{o}m}
method.
\item This method is a fourth order algorithm which is a general form
of the standard \emph{Runge-Kutta} method, used for solving second order
ordinary differential equations.
\item The equations were integrated from $z = 0$ to $z = 10$.
\item The dimming factor $(D_{II}/D_{III})^2$ was determined and plotted
as a function of redshift $z$.
\end{itemize}
\end{frame}

\begin{frame}{Solutions to the Dyer-Roeder Equation}
We solve the Dyer-Roeder
equation for the three different cosmologies resulting in three
solutions for the angular diameter distance $D$.
\begin{itemize}
    \item $D_{I}$   \texttt{d1.dat} (The top curve)
    \item $D_{II}$  \texttt{d2.dat} (The bottom curve)
    \item $D_{III}$ \texttt{d3.dat} (The centre curve)
\end{itemize}
    \begin{center}
        \includegraphics[width=0.5\textwidth]{results/cosmo_all.eps}
    \end{center}
\end{frame}

\begin{frame}{Increasing Redshift}
\begin{itemize}
\item According to the big bang model of the universe, objects with large
redshifts are further away.
\item Hence the results of solving the Dyer-Roeder
equation for the cases (1 and 3) when the ``clumpiness'' parameter
$\tilde{\alpha}$ is ignored the angular diameter distance $D$ increases
as the redshift increases.
\item In both cases we get asymptoting values:
\end{itemize}
\begin{equation}
    \lim_{z \rightarrow \infty} D_{I}(z) = 0.5
\end{equation}
\begin{equation}
    \lim_{z \rightarrow \infty} D_{II}(z) = 0.4
\end{equation}
\end{frame}
%In both these cases where $\tilde{\alpha}$ is not in the equation,
%the lens is the only matter along the line of sight to the source and
%hence we get an increase in the angular diameter distance as the redshift
%increases. Of particular interest is that in case $I$ ($\Omega = 0$),
%the value of $D$ increases more rapidly than in case $III$ ($\Omega = 1$).
%Hence we see that the density of the universe plays influences the results
%of any models chosen for gravitational lensing. Since the mean and
%critical densities of the universe are unknown (especially since it is
%believed that most of the matter of the universe is dark matter),
%there are limits to how accurately we can determined parameters from
%gravitational lensing observations.

%The second case is perhaps the most interesting of the three cosmologies
%in that we have a non-zero contribution from the ``clumpiness'' parameter
%$\tilde{\alpha}$. As a result is also produces the most interesting result
%as we have a maxiumum in $D$ at $z \simeq 1$.
%In this case where $\tilde{\alpha} = 1$, the gravitational lense
%make only a minor contribution to the total amount of smoothly distributed
%matter in the line of sight of the observer and the source. In this situation
%the matter is smootly distributed and we get gravitational lens being
%the matter between source and observer. We can then think of this
%as the light passing through a material of a different refractive index
%as in electrodynamics.

\begin{frame}{Dimming factor against source redshift}
    \begin{center}
        \includegraphics[width=0.7\textwidth]{results/dim.eps}
    \end{center}
\end{frame}

%Figure~\ref{fig:dimming} shows the dimming factor $(D_{II}/D_{III})^2$ plotted
%against source redshift $z$. In case $II$ has a smoothly distributed
%matter distribution meaning that we see more and brigter images than
%in case $III$ where the ``clumpiness'' parameter results in less
%light reaching the observer. As a result, a case $III$ universe appears
%to be dimmer than a case $II$ universe. The dimming ratio of factor
%is $(D_{II}/D_{III})^2$ and is plotted against red shift in the plot above.
%As we can see in the plot the higher the redshift the more dimness in
%a case $III$ universe.

%\section{Lensing an Extended Source by a Point Mass}
%    \subsection{Method}
%    A program was written to simulate the gravitational lensing
%    of a two dimensional grid. Two make the results of the simulation
%    more interesting the two dimensional grid chosen was represented
%    as a two dimensional array of integers ranging from 0 to 255.
%    The value at each array element represents an intensity value. As a result
%    the grid represents a two dimensional Portable Grey Map (PGM) image.
%    The program was written to accept the name of a PGM file on the command
%    line, load the image into memory,, apply the gravitational lensing calculations
%    and output a new ``lensed'' PGM image to standard output. The program
%    also takes the mass $M$ (in $kg$) of the lensing body, the effective
%    distance D specified in equation~\ref{eq:effd}, and the $(x,y)$
%    location of the lensing body.
%    The details of the lensing algorithm can be found in Algorithm~\ref{alg:lense}.

\begin{frame}{Lensing an Extended Source by a Point Mass}
        \begin{block}{Lensing Algorithm}
        \begin{algorithmic}[1]
            \Procedure {Lense}{}
                \State $img \leftarrow \textbf{load\_image}()$
                \State $x_c, y_c \leftarrow \textbf{find\_centre}(img)$
                \State $\theta_E \leftarrow \textbf{calc\_einstein\_radius}()$
                \For {$c \in img.coordinates$}
                    \State $b_x, b_y \leftarrow \textbf{calc\_location\_lensing\_body}()$
                    \State $\beta \leftarrow \textbf{impact\_radius}()$
                    \State $\textbf{calc\_angle\_for\_each\_quadrant}()$
                    \State $\theta_{\pm} \leftarrow \textbf{calc\_new\_angles}()$
                    \State $c \leftarrow \textbf{calc\_new\_deflected\_coordinate}()$
                \EndFor
                \State $\textbf{save\_image(img)}$
            \EndProcedure
        \end{algorithmic}
        \end{block}
\end{frame}

%    \begin{verbatim}
%        load the image
%        calculate the Einstein radius
%        for all coordinates in the image
%           calculate the location of the lensing body
%           calculate the impact radius beta
%           determine the correct angle for each quadrant
%           calculate the new angles for each point thetaPlus and thetaMinus
%           set the current coordinate to the new deflected coordinates
%        end for
%        output the image
%    \end{verbatim}

\begin{frame}{Lensing Images}
    A number of different images were used in the simulation. Two were used
    for testing purposes and to observe the effect, and three were images
    of astronomical interest. The images used include:
    \begin{itemize}
        \item A human face
        \item The starship Enterprise
        \item The Andromeda Galaxy
        \item The Milky Way
        \item The Pleiades Star Cluster
    \end{itemize}

    %The program used for the simulation called \texttt{image.c} can
    %be found in Appendix C. The programming language used in this case
    %was C instead of FORTRAN, for the following reasons:
    %\begin{itemize}
    %    \item Using a C \texttt{struct} was appropriate in representing
    %          the data stored in a single PGM image.
    %    \item Image files of varying widths and heights were used, and
    %          therefore the program made used of dynamic memory allocation
    %          to allocate only the memory that was required to store
    %          the images.
    %\end{itemize}

    In all cases the distance ratio used was $D = 1$ and just the mass
    was varied. This is because when calculating the Einstein radius,
    the quantity $DM$ appears as follows:
    \begin{equation}
        \theta_E = \sqrt{\frac{4GMD}{c^2}}
    \end{equation}
\end{frame}

%\subsection{Results and Discussion}

%\subsubsection{A Human Face}
%An image of a face was used to first test the program.
%The original image and the simulated lensing of the face is shown
%in Figure~\ref{fig:face}.  The image on the
%left is the original, and the image on the right is one that has been
%gravitationally lensed by a point mass in the lower right hand corner
%of mass, $M = 5 \times 10^{30}$ kg.

\begin{frame}{Gravitational lensing of image of a human face by a mass $M = 5 \times 10^{30}$ kg}
    \begin{center}
    \begin{tabular}{cc}
        \includegraphics[width=0.3\textwidth]{pics/kr.eps} &
        \includegraphics[width=0.3\textwidth]{pics/kr_5e30.eps} \\
    \end{tabular}
    \end{center}
\end{frame}

%Apart from the obvious distortion in the lensed image, of particular
%interest is the formation of two images, one small and one large.
%The smaller image is inside the Einstein radius of the lense.

%\subsubsection{The Enterprise}
%Figure~\ref{fig:enterprise} shows three images of the starship Enterprise
%from Star Trek.
%\begin{itemize}
%    \item The image on the left is the original
%    \item The center image has a gravitational lense of mass
%          $M = 2 \times 10^{30}$ kg with the lense placed in the
%          centre.
%    \item The image on the right has a gravitational lense of
%          mass $M = 1 \times 10^{31}$ in the lower right hand corner.
%\end{itemize}

%With the lense in the center of the image it is more difficult to
%see the two images. The image on the right resulting from a larger
%mass (and with the lens in the bottom right hand corner)
%clearly shows both the images and large distortions.
%In many cases of actual observations, the images inside the Einstein radius
%are two small to be resolved by even the most powerful telescopes.

\begin{frame}{Gravitational lensing of an image of the starship Enterprise}
    \begin{center}
        \begin{tabular}{ccc}
            \includegraphics[width=0.3\textwidth]{pics/1701.eps} &
            \includegraphics[width=0.3\textwidth]{pics/ent_2e30.eps} &
            \includegraphics[width=0.3\textwidth]{pics/ent_1e31.eps}
        \end{tabular}
    \end{center}
\end{frame}

%\subsubsection{The Andromeda Galaxy}
%%Figure~\ref{fig:andromeda} shows the effect of different masses gravitational lensing
%an image of the Andromeda galaxy.  A number different lenses
%of varying masses were used. In all the cases the lensing object
%is located in the upper left hand corner of the image [position $(0,0)$].

\begin{frame}{Gravitational Lensing of the Andromeda Galaxy}
\begin{center}
    \begin{tabular}{cc}
    \includegraphics[width=0.3\textwidth]{pics/andromeda.eps}  &
    \includegraphics[width=0.3\textwidth]{pics/1e30and.eps}  \\
    $M = 0 kg$ & $M = 1 \times 10^{30}$ kg, $D = 1$ \\
    \includegraphics[width=0.3\textwidth]{pics/1e31and.eps}  &
    \includegraphics[width=0.3\textwidth]{pics/4e31and.eps} \\
    $M = 1 \times 10^{31}$ kg, $D = 1$ &  $M = 4 \times 10^{31}$ kg, $D = 1$ \\
    \end{tabular}
\end{center}
\end{frame}

%As the mass of the lense is increased, the amount of distortion in the
%resulting image is increased. The secondary image is also clearly visible.

%\subsubsection{The Milky Way}
%Figure~\ref{fig:milky-way} shows the effect of simulated gravitational lensing
%on an image of the centre of the Milky Way galaxy.
%The image on left is looking towards the centre of the Milky Way in the
%infrared. The image on the right is gravitationally lensed by a
%$M = 3 \times 10^{30}$ kg mass. The lense is located slightly left and above
%from the center. This image of the Milky Way allows us to see the distortion
%of the image within the Einstein radius much easier.

\begin{frame}{Gravitational Lensing of the Milky Way}
    \begin{center}
            \includegraphics[width=0.45\textwidth]{Pics/milky.eps} \hspace{1mm}
            \includegraphics[width=0.45\textwidth]{Pics/milky3e30.eps}
    \end{center}
\end{frame}

%\subsubsection{The Pleiades Star Cluster}
%Figure~\ref{fig:pleiades} shows the effect of simulated gravitational lensing
%on a star cluster.
%The image on the left is that of the Pleiades star cluster. The image
%on the right is gravitationally lensed by a $M = 2 \times 10^{30}$ kg
%mass.

\begin{frame}{Gravitational Lensing of the Pleiades Star Cluster}
    \begin{center}
            \includegraphics[width=0.45\textwidth]{Pics/pl.eps} \hspace{1mm}
            \includegraphics[width=0.45\textwidth]{Pics/pl2e30.eps}
    \end{center}
\end{frame}

%\section{Caustics for the Chang-Refsdal Lens}
%    \subsection{Method}
%    A FORTRAN program was written (see Appendix A -- \texttt{caustics.f})
%    which calculated critical curves and caustics for given
%    values of $\gamma$ and $\epsilon$. As stated earlier, critical
%    curves are obtained when Jacobian determinant is zero ($\det A = 0$)
%    This gives:

\begin{frame}{Caustics for the Chang-Refsdal Lens}
    \begin{equation}
        \det A = 1 - \left( \gamma + \frac{\epsilon}{\bar{z}^2} \right)
                     \left( \gamma + \frac{\epsilon}{\bar{z}^2} \right) = 0
    \end{equation}.
    Using complex polar coordinates letting $z = x \cos \phi + ix \sin \phi$
    we get
    \begin{equation}
        x^4 (1 - \gamma^2) - 2 \gamma \epsilon x^2
            (\cos^2\phi - \sin^2\phi) - 1 = 0
    \end{equation}
    The equation can be parameterised by letting:
        \[ \lambda = \cos^2\phi - \sin^2\phi \]
        \[ u = x^2 \]
    giving
    \begin{equation}
        u^2(1 - \gamma^2) - 2\gamma \epsilon \lambda u - 1 = 0.
    \end{equation}
\end{frame}

\begin{frame}{Caustics for the Change-Refsdal Lens (continued)}
    Solving this quadratic we obtain:
    \begin{equation}
        u = \frac{\gamma\epsilon\lambda \pm \sqrt{\gamma^2(\lambda^2 - 1) + 1}}
                 {(1 - \gamma^2)}
    \end{equation}
    For a given values of $\gamma$, $\epsilon$ and for $0 < \phi < 2\pi$
    the program calculates $u$ and then $x$. The $(x,y)$ coordinates
    for the caustic curve is then given by $(x\cos\phi, x\sin\phi)$.
    The corresponding caustics are given by:
    \begin{equation}
        y_1 = \left[ (1 + \gamma)x - \frac{\epsilon}{x} \right]
              \sqrt{\frac{1 + \lambda}{2} }
    \end{equation}
    \begin{equation}
       y_2 = \left[ (1 - \gamma)x - \frac{\epsilon}{x} \right]
                     \sqrt{\frac{1 - \lambda}{2} }
    \end{equation}
\end{frame}

\begin{frame}{Selecting Values for $\epsilon$ and $\gamma$}
    In the program a value of $\epsilon = 0.5$ was chosen. Values
    of gamma were chosen for the four representative regions in which
    caustics are found. The four regions are:
    \begin{itemize}
        \item $\gamma < -1$
        \item $-1 < \gamma < 0$
        \item $0 < \gamma < 1$
        \item $\gamma > 1$
    \end{itemize}
    The four values of $\gamma$ selected are:
    \begin{itemize}
        \item $\gamma = -1.3$
        \item $\gamma = -0.4$
        \item $\gamma = +0.8$
        \item $\gamma = +1.6$
    \end{itemize}
\end{frame}

%    \subsection{Results}
%    The generated critical curves and caustics for the different values of $\gamma$ selected
%    are shown in the following figures:
%    \begin{itemize}
%        \item $\gamma = -1.3$ Figure~\ref{fig:critical-gamma-1-3}
%        \item $\gamma = -0.4$ Figure~\ref{fig:critical-gamma-0-4}
%        \item $\gamma = +0.8$ Figure~\ref{fig:critical-gamma08}
%        \item $\gamma = +1.6$ Figure~\ref{fig:critical-gamma1-6}
%    \end{itemize}

    \begin{frame}{Critical curves and caustics for $\gamma = -1.3$}
        \begin{center}
                \includegraphics[width=0.5\textwidth]{images/neg1-3-critical.eps}
                \includegraphics[width=0.5\textwidth]{images/neg1-3-caustic.eps}
        \end{center}
    \end{frame}

    \begin{frame}{Critical curves and caustics for $\gamma = -0.4$}
        \begin{center}
                \includegraphics[width=0.5\textwidth]{images/neg0-4-critical.eps}
                \includegraphics[width=0.5\textwidth]{images/neg0-4-caustic.eps}
        \end{center}
    \end{frame}

    \begin{frame}{Critical curves and caustics for $\gamma = 0.8$}
        \begin{center}
                \includegraphics[width=0.5\textwidth]{images/pos0-8-critical.eps}
                \includegraphics[width=0.5\textwidth]{images/pos0-8-caustic.eps}
        \end{center}
    \end{frame}

    \begin{frame}{Critical curves and caustics for $\gamma = 1.6$}
        \begin{center}
            \includegraphics[width=0.5\textwidth]{images/pos1-6-critical.eps}
            \includegraphics[width=0.5\textwidth]{images/pos1-6-caustic.eps}
        \end{center}
    \end{frame}

%    \subsection{Discussion}
%    As can be seen from the diagrams in the previouse section,
%    the caustic curves with values of $\gamma$ ranging between
%    $-1$ and $1$ are non overlapping curves, whereas the others
%    overlap in a ``petal'' pattern.

%    The caustics can be used to determine light curves for background
%    sources since they mark where intensity increases in the resulting
%    gravitationally lensed image occur.

    \begin{frame}{Gravitational Lensing by a Binary Star System}
        \begin{center}
            \includegraphics[width=0.7\textwidth]{images/lens_model.eps}
        \end{center}
    \end{frame}

%    The use of critical lines and caustics are even more important
%    when modelling graviational lenses especially when both the source
%    and the lens are more complicated objects such as galaxies.
%    For example Figure~\ref{fig:binary-star} shows the representation of
%    gravitational lensing for a distant star by a binary star system.
%    As can be seen from the diagram, a simple point mass model is
%    not sufficient to handle a system such as this, and hence
%    more detailed models such as the Chang-Refsdal lens model
%    are needed to simulate this situation so that it we can accurate
%    comparisons with observations.

%----------------------------------------------------------------------
% END DOCUMENT
%----------------------------------------------------------------------

\end{document}