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Spin Maps Concatenation

So if \bgroup\color{black}$T$\egroup is a spin map, it is then given by a couple of objects:

$\displaystyle T=
\left({m,S}\right)$     (11.8)

where \bgroup\color{black}$m$\egroup is an orbital map and \bgroup\color{black}$S$\egroup is an orbital dependent spin matrix. If a beam line #1 is followed by beam line #2, then the spin map for the full beam line is given by:
$\displaystyle {T}_{2}\circ {T}_{1}$ $\textstyle =$ $\displaystyle \left({{m}_{2},{S}_{2}}\right)\circ \left({{m}_{1},{S}_{1}}\right)$  
  $\textstyle =$ $\displaystyle \left({{m}_{2}\circ {m}_{1},{S}_{2}\circ {m}_{1}{S}_{1}}\right)$ (11.9)

The matrix \bgroup\color{black}${S}_{2}\circ {m}_{1}{S}_{1}$\egroup is simply the product of \bgroup\color{black}$S_2 S_1$\egroup where \bgroup\color{black}$S_2(\overline{z})$\egroup is evaluated at \bgroup\color{black}$\overline{z}=m_2(z)$\egroup with \bgroup\color{black}$z=(x,p_x,y,p_y,z_5,z_6)$\egroup.



Frank Schmidt 2010-10-15