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| Subject | ： Re: SAD Update. V1.0.10.3.10a04. Fringe of Solenoid. |
| Date | ： 2009/12/31(Thu) 19:26:03 |
| Contributor | ： K. Oide |
Dear Users,
1. There may be a factor of 2 difference. See below.
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\section{Model of Magnetic Field}
We consider a longitudinal solenoid field with an axial symmetry:
\begin{equation}
B_z(s)= \left\{\begin{array}{lc}
0, & s<-f/2\\
B_0(s/f+1/2), & -f/2\le s \le f/2\\
B_0, & f/2<s
\end{array}\right.\ ,
\end{equation}
where $f$ is the length of the fringe. The associated vector potentials for $s\le f/2$ are
\begin{eqnarray}
A_x&=&-\frac{B_0y}{2}{\rm max}(s+f/2,0)+\frac{B_0(x^2+y^2)y}{16f}\left(\delta(s+f/2)-\delta(s-f/2)\right)\ ,\label{ax}\\
A_y&=& \frac{B_0x}{2}{\rm max}(s+f/2,0)-\frac{B_0(x^2+y^2)x}{16f}\left(\delta(s+f/2)-\delta(s-f/2)\right)\ ,\label{ay}\\
A_z&=&0\ ,
\end{eqnarray}
where the terms with $\delta$-functions are necessary in order to satisfy the Maxwell equations, while keeping the axial symmetry.
\section{Solution}
The fringe field has at least two effects, linear and nonlinear. The linear effect is caused by the first terms in Eqs. (\ref{ax}) and (\ref{ay}) that are linear in $x$ and $y$. We can expect that such linear effects can be expressed by a model with hard edges sliced along $s$, if the number of slices is sufficiently large. Thus here we concentrate on the nonlinear effects that are caused by the $\delta$-function terms in Eqs. (\ref{ax}) and (\ref{ay}).
Let us obtain the transformation associated with the nonlinear terms up to the first order of $B_0$. It is expressed as
\begin{equation}
\exp(:-f/2:)\exp(:-\delta:)\exp(:f:)\exp(:\delta:)\exp(:-f/2:)\ ,\label{trans}
\end{equation}
where $\exp(:-f/2:)$ is a drift-back by a distance $-f/2$, and $\exp(:\delta:)$ is the nonlinear term at $s=-f/2$, etc. Then the transformation (\ref{trans}) is approximated as
\begin{equation}
\begin{pmatrix}
x_1\\p_{x1}\\y_1\\p_{y1}\end{pmatrix}=
\begin{pmatrix}
x_0+b\left(2p_{x0}x_0y_0-p_{y0}(x_0^2-y_0^2)\right)/8p^2\\
p_{x0}+b\left(2p_{x0}p_{y0}x_0-(p_{x0}^2-p_{y0}^2)y_0\right)/8p^2\\
y_0-b\left(2p_{y0}x_0y_0+p_{x0}(x_0^2-y_0^2)\right)/8p^2\\
p_{y0}-b\left(2p_{x0}p_{y0}y_0+(p_{x0}^2-p_{y0}^2)x_0\right)/8p^2
\end{pmatrix}\ ,\label{apptrans}
\end{equation}
where $b\equiv B_0/(B\rho)$, up to the first order of $B_0$. The transformation (\ref{apptrans}) is expressed as $\exp(:H:)$ with a Hamiltonian:
\begin{equation}
H=-\frac{b}{8p^2}(xp_y-yp_x)(xp_x+yp_y)\ .\label{hami}
\end{equation}
To solve Eq.~{\ref{hami}), it is convenient to use another set of variables:
\begin{equation}
\begin{pmatrix}
r\\ p_r\\ \varphi\\ p_\varphi\end{pmatrix}=\begin{pmatrix}
\log(x^2+y^2)/2\\xp_x+yp_y \\
\tan^{-1}(y/x)\\
xp_y-yp_x\end{pmatrix}\ ,
\end{equation}
which is generated by a generating function:
\begin{equation}
G(x,p_r,y,p_\varphi)=\frac{\log(x^2+y^2)}{2}p_r+\tan^{-1}\left(\frac{y}{x}\right)p_\varphi\ .
\end{equation}
Then the Hamiltonian (\ref{hami}) is rewritten as
\begin{equation}
H=-\frac{b}{8p^2}p_\varphi p_r\ .\label{hami1}
\end{equation}
The transformation with (\ref{hami1}) is simply written as:
\begin{equation}
\begin{pmatrix}
r_1\\ \varphi_1 \end{pmatrix}=\begin{pmatrix}r_0-bp_\varphi/8p^2\\ \varphi_0-bp_r/8p^2\end{pmatrix}\ ,
\end{equation}
where $p_r$ and $p_\varphi$ are constant.