[thread display] [new arrival display] [word search] [past log] [管理用]
| Subject | ： Re: SAD Update. V1.0.10.3.10a00. Fringe of Solenoid. |
| Date | ： 2009/12/30(Wed) 07:56:20 |
| Contributor | ： K. Oide |
Dear Users,
1. This version implements the fringe of solenoid only for tracking. The description was updated as below, correcting some signs in the previous description.
===============================
\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}(s/f+1/2)\theta(s+f/2)+\frac{B_0(x^2+y^2)y}{8f}\left(\delta(s+f/2)-\delta(s-f/2)\right)\ ,\label{ax}\\
A_y&=& \frac{B_0x}{2}(s/f+1/2)\theta(s+f/2)-\frac{B_0(x^2+y^2)x}{8f}\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)/4\\
p_{x0}+b\left(2p_{x0}p_{y0}x_0-(p_{x0}^2-p_{y0}^2)y_0\right)/4\\
y_0-b\left(2p_{y0}x_0y_0+p_{x0}(x_0^2-y_0^2)\right)/4\\
p_{y0}-b\left(2p_{x0}p_{y0}y_0+(p_{x0}^2-p_{y0}^2)x_0\right)/4
\end{pmatrix}\ ,\label{apptrans}
\end{equation}
where $b\equiv B_0p_0/(B\rho p$, up to the first order of $B_0$. The transformation (\ref{apptrans}) can be generated by a generating function:
\begin{equation}
G(\overline{x},p_x,\overline{y},p_y)=\overline{x}p_x+\overline{y}p_y-\frac{b}{4}\left[(p_x^2-p_y^2)\overline{x}\overline{y}-p_xp_y(\overline{x}^2-\overline{y}^2)\right]\ ,\label{genf}
\end{equation}
with the accuracy of the first order of $b$. An interesting thing is that (\ref{genf}) is independent on the length of fringe, $f$.
We can solve (\ref{genf}) by two parts
\begin{eqnarray}
G_1(\overline{x},p_x,\overline{y},p_y)&=&\overline{x}p_x+\overline{y}p_y-\frac{b}{4}(p_x^2-p_y^2)\overline{x}\overline{y}\label{genf1}\\
G_2(x,\overline{p_x},y,\overline{p_y})&=&x\overline{p_x}+y\overline{p_y}-\frac{b}{4}(x^2-y^2)\overline{p_x}\overline{p_y}\ ,\label{genf2}
\end{eqnarray}
as we are interested only in the first order of $b$. The generating function (\ref{genf1}) has the solution for $x$ and $y$:
\begin{eqnarray}
x&=&\frac{\partial G_1}{\partial p_x}=\overline{x}-\frac{b}{2}\overline{x}\overline{y}p_x\\
y&=&\frac{\partial G_1}{\partial p_y}=\overline{y}+\frac{b}{2}\overline{x}\overline{y}p_y
\end{eqnarray}
which are satisfied by
\begin{eqnarray}
\overline{x}&=&\frac{4x}{2-b(p_yx+p_xy)+\sqrt{4+b^2(p_yx+p_xy)^2+4b(p_yx-p_xy)}}\ ,\\
\overline{y}&=&\frac{4y}{2+b(p_yx+p_xy)+\sqrt{4+b^2(p_yx+p_xy)^2+4b(p_yx-p_xy)}}\ ,
\end{eqnarray}
assuming that the terms with $b$ are smaller than 1.
Then the transformations for momenta caused by $G_1$ are given by
\begin{eqnarray}
\overline{p_x}&=&\frac{\partial G_1}{\partial\overline{x}}=p_x-\frac{b}{2}(p_x^2-p_y^2)\overline{y}\ ,\\
\overline{p_y}&=&\frac{\partial G_1}{\partial\overline{y}}=p_y-\frac{b}{2}(p_x^2-p_y^2)\overline{x}\ .
\end{eqnarray}
The solution for $G_2$ is quite similar to above:
\begin{eqnarray}
\overline{p_x}&=&\frac{4p_x}{2-b(p_yx+p_xy)+\sqrt{4+b^2(p_yx+p_xy)^2-4b(p_yx-p_xy)}}\ ,\\
\overline{p_y}&=&\frac{4p_y}{2+b(p_yx+p_xy)+\sqrt{4+b^2(p_yx+p_xy)^2-4b(p_yx-p_xy)}}\ ,\\
\overline{x}&=&\frac{\partial G_2}{\partial\overline{p_x}}=x-\frac{b}{2}(x^2-y^2)\overline{p_y}\ ,\\
\overline{y}&=&\frac{\partial G_2}{\partial\overline{p_y}}=y-\frac{b}{2}(x^2-y^2)\overline{p_x}\ .
\end{eqnarray}