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<H2><A NAME="SECTION00023000000000000000">
Uniform structures</A>
</H2>
<P>
For uniform structures we don't need to integrate over longitudinal
direction of the structure as the fields don't depend on this so we
need only get the field itself.
<P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
\begin{array}{l}
w_l (\rho ,s) = - \frac{1}{q}\left. {E_z } \right|_{z = ct - s} \\
w_t (\rho ,s) = - \frac{1}{q}[E_t + z \times B]_{z = ct - s} \\
\end{array}
\end{displaymath}
-->
<IMG
WIDTH="242" HEIGHT="54" BORDER="0"
SRC="img9.gif"
ALT="\begin{displaymath}
\begin{array}{l}
w_l (\rho ,s) = - \frac{1}{q}\left. {E_z }...
...= - \frac{1}{q}[E_t + z \times B]_{z = ct - s} \\
\end{array}\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<P>
Wake fields are estimated over the unit length of the structure.
<P>
Different units of measurements for Wakes used here <IMG
WIDTH="17" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"
SRC="img10.gif"
ALT="$\frac{1}{L}$">.
Consequently to get usual loss or kick factors we need to multiply
by L.
<P>
Modes in axially symmetric structures
<P>
All the filed components depend on the angle by <IMG
WIDTH="24" HEIGHT="18" ALIGN="BOTTOM" BORDER="0"
SRC="img11.gif"
ALT="$e^{l\theta }$">.
<P>
If the bunch and the particle are on-axis then symmetry dictates
that there is no effect: the effect can be expanded in powers of
<IMG
WIDTH="13" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img12.gif"
ALT="$r$">. We use cylindrical co-ordinates <IMG
WIDTH="45" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
SRC="img13.gif"
ALT="$r,\theta,z$"> and consider the
effect on a particle, a distance <IMG
WIDTH="13" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img12.gif"
ALT="$r$"> from the centre, of a slice of
thickness <IMG
WIDTH="22" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
SRC="img14.gif"
ALT="$dz$"> of the bunch, preceding this particle by a distance
<IMG
WIDTH="13" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img15.gif"
ALT="$z$">. Then [<A
HREF="node27.html#Yokoya">4</A>,<A
HREF="node27.html#Stupakovone">5</A>,<A
HREF="node27.html#Stupakovtwo">6</A>]
<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{eqnarray*}
\overline F_\perp =& \\-e &\sum_{m=1}^\infty W_m(z) m r^{m-1}\Big[
\hat r\left(Q_m \cos m \theta + \tilde Q_m \sin m \theta \right)
\\&- \hat\theta \left( Q_m \sin m \theta + \tilde Q_m \cos m \theta \right)
\Big]
\end{eqnarray*}
-->
<IMG
WIDTH="429" HEIGHT="98" BORDER="0"
SRC="img16.gif"
ALT="\begin{eqnarray*}
\overline F_\perp =& \\ -e &\sum_{m=1}^\infty W_m(z) m r^{m-1}...
...left( Q_m \sin m \theta + \tilde Q_m \cos m \theta \right)
\Big]
\end{eqnarray*}"></DIV>
<BR CLEAR="ALL"><P></P>
where
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
Q_m=\int_0^{2 \pi} d \theta' \int_0^\infty dr' (r')^{m+1} cos m
\theta' \rho(r',\theta',z)
\end{displaymath}
-->
<IMG
WIDTH="344" HEIGHT="46" BORDER="0"
SRC="img17.gif"
ALT="\begin{displaymath}Q_m=\int_0^{2 \pi} d \theta' \int_0^\infty dr' (r')^{m+1} cos m
\theta' \rho(r',\theta',z) \end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
\tilde Q_m=\int_0^{2 \pi} d \theta' \int_0^\infty dr' r'^{m+1} sin m
\theta' \rho(r',\theta',z)
\end{displaymath}
-->
<IMG
WIDTH="331" HEIGHT="46" BORDER="0"
SRC="img18.gif"
ALT="\begin{displaymath}
\tilde Q_m=\int_0^{2 \pi} d \theta' \int_0^\infty dr' r'^{m+1} sin m
\theta' \rho(r',\theta',z) \end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <!-- MATH
$\rho(r,\theta,z)$
-->
<IMG
WIDTH="67" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"
SRC="img19.gif"
ALT="$\rho(r,\theta,z)$"> is the
density of the bunch We assume, as in [<A
HREF="node27.html#Zagorodov">7</A>], that only
<IMG
WIDTH="50" HEIGHT="20" ALIGN="BOTTOM" BORDER="0"
SRC="img20.gif"
ALT="$\theta'=0$"> need be considered, i.e. we consider
deflection/distortion in a particular direction. This then
simplifies to
<P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
F_{\perp,m}=-e W_m m r^{m-1} Q_m
\end{displaymath}
-->
<IMG
WIDTH="192" HEIGHT="32" BORDER="0"
SRC="img21.gif"
ALT="\begin{displaymath}F_{\perp,m}=-e W_m m r^{m-1} Q_m \end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
Q_m=\int_0^{2\pi} d \theta'
\int_0^\infty dr' {r'}^{m+1} \rho(r',\theta')=Q(z) \overline{r^m}
\end{displaymath}
-->
<IMG
WIDTH="344" HEIGHT="46" BORDER="0"
SRC="img22.gif"
ALT="\begin{displaymath}Q_m=\int_0^{2\pi} d \theta'
\int_0^\infty dr' {r'}^{m+1} \rho(r',\theta')=Q(z) \overline{r^m}\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
Where the moments <!-- MATH
$\overline{r^m}$
-->
<IMG
WIDTH="26" HEIGHT="17" ALIGN="BOTTOM" BORDER="0"
SRC="img23.gif"
ALT="$\overline{r^m}$"> are known from the bunch
distribution. <IMG
WIDTH="41" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"
SRC="img24.gif"
ALT="$Q(z)$"> is the charge of the slice, and the wake
function for a steeply tapered collimator, moving from aperture <IMG
WIDTH="12" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
SRC="img25.gif"
ALT="$b$">
to aperture <IMG
WIDTH="14" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img26.gif"
ALT="$a$">, is given by([<A
HREF="node27.html#Raimondi">8</A>])
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
W_m(z)=2 \left( {1 \over a^{2m}} - {1 \over b^{2m}} \right) e^{-mz/a} \Theta(z)
\end{displaymath}
-->
<IMG
WIDTH="283" HEIGHT="45" BORDER="0"
SRC="img27.gif"
ALT="\begin{displaymath}W_m(z)=2 \left( {1 \over a^{2m}} - {1 \over b^{2m}} \right) e^{-mz/a} \Theta(z)\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <IMG
WIDTH="40" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"
SRC="img28.gif"
ALT="$\Theta(z)$"> is a unit step function. The factor of 2 arises
because one has to consider both sides of the collimator
[<A
HREF="node27.html#Zimmermann">11</A>].
<P>
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<B> Next:</B> <A NAME="tex2html104"
HREF="node6.html">Types of wake potentials</A>
<B> Up:</B> <A NAME="tex2html100"
HREF="node2.html">General information</A>
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HREF="node4.html">Superposition</A>
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<ADDRESS>
german_kourevlev
2006-01-19
</ADDRESS>
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