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\title{UBC Phys560/UVic Phys522, lecture 3}
\author{Rick Baartman}
%\institute[TRIUMF]{\includegraphics[height=1.cm]{TRIUMF_Logo17_Blue.png}}
\institute[TRIUMF]{\textcolor{triumfcyan}{\tlogo A}}
\date{\today}
\begin{document}
\frame{\titlepage}
\frame{\frametitle{What is a beam?}
\small
{\color{myblue}
Answer: a group of particles traveling more or less in the same direction.}
{\color{myred}Q: Why can't they be perfect? I.e.\ why not travel in \underline{exactly} the
same direction?}
Look at an ion source. Atoms are ionized by hitting them hard enough. How
hard? Ionization energy is typically on order of 3\,eV. So a typical
energy before extraction is 3\,eV. After extraction, it is 30\,keV, but it
will still have the same transverse momentum.
\begin{center}\fbox{\includegraphics[width=.5\textwidth]{../TrainingImages/source.pdf}}\end{center}
\beq
{x'}={v_x\over v_z}=\sqrt{mv_x^2/2\over
mv_z^2/2}=\sqrt{3\,\mbox{eV}\over 30\,\mbox{keV}}=10\,\mbox{mrad.}
\eeq
\color{myred}
So a beam of such particles is 10\,mm in radius after drifting for 1 metre,
10\,cm after drifting 10\,m. Therefore need focusing.
}\frame{\frametitle{What is beam quality?}
\small
{\color{myred}Beam has high quality if it is small and its divergence (rate
of spreading) is also small.}
\begin{center}
\fbox{\includegraphics[width=.5\textwidth]{../TrainingImages/abquality.pdf}}
\end{center}
But beam may be high in quality and nevertheless violate these two
conditions.
\begin{center}
\fbox{\includegraphics[width=.5\textwidth]{../TrainingImages/cquality.pdf}}
\end{center}
So we need a better definition.
}\frame{\frametitle{`Phase' space}
\small{\color{myblue}
Particles can be characterized by position ($x$) and momentum, or direction
(${x'}$). So on an $(x,{x'})$ coordinate plane, every particle can be
assigned a point. Look at our 3 examples again:}
\begin{center}
\fbox{\includegraphics[width=.7\textwidth]{../TrainingImages/abcps.pdf}}
\end{center}
Now it is easy to see why case C is high quality even though it corresponds
to large, divergent beam. The area it occupies in phase space is zero.
This area is called `{\color{myred}emittance}'. It is related to entropy: with good
practices, it is conserved, but it can increase and it's difficult to make
it smaller.
{\color{myred}Liouville's theorem: Under the action of conservative forces, emittance is
conserved.} This is true even of space charge forces (particle-particle repulsion), though space charge is rarely linear.
}
\frame{\frametitle{Real life emittances}
Besides the $(x,{x'})$ coordinate plane, there is also the $(y,\phi)$
coordinate plane. The two emittances are not necessarily the same.
\begin{columns}\begin{column}{.5\textwidth}
\includegraphics[width=.75\textwidth,angle=90]{../TrainingImages/duos.pdf}
\end{column}
\begin{column}{.5\textwidth}
Also, notice the lack of a well-defined boundary. This means ``emittance''
is a fuzzy term; sometimes we say $\epsilon_{\rm
rms}$, sometimes $\epsilon_{\rm 90\%}$, etc.
\end{column}
\end{columns}
}
\frame{\frametitle{How many dimensions?}\small
Recall from QM, uncertainty relations between pairs of variables. These are the canonical pairs: \beq(x,p_x),\ (y,p_y),\ (z,p_z),\ (t,E)\eeq
There are others, depending upon how coordinates are defined; e.g.\ $\theta$ and angular momentum. But dynamical relations like $E^2=m^2c^4+p^2c^2$ tie these together. Of 8 coordinates, only 6 are dependent.
\\~\\
In general, there are two conditions that beam transport and accelerator designers strive to meet: (1) Keep the 3 phase space planes independent of one another (decoupled), (2) make sure the forces are as linear as possible.
\\~\\
In general, the momenta are usually normalized to the reference momentum $p_0$. Then e.g.\ the $x$ phase space is $(x,x')$, where $x'=p_x/p_0$ because $p_x/p_0=v_x/v_0=\frac{dx}{dt}/\frac{dz}{dt}=dx/dz$.
}
\frame{\frametitle{Examples: Drift and thin lens }
(apology: $\theta$ should be $x'$)
\includegraphics[width=.8\textwidth]{../TrainingImages/shearing}
}
\frame{\frametitle{Drift and thin lens in Matrix Formalism}
%\includegraphics[width=.9\textwidth]{../TrainingImages/matrixthin}
\small A handy way of calculating the effect of linear optics is with
\emph{transfer matrices}.
\begin{equation}
\label{tm}
\begin{pmatrix}x_{\rm f}\cr{x'}_{\rm f}\end{pmatrix}=
\begin{pmatrix}m_{11}&m_{12}\cr m_{21}&m_{22}\end{pmatrix}
\begin{pmatrix}x_{\rm i}\cr{x'}_{\rm i}\end{pmatrix}
\end{equation}
For a drift of length $d$, the matrix is
$\begin{pmatrix}1&d\cr 0&1\end{pmatrix}$, and for a focusing
thin lens, it is $\begin{pmatrix}1&0\cr {-1\over f}&1\end{pmatrix}$. So for 2 opposite thin lenses separated by
a drift, we have a total matrix
\begin{equation}
\label{tom}
\begin{pmatrix}1&0\cr {1\over f}&1\end{pmatrix}
\begin{pmatrix}1&d\cr 0&1\end{pmatrix}
\begin{pmatrix}1&0\cr{-1\over f}&1\end{pmatrix}=
\begin{pmatrix}1-{d\over f}&d\cr{-d\over f^2}&1+{d\over f}\end{pmatrix}.
\end{equation}
The crucial element here is $m_{21}$. If it is positive, the system
defocuses, if negative, it focuses. {\color{red}We note that this `doublet' focuses
irrespective of the sign of $f$.} I.e.\ it focuses in both $x$ and $y$. This is called {\color{myred}\bf Strong Focusing}. (More on this later.)
}
\frame{\frametitle{Simple DC accelerator}
Put a hole in one plate: (from J.D.Jackson {\it Classical Electrodynamics})
\\~\\
\includegraphics[width=\textwidth]{../2015PHYS/jackson.png}}
\frame{\frametitle{DC Accelerator Column Potential}
\includegraphics[width=.3\textwidth]{DCacc}\includegraphics[width=.7\textwidth]{AccCol}
}
\frame{\frametitle{Accelerator column entrance aperture as thin lens}\small
Write Laplace equation ($\nabla^2\Phi=0$) in polar coordinates $(\rho,\theta,z)$ for azimuthal symmetry (no $\theta$):
\beq \frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\Phi}{\partial\rho}\right)+\frac{\partial^2\Phi}{\partial z^2}=0
\eeq
Expand near axis,
\beq\Phi(\rho,z)=\Phi(0,z)-\frac{\rho^2}{ 4}\frac{d^2\Phi(0,z)}{ dz^2}+\ldots\eeq
By ``thin lens'' we mean that the motion due to focusing is negligible in the region where focusing takes place.
All we need is the function $\Phi(0,z)$; the on-axis potential $\phi(z)$.}
\frame{
So the focus force $F_\rho$ on the particle is $q{\cal{E}}_\rho=q\frac{\rho}{2}\phi''$. In $x$-direction, $F_x=m\frac{d^2x}{dt^2}=q\frac{x}{2}\phi''$. As $z=vt+z_0$, $dt=dz/v$, and \beq x''=q\frac{\phi''}{2mv^2}x\eeq
\\~\\
To get net effect, we ignore variation of $x$ through the aperture, and integrate: \beq\Delta x'=-\frac{q\Delta{\cal{E}}x}{2mv^2}=-\frac{q\Delta{\cal{E}}x}{4E_{\rm kin}}\eeq because the electric field ${\cal{E}}=-\phi'$.
As this should be $-x/f$, we can read off the focal length $f$.}
%\textcolor{myblue}{Exercise \stepcounter{exe}\arabic{exe}: By direct
% substitution, find the next term in the expansion above.}
\frame{\frametitle{Simple DC accelerator focuses!}
The parabolic potential dip means there is a transverse electric field that is linear with distance. If we integrate to find the focal length, \beq\frac{1}{f}=\frac{q\Delta{\cal{E}}}{4E_{\rm kin}}\eeq
N.B.: This result does not depend on the details of the aperture; not even its size!
\\~\\
But it is nonlinear; the next term in the expansion gives a cubic force.
%\includegraphics[width=.5\textwidth]{../TrainingImages/linear}
}
\frame{\frametitle{Effect of linear lens}
Note that the angle change in direction of a particle initially parallel to
axis is proportional to that particle's distance from the axis. That's what
makes an initially parallel beam come to an infinitesimal
point.
\begin{center}
\includegraphics[width=.6\textwidth]{../TrainingImages/linear}
\end{center}
$\Delta{x'}=x/f$, where $f$ is the focal length, or in transfer matrix
format,
\begin{equation}\nonumber
\label{lens}
\begin{pmatrix}x_{\rm f}\\ {x'}_{\rm f}\end{pmatrix}=
\begin{pmatrix}1 & 0\\{-1\over f} &1\end{pmatrix}
\begin{pmatrix}x_{\rm i}\\ {x'}_{\rm i}\end{pmatrix}
\end{equation}
}
\frame{\frametitle{Effect of non-linear lens}
Note that the angle change in direction of a particle initially parallel to
axis is not proportional to that particle's distance from the axis. The
initially-parallel beam cannot come to an infinitesimal
point.
\begin{center}
\includegraphics[width=.3\textwidth,angle=-90]{../TrainingImages/nonlinear}
\end{center}
$\Delta{x'}=x/f+Cx^3$, where $C$ is an aberration coefficient.}
\frame{\frametitle{Effective emittance increase}
\small So in phase space, a non-linear lens leads to an
effective emittance increase. Below are successive snapshots in phase
space of a beam acted upon by lenses with small nonlinearity. (Proceed left to right and top to bottom.)
\begin{center}
\includegraphics[width=.5\textwidth]{../TrainingImages/lawson}
\end{center}
Note that the initial elliptical emittance diagram filaments into a shape
to such an extent that the effective emittance has grown by 50\%.
}
\frame{\frametitle{Linear lenses}
\begin{columns}\begin{column}{.6\textwidth}
\small We want to design a linear lens, preferably focusing
on both directions. That means $F_x\propto x$ and $F_y\propto y$. Turns
out this is not possible in a region free from metal boundaries. Gauss'
Law: flux lines cannot all be pointing in without coming out somewhere
else.
Or, in mathematical language, $E_x=-kx$, $E_y=-ky$ means
the potential function is $V(x,y)=k(x^2+y^2)/2$, but this violates
Laplace's equation: $\nabla^2V={\partial^2V\over\partial
x^2}+{\partial^2V\over\partial y^2}=k+k\neq0$.
One can achieve such astigmatic focusing, but only with an electrode (a
wire) along the beam axis. This is \textbf{very} inconvenient.\end{column}\begin{column}{.4\textwidth}
\includegraphics[width=\textwidth]{../TrainingImages/wire}\end{column}\end{columns}
}
\frame{
Second choice: Linear, but focusing in one direction, defocusing in the
other. $E_x=-kx$, $E_y=ky$ means the potential function is
\begin{equation}
\label{qua}
{\color{mygreen}V(x,y)=k(x^2-y^2)/2}.
\end{equation}
This \underline{does} satisfy Laplace's equation:
$\nabla^2V={\partial^2V\over\partial x^2}+{\partial^2V\over\partial
y^2}=k-k=0$.
Equipotential lines are hyperbolae.
\begin{center}
\includegraphics[width=.35\textwidth]{../TrainingImages/hyperb2}
\end{center}
}
\frame{\frametitle{A real quadrupole}
\small Such a potential field can be set up with a set of 4
hyperbolic surfaces (hence the name quadrupole); 2 opposite surfaces being
kept at a $+V$ potential, and the other two opposite surfaces being kept at
a $-V$ potential with electrostatic power supplies. It is intuitively clear
that though hyperbolic surfaces extend to infinity, truncating them at some
point will have minimal effect in the region of interest between the
electrodes. In fact, the surfaces needn't be accurately hyperbolic either;
circular arcs will be sufficient provided the beam is confined to a region
somewhat smaller than the inside radius $a$ of the quadrupole. {\color{myred}The radius of the circular arcs which
most closely matches the hyperbola is 1.145 times the aperture radius $a$.}
\\~\\
An example is shown below. These are the dimensions of a standard TRIUMF quadrupole used up to beam eneries of $60$\,keV.}
\frame{
\begin{center}
\includegraphics[width=.7\textwidth]{../TrainingImages/esquad.jpg}
\end{center}
}
\frame{\frametitle{Quadrupole focal length}
\small Let us call the aperture radius $a$, and the
electrode voltages are (going clockwise) $+V_Q,-V_Q,+V_Q,-V_Q,$. Then the
potential is
\begin{equation}
V(x,y)={V_Q\over a^2}(x^2-y^2)
\end{equation}
The force in the $x$-direction is $F_x=q{\cal{E}}_x=-q\partial V/\partial
x=(2qV_Q/a^2)x$. But remember $F=ma$ and the change in direction ${x'}$
is ${x'}={\Delta v_x\over v_0}={a_xL\over v_0^2}={F_xL\over mv_0^2}$. Put
that together with the formula for the force and we get
\begin{equation}
{x'}=-{2q\over mv_0^2}{LV_Q\over a^2}x=-{V_Q\over V_0}{Lx\over a^2}
\end{equation}
But ${x'}$ is just $-x/f$, so the focal length is given by \color{black}
\begin{equation}\label{wof}
\color{red}{1\over f}={V_Q\over V_0}{L\over a^2}
\end{equation}
}
\frame{
Example: The typical ISAC LEBT quadrupole is 2 inches long, aperture radius
is 1 inch, and $V_Q= 1$\,kV, and the particles passing through come from an
ion source which is at 30 kV.
Then ${1\over f}={1\over 30}{2\over 1^2}\,\mbox{in}^{-1}$, or,
$f=15$\,inches.
\\~\\
\color{red}Notice that the
optics is independent of either the charge or the mass of the beam's
particles.
}
\frame{\frametitle{Thick Lens}\small
But we need not take the thin lens approximation. In that case for a quadrupole, we get \beq x''+Kx=0\mbox{ where }K=\frac{V_Q}{V_0a^2}\eeq
This is the familiar equation for {\color{myred}simple harmonic motion}, and so we can write immediately: for $K>0$: $x=C_1\cos(K^\frac{1}{2}s)+C_2\sin(K^\frac{1}{2}s)$ for $K>0$, and $x=C_1\cosh((-K)^\frac{1}{2}s)+C_2\sinh((-K)^\frac{1}{2}s)$ for $K<0$. But we find the constants from the initial coordinates $(x_{\rm i},x'_{\rm i})$, and set the final location $s=L$. This is then the integration through the complete quadrupole:
\begin{eqnarray}
x_{\rm f}&=&\cos(K^\frac{1}{2}L)x_{\rm i}+K^{-\frac{1}{2}}\sin(K^\frac{1}{2}L)x'_{\rm i}\\
x'_{\rm f}&=&-K^\frac{1}{2}\sin(K^\frac{1}{2}L)x_{\rm i}+\cos(K^\frac{1}{2}L)x'_{\rm f}
\end{eqnarray}
and for $K<0$:
\begin{eqnarray}
x_{\rm f}&=&\cosh((-K)^\frac{1}{2}L)x_{\rm i}+(-K)^{-\frac{1}{2}}\sinh((-K)^\frac{1}{2}L)x'_{\rm i}\\
x'_{\rm f}&=&(-K)^\frac{1}{2}\sinh((-K)^\frac{1}{2}L)x_{\rm i}+\cosh((-K)^\frac{1}{2}L)x'_{\rm f}
\end{eqnarray}
(Looks messy but only because of the ugly square root.)
}\frame{
We can put this into matrix form for $K>0$ (now with $k\equiv\sqrt{K}$):
\begin{equation}\label{tlensf}
\begin{pmatrix}x_{\rm f}\cr x'_{\rm f}\end{pmatrix}=
\begin{pmatrix}\cos kL&\frac{\sin kL}{k}\cr -k\sin kL&\cos kL\end{pmatrix}
\begin{pmatrix}x_{\rm i}\cr x'_{\rm i}\end{pmatrix} \end{equation}
For $K<0$ (now with $k\equiv\sqrt{-K}$):
\begin{equation}\label{tlensd}
\begin{pmatrix}x_{\rm f}\cr x'_{\rm f}\end{pmatrix}=
\begin{pmatrix}\cosh kL&\frac{\sinh kL}{k}\cr k\sinh kL&\cosh kL\end{pmatrix}
\begin{pmatrix}x_{\rm i}\cr x'_{\rm i}\end{pmatrix} \end{equation}
You will notice that the matrix determinant is always $1$. This is related to the conservation of emittance. Transformations can only stretch in an area-conserving manner, and/or rotate.
}
\frame{\frametitle{Magnetic Quadrupole}
\begin{columns}\begin{column}{.35\textwidth}
\includegraphics[width=\textwidth]{../TrainingImages/quad}\end{column}
\begin{column}{.65\textwidth}
Quadrupoles have magnetic fields that vary linearly with distance from axis. The gradient is $g=B_{\rm p}/a$ where $B_{\rm p}$ is the pole tip field and $a$ is the aperture radius. So the force equation is $m\frac{d^2x}{dt^2}=-qvB_y=-qvgx$ and again this is a simple harmonic motion equation, and all the solutions are the same as for electrostatic.
\end{column}
\end{columns}
The $K$ is re-defined as \beq K=\frac{qg}{mv}=\frac{B_{\rm p}}{aB\rho}\eeq
\\~\\
The thin lens focal strength is given by $\frac{1}{f}=\frac{gL}{B\rho}=\frac{B_{\rm p}L}{aB\rho}$
}
\frame{\frametitle{...about $B\rho$}
Recall radius of curvature $\rho$ in a magnetic field $B$ is given by \beq B\rho=p/q=mv/q,\eeq and called the ``rigidity''. For this reason, momentum and $B\rho$ tend to be used interchangably. For accelerator physicists, knowing momentum in ``Tesla-metres'' is more useful than knowing it in ``MeV/$c$''.
\\~\\
{\color{myblue}Exercise 9: For protons, calculate the conversion factor between $B\rho$ in Tesla-metres, and momentum in MeV/$c$.}
}
\frame{\frametitle{Which is better?}
Both for focusing and for redirecting or bending the beam path, we can choose either electrostatic or magnetic. To compare the two, look at the Lorentz force $\vec{F}=q(\vec{\cal{E}}+\vec{v}\times\vec{B})$. The highest DC electric fields are about $1$\,kV/mm. So at high energy ($\beta\sim1$) the equivalent $B={\cal{E}}/c=0.033$\,Tesla or $330$\,Gauss. It's not hard to make magnetic fields in the few Tesla range, so at high energy, magnets are about $100$ times stronger than electrostatics.
But at low speeds, for example a carbon ion at $60$\,keV has $\beta=0.0033$, electric fields are more effective. As well, electrostatic elements have two interesting advantages: (1) Electric fields are determinate whereas magnetic fields from ferromagnetic elements have some indeterminacy due to hysteresis; (2) electrostatic optics is mass-independent whereas equal-energy charged particles have mass-dependent trajectories in magnetic fields.
\\~\\
OTOH, electrostatics need vacuum or at least special environments to reach fields like 1\,kV/mm.
}
\frame{\frametitle{Summary: ${\cal{E}}$ vs.\ $B$}
\begin{table}[htp]
\caption{Magnetic beam optics elements vs.\ electrostatic}
\begin{center}
\begin{tabular}{|c||c|c|}\hline
{Property}&Magnetic&Electrostatic\\\hline
Energy&best for $\beta>\sim0.05$&\makecell{$\cal{E}$ is limited so not\\ used above about $500$\,keV}\\\hline
Vacuum&Does not need&\makecell{Must have\\or high pressure SF$_6$}\\\hline
Precision&\makecell{Hysteresis;\\field depends on history}&Exact\\\hline
Weight&\makecell{Heavy\\(large steel yokes)}&Very light\\\hline
Size&\makecell{Large elements; \\small beam pipe}&\makecell{Tiny elements inside\\ large beam pipe}\\\hline
\end{tabular}
\end{center}
\label{default}
\end{table}%
}
\frame{\frametitle{Strong focusing principle}
\small So if we can't avoid defocusing when we try to focus,
what do quadrupoles actually gain?
\\~\\
In a quadrupole transport channel, lenses alternate in sign, and $x$ and
$y$ are opposite: $x$;~f,d,f,d,f,... $y$;~d,f,d,f,d,...
\begin{center}
\includegraphics[angle=-90,width=.7\textwidth]{../TrainingImages/periodic}
\end{center}
The key point is that the beam is smaller in the defocusing lenses than in
the focusing lenses. Therefore, the forces don't cancel, and such a channel
always has a net focusing effect.
}
%\frame{\frametitle{Transfer Matrices}}
\frame{\frametitle{Periodic focusing}
\begin{center}
\includegraphics[angle=-90,width=.35\textwidth]{../TrainingImages/periodic}
\end{center}
\small The periodic focusing channel is the `backbone' of
particle beam transport. One can prove that there always exists a set of
initial conditions for which the beam envelope has the same periodicity
as that of the optics. (This is essentially the same as Bloch's theorem
in solid state physics. There are other parallels as well: a periodic
focusing channel is like a periodic potential in a crystal.) Under these
conditions, the beam is said to be matched. Particles in periodic
channels follow sinusoidal-like trajectories. A matched beam has
identical ellipses at a reference point in each period, but particles
move around the ellipse by an angle called the phase advance per cell.
Note in the above figure, the phase advance per cell is 60$^\circ$.
Interesting resonance effects can occur when the phase advance per cell
is $2\pi/n$ where $n$ is a small integer.
}
\frame{\frametitle{Dipoles as optical elements}
\includegraphics[width=\textwidth]{../TrainingImages/dipoles_sector_rect}\\
A sector magnet focuses horizontally but not vertically. \\
A rectangular magnet focuses vertically but not horizontally. \\
(Throughout, magnetic field is assumed to be vertical.)
}
\frame{\frametitle{Vertical focal strength}
\begin{columns}\begin{column}{.6\textwidth}
Slide 9, lecture 2: $m\frac{d^2z}{dt^2}=qv\frac{dB}{dr}z$ with ``field index'' being defined as the exponent power of the field variation: $B=B_0\left(\frac{r}{\rho}\right)^k$ so that $\frac{r}{B}\frac{dB}{dr}=k$, we get \beq z''-\frac{k}{\rho^2}z=0.\eeq (Recall: $B_0\rho=mv/q$.)
\end{column}
\begin{column}{.4\textwidth}\includegraphics[width=\textwidth]{slide9lecture2}\end{column}\end{columns}
\vspace{1cm}
This is {\bf again} a simple harmonic motion (SHM) equation just like those above. We just replace the $K$ above with $-k/\rho^2$.
}
\frame{\frametitle{Frenet-Serret}
Now do radial. But first, a complication:
To calculate paraxial rays in a bending system, we have to tranform to a non-inertial frame; a frame rotating with radius of curvature $\rho$. Why? Because we want rays only slightly away from the reference ray.
Start with \beq m\ddot{r}-\frac{mv^2}{r}=-qvB\eeq Divide by $mv^2$, remembering that $B_0\rho=mv/q$: \beq r''-\frac{1}{r}=-\frac{B}{B_0\rho}\eeq $x$ is the deviation of the particle's radius from the intended (reference) one: $r=\rho+x$. $\rho$ is constant so $r''=x''$. Remember the Taylor expansion and expand both $1/r$ and $B$: \beq \frac{1}{r}=\frac{1}{\rho}\left(\frac{1}{1+x/\rho}\right)\approx\frac{1}{\rho}\left(1-\frac{x}{\rho}\right)=\frac{1}{\rho}-\frac{x}{\rho^2}\eeq
}
\frame{\frametitle{}
and: \beq B=B_0\left(\frac{r}{\rho}\right)^k=B_0\left(1+\frac{x}{\rho}\right)^k\approx B_0\left(1+k\frac{x}{\rho}\right)\eeq Put these last three together and we find: \beq x''+(1+k)\frac{x}{\rho^2}=0\eeq {\bf Again} it's SHM and we can use all the above quadrupole results with $K=(1+k)/\rho^2$ to get the matrices.
\\~\\
Notice both in the vertical and horizontal cases we get sin and cosines of $L/\rho$. This is of course just the angle of rotation: $\theta=L/\rho$.}
\frame{\frametitle{Exercise}{\color{myblue}Exercise 10: For the very simplest dipole case of a flat field $k=0$, find the radial and vertical transfer matrices.}
}
\frame{\frametitle{Edge focusing}We derived the optics of dipoles assuming there is no issue with field when entering and exiting the magnet. This is not the case.
\\~\\
A particle a radial distance $x$ entering a slanted magnet adge, slant angle $\zeta$, is an extra distance $\delta s$ in the magnet, where $\delta s/x=\tan\zeta$. During this extra distance, it gains an extra turning in the magnet by angle $\Delta x'=\delta s/\rho=x\tan\zeta/\rho$. Thus we have a thin lens of focal strength: \beq\frac{1}{f_x}=\frac{\tan\zeta}{\rho}\eeq
}
\frame{\frametitle{Edge vertical effect}
\includegraphics[width=.9\textwidth]{edge_vert_effect}}
\frame{\frametitle{}
Vertically the effect is more subtle. It depends upon the ``bulging'' effect of the field falloff. Above and below the median plane, this causes a field component at right angles to the slanted edge, and because this edge is slanted not normal to the beam, this gives an $x$-component to the magnetic field, which gives a force in the vertical $z$ direction that is proportional to $z$. Not hard to work out from the magnetic field expansion; the result is a focal strength: \beq\frac{1}{f_z}=-\frac{\tan\zeta}{\rho}\eeq
\\~\\
Notice that for all 4 cases; both types of quadrupoles, dipoles, and edge focusing, there is a sum rule for $\frac{1}{f_x}+\frac{1}{f_z}$. Why is this?
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