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\title{\textcolor{myred}{\Large Classical Mechanics $\rightarrow$ Accelerator Physics $\rightarrow$ LHC}\\ \vspace{1em}
\textcolor{myblue}{\tlogo D}}
\author{Rick Baartman, TRIUMF}
%\date{January 25 -- February 3, 2011}
\begin{document}
\maketitle
\raggedright
\href{http://lin12.triumf.ca/text/Talks/2013TRISEP/}{\tt http://lin12.triumf.ca/text/Talks/2013TRISEP/}
\raggedright
\setlength{\rightskip}{0pt plus 200pt}
\newcounter{exe}
\foilheadc{Summary}
\begin{itemize}
\item some stuff on rotating frames
\item making reference orbit the unperturbed orbit
\item general, simplest motion (simple harmonic, or Hill's equation)
\item Strong Focusing
\item Calculating beam size
\item Separating the longitudinal motion
\item (LHC) Luminosity limit
\end{itemize}
\foilheadc{Example: Circular Reference, Field-Free}\small
Let us say we have chosen a circular path ($\rho=$constant) as reference
trajectory but that there are no fields: all $A_i$ and $\Phi$ are
zero. Then, ignoring motion in the $y$-direction,
\begin{equation}
H=-\left(1+\frac{x}{\rho}\right)\sqrt{\left(\frac{E}{c}\right)^2-m^2c^2-P_x^2}\label{eq:poop}
\end{equation}
Launch a particle along the circular path ($E\neq0$, $x=0$, $P_x=0$ at $s=0$). What do you think will happen? How will $x$ and $P_x$ and $E$ initially change?
%\textcolor{myblue}{Exercise \stepcounter{exe}\arabic{exe}: Answer the above question. You don't have to solve the differential equations, just find initial expressions for the rate of change of $x$ and $P_x$.}
This Hamiltonian has $s$ as independent variable; it is in fact $-P_s$. Therefore,
\begin{eqnarray}
\frac{dx}{ds}&=&\frac{\partial H}{\partial P_x},\ \frac{dP_x}{ds}=-\frac{\partial H}{\partial x},\\
\frac{dt}{ds}&=&-\frac{\partial H}{\partial E },\ \frac{dE}{ds}=\frac{\partial H}{\partial t},\\
\frac{dP_s}{ds}&=&-\frac{\partial H}{\partial s} \end{eqnarray}
$E$ is constant since $\frac{\partial H}{\partial t}=0$. Therefore let's introduce constant $P_0$: $P_0^2=\left(\frac{E}{c}\right)^2-m^2c^2$
\begin{eqnarray}\label{eo1}
\frac{dx}{ds}&=&\frac{\partial H}{\partial P_x}=\left(1+\frac{x}{\rho}\right)\frac{P_x}{\sqrt{P_0^2-P_x^2}}\\\label{eo2}
\frac{dP_x}{ds}&=&-\frac{\partial H}{\partial x}=\frac{1}{\rho}\sqrt{P_0^2-P_x^2}
\end{eqnarray}
Plugging the initial values $x=P_x=0$ at $s=0$, we find
\begin{eqnarray}
\left.\frac{dx}{ds}\right|_{s=0}&=&0\\
\left.\frac{dP_x}{ds}\right|_{s=0}&=&\frac{P_0}{\rho}\label{cq}
\end{eqnarray}
So $P_x$ begins to depart immediately from zero, and then $x$ will begin to grow as well.
Note that since $\frac{\partial H}{\partial s}=0$, $P_s=\left(1+\frac{x}{\rho}\right)\sqrt{P_0^2-P_x^2}$ is a constant, which, with our initial conditions, is $P_0$. This can be used to find $x$ once $P_x$ is known. \clearpage
\begin{wrapfigure}{l}{8cm}
\includegraphics[width=7cm]{../TrainingImages/circ_line}\vspace{-1.2cm}
\end{wrapfigure}
The solution of \ref{eo2} is simply \[P_x=P_0\sin(s/\rho).\] This gives \[x=\rho\left(\frac{1}{\cos(s/\rho)}-1\right).\]
Does this look sensible? Yes. $\cos(s/\rho)=\frac{\rho}{x+\rho}$
Since $s/\rho$ never gets beyond $\pi/2$ because the solution to the $dt/ds$ equation is \[\beta ct=\rho\tan(s/\rho):\] $s/\rho=\pi/2$ corresponds to $t=\infty$. Remember, the particle is simply traveling in a straight line. See drawing at left.
\clearpage
\foilheadc{Centrifugal/Coriolis}
It is instructive to delve into this example a bit more deeply to connect it to other known facts about rotating coordinate systems.
Let us go back to the time-based Hamiltonian without fields:
\begin{equation}
H_t=c\sqrt{m^2c^2+P_x^2+\left(\frac{P_s}{1+x/\rho}\right)^2}
\end{equation}
Reminder: this is in fact total energy $E$, and it is constant here, since $\frac{\partial H_t}{\partial t}=0$.
Equations of motion follow from the other partial derivatives:
\begin{eqnarray}
\dot{x}=\frac{c^2P_x}{E}&\mbox{ and }&\dot{P_x}=\frac{c^2}{\rho E}\frac{P_s^2}{(1+x/\rho)^3}\\
\dot{s}=\frac{c^2}{E}\frac{P_s}{(1+x/\rho)^2}&\mbox{ and }&\dot{P_s}=0
\end{eqnarray}
So both $E$ and $P_s$ are constant.
\textcolor{mygreen}{Note: Non-relativistic $\Longrightarrow E/c^2\rightarrow m$.
Thus, $P_x=m\dot{x}$, but $P_s\neq m\dot{s}$, not quite: $ P_s=m\dot{s}\left(1+\frac{x}{\rho}\right)^2$}
To find accelerations, differentiate with time once more and make a few substitutions:
\fbox{$\ddot{x}=\frac{\dot{s}^2}{\rho}\left(1+\frac{x}{\rho}\right)=\omega^2r$} Centrifugal! \fbox{$\ddot{s}=\frac{-2c^2}{E}\frac{P_s}{(1+x/\rho)^3}\frac{\dot{x}}{\rho}=-2\omega\frac{\dot{x}}{1+x/\rho}$} Coriolis!\\
where $\omega\equiv\dot{s}/\rho$ and $r\equiv x+\rho$.
%\mbox{\ \ Centrifugal!}mbox{\ \ Coriolis!}
Goldstein:\\
\includegraphics[width=25cm]{../TrainingImages/CentCoriolis.png}
\normalsize
\foilheadc{RF Cavities; time-dependent field example}
\color{mygreen}In the accelerating cavities, there is an rf electric field that can be described as a harmonically time varying term in $A_s$. This may come as a surprise, since we tend to think of electric fields as coming from a time-varying potential $\Phi(t)$. Recall
\[\vec{\cal{E}}=-\nabla\Phi-\frac{\partial\vec{A}}{\partial t}\mbox{ and }\vec{B}=\nabla\times\vec{A} .\] We can always choose a function $\Psi(\vec{x},t)$ and change $\vec{A}$ and $\Phi$ according to
\[ \vec{A} \rightarrow \vec{A} + \nabla \Psi\mbox{ and }
\Phi \rightarrow \Phi - \frac{\partial\Psi}{\partial t} \]
It is extremely handy to zero $\Phi$, as this simplifies the Hamiltonian. This can be done by choice of appropriate \textcolor{red}{gauge}. In this gauge, $ \vec{\cal{E}}=-\frac{\partial\vec{A}}{\partial t}$, so in a region of no magnetic field and electric field only in the $s$-direction ($A_x=A_y=0$),
\begin{equation}\label{deds}
dE/ds=\partial{H_s}/\partial{t}=-q\partial{A_s}/\partial{t}
\end{equation}
\begin{center}
\includegraphics{../TrainingImages/simp_cav2}
\end{center}
We characterize the cavity by its ``Voltage'' $V$ which is the energy gain per charge ($\Delta E/q$) for a particle crossing the gap at just the right time. By inspection of eqn.\,\ref{deds}, we find the vector potential to be \begin{equation}\label{cavAs}
A_s=-\frac{V}{\omega}\ \delta(s-s_{\bf o})\ \cos(\omega t+\phi_{\bf o}). \end{equation}
This is for an infinitesimally short gap (usu.\ a good approx.) at $s=s_{\bf o}$, and rf angular frequency $\omega$. $\phi_{\bf o}$ is the particle's phase.
\textcolor{myblue}{Check it out: $H_s=-qA_s-\sqrt{\left(\frac{E}{c}\right)^2-m_0^2c^2-P_x^2-P_y^2}$, $dE/ds=\partial{H_s}/\partial{t}=-qV\ \delta(s-s_{\bf o})\sin(\omega t+\phi_{\bf o})$, integrate to get $\Delta E=-qV\sin(\omega t+\phi_{\bf o})$.}
\textcolor{mygreen} {Close inspection of this will show that this vector potential violates Maxwell's equations. For the cylindrical gap shown, the voltage $V$ is not constant across the gap but rises with $r=\sqrt{x^2+y^2}$; one must multiply by a modified Bessel function $I_0$\footnote{\color{mygreen}See H.\ Hereward in {\it Linear Accelerators}, Septier and Lapostolle (eds., 1970), and M.M.\,Gordon in \textit{Particle Accelerators} {\bf 14} (1983) p.\,119.}:
\[A_s=-\frac{V}{\omega}\ I_0\left(\frac{\omega r}{\beta\gamma c}\right)\ \delta(s-s_{\bf o})\ \cos(\omega t+\phi_{\bf o}) \]}
{One can use this vector potential in the Hamiltonian \begin{equation}\label{ham13}
H_s=-qA_s-\sqrt{\left(\frac{E}{c}\right)^2-m_0^2c^2-P_x^2-P_y^2}\end{equation} to derive the focal power of an rf gap. Note that $r\ll\lambda$ where $\lambda$ is the rf wavelength $2\pi c/\omega$, so that an appropriate approximation can be made for the Bessel function. Note further that this will only give the $\cos\phi$ term, not the $\sin^2\phi$ term.
But the neat thing is that we find the focal power of an rf gap knowing nothing about the fields except their symmetry.}
\color{black}
\foilheadc{Change Longitudinal Coordinates}
The Hamiltonian $H$ in eqn.\,\ref{simsqH} is awkward because it mixes small dynamic quantities $x,y,P_x,P_y$ with a large one $E$.
We only care about particles with a small $\Delta E$ deviation from the reference energy $E_0$, and a small $\Delta t$ deviation from the reference time\footnote{We will drop the subscripts $0$ on $\beta$ and $\gamma$ else the notation is too cluttered. It should be remembered that they are constants as regards the Hamiltonian.} $t_0=s/(\beta_0c)$. We do this with a canonical transformation from $(t,-E)$ to $(\Delta t,-\Delta E)$. The generating function is \begin{equation}\label{gend}
F(t,-\Delta E)=\left(t-\frac{s}{\beta c}\right)(-\Delta E-E_0)
\end{equation}
\clearpage
\begin{quote}\small
{\bf Canonical Transformations}
Reminder: Time $t$ is now a position coordinate and the independent variable is $s$, not $t$.
$F$ is a function of the old position ($t$) and new ``momentum'' ($\Delta E$). Therefore, we must have derivative of $F$ w.r.t.\ old position is old momentum:
\[\frac{\partial F}{\partial t}=-E,\]
and derivative of $F$ w.r.t.\ new momentum is new position:
\[\frac{\partial F}{\partial(-\Delta E)}=\Delta t.\]
Check whether this is true. You can find a more general treatment on \href{http://en.wikipedia.org/wiki/Canonical_transformation}{Wikipedia}.
\end{quote}
The new Hamiltonian is
\begin{equation}\label{newh}
\tilde{H}_s=H_s+\frac{\partial F}{\partial s}=H_s+\frac{E_0+\Delta E}{\beta c}
\end{equation}
\foilheadc{Magnetic field with mid-plane symmetry}
Let us now restrict ourselves to only magnetic cases ($\Phi=0$), and
magnets with median plane symmetry. $A_x=A_y=0$ for the appropriate
choice of \href{http://en.wikipedia.org/wiki/Gauge_fixing}{Gauge}, but then $A_s$ is a function of $x,y$ only:
\begin{equation} \label{simsqH}
H_s=-qA_s-(1+x/\rho)\sqrt{E^2/c^2-m^2c^2-P_x^2-P_y^2}
\end{equation}
Recall: (with $h\equiv 1+x/\rho$)
\begin{equation}
\label{curl1}
\vec{B}=\nabla\times \vec{A}=\left(
\frac{1}{h}\,\frac{\partial A_s}{\partial y}-\frac{\partial A_y}{\partial s}\,,\,
\frac{\partial A_x}{\partial s}-\frac{1}{h}\,\frac{\partial A_s}{\partial x}\,,\,
\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}
\right)
\end{equation}
If $A_x=A_y=0$, this becomes:
\begin{equation} \label{curl2}
B_x= \frac{1}{h}\,\frac{\partial A_s}{\partial y}\,,\,
B_y=-\frac{1}{h}\,\frac{\partial A_s}{\partial x}\,,\,
B_s=0
\end{equation}
\foilheadc{Expand $H$}
Combine eqns.\ \ref{newh}, \ref{simsqH}, keeping only terms to second order (and dropping constants) to get the Hamiltonian:
\begin{equation}
\tilde{H}_s(P_x,x,P_y,y,\Delta E,\Delta t;s)=-qA_s-x\frac{P_0}{\rho}+\frac{P_x^2}{ 2P}+\frac{P_y^2}{ 2P}-\frac{x\Delta E}{\beta c\rho}+\frac{(\Delta E)^2}{2\beta^3\gamma^3mc^3}
\end{equation}
\textcolor{mygreen}{Exercise \stepcounter{exe}\arabic{exe}: Prove this. (Note: This is actually kind of tricky: since there are first order terms inside the square root, it has to be expanded to second order.)}
Note that there is one remaining first order term, $-(P_0/\rho)x$. It matches and will be cancelled by a term $-(P_0/(q\rho))x$ in $A_s$, as we shall see.
\foilheadc{Trick: Use $P$ instead of $E$}
It is much nicer and more intuitive to use a distance $\tau=\beta c\Delta t$ instead of $\Delta t$. This is the distance in the lab frame that the particle is ahead of the reference particle. The conjugate coordinate, $\Delta E$, must also change to $\Delta E/(\beta c)$, but since\footnote{This might not be obvious. Start with $E^2=m^2c^4+P^2c^2$, so $E\Delta E=c^2\,P\Delta P$.} $\vec{p}=\frac{E}{c^2}\ \vec{v}$, this is simply the total momentum increment $\Delta P=P-P_0$.
\begin{equation}\label{H3DA}
H=-\left(qA_s+x\frac{P_0}{\rho}\right)-\frac{x\Delta P}{\rho}+\frac{P_x^2}{ 2P}+\frac{P_y^2}{ 2P}+\frac{(\Delta P)^2}{2P\gamma^2}
\end{equation}
\foilheadc{But how to choose $\rho$?}
What does it mean that $s$ is along the unperturbed orbit? It means that when $x=y=P_x=P_y=0$, they will remain zero, in particular, $\frac{dP_x}{ds}|_{x=y=0}=0$. For this to be true, we need
\[0=\frac{dP_x}{ds}=-\frac{\partial H_s}{\partial x}=
q\frac{\partial A_s}{\partial x}+\frac{1}{\rho}\sqrt{\frac{E^2}{c^2}-m^2c^2},\]
or using the expression for $\vec{B}$ (\ref{curl2}), we find that on the reference orbit, \begin{equation}
B_{y0}\ \rho=\frac{P_0}{q}\end{equation}
(where $P_0\equiv\sqrt{\frac{E_0^2}{c^2}-m^2c^2}=\beta\gamma mc$ the kinetic momentum) as we previously found for the case of a uniform $B$. But the present case is much more general; $\rho$ and $B_{y0}$ are functions of $s$. It is for this reason that total momentum is often called, simply, $B\rho$, which is called the beam's \textcolor{red}{rigidity}.
So on the reference orbit, $B_y$ has a value given by the momentum and $\rho(s)$. Away from the orbit, both $B_x$ and $B_y$ can be expanded:
\begin{eqnarray}
B_x&=&B'(s)\ y+...\nonumber\\\label{linB}
B_y&=&B_{y0}(s)+B'(s)\ x+...
\end{eqnarray}
Notice that the coefficient $B'$ is the same in both cases. This is a consequence of $\nabla\times\vec{B}=0$, i.e.\ $\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}=0$.
What is the vector potential $A_s$ that is consistent with such a field? We need it for the Hamiltonian.
\foilheadc{Transverse focusing}
The main trick in accelerator design is to customize $A_s$ to control the particles' motion.
For transverse motion, the goal is linear restoring force. This means the optical elements that focus must result in quadratic terms in $A_s$. We ignore the rf cavities and changes in $E$ (although we allow different particles to have different $E$).
Then \begin{equation}\label{linAs}
A_s=-\frac{P_0}{q}\left[\frac{x}{\rho}+\left(\frac{1}{\rho^2}-K_1\right)\frac{x^2}{2}
+K_1\frac{y^2}{2}\right],\end{equation}
where $K_1\equiv -B'/(B\rho)$, $B'=\partial B_y/\partial x$.
The first term $x/\rho$ is, as we've already established, the one needed to cancel the linear term in the dynamic part of $H_s$. The second term $x^2/(2\rho^2)$ is focusing that comes automatically in a bend. If one imagines the reference orbit as a circular arc, then another particle starting at a slightly deviated angle also follows a circular arc of same radius. These circles cross each other twice per turn.
The remaining quadratic terms have equal and opposite coefficients for $x^2$ and $y^2$. This is a consequence of the Maxwell equation $\nabla\times\vec{B}=0$, as was stated above.
\textcolor{mygreen}{Exercise \stepcounter{exe}\arabic{exe}: Use the curl expression \ref{curl2} in \ref{linAs} to verify that this vector potential gives the magnetic field \ref{linB}. You will find leftover terms of higher order. We are ignoring these for the moment as we are concentrating on the linear motion only.}
\clearpage
The focusing devices that focus in one transverse direction defocus in the other. For straight beamlines ($1/\rho=0$) and transverse field, the only way to focus in both directions is to alternate the polarity of the focusing devices (quadrupoles). At first it might seem that then the net focusing would be zero. This was the assumption made for early accelerators, so designers set $K_1$ to approximately $1/(2\rho^2)$, making focusing in both directions the same sign. This is now called \textcolor{red}{weak focusing}.
\begin{wrapfigure}{l}{19cm}
\includegraphics[width=18cm]{../TrainingImages/quads}\caption{3 quadrupoles}\vspace{-4cm}
\end{wrapfigure}
\small During the 50's, it was discovered that by alternating the sign of $K_1$, far from canceling the focusing, it could be made very strong. This came to be called \textcolor{red}{strong -} or \textcolor{red}{alternating gradient - focusing}.
\clearpage\normalsize
\foilheadc{Multipoles}
For the moment, let's restrict ourselves to straight reference trajectory ($\rho=\infty$) and transverse fields. Then $A_s$ depends only upon $x$ and $y$, and as $\nabla \times \left( \nabla \times \vec{A} \right) = \nabla(\nabla \cdot \vec{A}) - \nabla^{2}\vec{A}= \vec{0}$ in the Coulomb gauge implies $\nabla^2A_s=0$, it means $A_s$ has the following form:
\begin{equation}\label{mpoles}
A_s=\Re\left[\sum_nC_n(x+iy)^n\right]
\end{equation} ($\Re$ means real part)
where $C$ is complex. The $n=1,\,2,\,3$, etc.\ are called \textcolor{red}{dipole}, \textcolor{red}{quadrupole}, \textcolor{red}{sextupole}, etc.
\textcolor{mygreen}{Exercise \stepcounter{exe}\arabic{exe}: Work out the first 3 multipole magnetic fields ($n=1,\,2,\,3$).}
\foilheadc{Higher Order}
\textcolor{mygreen}{For $n\geq 3$, the forces e.g.\ $P_x'=-\partial H/\partial x$ are ``higher order'' or nonlinear and beyond the scope of these lectures.}
\textcolor{mygreen}{But note as well that this $\vec{A}$ does not satisfy $\nabla\times\vec{B}=\vec{0}$ because e.g.\ $\frac{\partial B_x}{\partial s}\neq0$. This is an approximation in which we are ignoring effects of ``fringe fields''. Time does not permit for a proof, but these effects are ``higher order'', so when we expand the Hamiltonian only up to quadratic, they do not appear.}
\foilheadc{Linear Motion Hamiltonian for Magnetic field with Mid-Plane Symmetry}
Finally, for canonical variables $(x,P_x,y,P_y,\tau,\Delta P)$, independent variable $s$:
\begin{equation}\label{H3D}
H=\frac{P_x^2}{ 2P}+P_0\left(\frac{1}{\rho^2}-K_1\right)\frac{x^2}{2}
+\frac{P_y^2}{ 2P}+P_0K_1\frac{y^2}{2}-\frac{x\Delta P}{\rho}
+\frac{(\Delta P)^2}{2P\gamma^2}
\end{equation}
The cross-term $-x\Delta P/\rho$ interestingly shows the power and efficiency of the Hamiltonian approach, as it encapsulates two seemingly unrelated effects. (1)For $\Delta P>0$, $P_x$ will increase ($P_x'=\Delta P/\rho$). (2)Particles of higher momentum than the reference on average travel on an outer orbit $x>0$ and so counter-intuitively fall \textbf{behind} the reference particle when going through a bend ($\tau'=-x/\rho$).
\foilheadc{Note on longitudinal motion}
Notice only one term contains any trace of a ``relativistic effect''; the last one. As result of $P_x$, a particle deviates in the $x$ direction: $dx/ds=P_x/P$. Same for $P_y$. But in the longitudinal direction, the extra momentum causes a growing deviation that is scaled down by the relativistic factor $\gamma^2$: $d\tau/ds=\frac{1}{\gamma^2}\Delta P/P$. For the LHC ``energy'' of 7\,TeV, this factor is about $10^{-8}$, meaning that practically speaking, all are going at the same speed, they are locked in step: a proton at 1\% higher energy than reference is traveling only one part in $10^{10}$ faster. (You can check this directly by comparing $\beta$ for 7\,TeV and 7.07\,TeV.)
\color{mygreen}
\foilheadc{Digression: Cyclotrons}
Cyclotrons do not have a fixed reference trajectory. For example, the TRIUMF cyclotron has an injection energy of 0.3\,MeV which is at a radius of 0.25\,m, but top energy is 500\,MeV, radius 7.5\,m; a factor of 30!
At any energy, there is a closed orbit (Bloch's theorem) and this is called the \textcolor{red}{equilibrium orbit}. It makes little sense to use $s$ as independent variable. Instead, the azimuth, $\theta$ is used. The Canonical variables $(-E,t)$ and $(y,P_y)$ are as before, but in the median plane, we use $r$ and $\theta$ with conjugate coordinates:
\begin{align}
P_r=&p_r+qA_r\\
P_\theta=&rp_\theta+qrA_\theta
\end{align}
The Hamiltonian is $-P_\theta$:
{\small\begin{equation}
H(r,P_r,y,P_y,t,E;\theta)=-qrA_\theta-r\sqrt{E^2/c^2-m^2c^2-(P_r-qA_r)^2-(P_y-qA_y)^2} \end{equation}}
The equations of motion can finally be cast into the following form:
{\small\begin{align}
\frac{dp_r}{d\theta}=&Q-rB_y+\frac{r}{Q}\ p_yB_\theta\\
\frac{dr}{d\theta}=&\frac{r}{Q}\ p_r\\
\frac{dp_y}{d\theta}=&rB_r-\frac{r}{Q}\ p_rB_\theta\\
\frac{dy}{d\theta}=&\frac{r}{Q}\ p_y\\
\frac{dt}{d\theta}=&\frac{r}{Q}\ E
\end{align}
where $Q\equiv\sqrt{p^2-p_r^2-p_y^2}$.
(These are in cyclotron units: $B$ in units of central field $m\omega_0/q$, $t$ in units of $\omega_0^{-1}$, lengths in units of $c/\omega_0$, $E$ in units of $mc^2$, $p$ in units of $mc$.)
We Runge-Kutta integrate these through the measured magnetic field. The measured field is made on an $(r,\theta)$ grid (I-beam that pivots at centre) We integrate with the same steps of $\theta$, but we must interpolate radially. The tricky part is in devising an
interpolation scheme for $B$ consistent with Maxwell's equations.
For some starting $E$, with $p_r=0$ and $r$ set to some guess, we integrate over one sector of an $N$-sector machine. But of course the
orbit will not close on itself. To find the closed orbit, we track the
first order differentials of motion as well, then invert the resulting transfer marices, and iterate to find the values $(r,p_r)$ of the closed orbit
Finally, the motion is analyzed for isochronism and vertical focusing strength. Since the particles must stay in step with the rf accelerating field, and there are 1200 turns and 6000 rf periods from injection to extraction, the field must be correct to about $1:10^5$. In making the adjustment (with steel shims), if it's not done carefully, the vertical focusing frequency (tune) becomes imaginary. }
\includegraphics[height=\textheight]{triumf72.jpg}
\color{black}
\foilheadc{Back to Synchrotrons: Transverse}
We can renormalize the Hamiltonian by dividing it and all momenta by $P_0$. Ignoring the $\Delta P$ for the moment, we see the two transverse motions are independent and can be separated. Each has the same form:
\begin{equation}
H_y=\frac{P_y^2}{2}+K(s)\,\frac{y^2}{2}
\end{equation}
This is a harmonic oscillator equation, but with varying restoring force (\href{http://en.wikipedia.org/wiki/Hill_differential_equation}{Hill's equation}. We can combine the equations for $y$ and $P_y$:
\begin{equation}\label{eom}
y''+K(s)y=0
\end{equation}
For a ring, $K$ is periodic with period equal to the orbit length. But there is usually more periodicity than this, as it is convenient to make the ring of identical sectors. As well, for beamlines over long distances, it is advantageous, both for physics reasons and economically, to compose the line of identical \textcolor{red}{cells}.
Let us launch a particle with position $y=y_0$ and momentum $P_y=y'_0$. After one cell, it will have new coordinates $(y_1,y'_1)$, after $n$ cells, $(y_n,y'_n)$. Plotting these points in $(y,y')$ \textcolor{red}{phase space}, we find that they all lie on an ellipse, and that they progress a fixed angle for every cell. This angle is called the \textcolor{red}{phase advance per cell} or divided by $360^\circ$ it is the \textcolor{red}{cell tune}, or if the period is the whole ring, simply the \textcolor{red}{tune}. IOW, the tune of a ring is the number of oscillations in phase space for one revolution.
If $K$ is a constant, this is easily understood: $y$ and $y'$ oscillate as $\cos\sqrt{K}s$ and $\sin\sqrt{K}s$ and so the tune is $\nu_y=\sqrt{K}L/(2\pi)$, where $L$ is the ring circumference.
But as we've shown, the ring is composed of $K$'s that alternate in sign.
\foilheadc{``Strong Focusing''}
Focusing is thus achieved by using ``quadrupoles''. These focus in $x$, defocus in $y$, or vice versa. So would not the net effect cancel?
No.
The reason is an effect called ``strong focusing''. Roughly speaking, a series of lenses of opposite signs is net focusing because the beam ``size'' is larger in the focusing lenses than in the defocusing ones. This causes the focusing to have more effect than the defocusing.
Historical note: This seemingly trivial effect was discovered in the 1950s and not in full use until the 1960s.
\foilheadc{Thin lens limit}
An idealized case that (perhaps surprisingly) contains most of the physics is the limit of infinitesimally short lenses:\[K(s)=\pm\delta(s)/f\]where $f$ is the focal length. {\small (As we have seen, $K=B'/(B\rho)$, $\frac{1}{f}=B'l/(B\rho)$ where $l$ is the lens length. In this thin lens limit, $B'\rightarrow\infty,\,l\rightarrow 0$ as $B'l$ remains fixed.)} This is simple to integrate. Through the thin lens, $y_1=y_0,\,y'_1=y'_0\mp y_0/f$, or in matrix form:\footnote{Reminder: $y'=P_y/P_0$}\begin{equation}\label{tlens}
\begin{bmatrix}y_1\cr y'_1\end{bmatrix}=\begin{bmatrix}1&0\cr\mp\frac{1}{f}&1\end{bmatrix}
\begin{bmatrix}y_0\cr y'_0\end{bmatrix} \end{equation}
The drift $d$ between lenses is even simpler:
\begin{equation}\label{drift}
\begin{bmatrix}y_1\cr y'_1\end{bmatrix}=\begin{bmatrix}1&d\cr 0&1\end{bmatrix}
\begin{bmatrix}y_0\cr y'_0\end{bmatrix} \end{equation}
The simplest cell is one where we have Focus (F), drift (O), defocus (D), drift(O); for this reason called a \textcolor{red}{FODO} cell. Stack the matrices:
\[\begin{bmatrix}1&d\cr 0&1\end{bmatrix}
\begin{bmatrix}1&0\cr \frac{1}{f}&1\end{bmatrix}
\begin{bmatrix}1&d\cr 0&1\end{bmatrix}
\begin{bmatrix}1&0\cr-\frac{1}{f}&1\end{bmatrix}=
\begin{bmatrix}1-\frac{d}{f}-\frac{d^2}{f^2}&2d+\frac{d^2}{f}\cr -\frac{d}{f^2}&1+\frac{d}{f}\end{bmatrix}\]
Notice that the cell is always focusing: $M_{21}<0$ irrespective the sign of $f$ so a particle coming in with a certain $y'$ will have it decremented, meaning it is bent towards the axis.
\textit{(Note: Matrices of this kind apply to any kind of linear optical element and are called \textcolor{red}{transfer matrices}).}
\begin{figure}\centering
\includegraphics[width=20cm]{../TrainingImages/periodic.png}\caption{$60^\circ$ FODO cells}
\end{figure}
The reason for the net focusing is clear from the above figure. The beam is smaller in the defocusing lenses than in the focusing lenses. In this particular example, the phase advance per cell is $\mu=60^\circ$.
% As stated above the particle advances around an ellipse by an angle $\mu$ per cell. The general matrix describing this kind of motion is:
% \[\mbox{\textbf{M}}_{\bf cell}(\mu)=\mbox{\textbf{I}}\cos\mu+\mbox{\textbf{J}}\sin\mu\mbox{, where }\mbox{\textbf{I}}=\begin{bmatrix}1&0\cr 0&1\end{bmatrix}
% \mbox{, and }\mbox{\textbf{J}}=\begin{bmatrix}\alpha_{\rm cs}&\beta_{\rm cs}\cr -\gamma_{\rm cs}&-\alpha_{\rm cs}\end{bmatrix}\] and $\gamma_{\rm cs}=(1+\alpha_{\rm cs}^2)/\beta_{\rm cs}$. (I apologize that these symbols duplicate and have nothing to do with the relativistic parameters $\beta$ and $\gamma$.)
%
% \textcolor{myblue}{Exercise \stepcounter{exe}\arabic{exe}: Prove that $\mbox{\textbf{M}}_{\bf cell}(\mu_2)\mbox{\textbf{M}}_{\bf cell}(\mu_1)=\mbox{\textbf{M}}_{\bf cell}(\mu_1+\mu_2)$}.
%
% $\alpha_{\rm cs}$ and $\beta_{\rm cs}$ are known as the \textcolor{red}{Courant-Snyder parameters}. The parameter $\alpha_{\rm cs}$ is an asymmetry parameter related to the ellipse tilt. The parameter $\beta_{\rm cs}$ is the aspect ratio. For the thin-lens FODO, we find:
% \begin{align}
% \label{TLcosmu}\cos\mu&=1-\frac{d^2}{2f^2}\\
% \label{TLbeta}\beta_{\rm cs}&=2d\,\frac{1+\sin(\mu/2)}{\sin\mu}
% \end{align}
%
% Motion within the FODO system is not guaranteed to be stable: if $f