ANL-AAI-TN-2007-01

May 2007

TRAJECTORIES THROUGH DEFLECTING CAVITIES

Kwang-Je Kim

Advanced Photon Source, Argonne National Laboratory, Argonne, IL, 60439

Abstract

A deflecting cavity is required for transverse-longitudinal emittance
exchange scheme \[1,2\]. John Power found, by tracking calculation, that
a reference particle receives a transverse off-set after passing through
a deflecting cavity \[3\]. Don Edwards computed the transfer matrix of a
1-cell, pill-box deflecting cavity including the offset \[4\]. Here we
show that the off-set can be made to vanish either by adding a short
cavity at each end of the main cavity, or by choosing the size of the
beam hole properly.

TRAJECTORIES THROUGH DEFLECTING CAVITIES

1.  **Introduction**

A deflecting cavity is required for transverse-longitudinal emittance
exchange scheme \[1,2\]. John Power found, by tracking calculation, that
a reference particle receives a transverse off-set after passing through
a deflecting cavity \[3\]. Don Edwards computed the transfer matrix of a
1-cell, pill-box deflecting cavity including the offset \[4\]. Here we
show that the off-set can be made to vanish either by adding a short
cavity at each end of the main cavity, or by choosing the size of the
beam hole properly.

2.  **π-Mode Pill-Box Cavity**

Here we consider a cavity consisting of several pill-box cells in π
-mode configuration. For deflection, the mode in each cell is TM~011~.
The non-vanishing components of the electromagnetic field near the
cavity axis is

> $E_{z}\  = \ \varepsilon\ x\ \Theta(z)\ \text{cos}\ \omega\ t$, (1)
>
> $B_{y\ } = \ \varepsilon\ \Theta(z)\ \frac{1}{\omega}\ \text{sin}\ \omega\ t$.
> (2)

The coordinate system is such that the cavity axis is the z-direction,
while x and y are the two transverse coordinates. The step function Θ(z)
is necessary to represent the fact that the cells are connected in a π
-mode configuration. It is defined to be unity in magnitude and
alternating in signs from one λ/2 cell to the next. We choose z=0 be the
center of the middle cell and assume that the cells are placed
symmetrically about the middle cell. Thus,

$\begin{array}{r}
\Theta(z)\  = \ \text{1;}\ \frac{- \lambda}{4}\  \leq \ z\  \leq \frac{\lambda}{4}\ , \\
 \\
\  = \  - \text{1;}\ \frac{- 3}{4}\ x\  < \ z\  < \  - \frac{\lambda}{4},\ \text{or}\ \frac{\lambda}{4}\  < \ z\  < \ \frac{3}{4}\ \lambda\text{.}
\end{array}$ (3)

......

Here λ = 2π c/ω is the free space wavelength. The cell arrangement is
shown in Fig. 1.

The equation of motion is then

$\frac{d^{2}x}{\text{dz}^{2}}\  = \  - G\ \frac{1}{k}\ \Theta(z)\ \text{sin}\ \omega\ t$,
(4)

$\frac{\text{dδ}}{\text{dz}}\  = \ G\ \Theta\ (z)\ \left( x\ \text{cos}\ \omega\ t\  + \ \frac{1}{k}\ \frac{\text{dx}}{\text{dz}}\text{sin}\ \omega\ t \right)$
. (5)

Here

$G\  = \ \frac{\varepsilon}{E_{e}}$, (6)

where E~e~ is the electron energy. We set

$\text{kz}\  - \ \text{ωt}\  = \ \theta$, (7)

where

$k\  = \ \text{ω/c}\  = 2\text{π/λ}$, (8)

$\theta\  = \ k\ \Delta\ z$. (9)

Here Δz is the longitudinal position of the particle relative to the
reference particle. Throughout this note, we will regard *θ* as
constant, neglecting a small change due to acceleration.

By using Eq (7), we can integrate Eq. (4) with respect to z. For a
cavity extending from *-ℓ* to *ℓ*, we need to use integral:

$\begin{array}{r}
\ \int_{- \ell}^{\ell}{\Theta(z)\text{sin}\omega t\ \text{dz}}\  = \ \int_{- \ell}^{\ell}{\Theta(z)\text{sin}\ \left( \text{kz}\  - \ \theta \right)\ \text{dz}} \\
 \\
\  = \int_{- \ell}^{\ell}{\Theta(z)\left( \text{sin}\ \text{kz}\ \text{cos}\ \theta\  - \ \text{cos}\ \text{kz}\ \text{sin}\ \theta \right)}\ \text{dz} \\
 \\
\  = \  - 2\ \text{sin}\ \theta\ \int_{o}^{\ell}{\Theta(z)\ \text{cos}\ \text{kz}\ \text{dz}}\text{.}
\end{array}$ (10)

In the above we used the fact that *Θ(z) = Θ(-z)* to drop the terms in
the integrand containing the factor sin *kz* since it is an odd
function.

If

$\ell\  = \ \frac{\lambda}{4}\ \text{n,}\ n\  = \ 1,\ 2,\ \text{.}\text{.}$
(11)

Then

$\int_{o}^{\frac{\lambda n}{4}}{\Theta(z)}\ \text{cos}\ \text{kz}\ \text{dz}\  = \ \frac{n}{k}$.
(12)

$$$$

Therefore the transverse angular kick increases linearly as the number
of the cells, which is why a multi-cell cavity is used.

An odd integer n corresponds to the case of n-full cells. Note that the
number of the full cells can be only an odd number due to our symmetry
assumption. An even integer n, on the other hand, corresponds to the
case where a number of full cells are flanked at both ends by a
half-cell, a cell of length *λ/4*. See Fig. 1

As an example where Eq (11) is not satisfied, consider the case where a
single full cell is flanked by a cavity of length shorter than *λ/2*.
Then

$\frac{\lambda}{4}\  \leq \ \ell \leq \ \frac{\lambda}{2}$. (13)

Performing the integral Eq. (10), we obtain the change in the slope
$x^{'} \equiv \ \frac{\text{dx}}{\text{dz}}$ as follows:

$x_{2}^{'}\  = \ x_{1}^{'}\  + \ \frac{2G}{k^{2}}\ \left( 2\  - \text{sin}\ \varphi \right)\ \text{sin}\ \theta$.
(14)

Here and in the following, $\varphi$ is one half of the total length of
the cavity in radians:

$\varphi\  = \ k\ \ell$. (15)

The transverse displacement can be computed by integrating Eq. (4) by
means of manipulation such as follows:

$\begin{array}{r}
\int_{- \ell}^{\ell}\text{dz}\ \int_{- \ell}^{z}{\text{sin}\ \omega\ t\ \Theta(z)\ \text{dz}}\  \\
 \\
\  = \ \int_{- \ell}^{\ell}\text{dz}\ (\ell\  - \ z)\ \text{sin}\ \omega t\ \Theta(z)\ \text{dz} \\
 \\
\  = \ell\ \int_{- \ell}^{\ell}\text{dz}\ \text{sin}\ \omega t\ \Theta(z)\ \text{dz}\  - \ \int_{- \ell}^{\ell}{z\ \text{sin}\ \omega t\ \Theta(z)\ \text{dz}\text{.}}
\end{array}$ (16)

The first term involves the same integral as in Eq. (10), and the
evaluation of the integral in the second term is facilitated by
expanding the integrand into the odd and even functions in *z*. We then
obtain

$x_{2}\  = \ x_{1}\  + \ 2x_{1}^{'}\ell\  + \ \frac{2G\ell}{k^{2}}\ \left( 2\  - \ \text{sin}\ \varphi \right)\text{sin}\ \theta\  + \ 2G\ \frac{1}{k^{3}}\ \left( 2\  + \ \varphi\ \text{cos}\ \varphi\  - \ \text{sin}\ \varphi \right)\ \text{cos}\ \theta$.
(17)

Equations (14) and (17) reproduce the results obtained by D. Edwards for
the case $\varphi$ *= π/2* \[3\].

The last term in Eq. (17) is the transverse offset of the reference
particle for which *x~1~ = x~1~′ = θ = 0*. Guessing that the situation
here is similar to the case of the particle trajectory through a
periodic undulator, where a "half period" must be added at both ends to
avoid a net displacement, one may expect that the offset may be removed
if a cell of length *λ/4*, a "half cell" is introduced before and after
the main cell. This correspond to the case *ℓ = λ/2 or* $\varphi$ *= π*.
However, the trick does not work; the last term does not vanish when
$\varphi$ *= π*.

However, the off-set will vanish if $\varphi$ is chosen so that

$2\  + \ \varphi\ \text{cos}\ \varphi\  - \ \text{sin}\ \varphi\  = \ 0$.
(18)

The solution of this equation for *π /2 ≤* $\varphi$ *≤ π*
(corresponding to the range (13)) is

$\varphi\  = \ 2\text{.}\text{14}\ \text{or}\ \ell\  = \ \frac{\lambda}{4}\ \left( 1\  + \ 0\text{.}\text{36} \right)$.
(19)

Therefore, the length of the end cell should be very short, about one
third of the half cell.

The momentum kick can be obtained by inserting the transverse trajectory
expressions into Eq. (5) and integrating it. This was done by D. Edwards
for $\varphi$ *= π /2*. For a general case with an arbitrarily ℓ,
however, the computation is quite tedious and will not be reported here.

3.  **Hybrid EM Mode**

The pillbox cavity considered in the previous section is the limiting
case where the radius of the beam hole *ρ* vanishes. An approximate but
simple expression for the electromagnetic field taking into account the
non-vanishing *ρ* was "derived" and known as the hybrid electromagnetic
(HEM) mode \[5\]. Its components of the electromagnetic field for the
standing wave case are as follows:

> $\begin{array}{r}
> E_{x}\  = \ \varepsilon\ \frac{k}{4}\ \left( \rho^{2}\  + \ x^{2}\  - \ y^{2} \right)\ \text{sin}\ z\ \text{cos}\ \omega t, \\
>  \\
> E_{y}\  = \ \varepsilon\ \frac{k}{2}\ \text{xy}\ \text{sin}\ \text{kz}\ \text{cos}\ \omega t, \\
>  \\
> E_{z}\  = \ \varepsilon\ x\ \text{cos}\ \text{kz}\ \text{cos}\ \omega\ t, \\
>  \\
> \text{cB}_{x}\  = \ \varepsilon\frac{k}{2}\ \text{xy}\ \text{cos}\ \text{kz}\ \text{sin}\ \omega t, \\
>  \\
> \text{cB}_{y}\  = \  - \varepsilon\frac{1}{k}\ \left( \frac{(k\rho)^{2}}{4}\  - \ 1\  + \ \frac{k^{2}(x^{2}\  - \ y^{2})}{4} \right)\ \text{cos}\ \text{kz}\ \text{sin}\ \omega t, \\
>  \\
> \text{cB}_{z}\  = \  - \varepsilon\ y\ \text{sin}\ \text{kz}\ \text{sin}\ \omega t\text{.}
> \end{array}$ (20)

The factors sin *kz* and cos *kz* replace *Θ(z)* in the pill box case.
These field expressions result from the Maxwell equations with two
assumptions: First, each field components is a sum of terms, each of
which is a product of two functions, one a quadratic polynomial
involving the transverse coordinates x and y and another depending only
on the longitudinal coordinate z. Second, the azimuthal component of the
electric field vanishes at the beam hole radius
$\sqrt{x^{2}\  + \ y^{2}}\  = \ \rho$. Ref. 5 gave a derivation of the
HEM mode for traveling wave case. The derivation for the case of the
standing wave is given in Appendix.

The field computed by Microwave Studio for a realistic cavity with a
beam hole is shown in Fig. 2 and resembles closely those given by Eq.
(20), except that the field amplitudes decrease rapidly outside the
cavity.

Neglecting the quadratic terms, the trajectory equations for a HEM
cavity are:

$\frac{d^{2}x}{\text{dz}^{2}}\  = \ \frac{G}{k}\ \left\{ \xi\ \text{sin}\ (2\text{kz}\  - \ \theta)\  - \ \frac{1}{2}\ \left\lbrack \text{sin}\ (2\text{kz}\  - \ \theta)\  - \ \text{sin}\ \theta \right\rbrack \right\}$,
(21)

$$$\frac{d\delta}{\text{dz}}\  = \ \frac{G}{2k}\ \left\{ \text{kx}\ \lbrack\text{cos}\ (2\text{kz}\  - \ \theta)\  + \ \text{cos}\ \theta\rbrack\  + \ x^{'}\ (1\  - \ \xi)\ \lbrack\text{sin}\ (2\text{kz}\  - \ \theta) - \ \text{sin}\ \theta\rbrack \right\}$.
(22)

In the above

$\xi\  = \ \left( \frac{k\rho}{2} \right)^{2}$. (23)

Equation (21) can be integrated. Neglecting terms quadratic in x~1~,
x~1~′, and θ, the results are:

$x_{2}^{'}\  = \ x_{1}^{'}\  + \ \frac{G}{k^{2}}\ \theta\ \left( \varphi\  - \ \left( \xi\  - \ \frac{1}{2} \right)\ \text{sin}\ 2\ \varphi \right)$,
(24)

$$$\begin{array}{r}
x_{2}^{'}\  = \ x_{1}\  + \ 2\ell\ x_{1}\  + \frac{G}{2k^{3}}\ \left( \xi\  - \ \frac{1}{2} \right)\ \left( 2\varphi\ \text{cos}\ 2\varphi\  - \ \text{sin}\ 2\ \varphi \right)\ , \\
 \\
\  + \ \frac{G}{k^{3}}\theta\ \left\lbrack \varphi\  - \ \left( \xi\  - \ \frac{1}{2} \right)\ \text{sin}\ 2\ \varphi\  \right\rbrack\ \varphi\text{.}
\end{array}$ (25)

Equations (24) and (25) correspond to Eq. (14) and (17) for the pill box
cavity, respectively, except here $\varphi$ *= kℓ* is arbitrary, not
restricted by the inequality, Eq. (13).

We note from Eq. (25) that the transverse offset for the reference
particle, for which *x*~1~ *= x′*~1~ *= θ = 0*, vanishes if *2*$\varphi$
cos *2*$\varphi$ *-* sin *2*$\varphi$ *= 0*, which is satisfied if

$\ell\  = \ \frac{\varphi}{k}\  = \ \frac{\lambda}{4}\ \left( 1\text{.}\text{43},\ 2\text{.}\text{46},\ 3\text{.}\text{47},\ \text{.}\text{.} \right)$.
(26)

The first solution in the above, *ℓ = 1.43 λ/4*, is close to the
solution for the pillbox case, Eq. (19).

The offset also vanishes when $\xi\  = \ 12,\ i\text{.}e\text{.},$

$\ \rho\  = \ \frac{1}{\sqrt{2}\ \pi}\ \lambda$ (27)

Although the solution given above will be modified for a realistic
cavity with field vanishing gradually outside the cavity, it is
reasonable to expect that there will be solutions for a realistic cavity
resembling these solutions.

For the realistic cavity worked out by John Power \[3\], we have

$\begin{array}{r}
\lambda\  = \ \text{23}\text{.}\text{06}\ \text{cm}\ (L - \text{Band}) \\
 \\
\rho\  = \ 4\text{.}\text{392}\ \text{cm}
\end{array}$

Using these values, we obtain (ξ -- ½) = 0.142, a rather small value.
The corresponding transverse offset is significantly smaller than that
for the single cell pill box cavity. In such a case, the contribution
from the terms linear in *θ* are not negligible, and may also change the
sign of the transverse offset. John Power found such a small offset.

To find the momentum kick, it is necessary first to find expressions for
*x(z)* and *x′ (z)* within the cavity, insert them into Eq. (22), and
integrate the resulting expression. The calculation is laborious and was
done by using Mathematica. The result for a general case contains too
many terms. For the case *ξ -- ½ = 0*, the result simplifies
considerably and is as follows:

$\begin{array}{r}
\delta_{2}\  = \ \delta_{1}\  + \ \frac{G}{2k^{2}}\ \left( x_{1}k\  + \ x_{1}^{'}\varphi \right)\left( 2\varphi\  + \ \text{sin}\ 2\varphi \right) \\
 \\
\  + \ \frac{1}{2}\left( \frac{G}{k^{2}} \right)^{2}\ \theta\ \varphi^{2}\ \left( \frac{2}{3}\varphi\  + \frac{1}{2}\ \text{sin}\ 2\varphi \right)
\end{array}$ (28)

APPENDIX

We consider Maxwell equation in free space with the harmonic time
dependence e^-iωt^. The two curl equations are:

$\nabla\  \times \ E\  = \text{ikc}B$ (A-1)

$\nabla\  \times \ (cB)\  = \  - \text{ik}E$ (A-2)

We assume that each field component is of the following form:

$Τ_{i}(x,y)\ Ζ_{i}(z)$

Where T~i~(x,y) is polynomial in x and y containing up to quadratic
terms.

Now consider the modification of the field profile due to the circular
beam hole in a TM~011~ cavity as shown in Fig. 3. Due to the surface
current, charge will accumulate at the hole radius, inducing an electric
field in the hole region. From the symmetry of the field distribution,
we can then write down the most general expression for each field
component quadratic in the transverse variables x and y as follows:

$E_{x}\  = \ a\  + \ b\ \left( x^{2}\  + \ y^{2} \right)\  + \ c\ \left( x^{2}\  - \ y^{2} \right),$
(A.3)

$E_{y}\  = \ d\ x\ y,$ (A.4)

$E_{z}\  = \ e\ x,$ (A.5)

$\text{cB}_{x}\  = \ p\ x\ y,$ (A.6)

$\text{cB}_{y}\  = \ q\  + \ u\left( x^{2}\  + \ y^{2} \right)\  + \ v\left( x^{2}\  - \ y^{2} \right),$
(A.7)

$\text{cB}_{z}\  = \ w\ y$ (A.8)

Here the coefficients a, b,...are functions of z.

Each field component must satisfy the wave equation:

$\left( \nabla_{}^{2}\  + \ \frac{\partial^{2}}{\partial z^{2}}\  + \ k^{2} \right)\ \left( E_{i},\ B_{i} \right)\  = \ 0$
(A.9)

Inserting Eq. (A.3) into (A.9), and demanding this be satisfied for
arbitrary values of x and y, we find

$4b\  + \ \left( \frac{\partial^{2}}{\partial z^{2}}\  + \ k^{2} \right)\ a\  = \ 0$,
(A.10)

$\ \left( \frac{\partial^{2}}{\partial z^{2}}\  + \ k^{2} \right)\ b\  = \ 0$
(A.11)

$\ \left( \frac{\partial^{2}}{\partial z^{2}}\  + \ k^{2} \right)\ c\  = \ 0$.
(A.12)

From (A.10) and (A.11)

$\ \left( \frac{\partial^{2}}{\partial z^{2}}\  + \ k^{2} \right)^{2}\ a\  = \ 0$,
(A.13)

which then implies

$\ \left( \frac{\partial^{2}}{\partial z^{2}}\  + \ k^{2} \right)\ a\  = \ 0$.
(A.14)

From (A.14) and (A.15), it then follows:

$b\  = \ 0$ (A.15)

Thus, we see that E~x~ satisfies the transverse Laplace equation:

$\nabla_{}^{2}\ Ε_{x}\  = \ 0$. (A.16)

From Eqs. (A.4) and (A.5), it also follows that

$\nabla_{}^{2}\ \left( Ε_{y},\ Ε_{z} \right)\  = \ 0$ (A.17)

Similarly we can show that u=0, and therefore all components of magnetic
field satisfy

$\nabla_{}^{2}\ Β_{i}\  = \ 0$ (A.18)

It then follows that all coefficients satisfy (A.14) and is proportional
to

$e^{i\overline{k}z}$

Here $\overline{k}$ is equal to k = ω/c in magnitude but can be of
opposite sign. This distinction is necessary later to construct the
standing wave solution.

We now look at the curl equations component by component, using Eq.
(A.3) -- (A.8). We then find the following relations among the
coefficients:

$d\  + \ 2c\  + \ k\overline{k}q\  = \ k^{2}a$, (A.19)

$d\  + \ 2c\  = \  - i\overline{k}e$, (A.20)

$i\overline{k}a\  - \ e\  = \ \text{ikq}$, (A.21)

$w\  = \  - e\left( \frac{\overline{k}}{k} \right)$. (A.22)

There are no more relationships from Maxwell equations and the symmetry.

We now impose the condition that the tangential component of the
electric field at the radius *ρ* of the beam hole vanishes:

$- x\ E_{y}\  + \ y\ E_{x}\  = \ 0\ \text{for}\ x^{2}\  + \ y^{2}\  = \ \rho^{2}$.
(A.23)

Using Eqs. (A.3) and (A.4), this becomes

$\begin{array}{r}
 - {dx}^{2}y\  + \ y\ a\  + \ cy\left( x^{2}\  - \ y^{2} \right) \\
 \\
\  = \  - {dx}^{2}\  + \ a\  + \ c\left( 2x^{2}\  - \ \rho^{2} \right)\  = \ 0\ \text{.}
\end{array}$ (A.24)

Thus,

$a\  = \ {c\rho}^{2}$, (A.25)

$d\  = \ 2c$. (A.26)

Using Eqs. (A.19 -- A.22) and Eqs. (A.25, A.26), we determine all field
component in terms of the coefficient $\varepsilon$:

$Ε_{x}\  = \ \varepsilon\ \frac{- i\overline{k}}{4}\ \left( \rho^{2}\  + \ x^{2}\  - \ y^{2} \right)\ e^{i(\overline{k}z - \text{kct})}$
(A.27)

$Ε_{y}\  = \ \varepsilon\ \frac{- i\overline{k}}{2}\ xy\ e^{i(\overline{k}z - \text{kc}\ t)}$
(A.28)

$Ε_{z}\  = \ \varepsilon\ x\ e^{i(\overline{k}z - \text{kct})}$ (A.29)

${cΒ}_{x}\  = \ \varepsilon\ \frac{\text{ik}}{2}\ xy$ (A.30)

$$
${cΒ}_{y}\  = \ \varepsilon\left\lbrack \frac{1}{\text{ik}}\ \left( \frac{(k\rho)^{2}}{4}\  - \ 1 \right)\  - \frac{\text{ik}}{4}\ \left( x^{2}\  - \ y^{2} \right) \right\rbrack\ e^{i(\overline{k}z\  - \ \text{kc}t)}$
(A.31)

${cΒ}_{z}\  = \ \varepsilon\ \frac{- \overline{k}}{k}\ y\ e^{i(\overline{k}z - \text{kc}t)}$
(A.32)

This is the traveling wave form of the SEM mode, derived by Montague
\[5\]. The standing waveform is obtained by constructing

$\frac{1}{4}\left( Ε_{i}\ \left( \overline{k},\ k \right)\  + \ Ε_{i}\ \left( \overline{k},\  - k \right)\  + \ Ε_{i}\ \left( - \overline{k},\ k \right)\  + \ Ε_{i}\ \left( - \overline{k},\  - k \right) \right)$,

and similarly for the magnetic field. One then obtains field expression
for the standing wave HEM mode given by Eq. (20).

**References**

\[1\] P. Emma, et. al., Physical Review Special Topics -- Accelerator &
Beams, 9, 100702 (2006)

\[2\] M. Cornacchia and P. Emma, Physical Review Special Topics --
Accelerator & Beams, 5, 084001 (2002)

\[3\] J. Power, AWA Note, Feb. 2007

\[4\] D. Edwards, "Note on Transit in Deflecting Mode Pillbox Cavity",
Fermilab Note, Feb. 12, 2007

\[5\] B. W. Montague, "Particle Separation at High Energies..II
Radiofrequency Separation", Progress in Nuclear Techniques and
Instrumentation, Vol. III, North Holland, 1968.

Fig. 1. Cell Arrangement of π-mode Pill Box Cavity. The vertical blue
lines are ends of full (λ/2) cells and the vertical red lines are those
of half (λ/4) cells.

**E(V/m) vs. z (mm) at x=5mm and y=5mm**

![](media/image1.png){width="4.679861111111111in"
height="3.5118055555555556in"}

**Matlab plots B(tesla) vs. z (mm) at x=5mm and y=5mm**

![](media/image2.png){width="4.679861111111111in"
height="3.5118055555555556in"}

Fig. 2. Electric and Magnetic field components computed from Microwave
Studio for $\left( \frac{1}{2}\ 1\ \frac{1}{2} \right)$ cell cavity with
a beam hole.

Fig. 3. Field and Current Schematics of the HEM Mode. The lines on the
side wall are; green=magnetic field, red=current, blue=electric field.
Note the charge accumulation at the perimeter of the beam hole.