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14.2 An example of transportation matrix: five cell cavity

In this part, we will modelized a five cell cavity by five pillboxes and two drifts pipe on each side of the five cells. Such a cavity is present on a0 beam line. We will assume that the displacement isn't influenced by the cavity, which means that in this plane the beam exactly behave like in a drift. As a result of this we will work with a 4D matrix, and we will add the 2 -components at the end of our study.

14.2.1 Equation of motion in a pillbox

The following calculation where developped in a similar way by Donald A. Edwards [10]. We kept harmonious notations. Here we will use the S axis, which will be identical to the Z axis. The middle of the cell will correspond to

, and the cell length will be $\lambda/2$ . The r subscript will stands for the reference particle data. X will correspond to absolute transverse coordinate, and thus $x\equiv X-X_r$ . From the equations of Padamsee, Knobloch and Hays [35], page 41 , for a $TM_{110}$ mode pillbox, in the paraxial approximation, usual for linear dynamics, we can assume that the fields will be of the following form. As the longitudinal E field is linear with respect to X near the longitudinal axis we will use $E'=\frac{\partial E_z}{\partial X}$ assumed to be constant near the axis:

$\displaystyle E_s(X,t)$	$\displaystyle =$	$\displaystyle E'X\cos(\omega t)$	(14.12)
$\displaystyle B_y(X,t)$	$\displaystyle =$	$\displaystyle \frac{E'}{\omega}\cos(\omega t)$	(14.13)

The equations of motion, with the only Lorentz force $\vec{F}=e(\vec{E}+\vec{v})\times\vec{B}$ are:

$\displaystyle \frac{dP_X}{dt}$	$\displaystyle =$	$\displaystyle -eE'\frac{V_s}{\omega}\sin(\omega t)$	(14.14)
$\displaystyle \frac{dP_S}{dt}$	$\displaystyle =$	$\displaystyle eE'X\cos(\omega t)+eE'\frac{V_X}{\omega}\sin(\omega t)$	(14.15)

We will define without justification the transit time factor T:

$\displaystyle T=\frac{2eE'}{pck}$

14.2.2 The One cell pill box model

For a particle leading the reference particle by a distance z we have:

$\displaystyle E_s$	$\displaystyle =$	$\displaystyle E'X\cos(\omega t-kz) \approx E' X \cos(\omega t) + E'X kz \sin(\omega t)$	(14.16)
$\displaystyle B_y$	$\displaystyle =$	$\displaystyle \frac{E'}{\omega}\cos(\omega t-kz) \approx \frac{E'}{\omega} \sin(\omega t) - \frac{E'}{c} z \cos(\omega t)$	(14.17)

which can be written differently using $s=v_s t \approx ct$ and $\omega=kc$ . Let's write equations 14.14 and 14.15 for the reference particle, using derivative with respect to

now:

$\displaystyle \frac{dP_{X,r}}{ds}$	$\displaystyle =$	$\displaystyle \frac{-eE'}{kc}\sin(ks)$	(14.18)
$\displaystyle \frac{dP_{S,r}}{ds}$	$\displaystyle =$	$\displaystyle \frac{eE'}{c}X_r\cos(ks)+\frac{eE'}{kc^2}V_{X_r}\sin(ks)$	(14.19)

And for a particle, set at

, neglecting second orders terms:

$\displaystyle \frac{dP_X}{ds}$	$\displaystyle =$	$\displaystyle -\frac{eE'}{kc}\sin(ks)+\frac{eE'}{c}z\cos(ks)$	(14.20)
$\displaystyle \frac{dP_{S}}{ds}$	$\displaystyle =$	$\displaystyle \frac{eE'}{c}X\cos(ks)+\frac{eE'}{kc^2}V_X\sin(ks)$	(14.21)

Substracting those two sets of equations gives us

$\displaystyle \frac{dp_x}{ds}$	$\displaystyle =$	$\displaystyle \frac{eE'}{c}z\cos(ks)$	(14.22)
$\displaystyle \frac{dp_s}{ds}$	$\displaystyle =$	$\displaystyle \frac{eE'}{c}X\cos(ks)+\frac{eE'}{kc}\frac{dx}{ds}\sin(ks)$	(14.23)

Let's integrate 14.23 from $s=-\lambda/4$ to

, remembering that $k\lambda/4=\pi/2$ :

$\displaystyle p_x(s)=p_x(-\lambda/4)+\frac{eE'}{ck}z\left(sin(ks)+1\right)$

(14.24)

Now, as

we have:

$\displaystyle x'(s)=x'_{in}+\frac{Tp}{2}z\sin(ks)+\frac{Tp}{2}z$

(14.25)

Which we can integrate again:

$\displaystyle x(s)=x_{in}+\left(x'_{in}+\frac{Tp}{2}z\right)\left(s+\frac{\lambda}{4}\right)-\frac{Tp}{2k}z\cos(ks)$

(14.26)

The integration of 14.24 is now possible thanks to the two last expressions. We will provide details of the algebra here. Inserting expressions 14.26 and 14.27 in equation 14.24 yields to:

$\displaystyle \frac{dp_s}{ds}$	$\displaystyle =$	$\displaystyle \frac{Tkp}{2} \left[ x_{in}+\left(x'_{in}+\frac{Tp}{2}z\right)\left(s+\frac{\lambda}{4}\right)-\frac{Tp}{2k}z\cos(ks) \right] \cos(ks)$	(14.27)
	$\displaystyle +$	$\displaystyle \frac{Tp}{2} \left[ x'_{in}+\frac{Tp}{2}z\sin(ks)+\frac{Tp}{2}z \right] \sin(ks)$	(14.28)

After factorization:

$\displaystyle \frac{dp_s}{ds}$	$\displaystyle =$	$\displaystyle x_{in} \frac{T}{2}p\left[k\cos(ks)\right]$	(14.29)
	$\displaystyle \; +$	$\displaystyle x'_{in} \frac{T}{2}p \left[k\left(s+\frac{\lambda}{4}\right) \cos(ks) + \sin(ks)\right ]$	(14.30)
	$\displaystyle \; +$	$\displaystyle z \frac{T^2}{4}p \left[k\left(s+\frac{\lambda}{4}\right)\cos(ks)-\cos^2(ks)+\sin^2(ks)+\sin(ks)\right]$	(14.31)

We use the relation $-\cos^2(ks)+\sin^2(ks) = - \cos(2ks)$ , and the following integrals:

$\displaystyle \int_{-\lambda/4}^{s} \sin(ks)$	$\displaystyle =$	$\displaystyle -\frac{\cos(ks)}{k}$	(14.32)
$\displaystyle \int_{-\lambda/4}^{s} \cos(2ks)$	$\displaystyle =$	$\displaystyle \frac{sin(2ks)}{2k}$	(14.33)
$\displaystyle \int_{-\lambda/4}^{s} k\left(s+\frac{\lambda}{4}\right)\cos(ks)$	$\displaystyle =$	$\displaystyle \left(s+\frac{\lambda}{4}\right) \sin(ks) + \frac{\cos(ks)}{k}$	(14.34)

It yields:

$\displaystyle \frac{\delta p }{p}$	$\displaystyle =$	$\displaystyle x_{in}\;\frac{T}{2}\left[1+\sin(ks)\right]$	(14.35)
	$\displaystyle \; +$	$\displaystyle x'_{in}\;\frac{T}{2}\left[\left(s+\frac{\lambda}{4}\right)\sin(ks)\right]$	(14.36)
	$\displaystyle \; +$	$\displaystyle z \;\frac{T^2}{4}\left[\left(s+\frac{\lambda}{4}\right)\sin(ks)-\frac{1}{2k}\sin(2ks)\right]$	(14.37)

14.2.3 Cavity Transit matrix

To have the matrix of the entire cell we just have to evaluate the previous expression at $s=\lambda/4$ which corresponds to the exit of the cell.

$\displaystyle M_{cell}=\begin{pmatrix} 1 & \lambda/2&T\lambda/4 & 0\\ 0 &1 & T& 0\\ 0& 0& 1& 0\\ T&T\lambda/4 &T^2\lambda/8 & 1\\ \end{pmatrix}$

Then the matrix of the 5 cells is:

$\displaystyle M_{cell}^5=\begin{pmatrix} 1 & 5\lambda/2&25T\lambda\/ & 0\\ 0 &1 & 5T& 0\\ 0& 0& 1& 0\\ 5T&25T\lambda/4 &85T^2\lambda/8 & 1\\ \end{pmatrix}$

The matrix for a drift of d is:

$\displaystyle M_d=\begin{pmatrix} 1 & d & 0 & 0\\ 0 &1 & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ \end{pmatrix}$

Eventually, the matrix of the whole cavity with surrounding pipes is:

$\displaystyle M_{cav}=M_d.M^{5}_{cell}.M_d=\begin{pmatrix} 1 & 2d+5\lambda/2 & ... ...& 0\\ 0& 0& 0& 0\\ 5T& 5Td + 25T\lambda/4& 85T^2\lambda/8& 1\\ \end{pmatrix}$

Before evaluating this matrix let's write it in 6D, so including

and

components:

$\displaystyle M_{cav}=\begin{pmatrix} 1 & 2d+5\lambda/2 & 0&0&5Td+ 25T\lambda/4... ...& 0& 0&0&0& 0\\ 5T& 5Td + 25T\lambda/4& 0&0&85T^2\lambda/8& 1\\ \end{pmatrix}$

Numerically:

$\displaystyle M_{cav}=\begin{pmatrix} 1 & 0.300 & 0&0&-0.452& 0\\ 0 &1 & 0&0&-... ... 0& 0&1&0& 0\\ 0& 0& 0&0&1& 0\\ -3.016& -0.452& 0&0&0.297& 1\\ \end{pmatrix}$

with

$\displaystyle \begin{matrix} d&=&0.05385\:m\\ \lambda&=&0.07692\:m\\ T&=&-0.6031\\ \end{matrix}$

Emmanuel Branlard
http://emmanuel.branlard.free.fr/