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6.1 Fields for a cylindrical distribution

6.1.1 Theoretical expression of the potential

Given a homogeneous distribution of charges in a cylindrical volume with revolution symetry around the longitudinal axis

, the potential on any point

of this axis can be expressed using the Biot and Savart law:

$\displaystyle \Phi(0,z_0) = \frac{\rho}{4 \pi \epsilon_0} \int_r \int_z \frac{2\pi r }{\sqrt{r^2+(z-z_0)^2}}$ d $\displaystyle r$ d $\displaystyle z$

(6.1)

Where $\rho$ is the density of charges. For a cylinder of semi-axis

and radius

with

charges we have:

$\displaystyle \rho= \frac{N e}{2 b \pi a^2}$

(6.2)

We first integrate equattion 6.1 along

, between

and

, and then do the change of variable

$\displaystyle \displaystyle \Phi(0,z_0) = \frac{\rho}{2 \epsilon_0} \int_{z_0-b}^{z_0+b} \left( \sqrt{a^2+Z^2} -\vert Z\vert\right)$ d $\displaystyle Z$

(6.3)

The integration of the second term is:
$\begin{numcases}{\displaystyle \int_{z_0-b}^{z_0+b}\vert Z\vert\text{d}Z\ = } z_... ... & $\vert z_0\vert \leq b$\ \\ 2 z_0 b & $\vert z_0\vert \geq b$ \end{numcases}$
The first part is calculated eather by blindly applying a formula we provide in annex F or by simply writting $\sqrt{1+z^2}= 1+z^2\: /\: \sqrt{1+z^2}$ , leading to so a sum of two integrals, where the integration by part of the second one leads to the original integral:

$\displaystyle \int \sqrt{1+z^2} = \Big[ \arg\sinh z \Big] + \Big[z \sqrt{1+z^2} \Big] -\int \sqrt{1+z^2}$

(6.4)

Eventually, for a cylindrical distribution, the potentiall inside ( $\vert z_0\vert<b$ , subscribt ``int''), and outside( $\vert z_0\vert>b$ , subscribt ``ext'') the distribution are:

$\displaystyle \Phi(0,0,z_0)_{int}$	$\displaystyle =$	$\displaystyle \Phi_0+ \frac{\rho}{2 \epsilon_0} \left\{-\left(z_0^2+b^2 \right)... ...\left(\arg\sinh\frac{z_0+b}{a}-\arg\sinh\frac{z_0-b}{a} \right) \right. \right.$
	$\displaystyle +$	$\displaystyle \left. \left. \left(z_0+b \right)\sqrt{\left(z_0+b \right)^2+a^2} -\left(z_0-b \right)\sqrt{\left(z_0-b \right)^2+a^2} \ \right] \ \right\}$	(6.5)

$\displaystyle \Phi(0,0,z_0)_{ext}$	$\displaystyle =$	$\displaystyle \Phi_0+ \frac{\rho}{2 \epsilon_0} \left\{-2 b z_0 + \frac{1}{2} \... ...\left(\arg\sinh\frac{z_0+b}{a} -\arg\sinh\frac{z_0-b}{a}\right) \right. \right.$
	$\displaystyle +$	$\displaystyle \left. \left. \left(z_0+b \right)\sqrt{\left(z_0+b \right)^2+a^2} -\left(z_0-b \right)\sqrt{\left(z_0-b \right)^2+a^2} \ \right] \ \right\}$	(6.6)

The function argsinh is the inverse functions of sinh, the hyperbolic sine.

6.1.2 Theoretical expression of the Fields for the cylinder

From the previous expression of the potential, the field component $E_z=-\frac{d\Phi}{dz}$ on the axis can be expressed:

$\displaystyle E_{z,int}(0,z_0)$	$\displaystyle =$	$\displaystyle - \frac{\rho}{2 \epsilon_0} \left[ -2z_0 + \frac{1}{2} \left( \fr... ...a^2+2\left(z_0-b\right)^2 }{ \sqrt{\left(z_0-b \right)^2+a^2} } \right) \right]$	(6.7)

$\displaystyle E_{z,ext}(0,z_0)$	$\displaystyle =$	$\displaystyle - \frac{\rho}{2 \epsilon_0} \left[ -2 b\: \text{sign}\left(z_0\ri... ...a^2+2\left(z_0-b\right)^2 }{ \sqrt{\left(z_0-b \right)^2+a^2} } \right) \right]$	(6.8)

6.1.3 Graphical comparisons

On figure 6.1 one can see the good agreement between the numerical (solid colored line) and the theoretical(dashed lines) fields for different values of $\sigma _z$ , the RMS value of the longitudinal bunch half-length, and $\sigma_r$ , the rms value of the radius. Table 6.1 display the parameters used for the simulation.

**Figure 6.1:** Cylindrical distribution: Longitudinal field along z - comparison between simulation and theory. The theoretical fields are represented with black dashed lines. A perfect agreement is found, but the maximum amplitude of the simulated field is always smaller. This depends on the resolution of the grid, and the smoothing of the fields.

Table 6.1: Parameters used for the benchmarking of the space charge algorithm - cylinder distribution

Parameter	Value
Distribution type	Uniform cylinder
Distribution $\sigma_x$	$\left\{1 ; 2 ; 3 ; 4\right\}/\sqrt{2}$ mm
Distribution $\sigma_y$	$\left\{1 ; 2 ; 3 ; 4\right\}/\sqrt{2}$ mm
Distribution $\sigma _z$	$\left\{0.1 ; 0.2 ; 0.3 ; 0.4 \right\}$ mm
Distribution energy	15 MeV
Distribution charge	-0.25 nC
3D grid used	$N\ast_0=17$ , $N\ast_f$ =64, $N\ast_2$ =128

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