Commit 18d5cf22 authored by snuverink_j's avatar snuverink_j
Browse files

remove nonumber

parent 98531c5b
......@@ -593,17 +593,17 @@ Since the magnetic field data off the median plane field components is necessary
According to the approach given by Gordon and Taivassalo, by using a magnetic potential and measured $B_z$ on the median plane,
at the point $(r,\theta, z)$ in cylindrical polar coordinates, the 3$th$ order field can be written as
\begin{eqnarray}\label{eq:Bfield}
B_r(r,\theta, z) & = & z\diffp{B_z}{r}-\frac{1}{6}z^3 C_r, \nonumber\\
B_r(r,\theta, z) & = & z\diffp{B_z}{r}-\frac{1}{6}z^3 C_r, \\
B_\theta(r,\theta, z) & = & \frac{z}{r}\diffp{B_z}{\theta}-\frac{1}{6}\frac{z^3}{r} C_{\theta}, \\
B_z(r,\theta, z) & = & B_z-\frac{1}{2}z^2 C_z, \nonumber
B_z(r,\theta, z) & = & B_z-\frac{1}{2}z^2 C_z,
\end{eqnarray}
where $B_z\equiv B_z(r, \theta, 0)$ and
\begin{eqnarray}\label{eq:Bcoeff}
C_r & = & \diffp[3]{B_z}{r} + \frac{1}{r}\diffp[2]{B_z}{r} - \frac{1}{r^2}\diffp{B_z}{r}
+ \frac{1}{r^2}\diffp{B_z}{{r}{\theta^2}} - 2\frac{1}{r^3}\diffp[2]{B_z}{\theta}, \nonumber \\
+ \frac{1}{r^2}\diffp{B_z}{{r}{\theta^2}} - 2\frac{1}{r^3}\diffp[2]{B_z}{\theta}, \\
C_{\theta} & = & \frac{1}{r}\diffp{B_z}{{r}{\theta}} + \diffp{B_z}{{r^2}{\theta}}
+ \frac{1}{r^2}\diffp[3]{B_z}{\theta}, \\
C_z & = & \frac{1}{r}\diffp{B_z}{r} + \diffp[2]{B_z}{r} + \frac{1}{r^2}\diffp[2]{B_z}{\theta}. \nonumber
C_z & = & \frac{1}{r}\diffp{B_z}{r} + \diffp[2]{B_z}{r} + \frac{1}{r^2}\diffp[2]{B_z}{\theta}.
\end{eqnarray}
All the partial differential coefficients are on the median plane and can be calculated by interpolation. \textit{OPAL-cycl} uses Lagrange's 5-point formula.
......
......@@ -166,17 +166,17 @@ Since the magnetic field data off the median plane field components is necessary
According to the approach given by Gordon and Taivassalo, by using a magnetic potential and measured $B_z$ on the median plane,
at the point $(r,\theta, z)$ in cylindrical polar coordinates, the \engordnumber{3} order field can be written as
\begin{eqnarray}\label{eq:Bfield}
B_r(r,\theta, z) & = & z\diffp{B_z}{ r}-\frac{1}{6}z^3 C_r, \nonumber\\
B_r(r,\theta, z) & = & z\diffp{B_z}{ r}-\frac{1}{6}z^3 C_r, \\
B_\theta(r,\theta, z) & = & \frac{z}{r}\diffp{B_z}{\theta}-\frac{1}{6}\frac{z^3}{r} C_{\theta}, \\
B_z(r,\theta, z) & = & B_z-\frac{1}{2}z^2 C_z, \nonumber
B_z(r,\theta, z) & = & B_z-\frac{1}{2}z^2 C_z,
\end{eqnarray}
where $B_z\equiv B_z(r, \theta, 0)$ and
\begin{eqnarray}\label{eq:Bcoeff}
C_r & = & \diffp[3]{B_z}{r} + \frac{1}{r}\diffp[2]{B_z}{r} - \frac{1}{r^2}\diffp{B_z}{r}
+ \frac{1}{r^2}\diffp{B_z}{{r}{\theta^2}} - 2\frac{1}{r^3}\diffp[2]{B_z}{\theta}, \nonumber \\
+ \frac{1}{r^2}\diffp{B_z}{{r}{\theta^2}} - 2\frac{1}{r^3}\diffp[2]{B_z}{\theta}, \\
C_{\theta} & = & \frac{1}{r}\diffp{B_z}{{r}{\theta}} + \diffp{B_z}{{r^2}{\theta}}
+ \frac{1}{r^2}\diffp[3]{B_z}{\theta}, \\
C_z & = & \frac{1}{r}\diffp{B_z}{r} + \diffp[2]{B_z}{r} + \frac{1}{r^2}\diffp[2]{B_z}{\theta}. \nonumber
C_z & = & \frac{1}{r}\diffp{B_z}{r} + \diffp[2]{B_z}{r} + \frac{1}{r^2}\diffp[2]{B_z}{\theta}.
\end{eqnarray}
All the partial differential coefficients are on the median plane and can be calculated by interpolation. \textit{OPAL-cycl} uses Lagrange's 5-point formula.
......
......@@ -62,7 +62,7 @@ This gives the general form of the envelope equation
\begin{eqnarray}
\ddot\sigma_x + \left[\gamma^2\beta\dot\beta\right]\dot\sigma_x
+ \left[\frac{K}{m\gamma}\right]\sigma_x &=& \left[\frac{F_{x,s}^{(1)}}{m\gamma}\right]\sigma_x+\left(\frac{c\epsilon_n}{\gamma}\right)^2\frac{1}{\sigma_x^3},
\nonumber
\\
\sigma_x\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]\sigma_x\primed + \left[\frac{K}{mc^2\beta p_n}\right]\sigma_x &=& \left[\frac{F_{x,s}^{(1)}}{mc^2\beta p_n}\right]\sigma_x+\left(\frac{\epsilon_n}{p_n}\right)^2\frac{1}{\sigma_x^3}.
\end{eqnarray}
......@@ -81,7 +81,7 @@ These fields must then be boosted back to the laboratory frame according to
\begin{eqnarray}
\textbf{E}&=&\gamma(\textbf{E}\primed-\mathbf{\mathbf{\beta}}\times
c\textbf{B}\primed)-\frac{\gamma^2}{1+\gamma}\mathbf{\beta}(\mathbf{\beta}\cdot
\textbf{E}\primed),\nonumber
\textbf{E}\primed),
\\
\textbf{B}&=&\gamma(\textbf{B}\primed+\mathbf{\mathbf{\beta}}\times
\textbf{E}\primed/c)-\frac{\gamma^2}{1+\gamma}\mathbf{\beta}(\mathbf{\beta}\cdot
......@@ -111,7 +111,7 @@ The linearized force equation then takes the form
%
\begin{eqnarray}
\mathbf{F} &=& e(E_x-\beta c B_{y})\hat{\mathbf{x}} + e(E_y + \beta c B_{x})\hat{\mathbf{y}}
\nonumber
%\\
%&=&
= \frac{e\lambda}{\gamma^2}
......@@ -126,7 +126,7 @@ With this, the envelope equations become
\begin{eqnarray}
\ddot\sigma_i + \left[\gamma^2\beta\dot\beta\right]\dot\sigma_i
+ \left[\frac{K_i}{m\gamma}\right]\sigma_i &=& \left[\frac{e\lambda}{m\gamma^3}\left(\diffp{E_i\primed}{{x_i}{\lambda\primed}}\right)\right]\sigma_i+\left(\frac{c\epsilon_{n,i}}{\gamma}\right)^2\frac{1}{\sigma_i^3}.
%\nonumber
%
%\\
%\sigma_i\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]\sigma_i\primed + \left[\frac{K_i}{mc^2\beta p_n}\right]\sigma_i &=& \left[\frac{e\lambda}{mc^2\beta^2 \gamma^3}\left(\diffp{E_i\primed}{{x_i}{\lambda\primed}}\right)\right]\sigma_i+\left(\frac{\epsilon_{n,i}}{p_n}\right)^2\frac{1}{\sigma_i^3}.
\label{eq:GenEnv}
......@@ -142,7 +142,7 @@ Larmor frame. Working in the Larmor frame, the equations of motion for $\sigma_x
%
\begin{eqnarray}
\ddot\sigma_L + \left[\gamma^2\beta\dot\beta\right]\dot\sigma_L + \left[\frac{K}{m\gamma}+(\dot\theta_r)^2\right]\sigma_L &=& \left[\frac{e\lambda}{m \gamma^3}\left(\diffp{E_r\primed}{{r}{\lambda\primed}}\right)\right]\sigma_L+\left(\frac{c\epsilon_n}{\gamma}\right)^2\frac{1}{\sigma_L^3}.
\nonumber
%\\
%\sigma\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]\sigma\primed
%+ \left[\frac{K}{mc^2\beta p_n} + (\theta_r\primed)^2\right]\sigma &=&
......@@ -182,7 +182,7 @@ the envelope equation is written in its `standard' form \ref{bib:JBong}
\begin{eqnarray}
\ddot\sigma_L + \left[\gamma^2\beta\dot\beta\right]\dot\sigma_L + \left[\frac{K}{m\gamma}+(\dot\theta_r)^2\right]\sigma_L &=&
\left[\frac{c^2k_p}{\beta \gamma^3}\right]\frac{1}{\sigma_L}+\left(\frac{c\epsilon_n}{\gamma}\right)^2\frac{1}{\sigma_L^3},
\nonumber
%\\
%\sigma\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]\sigma\primed
%+ \left[\frac{K}{mc^2\beta p_n} + (\theta_r\primed)^2\right]\sigma &=&
......@@ -195,7 +195,7 @@ equations can be written in terms of the beam radius $R$:
\begin{eqnarray}
\ddot R + \left[\gamma^2\beta\dot\beta\right]\dot R + \left[\frac{K}{m\gamma}+(\dot\theta_r)^2\right]R&=&
\left[\frac{4c^2k_p}{\beta \gamma^3}\right]\frac{1}{R}+\left(\frac{4c\epsilon_n}{\gamma}\right)^2\frac{1}{R^3}.
%\nonumber
%
%\\
%R\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]R\primed
%+ \left[\frac{K}{mc^2\beta p_n} + (\theta_r\primed)^2\right]R &=&
......@@ -272,7 +272,7 @@ From these expressions the envelope equations can be derived using Equation~\ref
\begin{eqnarray}
\ddot\sigma_i + \left[\gamma^2\beta\dot\beta\right]\dot\sigma_i
+ \left[\frac{K_i}{m\gamma}\right]\sigma_i &=& \left[\frac{2c^2}{\beta\gamma^3}\right]\frac{k_p}{\sigma_x+\sigma_y}+\left(\frac{c\epsilon_{n,i}}{\gamma}\right)^2\frac{1}{\sigma_i^3}.
%\nonumber
%
%,\\
%\sigma_i\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]\sigma_i\primed + \left[\frac{K_i}{mc^2\beta p_n}\right]\sigma_i &=& \left[\frac{2}{p_n^3}\right]\frac{k_p}{\sigma_x+\sigma_y}+\left(\frac{\epsilon_{n,i}}{p_n}\right)^2\frac{1}{\sigma_i^3}.
\end{eqnarray}
......@@ -282,7 +282,7 @@ The equivalent equations for the ellipse axes are the given by
\begin{eqnarray}
\ddot X_i + \left[\gamma^2\beta\dot\beta\right]\dot X_i
+ \left[\frac{K_i}{m\gamma}\right]X_i &=& \left[\frac{8c^2}{\beta\gamma^3}\right]\frac{k_p}{X+Y}+\left(\frac{4c\epsilon_{n,i}}{\gamma}\right)^2\frac{1}{X_i^3}.
%\nonumber
%
%\\
%X_i\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]X_i\primed + \left[\frac{K_i}{mc^2\beta p_n}\right]X_i &=& \left[\frac{8}{p_n^3}\right]\frac{k_p}{X+Y}+\left(\frac{4\epsilon_{n,i}}{p_n}\right)^2\frac{1}{X_i^3}.
\end{eqnarray}
......@@ -348,7 +348,7 @@ infinitesimal field contribution in this case is
\begin{eqnarray}
d\mathbf{E} =
\frac{1}{4\pi\epsilon_0}\frac{(\mathbf{x}-\mathbf{x}\primed)}
{|\mathbf{x}-\mathbf{x}\primed|^3}\sigma dA\primed,\nonumber
{|\mathbf{x}-\mathbf{x}\primed|^3}\sigma dA\primed,
\end{eqnarray}
%
with $\mathbf{x}-\mathbf{x}\primed=
......@@ -369,7 +369,7 @@ charged disks from $0\leq z\primed\leq L$:
%\begin{equation}
%\mathbf{E}_z(r=0)=\frac{Q}{2\pi\epsilon_0R^2}
%\left(\sqrt{(1-z/L)^2+(R/L)^2}-\sqrt{z^2+R^2}-|L-z|+|z|\right)\hat{\mathbf{z}}.
%\nonumber
%
%\end{equation}
%
%This can be written more compactly as
......@@ -393,7 +393,7 @@ Gauss's law takes the form
r\left(\frac{\rho_0}{\epsilon_0}-\diffp{E_z}{z}\right),\label{eq:FindErho}
%\cong
r\left(\frac{\rho(r,z)}{\epsilon_0}-\diff{}{z}E_z(0,z)\right),
\nonumber
\end{eqnarray}
%
where the charge density $\rho = \rho_0[\theta_H(r)-\theta_H(r-R)]\cdot
......@@ -425,7 +425,7 @@ Thus the radial component is given to first order in $r$ as
%
\begin{eqnarray}
E_{r}=r\frac{\rho_0}{4\epsilon_0}
\left(\frac{L-z}{\sqrt{(L-z)^2+R^2}}+\frac{z}{\sqrt{z^2+R^2}}\right)r.\nonumber
\left(\frac{L-z}{\sqrt{(L-z)^2+R^2}}+\frac{z}{\sqrt{z^2+R^2}}\right)r.
\end{eqnarray}
%
This can be written as
......@@ -474,7 +474,7 @@ Following the previous analysis, the envelope equation then becomes
\begin{eqnarray}
\ddot\sigma + \left[\gamma^2\beta\dot\beta\right]\dot\sigma + \left[\frac{K}{m\gamma}+(\dot\theta_r)^2\right]\sigma &=&
\left[\frac{c^2k_p}{\beta \gamma^3}\left(\frac{G(\zeta,A)}{2}\right)\right]\frac{1}{\sigma}+\left(\frac{c\epsilon_n}{\gamma}\right)^2\frac{1}{\sigma^3},
\nonumber
%\\
%\sigma\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]\sigma\primed
%+ \left[\frac{K}{mc^2\beta p_n} + (\theta_r\primed)^2\right]\sigma &=&
......@@ -510,7 +510,7 @@ beam are equivalent to those of a circular beam with a effective radius $R^{*} =
charges (for emission from a cathode) are included. This is done by letting
%
\begin{eqnarray}
\frac{I(\zeta)g(\zeta)}{\gamma^2}\rightarrow I(\zeta)[(1-\beta^2)g(\zeta) - (1+\beta^2)g(\xi) ],\nonumber
\frac{I(\zeta)g(\zeta)}{\gamma^2}\rightarrow I(\zeta)[(1-\beta^2)g(\zeta) - (1+\beta^2)g(\xi) ],
\end{eqnarray}
%
where $\xi = z_s + z_h$. Effectively, this just includes a mirror bunch behind
......@@ -520,15 +520,15 @@ equations are given in the Larmor frame as:
%
\begin{eqnarray}
\lefteqn{\ddot\sigma_L + \left[\gamma^2\beta\dot\beta\right]\dot\sigma_L + \left[\frac{K}{m\gamma}+(\dot\theta_r)^2\right]\sigma_L=} & &
\nonumber
\\
& & \hspace{2cm}\frac{c^2k_p}{2\beta \gamma\sigma_L}[ \gamma^{-2}G(\zeta,A) - (1+\beta^2)G(\xi,A) ]+\left(\frac{c\epsilon_n}{\gamma}\right)^2\frac{1}{\sigma_L^3}.
%\nonumber
%
%\\
%\sigma\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]\sigma\primed
%+ \left[\frac{K}{mc^2\beta p_n} + (\theta_r\primed)^2\right]\sigma &=&
%\frac{k_p}{2\gamma\beta^3\sigma}[ \gamma^{-2}G(\zeta,A) - (1+\beta^2)G(\xi,A) ]+\left(\frac{\epsilon_n}{p_n}\right)^2\frac{1}{\sigma^3}.
%\nonumber
%
%\\
\end{eqnarray}
%
......@@ -537,13 +537,13 @@ For an elliptical beam in an uncoupled focusing channel the equations are
\begin{eqnarray}
\lefteqn{\ddot\sigma_i + \left[\gamma^2\beta\dot\beta\right]\dot\sigma_i
+ \left[\frac{K_i}{m\gamma}\right]\sigma_i =} &&
\nonumber
\\
&& \hspace{2cm}\frac{c^2k_p}{2\beta\gamma\sigma^*}[ \gamma^{-2}G(\zeta,A^*) - (1+\beta^2)G(\xi,A^*) ]+\left(\frac{c\epsilon_{n,i}}{\gamma}\right)^2\frac{1}{\sigma_i^3}.
%\nonumber
%
%\\
%\sigma_i\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]\sigma_i\primed + \left[\frac{K_i}{mc^2\beta p_n}\right]\sigma_i &=& \frac{k_p}{2\beta^3\gamma\sigma^*}[ \gamma^{-2}G(\zeta,A^*) - (1+\beta^2)G(\xi,A^*) ]+\left(\frac{\epsilon_{n,i}}{p_n}\right)^2\frac{1}{\sigma_i^3}.
%\nonumber
%
%\\
\end{eqnarray}
%
......@@ -553,10 +553,10 @@ This is done below:
%
\begin{eqnarray}
\lefteqn{\ddot R+ \left[\gamma^2\beta\dot\beta\right]\dot R + \left[\frac{K}{m\gamma}+(\dot\theta_r)^2\right]R=} &&
\nonumber
\\
& & \hspace{2cm} \frac{2c^2k_p}{\beta \gamma R}[ \gamma^{-2}G(\zeta,A) - (1+\beta^2)G(\xi,A) ]+\left(\frac{4c\epsilon_{n,x}}{\gamma}\right)^2\frac{1}{R^3},
%\nonumber
%
%\\
%R\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]R\primed
%+ \left[\frac{K}{mc^2\beta p_n} + (\theta_r\primed)^2\right]R &=&
......@@ -568,10 +568,10 @@ For an elliptical beam in an uncoupled focusing channeling the equations are
\begin{eqnarray}
\lefteqn{\ddot X_i + \left[\gamma^2\beta\dot\beta\right]\dot X_i
+ \left[\frac{K_i}{m\gamma}\right]X_i=} &&
\nonumber
\\
& & \hspace{2cm}\frac{2c^2k_p}{\beta\gamma R^*}[ \gamma^{-2}G(\zeta,A^*) - (1+\beta^2)G(\xi,A^*) ]+\left(\frac{4c\epsilon_{n,i}}{\gamma}\right)^2\frac{1}{X_i^3}.
%\nonumber
%
%\\
%\sigma_i\nprimed{2} + \left[\frac{\gamma\primed}{\beta^2\gamma}\right]\sigma_i\primed + \left[\frac{K_i}{mc^2\beta p_n}\right]\sigma_i &=& \frac{k_p}{2\beta^3\gamma$\sigma^*}[ \gamma^{-2}G(\zeta,A^*) - (1+\beta^2)G(\xi,A^*) ]+\left(\frac{\epsilon_{n,i}}{p_n}\right)^2\frac{1}{\sigma_i^3}.
\end{eqnarray}
......
......@@ -469,7 +469,7 @@ where $d(z)$ is the distance to the field boundary and $\lambda$ characterizes t
We consider the Frenet-Serret coordinate system $ ( \hat{\mathbf{x}}, \hat{\mathbf{s}}, \hat{\mathbf{z}} )$ with the radius of curvature $ \rho $ constant and the scale factor $ h_s = 1 + x/ \rho$. A conversion to these coordinates implies that
\begin{align}
\nabla \cdot \mathbf{B} & = \frac{1}{h_s} \left[ \frac{\partial (h_s B_x )}{\partial x} + \frac{\partial B_s}{\partial s} + \frac{\partial (h_s B_z )}{\partial z} \right] \\
\nabla \times \mathbf{B} & = \frac{1}{h_s} \left[ \frac{\partial B_z}{\partial s} - \frac{\partial (h_s B_s )}{\partial z} \right] \hat{\mathbf{x}} + \left[ \frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x} \right] \hat{\mathbf{s}} + \frac{1}{h_s} \left[ \frac{\partial (h_s B_s)}{\partial x} - \frac{\partial B_x}{\partial s} \right] \hat{\mathbf{z}} \nonumber
\nabla \times \mathbf{B} & = \frac{1}{h_s} \left[ \frac{\partial B_z}{\partial s} - \frac{\partial (h_s B_s )}{\partial z} \right] \hat{\mathbf{x}} + \left[ \frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x} \right] \hat{\mathbf{s}} + \frac{1}{h_s} \left[ \frac{\partial (h_s B_s)}{\partial x} - \frac{\partial B_x}{\partial s} \right] \hat{\mathbf{z}}
\end{align}
To simplify the problem, consider multipoles with mid-plane symmetry, i.e.
\begin{equation}
......@@ -512,10 +512,10 @@ The last equation from $\nabla \times \mathbf{B} = 0$ should be consistent with
\end{equation}
This results follows directly from (\ref{eq:11}) and (\ref{eq:12}); therefore the relations are consistent. Furthermore, the last required relations is obtained from the divergence of \textbf{B}
\begin{equation}
\sum_{i,k=0}^{\infty} \left[ \frac{a_{i,k} x^i z^{2k+1}}{\rho} + i a_{i,k} x^{i-1} z^{2k+1} + \frac{i a_{i,k} x^i z^{2k+1}}{\rho} + \partial_s c_{i,k} x^i z^{2k+1} + 2kb_{i,k}x^i z^{2k-1} \right] = 0 \nonumber
\sum_{i,k=0}^{\infty} \left[ \frac{a_{i,k} x^i z^{2k+1}}{\rho} + i a_{i,k} x^{i-1} z^{2k+1} + \frac{i a_{i,k} x^i z^{2k+1}}{\rho} + \partial_s c_{i,k} x^i z^{2k+1} + 2kb_{i,k}x^i z^{2k-1} \right] = 0
\end{equation}
\begin{equation}
\Rightarrow \partial_s c_{i,k} + \frac{2(k+1)}{\rho} b_{i-1,k+1} + 2(k+1) b_{i,k+1} + \frac{1}{\rho} a_{i,k} + (i+1) a_{i+1,k} + \frac{1}{\rho} a_{i,k} = 0 \nonumber
\Rightarrow \partial_s c_{i,k} + \frac{2(k+1)}{\rho} b_{i-1,k+1} + 2(k+1) b_{i,k+1} + \frac{1}{\rho} a_{i,k} + (i+1) a_{i+1,k} + \frac{1}{\rho} a_{i,k} = 0
\end{equation}
Using the relation (\ref{eq:11}) to replace the $a$ coefficients with $b$'s we arrive at
\begin{equation}
......
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