Commit 18d5cf22 by snuverink_j

### 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 $0\leq z\primed\leq L$: % %\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 % % % %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. ... ... @@ -512,10 +512,10 @@ The last equation from\nabla \times \mathbf{B} = 0$should be consistent with This results follows directly from (\ref{eq:11}) and (\ref{eq:12}); therefore the relations are consistent. Furthermore, the last required relations is obtained from the divergence of \textbf{B} \sum_{i,k=0}^{\infty} \left[ \frac{a_{i,k} x^i z^{2k+1}}{\rho} + i a_{i,k} x^{i-1} z^{2k+1} + \frac{i a_{i,k} x^i z^{2k+1}}{\rho} + \partial_s c_{i,k} x^i z^{2k+1} + 2kb_{i,k}x^i z^{2k-1} \right] = 0 \nonumber \sum_{i,k=0}^{\infty} \left[ \frac{a_{i,k} x^i z^{2k+1}}{\rho} + i a_{i,k} x^{i-1} z^{2k+1} + \frac{i a_{i,k} x^i z^{2k+1}}{\rho} + \partial_s c_{i,k} x^i z^{2k+1} + 2kb_{i,k}x^i z^{2k-1} \right] = 0 \Rightarrow \partial_s c_{i,k} + \frac{2(k+1)}{\rho} b_{i-1,k+1} + 2(k+1) b_{i,k+1} + \frac{1}{\rho} a_{i,k} + (i+1) a_{i+1,k} + \frac{1}{\rho} a_{i,k} = 0 \nonumber \Rightarrow \partial_s c_{i,k} + \frac{2(k+1)}{\rho} b_{i-1,k+1} + 2(k+1) b_{i,k+1} + \frac{1}{\rho} a_{i,k} + (i+1) a_{i+1,k} + \frac{1}{\rho} a_{i,k} = 0 Using the relation (\ref{eq:11}) to replace the$a$coefficients with$b\$'s we arrive at ... ...
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