opale.tex 26 KB
 snuverink_j committed Sep 07, 2017 1 \chapter{\textit{OPAL}-env}  snuverink_j committed Sep 06, 2017 2   snuverink_j committed Sep 07, 2017 3 The \textit{OPAL-e} Envelope Tracker is an algorithm used to solve the envelope equations  snuverink_j committed Sep 06, 2017 4 5 of a beam propagating through external fields and under the influence of its own space charge. The algorithm is based on the multi-slice analysis approach  snuverink_j committed Sep 08, 2017 6 used in the theory of emittance compensation \ref{bib:JBong}. The space charge  snuverink_j committed Sep 06, 2017 7 model used can be switched between an analytic model derived and used in HOMDYN  snuverink_j committed Sep 08, 2017 8 \ref{bib:HOMY} and a similar model developed at PSI called Beam Envelope Tracker  snuverink_j committed Sep 06, 2017 9 10 11 12 13 (BET). \subsection{Envelope Equation without Dispersion For Long Beams} The equation for the propagation of a general beam enveloped given here  snuverink_j committed Sep 08, 2017 14 follows the work of Sacherer \ref{Sach}. Using the variables $x$ and  snuverink_j committed Sep 08, 2017 15 $p_x$ as the phase space variables, the equation of motion for $\sigma_x =  snuverink_j committed Sep 06, 2017 16 17 18 \langle x^2\rangle^{1/2}$ is found by differentiating with respect to time: % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 19 20 \dot\sigma_x = \frac{\langle x\dot x \rangle}{\sigma},\hspace{1cm} \ddot\sigma_x = \frac{1}{\sigma_x^3}\left[\langle x^2\rangle \langle \dot x^2\rangle-\langle x\dot x\rangle^2\right]+\frac{\langle x\ddot x\rangle}{\sigma_x}  snuverink_j committed Sep 06, 2017 21 22 23 24 25 \end{eqnarray} % Noting that $\dot x = p_x/m\gamma$, the above equations become % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 26 \ddot\sigma_x = \frac{1}{\sigma_x^3}\left(\frac{c\epsilon_n}{\gamma}\right)^2+\frac{\langle x\ddot x\rangle}{\sigma_x},  snuverink_j committed Sep 06, 2017 27 28 29 30 31 32 33 34 \end{eqnarray} % where the normalized emittance is defined as $\epsilon_{n,x} = \frac{1}{mc}\sqrt{\langle x^2\rangle \langle p_x^2\rangle-\langle xp_x\rangle^2}$. The last term in this equation is expanded using $\ddot x = -\gamma^2\beta\dot\beta \dot x + F_x/m\gamma$: % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 35 \ddot\sigma_x = \left[-\gamma^2\beta\dot\beta\right]\dot\sigma_x + \frac{\langle xF_x\rangle}{m\gamma\sigma_x}+\left(\frac{c\epsilon_n}{\gamma}\right)^2\frac{1}{\sigma_x^3},  snuverink_j committed Sep 06, 2017 36 37 38 39 40 41 \end{eqnarray} % The force is split into a (linear) external part and the self-fields: $F_x(t,x) = -K_x(t)x + F_{x,s}$. Plugging this into the envelope equation gives % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 42 \ddot\sigma_x + \left[\gamma^2\beta\dot\beta\right]\dot\sigma_x + \left[\frac{K}{m\gamma}\right]\sigma_x = \left[\frac{\langle xF_{x,s}\rangle}{m\gamma}\right]\frac{1}{\sigma_x}+\left(\frac{c\epsilon_n}{\gamma}\right)^2\frac{1}{\sigma_x^3}.  snuverink_j committed Sep 06, 2017 43 44 \end{eqnarray} %  snuverink_j committed Sep 08, 2017 45 Following Sacherer \ref{Sach}, the term $\langle xF_{x,s}\rangle$ can be  snuverink_j committed Sep 06, 2017 46 47 48 49 50 interpreted as the linear part of the space charge field (in a least squares sense). The linear part of the field is defined by $F_{x,s}^{(1)}$, such that the quantity % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 51 D = \langle (F_{x,s}^{(1)}x-F_{x,s})^2\rangle = \int (F_{x,s}^{(1)}x-F_{x,s})^2\rho_x\;\mathrm{d}x,  snuverink_j committed Sep 06, 2017 52 53 \end{eqnarray} %  snuverink_j committed Sep 08, 2017 54 is minimized in a least squares sense. This is accomplished by setting $\pdifferential{}D / \pdifferential{} F_{x,s}^{(1)} = 0$, which implies \ref{Sach}:  snuverink_j committed Sep 06, 2017 55 56 % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 57 F_{x,s}^{(1)}x = \frac{\langle xF_{x,s}\rangle}{\sigma_x^2}x.  snuverink_j committed Sep 06, 2017 58 59 60 61 62 \end{eqnarray} % This gives the general form of the envelope equation % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 63 64 \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},  snuverink_j committed Sep 11, 2017 65   snuverink_j committed Sep 06, 2017 66 \\  snuverink_j committed Sep 08, 2017 67 \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}.  snuverink_j committed Sep 06, 2017 68 69 70 71 72 73 74 75 76 77 78 79 80 81 \end{eqnarray} % In this equation $p_n = \beta\gamma$ is the normalized momentum. A general form for the fields can be now introduced. For a long beam, the linear part of the fields (in the beam rest frame) are given by % \begin{eqnarray} E_x=\left( \left.\diffp{E_x}{x}\right|_{0}\right)x,\hspace{1cm}E_y= \left(\left.\diffp{E_y}{y}\right|_{0}\right)y. \end{eqnarray} % These fields must then be boosted back to the laboratory frame according to % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 82 83 \textbf{E}&=&\gamma(\textbf{E}\primed-\mathbf{\mathbf{\beta}}\times c\textbf{B}\primed)-\frac{\gamma^2}{1+\gamma}\mathbf{\beta}(\mathbf{\beta}\cdot  snuverink_j committed Sep 11, 2017 84 \textbf{E}\primed),  snuverink_j committed Sep 06, 2017 85 \\  snuverink_j committed Sep 08, 2017 86 87 \textbf{B}&=&\gamma(\textbf{B}\primed+\mathbf{\mathbf{\beta}}\times \textbf{E}\primed/c)-\frac{\gamma^2}{1+\gamma}\mathbf{\beta}(\mathbf{\beta}\cdot  snuverink_j committed Sep 06, 2017 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 \textbf{B}\primed),\label{eq:FieldTrans} \end{eqnarray} % For a beam moving along the $z$-axis with speed $c\beta$, the fields are given by % \begin{eqnarray} \mathbf{E} = \gamma E_x\primed\hat{\mathbf{x}} + \gamma E_y\primed\hat{\mathbf{y}}, \hspace{2cm} \mathbf{B} = \frac{\gamma\beta}{c}(-E_y\primed\hat{\mathbf{x}} + E_x\primed\hat{\mathbf{y}}). \end{eqnarray} % Here the primes label the field components in the lab frame. Assuming these components are proportional to the line charge density $\lambda\primed$ the fields can be written in a form that takes into account the Lorentz contraction of the current density: % \begin{eqnarray} E_x = \gamma E_x\primed = \gamma\lambda\primed\diffp{E_x\primed}{\lambda\primed}=\lambda\diffp{E_x\primed}{\lambda\primed}. \end{eqnarray} % The linearized force equation then takes the form %\begin{eqnarray} %\mathbf{F} = e(E_r-\beta c B_{\theta})\hat{\mathbf{r}} = \frac{e\lambda}{\gamma^2}\left(\diffp{E_r\primed}{{r}{\lambda\primed}}\right) r\hat{\mathbf{r}} %\end{eqnarray} % \begin{eqnarray} \mathbf{F} &=& e(E_x-\beta c B_{y})\hat{\mathbf{x}} + e(E_y + \beta c B_{x})\hat{\mathbf{y}}  snuverink_j committed Sep 11, 2017 114   snuverink_j committed Sep 06, 2017 115 116 117 118 119 120 121 122 123 124 125 126 %\\ %&=& = \frac{e\lambda}{\gamma^2} \left[ \left(\diffp{E_x\primed}{{x}{\lambda\primed}}\right) x \hat{\mathbf{x}}+ \left(\diffp{E_y\primed}{{y}{\lambda\primed}}\right) y \hat{\mathbf{y}} \right]. \end{eqnarray} % With this, the envelope equations become % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 127 128 \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}.  snuverink_j committed Sep 11, 2017 129 %  snuverink_j committed Sep 06, 2017 130 %\\  snuverink_j committed Sep 08, 2017 131 %\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}.  snuverink_j committed Sep 06, 2017 132 133 134 135 136 137 138 139 \label{eq:GenEnv} \end{eqnarray} So far this derivation has assumed that there is no coupling between $x$ and $y$ in both the external and self forces. In the presence of solenoids this is no longer the case for the external fields. If the beam and external fields are cylindrical symmetric then the previous analysis can be performed in the  snuverink_j committed Sep 08, 2017 140 Larmor frame. Working in the Larmor frame, the equations of motion for $\sigma_x  snuverink_j committed Sep 08, 2017 141 =\sigma_y =\sigma_L$ decouple and the envelope equation is given by  snuverink_j committed Sep 06, 2017 142 143 % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 144 \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}.  snuverink_j committed Sep 11, 2017 145   snuverink_j committed Sep 06, 2017 146 %\\  snuverink_j committed Sep 08, 2017 147 148 149 %\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 &=& %\left[\frac{e\lambda}{mc^2\beta^2 \gamma^3}\left(\diffp{E_r\primed}{{r}{\lambda\primed}}\right)\right]\sigma+\left(\frac{\epsilon_n}{p_n}\right)^2\frac{1}{\sigma^3}.  snuverink_j committed Sep 06, 2017 150 151 152 153 154 \end{eqnarray} % In this expression, $\theta_r$ is the Larmor angle, and is given by % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 155 \theta_r = -\int\left(\frac{eB_z}{2m\gamma }\right)\;\mathrm{d}t = -\int\left(\frac{eB_z}{2m\gamma \beta c}\right)\;\mathrm{d}z.  snuverink_j committed Sep 06, 2017 156 157 158 159 160 \end{eqnarray} \subsubsection{Long Uniform Cylindrical Beam} The envelope equation for a cylindrical beam is now explicitly derived. Assuming  snuverink_j committed Sep 08, 2017 161 cylindrical symmetric fields and working in the Larmor frame then $\sigma_x  snuverink_j committed Sep 08, 2017 162 163 = \sigma_y = \sigma_L$. It is important to distinguish this parameter from $\sigma_r = R/\sqrt{2}$. In fact, $\sigma_L = R/2$ for a circular beam. The  snuverink_j committed Sep 06, 2017 164 165 166 electric in the lab frame can be easily computed from Gauss's law: % \begin{eqnarray}  snuverink_j committed Sep 08, 2017 167 E_r\primed = \frac{1}{8\pi\epsilon_0}\frac{\lambda\primed}{\sigma^2}r \hspace{1cm} (r