### replace sb and intertext

parent 1ac56c43
 ... ... @@ -54,15 +54,15 @@ t[i] = t + (z[i] - z) / (c * b[i]) \end{verbatim} By doing so we assume that the kinetic energy, K, increases linearly and proportional to the maximal voltage. With this model for the progress of time we can calculate $\varphi$ according to Equation~\ref{rulelag}. Next a better model for the kinetic Energy can be calculated using \begin{alltt} K[i] = K[i-1] + q $$\Delta$$z[i](cos($$\varphi$$)(Ez[i-1]($$\Gamma\sb{21}$$[i] - $$\Gamma\sb{22}$$[i]) + Ez[i]$$\Gamma\sb{22}$$[i]) $$\,\,$$- sin($$\varphi$$)(Ez[i-1]($$\Gamma\sb{11}$$[i] - $$\Gamma\sb{12}$$[i]) + Ez[i]$$\Gamma\sb{12}$$[i])). K[i] = K[i-1] + q $$\Delta$$z[i](cos($$\varphi$$)(Ez[i-1]($$\Gamma_{21}$$[i] - $$\Gamma_{22}$$[i]) + Ez[i]$$\Gamma_{22}$$[i]) $$\,\,$$- sin($$\varphi$$)(Ez[i-1]($$\Gamma_{11}$$[i] - $$\Gamma_{12}$$[i]) + Ez[i]$$\Gamma_{12}$$[i])). \end{alltt} With the updated kinetic energy the time model and finally a new $\varphi$, that comes closer to the actual maximal kinetic energy, can be obtained. One can iterate a few times through this cycle until the value of $\varphi$ has converged. \section{Traveling Wave Structure} \index{Autophase Algorithm!Traveling Wave Structure} \begin{figure} \includegraphics[angle=0,width=0.9\linewidth]{figures/field_crop} \includegraphics[angle=0,width=0.9\linewidth]{figures/field_crop.png} \caption{Field map 'FINLB02-RAC.T7' of type 1DDynamic} \label{fig:tws} \end{figure} ... ... @@ -88,7 +88,7 @@ For this example we find V_\text{core} &= \frac{V_{0}}{\sin(2.0/3.0 \pi)} = \frac{2 V_{0}}{\sqrt{3.0}}\\ \varphi_\text{c1} &= \frac{\pi}{6}\\ \varphi_\text{c2} &= \frac{\pi}{2}\\ \varphi_\text{ef} &= - 2\pi \cdot(\text{\texttt{NUMCELLS}} - 1) \cdot \text{\texttt{MODE}} = 26\pi \varphi_\text{ef} &= - 2\pi \cdot(\mathbf{NUMCELLS} - 1) \cdot \mathbf{MODE} = 26\pi \end{align*} \subsection{Alternative Approach for Traveling Wave Structures} If $\beta$ doesn't change much along the traveling wave structure (ultra relativistic case) then $t(z,\varphi)$ can be approximated by $t(z,\varphi)=\frac{\omega}{\beta c}z + t_{0}$. For the example from above the energy gain is approximately ... ... @@ -142,7 +142,11 @@ where E_z^{(1)}(z) \qquad & 0 \le z \le 3\cdot s\\ 0 & \text{otherwise} \end{cases} \intertext{and} \end{align*} and \begin{align*} -\frac{\omega}{\beta c} \xi &= \varphi. \end{align*} Here we also used some trigonometric identities: ... ...
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