Nonlinear cooling of an annular beam distribution

In recent years, intense efforts have been devoted to studying how nonlinear effects can be used to shape the transverse beam distribution by means of an adiabatic crossing of nonlinear resonances. By this approach, it is possible to split the beams in the transverse plane so that the initial single-Gaussian beam is divided into several distinct distributions. This is at the heart of the multiturn extraction process that is successfully in operation at the CERN Proton Synchrotron. Nonlinear effects can also be used to cool a beam by acting on its transverse beam distribution. In this paper, we present and discuss the special case of a beam with an annular distribution, showing how its emittance can be effectively reduced by means of properly devised manipulations based on nonlinear effects.

Figure 1

Phase-space portrait of the Hamiltonian (13) with parameters $\lambda =0.1$, $\mu =0.01$. The red line represents the separatrix.

Figure 2

Cooling ratio $⟨J⟩/J_0$ for trapping in $G_1$ and $G_2$, with the variation of $\lambda$ according to Eq. (28), for three values of ${\mu }^{*}$.

Figure 3

Left: implicit solution ${\lambda }^{*}({\mu }^{*})$ of the equation $A_3({\lambda }^{*},{\mu }^{*})=2\pi J_0$. Center: expected cooling ratio $J/J_0$ for trapping particles via the coupled variation of $\lambda$ and $\mu$, as a function of ${\mu }^{*}$. Right: the required value of ${\mu }^{\prime }$ to achieve the cooling efficiency shown in the center plot, as a function of ${\mu }^{*}$. Thanks to the ratios of variables reported on the axes, the plotted functions are unique and independent from the value of $J_0$.

Figure 4

Simulated cooling ratio obtained by applying Protocol A for different values of $\mathrm{\Delta }\lambda$, as a function of the initial annular distribution action $J_0$. A comparison with the theoretical bound on the cooling efficiency given by Eq. (28) is presented. The Hamiltonian (4) has been used, with $k_3=1$, ${\omega }_0/(2\pi )=0.414$, ${\mathrm{\Omega }}_2=-0.3196$, ${\mu }^{*}=7.5\times {10}^{-3}$, $t_1=5\times {10}^4$.

Figure 5

Evolution of $\lambda (t)$ and $\mu (t)$ during the three phases of Protocol B.

Figure 6

Expected and simulated cooling ratio for trapping in $G_2$ using Protocol B as a function of ${\lambda }^{*}$. The initial distribution is ${\rho }_{0.05}$. The Hamiltonian (4) has been used, with $k_3=1$, ${\omega }_0/(2\pi )=0.414$, ${\mathrm{\Omega }}_2=-0.3196$, ${\mu }_1=0.02$, $t_1=1/\dot{\lambda }=5\times {10}^4$.

Figure 7

Distributions at the end of the first (left), second (center), and third (right) phase for an initial distribution ${\rho }_{0.05}$ for Protocol B for three values of ${\lambda }^{*}$. The color scale represents the initial angle ${\varphi }_0$ and the initial distribution is the same as that shown in the left plots of Fig. 11. Note that for ${\lambda }^{*}>4J_0=0.2$, the angular dependence of the final action is reversed w.r.t. ${\lambda }^{*}<0.2$. The Hamiltonian of Eq. (4) has been used, with $k_3=1$, ${\omega }_0/(2\pi )=0.414$, ${\mathrm{\Omega }}_2=-0.3196$, ${\mu }_1=0.02$, and $t_1=1/\dot{\lambda }=5\times {10}^4$.

Figure 8

Top: cooling ratio for trapping in $G_2$ using Protocol B, at different values of ${\lambda }^{*}$, as a function of the initial action of the annular distribution $J_0$. Bottom: cooling ratio for the same values of ${\lambda }^{*}$ as a function of ${\sigma }_0$, using the initial distribution ${\rho }_{\zeta ,{\sigma }_0}^{\mathrm{a}}(J_0,\varphi )$ defined in Eq. (45) with $\zeta =0.05$. ${⟨J⟩}_{\mathrm{i}}$ is the average of the initial actions. The Hamiltonian (4) has been used, with $k_3=1$, ${\omega }_0/(2\pi )=0.414$, ${\mathrm{\Omega }}_2=-0.3196$, ${\mu }_1=0.02$, and $t_1=1/\dot{\lambda }=5\times {10}^4$.

Figure 9

Evolution of $\lambda (t)$ and $\mu (t)$ during the two phases of Protocol C. The two values ${\lambda }_{\max }$ and ${\mu }_{\max }$ have expressions in function of the computed ${\lambda }^{*}$, ${\mu }^{*}$, and ${\mu }^{\prime }$, i.e., ${\mu }_{\max }={\mu }^{*}+\lambda |{\mu }^{\prime }|$, ${\lambda }_{\max }={\lambda }^{*}+{\mu }^{*}/|{\mu }^{\prime }|$.

Figure 10

Expected and simulated cooling ratio for trapping in $G_1$ using Protocol C as a function of ${\mu }^{*}$ for an initial distribution ${\rho }_{0.05}$. The Hamiltonian (4) has been used, with $k_3=1$, ${\omega }_0/(2\pi )=0.414$, ${\mathrm{\Omega }}_2=-0.3196$, and $t_1=1\times {10}^5$.

Figure 11

Particle distributions when applying Protocol C at the beginning (left), after the first phase (middle), and at the end of the second phase (right), for two values of ${\mu }^{*}$. The hue encodes the initial angle of the action distribution. The Hamiltonian (4) has been used, with $k_3=1$, ${\omega }_0/(2\pi )=0.414$, ${\mathrm{\Omega }}_2=-0.3196$, $t_1=1\times {10}^5$.

Figure 12

Top: simulated cooling ratio $⟨J⟩/J_0$ for initial distributions ${\rho }_{J_0}$ using Protocol C at different values of ${\mu }^{*}$, having computed ${\mu }^{\prime }$ and ${\mu }_{\max }$ for $J_0={\widehat{J}}_0=0.05$ (indicated by a vertical dotted line in the plot). Bottom: cooling ratio for the same values of ${\mu }^{*}$ as a function of ${\sigma }_0$, using the initial distribution ${\rho }_{\zeta ,{\sigma }_0}^{\mathrm{a}}(J_0,\varphi )$ defined in Eq. (45) with $\zeta =0.05$. ${⟨J⟩}_{\mathrm{i}}$ is the average of the initial actions. The Hamiltonian (4) has been used, with $k_3=1$, ${\omega }_0/(2\pi )=0.414$, ${\mathrm{\Omega }}_2=-0.3196$, and $t_1=1\times {10}^5$.

Figure 13

Final particle distributions after applying Protocol C, for initial distributions ${\rho }_{0.045}$ (left) and ${\rho }_{0.05}$ (right), having computed ${\mu }^{\prime }$ for ${\widehat{J}}_0=0.05$. The hue indicates the initial angle ${\varphi }_0$. We see that, although the process has been optimized for ${\widehat{J}}_0=0.05$, for $J_0<{\widehat{J}}_0$ (and close to the minimum shown in Fig. 12), the resulting cooling is better. The Hamiltonian (4) has been used, with $k_3=1$, ${\omega }_0/(2\pi )=0.414$, ${\mathrm{\Omega }}_2=-0.3196$, $t_1=1\times {10}^5$, and ${\mu }^{*}=0.01$.