Operation with passive RF

June 12, 2003

In a scenario where the west cavities are driven by the klystron and the east cavities are driven by the beam, we depend on the beam to transfer energy from west to east. The power required to maintain a particular accelerating voltage in the passive cavity P = V_acc^2/(R/Q)/Q_ext and is independent of the beam current. The electron(positron) energy that must be transferred from active to passive cavity is E = P/I and results by displacing the bunch by phase phi from the zero crossing

If Q_ext = 1e6, R/Q=89ohms, V_acc=2.5MV, and I=50mA the

Consider the effect of adding 1.4MV at the west cavities and then subtracting 1.4MV in the east. Note that the energy loss due to synchrotron radiation is negligible in comparison to energy shifts from the RF. The results are shown in the plot. The pair on the left show the closed orbits with phi=0, that is no energy gain at either cavity. The pair on the right show closed orbits with phi=0.6radians for the west cavities and phi=-0.6radians for the east. The upper pair show energy vs s and the lower pair horizontal displacement vs s. Separators are off. Red is positrons and green is electrons.

Alternatively, if Q_ext = 3e6, R/Q=89ohms, V_acc=2.5MV, and I=50mA then