Monopole Mode Excitation

The voltage induced by a point-charge, $q$, according to the fundamental theorem of beam loading given by

  \Delta V_{q,n} = - q \frac{\omega_n}{2}(R/Q)_n (\beta) (1)

where $\omega_n$ is the angular frequency of mode, $n$, $\beta$ is the velocity of the particle as a fraction of the speed of light and $(R/Q)_n (\beta)$ is given by

  (R/Q)_n (\beta)  = \frac{\left| \int_{-\infty}^{\infty}E_{z,n}(r=0,z)e^{i\omega_n \frac{z}{\beta c}}dz\right|^2}{\omega_n U_n}  (2)

where $U_n$ is the stored energy in the mode and $E_{z,n}(r = 0,z)$ the electric field along the beam axis. Due to losses, the amplitude of the voltage will decay according to

  V_n(t) = \Delta V_{q,n}e^{-t/T_{d,n}}e^{i\omega_n t} (3)

where $T_{d,n} = 2Q_{L,n}/\omega_n$ is the voltage decay time constant with $Q_{L,n}$ as the loaded quality factor of the complete system and is the inverse sum of all the subsystems

  \; \frac{1}{Q_L} = \frac{1}{Q_{ex}} + \frac{1}{Q_0}. (4)

For a superconducting cavity, the natural $Q_0$ is of the order of $10^{10}$ meaning that that $Q_L$ is dominated by $Q_{ex}$ and so the approximation that $Q_L \approx Q_{ex}$ is made.

In pulsed operation the voltage after one pulse with $N$ bunches and no HOM voltage present before the arrival of the first bunch, is given by

  V_{p,n} = \Delta V_{q,n} \sum^N_{k=1} e^{-kT_b/T_{d,n}+ik\omega_nT_b} = \Delta V_{q,n} \frac{1-e^{-NT_b/T_{d,n}+iN\omega_nT_b}}{1-e^{-T_b/T_{d,n}+i\omega_nT_b}} (5)

where $T_b$ is the time between bunches. After $m$ pulses with pulse period $T_p$, the voltage present in the cavity is given by:

  V^m_{p,n} =  \Delta V_{q,n}  \frac{1-e^{-NT_b/T_{d,n}+iN\omega_nT_b}}{1-e^{-T_b/T_{d,n}+i\omega_nT_b}} \frac{1-e^{-mT_p/T_{d,n}+im\omega_nT_p}}{1-e^{-T_p/T_{d,n}+i\omega_nT_p}} (6)

Monopole Mode Interaction

  dU_n = q(\Re(V_n)\cos(\omega_n dt) - \Im(V_n)\sin(\omega_n dt)) - \frac{1}{2}\Delta V_{q,n} (7)

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Topic revision: r4 - 06 Jun 2012 - TomCraneAdmin

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