Beam Excitation and Resonances

As already mentioned in chapter 2.3 there is a relationship between spin motion and orbital motion; this shall be examined in this section. Assume an electron performing vertical betatron oscillations according to Hill's Equation and consider these oscillations damped weakly:

$$\frac{d^{2}z}{dt^2} + \delta \frac{dz}{dt} + {\omega_{\beta,z}}^2z = 0$$ (26)

where $\delta$ is the damping constant and $\omega_{\beta,z}$ is the vertical betatron oscillation frequency. Now a vertical kicker magnet shall be used to excite the beam. Compared to the perturbation of the beam through the kicker magnet damping is negligible:

$$\frac{d^{2}z}{dt^2} + {\omega_{\beta,z}}^2z = F(t)$$ (27)

The radial field $F(t)$ is time dependent, but it is more complicated than simply $\propto \cos(2\pi\nu_{kick}t)$ because the electron only sees the excitation during the small period of time in which it travels through the kicker magnet. Therefore $F(t)$ can be arranged as a cosine multiplied with a step function [13]:

$$F(t) = \left[a_0 + \sum_{n=1}^{\infty} a_n\cos(n2\pi\omega_0t)\right]\cdot \cos(2\pi\omega_{kick}t)$$ (28)

where $\omega_0$ is the revolution frequency. The solution to equation 27 is:

$$z(t) = \sum_{n=0}^{\infty} A_n\cos\left[2\pi(\omega_{kick}\pm n\omega_0)\right]$$ (29)

and the amplitudes are determined by:

$$A_n \propto \frac{1}{{\omega_{\beta,z}}^2 - (2\pi\omega_{kick} \pm n2\pi\omega_0)^2}$$ (30)

The resonance condition leads to:

$$\omega_{res} = ~\mid\frac{\omega_{\beta,z}}{2\pi\omega_0}\mp... ...d\cdot\omega_0 = ~\mid Q_z\mp n\mid\omega_0 \qquad n=0,1,2,...$$ (31)

(where $Q_z$ is the vertical tune) which means that if $\omega_{kick}=\omega_{res}$ the vertical beam oscillations will be excited to a point where the beam is lost. The same can be shown for the horizontal betatron oscillations ($Q_x$) as well as for synchrotron motion ($Q_s$). Such resonances must be avoided during the whole experiment.

The motion of the spin vector can be described in a very similar way: A solution to the Thomas-BMT equation (equation 11) is given by (neglecting phase)⁵:

$$S_y(t) = A\sin(\Omega t)\qquad\textrm{where}\qquad A = \sqrt{1-{S_z}^{2}} \ll 1$$ (32)

Now the spin precession is excited with the kicker magnet:

$$\frac{d^2S_y}{dt^2} + \Omega^2S_y = F(t)$$ (33)

Taking into account again that $F(t)$ is not a pure cosine, but a cosine multiplied with a step function, one derives:

$$S_y(t) = \sum_{n=0}^{\infty} A_n\cos\left[2\pi(\omega_{kick}\pm n\omega_0)\right]$$ (34)

$$A_n\propto\frac{1}{{\Omega^2 - (2\pi\omega_{kick} \pm n2\pi\omega_0)^2}}$$ (35)

Keeping in mind equations 11, 12 and 14 an recalling that

$$\Omega = 2\pi\omega_0(a\gamma+1)$$ (36)

the resonant depolarizing frequencies are derived:

$$\omega_{depol} = \mid a\gamma + 1 \mp n~\mid \omega_0 \qquad n = 0,1,2,...$$ (37)

This shows that the resonant depolarizer frequencies are sidebands of the revolution frequency. The case where $n=a\gamma$ is called the integer spin tune. If the kicker frequency is in resonance with the depolarizing frequency ( $\omega_{kick}=\omega_{depol}$) it is possible to tilt the mean spin vector of the electrons in the beam into the horizontal plane; in presence of spin diffusion effects (with a decoherence time much smaller than $\tau_{depol}$) this depolarizes the beam.