Cyclotrons and Synchrotrons — Question 5

Question 5

Relativistic slowing of the orbital frequency creates RF phase slip

Use $\gamma=1+K/(mc^2)$ and the relativistic period $T=2\pi\gamma m/(qB)$. The RF remains set at the classical angular frequency $\omega_0=qB/m$. Consider a particle coasting at K=.0500mc squared with no energy change during the interval considered.

1. Find the RF phase advance during one particle semicircle and its excess over pi.

2. Find the accumulated excess after ten semicircles and the RF frequency required to restore synchronism at this energy.

Question 5 - Solution

Relativistic slowing of the orbital frequency creates RF phase slip

(a) Method and calculation

$$t_{\mathrm{half}}=\pi\gamma/\omega_0,\quad\Delta\phi_{\mathrm{RF}}=\omega_0t_{\mathrm{half}}=\pi\gamma.$$$$\delta\phi=\pi(\gamma-1)=.0500\pi=9.00^\circ.$$

(b) Interpretation and further calculation

$$10\delta\phi=.500\pi=90.0^\circ.$$$$\omega_{\mathrm{RF,new}}=\omega_0/\gamma=\omega_0/1.05.$$

Physical check

The constant-energy assumption makes the phase slips equal; during acceleration their values must be summed at the changing energy.