Cyclotrons and Synchrotrons — Question 8

Question 8

Synchrotron RF timing and bending-field ramps obey separate constraints

A synchrotron has circumference C=300 m and RF harmonic number h=20, meaning $f_{\mathrm{RF}}=h f_{\mathrm{rev}}$. A particle accelerates from .600c to .800c at fixed orbit geometry. Use c=$3.00\times10^8$ m/s and $p=\gamma mv$.

1. Find the RF frequency at each speed.

2. Find the ratio of final to initial bending field and explain why it differs from the RF-frequency ratio.

Question 8 - Solution

Synchrotron RF timing and bending-field ramps obey separate constraints

(a) Method and calculation

$$f_{\mathrm{RF}}=h\frac{\beta c}{C}.$$$$f_1=12.0\,\mathrm{MHz},\quad f_2=16.0\,\mathrm{MHz},\quad f_2/f_1=4/3.$$

(b) Interpretation and further calculation

$$B_2/B_1=p_2/p_1=\frac{\gamma_2\beta_2}{\gamma_1\beta_1}=\frac{(5/3)(4/5)}{(5/4)(3/5)}=16/9.$$$$f_{\mathrm{RF}}\propto v,\quad B\propto\gamma v\quad\text{at fixed geometry}.$$

Physical check

RF timing follows the time around the orbit; magnetic bending follows momentum.