On slip factor and phase transitions in particle accelerators In a uniform magnetic field $B$, a particle with mass $m$, charge $q$ and initial velocity $v$, undergoes a centripetal force (Lorentz force) which makes it travel on a circular orbit, with angular frequency (also called cyclotron frequency) $f_c$ given by:
$$
f_c = \frac{qB}{2\pi m}
$$
Notice this frequency does not depend on initial velocity $v$, as long as $v/c=:\beta\ll1$.
In these lectures, Introduction to particle accelerators, given by Professor E. Tsesmelis, however, there is a clear relation between frequency and momentum at every velocity:


So why even at low velocities the frequency increases as momentum (and hence velocity) increases?
 A: As you note at the start of your question, the cyclotron frequency equation holds for uniform magnetic fields. In this case, the circumference of a particle's orbit is proportional to its momentum. Since, for $\beta \ll 1$, its velocity is also proportional to its momentum, its revolution time is independent of momentum.
Professor Tsemelis treats the general case of an accelerator with a potentially non-uniform magnetic field, which you'd need if you want quadrupoles, long dipole-free straight sections for RF cavities or detectors, etc. Here, the proportionality between particle momentum and orbital circumference no longer holds, and you need to use the momentum compaction factor ($\alpha_p$) to describe how a change in momentum causes a change in orbital circumference. If you have $\alpha_p < 1$, then the fractional increase in orbital circumference is less than fractional increase in momentum, while the fractional increase in speed is equal to the fractional increase in momentum for non-relativistic particles. This results in higher-momentum particles completing one revolution faster than low-momentum ones.
