Cyclotron cannot accelerate particles beyond certain KE due to relativistic mass gain. I am from Chemistry background.

  • $\begingroup$ Can you clarify what you are asking? In synchrotrons we are normally interested in the kinetic energy of the particles rather than the velocity. So it's the job of the synchrotron to pump in as much KE as possible. $\endgroup$ – John Rennie Nov 5 '19 at 17:30
  • $\begingroup$ The answer is fairly clear if you know how a cyclotron works. That is, do you know the mathematics for a charged particle orbitting in a uniform magnetic field and about the "Dees" and how they give a energy kick to the particle on each half-orbit? $\endgroup$ – dmckee --- ex-moderator kitten Nov 5 '19 at 18:05
  • $\begingroup$ That question is too short. It's not clear what you want to know. $\endgroup$ – Cham Nov 5 '19 at 18:06
  • $\begingroup$ I know how cyclotron works. Limitation of cyclotron is that it can accelerate particle to a certain extent especially for electron. How this problem is solved in synchrotron? $\endgroup$ – ggs Nov 5 '19 at 18:25
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    $\begingroup$ Relativity poses a limitation on speed not on KE. There is no bound on KE. It can grow as much as you will; depending only on the amount of energy you have to spend. Near the speed of light doubling the KE will result in almost no increase in speed. $\endgroup$ – J. Manuel Nov 5 '19 at 19:06

The radius of curvature of a particle in a uniform magnetic field of strength $B$ is $$ r_c = \frac{p_\perp}{qB} \;, $$ where $q$ is the charge of the particle, $p$ is momentum and the $\perp$ indicates that we care only about the componenet of momentum perpendicular to the field. From here we'll pretend there is no axial momentum so that $p = p_\perp$.

Now, the time to go one time around the path (the period) is $$ T = \frac{2 \pi r_c}{v}\;$$ so substituting in for the radius of curvature and using $p = \gamma m v$ we find that the cyclic frequency is $$ f = \left( \frac{qB}{2 \pi m} \right) \frac{1}{\gamma}$$ Here $m$ is the invariant (rest) mass of the particle and $\gamma = [1 - (v/c)^2]^{-1/2}$ is the Lorentz factor. The values in the parentheses are constant, and the Lorentz factor is effectively constant at low speed which is why a plain cyclotron works with a constant frequency.

But as the speed increases $\gamma$ begins to increase and it takes longer for the particle to complete each half cycle. If you continued to operate the dees with the same driving frequency they would get out of synch with the motion of the particle and would no longer be consistently giving it a energy boost.

To compensate you control the driving frequency so that it drops as the particle's orbital frequency does, keeping the acclerating potentials in synch with the motion of the particle and allowing you to add more energy.


A synchrotron works works with short pulses of particles. The room sized machine I worked with (at MIT back in the fifties) started with particles at about 1 Mev from a Van deGraph generator and used a magnetic field that increased as the particles in a pulse gained mass, in order to keep them moving in a circular evacuated tube. (The magnet was half of an LC circuit. The C was a room full of capacitors next door.) As I recall, the particles were electrons, and gained energy (and mass) but not speed after entering the magnetic field.


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