1

I was trying to compute what is the maximum energy reached by the LHC assuming it has a circunference of 27 km length and the magnetic field of the magnets is ~8 T. To do that, I tried comparing the centripetal force suffered by protons $mv^2/R$ where $m$ is the mass of the proton, $v$ the speed of the particle and $R$ the radius of the circunference with the electromagnetic force $qvB$, with $q$ the charge of a proton and $B$ the magnetic field strength. If I do this, I'm getting a speed larger than the speed of light, so I guess I'm screwing up in some reasoning...

Juanjo
  • 900
  • Likely because you aren't including relativistic effects and only treating it with the classical equation – Triatticus May 25 '18 at 05:33
  • so both forces cannot be equaled? – Juanjo May 25 '18 at 05:59
  • No it's just you cannot use classical mechanics in a regime where something is traveling with near speed of light velocity – Triatticus May 25 '18 at 06:06
  • then what do you suggest compute the energy of the accelerator? I even found the approximation $p_{max}=0.3BR$ (with $p$ in GeV/c, $B$ in Tesla and $R$ in meters), but I do not know how to derive it – Juanjo May 25 '18 at 06:10
  • You may want to look at this related question: https://physics.stackexchange.com/questions/20919/relativistic-centripetal-force – mng May 25 '18 at 20:03

1 Answers1

1

The proton is relativistic so the centrifugal force term is different. The correct relativistic relation is amazingly simple: $p=0.3 BR$. (Units GeV/c, Tesla, meters).

Your estimate will be a bit high as the ring is not fully occupied by magnets. There are gaps for RF, collimators, quadrupoles, and other components.

RogerJBarlow
  • 10,063
  • Thanks for the answer. I already had that relation but I wanted to understand how to derive it (don't like to apply formulae without knowing their origin...) – Juanjo May 25 '18 at 06:12
  • 2
    Good policy. Should have added that 0,3 is $c$ times $10^{-9}$ because of using Giga eV. – RogerJBarlow May 25 '18 at 08:03