The thought randomly occurred to me that a circular particle accelerator would have to exert a lot of force in order to maintain the curvature of the trajectory. Many accelerators move particles at fully relativistic speeds, and I want to ask how that affects things.
Why does this matter? Well, if I understand correctly, a particle in the LHC moving at $0.999 c$ would be dramatically more difficult to keep moving in the circle than a particle moving at $0.99 c$. Reading Wikipedia, I was delighted to find a fully specified problem within a single paragraph.
http://en.wikipedia.org/wiki/Large_Hadron_Collider
The LHC lies in a tunnel 27 kilometres (17 mi) in circumference, as deep as 175 metres (574 ft) beneath the Franco-Swiss border near Geneva, Switzerland. Its synchrotron is designed to collide opposing particle beams of either protons at up to 7 teraelectronvolts (7 TeV or 1.12 microjoules) per nucleon, or lead nuclei at an energy of 574 TeV (92.0 µJ) per nucleus (2.76 TeV per nucleon).
I can (or Google can) calculate the proton case to come to $0.999999991 c$. In order to correctly calculate the force that the LHC must exert on it, do I need to start from $F=dp/dt$, or can I get by with $v^2/r$ times the relativistic mass? I'm not quite sure how to do the former.
Question: Say I have 2 protons, both moving so fast they're going almost the speed of light but one has twice the energy of the other. They move side-by-side and enter either an electric or magnetic field that causes them to accelerate perpendicular to the velocity vector. Is the radius of curvature mostly the same for the two, or is it different by a major factor (like 0.5x, 1x, or 2x)?
I want to read some comments from people who understand these physics well. The particles beyond a certain energy are all going almost exactly the same speed. Is it still okay for me to say they have the same "acceleration" too? They just happen to have absurdly huge inertia compared to the same particle at a more modest fraction of the speed of light. Is this the correct perspective?
Best Answer
Because we use magnetic fields to bend the path of particles in accelerators and E&M is Lorentz invariant by construction, we just apply the bending radius in a magnetic field equation in the lab frame and never bother to compute the force. The radius of curvature is
$$ R = \frac{p}{qB} .$$
Note that for a ring like the LHC, the bending is not actually uniform, but only in the bending magnets, but not in the quadrupoles or cavities (if any), so it is locally tighter than you would get by naively apply the above equation with a 27 mile radius.
If you insist on finding the force you'll note that over the course of one cycle the momentum changes by $2\pi |p|$, and it takes $(27\text{ km})/c$ seconds to get there so the mean force is
$$ F = \frac{p c}{r} = \frac{2 \pi p c}{27 \,\mathrm{Km}}. $$
Again, in any given magnet it will be larger by a factor of less than 10 because the magnets don't cover the whole beam line.
Aside: If you spend much time doing particle physics you'll come to love ultra-relativistic mechanics: it's even easier than non-relativistic mechanics.