special relativity - Relativistic centripetal force

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?

Answer

I'm not someone who inderstands the physics well, so I'm not 100% sure of this. Comments on this will be greatly appreciated. (Update: This approach seems to be correct)

F=dp/dt works as well as $v^2/r$. As long as you want to spin it at a constant velocity, your lorentz factor will be constant, and since $p=\gamma m_0v$, your force will become $m_0\gamma d\vec{v}/dt$. Then you can solve it normally using vectors (exact same proof a the classical one for CPF). Your end result will be $\gamma m_0 v^2/r$.

When dealing with accelerations due to forces, again use $F=d(\gamma m_0 v)/dt$. If $m_0$ is constant, then we get $F=\frac{m_0a}{(1-\frac{v^2}{c^2})^\frac{3}{2}}=\gamma^3m_0a$. Since we still have $\gamma$, the accelerations will be very much different.

So if a proton has twice the energy of another, the proton has twice the $\gamma$ (from $E=\gamma m_0c^2$). So, the acceleration will be eight times less.