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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?

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I think if you want to do calculations for a single proton, you probably need some 4 vectors. I know there are some wierd effects as far as relativistic circular motion. It can be approximated by alot of small perpendicular lorentz boosts. this is beyond me, but it is great question –  Timtam Feb 13 '12 at 3:08

2 Answers 2

up vote 5 down vote accepted

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 $F=\frac{m_0a}{(1-\frac{v^2}{c^2})^\frac{3}{2}}=\gamma^3m_0v$. 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.

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It's not really an opinion if you back it up with equations... FWIW this looks like it may be correct. –  David Z Feb 13 '12 at 3:54
    
This is the correct approach, but there is no point in computing $v$ at all. Just go with the time derivative of momentum. –  dmckee Feb 13 '12 at 3:57
    
@dmckee Exactly where should I not compute v? I don't see any place where I have computed it, I've just kept the momentum terms separate so that I get acceleration. –  Manishearth Feb 13 '12 at 4:26
    
@dmckee If you're talking about the radius of curvature, I didn't use $R=p/qB$ as I myself was not sure if it was applicable. After all, the formula is derived using classical $F=ma$. Turns out that it is applicable, I just had to derive it (partially to convince myself). –  Manishearth Feb 13 '12 at 4:28
    
@DavidZaslavsky Yes, but each equation has a limit of application. For example, one cannot apply normal time dilation formulae for accelerating bodies. Similarly, using the vectorial approach to prove that CPF$=v^2/r$ could have some flaw. I don't like giving wrong answers, so I marked it as an 'opinion' (yes, bad choice of words here =P ). –  Manishearth Feb 13 '12 at 4:32

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 klystrons (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.

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Nice $F$ equation. If something is "ultra-relativistic", then can I say $E=pc$? In that case, the physics picture is almost stupidly simple. In the LHC, they send the protons in "packets" right? If you had 2 different energies they'd have to manage 2 different electronic signals (through the superconducting magnets), even though an ultra-relativistic particle would take almost forever to lap another ultra-relativistic particle? This is wacky! –  Alan Rominger Feb 13 '12 at 4:33
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It's $E^2=(pc)^2+(m_0c^2)^2$. I guess the $m_0c^2$ becomes negligible at some point (nice exchange of roles, it's usually the other way around!). Then you'll get $E=pc$. –  Manishearth Feb 13 '12 at 4:48
    
Yeah, $E = pc$ (a.k.a. assuming a massless particle) is quite common in high energy physics. –  David Z Feb 13 '12 at 5:01
    
You've got it. The mass of a proton is about 1 GeV, so when the beam energies are at TeV scale, the mass becomes negligible. Electrons have a mass of about 1/2 MeV, so it can be neglected at GeV beam energies, and so on. –  dmckee Feb 13 '12 at 5:10

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