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The LHC in Geneva is a circular accelerator, 27 km long, why is it like that ?

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Could someone please edit the title to include a verb, I don't have enough rep yet: "Why is the LHC circular and 27km long?" –  ihuston Nov 2 '10 at 23:04
    
@ihuston: oops ! –  Cedric H. Nov 2 '10 at 23:31
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4 Answers 4

When you accelerate charged particles, they lose energy by emitting photons (a process called "Bremsstrahlung" or "braking radiation"). This is a nuisance in particle accelerators, because (1) you want to impart as much energy as possible to the particles being accelerated (that's the point!) and this is a loss; and (2) the bremsstrahlung can be in the form of harmful (to people and machines) ionizing radiation.

Stronger magnets are also necessary to produce the tighter turns.

Acceleration in the beam direction is, of course, unavoidable. But any time you have motion along a curved path, you also need acceleration perpendicular to the direction of motion in order to curve the trajectory.

In order to minimize this acceleration perpendicular to the direction of motion, it is desirable to have the accelerator be as straight as possible. If you are building a machine that accelerates particles along a closed path, this means you want to make the radius as big as possible. (I assume that, if you do the math, it turns out that a circular path (with uniform curvature) is better than an elliptical one, or one with long straightaways followed by tight turns.)

An alternative is to build a linear accelerator, which simply accelerates particles in a straight line. The Stanford Linear Accelerator (SLAC) is one such accelerator; and the International Linear Collider is in the planning phases.

Why is the LHC 27 km in circumference? Because they are re-using the tunnels from the Large Electron–Positron Collider (LEP), which was 27 km around.

Why was LEP 27 km around? It was almost certainly a balance between the science goals and the money available (longer tunnels = more expensive).

Check out the Superconducting Super Collider (SSC) on Wikipedia, the doomed American successor to LEP. It was going to be 87 km in circumference.

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synchrotron radiation goes as 1/m^4, thus being negligible for protons. It is usually a design constraint for electron accelerator but it is not as important for a proton machine. –  Cedric H. Nov 2 '10 at 22:40
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A couple thoughts: it's mostly the smallest radius of curvature anywhere along the path that constrains the design. So if you're going to build the tunnel with a 27 km radius at one point, you might as well make it 27 km all around, i.e. a circle. Plus, with an ellipse, you'd have a varying radius of curvature so you'd need different magnet strengths all around the track. I think those are some of the reasons these things are circular. –  David Z Nov 3 '10 at 5:42
    
Points of fact that may be of interest: (1) SLAC has a pair of curved section as the end of the straight run so that the accelerators can share a tunnel, but still collide. (2) CEBAF (at JLAB) is an oval electron machine (modest energies, space constraints and the multiple interlaced beams with N-pass pickoff for different Ns). I other words, nibot's discussion gets at some of the big engineering factors, but there are others that sometimes come into play. –  dmckee Nov 17 '10 at 18:49
    
Does it help for it to be a closed loop, or is it merely for space-saving reasons? I.e. do the particles get accelerated more and more on each subsequent "lap", thus making a 27km circular accelerator potentially more powerful than a 27km long straight line? –  romkyns Jan 12 '11 at 15:13
    
@romkyns: It's about money. Each klystron can only add so much energy to the beam per pass and klystrons cost a lot. A closed path means you don't have to buy so many. –  dmckee Mar 6 '11 at 0:56
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The LHC is a synchrotron, that is, a accelerator with a magnetic field confining the orbit on a circular path and using RF accelerating cavities to accelerate the particles.

The voltage provided by the cavities is limited (the order of MV) and thus a linear accelerator cannot achieve such high energies (of the order of the TeV) (although some projects of TeV linear collider are in development, CLIC and the ILC) because it would be extremely long. The idea is thus to have a circular path, the particle going through the cavities at each turn and gaining a small amount of energy each turn.

To have this circular path, we use a magnetic field, it does not accelerate the particles, but it provides a force perpendicular to the motion, thus allowing to bend the trajectory and to obtain a circular orbit.

Why does it have to be that long ? A fundamental relation for the synchrotrons is:

p = q B r

where p is the particle momentum, q is the charge of the particle, B is the magnetic field and r is the radius of curvature.

We can then see that to have a high momentum (and energy) we need a high magnetic field and a large radius.

In the LHC, the magnetic field is already at the limit of what a superconducting magnet can achieve (almost 8.5 T).

The LHC then needs a very large tunnel. For that it reuses the tunnel of the LEP which was also a synchrotron, but for electrons.

In that case the size of the tunnel is not really given by the same reasoning. We need to take into account the synchrotron radiation: any accelerated charge radiates energy in the form of a EM radiation: "light".

But the amount of radiation goes with the inverse of the fourth power of the mass, the electron being very light they emit a large amount of radiation. For protons, this effects is almost negligible, that's why it does count for the LHC.

But for LEP (and LEP gave it's tunnel to the LHC) this was the main limitation to the achieved energy. And to obtain a high energy, the larger the tunnel the better, because the amount of radiation decreases with the bending angle of the dipole magnets, meaning that a large radius leads to lower radiation.

Finally, the size of precisely 27 km was chosen for geographic consideration: the tunnel is between the Jura Mountains and the Leman lake, this implies strict constraints in the civil engineering.

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As we are in the beta phase, it took the liberty to ask/answer a question. –  Cedric H. Nov 2 '10 at 22:25
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Cyclotrons, while generally economic because of the reduced space and cost to build (compared to linear accelerators), suffer from two notable problems.

Note that the LHC is in fact a synchotron, which is notably improved over a cyclotron for relativistic (high energy) particles. The following principles still apply pretty well, however.

  1. The magnetic field required to impart the centripetal acceleration on the particle to keep it in orbit is inversely proportional to the radius of orbit. Creating stronger magnetic fields requires larger and much more costly magnets. Maximising the radius of orbit helps reduce the required strength of magnetic field.

  2. Maxwell's equations of electromagnetism indicate that an accelerating charge loses energy by emitting radiation. In fact, if you do the maths, this introduces the limit on the maximum velocity/energy of any accelerated particle. The larger the radius, the lower the centripetal force/acceleration, hence the lower the rate of energy loss.

To illustrate point 1, the force on a charged particle is given as

$$F = Bqv$$

where B is the magnetic field strength. And the centripetal force on a charge is given as

$$F = \frac{mv^2}{r}$$

where r is the radius of the particle accelerator tunnel from the centre of the orbit.

Equating the two gives

$$B = \frac{mv}{qr} = \frac{m\omega}{q}$$

Clearly, increasing the radius $r$ requires a lower $B$ (magnetic field).

Point 2 is slightly more involved to show, as one has to apply Maxwell's equations and calculate the energy flux emitted. Hopefully you get the point however.

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Nice simple explanation, but the LHC is not a cyclotron, but a synchrotron. And synchrotron radiation goes as 1/m^4, thus being negligible for protons. –  Cedric H. Nov 2 '10 at 22:09
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True, but the LHC is a synchrotron, not a cyclotron. –  nibot Nov 2 '10 at 22:11
    
@Credic, nibot: That's a fair point. I originally meant to put that, but left it out. (It's edited now.) The same principles still apply to lesser degrees however. I believe you're right about Maxwell's equations applied relativistically reducing the radiation flux, but I'm not sure it's totally negligible. Would need to think about it a bit more, really. –  Noldorin Nov 2 '10 at 22:19
    
@Cedric: Note that point one in particular still applies very much. –  Noldorin Nov 2 '10 at 22:20
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About synchrotron radiation: the LHC is actually the first proton accelerator where this is not a completely negligible effect, but it is a second order effect: it is taken into account in the heat load of the cryo system and it is used for beam characterization, but from a pure synchrotron motion/acceleration point of view it is negligible. –  Cedric H. Nov 2 '10 at 22:28
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To accelerate particles to very high energies, you need a long path, because the force is limited (just as in case of an airplane runway).

You run them in circles to have it confined, or you would have to build something the size of a solar system.

There are, however, what is called Linear Accelerators, like SLAC. These are used to accelerate electrons, which are too light to be run in circles. They radiate a lot of energy when pushed sideways.

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Amusingly, circular runways have also been tried, usually so that there will always be a place where you can take-off/land into the wind, but at least once for acceleration, as mentioned in this old Time magazine: time.com/time/magazine/article/0,9171,936623,00.html –  nibot Nov 2 '10 at 22:19
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