## Monday, 9 April 2012

### Going Round in Circles

Why is the Large Hadron Collider a big ring?

One answer is that is uses the same tunnel as the Large Electron Positron collider, the previous CERN high energy experiment.  When LEP was shut down, the LHC made use of the facilities, rather than dig new ones.  However, this ultimately only kicks the question back a step to ask why LEP was built that way.  And neither machine is an oddity; the recently closed Tevatron also used a circular design, as did BaBar, CLEO, SppS and most high energy machines in at least my lifetime.

There are ultimately two reasons for this, both of which are about the economics of getting the most information within the budget.  The simpler reason is that by sending particles around in a circle, they can pass through the same accelerating element multiple times.  Thus we don't have to build as many such components for the same energy.

As an aside, you may be wondering what stops us getting arbitrarily high energies by going through the same accelerator time and again.  There are two limits, one theoretical and one practical.  The practical limit is that no machine is perfect, and you will lose some of the particles in your beam in each loop, till they are all gone.  The theoretical limit is that the all these machines work with charged particles, in order to accelerate them.  But a charged particle going in a circle will radiate away electromagnetic energy.  The amount of energy lost this way is proportional to the kinetic energy (and inversely proportional to the mass).  When the total energy lost each cycle equals what you add, you hit a hard limit that cannot be overcome.  The impressive thing is that in all these modern machines, it is the latter theoretical limit that is most important.

The other thing that factors into these decisions is that when two particles (or, indeed, anything) collide, the available energy is largest if the collision is somewhat symmetric.  What I mean by that is that the total momentum of the collision should be zero.

For small speeds, the momentum of an object is equal to its speed times its mass, and it also carries a direction (it is a vector).  This simple expression receives relativistic corrections close to the speed of light:
$p = \gamma m c$,
where p is the momentum; m the mass; c the speed of light; and the Lorentz gamma factor is
$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$.

The important thing about momentum is that it is a conserved quantity, once you remember to include the direction.  This is actually something we all have some intuition for.  Imagine a two ton truck hitting the back of a one ton car; you would expect the car to increase its speed by twice the speed decrease of the truck.  That instinct is essentially correct, and conservation of momentum is simply a formal way to state it.

Returning to our particle collision, we can finally make the main point.  If the two colliding particles have equal and opposite momenta, then they could potentially produce a particle at rest.  If they have unequal momenta, any final state particle would instead by moving.  Thus at least some of the energy in the initial state must go to kinetic energy in the final state, leaving less energy available to probe new physics.

Note that this does not mean that experiments with non-zero momentum are bad.  Many such experiments have been done, for a variety of reasons.  However, an experiment like the LHC, which aims to discover and examine new particles, wants to use all of its energy for that purpose.