# How's the energy of particle collisions calculated?

Physicists often refer to the energy of collisions between different particles. My question is: how is that energy calculated? Is that kinetic energy?

Also, related to this question, I know that the aim is to have higher and higher energy collisions (e.g to test for Higgs Boson). My understanding is that to have higher energy you can either accelerate them more, or use particles with higher mass. Is this correct?

## 3 Answers

I think your question is divided in two parts.

1. When talking about energy, in the field of HEP or accelerator physics we can talk about

• total energy
• kinetic energy
• momentum

As for elementary particles relativistic effects manifest themselves almost all the time, you need to use a relativistic form for the energy:

$$E = \sqrt{p^2c^2+m^2c^4}$$

Instead of a bare $E = p^2/2m$ (as in classical mechanics, where $p = m v$).

This relation as two parts: one depending on the momentum and one (constant) given by the mast of the particle.

It should be also noted that for ultra-relativistic cases, where $E \gg E_0$, we have $E = p c$.

Usually for low energy applications, like linear accelerators or low energy experiments we talk about the kinetic energy, which is $E_k = \sqrt{p^2c^2+m^2c^4} - mc^2$. For example if you talk about protons of 160 MeV obviously it is kinetic energy, as the rest mass of the proton is rhougly 1 GeV.

For higher energy application usually you can make the ultra-relativistic approximation and you then talk about total energy (in eV) or about the momentum in $\mathrm{eV}/c$; taking $c=1$, both are numerically equal.

When you are not sure about which approximation you can take, it is better to explain which one you take.

Example: For a proton in the LHC with a momentum of 3.5 TeV/c you can calculate its total energy which is ...

2. How is this energy "calculated"? (I assume you meant "experimentally" or something like that.)

In HEP physics we use what we call electron-volt as a unit of energy. A particle of unit charge will have an energy of 1 eV if it descends from rest a potential difference of 1 V.

So for example, when you accelerate protons in the LHC, if you have cavities giving you 10 MV, the particle will gain 10 MeV every turns.

• I was referring to the case of LHC. So in your example of a proton with an energy of 3.5 TeV, what speed should the proton have? – Albert Nov 3 '10 at 17:11
• The reasoning to have the speed of the protons at such high energy is similar to that case: physics.stackexchange.com/questions/139/speed-of-neutrinos/… . Basically: E/E0 = gamma =~ 3500 and gamma = 1 / sqrt(1- (v/c)^2) , you can then extract v (should be 0.some9s * c ). – Cedric H. Nov 3 '10 at 17:18
• OK, so if you use another particle, let's say with a rest mass twice as that of the proton, you achieve a bigger energy right? – Albert Nov 3 '10 at 17:49
• No, because you won't be able to accelerate it to higher energy, because this other particle is heavier and then if you accelerate it with say the LHC, you would need a higher magnetic field, which you don't have. – Cedric H. Nov 3 '10 at 17:51
• Partly the answer is that the energies we're talking about are mostly kinetic (much bigger than the mass), so the mass is relatively unimportant. Another part of the answer is that a nucleus is a big object consisting of many protons and neutrons loosely held together. When you collide nuclei, the mass of the protons and neutrons never gets entirely funneled into the creation of one heavy object (like a Higgs boson), because they're all moving quasi-independently. Even when protons are collided, heavy objects like a Higgs are made from only two gluons, or two quarks, not the whole proton. – Matt Reece Nov 4 '10 at 23:04

In experimental high energy physics, the beam energy is well known. For example, if you have a beam of protons, you know the energy because the engineers controlling the accelerator make sure that the beam is very well collimated, and goes on track, etc. If it weren't so, the beam would hit the pipe walls and you'd lose it... There are also instruments along the beam pipe which measure the current, so all that is used to control the total energy of the beam.

Secondly, experimental particle physicists seldomly look into individual collisions — because there are millions of collisions per second, with millions of electronic channels, the pile up of collision events is not negligible (typically 10-20 events per "frame"). It's simply too complicated (and error prone) to look at individual events. To make stand out the potentially interesting events from the manifestly uninteresting background events (those too common and already studied, such as low-mass-particle decays), they make specific cuts in quantities that they know (from numerical simulations) will exclude manifestly uninteresting events, and in the end you have events which are likely to be the "type" you're looking for. On example is this: if you only accept reconstructed tracks which have a linear momentum higher than a certain value, you exclude a lot of particles ("good" and "bad") in the direction of the collision (beam-wise) but those who are scattered perpendicularly to that direction (i.e. away from the beam), and have a high energy, are likely to be interesting (as expected from numerical simulations, i.e. "Monte Carlo" as it is called).

Bottom line (to make this short):

They know very well what goes in, but they know not too well what goes out for individual events. When you only count high-energy particles (i.e. those who don't bend very much under the detector's strong magnetic field) and you start superimposing (piling up!!) all the events which have a couple (in about 100 or so) of "promissing" particle tracks, they start accumulating around the "correct value". That's how they know that "when two particles collided, a heavy one was created momentarily and it decayed into lighter ones".

You can have a clue about how piling up individual events might give the approximate good answer from this illustration:

Suppose you have a glass of sand and you let it drop slowly into the floor. You then ask a friend to come into the room and tell him to make an estimation of where in the room you dropped the sand from. That should be easy. He might even make a more or less good prediction of how high you dropped the sand by how much it spreads on the floor (it should spread wider if dropped higher). Even if your friend knows exactly how much sand was dropped (from knowing how much sand was in the glass), he only has a "good enough" hint at what happened (where in the room, how high) when the sand was dropped from the glass.

• Not a bad discussion, and you get my vote, but to say that particle physicist rarely look at single collisions is to mistake the collider guys for all of particle physics. When we're talking about neutrinos or ultra-high energy cosmic rays or the non-perturbative energy regime (as at JLAB) we tend to look at one event at a time. – dmckee --- ex-moderator kitten Mar 6 '11 at 1:16

It is worth remembering that in at velocities close to c. A particle kinetic energy is inter-wind with its rest mass. So the actual energy equation is

$E = \sqrt{p^2c^2+m^2c^4}$

So the energy of a collision is the sum of the above energy for the two colliding particles.

That is why you construct improved accelerators. As it is the only way to tune up the energy. To the extend of my knowledge we don't how to tune up mass.

• Isn't p = m*c ? – Albert Nov 2 '10 at 20:21
• sorry... I mean p = m * v – Albert Nov 2 '10 at 20:25
• "To the extend of my knowledge we don't how to tune up mass." : just by taking another particle – Cedric H. Nov 2 '10 at 20:53
• @Albert: No when the speed of particle is very fast. (p = mv/√(1-v^2/c^2), m is rest mass.) – kennytm Nov 2 '10 at 20:57