Accelerators, Detectors and the Standard Model – Nature of Particles and Their Interactions (Senior 6 Physics) | Ganzaa

Nature of Particles and Their Interactions

Lessons

Senior 6 · Physics

Nature of Particles and Their Interactions

5 lessons

Lesson 5 of 5

Accelerators, Detectors and the Standard Model

25 min

A 27-kilometre microscope

On 4 July 2012, physicists at CERN, the European laboratory for particle physics near Geneva, announced that two huge detectors, ATLAS and CMS, had each found a new particle with a mass of about 125 GeV: the Higgs boson, predicted almost fifty years earlier. They found it by smashing protons together in the Large Hadron Collider (LHC), a ring 27 km round, buried about 100 m underground, where since 2022 two beams of protons have collided at a total energy of 13.6 TeV. Why does particle physics need such enormous machines, and how can anyone see a particle that lives for less than $10^{-21}$ s? This lesson answers both questions and puts the whole unit together in the Standard Model.

Why high energies?

To see small things. A probe can resolve details about as small as its de Broglie wavelength, $\lambda = \frac{h}{p}$ . For a very fast particle, $p \approx \frac{E}{c}$ , so $\lambda \approx \frac{hc}{E}$ . Higher energy means a shorter wavelength and finer detail. At 17 GeV, the top energy of the SLAC experiment in Lesson 1, $\lambda \approx 7 \times 10^{-17}$ m, much smaller than a proton.

To make heavy things. By $E = m c^{2}$ , new particles can be created only if enough energy is available in the collision: at least 125 GeV for a Higgs boson, 173 GeV for a top quark.

Colliding beams. When a moving particle hits a stationary target, much of its energy goes into the forward motion of the products, because momentum must be conserved. When two beams of equal energy collide head on, the total momentum is zero and ALL the energy is available to make new particles. That is why the LHC collides two beams.

Accelerators

Accelerators use electric fields to give charged particles energy and magnetic fields to steer them.

Linear accelerator (linac). Particles pass through a line of hollow metal tubes connected alternately to an alternating supply. They are accelerated in each gap and shielded inside the tubes while the voltage reverses. The tubes get longer along the line because the particles go faster.
Cyclotron. A magnetic field bends the particles into semicircles inside two hollow "dees", and an alternating voltage across the gap accelerates them each half-turn.
Synchrotron. The radius is kept fixed by increasing the magnetic field as the particles gain momentum, so the magnets need fill only a narrow ring. The LHC is a synchrotron with superconducting magnets of about 8 T.

Figure 5.1: A cyclotron.

The cyclotron equations

Inside a dee the magnetic force provides the centripetal force: $qvB = \frac{mv^2}{r}$ , so $r = \frac{mv}{qB}$ . The time for a half-turn is $\frac{\pi r}{v} = \frac{\pi m}{qB}$ , which does not depend on the speed. So a supply of fixed frequency

f = \frac{qB}{2\pi m}

stays in step with the particles. They leave at the outer radius $R$ with speed $v = \frac{qBR}{m}$ and kinetic energy

E_k = \frac{1}{2}mv^2 = \frac{q^2B^2R^2}{2m}

These results assume the particles are much slower than light. At higher speeds their effective mass grows, they fall out of step, and a synchrotron is needed.

Worked example 1: a medical cyclotron

Hospitals use small cyclotrons to make positron emitters for PET scans. A cyclotron accelerates protons in a field of 1.5 T, and its dees have a radius of 0.50 m. Find (a) the frequency of the AC supply, (b) the final kinetic energy of the protons in MeV. ( $m_p = 1.67 \times 10^{-27}$ kg, $e = 1.60 \times 10^{-19}$ C.)

Solution

(a) $f = \frac{qB}{2\pi m} = \frac{(1.60 \times 10^{-19})(1.5)}{2\pi(1.67 \times 10^{-27})} = 2.3 \times 10^{7}\,\mathrm{Hz} = 23\,\mathrm{MHz}$ .

(b) $E_k = \frac{q^2B^2R^2}{2m} = \frac{(1.60 \times 10^{-19})^2(1.5)^2(0.50)^2}{2(1.67 \times 10^{-27})} = 4.3 \times 10^{-12}\,\mathrm{J}$ . Dividing by $1.60 \times 10^{-13}\,\mathrm{J\,MeV^{-1}}$ : $E_k = 27\,\mathrm{MeV}$ .

Check the assumption: $v = \frac{qBR}{m} = 7.2 \times 10^{7}\,\mathrm{m\,s^{-1}}$ , about 0.24c, so the non-relativistic result is only a few per cent off.

Detectors

Most particles made in a collision decay almost at once, so physicists detect their longer-lived products and work backwards, using the conservation laws of Lesson 4. A modern detector is built in layers around the collision point, like an onion:

Figure 5.2: How each kind of particle shows up in the layers of a detector.

The tracker, in a strong magnetic field, records the curved paths of charged particles. From $r = \frac{p}{qB}$ the curvature gives the momentum, and the direction of bending gives the sign of the charge. A useful form: $p\,(\text{in GeV}/c) \approx 0.3\,B\,(\text{T})\,r\,(\text{m})$ for a particle of charge $e$ .
Calorimeters absorb particles completely and measure their energy: the electromagnetic one stops electrons and photons, the hadron one stops protons, neutrons and pions.
Muon chambers on the outside detect muons, the only charged particles that pass through everything else.
Neutrinos escape unseen; they are spotted as "missing" momentum when the visible momenta across the beam do not add up to zero.

Earlier detectors included the cloud chamber of Unit 5 and the bubble chamber, in which charged particles left trails of bubbles in a superheated liquid, as in Figure 2.1.

Worked example 2: momentum from a curved track

In the CMS detector, whose magnet produces 3.8 T, a muon's track has a radius of curvature of 2.0 m. Find its momentum in GeV/c.

Solution

$p = qBr = (1.60 \times 10^{-19})(3.8)(2.0) = 1.22 \times 10^{-18}\,\mathrm{kg\,m\,s^{-1}}$ .

Then $pc = (1.22 \times 10^{-18})(3.00 \times 10^{8}) = 3.65 \times 10^{-10}\,\mathrm{J}$ , and $\frac{3.65 \times 10^{-10}}{1.60 \times 10^{-10}\,\mathrm{J\,GeV^{-1}}} = 2.3\,\mathrm{GeV}$ , so $p = 2.3\,\mathrm{GeV}/c$ . (Check with the shortcut: $0.3 \times 3.8 \times 2.0 = 2.3$ .) A straighter track would mean a larger momentum.

The Standard Model

Everything in this unit fits into one theory, the Standard Model of particle physics:

Figure 5.3: The particles of the Standard Model, with masses from the Particle Data Group's 2026 Review.

Twelve matter particles (fermions): six quarks and six leptons in three generations. Ordinary matter needs only the first: up and down quarks (in protons and neutrons) and electrons.
Force carriers (gauge bosons): gluons (strong), the photon (electromagnetic), W and Z (weak).
The Higgs boson, the visible sign of the Higgs field that fills space. Particles that interact with this field acquire mass; the W and Z are heavy for this reason, while the photon, which does not interact with it, is massless.

The Standard Model has passed every precision test for fifty years, but it is not complete. It does not include gravity. It says nothing about the dark matter that astronomers find makes up most of the mass of galaxies. It had to be extended when neutrinos were found to have mass. And it does not explain why the universe contains matter but almost no antimatter (Lesson 2).

Uses of particle physics

Application	How particle physics is used
PET scanning	positron emitters made in cyclotrons; annihilation photons detected in coincidence (Lesson 2)
cancer treatment	beams of protons or carbon ions from accelerators deposit most of their energy at a chosen depth, sparing healthy tissue
synchrotron light sources	electrons circling in a synchrotron emit intense X-rays, used to find the structures of proteins and new materials
the World Wide Web	invented at CERN by Tim Berners-Lee in 1989 so that physicists could share data
muon imaging	cosmic-ray muons passing through large objects reveal hidden spaces; in 2017 this found a large void inside the Great Pyramid of Giza
industry and security	electron accelerators sterilise medical supplies; particle detectors scan cargo

Warning

Common mistakes. (1) Thinking the cyclotron frequency depends on the particle's speed: $f = \frac{qB}{2\pi m}$ does not (until relativity matters). (2) Forgetting that a cyclotron's AC frequency must match the HALF-turns: the field reverses twice per revolution. (3) Saying the Higgs boson gives all mass: most of the mass of a proton comes from the energy of confined quarks and gluons (Lesson 1). (4) Believing neutrinos are detected in the calorimeters: they pass through and are inferred from missing momentum. (5) Quoting the Standard Model as complete: gravity and dark matter are not in it.

Tip

To identify a particle in a detector, ask three questions: Does it leave a track? (charged or not) Where does it stop? (EM calorimeter: electron or photon; hadron calorimeter: hadron; nowhere: muon or neutrino) Which way does it bend? (sign of charge).

The Standard Model of Particle Physics: A Triumph of Science (Quanta Magazine). Watch how quarks, leptons, force carriers and the Higgs field fit together, and which questions the model leaves open.

✦

Practice

Give two reasons why particle physics experiments need very high energies.
Find the de Broglie wavelength of a 6.8 TeV proton ( $hc = 1.24 \times 10^{-6}\,\mathrm{eV\,m}$ ), and compare it with the size of a proton.
Show that the time a particle spends in one dee of a cyclotron is independent of its speed.
A cyclotron with B = 0.80 T accelerates deuterons ( $q = e$ , $m = 3.34 \times 10^{-27}$ kg). Find the frequency of its AC supply.
Why does a synchrotron, not a cyclotron, accelerate protons to TeV energies?
In a detector, a particle leaves a curved track in the tracker and is absorbed in the electromagnetic calorimeter. Its track bends the same way as a known positive particle. Identify it.
A track in a 2.0 T field has a radius of 0.50 m. Find the particle's momentum in GeV/c (charge $e$ ).
Gatete says: "The Standard Model explains every force in nature." Evaluate his statement.

Show answersHide answers

A shorter de Broglie wavelength resolves smaller structures ( $\lambda \approx \frac{hc}{E}$ ); and $E = m c^{2}$ means heavy particles (W, Z, top, Higgs) can only be made with enough energy.
$\lambda \approx \frac{1.24 \times 10^{-6}}{6.8 \times 10^{12}} = 1.8 \times 10^{-19}\,\mathrm{m}$ , about $10^{4}$ times smaller than a proton (about $1.7 \times 10^{-15}$ m).
Half-turn time $= \frac{\pi r}{v}$ , and $r = \frac{mv}{qB}$ , so time $= \frac{\pi m}{qB}$ : $v$ cancels.
$f = \frac{(1.60 \times 10^{-19})(0.80)}{2\pi(3.34 \times 10^{-27})} = 6.1 \times 10^{6}\,\mathrm{Hz}$ (6.1 MHz).
At high energies the protons are relativistic: their period grows and they fall out of step with a fixed-frequency supply, and the radius would become enormous. A synchrotron raises B (and adjusts the frequency) to keep them on a fixed ring.
A positron: charged (track), positive (bending), and stopped in the EM calorimeter like an electron.
$p \approx 0.3 \times 2.0 \times 0.50 = 0.30\,\mathrm{GeV}/c$ .
Wrong. It describes the strong, weak and electromagnetic interactions very successfully, but it does not include gravity, and it does not explain dark matter or the matter–antimatter imbalance.

Key points

High energy gives short wavelengths ( $\lambda \approx \frac{hc}{E}$ ) and enough $m c^{2}$ to make heavy particles; colliding beams make all the energy available.
Cyclotron: $f = \frac{qB}{2\pi m}$ , $E_k = \frac{q^2B^2R^2}{2m}$ ; synchrotrons raise B to keep the radius fixed.
Detectors: tracker (charge and momentum, $p = q B r$ ), calorimeters (energy), muon chambers; neutrinos inferred from missing momentum.
Standard Model: 6 quarks, 6 leptons, gauge bosons and the Higgs boson (125.13 GeV, PDG 2026); gravity and dark matter lie outside it.

1A 27-kilometre microscope

2Why high energies?

3Accelerators