University Physics IV · Particle Physics · 14.7
Colour, Confinement & Asymptotic Freedom
Colour is a charge nobody can see directly: it is counted from a ratio of cross-sections and forced by a statistics violation. This lesson follows the one number that carries the whole story — how αₛ makes TeV collisions calculable, why it leaves the proton uncomputable by hand, and why quarks arrive as jets.
Build the model
Connect the measurement to the mechanism.
Colour was not invented for elegance. The Δ⁺⁺ is three up quarks with aligned spins and no orbital angular momentum — symmetric in space, spin and flavour at once, which Fermi statistics forbid unless a further label distinguishes them, and that label needs three values for a totally antisymmetric combination to exist. Promote the resulting SU(3) to a local symmetry and eight gluons are forced, and unlike photons they carry the charge they mediate.
That one fact inverts the vacuum. Quark loops screen a colour charge as electron loops screen an electric one, but gluon loops smear it outward, and the competition reads 11Nc against 2nf: with three colours and six flavours antiscreening wins by 33 to 12, so αₛ falls as the probing energy rises. High-energy QCD is therefore calculable — that is asymptotic freedom, and it is the entire licence behind jet-rate and cross-section predictions good to a few per cent.
The price is charged at the other end. The same running drives αₛ past 1 near the hadron scale, where a series in αₛ is worthless, so the theory that gets a TeV collision right cannot prove on paper the most obvious fact about the strong force: that no quark has ever been seen alone. Confinement is an experimental result and a lattice computation, not a theorem.
- Simple definition
- Colour is the three-valued charge of the strong interaction, and its coupling αₛ shrinks as the probing energy rises — asymptotic freedom — while growing without bound as quarks separate, which is why quarks are only ever found inside colour-neutral hadrons.
- Example
- αₛ = 0.118 at Q = 91.2 GeV, where the probe resolves ħc/Q = 0.0022 fm; at Q = 1 GeV, resolving 0.20 fm, the same one-loop formula returns 0.34, and below that it diverges.
2 below charm, 10/3 above it, 11/3 above bottom — and one third of each without colour. The measurement counts Nc.
Sum over quarks with 2mq c² < √s; Qq in units of e. R is dimensionless.
CF sets the −4/3 attraction inside a colour-singlet pair; CA ≠ 0 is the self-coupling QED has no analogue of.
Dimensionless group factors: CF weights a quark–gluon vertex, CA a gluon–gluon one.
The bracket is 23 at the Z. Positive means asymptotic freedom, and it stays positive up to nf = 16.
11Nc from gluon loops (antiscreening), 2nf from quark loops (screening); nf counts flavours with 2mq < Q.
Carries the world average to any scale without quoting Λ at all: 0.203 at 5 GeV, 0.088 at 1 TeV.
αₛ(MZ) = 0.1180 ± 0.0009 at MZ = 91.19 GeV; both sides dimensionless.
One-gluon exchange dominates inside 0.1 fm; beyond that the linear term takes over and never weakens.
r in fm, ħc = 0.1973 GeV⋅fm, string tension k ≈ 0.90 GeV/fm = 1.4 × 10⁵ N.
Pull a quark that far and the tube buys a new pair instead of stretching — which is why a jet appears, not a quark.
mq ≈ 0.33 GeV/c² constituent mass, k ≈ 0.90 GeV/fm, r* in fm.
Colour: a charge invented to rescue Fermi statistics
The Δ⁺⁺ is uuu with spin 3/2 and no orbital angular momentum. Its space part is symmetric because the state is a ground state, its spin part is symmetric because all three spins are aligned, and its flavour part is symmetric because all three quarks are up. Three symmetric factors give a symmetric total, which the exclusion principle forbids for identical spin-½ particles. The only escape is a further label. It must carry at least three values, so that the totally antisymmetric combination εᵢⱼₖ exists and can absorb the required sign; and exactly three, because every observed hadron is a colour singlet of SU(3) — three-quark baryons, quark–antiquark mesons, nothing carrying net colour. An independent check comes from π⁰ → γγ, whose amplitude runs through a quark loop and therefore scales as Nc, making the rate scale as Nc². The measured width of 7.8 eV is nine times the 0.87 eV that a single-coloured quark would give.
Counting the colours: the R ratio
In e⁺e⁻ annihilation through a virtual photon, making μ⁺μ⁻ and making a quark pair use the same QED vertex; only the fractional charge and the number of available copies differ. Once a quark pair exists it hadronises with probability one, so the hadronic cross-section is settled at the quark level and the ratio is clean: R = σ(hadrons)/σ(μ⁺μ⁻) = Nc Σq Qq², summed over the flavours light enough to pair-produce at that √s. Below open charm the sum is 4/9 + 1/9 + 1/9 = 2/3, giving R = 2. Above the charm threshold add 4/9 for R = 10/3; above bottom add 1/9 for R = 11/3. The data step up at exactly those energies and sit near 2.2, 3.6 and 3.9 once the small (1 + αₛ/π) correction is included — three times what a colourless quark model predicts. No fit and no free parameter: a ratio of two cross-sections counts colours.
Screening, antiscreening, and the sign that decides
In QED a bare charge polarises the vacuum: virtual e⁺e⁻ pairs align and partly cancel it, so a closer probe sees more charge, and α runs from 1/137 at zero momentum to about 1/128 at the Z. QCD has the same fermion loops — nf of them — screening in the same direction. It also has something QED does not. Gluons carry colour, so gluon loops spread the colour charge outward instead of shielding it, and a closer probe finds less of it. The one-loop beta function weighs the two effects against each other, Q dαₛ/dQ = −[(11Nc − 2nf)/6π] αₛ², where 11Nc comes from gluon loops and 2nf from quark loops. With Nc = 3 and nf = 6 the bracket is 33 − 12 = 21, comfortably positive, so antiscreening wins and αₛ falls as Q rises. Seventeen flavours would be needed to flip the sign — the physical world is nowhere near that.
Asymptotic freedom is a licence to calculate
Integrate the beta function and anchor it on a measurement rather than on Λ: 1/αₛ(Q) = 1/αₛ(MZ) + [(11Nc − 2nf)/6π] ln(Q/MZ). Starting from αₛ(MZ) = 0.118 with nf = 5, this gives 0.203 at 5 GeV and 0.088 at 1 TeV — a factor of 2.1 across two decades of energy, which is how slowly a logarithm moves. What it buys is a perturbation series that works: at 1 TeV each further order costs roughly a factor of 0.088, so a few terms give a per-cent answer, and jet rates, Drell-Yan spectra and Higgs cross-sections are all obtained this way. Run the same formula backwards and it prices its own expiry. At 1 GeV it returns 0.34, at 0.5 GeV 0.47, it passes 1 near 0.2 GeV, and it diverges at Λ. Long before the divergence the series has stopped converging, because each new term is as large as the last.
What comes out is a jet, not a quark
Knock a quark out of a hadron and the colour field between it and the remnant does not fan out like a Coulomb field. The gluons' mutual attraction collapses the flux into a tube of roughly fixed cross-section, so the stored energy grows linearly: V(r) → k r with k ≈ 0.90 GeV per fm, a constant 1.4 × 10⁵ N — the weight of about fifteen tonnes — that never falls off with distance. After roughly 0.7 fm the tube holds enough energy to create a light quark–antiquark pair, so it breaks instead of stretching, and the process repeats down the chain. What emerges is a collimated spray of colour-singlet hadrons following the original parton's direction, with only a few hundred MeV/c of transverse momentum about it; jets therefore narrow as 1/E. A hard radiated gluon makes a third jet at a rate proportional to αₛ, and the three-jet events seen at PETRA in 1979 were the first direct sight of the gluon.
Confinement is measured and simulated, not proved
Nothing above derives confinement. The running formula says only that the perturbative expansion fails below about 1 GeV, and the failure of a method is not a theorem about the theory. What stands in its place is evidence of two kinds. Experimentally, no free quark has ever been isolated: Millikan-style searches for fractional charge bound their abundance below roughly one per 10²⁰ nucleons. Computationally, lattice QCD places the fields on a Euclidean grid of spacing a ≈ 0.05–0.1 fm and samples the path integral numerically; Wilson loops return a static quark–antiquark potential with a linear tail of tension k ≈ 0.9 GeV/fm, and the hadron spectrum comes out right to a few per cent. Those results carry their own extrapolations — a → 0, infinite volume, physical quark masses — and the analytic statement behind them, that pure SU(3) Yang–Mills has a mass gap, remains one of the Clay Millennium problems.
Change one variable at a time
Make the relationship visible.
Set Nc = 3, nf = 3 and the curve dives from αₛ = 0.50 at 1 GeV to 0.084 at 1 TeV. Now pull Nc to 2 and nf to 18: 11Nc − 2nf = −14, the sign flips, and the curve tilts the other way — rising with energy exactly as QED's α does.
PROBE SCALE Q91.2 GeV
COUPLING αₛ(Q)0.118
PROBE SIZE ħc/Q0.0022 fm
11Nc − 2nf23
Live interpretationPROBE SCALE Q: 91.2 GeV. COUPLING αₛ(Q): 0.118. PROBE SIZE ħc/Q: 0.0022 fm. 11Nc − 2nf: 23
Catch the common trap
Explain before calculating.
Electron–positron annihilation at √s = 6 GeV — above the charm threshold, below the bottom threshold near 9.5 GeV. Ignoring the small QCD correction, what does the parton model predict for R = σ(e⁺e⁻ → hadrons)/σ(e⁺e⁻ → μ⁺μ⁻)?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAt √s = 2.5 GeV, e⁺e⁻ annihilation is above the strange threshold and below open charm, which needs 3.73 GeV to make a D pair. Predict R = σ(hadrons)/σ(μ⁺μ⁻) with and without colour, add the first QCD correction using αₛ(2.5 GeV) ≈ 0.30, and compare with the measured R ≈ 2.2.
- Only pairs light enough to be produced contribute. At 2.5 GeV that is u, d and s; charm is absent because open charm needs 3.73 GeV.
- Sum the squared charges in units of e: (2/3)² + (−1/3)² + (−1/3)² = 4/9 + 1/9 + 1/9 = 2/3.
- Without colour each flavour appears once, so R = 2/3 = 0.67. With Nc = 3 there are three identically charged copies of every quark, so R = 3 × 2/3 = 2.
- Apply the first QCD correction: R → R(1 + αₛ/π) = 2 × (1 + 0.30/3.1416) = 2 × 1.0955 = 2.19.
- Compare with data: measured R ≈ 2.2. The colourless prediction is low by a factor of three, far outside any experimental uncertainty, while the αₛ/π term is a 10% adjustment on top of a correct answer.
AnswerR = 2 at the parton level and 2.19 with the αₛ/π correction, against a measured 2.2. A single-coloured quark model gives 0.67 and is excluded by a factor of three.
MediumTake αₛ(MZ) = 0.118 at MZ = 91.2 GeV and run it at one loop with Nc = 3 and nf = 5. Find αₛ at Q = 5 GeV and at Q = 500 GeV, give the distance ħc/Q that each probe resolves, and say what the ratio of the two couplings shows.
- Beta coefficient: (11Nc − 2nf)/6π = (33 − 10)/(6π) = 23/18.850 = 1.2202.
- Anchored form: αₛ(Q) = 0.118 / [1 + 0.118 × 1.2202 × ln(Q/91.2)] = 0.118 / [1 + 0.1440 ln(Q/91.2)].
- Q = 5 GeV: ln(5/91.2) = ln 0.05482 = −2.9036, so the bracket is 1 − 0.1440 × 2.9036 = 0.5819 and αₛ = 0.118/0.5819 = 0.203.
- Q = 500 GeV: ln(500/91.2) = ln 5.4825 = +1.7016, the bracket is 1 + 0.1440 × 1.7016 = 1.2450, and αₛ = 0.118/1.2450 = 0.0948.
- Probe distances: ħc/Q = 0.1973 GeV⋅fm ÷ Q gives 0.0395 fm at 5 GeV and 3.9 × 10⁻⁴ fm at 500 GeV.
- The resolved distance falls by a factor of 100 while the coupling falls only by 0.203/0.0948 = 2.1 — a logarithm, not a power. One loop is about 3% low at 5 GeV, where the measured value is near 0.21.
Answerαₛ(5 GeV) = 0.203 and αₛ(500 GeV) = 0.095. Cutting the probed distance a hundredfold weakens the coupling only by a factor of 2.1.
HardA 20 GeV quark is knocked out of a hadron. Model its colour field as a flux tube of tension k = 0.90 GeV/fm, take a light quark's constituent mass as 0.33 GeV/c², and take ⟨pT⟩ ≈ 0.35 GeV/c about the jet axis. Find (a) the separation at which the tube can pay for a new quark–antiquark pair, (b) the time that takes at speeds near c, (c) the force the tube exerts, and (d) the angle at which a hadron carrying a tenth of the jet's energy emerges.
- (a) The tube stores k r, and it can create a pair when k r* = 2 mq c² = 0.66 GeV. So r* = 0.66/0.90 = 0.73 fm — about one hadron radius, which is the whole point.
- (b) The separation grows at nearly c, so t = r*/c = 0.73 × 10⁻¹⁵ m ÷ 3.00 × 10⁸ m s⁻¹ = 2.4 × 10⁻²⁴ s. Hadronisation is over long before anything reaches a detector element.
- (c) Convert the tension: 0.90 GeV/fm = 0.90 × 1.602 × 10⁻¹⁰ J ÷ 10⁻¹⁵ m = 1.4 × 10⁵ N, the weight of about 15 tonnes — and unlike a Coulomb force it does not fall off with r.
- (d) A hadron with z = 0.10 of 20 GeV carries p∥ ≈ 2.0 GeV/c against pT ≈ 0.35 GeV/c, so θ = arctan(0.35/2.0) = arctan 0.175 = 9.9°, and the angle shrinks as the jet energy rises.
- Every line says the same thing: the quark's energy is spent locally, on making hadrons, inside a region the size of a proton. The detector records a spray about 10° wide, and the '20 GeV quark' is an inference from that spray's total momentum, never a track.
Answerr* = 0.73 fm, reached in 2.4 × 10⁻²⁴ s against a constant 1.4 × 10⁵ N; a hadron carrying a tenth of the energy leaves at about 10°.