Running couplings and asymptotic freedom

Write a cross section in perturbation theory and the loop integrals diverge; renormalisation subtracts the divergence at some scale μ and hands you a finite coupling αₛ(μ) in exchange. But μ was your choice, and no observable can know it: μ dR/dμ = 0 for every physical R. Since R depends on μ both explicitly, through logarithms ln(Q²/μ²), and implicitly, through αₛ(μ), the only way the total derivative can vanish is for αₛ to change with μ in a fixed way — that is the renormalisation-group equation, and the function on its right-hand side is the beta function. At one loop, μ² dαₛ/dμ² = −(β₀/4π)αₛ². Setting μ = Q, the scale of the process, removes the large logarithms and gives αₛ(Q): the coupling seen by a probe of momentum transfer Q. The convention matters: the derivative is with respect to ln Q², not ln Q, and the 4π is a normalisation that keeps β₀ a tidy combination of integers — confuse ln Q with ln Q² and every number below is off by a factor of two; drop the 4π and it is off by 4π.

β₀ = (11/3)C_A − (4/3)T_F n_f, and both group invariants come from the SU(3) trace identities of the previous topic. The adjoint Casimir (f^acd f^bcd = C_A δ^ab) gives C_A = 3; the fundamental normalisation Tr(T^a T^b) = T_F δ^ab gives T_F = ½. So β₀ = 11 − 2n_f/3: 9 for the three light quarks, 25/3 with charm, 23/3 with bottom, 7 above the top. In QED there is no photon self-coupling, so the C_A term is absent and only the screening term survives — virtual e⁺e⁻ pairs shield a charge, and a harder probe, reaching inside the cloud, sees more of it: α(M_Z) ≈ 1/128 against 1/137 at low Q. In QCD the gluons carry colour themselves and, in the usual picture, the virtual gluon cloud anti-screens by spreading the colour charge out, so a shorter-distance probe sees less of it. The +11 beats the −2/3 per flavour until n_f = 17 would flip the sign; with six flavours the strong coupling gets weaker as Q grows, and that is asymptotic freedom.

Separate variables: dαₛ/αₛ² = −(β₀/4π) d ln Q², and integrating from μ² to Q² gives 1/αₛ(Q²) = 1/αₛ(μ²) + (β₀/4π) ln(Q²/μ²). That is the working formula, and it says that 1/αₛ, not αₛ, is the thing to plot: a straight line in ln Q² whose slope is fixed by the group and whose intercept is the single measured input. Anchor at M_Z: αₛ(M_Z) = 0.118 gives 1/αₛ = 8.475, and with n_f = 5 the slope β₀/4π = 0.610 per unit of ln Q², or 2.81 per decade of Q. Run down to 10 GeV: ln(10²/91.19²) = −4.421, the shift is −2.697, so 1/αₛ = 5.778 and αₛ(10 GeV) = 0.173. Run up to 1 TeV, pretending n_f stays 5: the shift is +2.922, 1/αₛ = 11.40 and αₛ = 0.088. A factor of 100 in Q has moved αₛ by a factor of two, and it is that logarithmic slowness which makes jet cross sections at the LHC computable to a few per cent and hadron masses not computable this way at all.

Extend the straight line downward and 1/αₛ hits zero at a finite Q. Call it Λ and the running collapses to αₛ(Q²) = 4π/[β₀ ln(Q²/Λ²)], with Λ = μ exp[−2π/(β₀ αₛ(μ²))]. Classical QCD with massless quarks has no scale in it at all, only a pure number g; quantisation has turned that number into a length, ħc/Λ, and this is dimensional transmutation. Put in the anchor: 2π/(7.667 × 0.118) = 6.945, so Λ = 91.19 GeV × e^(−6.945) = 88 MeV. That is a one-loop, five-flavour number. Match through the b and c thresholds down to three flavours and it becomes about 140 MeV; do the running at four loops in the MS-bar scheme and the tabulated Λ⁽⁵⁾ is about 210 MeV, with Λ⁽³⁾ near 330 MeV. Λ is exponentially sensitive to αₛ(M_Z) — raising the anchor from 0.118 to 0.130 nearly doubles it — and it depends on scheme, loop order and n_f. So it is not a constant of nature; only its order of magnitude, a few hundred MeV or about 1 fm, is physical.

n_f is not a constant either: it counts the quarks light enough to appear in the loop, and it steps at m_c ≈ 1.27, m_b ≈ 4.18 and mₜ ≈ 173 GeV. At one loop the matching is trivial — αₛ is continuous across each threshold and only β₀ changes — so the run is piecewise: apply the closed form segment by segment, or hand the ODE to SciPy. In t = ln Q² the right-hand side is f(t, α) = −(11 − 2n_f/3) α²/4π with n_f looked up from Q = e^(t/2), and SciPy's initial-value solver, integrating f from t₀ to t₁ with α = 0.118 at t₀, reproduces the analytic answer to any tolerance you set. Running down from M_Z: αₛ(m_b) = 0.212 with n_f = 5, then αₛ(m_τ) = 0.279 with n_f = 4, against a τ-decay measurement of about 0.31. That 10% shortfall is not a bug; it is the two-loop term, whose size relative to one loop is β₁αₛ/(4πβ₀) = 14% at m_τ and 5% at M_Z. Running up through the top threshold: αₛ(mₜ) = 0.108, then with β₀ = 7, αₛ(1 TeV) = 0.0892 — 1.7% above the 0.0877 you get by forgetting the threshold. Both effects are small at high Q and grow as you descend: one loop is a percent-level tool above 100 GeV and a 10% tool at 2 GeV.

Below about 1 GeV, αₛ passes 0.5 and the series stops being useful: the two-loop term of the beta function is already more than a quarter of the one-loop term, and in the series for an observable such as the R ratio the αₛ⁴ correction is no smaller than the αₛ³ one. The one-loop pole at Λ is therefore not a prediction that the force becomes infinite; it is the equation announcing that it no longer describes QCD. The rest is known from elsewhere. Lattice QCD evaluates the path integral numerically on a grid and finds Wilson loops obeying an area law — a linear potential V(r) ≈ −(4/3)αₛ ħc/r + σr with string tension σ ≈ 0.9 GeV per fm: at 1 fm the linear term alone stores about 0.9 GeV, and near 1.2 fm it becomes cheaper to create a light q q̄ pair than to stretch further, so the string breaks into two mesons. Experiment sees the same thing as jets: a quark struck at 100 GeV never arrives, only a collimated spray of hadrons, and the jet counts, the R ratio and the π⁰ → γγ rate all demand N_c = 3. No analytic proof that Yang–Mills theory has a mass gap exists; it is a Clay Millennium problem. Confinement is established as most physics is, by computation and measurement, and an honest lesson says so.