Quarks, leptons and generations

Twelve fermions, and the useful way to hold them is by what they are charged under rather than by how heavy they are. Six quarks — u, d, c, s, t, b — each carry colour and sit in a triplet of SU(3); the up-type three have charge +2/3, the down-type three −1/3, and every one carries baryon number 1/3. Six leptons — e, μ, τ and one neutrino each — are colour singlets, so the strong force never touches them: charge −1 for the charged three, 0 for the neutrinos, and a lepton-family number Lₑ, L_μ or L_τ apiece. Antiparticles double the list with every sign flipped. Counting two-component Weyl fields rather than particles, one generation is fifteen: six for the coloured left-handed quark doublet, three each for u_R and d_R, two for the lepton doublet, one for e_R — and no ν_R. Three generations makes forty-five. That total is the ledger any proposed new fermion has to be added to, and the colour label in it is inferred from spectroscopy and jet counting, never read directly off a detector.

Apply the projectors P_L = ½(1 − γ⁵) and P_R = ½(1 + γ⁵) to a Dirac field and you get two independent pieces, and the Standard Model treats them as different objects. ψ_L goes into a weak-isospin doublet — (u, d)_L and (ν, e)_L — with T₃ = +½ on the upper member and −½ on the lower; ψ_R is an SU(2) singlet with T₃ = 0. The W couples to the doublet index and therefore to left-handed chirality alone, which is why the charged current violates parity maximally and why a right-handed electron cannot emit a W. Do not read chirality as a spin direction: chirality coincides with helicity only in the massless limit, and the mismatch is of order m/E — about 5×10⁻⁴ for a 1 GeV electron, but of order 1 for one at rest. The split carries a price. A Dirac mass term m(ψ̄_L ψ_R + ψ̄_R ψ_L) joins a doublet to a singlet, so it is not gauge invariant, and no fermion in this model is allowed a mass written in by hand.

Weak hypercharge is fixed, not chosen: Y = 2(Q − T₃). For the left-handed quark doublet the up member gives 2(2/3 − 1/2) = 1/3 and the down member 2(−1/3 + 1/2) = 1/3 — the same value, as it must be, since Y labels the doublet rather than its members. The right-handed quarks are singlets with T₃ = 0, so Y = 2Q: 4/3 for u_R and −2/3 for d_R. The lepton doublet gets Y = −1 and e_R gets −2. Nothing so far explains why the charges take these particular fractions, and this is where the arithmetic earns its place: quantum consistency demands that certain sums over a generation vanish. Add the electric charges of one generation with colour counted — 3(+2/3) + 3(−1/3) + 0 + (−1) — and the total is exactly zero, which happens only because quarks come in three colours and carry thirds. That is why the proton's charge cancels the electron's; experiment holds the residue below 10⁻²¹ of e.

The second and third generations duplicate the first slot for slot: (c, s, μ, ν_μ) and (t, b, τ, ν_τ) carry exactly the charges (u, d, e, νₑ) carry and differ only in mass. Because the gauge charges match, every gauge interaction rate matches once phase space is accounted for — that is lepton universality, and Γ(Z→e⁺e⁻), Γ(Z→μ⁺μ⁻) and Γ(Z→τ⁺τ⁻) agree at 84.0 MeV to a few parts in a thousand. What does not carry across is flavour. The three lepton-family numbers are separately conserved in every charged-lepton process seen so far, so μ⁻ → e⁻γ is absent below a branching ratio of 3.1×10⁻¹³, and the muon must instead decay as μ⁻ → e⁻ + ν̄ₑ + ν_μ, exporting both family labels in neutrinos. Quarks behave differently: the CKM matrix mixes their generations, so s → u proceeds with amplitude |Vᵤₛ| = 0.2243 and strange particles decay weakly in around 10⁻¹⁰ s.

Nothing in the model says how many copies there should be, so the number is measured. The Z couples to every fermion pair light enough to produce, neutrinos included, and neutrinos leave no signal — so they appear as width that no visible channel accounts for. LEP scanned e⁺e⁻ collisions across √s = 88–94 GeV and fitted the resonance: Γ_Z = 2495.5 ± 2.3 MeV, of which 1744.4 MeV is hadronic and 3 × 83.99 = 252.0 MeV is charged leptons. The remainder, 499 MeV, is invisible. Dividing by the predicted width per species, Γ_νν = 1.9912 Γ_ℓℓ = 167.2 MeV, gives N_ν = 2.984 ± 0.008. Four light species would have put Γ_Z at 2665 MeV, seventy-four standard deviations high, and pulled the peak hadronic cross-section down from 41.5 nb to 36.4 nb. The result is sharp but narrow: it counts only neutrinos below M_Z/2 = 45.6 GeV that couple to the Z with full strength.

Because no fermion may carry a bare mass, each one is generated by a Yukawa coupling to the Higgs field, m_f = y_f v/√2 with v = 246.22 GeV. The coupling is a free parameter, one per fermion, fitted to the measured mass and predicted by nothing. The numbers it has to take are absurd. The top quark needs yₜ = 0.991, essentially unity; the electron needs 2.94×10⁻⁶; a neutrino at the 0.05 eV scale that oscillation demands would need about 3×10⁻¹³. The charged fermions alone spread over 5.5 decades, since mₜ/mₑ = 3.4×10⁵, and the full range reaches 12.5 decades. Counting the neutrino sector in, the model carries roughly twenty-six measured parameters, and some twenty of them are the masses and mixings of this one table. So the inventory is predictive about how fermions interact and silent about why any of them weighs what it does. That gap is the flavour problem, and it is open.