§4.10The Third Family

Ordinary matter needs four particles: the electron, its neutrino, and the u and d quarks inside nucleons. That is the first family family (generation) a quark doublet plus a charged lepton and its neutrino: (u,d,e,νₑ), (c,s,μ,ν_μ), (t,b,τ,ν_τ). Three exist, identical in structure and differing only in mass. defined in §4.10 — open in glossary . The muon, its neutrino, and the s and c quarks repeat the pattern exactly — same charges, same spins, same structure, different masses. That is the second.

By the mid-1970s a third charged lepton, the τ, had already turned up. Nature does not appear to build half a family, so two more quarks were expected: beauty (or bottom) b with charge −1/3 and $B = -1$, and top t with charge +2/3 and $T = +1$. Like every other flavour, both are conserved by the strong and electromagnetic interactions and violated by the weak one.

Finding them took 1977 and 1995 respectively, and the eighteen years between is the story of this section.

Beauty: Lederman at Fermilab, 1977

import numpy as np

for L in (6, 12, 18):
    print(f"  {L:2d} interaction lengths -> hadrons surviving: exp(-{L}) = {np.exp(-L):.2e}")
print("  muons feel no strong force, so they walk through essentially untouched")

n_ext, n_tot, n_pairs = 1e11, 1.6e16, 9000
print(f"\n  {n_ext:.0e} protons per extraction, {n_tot:.1e} on target in total")
print(f"  -> {n_pairs} dimuon events above 5 GeV, i.e. one per {n_tot/n_pairs:.1e} protons")

   6 interaction lengths -> hadrons surviving: exp(-6) = 2.48e-03
12 interaction lengths -> hadrons surviving: exp(-12) = 6.14e-06
18 interaction lengths -> hadrons surviving: exp(-18) = 1.52e-08
muons feel no strong force, so they walk through essentially untouched

1e+11 protons per extraction, 1.6e+16 on target in total
-> 9000 dimuon events above 5 GeV, i.e. one per 1.8e+12 protons

The spectrum, before and after subtracting the continuum

m = {"1S": 9460, "2S": 10023, "3S": 10355}
for a, b in [("1S", "2S"), ("2S", "3S")]:
    d = m[b] - m[a]
    sig = 0.02 * m[a]
    print(f"  m({b}) - m({a}) = {d:4d} MeV, resolution sigma = {sig:.0f} MeV -> {d/sig:.1f} sigma apart")
print("  three states, separations of 1.7 and 3.0 sigma:"
      " 'three barely resolved resonances'")

  m(2S) - m(1S) =  563 MeV, resolution sigma = 189 MeV -> 3.0 sigma apart
m(3S) - m(2S) =  332 MeV, resolution sigma = 200 MeV -> 1.7 sigma apart
three states, separations of 1.7 and 3.0 sigma: 'three barely resolved resonances'

Precision: CLEO at CESR

At an $e^+e^-$ collider the mass is not measured at all — it is set. Tune the beams to $\sqrt s$, count hadronic events, and step. The resonance position is then known as well as the machine energy is known, which is a completely different systematic from a spectrometer resolution.

CLEO at Cornell did exactly that around each of the three states.

All three are $J^{PC} = 1^{--}$ with $I = 0$: they are the $^3S_1$ levels of the $b\bar b$ “atom”, bottomonium charmonium / bottomonium the cc̄ and bb̄ bound systems, treated as atoms with the level structure of hydrogen. J/ψ is 1³S₁, ψ′ 2³S₁, ψ(3770) 3³S₁; η_c is 1¹S₀. defined in §4.10 — open in glossary , in the same spectroscopic scheme charmonium used. And they are narrow for the same reason charmonium was.

The threshold argument, a third time

Here is the payoff of building the §4.9 figure as a reusable component rather than a picture: the argument is identical, only the numbers change.

mB, mBs = 5279.3, 5366.9          # Table 4.4
th_BB, th_Bs = 2*mB, 2*mBs
print(f"  2 m(B+) = {th_BB:.1f} MeV      2 m(Bs0) = {th_Bs:.1f} MeV")
for name, mm in [("1S", 9460), ("2S", 10023), ("3S", 10355), ("4S", 10579)]:
    d = mm - th_BB
    state = "OPEN -> falls apart" if d > 0 else "closed -> must annihilate bbbar"
    print(f"  Y({name}) at {mm:5d}: m - 2m(B) = {d:+8.1f} MeV   {state}")
print("  the 4S clears the bar by 20 MeV and is 20 MeV wide")

  2 m(B+) = 10558.6 MeV      2 m(Bs0) = 10733.8 MeV
Y(1S) at  9460: m - 2m(B) =  -1098.6 MeV   closed -> must annihilate bbbar
Y(2S) at 10023: m - 2m(B) =   -535.6 MeV   closed -> must annihilate bbbar
Y(3S) at 10355: m - 2m(B) =   -203.6 MeV   closed -> must annihilate bbbar
Y(4S) at 10579: m - 2m(B) =    +20.4 MeV   OPEN -> falls apart
the 4S clears the bar by 20 MeV and is 20 MeV wide

The B mesons, and a lifetime puzzle

Now the puzzle. The B mesons weigh nearly three times what the D mesons weigh, and weak decay rates grow steeply with mass. So the B mesons should be shorter-lived than the D mesons. They are longer-lived: 1.5 ps against 0.4 ps.

Top: too heavy to be a hadron

Eighteen more years. The top is heavy — over 170 GeV — and it is produced predominantly in pairs by the strong interaction, so the collision needs to supply twice that. Only the Tevatron, colliding protons and antiprotons at $\sqrt s = 2$ TeV, could do it, and even there top production happens once in 10¹⁰ collisions.

The top is the exception to all of it, and the reason is its width.

hbar = 6.582119569e-25          # GeV s
G_t = 1.3                       # GeV, the Standard Model width of the top
tau_t = hbar / G_t
d_had = 1e-15                   # m, roughly one hadron diameter
t_had = d_had / 2.99792458e8

print(f"  top width 1.3 GeV -> tau = hbar/Gamma = {tau_t:.2e} s")
print(f"  hadronization time  d/c with d = 1 fm = {t_had:.2e} s")
print(f"  ratio = {t_had/tau_t:.1f} : the top decays about {t_had/tau_t:.0f}"
      f" times faster than it can dress")
print("  so there are no top hadrons -- the only quark ever seen as a free particle")

  top width 1.3 GeV -> tau = hbar/Gamma = 5.06e-25 s
hadronization time  d/c with d = 1 fm = 3.34e-24 s
ratio = 6.6 : the top decays about 7 times faster than it can dress
so there are no top hadrons -- the only quark ever seen as a free particle

The signature

from math import lgamma, log, exp, erfc, sqrt

def pois_ge(k, mu):                       # P(N >= k) for a Poisson of mean mu
    return 1.0 - sum(exp(-mu + i*log(mu) - lgamma(i+1)) for i in range(k))

def sigma_of(p):                          # invert the one-sided Gaussian tail
    lo, hi = 0.0, 40.0
    for _ in range(200):
        mid = (lo + hi) / 2
        lo, hi = (mid, hi) if 0.5*erfc(mid/sqrt(2)) > p else (lo, mid)
    return lo

n = 27
print("  background expectation  b = 6.7 +- 2.1")
print(f"  observed                n = {n}")
p = pois_ge(n, 6.7)
print(f"  P(N >= {n} | b = 6.7) = {p:.2e}  ->  {sigma_of(p):.1f} sigma (Gaussian equivalent)")
print(f"  naive n/sqrt(b) would claim {n/6.7**0.5:.1f} sigma -- Poisson is the honest number")

p2 = pois_ge(n, 8.8)
print("\n  but b itself is uncertain. Shift it up by one standard deviation:")
print(f"  P(N >= {n} | b = 8.8) = {p2:.2e}  ->  {sigma_of(p2):.1f} sigma")
print("  the +-2.1 costs a full sigma, and that is BEFORE the look-elsewhere effect")

  background expectation  b = 6.7 +- 2.1
observed                n = 27
P(N >= 27 | b = 6.7) = 2.98e-09  ->  5.8 sigma (Gaussian equivalent)
naive n/sqrt(b) would claim 10.4 sigma -- Poisson is the honest number

but b itself is uncertain. Shift it up by one standard deviation:
P(N >= 27 | b = 8.8) = 6.36e-07  ->  4.8 sigma
the +-2.1 costs a full sigma, and that is BEFORE the look-elsewhere effect

Because the top decays before it hadronizes, its mass can be reconstructed from its decay products the way you would reconstruct any free particle — no binding energy, no confinement corrections, no theoretical scheme required to say what “mass” means:

That is a 0.17 % measurement of the mass of a quark, and it is possible only for this one. §9.11 is how it is done to that precision, and why the Standard Model cares so much about the answer.

Where this goes next

• §4.11 closes the chapter with the full cast: six flavours, three families, and the generalisation of Gell-Mann–Nishijima to all of them.
• §6.2 is hadronization and jets done properly, instead of the sketch this section needed.
• §7.9 turns “crossing a family is suppressed” into the CKM matrix, and explains the B lifetime quantitatively.
• §9.11 measures the top mass properly, and shows why a precise m_t plus a precise m_W is a test of the whole Standard Model.
• Chapter 8 is what the ϒ(4S) is for: a machine parked on that resonance makes BB̄ pairs and nothing else, which is how CP violation in B mesons was measured.