What the W and Z Must Be, and How to Make One · Elementary Particle Physics

The numbers were on paper before the machine existed. This section predicts masses, widths and rates from one measured angle, and then builds the collider that will check them.

🎯 Why this matters

One measured angle goes in and a dozen numbers come out. That ratio between what is put in and what is predicted is what separates a theory from a description, and it is why a single disagreement would have been fatal.

§9.4 measured the angle. Everything in this section is a consequence — and the point is how little is put in. One partial width, the colour factor 3, the CKM elements of §7.11 and the $c_Z$ table you already have, and out come both masses, both total widths, every branching ratio and both production cross-sections.

Then §9.6 asks the practical question. If the $W$ weighs 80 GeV, how do you make one in 1983?

§9.5 One number, and the rest is arithmetic

The leptonic width of the $W$ requires quantum field theory and the book does not derive it. But it only has to be quoted once:

\htmlClass{t-g}{\Gamma_{e\nu}} = \frac{g^2}{4\pi}\frac{\htmlClass{t-m}{M_W}}{12} = \frac{\htmlClass{t-gf}{G_F}\,\htmlClass{t-m}{M_W^3}}{6\pi\sqrt2} \simeq 225\;\text{MeV}

(9.41)

Bettini p. 366. The only dynamical input in the whole section — everything below is this number times a ratio of couplings.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

Now every other channel is that number times something you already know. Quarks bring two factors: the colour multiplicity 3, and $|V_{qq'}|^2$ from the mixing matrix. And three channels are simply shut — $t\bar d$ , $t\bar s$ , $t\bar b$ — because $M_W < m_t$ , which nobody knew when the $W$ was found.

Bettini Eqs. (9.42)–(9.46). The last column is 225 MeV × the factor in the third — no new physics enters anywhere.

channel↕	how many↕	factor × $\Gamma_{e\nu}$ ↕	width (MeV)▼
$e\nu$ , $\mu\nu$ , $\tau\nu$	3	1 each	225 each, 675 total
$c\bar s$	1	$3\|V_{cs}\|^2 = 3(0.99)^2 = 2.94$	660
$u\bar d$	1	$3\|V_{ud}\|^2 = 3(0.974)^2 = 2.85$	640
$u\bar s$	1	$3\|V_{us}\|^2 = 3(0.224)^2 = 0.151$	34
$c\bar d$	1	$3\|V_{cd}\|^2 = 3(0.22)^2 = 0.145$	33
$t\bar d$ , $t\bar s$ , $t\bar b$	3	closed: $M_W < m_t$	0
$u\bar b$ , $c\bar b$	2	$3\|V_{ub}\|^2$ , $3\|V_{cb}\|^2$ — both $\sim10^{-3}$	≈ 0
total			2040 = 2.04 GeV

💡 What this really says — the branching ratios are Cabibbo angles in disguise

Look at the two large quark channels and the two small ones. $u\bar d$ and $c\bar s$ are each worth about 2.9 lepton channels; $u\bar s$ and $c\bar d$ are worth 0.15. The ratio is $\tan^2\theta_C$ , the same suppression that §7.9 extracted from kaon decays.

So the $W$ width is a weighted sum over the CKM matrix, and because each row of a unitary matrix has unit norm, the total is almost exactly $3 \times (1 + 3 + 3) = 3 + 6 = 9$ lepton units — three lepton families plus two open quark families times three colours. The measured $\text{BR}(W\to e\nu) = 10.7\,\%$ against a naive $1/9 = 11.1\,\%$ is that counting, and the small gap is QCD corrections.

The W width is a counting argument: 3 + 3 + 3 lepton units, and one row switched off

eν, μν, τν675 MeV = 3 unitsud̄ + us̄ + ub̄674 MeV = 3.00cd̄ + cs̄ + cb̄693 MeV = 3.08t-rowclosed0369width, in units of Γ(eν) = 225 MeVEach CKM row is worth 3 because unitarity makes its squared moduli sum to 1, times 3 colours.So the answer never needed the individual matrix elements — only that the rows are normalised.BR(W → eν) = 1/9.08 = 11.0 %, against a naive 1/9 = 11.1 % and a measured 10.7 %.The remaining 3 % is QCD: the quark channels get a (1 + α_s/π) enhancement the leptons do not.

Supplied — the table above has all the channels and the counting only becomes visible when they are grouped by CKM row. The three blocks are the same size for a reason that has nothing to do with their contents: each quark row contributes $3\sum_j |V_{ij}|^2 = 3$ by unitarity alone, so a W does not need to know the Cabibbo angle to decide how often it decays to an electron.

The individual elements only redistribute width within a block — which is why ud̄ is 640 MeV and us̄ is 34, in the ratio $\tan^2\theta_C$ , while their sum is fixed. The third row is worth zero not because its elements are small but because $M_W < m_t$ , a fact nobody knew when the W was discovered.

This is why problem 9.11 asks for the branching ratio and expects you to answer without computing a single width: count the channels, weight the quarks by three for colour, and let unitarity handle the mixing. The individual CKM elements never appear.

The $Z$ works the same way with a different master constant, and now the couplings are the $c_Z$ of §9.3:

Bettini Eqs. (9.48)–(9.54), with the master constant G_F M_Z³/(3√2 π) = 660 MeV. The third column is the sum of c_Z² over the chiralities — read straight off §9.3's table.

channel↕	how many↕	$\sum c_Z^2$ (at $s^2 = 0.232$ )↕	colour↕	width (MeV)↕
$\nu\bar\nu$	3	$(1/2)^2 = 0.250$	1	165 each — 495 invisible
$\ell^+\ell^-$	3	$(-\tfrac12+s^2)^2 + (s^2)^2 = 0.126$	1	83 each — 249
$u\bar u$ , $c\bar c$	2	$(\tfrac12-\tfrac23 s^2)^2 + (\tfrac23 s^2)^2 = 0.143$	3	280 each
$d\bar d$ , $s\bar s$ , $b\bar b$	3	$(-\tfrac12+\tfrac13 s^2)^2 + (\tfrac13 s^2)^2 = 0.185$	3	370 each
$t\bar t$	1	closed: $M_Z < 2m_t$	3	0
hadrons, total				1670
total				2420 = 2.42 GeV

both bosons, computed from scratch

import numpy as np
GF = 1.1663788e-5
MW, MZ, s2 = 80.0, 91.0, 0.232

Ge = GF*MW**3/(6*np.pi*np.sqrt(2))
print("the W, from ONE dynamical number")
print(f"  Gamma(W -> e nu) = G_F M_W^3 / (6 pi sqrt2) = {Ge*1e3:.1f} MeV     book: 225\n")
print(f"  leptons   3 x {Ge*1e3:.1f}                        = {3*Ge*1e3:4.0f} MeV")
tot = 3*Ge
for nm, V in (('u dbar', 0.974), ('c sbar', 0.990), ('u sbar', 0.224), ('c dbar', 0.220)):
    f = 3*V**2; w = f*Ge; tot += w
    print(f"  {nm}    3 x {V:.3f}^2 x Gamma_e = {f:5.3f} x  = {w*1e3:4.0f} MeV")
print("  " + "-"*54)
print(f"  Gamma_W                                    = {tot:.3f} GeV      book: 2.04")
print( "  measured (9.92):                             2.085 +- 0.042\n")
print(f"  BR(W -> e nu) = {Ge/tot*100:.1f}%   naive channel count 1/9 = 11.1%")
print( "  measured 10.7% -- the small gap is QCD corrections")

K = GF*MZ**3/(3*np.sqrt(2)*np.pi)
cz = lambda i3, q: i3 - s2*q
print(f"\nthe Z, from ONE constant and the c_Z table")
print(f"  G_F M_Z^3 / (3 sqrt2 pi) = {K*1e3:.0f} MeV                          book: 660\n")
rows = [('nu nubar', 3, [(0.5, 0)], 1), ('l+ l-', 3, [(-0.5, -1), (0, -1)], 1),
        ('u ubar', 2, [(0.5, 2/3), (0, 2/3)], 3), ('d dbar', 3, [(-0.5, -1/3), (0, -1/3)], 3)]
w = {}
for nm, n, cs, col in rows:
    b = sum(cz(i, q)**2 for i, q in cs); w[nm] = col*K*b
    print(f"  {nm:12s} x{n}   sum c_Z^2 = {b:.4f}  colour {col}  -> {w[nm]*1e3:6.1f} MeV")
print("  " + "-"*54)
print(f"  Gamma_inv = 3 x {w['nu nubar']*1e3:.1f}                      = {3*w['nu nubar']*1e3:4.0f} MeV       book:  495")
had = 2*w['u ubar'] + 3*w['d dbar']
print(f"  Gamma_had = 2 x {w['u ubar']*1e3:.1f} + 3 x {w['d dbar']*1e3:.1f}          = {had*1e3:4.0f} MeV       book: 1670")
GZ = 3*w['nu nubar'] + 3*w['l+ l-'] + had
print(f"  Gamma_Z                                    = {GZ:.3f} GeV      book: 2.42")
print( "  measured (9.80):                             2.4952 +- 0.0023")
print("\nwhy the Z is wider than the W despite weaker couplings per channel:")
print(f"  M_Z^3 / M_W^3 = {(MZ/MW)**3:.2f}   -- the mass cubed does most of the work")
print( "  and the Z has 21 open channels against the W's 9")

prints

the W, from ONE dynamical number
Gamma(W -> e nu) = G_F M_W^3 / (6 pi sqrt2) = 224.0 MeV     book: 225

leptons   3 x 224.0                        =  672 MeV
u dbar    3 x 0.974^2 x Gamma_e = 2.846 x  =  638 MeV
c sbar    3 x 0.990^2 x Gamma_e = 2.940 x  =  659 MeV
u sbar    3 x 0.224^2 x Gamma_e = 0.151 x  =   34 MeV
c dbar    3 x 0.220^2 x Gamma_e = 0.145 x  =   33 MeV
------------------------------------------------------
Gamma_W                                    = 2.035 GeV      book: 2.04
measured (9.92):                             2.085 +- 0.042

BR(W -> e nu) = 11.0%   naive channel count 1/9 = 11.1%
measured 10.7% -- the small gap is QCD corrections

the Z, from ONE constant and the c_Z table
G_F M_Z^3 / (3 sqrt2 pi) = 659 MeV                          book: 660

nu nubar     x3   sum c_Z^2 = 0.2500  colour 1  ->  164.9 MeV
l+ l-        x3   sum c_Z^2 = 0.1256  colour 1  ->   82.9 MeV
u ubar       x2   sum c_Z^2 = 0.1432  colour 3  ->  283.3 MeV
d dbar       x3   sum c_Z^2 = 0.1846  colour 3  ->  365.3 MeV
------------------------------------------------------
Gamma_inv = 3 x 164.9                      =  495 MeV       book:  495
Gamma_had = 2 x 283.3 + 3 x 365.3          = 1662 MeV       book: 1670
Gamma_Z                                    = 2.405 GeV      book: 2.42
measured (9.80):                             2.4952 +- 0.0023

why the Z is wider than the W despite weaker couplings per channel:
M_Z^3 / M_W^3 = 1.47   -- the mass cubed does most of the work
and the Z has 21 open channels against the W's 9

📏 Γ_inv is a counting experiment disguised as a width

The three neutrino channels contribute 495 MeV that no detector can see. The book defines the invisible width as the total width in undetectable channels, and then makes the observation this whole chapter has been building towards: if that width is due to $N_\nu$ neutrino species,

N_\nu = \frac{\Gamma_{\text{inv}}}{\Gamma_{\nu\bar\nu}}

and $\Gamma_{\nu\bar\nu}$ is predicted, exactly, with no free parameters — the neutrino’s $c_Z = 1/2$ does not even depend on $\sin^2\theta_W$ (§9.4). So measuring a width counts particles you cannot detect, including any that have not been thought of, provided only that they are lighter than $M_Z/2$ and couple to the $Z$ .

§9.9 does it and gets $N_\nu = 2.984 \pm 0.008$ . Note what that number forecloses: not just a fourth neutrino, but — if families are universal — a fourth family.

Making one: the resonance cross-section

Both bosons can be formed by quark–antiquark annihilation, so the machine to build is a quark–antiquark collider. The nearest available thing is a proton–antiproton collider, and near resonance the cross-section is §4.1’s Breit–Wigner:

\sigma(u\bar d \to e^+\nu_e) = \htmlClass{t-c}{\frac{1}{9}}\,\frac{3\pi}{\hat s} \;\frac{\htmlClass{t-in}{\Gamma_{ud}}\,\htmlClass{t-out}{\Gamma_{e\nu}}} {\left(\sqrt{\hat s}-M_W\right)^2 + \left(\Gamma_W/2\right)^2}

(9.57)

Bettini p. 368. Eq. (4.67) for a spin-1 resonance formed from two spin-½ particles, with one extra factor the colour makes necessary.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

At the peak this gives about 8.8 nb for the $W$ and about 0.8 and 1 nb for the $Z$ from $u\bar u$ and $d\bar d$ . The $Z$ numbers are an order of magnitude smaller for two compounding reasons: its partial widths are smaller and its mass is larger, and both enter squared or inverse-squared.

Erratum — Eq. (9.58) does not evaluate to the number it states

The equation is printed with its inputs visible:

\sigma_{\max} = \frac{4\pi}{3}\,\frac{1}{\mathbf{81^2}}\, \frac{0.640\times0.225}{2.04^2}\times 388\;[\mu\text{b}\cdot\text{GeV}^2] \approx 8.8\;\text{nb}

Evaluate it: $4\pi/3 = 4.1888$ , divided by $81^2 = 6561$ gives $6.385\times10^{-4}$ ; times $0.144/4.1616 = 0.03460$ gives $2.209\times10^{-5}$ ; times 388 gives $8.57\times10^{-3}$ μb = 8.6 nb.

Getting 8.8 requires $M_W = 80$ , not 81 — and 80 GeV is the value the chapter itself predicts in Eq. (9.29) and uses everywhere else. So the $81$ is the odd one out. A 2.5 % slip that changes nothing, but the page invites you to check the arithmetic and the arithmetic does not close.

Worth noting alongside it: the conversion constant here is 388 μb·GeV², while §9.9’s Example 9.4 uses 339 for the same quantity. The correct value is $\hbar^2c^2 = 389.4$ μb·GeV², so 388 is rounding and 339 is a transposition — see that section’s erratum. Both confirmed on the renders of PDF pp. 387 and 401.

Aside — Eq. (9.42)‘s 35 MeV is rounding, not an error

Recomputing (9.42) with the book’s own inputs, $3\times0.224^2\times225 = 33.9$ MeV, which rounds to 34 rather than the printed 35 — while (9.43), $3\times0.22^2\times225 = 32.7$ , rounds correctly to the printed 33.

The inconsistency is only apparent. Any $\Gamma_{e\nu}$ between about 228 and 232 MeV makes both round as printed, and the book only ever commits to ” $\simeq 225$ MeV”; the exact value depends on which $M_W$ you put in the cube. So this is rounding slack, not a mistake, and the total (9.46) = 2.04 GeV comes out right either way. Recorded because the discrepancy is visible to anyone who checks, and it is worth knowing that it is benign.

§9.6 A collider for quarks, and one event in ten billion

Rubbia, Cline and McIntyre proposed in 1976 to rebuild the CERN SPS as a proton–antiproton storage ring. The physics case is the resonance above; the engineering case is that antiprotons are hard. Van der Meer’s stochastic cooling is what made a dense enough antiproton beam possible, reaching $L = 10^{28}\;\text{cm}^{-2}\text{s}^{-1}$ .

🔢 Worked example — Example 9.3, and what a year buys

At the design luminosity, with $\sigma(p\bar p\to W\to e\nu) = 530$ pb and $\sigma(p\bar p\to Z\to ee) = 35$ pb, a 50 % detection efficiency and a running year of $10^7$ s:

N_W = 530\times10^{-36}\,\text{cm}^2 \times 10^{28}\,\text{cm}^{-2}\text{s}^{-1} \times 10^{7}\,\text{s} \times 0.5 = \mathbf{26}

N_Z = 35\times10^{-36} \times 10^{28} \times 10^{7} \times 0.5 = \mathbf{2}

Twenty-six $W$ s and two $Z$ s in a year. That is the entire experiment, and it explains the history exactly: the $W$ was announced several months before the $Z$ , not because it was easier to identify but because there were thirteen times as many.

It also explains why the “mnemonic rule” that a year is $\pi\times10^7$ seconds gets rounded down to $10^7$ — machine filling, maintenance and downtime eat the difference, and the book says so.

The hard part is not the rate; it is the background. The cross-sections above are eight to nine orders of magnitude below the total proton–antiproton cross-section of 60 mb, so the detector needs a discriminating power of at least $10^{10}$ . Hadronic $W$ and $Z$ decays are more frequent and completely hopeless — buried under $gg\to gg$ , $gq\to gq$ and the rest. Only the leptonic channels can be used, and finding them needs three ideas: cutting on transverse momentum , demanding isolation within a cone measured in pseudorapidity , and reading the neutrino off the missing transverse energy .

Bettini pp. 371–372. Each is a way of exploiting something the signal has and the background does not.

variable↕	definition↕	why it separates↕
transverse momentum $p_T$	the momentum component perpendicular to the beams	a lepton from an 80 GeV boson at rest carries $\sim M/2 = 40$ GeV of $p_T$ . A lepton from a charm or beauty decay carries at most a fraction of the parent mass — under 1 GeV — and the boost along the beam does not change the transverse component
isolation	no other track within $\Delta R = \sqrt{\Delta\eta^2+\Delta\varphi^2}$ , typically 0.3–0.7	the dangerous background is a high- $p_T$ lepton from a heavy-flavour decay inside a jet. That lepton has company; a lepton from a $W$ does not
missing transverse energy $E_T^{\text{miss}}$	the vector sum of all calorimeter cell energies, transverse components only	the neutrino from $W\to e\nu$ is invisible, so it shows up as an imbalance. Requires a hermetic detector — and only the transverse projection is usable, because the beam pipe is open

⚙️ Engineer’s bridge — why everything at a hadron collider is transverse

The variables above share a feature that looks like a limitation and is actually the design principle: all three are transverse, and the longitudinal direction is thrown away. It is worth understanding why, because it governs every plot in the rest of the chapter.

The colliding objects are not the protons; they are partons carrying unknown fractions $x_q$ and $x_{\bar q}$ of the beam momenta. So the parton–parton centre of mass is moving along the beam with a velocity you do not know, and cannot measure. Every longitudinal quantity is therefore contaminated by an unknown boost.

But a boost along $z$ leaves the transverse components untouched. So $p_T$ is the same in the parton frame and the lab, and a transverse momentum balance holds in both. Pseudorapidity is chosen for exactly the same reason: $\eta = -\ln\tan(\theta/2)$ has the property that differences in $\eta$ are boost-invariant, which is why $\Delta R$ is built from $\Delta\eta$ rather than $\Delta\theta$ .

The engineering analogue is close to exact. You are measuring a signal riding on an unknown DC offset, so you AC-couple: throw away the component the interference lives in, and work entirely in the one that is clean. You lose information — the longitudinal momentum of the $W$ genuinely cannot be reconstructed when a neutrino escapes — and you accept the loss because the alternative is unusable data. §9.7’s Jacobian peak is what you do once you have made that choice: extract the mass from a transverse distribution alone.

Where it breaks: projecting onto the transverse plane is lossy in a way that choosing a coordinate system is not, and the loss is not recoverable by better instrumentation. With one escaping neutrino you can still solve for the longitudinal component using a mass constraint; with two — as in $H \to WW \to \ell\nu\ell\nu$ — the event is underdetermined no matter what you build, which is why that channel yields a broad transverse-mass shape and never a peak. And missing transverse momentum is a balance, so it inherits every mismeasurement in the event: one badly calibrated jet manufactures a neutrino that was never there.

The detector that follows from those three requirements is UA1, and its structure is the one every collider detector has used since:

🧅 UA1 — the layered design every collider detector has used since

light up:

Pick a particle above to light up the layers it touches, and click any layer for what it measures. No single layer identifies anything — the pattern across all five does.

UA1 in cross-section (Bettini Figs. 9.12–9.13), and the origin of the layered design. The beams enter along the axis; a particle produced at the centre meets each layer in turn. Click a layer to see what it measures, or pick a particle to see its signature.

Reading the layers outward, and noting what each one is for in this measurement:

the central detector — a large cylindrical time projection chamber in a horizontal magnetic field, giving an electronic image of every charged track and its momentum from curvature;
the electromagnetic calorimeter — lead and scintillator, in which electrons and photons deposit all their energy. Comparing this energy with the momentum from curvature is what identifies an electron;
the hadron calorimeter — iron and scintillator, whose iron also serves as the magnet’s return yoke. Hermetic, because $E_T^{\text{miss}}$ requires that nothing escapes sideways;
iron absorbers, for the tails of forward hadronic showers;
muon chambers outside everything, because muons are what survive.

Bettini Eq. (9.67). Each row is a complete event description — this is what the trigger was built to find.

channel↕	signature↕
$W\to e\nu$	one isolated electron at high $p_T$ , and high $p_T^{\text{miss}}$
$W\to\mu\nu$	one isolated muon at high $p_T$ , and high $p_T^{\text{miss}}$
$Z\to e^+e^-$	two isolated electrons, opposite sign, both at high $p_T$ — and no missing energy
$Z\to\mu^+\mu^-$	two isolated muons, opposite sign, both at high $p_T$

💡 What this really says — colour counted twice, in opposite directions

Worth collecting, because it is easy to think one of them is a mistake.

In the widths, colour multiplies by 3. A $W$ can decay to $u\bar d$ in three ways — red, green, blue — and they are distinct final states, so the rate is three times a single one.

In the formation cross-section, colour divides by 9. To make a colourless $W$ , the incoming quark and antiquark must carry matching colour and anticolour. Given a random quark and a random antiquark, that is one combination in nine.

Both are correct, and they are the same counting seen from two ends: summing over final states multiplies, averaging over initial states divides. Eq. (9.57)‘s $\tfrac19$ and Eq. (9.44)‘s $\times 3$ appear four pages apart and the book does not connect them.

The general rule is worth internalising, because it is the commonest arithmetic error in this subject: sum over what you cannot distinguish in the final state, average over what you did not choose in the initial state.

🔑 If you remember only three things

Three cuts turn one in ten billion into a countable sample. The selection is the experiment; the detector only has to survive what the selection demands of it.
How often you make one follows from how it decays. Nothing new is measured to predict the production rate — the same widths do both jobs.
A collider for quarks is a machine nobody can aim. You collide protons and take whatever the partons bring, which is why the cuts matter more than the beam does.

Where this goes next

Everything above is a prediction. §9.7 is the year those predictions met the data: UA1 and UA2 found both bosons at the predicted masses, measured the $W$ mass from the Jacobian peak, confirmed its spin from the decay angular distribution, and extracted $\sin^2\theta_W$ from the mass ratio — where, as §9.4 noted, the calorimeter calibration cancels.

The widths computed here are tree-level, and all three come out a few per cent below the measured values — $\Gamma_W$ 2.04 against 2.085, $\Gamma_Z$ 2.42 against 2.4952. That is not a defect. §9.9 measures those same widths to four significant figures at LEP, and the gap between prediction and measurement is where the radiative corrections live — including, eventually, the top quark and the Higgs.

✅ Check yourself — the W, the Z, and the machine that found them

0/6 answered · 0 correct

1.Colour multiplies the W's hadronic widths by 3 but divides the formation cross-section (9.57) by 9. Is one of them wrong?
2.Why is the Z wider than the W (2.42 against 2.04 GeV) even though every one of its couplings is weaker than the W's?
3.Γ_inv is the Z's width into everything invisible, and N_ν = Γ_inv/Γ_νν̄. Why is this a strong measurement rather than a circular one?
4.Example 9.3 finds 26 W and 2 Z events per year at the design luminosity. What does that explain about the history?
5.Why is every observable at a hadron collider built from TRANSVERSE quantities — p_T, E_T^miss, Δη rather than Δθ?
6.Evaluating Eq. (9.58) with its own printed inputs — 4π/3, 1/81², 0.640 × 0.225/2.04², × 388 — gives 8.6 nb, but the equation states 8.8 nb. What is going on?