The Discovery of the W and the Z · Elementary Particle Physics

The discovery rested on a handful of events, and what made them convincing was not how many there were but that each could be checked against a prediction made in advance.

🎯 Why this matters

This only works when a theory predicts a shape and not merely a rate. A peak at a computable location can be established by a few events, where an excess of the same size would have persuaded nobody.

§9.5 predicted two masses, two widths and two cross-sections from one number and an angle. §9.6 built the machine and the three cuts. This section is 1983, when the predictions met the data — and what makes it worth reading in detail is that every prediction was checked, not just the masses.

One event, and what makes it unambiguous

A raw $W\to e\nu$ event is a mess: dozens of tracks, most from the rest of the event — the other partons in the same proton and antiproton, which took no part in the collision that mattered. The cut that clears it is almost embarrassingly simple. Discard everything with $p_T < 1$ GeV, a few times $\Lambda_{QCD}$ , and what remains is a single charged track.

The calorimeter’s view of the same event is the lego plot below. That track is then confirmed as an electron by a coincidence the detector was built for: its momentum from curvature in the central tracker and its energy from the calorimeter agree within errors. A hadron would deposit its energy differently; a muon would not stop at all.

Figs. 9.15, 9.20 and 9.22 — the same plot, three processes

tallest tower: 34 GeV of transverse energy

One tall, narrow tower and nothing opposite it. The electron carries ~M_W/2 of transverse energy into a single calorimeter cell; the neutrino carries a comparable amount away invisibly, which is the arrow. An imbalance THIS large cannot come from mismeasurement — that is the whole signature.

Calorimeter transverse energy over the (η, φ) plane, redrawn. Switch between the three: the W is one tower and an imbalance, the Z is two towers that balance, and the dijet event is two towers that look almost the same until you notice their width. The soft background is the 'rest of the event' and is drawn from a fixed seed, so it does not reshuffle when you switch.

💡 What this really says — the lego plot is the reason quarks were believed

Flip between the last two presets. A $Z\to e^+e^-$ event and a two-jet event produce the same picture — two localised energy deposits, back to back in azimuth, nothing else of consequence.

The book states the difference in one sentence: the quark peaks are “wider and more activity is present outside them”. That is the whole observable signature of hadronization. A quark cannot arrive at the calorimeter; what arrives is a spray of hadrons collimated around the original direction, and the width of that spray is set by $\Lambda_{QCD}$ over the jet energy (§6.1’s cone angle).

Two consequences, and both matter later:

you can measure a quark’s momentum by adding up a cone of calorimeter cells, which is what makes §9.11’s top-mass reconstruction from six jets possible at all;
and you cannot make the isolation cut without this. §9.6’s requirement that a lepton have no track within $\Delta R \sim 0.5$ works because jets are wide and leptons are not. The picture is the justification for the cut.

The comparison also explains why UA2 saw two-jet events before anyone was looking for them: at a hadron collider they are overwhelmingly the commonest hard process, and they look like the signal.

The Jacobian peak: a mass from a transverse distribution

Here is the problem. The $W$ decays to $e\nu$ , and the neutrino is invisible. You know its transverse momentum from the imbalance, but not its longitudinal component, so you cannot reconstruct the invariant mass. Yet UA1 measured $M_W$ to 4 %, using the Jacobian peak .

🪜 Where the peak comes from — Eqs. (9.69)–(9.72)

Step 1 of 4 — the one component that survives

p_T = \frac{M_W}{2}\sin\theta^*

Why you may do this: In the W rest frame the two-body decay gives each particle momentum M_W/2. The component transverse to the W's motion is the same in that frame and in the laboratory — a longitudinal boost does not touch it — so p_T is measurable even though the W's own momentum is not.

θ* is the decay angle in the W rest frame. The W is produced almost along the beams, so "transverse to the W" and "transverse to the beam" are nearly the same thing.

Bettini pp. 375–376. The trick is to find a variable whose distribution has a feature at a known place, rather than a variable you can compute event by event.

⚙️ Engineer’s bridge — look for a feature at a known place, not a number you can compute

This is a genuinely transferable move and it is worth naming, because it recurs throughout experimental physics and has an exact engineering counterpart.

You want $M_W$ . The direct route — reconstruct the invariant mass event by event — is blocked, because one particle is invisible along the beam. So instead of computing the quantity you want for each event, you form the distribution of a quantity you can measure, and you look for a feature in it whose position encodes what you wanted.

The reason it works is the change of variables. A smooth distribution in $\theta^*$ , pushed through a transformation whose derivative is singular at the endpoint, produces a sharp feature at that endpoint regardless of the input shape. You do not need to know the decay angular distribution to find the edge. That is robustness of exactly the kind you want when the dynamics is what you are trying to test.

An engineer meets the same structure constantly. It is a caustic — light through a curved surface concentrating at a focus whose position depends on the geometry and not on the illumination. It is the pile-up of a swept sine at the turning points of the sweep. It is the edge in the histogram of $\sin$ of a uniform variable. In every case a singular Jacobian converts a smooth distribution into a sharp landmark, and the landmark is where the information is.

Two practical corollaries the book leaves implicit. First, the edge is one-sided: events above $M_W/2$ come only from the $W$ ‘s own transverse momentum and from resolution, so the shape above the edge measures your smearing. Second, this is why the measurement is systematics-limited from the start. UA1 quoted $M_W = 83 \pm 3$ GeV with the uncertainty “substantially determined by the systematic uncertainty on the energy calibration” — statistics was never the problem.

Where it breaks: an edge is a landmark only while the smearing is smaller than the feature. Detector resolution rounds the Jacobian edge, and the shape of that rounding — not merely its width — is what maps onto the mass, so the measurement depends on a resolution model rather than on a measured position. It also assumes a production model: the W’s transverse momentum is not zero, and the recoil spectrum that shifts the edge has to come from theory or from a control sample. Looking for a feature at a known place converts a statistics problem into a modelling problem; it does not remove the problem.

the ideal Jacobian, 1/√((M_W/2)² − p_T²)
smeared by the W transverse momentum and the resolution
the edge at M_W/2 = 40 GeV

Fig. 9.17 redrawn (data: Albajar et al. 1989). The ideal distribution has an integrable divergence at M_W/2; the measured one is that convolved with the W's own transverse momentum and the calorimeter resolution. The peak survives the smearing — which is the point — but its width is now a systematic, not a measurement.

the edge, and the mass ratio it feeds

import numpy as np
MW = 80.0
print(f"the Jacobian factor 1/sqrt((M_W/2)^2 - pT^2), with M_W = {MW:.0f} GeV:")
for x in (0.50, 0.80, 0.90, 0.95, 0.99):
    p = x*MW/2
    print(f"  pT = {p:4.1f} GeV = {x:.2f} x M_W/2  ->  {1/np.sqrt(1-x**2):.1f}x its value at pT = 0")
print( "  the divergence is integrable, so the COUNT stays finite --")
print(f"  it is the density that spikes, and the spike is at M_W/2 = {MW/2:.1f} GeV")

print("\nthe two 1983 measurements, and the angle from their ratio:")
for nm, mw, dw, mz, dz, book in (('UA1', 83, 3, 93, 3, '0.211 +- 0.025'),
                                 ('UA2', 80, 1.5, 91.5, 1.7, '0.232 +- 0.027')):
    print(f"  {nm}: M_W = {mw} +- {dw}     M_Z = {mz} +- {dz}")
    print(f"       sin^2(theta_W) = 1 - (M_W/M_Z)^2 = {1-(mw/mz)**2:.4f}     book: {book}")
print("\n  (the small differences are because the published ratios were not")
print("   formed from these rounded central values)")

print("\ncompare with the independent low-energy determination of Sec. 9.4:")
print( "  CHARM2, neutrino-electron scattering:  0.2324 +- 0.0083")
print(f"  the modern masses:  1 - (80.377/91.1875)^2 = {1-(80.377/91.1875)**2:.4f}")
print("\n  a purely leptonic measurement at |Q| ~ 0.1 GeV and a mass ratio at")
print("  90 GeV agree.  THAT is the test, and it is why both are quoted.")

prints

the Jacobian factor 1/sqrt((M_W/2)^2 - pT^2), with M_W = 80 GeV:
pT = 20.0 GeV = 0.50 x M_W/2  ->  1.2x its value at pT = 0
pT = 32.0 GeV = 0.80 x M_W/2  ->  1.7x its value at pT = 0
pT = 36.0 GeV = 0.90 x M_W/2  ->  2.3x its value at pT = 0
pT = 38.0 GeV = 0.95 x M_W/2  ->  3.2x its value at pT = 0
pT = 39.6 GeV = 0.99 x M_W/2  ->  7.1x its value at pT = 0
the divergence is integrable, so the COUNT stays finite --
it is the density that spikes, and the spike is at M_W/2 = 40.0 GeV

the two 1983 measurements, and the angle from their ratio:
UA1: M_W = 83 +- 3     M_Z = 93 +- 3
     sin^2(theta_W) = 1 - (M_W/M_Z)^2 = 0.2035     book: 0.211 +- 0.025
UA2: M_W = 80 +- 1.5     M_Z = 91.5 +- 1.7
     sin^2(theta_W) = 1 - (M_W/M_Z)^2 = 0.2356     book: 0.232 +- 0.027

(the small differences are because the published ratios were not
 formed from these rounded central values)

compare with the independent low-energy determination of Sec. 9.4:
CHARM2, neutrino-electron scattering:  0.2324 +- 0.0083
the modern masses:  1 - (80.377/91.1875)^2 = 0.2231

a purely leptonic measurement at |Q| ~ 0.1 GeV and a mass ratio at
90 GeV agree.  THAT is the test, and it is why both are quoted.

Three more predictions, checked

The masses are the headline. The other three checks are what turn a bump into an identification.

Bettini pp. 375–378. Each row tests a different part of §9.3, and each would have been a way for the theory to fail.

prediction↕	how it was checked↕	result↕
lepton universality — the $W$ couples equally to $e$ , $\mu$ , $\tau$	compare the three cross-sections $p\bar p \to W \to \ell\nu$	$g_\mu/g_e = 1.00 \pm 0.07 \pm 0.04$ , $g_\tau/g_e = 1.01 \pm 0.10 \pm 0.06$ — and note this tests the charged current, where §7.8 tested it at low energy
spin 1 — the $W$ is a vector boson	the decay angular distribution in the $W$ rest frame	$(1+\cos\theta^*)^2/4$ , in perfect agreement; a spin-0 $W$ would give a flat line, which the data exclude outright
parity violation in the production and decay	the same distribution — it is asymmetric	confirmed. But see the ⚠️ below: this does not establish V−A
the mass ratio gives the angle	$\sin^2\theta_W = 1 - (M_W/M_Z)^2$	0.211 ± 0.025 (UA1) and 0.232 ± 0.027 (UA2), agreeing with the low-energy determinations of §9.4

The angular distribution is worth doing properly, because it is a clean piece of angular-momentum bookkeeping and the answer is forced:

\frac{d\sigma}{d\Omega} \propto \left|\htmlClass{t-d}{d^{\,1}_{1,1}(\theta^*)}\right|^2 = \left[\frac{1}{2}\left(1 + \htmlClass{t-c}{\cos\theta^*}\right)\right]^2

(9.73)

Bettini p. 377. A Wigner d-function with both indices equal to 1 — the initial and final states both have J_z = −1 along their own axes, which is what V−A forces.

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.

J = 1, Eq. (9.73): [(1+cos θ*)/2]² — and V+A gives the same curve
J = 0 — flat, and excluded by the data
UA1

Fig. 9.18(b) redrawn. The curve is Eq. (9.73), normalised to unit average. The flat dashed line is the J = 0 hypothesis. Note what the asymmetry does and does not prove — see below.

⚠️ The asymmetry proves parity violation, not V−A — Question 9.1

The book states it and then asks you to prove it: “the observed asymmetry shows that parity is violated but does not prove that the CC structure is V−A. The V+A structure predicts the same angular distribution.”

Here is why. Flip every chirality: the quark that was left becomes right, the antiquark likewise, and the electron’s helicity flips too. The initial state now has $J_z = +1$ along the beam instead of $-1$ , and the final state has $J_z = +1$ along the electron direction instead of $-1$ . The relevant rotation matrix element is $d^{\,1}_{-1,-1}$ , and

d^{\,1}_{-1,-1}(\theta^*) = d^{\,1}_{1,1}(\theta^*) = \tfrac12(1+\cos\theta^*)

identically. Both projections flipped together, and the $d$ -functions satisfy $d^{\,j}_{-m,-m'} = d^{\,j}_{m',m}$ , so the observable is unchanged.

The general lesson is worth carrying: flipping every chirality in a process at once is invisible to any observable that does not measure a chirality. The angular distribution compares two directions, and a mirror image of the whole process gives the same angle between them. To tell V−A from V+A you must measure a polarization — the actual helicity of a final-state particle — which is what §7.4’s pion-decay experiments did, and why they had to be done at all.

So 1983 established the $W$ ‘s existence, mass and spin, and that parity is violated. It did not re-establish V−A; that was already known, and this measurement is consistent with it rather than evidence for it.

The Z, and the ratio that gives the angle

The $Z$ is cleaner and rarer: two isolated tracks of opposite sign, both with calorimeter energy matching their momentum, and no missing energy. Its mass needs no Jacobian trick at all — with both decay products visible, the invariant mass is reconstructed directly from the two energies and the angle between them.

the first 24 UA1 Z → e⁺e⁻ events
mean: M_Z = 93 GeV, systematic ±3

Fig. 9.21 redrawn. Twenty-four events — and that was enough, because the background under a two-isolated-lepton peak at 90 GeV is essentially nil. Compare the W measurement, which needed hundreds of events because its observable is an edge in a distribution rather than a peak.

Erratum — Eq. (9.74) has the wrong trigonometric function, and cites the wrong equation

The relation used to extract the angle from the two masses is printed as

\cos^2\theta_W = 1 - \left(M_W/M_Z\right)^2

This cannot be right, and it contradicts the chapter’s own Eq. (9.28), $M_W/M_Z = \cos\theta_W$ . Substituting that in gives $\cos^2\theta_W = 1 - \cos^2\theta_W$ , i.e. $\cos^2\theta_W = 1/2$ and $\theta_W = 45°$ — for every possible pair of masses. The correct relation is

\sin^2\theta_W = 1 - \left(M_W/M_Z\right)^2

and the book uses the correct one immediately: (9.75) quotes $\sin^2\theta_W = 0.211$ and $0.232$ , and $1 - (80/91.5)^2 = 0.236$ reproduces the second. So only the printed equation is wrong, not the physics or the numbers that follow it.

The same sentence also cites Eq. (9.27) as the source. (9.27) is $M_W = 37.3\;\text{GeV}/\sin\theta_W$ ; the mass-ratio relation is (9.28).

Both confirmed on the render of PDF p. 397.

🔬 Experiment card — UA1 and UA2, CERN SPS proton–antiproton collider, 1983

Apparatus

The SPS converted to a

p\bar p

storage ring at

\sqrt{s} = 540

GeV, later 630, running at

L \approx 10^{28}\;\text{cm}^{-2}\text{s}^{-1}

— made possible by van der Meer’s stochastic cooling of the antiproton beam. Two experiments: UA1, a hermetic detector built around a large time projection chamber in a dipole field, with electromagnetic and hadronic calorimetry and external muon chambers; and UA2, differently designed and with comparable results.

What is measured

Events with an isolated high- $p_T$ lepton. For the

W

: one lepton plus large missing transverse momentum, and the observable is the distribution of the lepton’s $p_T$ — not any individual event’s mass. For the

Z

: two isolated opposite-sign leptons, and the observable is their invariant mass, from two calorimeter energies and the angle between the tracks.

The result

A Jacobian edge at

p_T \approx 40

GeV and a mass peak near 93 GeV:

$M_W = 83 \pm 3\;\text{(UA1)}, \quad 80 \pm 1.5\;\text{(UA2)}$ $M_Z = 93 \pm 3\;\text{(UA1)}, \quad 91.5 \pm 1.7\;\text{(UA2)}$

with the decay angular distribution following $(1+\cos\theta^*)^2$ and the three leptonic couplings equal to within 7 %.

What it proved

That the electroweak theory’s central prediction is correct — two massive vector bosons, at the masses that

G_F

and

\sin^2\theta_W

demand, with the predicted spin and universal couplings. The mass ratio then returns the weak mixing angle and it agrees with the low-energy determinations of §9.4, which is the unification claim tested across three orders of magnitude in one experiment.

Rubbia and van der Meer shared the 1984 Nobel Prize, one year later — among the shortest gaps in the prize’s history, and a fair reflection of how completely the result was anticipated.

🔑 If you remember only three things

A transverse distribution gives a mass. The edge sits where the kinematics puts it, so a quantity nobody measured directly falls out of one that was.
The machine was rebuilt around a number. The SPS became a proton–antiproton collider because the masses were known in advance, which is designing an accelerator to a prediction.
One run settled several claims at once. The same data tested the widths, the production ratio and the angle, and each had been fixed before the beam turned on.

Where this goes next

The masses are now known to a few per cent, and the mass ratio is already the best determination of $\sin^2\theta_W$ . §9.8 asks what “the” weak mixing angle even means once radiative corrections are included — it runs with $Q$ , and its slope changes sign at $M_W$ for a reason worth waiting for.

Then §9.9 replaces every number on this page. LEP put four experiments on the $Z$ peak and measured its mass to 20 parts per million, against UA1’s 3 %: a factor of fifteen hundred. What that precision buys is the subject of the rest of the chapter — including a count of the neutrino species and, eventually, the mass of a quark nobody had yet seen.

✅ Check yourself — the 1983 discovery

0/6 answered · 0 correct

1.The neutrino from W → eν escapes, so the invariant mass cannot be reconstructed event by event. How was M_W measured anyway?
2.Why is the Jacobian-peak measurement systematics-limited from the very first events?
3.The decay angular distribution came out as [(1+cos θ)/2]², asymmetric in cos θ. What does that establish?
4.Twenty-four Z events sufficed to measure M_Z, while the W measurement needed hundreds. Why the difference?
5.Eq. (9.74) is printed as cos²θ_W = 1 − (M_W/M_Z)². What is wrong with it?
6.In the lego plot, a Z → e⁺e⁻ event and a two-jet event give almost the same picture. What are the differences, and why do they matter?