Measuring the Weak Mixing Angle · Elementary Particle Physics

One number is extracted from processes eight orders of magnitude apart, and the theory’s whole claim is that they all give the same answer.

🎯 Why this matters

That makes the angle a consistency test rather than a parameter. A disagreement anywhere would not shift a fitted value — it would say the compression itself is wrong, and there is no version of this theory carrying two angles.

§9.3 reduced every neutral-current coupling in nature to $c_Z = I_{Wz} - Q\sin^2\theta_W$ . That is a spectacular compression, and it is worth nothing at all unless the angle you extract from one process agrees with the angle you extract from another.

So this section is a measurement, done in detail, of a single number. The list of ways to measure it is the actual test; CHARM2 is one entry, chosen because it is the cleanest.

Eight orders of magnitude, one number

Bettini §9.4. The theory says every row must give the same answer once radiative corrections are applied. Nothing but the theory connects them — different beams, different targets, different decades.

method↕	typical $\|Q\|$ ↕	how the angle enters↕	what limits it↕
Boson mass ratio $M_W/M_Z$	~90 GeV	$\cos\theta_W = M_W/M_Z$ directly	nothing much — this is the most precise determination, because a ratio measured in one calorimeter cancels the energy scale (§9.7)
Atomic parity violation	keV–MeV	γ–Z interference shifts nothing measurably, but interferes observably at the ppm level	atomic-structure calculations — many-electron wave functions to ppm
Polarized $e$ –deuteron scattering	1.3 GeV	a few-ppm asymmetry between the two beam polarizations	QCD: the quarks are inside nucleons inside a nucleus
Polarized Møller scattering $ee\to ee$	0.16 GeV	the same interference, but purely leptonic	the asymmetry is a fraction of a ppm, and $ep$ events swamp $ee$ ones
Forward–backward asymmetry, $e^+e^-\to f\bar f$	10–200 GeV	γ–Z interference makes the angular distribution asymmetric about 90°	statistics; large near $M_Z$ , which is why LEP owns it (§9.9)
Deep inelastic $\nu N$	several GeV	the NC/CC cross-section ratio, no interference needed	QCD again — a complex hadronic target
Elastic $\nu e$ — CHARM2	~0.1 GeV	the ratio $\sigma(\nu e)/\sigma(\bar\nu e)$	rate. The cross-section is $10^4$ times smaller than on a nucleus — the price of having no hadrons at all

💡 What this really says — the clean measurement is the rare one, and that is not a coincidence

Read the last column. Every entry that is easy is limited by hadronic theory, and every entry that is theoretically clean is limited by rate.

That trade is structural, not bad luck. The weak neutral current couples to everything, so the abundant targets are nuclei — and the moment a nucleus is involved you need to know what the quarks inside it are doing, which means QCD, which means §6.10’s lattice or a parton model with its own uncertainties. Choosing a leptonic target removes all of that, at a price the book states plainly: a neutrino–electron cross-section goes as the target mass, and swapping a nucleus for the handful of electrons bound to it costs four orders of magnitude. Neutrinos interact with the nuclei of CHARM2 and with its electrons, and the process wanted here happens once in ten thousand.

You are choosing which uncertainty to have. Statistical error shrinks as $1/\sqrt{N}$ and you can always buy more of it with beam time and mass. A theoretical uncertainty in the QCD correction does not shrink at all, no matter how long you run.

That is why CHARM2 is 792 tonnes of glass: it is buying back, with brute mass, the rate it gave up to get rid of the hadrons. And the choice paid — the book picks this experiment, out of seven, to describe in full.

The calculation, from the Z-charge factors alone

Both reactions run through a single $Z$ exchange, and the upper vertex — the neutrino’s — is the same in every diagram. So it cancels in the ratio, and everything reduces to the electron vertex, whose coupling §9.3 already gave us.

Figs. 9.3 and 9.4 — four diagrams, one vertex that matters

Click a vertex or an internal line.

Bettini Figs. 9.3 and 9.4, redrawn together because the comparison is the argument. The Z-charge factor at each lower vertex is the only thing that differs; the neutrino vertex is identical throughout and cancels in the ratio.

The two contributions to each reaction do not interfere — they differ in the electron’s helicity, which is in principle measurable — so you add the squares, not the amplitudes. And they are not weighted equally:

🪜 Why L+R costs a factor of 3 — Fig. 9.5

Step 1 of 4 — set up the two cases in the centre of mass

\nu_L + e^-_L \quad\text{versus}\quad \nu_L + e^-_R

Why you may do this: At these energies both particles are effectively massless, so negative chirality means negative helicity: spin antiparallel to momentum. The two particles fly at each other along a common axis, which we take as the quantisation axis.

Bettini pp. 361–362. Pure angular momentum, no dynamics. The result is general and is quoted again in §9.5 for the Z widths.

Putting the two together — each contribution weighted by $c_Z^2$ and by 1 or 1/3:

\sigma_{\nu_\mu e} \propto G_F^2 m_e E_\nu\left[\left(-\tfrac12 + \sin^2\theta_W\right)^2 + \tfrac13\sin^4\theta_W\right]

\sigma_{\bar\nu_\mu e} \propto G_F^2 m_e E_{\bar\nu}\left[\tfrac13\left(-\tfrac12 + \sin^2\theta_W\right)^2 + \sin^4\theta_W\right]

The antineutrino is right-handed, so the two roles swap: the 1/3 moves from the $e_R$ term to the $e_L$ term. That swap is the entire angular dependence of the measurement — without it the two cross-sections would have the same shape and their ratio would be 1 for every angle.

\htmlClass{t-r}{R} = \frac{\sigma_{\nu_\mu e}}{\sigma_{\bar\nu_\mu e}} = 3\,\frac{\htmlClass{t-n}{1 - 4\sin^2\theta_W + \tfrac{16}{3}\sin^4\theta_W}} {\htmlClass{t-d}{1 - 4\sin^2\theta_W + 16\sin^4\theta_W}}

(9.31)

Bettini p. 362. Every common factor is gone — G_F, the electron mass, the flux, the detector efficiency. What is left is a pure function of one angle.

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.

💡 What this really says — R = 1 exactly at sin²θ_W = ¼, so this is a null measurement

Put $\sin^2\theta_W = 1/4$ into (9.31). The numerator becomes $3(1 - 1 + \tfrac{16}{3}\cdot\tfrac1{16}) = 3\cdot\tfrac13 = 1$ and the denominator becomes $1 - 1 + 16\cdot\tfrac1{16} = 1$ . So

R = 1 \quad\Longleftrightarrow\quad \sin^2\theta_W = \tfrac14 \quad\text{exactly}

and differentiating at that point gives a slope of exactly $-8$ :

R \simeq 1 - 8\left(\sin^2\theta_W - \tfrac14\right)

which is good to 1 % at the measured value. The book does not point this out, and it changes how you should read the experiment.

CHARM2 is not measuring an angle; it is measuring a departure from a symmetric point. At $\sin^2\theta_W = 1/4$ the electron’s two Z-charge factors are $\mp 1/4$ — equal in magnitude, opposite in sign — and the neutrino and antineutrino see mirror-image situations. Nature sits a few per cent below that point, and $R = 1.15$ rather than 1 is the whole signal.

The engineering instinct is right here: a null measurement is worth far more than an absolute one, because everything multiplicative cancels near the null and you are left measuring a small difference against a known zero. The same reasoning built the Wheatstone bridge, the lock-in amplifier, and the double ratio of §8.8. It also explains the leverage — the slope −8 means a 6.6 % measurement of $R$ becomes a 3.6 % measurement of $\sin^2\theta_W$ , and that is exactly what CHARM2 achieved.

Eq. (9.31), exact
the linearisation R = 1 − 8(sin²θ_W − ¼)
the null: R = 1 at sin²θ_W = ¼ exactly
CHARM2: sin²θ_W = 0.2324 ± 0.0083, i.e. R = 1.151

The ratio against the angle. The curve passes through (¼, 1) exactly, and is straight to within 1 % across the whole measured region — so the experiment is best read as a measurement of how far below ¼ the angle sits.

Making the measurement

Two exposures, one to a $\nu_\mu$ beam and one to a $\bar\nu_\mu$ beam, and you count. The one thing that does not cancel is the beam: the two fluxes are different and both have a wide energy spread, and since the cross-sections are proportional to $E$ , what is needed is the ratio of energy-weighted fluxes:

F = \frac{\int \Phi_\nu(E)\,E\,dE}{\int \Phi_{\bar\nu}(E)\,E\,dE}, \qquad R_{\text{exp}} = \frac{N(\nu_\mu e)}{N(\bar\nu_\mu e)}\,F

⚙️ Engineer’s bridge — everything cancels except the one thing you must measure separately

It is worth listing what has just disappeared from the problem, because the design of the experiment is this list.

Gone, because the same detector sees both exposures: the fiducial mass, the electron-finding efficiency, the angular acceptance, the trigger threshold, the energy scale, the reconstruction bias. Gone, because the ratio is taken at the same energy: $G_F^2$ , $m_e$ , and the linear $E$ dependence.

Left over: the flux ratio $F$ . Which is now the dominant systematic, and the reason CHARM2 used a narrow band beam whose energy spectrum could be computed from the parent pion kinematics (problem 9.3) rather than a wide-band one.

This is the standard shape of a good ratio measurement, and an engineer builds it deliberately: arrange the experiment so that everything hard cancels, then spend the entire effort budget on the one quantity that does not. The residual is always something about the source, because the source is the only thing not shared between the two configurations — the same reason a differential amplifier is limited by its reference and a two-point calibration by its standards.

Where it breaks: the cancellation assumes the two configurations differ only in the bit you flipped. A ν and a ν̄ beam are not one source with a sign reversed — they are made by flipping a focusing horn, which selects different parent mesons with different momentum spectra, so the two beams have genuinely different fluxes and energy distributions. The quantity that fails to cancel is precisely the one you assumed away. That is the standing failure mode of differential measurement: the control bit is never perfectly orthogonal to everything else, and the cleaner the subtraction looks, the less anyone examines what rides along with it.

The signal is a single electron and nothing else, which is a topology that backgrounds imitate easily. The rescue is kinematic, and it is a one-line consequence of the electron being light:

🪜 Why the electron comes out almost straight — Eqs. (9.36)–(9.37)

Step 1 of 4 — conservation, three equations

E_i + m_e = E_e + E_\nu, \qquad 0 = E_\nu\sin\theta_\nu - E_e\sin\theta_e,

Why you may do this: The incoming neutrino has energy E_i, the outgoing electron E_e at angle θ_e and the outgoing neutrino E_ν at θ_ν. Longitudinal momentum gives a third equation, E_i = E_ν cos θ_ν + E_e cos θ_e.

Bettini pp. 363–364. Energy and momentum conservation for ν + e → ν + e, with the electron initially at rest.

the angle bound, and where the precision comes from

import numpy as np
me = 0.5109989e-3                                  # GeV
print("the kinematic bound is a pure number, with no free parameters:")
print(f"  E_e theta_e^2 <= 2 m_e = {2*me*1e3:.4f} MeV")
print("\nso the largest possible electron angle falls as 1/sqrt(E):")
for E in (5, 10, 20, 50):
    print(f"  E_e = {E:3d} GeV  ->  theta_max = {np.sqrt(2*me/E)*1e3:5.2f} mrad")

print("\nthat is what sets every detector requirement: to place an event inside")
print("that bound you must know the direction of a SINGLE electron track to a")
print("fraction of a few mrad, and the target is also the scatterer.  hence a")
print("low-Z medium (glass, not iron) and many position samples per module.")

R = lambda s: 3*(1 - 4*s + (16/3)*s**2) / (1 - 4*s + 16*s**2)
print("\nnow the sensitivity.  R is 1 exactly at sin^2 = 1/4, with slope -8:")
tags = {0.2324: "      <- CHARM2", 0.25: "      <- the null, exactly"}
for s in (0.20, 0.2231, 0.2324, 0.25, 0.30):
    print(f"  sin^2 = {s:.4f}  ->  R = {R(s):.4f}{tags.get(s, '')}")

s0, h = 0.2324, 1e-5
d = (R(s0+h) - R(s0-h)) / (2*h)
print(f"\n  dR/d(sin^2) at the measured point = {d:.2f}")
print(f"  so sigma(R)/R = {abs(d)*0.0083/R(s0)*100:.1f}%  ->  sigma(sin^2) = 0.0083, which is 3.6%")
print("\n  the measurement is LEVERAGED: a 6.6% number becomes a 3.6% one,")
print("  because R changes fast near the null.  that is the whole reason to")
print("  measure a ratio that happens to pass through 1 nearby.")

prints

the kinematic bound is a pure number, with no free parameters:
E_e theta_e^2 <= 2 m_e = 1.0220 MeV

so the largest possible electron angle falls as 1/sqrt(E):
E_e =   5 GeV  ->  theta_max = 14.30 mrad
E_e =  10 GeV  ->  theta_max = 10.11 mrad
E_e =  20 GeV  ->  theta_max =  7.15 mrad
E_e =  50 GeV  ->  theta_max =  4.52 mrad

that is what sets every detector requirement: to place an event inside
that bound you must know the direction of a SINGLE electron track to a
fraction of a few mrad, and the target is also the scatterer.  hence a
low-Z medium (glass, not iron) and many position samples per module.

now the sensitivity.  R is 1 exactly at sin^2 = 1/4, with slope -8:
sin^2 = 0.2000  ->  R = 1.4762
sin^2 = 0.2231  ->  R = 1.2381
sin^2 = 0.2324  ->  R = 1.1507      <- CHARM2
sin^2 = 0.2500  ->  R = 1.0000      <- the null, exactly
sin^2 = 0.3000  ->  R = 0.6774

dR/d(sin^2) at the measured point = -9.11
so sigma(R)/R = 6.6%  ->  sigma(sin^2) = 0.0083, which is 3.6%

the measurement is LEVERAGED: a 6.6% number becomes a 3.6% one,
because R changes fast near the null.  that is the whole reason to
measure a ratio that happens to pass through 1 nearby.

🛠️ Figs. 9.6 and 9.8 — CHARM2, a fine-grained calorimeter that is also the target

Click a numbered marker for what that piece does.

Bettini Figs. 9.6 and 9.8, redrawn. The detector is a stack of ~100 identical modules; the right-hand side expands one of them. Its material is the target, which is why the choice of glass is a physics decision and not an engineering one.

🔬 Experiment card — CHARM2, CERN SPS, 1980s–1990s

Apparatus

A narrow-band

\nu_\mu

\bar\nu_\mu

beam from the SPS, and a 792 t fine-grained calorimeter 4 m × 4 m in cross-section and 33 m long. It is built from about a hundred identical modules, each a 48 cm slab of glass — a low-

Z

material chosen to limit multiple scattering — followed by two tracking chambers measuring one coordinate each, and a plane of plastic scintillator counters that both trigger the read-out and measure the energy.

What is measured

For each single-electron event, the energy

E_e

and the angle

\theta_e

to the beam. The observable is the distribution of the product $E_e\theta_e^2$ , which kinematics bounds at

2m_e = 1.022

MeV for genuine elastic scattering, and which is unbounded for everything else. The whole measurement is then the ratio of the signal counts in the neutrino and antineutrino exposures, corrected by the ratio

F

of energy-weighted fluxes.

The result

A peak at small

E_e\theta_e^2

in both exposures, sitting on a background that is fitted at large angles — where the signal cannot be — and extrapolated underneath. Subtracting it gives

$\sin^2\theta_W = 0.2324 \pm 0.0083$

What it proved

That the angle governing a purely leptonic neutral-current process at

|Q|\sim 0.1

GeV is the same angle that governs the boson masses at 90 GeV. That is the electroweak theory’s central claim, tested with no hadronic input whatsoever — no parton distributions, no form factors, no lattice.

The precision is modest by later standards, and deliberately so: this measurement’s value is that its systematic uncertainties have nothing in common with anyone else’s.

The two backgrounds, and how kinematics kills both

The signal is one electron track and nothing else, occurring once per ten thousand neutrino interactions. Two things imitate it.

Bettini p. 363. Both are rejected by the same variable that defines the signal, which is why the experiment works at all.

background↕	why it happens↕	how big↕	why it fails the cut↕
$\nu_e N \to e^- X$	the $\nu_\mu$ beam carries an irreducible ~1 % $\nu_e$ contamination, from $K$ and $\mu$ decays in the beam line	1 % of the flux × $10^4$ times the cross-section = 100× the signal	a charged-current scatter off a nucleus has no $E\theta^2$ bound — the target is 2000× heavier, so the electron can come out at any angle
$\nu N \to \nu\pi^0 X$ , $\pi^0\to\gamma\gamma$	a neutral-current interaction where the hadronic system $X$ is too soft to see, and one photon converts to an $e^+e^-$ pair read as one track	comparable	same reason — and the fine granularity separates two overlapping showers from one

The procedure is the sideband method of §8.5 again: fit the background where the signal cannot be — at large $E_e\theta_e^2$ , beyond the kinematic limit — check that the prediction agrees there, then extrapolate under the peak and subtract.

ν exposure — signal peak on background
ν̄ exposure — a smaller peak, same background shape
background, fitted at large Eθ² and extrapolated
the kinematic limit 2m_e/E at E = 20 GeV

Fig. 9.9 redrawn schematically (data: Geiregat et al. 1991). The signal is the excess in the first bin only. Note how small the kinematic limit is on this scale — the whole signal lives inside the leftmost sliver, which is precisely why the background can be measured so confidently everywhere else.

🔑 If you remember only three things

Seven hundred tonnes is the price of a small coupling. The detector’s size is set by how rarely the process happens, not by what it has to resolve.
The backgrounds die by kinematics rather than by quality cuts. Both fakes are removed by where they sit, which is why the selection survives having no particle identification.
A free parameter can still be falsifiable. Nobody predicts what the angle is, and the theory is nonetheless at risk on every row of the table.

Where this goes next

The angle is now a measured quantity, and §9.5 spends it. Knowing $\theta_W$ and $G_F$ fixes $M_W$ and $M_Z$ ; knowing $c_Z$ for every fermion fixes every partial width of both bosons. The whole of §9.5 is arithmetic on one number — the leptonic $W$ width — plus the colour factor 3 and the table you have already met.

§9.8 returns to this section’s list and plots all of it at once, as $\sin^2\theta_W$ against $Q$ . The variation across eight orders of magnitude turns out to be a few per cent — and the slope changes sign at $M_W$ , for a reason that is genuinely surprising.

✅ Check yourself — measuring the weak mixing angle

0/6 answered · 0 correct

1.Why did CHARM2 choose neutrino–ELECTRON scattering when neutrino–nucleus scattering is ten thousand times more frequent?
2.Where does the factor 1/3 in the L+R cross-section come from?
3.Substituting sin²θ_W = 1/4 into Eq. (9.31) gives R = 1 exactly. What does that tell you about the measurement?
4.The signal is a single electron. Both principal backgrounds also give a single electron. Why does the cut on E_e θ_e² reject them?
5.Taking the ratio of the two exposures cancels the detector mass, the efficiency, the acceptance and the energy scale. What does NOT cancel, and what did the experiment do about it?
6.CHARM2 got sin²θ_W = 0.2324 ± 0.0083, far less precise than the LEP determinations. Why does the book describe it in full anyway?