Charged Leptons and Neutrinos · Elementary Particle Physics

No two of these particles were established the same way, which is why the chapter reads as a catalogue of arguments rather than a sequence of discoveries.

🎯 Why this matters

The neutrino is the case worth carrying. Its cross-section was computed correctly and the conclusion drawn from it was wrong, because “undetectable” quietly assumed the flux available in 1934. A reactor changed the answer without changing any physics.

Six leptons: the $e^-$ , the $\mu^-$ and the $\tau^-$ , and three neutral partners. The charged ones are identical except for their masses; the neutral ones were, for a quarter of a century, a bookkeeping device that nobody expected to see. This section is the story of how each was found, and it contains the best example in the book of a correct calculation leading to a wrong conclusion.

Table 2.3 — the charged leptons. Identical in every respect but mass.

Lepton	m (MeV)↕	τ↕	cτ↕	Found↕	How
$e$	0.511		—	1897	Thomson: cathode rays deflected by crossed E and B fields
$\mu$	105.6	2.2 μs	659 m	1937	Anderson & Neddermeyer; Street & Stevenson — and identified as a lepton only in 1946
$\tau$	1777	290 fs		1975	Perl at SPEAR, by a signature that looked like a violated conservation law

Three copies of one particle, spread over a factor of 3500 in mass and 10²⁰ in lifetime. Nothing in the Standard Model explains why there are three, or why these masses. The name τ is from the Greek <em>triton</em>, 'the third' — a name chosen before anyone knew whether there would be a fourth.

Thomson’s electron: a null measurement

The first elementary particle was found by refusing to accept a deflection. Cathode rays were thought to be waves in the ether. Thomson (1897) showed they bend in an electric field as well as a magnetic one, and then did the decisive thing: he set the two fields against each other and tuned them until the beam did not move at all.

⚙️ Engineer’s bridge — why a null is worth two measurements

With crossed fields, balance means $qE = qvB$ , so

v = \frac{E}{B},

and the charge has cancelled. Now switch the electric field off and measure the magnetic deflection alone, which depends on $q/m$ and on the $v$ you just determined. Two runs, one ratio.

Two things make this a null measurement rather than an ordinary one, and both are reasons you would design it the same way today:

The answer is read off the settings, not the response. You do not need to know the detector’s gain, the beam’s intensity, or how far the spot moves per volt — only that it moved zero. Every calibration downstream of the balance point drops out.
The zero is where the sensitivity is best. Near balance the residual deflection is a small difference of two large forces, so a small error in $E/B$ shows up loudly. You are working at the steepest part of the curve.

Wheatstone bridges, lock-in nulls and phase-comparison methods all trade on the same two properties. Thomson’s result — that $m/q$ came out universal, independent of the cathode material and the gas — is what made it a particle rather than a property of the apparatus.

Where it breaks: a null is worth two measurements only when the alternative predicts a non-null. Thomson’s universality is strong because a property of the cathode material would have varied with the cathode material — the alternative was specific. A null against a vague alternative constrains nothing, and no null can separate “no effect” from “an effect below my sensitivity”; converting the first into the second is the entire work, and it is the same problem the FCNC limits of §7.13 have to solve.

The τ: find a conservation law being violated

By 1967 the pattern (e, ν_e) and (μ, ν_μ) was clear enough for A. Zichichi to ask whether there was a third pair, and to design a way to look for it at ADONE. There was, but it was too heavy for that machine. Perl found it at SPEAR in 1975.

The method is worth studying because it hunts for a particle by looking for something impossible.

🪜 How to discover a lepton you cannot see

Step 1 of 6 — Start from what is forbidden

e^+ e^- \not\to e^\pm \mu^\mp

Why you may do this: Lepton flavour is separately conserved (below): the electronic count L_e and the muonic count L_μ are each fixed. An e μ pair has L_e = ±1 and L_μ = ∓1, so it cannot come from a state with L_e = L_μ = 0.

This is a strict selection rule, not a suppression. If you see such events at any rate, something new is producing them.

Bettini Eqs. (2.19a)–(2.19b). Note that the τ itself is never observed — only the impossibility of the events it leaves behind.

The neutrino: a correct calculation, a wrong conclusion

Pauli proposed the neutrino in 1930 as a “desperate remedy” — beta decay appeared to violate the conservation of energy, momentum and angular momentum, and one invisible spin-½ particle in the final state fixed all three at once. Fermi built it into a theory in 1933 (Ch. 7) and proposed detecting it by inverse beta decay on a nucleon.

The following year Bethe and Peierls computed what that would take.

💡 What this really says — the most instructive mistake in the book

Their number was $\sigma \approx 10^{-44}\ \text{cm}^2$ at a few MeV, which corresponds to a mean free path of a thousand light years in solid matter. Their conclusion:

it is therefore absolutely impossible to observe processes of this kind with the neutrinos created in nuclear transformations.

The calculation is right. The conclusion is wrong, and the reason is a lesson about what a detector actually fights. A rate is

W = \sigma\,\Phi\,N ,

three factors, not one. Bethe and Peierls could bound $\sigma$ ; they had no way to imagine what $\Phi$ and $N$ would become. The cross-section is a property of nature and cannot be improved. The flux and the target mass are engineering, and engineering moved. A gigawatt reactor — eight years in the future when they wrote — delivers $10^{17}$ neutrinos per square metre per second, and a tonne of water is not hard to obtain.

Note that this is not a criticism of their physics. It is an observation about which quantities in a problem are fixed and which are yours to choose, and it is worth asking of any “impossible” measurement you meet.

\htmlClass{t-sig}{\sigma}\left(\bar\nu_e + p \to e^+ + n\right) = 10^{-47} \left(\frac{\htmlClass{t-E}{E}}{\text{MeV}}\right)^{\htmlClass{t-two}{2}}\ \text{m}^2

(2.20)

Eq. (2.20). At low energy the weak cross-section rises as the square of the energy — a consequence of the contact-interaction limit of §2.1–2.2's bridge, where the propagator is a constant and all the energy dependence comes from phase space.

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.

reactor thermal power0.7 GWdistance from the core12 mmass of water200 kgrepresentative ν̄ energy1 MeV

Bethe & Peierls, 1934

σ(1.0 MeV) = 1.0e-47 m²
mean free path in water
  = 1/(nσ) = 1.5e+18 m
  = 1.6e+2 light years

“It is therefore absolutely impossible to observe processes of this kind.” The arithmetic is correct.

Reines & Cowan, 1956

ν̄ from the core  = 1.3e+20 /s
flux at 12.0 m     = 7.2e+16 /m²/s
free protons     = 1.3e+28
rate = σ Φ N     = 9.7e-3 /s

34.9 per hour

Both numbers are true at once. A single neutrino would indeed cross 1.6e+2 light years of water. But a detector does not follow one neutrino — it waits for one of 1.3e+20 per second to interact with one of 1.3e+28 protons. What Bethe and Peierls could not have known is that the flux would one day be a free parameter — the first nuclear reactor was still eight years in the future.

Reines actually observed 3.0 ± 0.2 events per hour with 200 kg at Savannah River. Set the preset and compare: the ideal calculation gives more, and the ratio is the detector's overall efficiency — the fraction of true interactions that survive the prompt-plus-delayed coincidence, the shielding and the analysis cuts.

σ = 10⁻⁴⁷ (E/MeV)² m² is Eq. (2.20) — note that it grows as the SQUARE of the energy, which is why the higher-energy end of the reactor spectrum does most of the work. Free protons only: a mole of water carries 2N_A of them, the ones bound in oxygen do not count for inverse beta decay.

🔬 Experiment card — Reines and Cowan, Savannah River, 1956

Apparatus

Two 100-litre tanks of water, each sandwiched between liquid-scintillator chambers viewed by photomultipliers, buried a dozen metres below a 0.7 GW reactor core. Forty kilograms of cadmium chloride dissolved in the water. Shielding on all sides.

What is measured

Not a particle — a pattern in time. See the walkthrough below: a prompt double pulse, then a second one microseconds later. Nothing else in nature produces that sequence at that spacing.

The result

W = 3.0 \pm 0.2

events per hour, and control measurements showing the rate could not be background. One quarter of a century after Pauli’s hypothesis.

What it proved

The neutrino exists — specifically the electron antineutrino, since that is what fission produces. Pauli’s reply on hearing: “Thanks for the message. Everything comes to him who knows how to wait.”

🛠️ Figs. 2.6–2.7 — the Savannah River detector, a sandwich

Click a numbered marker for what that piece does.

Redrawn from Figs. 2.6 and 2.7 (p. 83). Two 100-litre water tanks between three scintillator layers; 40 kg of cadmium chloride dissolved in the water.

🪜 The Savannah River signature, microsecond by microsecond

Step 1 of 5 — t = 0 — the interaction

\bar\nu_e + p \to e^+ + n

Why you may do this: A free proton in the water absorbs an antineutrino. Two particles come out, and — crucially — they are of two completely different kinds, which is what makes a two-part signature possible.

Expected rate: a few per hour in 200 kg. Every background source in the room is far more frequent.

Redrawn from Figs. 2.6 and 2.7 (pp. 83). The whole experiment is one idea: make the signal a sequence that background cannot imitate.

⚙️ Engineer’s bridge — signal design when the SNR is hopeless

Three per hour against a reactor’s neutron flux, cosmic rays and ambient radioactivity is not a signal-to-noise ratio you improve by amplifying. Reines and Cowan did what you would do:

The problem	What they did	The general move
background is everywhere	12 m underground, heavy shielding	reduce the noise at the source before touching the signal
single pulses are indistinguishable	require prompt pair then delayed capture	look for a code, not an amplitude
the code needs a clock	tune the delay with dissolved Cd	design the signal so its timing is a controllable parameter
you cannot trust one number	run control measurements with the reactor off	measure your background in situ, not from a model

The second row is the important one, and it is matched-filtering in everything but name: instead of asking “is this pulse big enough?”, ask “does this sequence match the template?”. A correlated pattern spread over microseconds carries information that no single sample does, and uncorrelated background cannot forge it. Radar, GPS, ultrasound, optical time-domain reflectometry — all the same trick, and all for the same reason.

And they engineered the template. The 5 μs delay is not a property of the neutrino; it is set by how much cadmium you dissolve. Being able to choose where your signal sits in a background spectrum is the strongest position an experiment can be in.

Where it breaks: you choose where the signal sits; you do not choose the background. Loading the target with cadmium places the capture gamma at 9 MeV, but nothing lets you move the reactor’s own gamma flux, the cosmic-ray rate or the natural radioactivity of the tank — and the strategy only pays if the background happens to be smooth and small at the place you picked. Reines and Cowan still needed a delayed coincidence, metres of shielding and a reactor-off subtraction on top of the design. Choosing your signal’s location is a large advantage over having none; it is not the same as choosing your SNR.

Two neutrinos: the Brookhaven experiment

By 1962 there was a second question: is the neutrino that comes with a muon the same particle as the one that comes with an electron? Lederman, Schwartz and Steinberger answered it at the AGS with an experiment whose design is almost brutally direct.

🛠️ Fig. 2.8 — the first accelerator neutrino beam

Click a numbered marker for what that piece does.

Redrawn from Fig. 2.8 (p. 84). Read it as three filters in series: make everything, remove everything that is not a neutrino, and then let the neutrinos tell you what they are by what they turn into.

🔢 Worked example — how thick must the filter be?

The iron has two independent jobs, and it is worth checking that one thickness does both.

Stop the muons. 13.5 m of iron is $13.5\ \text{m} \times 7.874\ \text{g cm}^{-3} = 10\,630\ \text{g cm}^{-2}$ . At a minimum-ionising $\mathrm{d}E/\mathrm{d}x$ of about 1.45 MeV cm² g⁻¹ (§1.11) that absorbs

10\,630 \times 1.45\ \text{MeV} \approx 15\ \text{GeV},

and the AGS ran at 15 GeV, so no secondary muon can carry more. The filter was sized to the machine.

Remove the hadrons. With $\lambda_0 = 16.8$ cm for iron (§1.13d), 13.5 m is 80 interaction lengths — an attenuation of $e^{-80}$ , which is nothing at all.

Two different physical processes, two different length scales, one thickness that satisfies both. Everything downstream is a neutrino by construction, which is what licenses the claim that what interacted in the spark chambers arrived as one.

📐 Physics you need first — the spark chamber

Invented by Conversi and Gozzini (1955) and developed by Fukui and Miyamoto (1959), and worth a paragraph because it is the missing link between the chambers of §1.13b–c and modern electronics.

Two parallel metal plates a few millimetres apart, with gas between them. After a particle has crossed, a high voltage — of order $10^6$ V/m — is applied suddenly. The gas breaks down, but only where the ionisation trail already is, because those free electrons seed the discharge. The result is a bright visible spark sitting on the particle’s path, which is photographed.

Compare with the family of §1.13c:

	bubble chamber	spark chamber	MWPC
triggerable?	no	yes	yes
readout	film	film	electronic
the target	the liquid	the plates	separate
position resolution	0.1 mm	~1 mm	0.6 mm

The spark chamber is triggerable and massive, and in 1962 that combination existed nowhere else. It is why this experiment happened when it did.

💡 What this really says — what “lepton flavour” actually asserts

The result is stated in one line — neutrinos made with muons produce muons, never electrons — and it means lepton flavour is two separately conserved counts, not one. Assign:

	$L_e$	$L_\mu$
$e^-$ , $\nu_e$	+1	0
$e^+$ , $\bar\nu_e$	−1	0
$\mu^-$ , $\nu_\mu$	0	+1
$\mu^+$ , $\bar\nu_\mu$	0	−1
everything else	0	0

Both are additive and both are conserved separately. Check the chain that made the beam and then interacted:

\pi^+ \to \mu^+ + \nu_\mu \qquad (L_\mu: 0 \to -1 + 1 = 0) \ ✓

\nu_\mu + n \to \mu^- + p \qquad (L_\mu: +1 \to +1) \ ✓

\nu_\mu + n \not\to e^- + p \qquad (L_e: 0 \to +1,\ L_\mu: +1 \to 0) \ \times

The third line is what the photographs did not contain. A third count $L_\tau$ was added when the τ was found — and Chapter 10 is about the discovery that these counts are, very slightly, not conserved after all, which is what neutrino oscillation means.

Aside — the experiment that found nothing, and was right

Between Pauli and Reines there is a third attempt worth knowing about. Pontecorvo (1946) proposed a radiochemical method: expose chlorine, wait for $\nu + {}^{37}\text{Cl} \to {}^{37}\text{Ar} + e^-$ , then extract the argon and count its decays. He overestimated the cross-section by a hundred, and Alvarez (1949) corrected it and identified the real difficulty — the argon atoms to be extracted are about one in $10^{30}$ chlorine atoms.

Davis tried it in 1958 at Savannah River with 18.5 tonnes of C₂Cl₄, calculated to be twenty times more sensitive than needed. He saw nothing.

That null result is a real measurement, and an important one: the particle produced in β⁻ decay — which we now call $\bar\nu_e$ — does not induce the chlorine–argon reaction. Antineutrinos and neutrinos are different, exactly as lepton number requires. Davis had built the right instrument and pointed it at the wrong source; a decade later he pointed it at the Sun and found something nobody expected (Ch. 10).

Reproduce it

import numpy as np
NA = 6.02214076e23
E_FIS = 200e6*1.602176634e-19            # J per fission

sig_bp = 1e-44*1e-4                       # cm^2 -> m^2
n_p = 2*NA/18.0*1e6                       # free protons per m^3 of water
print(f"Bethe & Peierls 1934: sigma = 1e-44 cm2 = {sig_bp:.1e} m2")
print(f"   water holds {n_p:.2e} free protons/m3 -> mean free path "
      f"{1/(n_p*sig_bp):.1e} m = {1/(n_p*sig_bp)/9.4607e15:.0f} light years")

P, R = 1e9, 10.0
nu = P/E_FIS*6
print(f"1 GW reactor: {P/E_FIS:.2e} fissions/s x 6 nu = {nu:.2e} nu/s")
print(f"   flux at {R:.0f} m = {nu/(4*np.pi*R**2):.1e} /m2/s   (the book's 1e17)")
phi = 1e17
print(f"   1 m3 of water: rate = sigma*phi*N = {sig_bp*phi*n_p:.1e} /s"
      f" -> one every {1/(sig_bp*phi*n_p):.0f} s  (book: 100 s)")

sig, target = 1e-47, 1e-3                 # m^2 at 1 MeV; wanted rate in Hz
print(f"Reines' target estimate at E = 1 MeV, sigma = {sig:.0e} m2:")
print(f"   per-proton rate = {sig*phi:.0e} /s -> for {target:.0e} Hz need {target/(sig*phi):.1e} protons")
moles = target/(sig*phi)/(2*NA)
print(f"   = {moles:.0f} moles of water = {moles*18e-3:.0f} kg   "
      f"(the book rounds to 1000 moles, 18 kg)")
ideal = sig*(0.7e9/E_FIS*6/(4*np.pi*12**2))*(200e3/18*2*NA)*3600
print(f"   with the 200 kg actually used: {ideal:.0f} /hour ideal vs 3.0 +- 0.2 observed"
      f" -> efficiency {3.0/ideal*100:.0f} %")

nCl = 18.5e3*1e3/165.83*4*NA
print(f"Davis 1958: 18.5 t of C2Cl4 = {nCl:.1e} Cl atoms; 1 Ar per 1e30 Cl is {nCl/1e30:.1f} atoms to extract")

X = 13.5*100*7.874
print(f"Brookhaven filter: 13.5 m of Fe = {X:.0f} g/cm2")
print(f"   stops muons up to ~{X*1.45/1000:.0f} GeV by ionisation — and the AGS ran at 15 GeV")
print(f"   = {13.5/0.168:.0f} nuclear interaction lengths, so hadrons are gone many times over")
print(f"spark chambers: one 1.1x1.1 m2 x 2.5 cm Al plate = {110*110*2.5*2.70/1000:.0f} kg"
      f" -> 10 t is {10000/(110*110*2.5*2.70/1000):.0f} plates")
print(f"tau: m = 1777 MeV, c*tau = {2.99792458e8*290e-15*1e6:.0f} um")

prints

Bethe & Peierls 1934: sigma = 1e-44 cm2 = 1.0e-48 m2
 water holds 6.69e+28 free protons/m3 -> mean free path 1.5e+19 m = 1580 light years
1 GW reactor: 3.12e+19 fissions/s x 6 nu = 1.87e+20 nu/s
 flux at 10 m = 1.5e+17 /m2/s   (the book's 1e17)
 1 m3 of water: rate = sigma*phi*N = 6.7e-03 /s -> one every 149 s  (book: 100 s)
Reines' target estimate at E = 1 MeV, sigma = 1e-47 m2:
 per-proton rate = 1e-30 /s -> for 1e-03 Hz need 1.0e+27 protons
 = 830 moles of water = 15 kg   (the book rounds to 1000 moles, 18 kg)
 with the 200 kg actually used: 35 /hour ideal vs 3.0 +- 0.2 observed -> efficiency 9 %
Davis 1958: 18.5 t of C2Cl4 = 2.7e+29 Cl atoms; 1 Ar per 1e30 Cl is 0.3 atoms to extract
Brookhaven filter: 13.5 m of Fe = 10630 g/cm2
 stops muons up to ~15 GeV by ionisation — and the AGS ran at 15 GeV
 = 80 nuclear interaction lengths, so hadrons are gone many times over
spark chambers: one 1.1x1.1 m2 x 2.5 cm Al plate = 82 kg -> 10 t is 122 plates
tau: m = 1777 MeV, c*tau = 87 um

🔑 If you remember only three things

A conservation law seen to fail is a discovery either way. The energy missing from beta decay meant either the end of conservation or a new particle — and the second took twenty-six years to collect on.
When the background is orders of magnitude larger, the design problem is rejection. No amount of gain helps, because the noise is amplified with the signal.
The second neutrino was settled by one yes-or-no question. Do neutrinos made alongside muons ever produce electrons? Everything in that apparatus exists to make the answer unambiguous.

Where this goes next

§2.5 is the Dirac equation — the theory that predicted, for each of these fermions, an antiparticle.
Ch. 7 is the weak interaction that every process on this page runs on, and §7.4–7.5 is why neutrinos have only one helicity.
Ch. 10 is the sequel: the lepton flavours established here turn out not to be exactly conserved, and neutrinos are not massless after all.

✅ Check yourself — leptons, and the particle that was impossible to detect

0/6 answered · 0 correct

1.Thomson balanced an electric field against a magnetic one until the cathode-ray beam did not move. Why is that better than measuring the deflection?
2.Perl found the τ by looking for e±μ∓ events, which lepton-flavour conservation forbids. How is that not a contradiction?
3.Bethe and Peierls calculated correctly that a neutrino crosses a thousand light years of matter, and concluded detection was 'absolutely impossible'. Where does the reasoning fail?
4.Reines and Cowan required a prompt double pulse followed by a delayed one a few microseconds later. What does that design buy?
5.The Brookhaven neutrino beam used a 13.5 m iron filter. What sets that number?
6.Davis's 1958 chlorine experiment at Savannah River saw nothing, and it was calculated to be twenty times more sensitive than needed. What did the null result establish?