The Positron; The Antiproton · Elementary Particle Physics

The same prediction took four years to confirm for the electron and twenty-three for the proton. The difference is not physics but how many wrong events you must discard to keep one right one.

🎯 Why this matters

Between the two discoveries lies the moment the field stopped waiting for nature to supply its particles and began commissioning them. The positron arrived in a cosmic-ray photograph; the antiproton was ordered, and a machine was built to fill the order.

§2.5 ended with a prediction nobody asked for: two of the four components of the Dirac bispinor describe a particle of the opposite charge — the positron and the antiproton . This page is the two experiments that found them — separated by twenty-three years, and by a complete change in what “an experiment” meant.

Anderson’s positron: four measurements on one photograph

Anderson built a cloud chamber $17 \times 17 \times 3$ cm³ in a field of up to 2 T and pointed it at cosmic rays, hunting for the $e^-$ ‘s mirror image. It had no trigger — the Rossi coincidence of §1.13b existed but he did not use it — so most photographs contained nothing at all.

🔬 Experiment card — Anderson, Caltech, 2 August 1932

Apparatus

A cloud chamber

17 \times 17 \times 3

cm³ in a vertical field of up to 2 T, with a 6 mm lead plate across the middle, pointed at cosmic rays. No trigger, and therefore only a few per cent of frames worth developing.

What is measured

Four quantities on one photograph: the curvature below the plate, the curvature above it, the droplet density along the track (that is

dE/dx

, §1.11), and the length of track after the plate. The plate is there to break a genuine degeneracy — a curved track alone cannot tell charge from direction of travel.

The result

p = 63

MeV/c below the plate and 23 MeV/c above it, so the particle was going up and is therefore positive; minimum-ionising throughout; and 50 mm of track after the plate, where a proton of 23 MeV/c would have stopped in about 5 mm.

What it proved

A positive particle with the electron’s mass to within 20 % — the positron, four years after Dirac’s equation predicted it. Twenty per cent was precise enough: the only competing hypothesis was 1836 times heavier.

🛠️ Anderson's chamber, and the plate that settled it

Click a numbered marker for what that piece does.

Redrawn from the description on p. 90 and Fig. 2.9. One chamber, one plate, one photograph — and four independent measurements on it: curvature twice, ionisation density, and range.

🪜 Anderson's chain of elimination

Step 1 of 6 — A positive track at minimum ionisation

Why you may do this: Droplet density per unit length is proportional to dE/dx (§1.11), and this track sits at the minimum. That already says something: a slow heavy particle would ionise far more.

In 1932 the only known positive particle was the proton. So the first question is whether it can be one.

Every step removes a hypothesis using a quantity already visible on the same photograph. Nothing was measured twice, and nothing was assumed.

⚙️ Engineer’s bridge — the lead plate is a sign convention made physical

The core difficulty is a genuine degeneracy: charge and direction of travel are not separately observable from a curved track. One measurement, two unknowns. Reverse the assumed direction and the inferred sign flips, and no amount of care in measuring the curvature helps.

Anderson’s solution is to inject a known irreversibility. Energy loss has an arrow — a particle always ends up with less than it started — so an absorber converts “which end came first” into “which end is more curved”, a quantity the photograph already contains.

You have done this. It is why a quadrature encoder has two channels offset by 90° rather than one: a single channel gives you speed and not direction, and the second one breaks the symmetry. It is why dither is added before quantisation, and why a pilot tone is inserted to recover a phase. When a measurement has a sign ambiguity, add a known asymmetry rather than a better sensor.

And the cost is visible: the plate also destroys 40 MeV of the thing you are measuring. Every symmetry-breaking probe perturbs its subject, and the design question is whether what you learn exceeds what you spoil. Here it does, easily.

Where it breaks: the trade is only acceptable because the quantity wanted is a sign. Anderson needed one bit — which end came first — and could pay 40 MeV for it. Run the same probe when you want the energy and it is fatal: the plate destroys precisely the quantity being measured, which is why the antiproton experiment of §2.7 could not use one and had to separate the sign with a spectrometer instead. “Perturb the subject to break the ambiguity” scales with how much of the answer the perturbation destroys, and that ratio is different for every measurement.

🔢 Worked example — a consistency check the book does not make

The particle entered the plate at 63 MeV/c and left at 23 MeV/c. Is that the right amount of energy for a 63 MeV electron to lose in 6 mm of lead?

Lead has $X_0 = 5.6$ mm (§1.13d), so the plate is

\frac{6\ \text{mm}}{5.6\ \text{mm}} = 1.07\ X_0 .

Above the critical energy — 7.3 MeV in lead, and 63 MeV is well above it — an electron loses energy predominantly by radiation, and its mean energy falls as $E(t) = E_0\,e^{-t/X_0}$ . So

E = 63\ \text{MeV} \times e^{-1.07} = 21.6\ \text{MeV},

against the measured 23 MeV — 6 % apart.

Two caveats worth stating. Bremsstrahlung is stochastic, so a single event can depart from the mean by a great deal; and the exponential law describes the mean of an ensemble, not one track. This is a consistency check, not a measurement. But it is the right consistency check, and a proton would have failed it by orders of magnitude — a 63 MeV/c proton has 2 MeV of kinetic energy and would not have crossed the plate at all.

Aside — Blackett and Occhialini, the same year, done properly

While Anderson was working, Blackett and Occhialini ran a Wilson chamber that was triggered by a Rossi coincidence of Geiger counters (§1.13b) and photographed by two cameras, so tracks could be reconstructed in space rather than in projection.

The consequence is visible in what they found. Instead of isolated positive tracks they saw pairs of opposite sign, both at minimum ionisation, originating from a common point — and by measuring curvature and droplet density on both, they got the electron mass for each. So they did not merely confirm the positron: they discovered pair production , $\gamma \to e^+e^-$ , the process that §1.11 makes the second half of every electromagnetic shower.

Anderson published first and shared the Nobel Prize. But the comparison is a fair illustration of what a trigger buys: Anderson’s untriggered chamber recorded a few per cent of useful frames, and Blackett’s recorded about 80 %.

The antiproton: twenty-three years of nothing

The positron arrived four years after Dirac’s equation. The antiproton took until 1955, and the delay is not an accident.

💡 What this really says — why the equation did not settle it

The Dirac equation predicts an antiparticle for a point-like spin-½ particle. The electron is one. The proton, as the book points out, is not:

its magnetic moment is not the $g = 2$ the equation demands — it is about 5.6;
and the neutron, which is electrically neutral, has a magnetic moment at all, which a point-like neutral Dirac particle could not.

Both anomalies say the same thing in retrospect: nucleons have internal structure (Ch. 6). At the time they simply meant the theoretical argument did not apply, so the question was genuinely open and had to be settled experimentally.

Cosmic-ray searches had failed — antiprotons exist there but are very rare — so what was needed was a machine.

The reaction is fixed by baryon-number conservation (§3.1–3.2): you cannot make one antiproton, you must make a proton with it.

p + p \to p + p + \bar p + p, \qquad E_p^{\text{thr}} = 7 m_p = 6.6\ \text{GeV}

That threshold — computed in Problem 1.9 — is why the Bevatron was built to 7 GeV. Chamberlain, Segrè, Wiegand and Ypsilantis extracted a 7.2 GeV beam onto a target in 1955, and then faced the real problem: only about one antiproton per 100 000 pions.

🔬 Experiment card — Chamberlain, Segrè, Wiegand and Ypsilantis, Bevatron, 1955

Apparatus

A 7.2 GeV proton beam on a target, feeding a two-stage spectrometer: dipole plus quadrupole imaging the target onto a slit that accepts negatives at

p = 1.19

GeV/c, then a second dipole and quadrupole. Downstream, two scintillators 12 m apart for time of flight, and two Cherenkov counters — C1 a threshold counter blind below

\beta = 0.99

, C2 a mirror counter that images the emission angle onto a photomultiplier.

What is measured

A mass, through

m = p/v \cdot \sqrt{1 - v^2/c^2}

(Eq. 2.53). The magnets fix

p

; everything after the slit is a velocity measurement, made three independent ways — time of flight, a threshold, and an angle.

The result

Flight times of 40.3 ns for pions and 51.0 ns for the candidates: 10.7 ns apart against ±1 ns resolution, about 11σ. Roughly fifty antiprotons, each tagged

(\text{not C1})\cdot\text{C2}

, in the slow peak.

What it proved

The antiproton exists, with the proton’s mass — so Dirac’s prediction survives for a particle that is not point-like (

g \neq 2

), which is exactly why the question had stayed open for twenty-three years. Nobel Prize, 1959.

\htmlClass{t-m}{m} = \frac{\htmlClass{t-p}{p}}{\htmlClass{t-v}{v}}\sqrt{1 - \frac{\htmlClass{t-v}{v}^2}{c^2}}

(2.53)

Eq. (2.53) — the same two-measurements-for-a-mass principle as §2.3's pion, and as every mass determination in the book. Fix p, measure v, and the mass follows.

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.

🛠️ Fig. 2.12 — the antiproton spectrometer, in two stages

Click a numbered marker for what that piece does.

Redrawn from Figs. 2.10–2.12 (pp. 93–94). Four independent handles on one particle: momentum from the magnets, velocity from time of flight, and velocity again — twice, differently — from the two Cherenkov counters.

⚙️ Engineer’s bridge — two optical ideas doing physics

Piccioni’s objection is about étendue. A dispersing element sorts by momentum, but the rate you get through a slit of width $d$ at distance $L$ goes as the solid angle $\sim (d/L)^2$ . Make the slit narrow enough for good resolution and the intensity collapses. Adding a focusing element does not violate anything — it re-shapes the accepted region of phase space rather than enlarging it — but re-shaping is exactly what you need: collect over a wide angle at the source, and deliver a narrow spot at the slit. This is the same trade a monochromator makes with its collimating mirrors, and the same reason a spectrograph has a camera lens and not just a grating.

C2’s mirror is a Fourier-transform lens. Cherenkov light comes off at an angle set by the speed, $\cos\theta = 1/\beta n$ , so the information you want is angular. A spherical mirror maps angle at the input to position in the focal plane — the standard optical Fourier transform — and once the angle has become a position, a stop or a well-placed photomultiplier is an angular filter, i.e. a velocity filter.

Put an illustrative $n = 1.33$ into the formula and the two species come out $40.8°$ and $16.8°$ apart. Twenty-four degrees is an enormous separation in a focal plane. Selecting on it is spatial filtering, and it is the same operation as picking a spatial frequency in an optical correlator or a bin in a filter bank.

Where it breaks: spatial filtering in optics is linear, lossless and reversible — the light you reject is still there, and you can put the filter back. Selecting particles on an angle is destructive: what misses the aperture is gone, and with it the statistics. That is why the antiproton search needed a beam intensity argument as well as an optics one — the selection that made the signal identifiable also threw away most of what it was selecting from, at a rate of roughly one antiproton per 50 000 pions.

💧 Cherenkov: cos θ = 1/βn

n = 1.33 → β must exceed 1/n = 0.751880, i.e. γ > 1.5168
maximum ring angle (β → 1): θ = 41.25°

particle	mass (MeV)	threshold E (MeV)	threshold p (MeV)	θ at p = 1 GeV
e	0.511000	0.8	0.6	41.25°
μ	105.658	160.3	120.5	40.88°
π	139.570	211.7	159.2	40.61°
K	493.677	748.8	563.0	33.02°
p	938.272	1423.1	1070.0	no light

Each curve starts abruptly at that particle's threshold and saturates at θ_max. The gaps between the curves are the separation power: pick n so that the particles you must distinguish sit on opposite sides of a threshold, or far apart in angle. Note the contrast with dE/dx in §1.11 — there the bands MERGE at high momentum, here they separate cleanly at low momentum and converge only well above threshold.

📏 Why two counters and a clock, for one measurement

Time of flight alone separates the two species by 11 σ. So why add two Cherenkov counters?

Because 11 σ of separation is not 10⁵ of rejection. The pions outnumber the antiprotons by a hundred thousand to one, and a Gaussian tail is not the enemy — random coincidences are. Two uncorrelated pions, one in S1 and a different one in S2, can produce an apparent 51 ns interval no matter how good your timing is, and the rate of such accidentals scales with the square of the beam intensity.

The counters attack that specifically. C1 says “not fast”; C2 says “slow”, by an independent mechanism. A random coincidence would have to be accompanied by the right Cherenkov pattern as well, and the probability of that is the product.

When your background exceeds your signal by orders of magnitude, add independent handles rather than sharpening one. Reines and Cowan reached the same conclusion by a different route in §2.4.

π⁻, tagged C1·C2
p̄, tagged (not C1)·C2

Fig. 2.13 redrawn — the two time-of-flight peaks, at the 40.3 ns and 51.0 ns computed from p = 1.19 GeV/c and the two masses, with the ±1 ns resolution as the width. About fifty antiprotons in the right-hand peak. Note that the peaks are labelled by the CHERENKOV tag, not by the timing: each of the two measurements confirms the other, which is the point of building both.

Reproduce it

import numpy as np
mp, mpi, me, c = 938.27209, 139.57039, 0.51099895, 2.99792458e8

B, X0 = 1.5, 5.6                                   # tesla, mm (lead)
print(f"Anderson, 2 August 1932 (B = {B} T):")
for label, p in (("before", 63.0), ("after ", 23.0)):
    print(f"   {label} the plate  p = {p:.0f} MeV -> radius {p/1000/(0.3*B)*100:4.1f} cm"
          + ("  (the chamber is 17 cm across)" if label == "before" else ""))
print(f"   as a proton:   T = {(np.hypot(23., mp)-mp)*1000:.0f} keV"
      f" -> stops in ~5 mm of gas, but 50 mm of track was seen")
print(f"   as a positron: T = {np.hypot(23., me)-me:.1f} MeV -> crosses the chamber easily")
t = 6.0/X0
print(f"   the 6 mm lead plate is {t:.2f} X0, so an electron should leave with")
print(f"      63 x exp(-{t:.2f}) = {63*np.exp(-t):.1f} MeV; measured 23 MeV,"
      f" {abs(63*np.exp(-t)-23)/23*100:.0f} % apart")

p, L = 1.19, 12.0                                  # GeV/c, m
print(f"antiproton spectrometer at p = {p} GeV/c:")
betas = {}
for name, m in (("pi  ", mpi/1000), ("pbar", mp/1000)):
    b = p/np.hypot(p, m); betas[name.strip()] = b
    print(f"   {name} : beta = {b:.4f}, time of flight over {L:.0f} m = {L/(b*c)*1e9:.1f} ns")
dt = L/(betas['pbar']*c)*1e9 - L/(betas['pi']*c)*1e9
print(f"   difference {dt:.1f} ns against +-1 ns resolution -> {dt:.0f} sigma")
print(f"   threshold for p p -> p p pbar p is 7 m_p = {7*mp/1000:.2f} GeV; the Bevatron reached 7.2")
print(f"   C1 must be blind above beta = 0.99, so n = {1/0.99:.4f}")
n = 1.33
a = [np.degrees(np.arccos(1/(n*betas[k]))) for k in ('pi', 'pbar')]
print(f"   C2 with n = {n}: pions radiate at {a[0]:.1f} deg, antiprotons at {a[1]:.1f} deg"
      f" ({a[0]-a[1]:.1f} deg apart)")

prints

Anderson, 2 August 1932 (B = 1.5 T):
 before the plate  p = 63 MeV -> radius 14.0 cm  (the chamber is 17 cm across)
 after  the plate  p = 23 MeV -> radius  5.1 cm
 as a proton:   T = 282 keV -> stops in ~5 mm of gas, but 50 mm of track was seen
 as a positron: T = 22.5 MeV -> crosses the chamber easily
 the 6 mm lead plate is 1.07 X0, so an electron should leave with
    63 x exp(-1.07) = 21.6 MeV; measured 23 MeV, 6 % apart
antiproton spectrometer at p = 1.19 GeV/c:
 pi   : beta = 0.9932, time of flight over 12 m = 40.3 ns
 pbar : beta = 0.7853, time of flight over 12 m = 51.0 ns
 difference 10.7 ns against +-1 ns resolution -> 11 sigma
 threshold for p p -> p p pbar p is 7 m_p = 6.57 GeV; the Bevatron reached 7.2
 C1 must be blind above beta = 0.99, so n = 1.0101
 C2 with n = 1.33: pions radiate at 40.8 deg, antiprotons at 16.8 deg (24.0 deg apart)

🔑 If you remember only three things

A threshold calculation became a machine specification. 7mₚ = 6.6 GeV is why the Bevatron was built to 7 — a kinematic invariant setting a budget.
Charge and direction of travel are one measurement, not two. A curved track cannot separate them, and the lead plate resolves the degeneracy by making the particle lose energy in a known direction.
Three velocity measurements were used where one would have sufficed. Independent handles, not a better single one, are what a rare-event search is built from.

Where this goes next

§2.8–2.9 returns to the Dirac bispinor and asks what its two halves really mean — helicity, chirality, and the possibility that a fermion is its own antiparticle.
§3.1–3.2 is baryon number, the conservation law that forced $p\bar p$ production in pairs and therefore set the Bevatron’s energy.
Ch. 6 is why the proton’s magnetic moment is not 2 — the structure that made the antiproton an open question in the first place.

✅ Check yourself — finding antimatter twice

0/6 answered · 0 correct

1.Why did Anderson need a lead plate across the middle of his chamber?
2.After the plate the track had p = 23 MeV/c and continued for 50 mm. Why does that rule out a proton?
3.The track lost momentum from 63 to 23 MeV/c crossing 6 mm of lead. Is that the right amount for an electron?
4.Piccioni pointed out that the first version of the spectrometer would not work. What was wrong with a dipole and a slit?
5.C2 has a low enough threshold that both pions and antiprotons radiate. How does it tell them apart?
6.Time of flight already separates pions from antiprotons by 11σ. Why add two Cherenkov counters as well?