The Quantum Numbers of the Charged Pion · Elementary Particle Physics

None of these three measurements built an instrument. The accelerator is the spectrometer, a coil of cable is the stopwatch, and a second experiment is what turns the first into a measurement of spin.

🎯 Why this matters

Every one of these tricks is still in use. Reaching for the machine you already have, delaying a signal so it arrives after the decision to record it, and forming a ratio so that what you cannot calculate cancels — none has been improved on, only re-implemented.

Discovering a particle is the easy half. Once it exists you have to measure everything about it — mass, lifetime, charge, spin, parity, branching ratios — and each of those is a separate experiment with its own idea in it. The book picks the charged $\pi^\pm$ and does three of them properly. They are worth studying not for the answers, which you already know, but because each measurement is a different trick for getting at something you cannot observe directly.

The mass: let the accelerator be the spectrometer

Mass is never measured. You measure two of the six observable quantities and combine them through $m^2 = E^2 - p^2$ (§1.13a). Burfening and collaborators (1951) chose momentum and energy, and got both from a single emulsion track.

🔬 Experiment card — the pion mass, Berkeley 184-inch cyclotron, 1951

Apparatus

No apparatus was built. A small target sits inside the cyclotron vacuum chamber, clipped by 380 MeV α particles on their final orbit; a copper shield stops everything else; two emulsion stacks lie below the orbit plane, one on each side, because the machine’s own field sends the two charge signs opposite ways.

What is measured

Two quantities on the same track. Momentum from geometry — the entry point and direction in the emulsion, plus the known target position and known field, fix the trajectory. Energy from range — the same track then stops, and its length is its energy through Bethe–Bloch (§1.11).

The result

m_{\pi^-} = 141.5 \pm 0.6

MeV and

m_{\pi^+} = 140.8 \pm 0.7

MeV (Eq. 2.5), against today’s

139.570\,39 \pm 0.000\,18

MeV (Eq. 2.6).

What it proved

The two charge states have the same mass — they differ by

0.8\sigma

, which is the CPT test, and it passes. Both sit high of the modern value, by

+3.2\sigma

and

+1.8\sigma

: the quoted errors were statistical, and the real limit was the emulsion range–energy calibration. A first measurement, behaving like one.

🛠️ Fig. 2.4 — the pion mass measurement inside the 184-inch cyclotron

Click a numbered marker for what that piece does.

Redrawn from Fig. 2.4 (p. 77). The elegance is that nothing was built: a target, a copper shield and two stacks of film, dropped into a machine designed for something else.

The result, and the modern value:

m_{\pi^-} = 141.5 \pm 0.6\ \text{MeV}, \qquad m_{\pi^+} = 140.8 \pm 0.7\ \text{MeV} \tag{2.5}

m_{\pi^\pm} = 139.570\,39 \pm 0.000\,18\ \text{MeV} \tag{2.6}

📏 What the 1951 numbers actually tell you

The book says the two values are “equal within the errors”, and they are: they differ by $0.7 \pm 0.9$ MeV, which is $0.8\sigma$ . That is the interesting comparison, because CPT requires them to be exactly equal (§2.1–2.2) — a particle and its antiparticle cannot have different masses. The measurement is a test of CPT, and it passes.

But compare them with the modern value instead and something else appears:

	1951	modern	discrepancy
$m_{\pi^-}$	141.5 ± 0.6	139.5704	+3.2σ
$m_{\pi^+}$	140.8 ± 0.7	139.5704	+1.8σ

Both are high, and one of them badly so. The quoted errors are statistical; the real limit was a systematic one — the range–energy relation for emulsion, which had to be calibrated and was not yet well known. This is the ordinary condition of a first measurement, and it is why “3σ” from a single experiment has never been enough to claim anything in this field.

The lifetime: build a stopwatch out of a delay line

26 nanoseconds is a hard interval to measure in 1950. Chamberlain and collaborators did it with scintillators, a coincidence circuit and an oscilloscope that was photographed.

🔬 Experiment card — the charged-pion lifetime, 1950

Apparatus

A 340 MeV photon beam on paraffin makes pions through

\gamma + p \to \pi^+ + n

. Two scintillators in coincidence demand that the pion cross the first and stop in the second. That coincidence starts an oscilloscope sweep; 0.5 μs of coaxial cable delays the counter signal until the sweep is running; a 0.5–2.5 μs gate lights a lamp beside the screen, and one camera photographs waveform and lamp together.

What is measured

The interval between two pulses from the same counter — the pion arriving, and the pion decaying — read off a photographed trace. The lamp in the frame is the validity bit: it says the muon decayed later too, so the stopped particle really was a pion.

The result

\tau = 26.5 \pm 1.2

ns from 554 events. Two pulses were separable only beyond 22 ns, so 56 % of the decays were unusable — and the fit did not care, because a truncated exponential keeps its slope.

What it proved

The charged pion lives 26 ns, which on the

10^{-23}

s scale of the strong interaction is an eternity: its decay cannot be strong. The value sits

0.4\sigma

from today’s

26.033

ns, and the error was pure statistics — they were fighting run time, not apparatus.

🛠️ Fig. 2.5 — Chamberlain's pion-lifetime experiment as a signal chain

Click a numbered marker for what that piece does.

Redrawn from Fig. 2.5 (p. 78). Read left to right it is a signal chain: source, sensor, trigger, delay, gate, display, record.

⚙️ Engineer’s bridge — the delay line, and why every scope still has one

The interesting element in that chain is 0.5 μs of cable, and the reason it is there is a genuine causality problem.

The scope sweep is triggered by the very pulse you want to look at. By the time the trigger circuit has decided, the pulse has already passed the display. So you must delay the signal relative to its own trigger — and in 1950 that means physically routing it through 150 m of cable, because a metre of coax buys about 5 ns.

Every oscilloscope since has the same block. Analogue scopes had a literal delay line in the vertical amplifier; a digital scope replaces it with a circular buffer that is always recording, so that when the trigger fires the samples before it are still in memory — which is why “pre-trigger” is a knob you can set to a negative time. Same problem, same solution, different technology:

Chamberlain, 1950	a modern DSO
0.5 μs coaxial delay line	circular sample buffer
fast coincidence starts the sweep	trigger comparator arms the capture
gate lamp beside the screen	per-event flags in the record header
camera photographs screen + lamp	waveform written to file with metadata

The physics moved on; the architecture did not.

Where it breaks: the architecture survived; the guarantees did not. A scope’s delay line is lossless and its trigger deterministic, so what you see is what arrived. Here the “record” is a photographic emulsion with finite grain and fading, the trigger is a coincidence with real inefficiency, and — the part with no software analogue — an event cannot be re-taken. You do not get to re-run the input with the trigger widened. Every decision about what to record is made before you know what you threw away, which is why the whole design effort goes into the trigger rather than into the analysis.

🔢 Worked example — 554 events, and what they can and cannot tell you

The error is pure statistics. For an exponential, the fractional error on the lifetime from $N$ events is $1/\sqrt N$ . With $N = 554$ that is 4.25 %, so

\delta\tau = \frac{26.5\ \text{ns}}{\sqrt{554}} = 1.13\ \text{ns},

against the published $\pm 1.2$ ns. The experiment was statistics-limited — they were not fighting their apparatus, they were fighting the run time.

The resolution problem is more interesting. Two pulses on the film were only separable if they were more than 22 ns apart. On a 26.5 ns lifetime that means

e^{-22/26.5} = 0.44,

so 56 % of the decays were unusable. Yet the measurement still works, and the reason is worth being explicit about: a truncated exponential still has the right shape. Fitting $e^{-t/\tau}$ over $t > 22$ ns determines $\tau$ from the slope, and the missing normalisation never enters. You have thrown away more than half the data and lost none of the information about the quantity you wanted.

That is a general and useful fact. Efficiency that is constant in the fitted variable cancels out of a shape fit. What would kill you is an efficiency that varies with $t$ — and here it does not, above the cut.

true distribution, e^(−t/26.5 ns)
resolvable: t > 22 ns (44 % of decays)

On a log axis an exponential is a straight line, and its slope is −1/τ. The cut removes the left-hand 56 % of the events and leaves the slope untouched — which is why a measurement with a resolution comparable to the quantity being measured still works. The published value, 26.5 ± 1.2 ns, is within 0.4σ of today's 26.033 ± 0.0005 ns.

The spin: measure a reaction, and then measure it backwards

The third measurement is the cleverest, and it is the one whose idea is worth carrying forward. You cannot see a spin. What you can see is that a reaction’s probability depends on how many spin states are available — and that dependence can be isolated by comparing a reaction with its own inverse.

🔬 Experiment card — the pion spin from π⁺d ⇌ pp, 1951

Apparatus

Two separate experiments, and the pairing is the instrument. One absorbs pions of

T_\pi = 24

MeV on deuterium,

\pi^+ d \to pp

; the other produces them with protons of

T_p = 341

MeV,

pp \to \pi^+ d

. Nothing links them except the requirement that both sit at the same CM energy —

\sqrt s = 2037

and 2040 MeV, 0.12 % apart.

What is measured

The two integrated cross-sections, and nothing else. The CM momenta

p^*_\pi = 78.5

MeV/c and

p^*_p = 396.7

MeV/c are not measured at all — they follow from the masses and

\sqrt s

The result

A ratio of

17.6 \pm 32\ \%

, against a kinematic factor

(p_p/p_\pi)^2 = 25.5

. Inverting Eq. (2.15) gives

2s_\pi + 1 = 0.97 \pm 0.31

What it proved

The charged pion has spin 0 — six standard deviations from the 3 that spin 1 would demand. And the strong-interaction matrix element, incalculable then and hard now, never had to be known: detailed balance made it cancel.

📐 Physics you need first — why 2s+1 appears in a rate

A particle of spin $s$ has $2s+1$ internal states (the possible values of the spin projection). Nothing in a cross-section distinguishes them, so:

Final states are summed over. Every accessible spin configuration is a separate way for the reaction to end, and they all count. More final spin states means a bigger cross-section.
Initial states are averaged over. You did not prepare a particular spin, so the cross-section you measure is the mean over the $(2s_a+1)(2s_b+1)$ possible initial configurations. More initial spin states means a smaller measured cross-section, because the same total is divided among more cases.

That asymmetry — sum the final, average the initial — is why the multiplicities appear upside down on the two sides of Eq. (2.12), and it is the entire content of the spin measurement.

The other ingredient is detailed balance , a consequence of time-reversal invariance (which the strong interaction respects, Ch. 3): the summed squared matrix element is the same forwards and backwards, $\sum |M_{fi}|^2 = \sum |M_{if}|^2$ . The dynamics is symmetric even though the counting is not.

\frac{d\sigma}{d\Omega}(a + b \to c + d) \propto \frac{\htmlClass{t-pf}{p_f}}{\htmlClass{t-pi}{p_i}}\;\frac{1}{\htmlClass{t-av}{(2s_a+1)(2s_b+1)}}\sum_{f,i}\left|\htmlClass{t-M}{M_{fi}}\right|^2

(2.12)

Eq. (2.12), from the golden rule of §1.6–1.7 with the common factors dropped. Everything hard is in the last term — which is exactly what the ratio removes.

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 — what Eq. (2.12) is actually asserting

A cross-section is how many ways the reaction can finish, divided by how many ways it could have started, times how hard each way is. Read the three factors in that order:

$p_f/p_i$ — how many ways to finish, in momentum space. This is the density of final states from the golden rule (§1.6–1.7), with the incident flux divided out.
$1/[(2s_a+1)(2s_b+1)]$ — how many ways it could have started. You did not prepare a spin state, so what you measure is the average over all of them.
$\sum|M_{fi}|^2$ — how hard each way is, summed over everything you did not observe. All of the physics is here, and none of it is known.

The measurement works because the first two factors are pure counting and the third is the same in both directions. Divide one reaction by its inverse and what is left is arithmetic.

🪜 From two measured cross-sections to a spin

Step 1 of 6 — Pick a reaction and its inverse(2.10, 2.11)

\pi^+ + d \to p + p \qquad\text{and}\qquad p + p \to \pi^+ + d

Why you may do this: Both must be measurable, and they must be run at the SAME CM energy — otherwise the matrix elements are not the same object and nothing cancels.

The absorption was measured at T_π = 24 MeV, the production at T_p = 341 MeV. Those look unrelated until you compute √s for each: 2037 and 2040 MeV, agreeing to 0.12 %.

Bettini Eqs. (2.10)–(2.15). Notice that the only physics input beyond kinematics is time-reversal invariance; everything else is bookkeeping.

π⁺ kinetic energy in the absorption experiment24 MeVmeasured σ(π⁺d→pp) / σ(pp→π⁺d)17.6its relative error32 %

absorption  π⁺d → pp  at T_π = 24 MeV
production  pp → π⁺d   needs T_p = 335 MeV to reach the same √s
                              √s = 2037.4 MeV

CM momenta at that energy   p*_π = 78.5 MeV/c
                            p*_p = 396.7 MeV/c
kinematic factor      (p_p/p_π)² = 25.53

predicted ratio   s_π = 0 → 17.02   ← the answer
                  s_π = 1 → 5.67
                  s_π = 2 → 3.40

measured 17.6 ± 5.6  ⇒  2s_π + 1 = 0.97 ± 0.31

Consistent with 2s_π + 1 = 1, i.e. spin 0. Spin 1 would need 3, which is 6.6σ away. Note how much of the work the kinematics does: the momentum ratio alone is 25.5, so the three spin hypotheses are spread over a factor of 5.0 and are easy to tell apart.

if s_π = 0
if s_π = 1
if s_π = 2

The three curves are the same physics with three different spin assignments; they never cross, so a single measurement at any energy separates them. The rise with energy is pure kinematics — p_p grows much faster than p_π because the deuteron is heavy and soaks up momentum.

Eq. (2.15): σ(π⁺d→pp)/σ(pp→π⁺d) = (2s_p+1)² p_p² / [2(2s_π+1)(2s_d+1) p_π²]. With s_p = ½ and s_d = 1 the counting factors collapse to 2/[3(2s_π+1)], and the only other ingredient is a ratio of CM momenta you can compute exactly.

⚙️ Engineer’s bridge — the ratio that cancels what you cannot model

The unknown here is $|M|^2$ : all of the strong-interaction dynamics, incalculable in 1951 and difficult today. The method does not estimate it, bound it or model it. It arranges for it to appear identically in the numerator and the denominator, and then divides.

You have done this. It is a differential measurement, a bridge circuit, a ratiometric ADC reading, a common-mode rejection: whenever the quantity you can measure is contaminated by a large factor you cannot characterise, look for a second measurement contaminated by the same factor and take the ratio. The error on the ratio is then set only by the things that differ.

The conditions are the same too, and they are the demanding part:

the contaminating factor must be genuinely identical, which here means the two experiments must sit at the same $\sqrt s$ — hence the fuss about 24 MeV pions matching 341 MeV protons;
the difference you want must survive the division, which here it does, because the spin multiplicities enter the two directions asymmetrically.

Get either wrong and you have measured nothing very precisely.

Where it breaks: the cancellation needs both directions measured with the same apparatus and the same acceptance, which is why the comparison is made within one experiment rather than between two. It also assumes the two matrix elements are equal — that is detailed balance, which holds only if time reversal is a good symmetry. Chapter 3 shows T is violated, and §8.5 measures it. The violation is far too small to matter here, but the ratio is not free of assumptions: it trades a hard measurement for a symmetry, and the symmetry is only approximately true.

The neutral pion, briefly

The three pions, side by side

Particle	m (MeV)↕	τ↕	cτ↕	Dominant decay	Interaction
$\pi^\pm$	139.570 39	26.033 ns	7.80 m	μ⁺ν_μ (99.99 %)
$\pi^0$	134.976 8	8.43 × 10⁻¹⁷ s		γγ (99.8 %)	electromagnetic

The mass difference is only 4.59 MeV — under 3.3 % — and the lifetimes differ by a factor of <strong>3 × 10⁸</strong>. Nothing about the masses explains that; it is entirely which interaction is allowed to run the decay. The π⁰ is its own antiparticle and can go to two photons; the π± must change a quark flavour and cannot.

Reproduce it

import numpy as np
mpi, mpi0, mp, md = 139.57039, 134.9768, 938.27209, 1875.61294
pstar = lambda M, a, b: np.sqrt((M**2-(a+b)**2)*(M**2-(a-b)**2))/(2*M)

print("Burfening 1951 vs modern:")
for n, v, e in (("pi-", 141.5, 0.6), ("pi+", 140.8, 0.7)):
    print(f"   m({n}) = {v} +- {e} -> {(v-mpi)/e:+.1f} sigma from {mpi:.4f}")
d, ed = abs(141.5-140.8), np.hypot(0.6, 0.7)
print(f"   the two 1951 values differ by {d:.1f} +- {ed:.2f} = {d/ed:.2f} sigma"
      f" -> equal, as CPT requires")

N, tau = 554, 26.5
print(f"lifetime: {N} events give a statistical error of {tau}/sqrt({N}) = {tau/np.sqrt(N):.2f} ns")
print(f"   published +- 1.2 ns, so the measurement was statistics-limited")
print(f"   22 ns two-pulse resolution on a {tau} ns lifetime -> only "
      f"{np.exp(-22/tau)*100:.0f} % of decays resolvable")
print(f"   {tau} +- 1.2 vs today's 26.033 -> {(tau-26.033)/1.2:.2f} sigma")

Tpi = 24.0
s1 = mpi**2 + md**2 + 2*(mpi+Tpi)*md
s2 = 2*mp**2 + 2*(mp+341.)*mp
rs = np.sqrt(s1)
ppi, pp = pstar(rs, mpi, md), pstar(rs, mp, mp)
print(f"spin, at T_pi = {Tpi:.0f} MeV:")
print(f"   sqrt(s) = {rs:.1f} MeV (absorption), {np.sqrt(s2):.1f} MeV (production at T_p = 341)"
      f" -> {abs(np.sqrt(s2)-rs)/rs*100:.2f} % apart")
print(f"   p*_pi = {ppi:.1f} MeV/c, p*_p = {pp:.1f} MeV/c, (p_p/p_pi)^2 = {(pp/ppi)**2:.2f}")
print("   predicted ratio: " + ", ".join(
    f"s={j} -> {2/(3*(2*j+1))*(pp/ppi)**2:.2f}" for j in (0, 1, 2)))
R = 17.6
print(f"   measured {R} +- 32 % -> 2s+1 = {2*(pp/ppi)**2/(3*R):.2f} "
      f"+- {2*(pp/ppi)**2/(3*R)*0.32:.2f}   (the book's value)")

print(f"pi+/pi0: mass difference {mpi-mpi0:.3f} MeV, "
      f"lifetime ratio {26.033e-9/8.43e-17:.2e}")

prints

Burfening 1951 vs modern:
 m(pi-) = 141.5 +- 0.6 -> +3.2 sigma from 139.5704
 m(pi+) = 140.8 +- 0.7 -> +1.8 sigma from 139.5704
 the two 1951 values differ by 0.7 +- 0.92 = 0.76 sigma -> equal, as CPT requires
lifetime: 554 events give a statistical error of 26.5/sqrt(554) = 1.13 ns
 published +- 1.2 ns, so the measurement was statistics-limited
 22 ns two-pulse resolution on a 26.5 ns lifetime -> only 44 % of decays resolvable
 26.5 +- 1.2 vs today's 26.033 -> 0.39 sigma
spin, at T_pi = 24 MeV:
 sqrt(s) = 2037.4 MeV (absorption), 2039.9 MeV (production at T_p = 341) -> 0.12 % apart
 p*_pi = 78.5 MeV/c, p*_p = 396.7 MeV/c, (p_p/p_pi)^2 = 25.53
 predicted ratio: s=0 -> 17.02, s=1 -> 5.67, s=2 -> 3.40
 measured 17.6 +- 32 % -> 2s+1 = 0.97 +- 0.31   (the book's value)
pi+/pi0: mass difference 4.594 MeV, lifetime ratio 3.09e+08

🔑 If you remember only three things

The instrument is an arrangement, not a device. Two reactions run in opposite directions give a spin that neither of them gives alone.
These measurements remove unknowns rather than compute them. Ratios, nulls and reversals all work by arranging for the incalculable quantity to cancel.
A first measurement’s real job is to be checkable. Two independent 1951 values of the pion mass differ by 0.8σ, which is why the pair is quoted and not the better one.

Where this goes next

§2.4 does the same job for the charged leptons and the neutrinos — and the neutrino measurements are harder by a factor no instrument on this page could have bridged.
§3.3–3.4 is time-reversal invariance, the assumption the spin measurement rests on, taken apart properly.
§4.1 reuses detailed balance on resonances, where the reaction and its inverse are the two halves of a single peak.

✅ Check yourself — three measurements, three tricks

0/6 answered · 0 correct

1.Burfening's experiment measured the pion mass with no spectrometer, no beamline and no separate magnet. How?
2.Chamberlain's chain contains 0.5 μs of coaxial cable between the second scintillator and the oscilloscope. Why is it there?
3.Two pulses on Chamberlain's film were only separable above 22 ns, and the lifetime is 26.5 ns — so 56 % of the decays were unusable. Why does the measurement still work?
4.In the detailed-balance argument, an extra factor of ½ appears in the absorption cross-section but not in the production one. What is it?
5.In the widget, the predicted ratio at T_π = 24 MeV is 17.0 for spin 0, 5.7 for spin 1 and 3.4 for spin 2 — but the kinematic factor (p_p/p_π)² alone is 25.5. What does that tell you about the method?
6.The π⁰ is only 4.6 MeV lighter than the π±, yet it lives 3 × 10⁸ times less long. Why?