Quark Flavours and Baryonic Numbers; Leptonic Flavours

A lifetime limit quoted in years is really a statement about a population: 10³⁴ years means 10³⁴ protons watched for a decade with nothing happening to any of them.

🎯 Why this matters

Reading any of these bounds therefore requires knowing the exposure behind it. That is why every limit in this book arrives with a kt·yr or a count of muons attached — without one, the years are uninterpretable.

P, C and T were multiplicative. These two sections are about the additive quantum numbers — the ones you count rather than multiply — and about a distinction §3.1 drew and this page cashes in: electric charge is conserved because a gauge symmetry forces it, while baryon and lepton number are conserved only because nobody has ever seen them fail. That is not a philosophical difference. It is the reason a 50 000-tonne detector sits a kilometre under a Japanese mountain, watching.

3.6 Baryon number, and the search for its failure

\htmlClass{t-B}{\mathcal B} = \htmlClass{t-Nb}{N(\text{baryons})} - \htmlClass{t-Na}{N(\text{antibaryons})}

(3.28)

Bettini p. 116. An additive count, like electric charge — a system's baryon number is the sum over its parts.

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 — everything conserves baryon number and nothing explains why

Within the limits of experiment, every known interaction conserves baryon number . Nothing in the Standard Model explains why. There is no gauge symmetry behind $\mathcal B$ the way U(1) invariance stands behind electric charge, so its conservation is not a theorem — it is a very well-tested observation.

That gap has a consequence you can act on: if $\mathcal B$ is only accidentally conserved, it should eventually fail, and the cleanest place to look is the lightest baryon. A $p$ has nothing lighter to decay into while keeping $\mathcal B$ , so if it decays at all, baryon number is not conserved. The whole of this section is one experiment built on that sentence.

⚙️ Engineer’s bridge — a checked invariant versus one you merely hope for

§3.3 drew this distinction for CPT; here it is again, and the consequences are practical rather than philosophical.

Electric charge is guaranteed by the construction. U(1) gauge invariance makes conservation a theorem, so no experiment is designed to test it — a violation would mean the framework is wrong, not that a parameter is small.
Baryon number is an invariant nobody enforces. It holds in every process ever recorded, and there is no structural reason it must. In code terms it is a property that happens to be true of every execution you have logged, not one the type system prevents you from breaking.

The engineering response to an unenforced invariant is not to assume it — it is to instrument it, and to quote how hard you looked. That is precisely what a proton-decay limit is: a number saying “we watched $10^{34}$ protons for twenty years and the assertion never fired”.

And the payoff if it ever does fire is enormous, which is why the search continues. Grand unified theories predict exactly this failure, and the observed excess of matter over antimatter in the universe requires some baryon-number violation to have happened (Chapter 12).

Where it breaks: an invariant checked to $10^{34}$ years is not an invariant proved, and here the distinction has consequences. Baryon number is not protected by a gauge symmetry — it is accidental, holding because no renormalizable term in the Standard Model violates it — so the theory itself offers no reason for it to be exact, and cosmology positively requires that it was violated at some point (ch12). An assertion checked very hard is still an assertion, and this one is expected to fail.

The channel searched hardest is

\htmlClass{t-p}{p} \to \htmlClass{t-e}{e^+} + \htmlClass{t-pi}{\pi^0}

(3.29)

Bettini p. 116. The most plausible channel, and the one with the cleanest signature in a water Cherenkov detector.

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 the detector must resolve, in numbers

The kinematics is completely fixed, because the proton is at rest and the final state has two bodies. In the proton rest frame each carries $p^* = 459.4$ MeV/c, the total energy is $938.27$ MeV by construction, and the π⁰ flies with $\gamma = 3.55$ — so its two photons open by at least $2\arcsin(1/\gamma) = 32.7°$ .

Those three numbers are the three cuts:

total energy = the proton mass, which no atmospheric-neutrino interaction reproduces except by coincidence;
two of the rings must reconstruct to $m_{\pi^0}$ ;
the rings must be separated, and a 33° minimum opening is comfortably more than the ring-fitting resolution.

For scale, the Cherenkov cone in water ( $n = 1.333$ ) opens at $41.4°$ for anything ultra-relativistic, and the momentum thresholds are 0.6 MeV/c for an electron, 120 MeV/c for a muon and 158 MeV/c for a pion — which is why a water detector sees the positron and the photon-induced electrons and never sees a slow pion at all.

From a mass of water to a lifetime limit

The book runs the arithmetic once, for Super-Kamiokande: 50 000 t of water 1000 m under the Japanese Alps, of which the central, best-shielded fiducial mass is 22 500 t.

\htmlClass{t-Np}{N_p} = \htmlClass{t-M}{M}\times 10^3 \times \htmlClass{t-NA}{N_A}\left(\htmlClass{t-frac}{10/18}\right) = 2.25\times10^7 \times 10^3 \times 6\times10^{23}\,(10/18) = 7.5\times10^{33}

Bettini p. 117. Four factors, and the last one is doing two jobs at once.

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 — 10/18 means two things at once, which is why it is worth pausing on

The number 10/18 appears once and means two things at the same time, which is why it is worth pausing on. In H₂O the molar mass is 18 g/mol and there are 10 protons per molecule (2 from the hydrogens, 8 from the oxygen). So

\frac{N_A}{18\ \text{g/mol}} \times 10\ \frac{\text{protons}}{\text{molecule}} = 3.35\times10^{23}\ \text{protons per gram}.

Water is a good target for exactly this reason: it is cheap, transparent, and more than half its nucleons are free or nearly-free protons. The same coincidence — 10 protons out of 18 nucleons — makes the proton fraction 56 %, which is much better than any heavy material would give.

Multiply by the fiducial mass and you have $N_p$ . Multiply by the running time and you have the exposure , which is the figure of merit for any rare search: 450 kt·yr, or $1.5\times10^{35}$ proton-years.

Rare-process search · mass → protons → exposure → limit

fiducial mass M = 22.5 ktrunning time Δt = 20 yrdetection efficiency ε = 44 %background rate = 0.002 events / kt·yr

1	protons in the fiducial volume	7.53 × 10³³	water is 10 protons per 18 g, so N_p = M × (N_A/18) × 10
2	exposure M Δt	450 kt·yr	the figure of merit — mass and time are interchangeable
3	protons-years N Δt	1.51 × 10³⁵	the number that actually enters the limit
4	expected background b	0.90	small — the search is still background-free
5	90 % CL limit on the signal	2.30	the Poisson rule of 2.3 for zero observed events
6	lifetime limit τ/B	2.88 × 10³⁴ yr	N Δt ε / s, at 90 % confidence

with background, 0.002 events / kt·yr
background-free ideal (∝ exposure)

The two regimes. While the expected background stays below a few events the limit grows in proportion to the exposure — the dashed line. Once background dominates it grows only as its square root, and the curve bends away permanently. Doubling the run time then buys 40 %, not 100 %.

The signal limit is approximated as s ≈ max(2.30, 1.28√b): the Poisson rule of 2.3 for zero observed events, crossing over to the Gaussian one-sided 90 % point once the background dominates. On the Super-K preset this gives 2.9 × 10³⁴ yr against the published 2.4 × 10³⁴ — the real analysis is a likelihood fit carrying systematic uncertainties, so read this as the arithmetic of a search, not as a reproduction of one.

⚙️ Engineer’s bridge — the moment a search stops paying linearly

The plot in that widget carries the most useful idea in this section, and it is not a physics idea.

While a search is background-free, its reach grows in proportion to the exposure. Zero events observed means the 90 % limit on the signal is a constant — the Poisson rule of 2.3 — so the lifetime limit is just $N_p \Delta t\, \varepsilon / 2.3$ , linear in running time. Double the time, double the reach.

Once background dominates, the reach grows only as the square root. The expected background $b$ itself grows with exposure, and what you can exclude above a known background scales as $\sqrt b$ . The limit then goes as $\Delta t/\sqrt{\Delta t} = \sqrt{\Delta t}$ , and doubling the run buys 40 %.

You have met this exchange rate before — it is the same $\sqrt N$ that made KTeV need 272 times Samios’ events to gain a factor 16 in §3.5. The engineering consequence is sharp: spending on background rejection buys more than spending on mass, right up until the background is gone, and nothing after that. Which is why the section devotes its middle third to going 1000 m underground, defining a fiducial volume, and identifying event topologies — before it ever mentions how big the tank is.

Where it breaks: “stop paying for statistics and start paying for background rejection” is the right instinct and it has a floor no engineering removes. Below some rate the irreducible background is the signal channel of another process — atmospheric neutrinos mimicking proton decay — and no depth, fiducial cut or topology requirement eliminates it, because it is genuinely indistinguishable event by event. Past that point the only remaining lever is exposure again, which is why Hyper-K’s answer to Super-K is twenty times the mass rather than a cleverer cut.

What one decay would look like

three fuzzy rings · total energy 938 MeV · two of them reconstructing to 135 MeV

💡 What this really says — topology is the signal

No single ring says “ proton decay ”. The event topology does: how many rings, of which type, with what total energy and what invariant masses among them.

Ring type is the part worth knowing. In water, an electron or a photon showers and multiple-scatters, so its Cherenkov light arrives smeared over a range of angles and the ring is fuzzy; a muon travels in a straight line and leaves a sharp ring. All three rings of $p \to e^+\pi^0$ are e-like and fuzzy — the positron makes one directly, and each photon from the $\pi^0$ converts and showers into another. The dominant background, atmospheric-neutrino interactions, frequently produces a muon and therefore a sharp ring, so ring type alone removes a large fraction of it before any energy cut is applied.

This is the same idea as the §2.6–2.7 antiproton search, where no single measurement identified the particle and the pattern across four did. Against a background you cannot make smaller, add an independent handle.

🔬 Experiment card — Super-Kamiokande, Abe et al. 2017 / Takenaka 2020

Apparatus

50 000 t of ultrapure water in the Kamioka Observatory, about 1000 m below the Japanese Alps — the rock is the shield against cosmic rays. The inner surface is lined with photomultipliers; the innermost 22 500 t, where the background is lowest, is the fiducial mass. Nuclear radioactivity is irrelevant here because its spectra end at 10–15 MeV while the signal is at a GeV.

What is measured

Ring topology, then kinematics. Each ring’s radius and centre give the particle’s velocity, and the total photon count gives its energy. An event survives only if the rings are e-like, two of them reconstruct to

m_{\pi^0}

, and the total energy equals the proton mass.

The result

No event satisfying those conditions, over an exposure of 450 kt·yr —

1.5\times10^{35}

proton-years — with a calculated detection efficiency of 44 %.

What it proved

\tau \ge \mathcal{B}(p\to e^+\pi^0)\times 2.4\times10^{34}

yr at 90 % confidence: more than

10^{24}

times the age of the universe. Baryon number survives its most sensitive test. Somewhat weaker limits exist for

\mu^+\pi^0

and

K^+\nu

\htmlClass{t-tau}{\tau} \ge \htmlClass{t-BR}{\mathcal B(p\to e^+\pi^0)} \times \htmlClass{t-lim}{2.4\times10^{34}}\ \text{yr}

(3.30)

Bettini p. 117, at 90 % confidence. Note what the limit is on: not the lifetime, but the lifetime divided by an unknown branching ratio.

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.

🔢 Worked example — every number in the Super-K chain

The whole section is four multiplications and one division. Running them shows where the published limit does and does not come from the arithmetic.

Reproduce it

import numpy as np
NA = 6.022e23
M, yrs, eff = 22.5, 450 / 22.5, 0.44          # kt, yr, detection efficiency
g = M * 1e9                                    # 1 kt = 1e9 g
per_g = NA / 18 * 10                           # molecules per gram x 10 protons each
Np = g * per_g
print("Super-Kamiokande, p -> e+ pi0")
print(f"  fiducial mass          {M} kt  = {g:.3e} g of water")
print(f"  protons per gram       (N_A/18) x 10 = {per_g:.3e}")
print(f"  protons N_p            {Np:.2e}         (book: 7.5e33)")
print(f"  running time           450 kt yr / {M} kt = {yrs:.0f} yr")
print(f"  proton-years N_p dt    {Np*yrs:.2e}         (book: 1.5e35)")
print(f"  efficiency             {eff*100:.0f} %  (the pi0 is often lost to charge exchange in oxygen)")
naive = Np * yrs * eff / 2.303                 # 90% CL, zero events, no background
print(f"  zero-background limit  N_p dt eps / 2.303 = {naive:.2e} yr")
print(f"  published limit        2.40e+34 yr  -> ratio {2.4e34/naive:.2f},"
      f" i.e. a real background is expected")
print(f"  vs the age of the universe (1.38e10 yr): {2.4e34/1.38e10:.1e} times")

mp, me, mpi = 938.27208816, 0.51099895, 134.9768
ps = np.sqrt((mp**2-(me+mpi)**2)*(mp**2-(me-mpi)**2))/(2*mp)
Ee, Epi = np.hypot(ps, me), np.hypot(ps, mpi)
gam = Epi / mpi
print("kinematics of the decay at rest:")
print(f"  p* = {ps:.1f} MeV/c   E(e+) = {Ee:.1f}   E(pi0) = {Epi:.1f}   sum = {Ee+Epi:.2f} MeV = m_p")
print(f"  pi0 gamma = {gam:.3f} -> the two photons open by at least "
      f"{2*np.degrees(np.arcsin(1/gam)):.1f} deg")

prints

Super-Kamiokande, p -> e+ pi0
fiducial mass          22.5 kt  = 2.250e+10 g of water
protons per gram       (N_A/18) x 10 = 3.346e+23
protons N_p            7.53e+33         (book: 7.5e33)
running time           450 kt yr / 22.5 kt = 20 yr
proton-years N_p dt    1.51e+35         (book: 1.5e35)
efficiency             44 %  (the pi0 is often lost to charge exchange in oxygen)
zero-background limit  N_p dt eps / 2.303 = 2.88e+34 yr
published limit        2.40e+34 yr  -> ratio 0.83, i.e. a real background is expected
vs the age of the universe (1.38e10 yr): 1.7e+24 times
kinematics of the decay at rest:
p* = 459.4 MeV/c   E(e+) = 459.4   E(pi0) = 478.8   sum = 938.27 MeV = m_p
pi0 gamma = 3.548 -> the two photons open by at least 32.7 deg

The 44 % efficiency is the physically interesting number, and it is not an instrumental one. Most protons in the tank are bound inside an oxygen nucleus, so the π⁰ produced by a decay has to escape past the other nucleons — and if it charge-exchanges on the way out, the final state carries a π⁺ or π⁻ instead and the event no longer looks like (3.29). More than half of all decays would be lost this way. The target material sets the efficiency as much as the electronics does.

The flavour numbers, and a notation trap

Baryon number generalises: each quark type gets its own additive count.

\begin{aligned} \htmlClass{t-plain}{N_d} &= N(d) - N(\bar d), & \htmlClass{t-plain}{N_u} &= N(u) - N(\bar u) \\[2pt] \htmlClass{t-neg}{-S} &= N_s = N(s) - N(\bar s), & \htmlClass{t-pos}{C} &= N_c = N(c) - N(\bar c) \\[2pt] \htmlClass{t-neg}{-B} &= N_b = N(b) - N(\bar b), & \htmlClass{t-pos}{T} &= N_t = N(t) - N(\bar t) \end{aligned}

(3.31)

Bettini p. 118. Six counts, one per quark flavour — and three different sign conventions hiding in them.

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.

The six flavour numbers, their signs, and the symbols they collide with

quark	flavour number	value for the quark	the symbol also means…
$d$	down number N_d	+1	—
$u$	up number N_u	+1	—
$s$	strangeness S		spin (s), and total spin of a pair (s) in §3.3
$c$	charm C	+1
$b$	beauty B	−1
$t$	topness T	+1

Strong and electromagnetic interactions conserve every flavour number; the weak interaction violates them all. That single line is the engine of chapters 7 and 8: changing a quark's flavour is something only the weak interaction can do, and it is why heavy quarks decay at all.

3.7 Lepton number and the lepton flavours

The same construction, on the other half of the matter content.

Aside — the same script letter, twice

$\mathcal{L}$ below is the lepton number. On §3.1 the same script letter was the Lagrangian density, and it will be again from Chapter 5 onwards. The book uses one symbol for both and so does this site, because deviating from the book’s notation costs more than the collision does — but it is worth naming, since the two appear within twenty pages of each other.

Context separates them cleanly: the lepton number is an integer you count, and the Lagrangian is a function you vary. If a formula adds $\mathcal{L}$ to a quark flavour number , it is the count.

\begin{aligned} \htmlClass{t-L}{\mathcal L} &= N(\text{leptons}) - N(\text{antileptons}) \\[3pt] \htmlClass{t-Le}{\mathcal L_e} &= N(e^- + \nu_e) - N(e^+ + \bar\nu_e) \\[2pt] \htmlClass{t-Lmu}{\mathcal L_\mu} &= N(\mu^- + \nu_\mu) - N(\mu^+ + \bar\nu_\mu) \\[2pt] \htmlClass{t-Ltau}{\mathcal L_\tau} &= N(\tau^- + \nu_\tau) - N(\tau^+ + \bar\nu_\tau) \\[3pt] \htmlClass{t-sum}{\mathcal L} &= \mathcal L_e + \mathcal L_\mu + \mathcal L_\tau \end{aligned}

(3.32–3.36)

Bettini p. 118. One total and three flavours, with the charged lepton and its own neutrino counted together.

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 — four laws here, not one, and they are not equally solid

There are four laws here, not one, and they are not equally solid.

Total $\mathcal L$ is conserved in everything ever observed.
Each flavour $\mathcal L_e, \mathcal L_\mu, \mathcal L_\tau$ is conserved in every collision and every decay ever observed.
But neutrinos change flavour in flight, which violates the flavour laws without touching the total.

The Standard Model as written forbids all three flavour violations, so that last item is not a detail — the book calls it the only phenomenon so far observed in contradiction of the Standard Model. Everything else in this book is the Standard Model working.

What each interaction conserves — the answer to Problem 3.1

quantum number	strong	electromagnetic	weak
Iisospin
I_zthird component of isospin
Sstrangeness
flavourthe other quark flavours (C, B̃, T)
Bbaryon number
Llepton number (total)
L_e, L_μ, L_τlepton flavours
Pparity
Cparticle–antiparticle conjugation
Ttime reversal
Jangular momentum
J_zthird component of angular momentum
Qelectric charge

lepton flavours · weak · conserved, with one exception

Conserved in every collision and decay ever measured — but NOT by neutrinos in flight. Atmospheric ν_μ disappear over thousands of kilometres and solar ν_e arrive as something else. The book calls this the only observed contradiction of the Standard Model (§3.7, Ch. 10).

Y* marks a law that holds in every collision and decay yet fails somewhere else — lepton flavour fails for neutrinos in flight, and T fails only in the neutral-kaon system. Click any cell for the reason and the measurement behind it.

💡 What this really says — reading the Y* row

The table opens on the lepton-flavour row, and its three cells are marked Y* rather than Y: conserved in every collision and decay, and violated anyway, somewhere the word “reaction” does not reach.

That verdict is worth its own symbol precisely because a two-valued table would have to lie in one direction or the other. Writing N would suggest you can build a collider experiment that produces a muon and detects an electron; writing Y would hide the single most important open crack in the Standard Model. The violation happens in flight, over thousands of kilometres, to a particle that interacts with nothing on the way — a mode of failure the table’s rows and columns were never designed to express.

Testing the flavour laws: forbidden muon decays

\frac{\Gamma(\htmlClass{t-a}{\mu^- \to e^- + \gamma})}{\Gamma_{\text{tot}}} \le 4.2\times10^{-13}, \qquad \frac{\Gamma(\htmlClass{t-b}{\mu^- \to e^- + e^- + e^+})}{\Gamma_{\text{tot}}} \le 1.0\times10^{-12}

(3.37)

Bettini p. 118, from Workman et al. (2022). Both final states are kinematically wide open and both are simply never seen.

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.

🔢 Worked example — why $4.2\times10^{-13}$ is not about making muons

A branching-ratio limit inverts into a count, exactly as the C-violation limits did in §3.3: to reach $4.2\times10^{-13}$ you must watch of order $1/(4.2\times10^{-13}) = 2.4\times10^{12}$ muon decays.

That sounds enormous, and it is not. A modern continuous muon beam delivers $\sim 10^8$ muons per second, so

\frac{2.4\times10^{12}}{10^{8}\ \text{s}^{-1}} = 2.4\times10^{4}\ \text{s} \approx 6.6\ \text{hours}.

Six and a half hours of beam. The limit is therefore not set by how many muons anyone can make — it is set entirely by background rejection: a normal muon decay $\mu \to e\nu\bar\nu$ happening at the same instant as an unrelated photon fakes $\mu \to e\gamma$ perfectly well, and the rate of such accidental coincidences grows as the square of the beam intensity while the signal grows only linearly.

Reproduce it

lim, beam = 4.2e-13, 1e8                      # branching limit, muons per second
N = 1 / lim
print(f"mu -> e gamma limit {lim:.1e}")
print(f"   decays that must be watched: {N:.1e}")
print(f"   at {beam:.0e} muons/s that is {N/beam:.1e} s = {N/beam/3600:.1f} hours of beam")
print("   so the search is background-limited, not exposure-limited")
print("   accidentals ~ rate^2 while signal ~ rate: doubling the beam")
print("   doubles the signal and QUADRUPLES the accidental background")
print("proton decay, for contrast: 1.5e+35 proton-years and 20 years of running")
print("   -> exposure-limited, and no beam exists that could shorten it")

prints

mu -> e gamma limit 4.2e-13
 decays that must be watched: 2.4e+12
 at 1e+08 muons/s that is 2.4e+04 s = 6.6 hours of beam
 so the search is background-limited, not exposure-limited
 accidentals ~ rate^2 while signal ~ rate: doubling the beam
 doubles the signal and QUADRUPLES the accidental background
proton decay, for contrast: 1.5e+35 proton-years and 20 years of running
 -> exposure-limited, and no beam exists that could shorten it

Two rare searches, two opposite bottlenecks. Proton decay is exposure-limited: you cannot buy protons faster than by building a bigger tank and waiting. $\mu \to e\gamma$ is background-limited: the events are cheap and the difficulty is entirely in not being fooled. Recognising which of the two regimes you are in is the first question to ask of any rare-process measurement — and the widget above draws exactly that boundary.

The exception: neutrinos change flavour in flight

The lepton flavour laws hold in every collision and decay, and fail anyway. The book gives the two phenomena in which it was found:

The only observed contradiction of the Standard Model

source	what is observed	why it cannot be absorption
atmospheric — ν_μ made by cosmic rays in the air	the flux falls to about 50 % over several thousand km, i.e. after crossing part of the Earth	Neutrino cross-sections are far too small for the Earth to absorb half of them (§2.4 put the mean free path in light years). The missing half has turned into another flavour, mainly ν_τ.
solar — ν_e from thermonuclear reactions in the Sun	only about half leave the surface as ν_e, and less than that at some energies	Here the mechanism is different again: the ν_e interact coherently with the electrons of dense solar matter and are partially converted into a superposition of ν_μ and ν_τ. Energy dependence is the signature.

Both are propagation phenomena, not collisions or decays — which is exactly why the flavour laws can hold in every reaction ever measured and still fail. Chapter 10 is this subject, from the mixing matrix to the experiments.

Testing the total: two much rarer processes

Total lepton number has never been seen to fail. Two searches probe it directly.

Limits on total lepton-number violation

process	$\Delta \mathcal{L}$	limit	what it means
μ⁻ Ti → e⁺ Ca in a muonic atom	2	< 3.6 × 10⁻¹¹
neutrinoless double β decay, (Z, A) → (Z+2, A) + 2e⁻	2	T½ > 2.3 × 10²⁶ yr (¹³⁶Xe)	The most comprehensive test. Two neutrons turn into two protons and two electrons with NO neutrinos — possible only if the neutrino is its own antiparticle (§2.9, §10.7).
the same, in germanium	2	T½ > 1.8 × 10²⁶ yr (⁷⁶Ge)	A different isotope with different backgrounds and a comparable reach — which is how a null result in a field of one experiment becomes a null result in physics.

For scale: one tonne of ¹³⁶Xe contains 4.4 × 10²⁷ nuclei, so a half-life of 2.3 × 10²⁶ yr corresponds to about 13 decays per year — in a detector that must also reject everything else the universe does to a tonne of xenon.

Erratum — the target is titanium, not thallium

Page 119 prints the μ-to-e conversion limit as $\sigma(\mu^-\mathrm{Tl} \to e^+\mathrm{Ca})/\sigma(\mu^-\mathrm{Tl}\to\text{all})$ , with Tl — thallium, $Z = 81$ . It should be Ti, titanium, $Z = 22$ : the process converts a bound μ⁻ into an e⁺ while the nucleus loses two units of charge, and $22 - 2 = 20$ is calcium. Thallium would have to land on $Z = 79$ , gold, not calcium. The cited measurement (Kaulard et al. 1998, SINDRUM II) is on titanium. Nothing else in the argument is affected.

⚙️ Engineer’s bridge — a conserved difference outlives its parts

Both broken laws on this page break in a specific, correlated way, and the pattern is one you would recognise from any system of checksums.

$p \to e^+\pi^0$ has $\Delta\mathcal B = -1$ and $\Delta\mathcal L = -1$ . Each count fails, and their difference $\mathcal B - \mathcal L$ survives. That is not a coincidence of the channel — it is why this channel is the one everyone searches for. Grand unified theories put quarks and leptons in the same multiplet, so transitions between them are allowed; what such a theory conserves is not $\mathcal B$ and $\mathcal L$ separately but the combination that its symmetry protects.

The engineering reading: when a system has several redundant counters and you suspect one may not be enforced, the informative question is not “which counter holds?” but “which linear combination of counters is the one the architecture actually protects?” A single parity bit can fail while the total checksum holds; a per-field invariant can break while an aggregate stays exact.

Where it breaks: $\mathcal B - \mathcal L$ is also only conjectured, not derived. It is the best-motivated combination, not a guaranteed one — which is why the searches continue in several channels at once rather than in the theoretically favoured one alone.

Where it breaks: the argument tells you which combinations survive and nothing about the rate. B − L is conserved by everything in the Standard Model including sphalerons, so it is the robust label — but knowing a quantity is conserved in the theory you have gives no guidance about how fast the theory-you-do-not-have violates it. That is why the experimental programme searches several channels at once rather than concentrating on the one a favoured grand unified theory prefers.

🔑 If you remember only three things

A null result is a quantitative result. Seeing nothing, for long enough and in enough material, is the measurement — there is no other kind of answer this page can give.
The two halves test their laws in opposite postures. Baryon number is tested by waiting for something to happen; lepton flavour by making as many muons as possible and watching what they never do.
Some conservation laws come with a guarantee and some come with data. Charge is protected by a symmetry of the theory; baryon number is protected by nobody having seen it break.

Where this goes next

§3.8–3.10 is the last quantum number of the chapter, isospin — the one that is not merely counted but rotated, and the one that predicts cross-section ratios with no dynamics at all.
Chapter 4 is the flavour numbers put to work: charm, beauty and topness as the organising principle of the hadron tables.
Chapter 7 is why the weak interaction alone violates the flavour numbers, and by exactly how much — the quark mixing matrix.
Chapter 10 is the neutrino exception taken seriously: mixing, masses, oscillation in flight, and whether the neutrino is its own antiparticle.
Chapter 12 returns to baryon-number violation as one of the open questions, since the universe’s excess of matter over antimatter requires it to have happened at least once.

✅ Check yourself — baryon number, lepton number, and how to look for their failure

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1.Electric charge and baryon number are both additive and both conserved in every process ever recorded. Why does anyone spend twenty years watching for proton decay and nobody watches for charge non-conservation?
2.Super-Kamiokande's fiducial mass is 22 500 t out of a 50 000 t total. Why is the rest of the water not counted?
3.In the ExposureLab, push the background rate up until the red curve visibly bends away from the dashed one. What has changed about the value of running longer?
Hint: Compare how the limit scales with time on the two sides of the bend.
4.The limit on $\mu^- \to e^- \gamma$ is $4.2\times10^{-13}$ , which needs about $2.4\times10^{12}$ muon decays — roughly six hours of a modern muon beam. So what is actually limiting that measurement?
5.Lepton flavour is conserved in every collision and every decay ever measured, and the Standard Model is nonetheless contradicted. How?