The Elements of the Standard Model · Elementary Particle Physics

The three interactions carry charges with nothing structurally in common: one signed number, three colours that must cancel, and a doublet only left-handed states are allowed to join.

🎯 Why this matters

The tidiest row of the table is the one that is wrong. Neutrinos are listed by the flavour they are made and detected with, and those are not the states that travel — which is the whole reason lepton flavour fails to be conserved.

Four sections ago this chapter had a hundred hadrons and no explanation. It now has six quarks. This last section is the audit: what the census actually says, what the three interactions look like side by side, and which questions the Standard Model answers by not answering them.

Six flavours, one relation

The hadrons carry six additive quantum numbers, all of them called flavours : the two values of $I_z$ , then strangeness $S$ , charm $C$ , beauty $B$ and top $T$ . Every one is conserved by the strong and electromagnetic interactions and violated by the weak one — a pattern that has held since §2.2 and never once wavered.

Chapter 3 defined hypercharge with strangeness alone. Now it needs all of them.

\htmlClass{t-Y}{Y} = \htmlClass{t-B}{\mathcal{B}} + \htmlClass{t-S}{S} + C + B + T

(4.81)

Bettini p. 176, generalising Eq. (3.38). Every flavour a hadron carries adds to its hypercharge.

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.

\htmlClass{t-Q}{Q} = \htmlClass{t-Iz}{I_z} + \frac{Y}{2}

(4.82)

Bettini p. 176 — Gell-Mann–Nishijima, in its final form. Six flavours in, one charge out.

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.

⚙️ Engineer’s bridge — this is a property-based test, and this site runs it

Eq. (4.82) is not a formula you evaluate to get an answer. It is an invariant: a statement that must hold for every row of the particle table, forever, or something upstream is wrong.

That is exactly the shape of a property-based test. You do not check Gell-Mann–Nishijima on the example in the book and move on; you assert it over the whole dataset and let it fail loudly if anyone mistypes a strangeness. So this site does — npm run check recomputes $Y = \mathcal{B} + S + C + B + T$ and $I_z = Q - Y/2$ for every hadron in Appendix 3, and requires that $I_z$ come out an integer or half-integer and never exceed the multiplet’s $I$ . Sixty hadrons, zero failures, checked on every build.

Where the analogy breaks: a property-based test can only find inconsistencies in the data, not errors in the physics. If the whole table shared a wrong convention the check would pass happily. An invariant tells you that your bookkeeping is self-consistent, never that it is true — which is why the book derives the relation from isospin and hypercharge rather than fitting it.

🔢 Worked example — six for six, in exact fractions

Run Eq. (4.82) across every row of Table 4.5. Use exact rationals, because the whole point is that thirds cancel exactly rather than nearly.

Reproduce it

from fractions import Fraction as F

B = F(1, 3)                                  # baryon number of any quark
#         name    Q        I       Iz       S   C   B'  T
rows = [("d", F(-1,3), F(1,2), F(-1,2),  0,  0,  0,  0),
        ("u", F( 2,3), F(1,2), F( 1,2),  0,  0,  0,  0),
        ("s", F(-1,3),      0,       0, -1,  0,  0,  0),
        ("c", F( 2,3),      0,       0,  0, +1,  0,  0),
        ("b", F(-1,3),      0,       0,  0,  0, -1,  0),
        ("t", F( 2,3),      0,       0,  0,  0,  0, +1)]

print("  q    Q      Iz     Y = B+S+C+B'+T    Iz + Y/2    Q ?")
for n, Q, I, Iz, S, C, Bf, T in rows:
    Y = B + S + C + Bf + T
    lhs = F(Iz) + Y/2
    print(f"  {n}  {str(Q):>5}  {str(Iz):>5}   {str(Y):>6}          "
          f"{str(lhs):>5}     {'OK' if lhs == Q else 'FAIL'}")
print("  six for six -- and the same relation holds for every hadron they build")

prints

  q    Q      Iz     Y = B+S+C+B'+T    Iz + Y/2    Q ?
d   -1/3   -1/2      1/3           -1/3     OK
u    2/3    1/2      1/3            2/3     OK
s   -1/3      0     -2/3           -1/3     OK
c    2/3      0      4/3            2/3     OK
b   -1/3      0     -2/3           -1/3     OK
t    2/3      0      4/3            2/3     OK
six for six -- and the same relation holds for every hadron they build

Note what carries the weight in each row. For u and d, $Y$ is just $\mathcal{B}$ and all the structure is in $I_z$ . For s, c, b and t, $I_z$ is zero and all the structure is in $Y$ . The relation is the same one either way — which is the content of the phrase “a generalisation of (3.38)”.

Table 4.5 — quantum numbers and masses of the six quarks

quark	$Q$	$I$	$I_z$	$S$	$C$	$B$	$T$	$\mathcal{B}$	$Y$	mass
d	−1/3	1/2	−1/2	0	0	0	0	1/3	1/3	4.67 ⁺⁰·⁴⁸₋₀.₁₇ MeV
u	+2/3	1/2	+1/2	0	0	0	0	1/3	1/3
s	−1/3	0	0		0	0	0	1/3	−2/3	93.4 ⁺⁸·⁶₋₃.₄ MeV
c	+2/3	0	0	0	+1	0	0	1/3	4/3
b	−1/3	0	0	0	0	−1	0	1/3	−2/3	4.18 ⁺⁰·⁰³₋₀.₀₂ GeV
t	+2/3	0	0	0	0	0	+1	1/3	4/3

<strong>Only u and d have isospin.</strong> Every heavier quark is an isosinglet, so for them Gell-Mann–Nishijima reduces to Q = Y/2. The masses span a factor of 80 000 across the six, and the meaning of a quark mass — for a particle that is never free — needs Chapter 6.

Erratum — the uncertainty on the charm mass

Table 4.5 gives $m_c = 1.27 \pm 0.002$ GeV. That is a claim of two parts in a thousand, and it is not right: the 2022 Particle Data Group value is $1.27 \pm 0.02$ GeV, ten times looser. A zero has crept in.

You can see it is wrong without looking anything up, by reading down the table’s own uncertainty column. The d is known to about $+10\,\%/-4\,\%$ , the u to $+23\,\%/-12\,\%$ , the s to $+9\,\%/-4\,\%$ and the b to about $\pm 0.6\,\%$ . A charm mass ten times better determined than the bottom mass would be a strange inversion — heavier quarks are the better-determined ones, because their masses are large compared with the QCD scale that makes the definition ambiguous in the first place.

Three interactions, three completely different charge structures

The book anticipates the next five chapters with a summary, and it is worth reading as a comparison rather than as three paragraphs.

What each interaction is made of — the shape of Chapters 5 to 9 in one table

property	electromagnetic	strong	weak
the charge
can you predict attraction from the signs?	yes		—
mediators	1 photon	8 gluons	3: W⁺, W⁻, Z⁰
mediator mass	0	0
do the mediators carry the charge?
who feels it	everything electrically charged		every fundamental fermion, and the bosons
where it is done properly	Chapter 5 (QED)	Chapter 6 (QCD)	Chapters 7, 8, 9

Gravity is missing from this table on purpose: it has no charge in this sense and no established mediator, which is the difference Bettini draws before listing the other three. Chapter 11 is what it does have.

The three columns of the table above, drawn. The bottom line of each panel is the consequence, and the three consequences are completely different in kind: an infinite range from a neutral mediator, a short range from a massless but self-coupled one, and a short range from a heavy one. Only the third is the elementary “heavy mediator ⇒ short range” argument of §1.10.

The census

Counting everything

flav, col = 6, 3
quarks = flav * col
lep, nu = 3, 3
ferm = quarks + lep + nu

print(f"  {flav} flavours x {col} colours                = {quarks:2d} quarks")
print(f"  + {lep} charged leptons + {nu} neutrinos     =  {lep+nu} leptons  (no colour)")
print(f"  = {ferm} fundamental fermions, and {ferm} antifermions -> {2*ferm} in Table 4.6")

gauge = 8 + 1 + 2 + 1
print(f"\n  gauge bosons: 8 gluons + 1 photon + W+ W- + Z0 = {gauge}")
print(f"  + the Higgs                                    =  1")
print(f"  everything the Standard Model contains         = {2*ferm + gauge + 1}")

prints

  6 flavours x 3 colours                = 18 quarks
+ 3 charged leptons + 3 neutrinos     =  6 leptons  (no colour)
= 24 fundamental fermions, and 24 antifermions -> 48 in Table 4.6

gauge bosons: 8 gluons + 1 photon + W+ W- + Z0 = 12
+ the Higgs                                    =  1
everything the Standard Model contains         = 61

Press count the colour degree of freedom below. Each quark tile splits into three, and the familiar twelve-fermion chart becomes Bettini’s twenty-four.

🧩 The whole cast — 17 particles

generation I

generation II

generation III

gauge bosons

scalar

Click a tile for its properties. Turn on the force overlay and look at the third row: the neutrinos are the only particles that feel just one interaction — which is exactly why they pass through the Earth unhindered, and why detecting them takes a thousand tonnes of target.

Table 4.6 — the 24 fundamental fermions and their antiparticles; every column is a family

	family I	family II	family III	anti I	anti II	anti III
down-type, R	dᴿ	sᴿ	bᴿ	d̄ᴿ	s̄ᴿ	b̄ᴿ
down-type, G	dᴳ	sᴳ	bᴳ	d̄ᴳ	s̄ᴳ	b̄ᴳ
down-type, B	dᴮ	sᴮ	bᴮ	d̄ᴮ	s̄ᴮ	b̄ᴮ
up-type, R	uᴿ	cᴿ	tᴿ	ūᴿ	c̄ᴿ	t̄ᴿ
up-type, G	uᴳ	cᴳ	tᴳ	ūᴳ	c̄ᴳ	t̄ᴳ
up-type, B	uᴮ	cᴮ	tᴮ	ūᴮ	c̄ᴮ	t̄ᴮ
neutrino		ν_μ	ν_τ	ν̄_e	ν̄_μ	ν̄_τ
charged lepton	e⁻	μ⁻	τ⁻	e⁺	μ⁺	τ⁺

Eight rows × three families × two (particle / antiparticle) = 48 entries, of which 24 are the fermions. The colour label is written as a superscript here; Bettini prints it as a left apex. <strong>Leptons have no colour row triplication</strong> — that missing factor of three is the whole difference between a particle that feels the strong force and one that does not.

The Higgs boson completes the list. Its role in the Standard Model is to give mass — to the W and Z, to the quarks and to the charged leptons — through a symmetry-breaking mechanism that is Chapter 9’s subject. In the theory as written, neutrinos get none, and remain rigorously massless.

💡 What this really says — where the arbitrariness lives

Look at what the Standard Model fixes and what it merely records.

Fixed, and non-negotiable: the charge structure. Once you say there are three colours , the gluon count is 8 and not 9. Once you say the weak interaction acts on chirality, only left-chiral neutrinos can exist. Once quarks carry a third of a baryon number, Gell-Mann–Nishijima follows. None of this is adjustable.

Recorded, and unexplained: every mass, and the number three. Why three families rather than one or seventeen? Why does the top weigh 80 000 times what the up quark weighs, when they are structurally the same object? The theory takes those numbers as input, and this chapter’s central achievement — reducing a hundred hadrons to six quarks — has not reduced them at all. It has moved the arbitrariness from the spectrum into the parameter list, where it is smaller and much more visible.

That relocation is the honest description of what a good theory does, and it is also why Chapter 12 is called an epilogue rather than a conclusion.

up-type quarks
down-type quarks
charged leptons
neutrinos (upper limits)

The twelve fundamental fermions on one logarithmic mass axis, one family per row. Every row has the same four entries and the same charges; only the positions differ, and they differ by eleven orders of magnitude. Notice the ordering is not even tidy — the s quark outweighs the muon, and the b outweighs the tau. Neutrino points are drawn at an indicative 10⁻⁹ GeV: only upper limits exist, and the Standard Model as written says zero.

The one thing the structure does not explain

import math

m = {"nu (limit)": 1e-9, "e": 0.511e-3, "u": 2.16e-3, "d": 4.67e-3, "s": 93.4e-3,
     "mu": 105.66e-3, "c": 1.27, "tau": 1776.86e-3, "b": 4.18, "t": 172.69}
for k, v in sorted(m.items(), key=lambda kv: kv[1]):
    print(f"  {k:11s} {v:10.4g} GeV")

lo, hi = min(m.values()), max(m.values())
print(f"  span = {hi/lo:.1e}, i.e. {math.log10(hi/lo):.1f} orders of magnitude")
print("  three copies of one structure, differing only in this")

prints

  nu (limit)       1e-09 GeV
e             0.000511 GeV
u              0.00216 GeV
d              0.00467 GeV
s               0.0934 GeV
mu              0.1057 GeV
c                 1.27 GeV
tau              1.777 GeV
b                 4.18 GeV
t                172.7 GeV
span = 1.7e+11, i.e. 11.2 orders of magnitude
three copies of one structure, differing only in this

Two observations on neutrinos

The book closes the chapter with two warnings, and both are about the row of Table 4.6 that does not behave like the others.

⚠️ The neutrino row is not a row of stationary states

$\nu_e$ , $\nu_\mu$ and $\nu_\tau$ are the states of definite lepton flavour — the states the weak interaction produces and the states a detector identifies, because production and detection are both weak processes. Every other entry in Table 4.6 is also a state of definite mass. These are not.

The stationary states are $\nu_1$ , $\nu_2$ and $\nu_3$ : quantum superpositions of the three flavour states, with definite masses and no definite flavour at all. They cannot be assigned to a family, so the tidy three-column structure of Table 4.6 breaks down in exactly one row.

An engineer has met this before: the eigenbasis is not the measurement basis. A unitary change of basis relates them, the state propagates in the eigenbasis and is measured in the other, and the relative phases accumulated in flight turn into beats — the flavour you detect is not the flavour you made. That is Chapter 10, and it is the same mathematics as the K⁰ system of Chapter 8.

⚠️ Therefore lepton flavour number is not conserved

It follows immediately. If the state that propagates is not a state of definite flavour, the flavour you count at the end need not be the flavour you counted at the start, and $L_e$ , $L_\mu$ , $L_\tau$ are not separately conserved — a conservation law of §2.4 demoted by the last paragraph of the chapter.

The total lepton number $L$ is a different matter: no violation has ever been observed, and the Standard Model conserves it. But it cannot be completely excluded, and the stakes are unusually high, because $L$ is the only quantum number that distinguishes a neutrino from an antineutrino. Every other label they might differ by is zero for both. So if $L$ is violated, neutrino and antineutrino may be two states of one particle, described by the Majorana equation of §2.9 rather than the Dirac equation.

That is not the Standard Model’s assumption. Whether nature agrees is §10.7, and the experiment that would settle it — neutrinoless double beta decay — has been running, without a signal, for decades.

🔑 If you remember only three things

The relation holds in exact thirds. The fractions cancel exactly rather than nearly, which is what makes it a constraint on the table instead of a coincidence within it.
A hundred particles became six quarks and a set of rules. What replaced the count of hadrons is a count of free parameters, and that number is the honest measure of what is understood.
Three generations, and no reason for three. The pattern repeats exactly twice and then stops, and nothing anywhere in the theory says why it should.

Where this goes next

Chapter 5 starts the second half of the book: the same particles, now with dynamics. QED first, because it has the simplest charge structure and because the tools built there are used for all three interactions.
Chapter 6 is colour taken seriously — where the antisymmetric factor of §4.8 comes from, and where a proton keeps 99 % of its mass.
§9.7 is the experimental proof that there are exactly three families with a light neutrino, measured from the width of the Z⁰.
Chapter 10 is the neutrino row of Table 4.6, and what happens when a table entry turns out not to be a stationary state.

✅ Check yourself — the elements of the Standard Model

0/6 answered · 0 correct

1.Why does Gell-Mann–Nishijima, $Q = I_z + Y/2$ , still work once four more flavours exist?
2.The s quark has $S = -1$ while the c quark has $C = +1$ . Why the different signs?
3.Press 'count the colour degree of freedom' on the chart. Why does the fermion count go from 12 to 24?
4.Which line of the interactions table is the deepest difference between electromagnetism and the strong force?
5.The neutrino row of Table 4.6 does not behave like the other seven rows. In what way?
6.What is at stake in whether total lepton number $L$ is conserved?