§6.2Nucleon Structure

Part II Bettini pp. 232–239 · ~25 min read

  • deep inelastic scattering
  • Bjorken scaling
  • parton distribution function
  • momentum sum rule

Scaling was the surprise that revealed point-like constituents, and its slow failure is what proved the theory right. The same observation had to be true and then eventually false.

🎯 Why this matters

A quantity that stops being constant is not always a defeat. Here the rate at which scaling breaks is what measures the coupling, so the deviation carries more information than the effect that made the discovery.

§6.1 looked at quarks from the outside: make a pair and watch it turn into jets. This section looks from the inside — fire an electron into a proton and see what it bounces off.

It is Rutherford’s experiment, run sixty years later on a target a hundred thousand times smaller, and it produced the same kind of surprise: something hard and point-like inside something that had looked smooth.

📐 Physics you need first — elastic, inelastic, deep inelastic

Three regimes, distinguished by what happens to the target.

  • Elastic: the proton survives. epepep \to ep. One measured quantity (the angle) determines everything, and the cross-section falls steeply with Q2Q^2 because a proton is an extended object — that fall is its form factor.
  • Inelastic: the proton is excited to a resonance, epeΔep \to e\Delta, which then decays. Still a definite final state.
  • Deep inelastic (DIS): the proton is smashed into a hadronic mess of large invariant mass WW, and nobody looks at the mess. Only the scattered electron is measured. That is what “inclusive” means: epeXep \to eX, where XX is whatever it likes.

The last one is the useful one, and the reason is counter-intuitive: by throwing away information about the final state, you gain a measurement that depends only on the target’s internal structure and not on the messy business of what the fragments do afterwards. It is the same bargain §6.1 struck with jets.

The apparatus

🛠️ Fig. 6.8 — the SLAC spectrometers, and what each part is for
e⁻, energy EθX — never measuredthe 2-mile LINACup to 20 GeV, monochromatictargetliquid H₂ or D₂8 GeV spectrometervertical bending20 GeV spectrometerBrown–Richter opticsdetectorsmeasure E′detectorsmeasure E′1234

Click a numbered marker for what that piece does.

Friedman, Kendall and Taylor, 1967–1969; the Nobel Prize followed in 1990. Schematic.

🔬 Experiment card — Friedman, Kendall and Taylor, SLAC 1967–1969

Apparatus
A two-mile linear accelerator delivering a collimated, near-monochromatic electron beam of known energy up to 20 GeV; a liquid hydrogen target for the proton and liquid deuterium for the neutron; and two magnetic spectrometers on rails, of 8 and 20 GeV, that rotate about the target to select the scattering angle. Nothing about the hadronic debris is instrumented at all.

What is measured
Two numbers per event — the scattered electron’s energy EE' and its angle θ\theta — from which Q2=4EEsin2(θ/2)Q^2 = 4EE'\sin^2(\theta/2) and ν=EE\nu = E - E' follow. Each spectrometer setting samples one point of the (Q2,ν)(Q^2, \nu) plane, so the result is built by sweeping both over many runs. What is deliberately not measured is everything the proton turned into.

The result
In the deep inelastic region the cross-section divided by the point-like one is almost independent of Q2Q^2, and independent of WW as well. Set against the elastic cross-section on the same axes — which falls by four decades over the same range — the contrast is unmissable.

What it proved
That the proton contains hard, point-like constituents. A form factor is the Fourier transform of a charge distribution, so falling means extended and flat means point-like; the elastic curve in the same figure is the control proving the apparatus could have seen a size had there been one. Quarks stopped being a classification scheme and became objects — Gell-Mann had until then treated them as mathematical. The Nobel Prize followed in 1990.

Two invariants, from two measured numbers

Q2=4EEsin2 ⁣θ2,ν=EE,W2=mp2+2mpνQ2\htmlClass{t-Q}{Q^2} = 4E E' \sin^2\!\frac{\htmlClass{t-th}{\theta}}{2}, \qquad \htmlClass{t-nu}{\nu} = E - E', \qquad \htmlClass{t-W}{W^2} = m_p^2 + 2m_p\htmlClass{t-nu}{\nu} - \htmlClass{t-Q}{Q^2}
(6.11, 6.14, 6.12)

Bettini pp. 233–234. Everything in this section is built from these three, and all three come from the two numbers the spectrometer reports.

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 — a swept measurement, and choosing your axes

Each setting of the spectrometer — one angle, one momentum bite — returns one point. The structure function is the map you build by sweeping both and repeating. That much is ordinary experimental practice, and it is a frequency sweep in all but name.

The interesting part is the choice of coordinates. The raw knobs are EE' and θ\theta; the natural invariants are Q2Q^2 and ν\nu; the variable that makes the data collapse is x=Q2/2mpνx = Q^2/2m_p\nu. Same data, three parametrisations, and only the third one shows the structure.

This is the move that pays off everywhere in engineering: a measurement that looks like a two-dimensional mess in the variables you happened to control often collapses onto a single curve in the right dimensionless combination. Reynolds number does it for flow; L/EL/E does it for neutrino oscillations in §8.1; xx does it here. Finding that variable is the result — Bjorken got the scaling law named after him for proposing which combination to plot against, before anyone knew what was inside.

Where it breaks: choosing the variable is only a discovery if the variable is not already the answer. xx was defined from the assumption that the electron strikes a free point-like constituent — so plotting against it and finding scaling does not independently prove partons exist; it shows the data are consistent with the hypothesis the variable encodes. Had the proton been a smooth charge cloud, xx would have had no special meaning and no curve would have collapsed. A well-chosen abscissa is a hypothesis in disguise, which is what makes it powerful and what stops it being a proof.

what one spectrometer setting selects

import numpy as np
mp, E = 0.9382720882, 17.0            # proton mass, and the SLAC beam energy

print("  theta     E'    Q^2 GeV^2     nu     x       W GeV    regime")
for th, Ep in [(6, 15.5), (6, 12.0), (10, 12.0), (10, 6.0), (18, 6.0), (26, 4.0), (34, 2.5)]:
    t = np.radians(th); nu = E - Ep
    Q2 = 4 * E * Ep * np.sin(t / 2)**2      # Eq. (6.11)
    x = Q2 / (2 * mp * nu)                  # Eq. (6.16)
    W2 = mp**2 + 2 * mp * nu - Q2           # Eq. (6.12)
    W = np.sqrt(W2) if W2 > 0 else float('nan')
    reg = ("UNPHYSICAL x>1" if x > 1 else "quasi-elastic" if W < 1.3
           else "resonance region" if W < 2.0 else "deep inelastic")
    print(f"  {th:5.0f}  {Ep:6.1f}  {Q2:9.2f}  {nu:7.2f}  {x:6.3f}  {W:7.3f}   {reg}")
print()
print("x = 1 is the ELASTIC limit: there Q^2 = 2 m_p nu exactly and W = m_p.")
print("x > 1 is forbidden -- nothing inside a proton is heavier than the proton,")
print("so no constituent can absorb more momentum than the whole of it.")
prints
  theta     E'    Q^2 GeV^2     nu     x       W GeV    regime
    6    15.5       2.89     1.50   1.026    0.899   UNPHYSICAL x>1
    6    12.0       2.24     5.00   0.238    2.833   deep inelastic
   10    12.0       6.20     5.00   0.661    2.016   deep inelastic
   10     6.0       3.10    11.00   0.150    4.292   deep inelastic
   18     6.0       9.98    11.00   0.484    3.397   deep inelastic
   26     4.0      13.76    13.00   0.564    3.393   deep inelastic
   34     2.5      14.53    14.50   0.534    3.682   deep inelastic

x = 1 is the ELASTIC limit: there Q^2 = 2 m_p nu exactly and W = m_p.
x > 1 is forbidden -- nothing inside a proton is heavier than the proton,
so no constituent can absorb more momentum than the whole of it.

Scaling: the surprise

The cross-section from a target with structure is described by two structure functions W1(Q2,ν)W_1(Q^2,\nu) and W2(Q2,ν)W_2(Q^2,\nu) — one sensitive to the current distribution, one to the charge distribution. In the SLAC kinematics W1W_1 is negligible, so Eq. (6.15) reduces to

dσdΩdE=(dσdΩ) ⁣pointW2(Q2,ν)\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega\,\mathrm{d}E'} = \htmlClass{t-pt}{\left(\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}\right)_{\!\text{point}}} \htmlClass{t-w}{W_2(Q^2,\nu)}
(6.15 (reduced))

Deep inelastic scattering, factorised: a cross-section you can calculate exactly, multiplied by a function of two variables that carries everything you do not know about the target.

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.

Everything hangs on how W2W_2 behaves. And what SLAC found is that in the deep inelastic region it barely depends on Q2Q^2 at all.

024610⁻⁴10⁻³0.010.11Q² (GeV²)dσ/dQ² ÷ (dσ/dQ²)_point
  • elastic — a form factor, falling like Q⁻⁸
  • W = 2 GeV
  • W = 3 GeV
  • W = 3.5 GeV
Fig. 6.10, redrawn from the Friedman Nobel lecture. Note the vertical axis is logarithmic and spans four decades. The elastic curve falls off a cliff — that is what an extended charge distribution does. The deep inelastic points are almost flat. Whatever the electron is hitting in the second case has no form factor, which is to say no size.

💡 What this really says — why “flat in Q²” means “point-like”, in one sentence

A form factor is the Fourier transform of a charge distribution. The transform of a spread-out blob falls with Q2Q^2; the transform of a point is a constant.

So the elastic curve plunging by four decades says the proton as a whole is about a fermi across, and the deep-inelastic data sitting flat says that whatever the electron struck inside it has no measurable extent at all.

This is precisely Geiger and Marsden in 1909: most alpha particles went through, a few came straight back, and the ones that came back said there was something small and hard in there. Rutherford needed backward scattering; Friedman, Kendall and Taylor needed a cross-section that refused to fall. Both are the same signature — structure shows up as a departure from the point-like law, and none was seen.

Feynman’s reading: partons and x

Feynman’s move was to look at the proton in a frame where it moves very fast. Then the transverse momenta of its constituents are negligible, time dilation freezes their interactions with each other, and the electron scatters off one free constituent — the impulse approximation. He called them partons .

Let that parton carry a fraction xx Bjorken x — of the proton’s four-momentum. If it is massless and stays massless after being struck,

(xP+q)2=0    x=Q22Pq=Q22mpν\htmlClass{t-lhs}{(\htmlClass{t-x}{x}P + q)^2 = 0} \;\Longrightarrow\; \htmlClass{t-x}{x} = \frac{\htmlClass{t-q}{Q^2}}{2P\cdot q} = \frac{\htmlClass{t-q}{Q^2}}{2m_p\htmlClass{t-nu}{\nu}}
(6.16)

Bjorken x, derived in one line from a single assumption: the struck constituent was massless before the collision and is massless after it.

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.

which is Eq. (6.16). And now the payoff: if the parton is point-like, the structure function cannot depend on Q2Q^2 — only on xx. Define the dimensionless version, Eq. (6.17):

F2(x,Q2)=νW2(Q2,ν)\htmlClass{t-f}{F_2(x, Q^2)} = \htmlClass{t-nu}{\nu}\,\htmlClass{t-w}{W_2(Q^2,\nu)}
(6.17)

The dimensionless structure function. Multiplying by ν is not cosmetic: it is what makes the statement 'depends on x alone' a statement about physics rather than about units.

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.

and Bjorken’s scaling law says F2F_2 is a function of xx alone. The data agree, and that is why quarks stopped being bookkeeping and became objects.

⚠️ Three symbols mean two things each in this section

The book uses ν for the energy transfer EEE - E' throughout §6.2, and also writes ν and ν̄ for neutrino beams, in Eqs. (6.24)–(6.27), two pages later. They appear within a few lines of one another and the book does not flag the collision.

On this site the energy transfer is always written ν\nu in plain italic and the neutrino always as νμ\nu_\mu or νˉμ\bar\nu_\mu with its flavour subscript. If a symbol has no subscript in §6.2, it is an energy.

W is worse, with three meanings in play. Here WW is the invariant mass of the hadronic final state. On the same page W1W_1 and W2W_2 are the two structure functions — a completely unrelated object, and dimensionful where F2=νW2F_2 = \nu W_2 is not. And from Chapter 7 onwards W±W^\pm is the charged weak boson. The subscripted ones are always structure functions; a bare WW in this chapter is always a mass.

And g now has a fourth meaning. On this page g(x)g(x) is the gluon parton distribution. Elsewhere in the book gg has been the metric tensor (§5.1), the gyromagnetic ratio (§5.9a) and the strong charge gsg_s (§6.3). The rule §5.9a set still applies: g means whatever the current page says it means, and the page must say.

What is actually inside

The distributions split into valence and sea quarks, and the measurements — electrons on hydrogen and deuterium, then neutrinos and antineutrinos, which see different quarks because the weak charged current turns dud \to u but not udu \to d — determine the distributions separately.

probewhat it couples towhat it gives
electronelectric charge, weighted z²F₂^{ep}/x = (4/9)(u+ū) + (1/9)(d+d̄+s+s̄) — Eq. (6.23a). Cannot tell a quark from an antiquark, because the charge is squared
neutrino ν_μweak charge, and only one chirality2[d(x) + ū(x)] — Eq. (6.27a). It turns d → μ⁻u and ū → μ⁻d̄, and nothing else, so it separates quarks from antiquarks
antineutrino ν̄_μthe mirror set2[u(x) + d̄(x)] — Eq. (6.27b). The complement, which closes the system
nothing at allthe gluons. No electric charge, no weak charge, and therefore invisible to every probe in this table

The factor 2 in the neutrino rows comes from the V−A structure of the weak interaction (ch. 7) — the neutrino couples to one chirality only, so it gets no averaging factor of 1/2. And the last row is the point of the whole section: the most abundant thing in the proton is the one nothing in the table can see.

parton distributions, and where the proton keeps its momentum

10⁻⁴10⁻³0.010.1100.250.50.751x (momentum fraction)x · f(x)
  • u valence — carries 23%
  • d valence — carries 9%
  • sea (ū, d̄, s, s̄) — carries 18%
  • gluon — carries 50%
u valence
22.7%
d valence
9.3%
sea (ū, d̄, s, s̄)
18.0%
gluon
50.0%
∫ x[q(x) + q̄(x)] dx summed over every flavour an electron or a neutrino can see0.50
…so the rest, carried by something with no electric and no weak charge:0.50

Half the proton is missing, and that is the measurement — Eq. (6.28). Move the Q² slider and watch the curves shift towards small x while those percentages barely move: raising the resolving power resolves one parton into two softer ones, which is scaling violation, but it cannot create or destroy momentum. The shapes here are an illustrative parametrisation; the three things that are not illustrative are the counting rules ∫u_v dx = 2 and ∫d_v dx = 1, which fix the valence normalisations exactly, and the measured 0.50, which fixes the rest.

Fig. 6.12, with the integral of Eq. (6.28) done live. The valence quarks peak around x ≈ 0.15–0.3 and vanish at both ends; the sea piles up at small x. The shapes are an illustrative parametrisation, but the three constraints are not: ∫u_v dx = 2 and ∫d_v dx = 1 are exact counting rules for a proton, and the 0.50 is measured.

🔢 Worked example — half of the proton is missing

Integrate the measured distributions — the momentum sum rule — over every quark and antiquark an electron or neutrino can see, Eq. (6.28):

01x[u+d+uˉ+dˉ+s+sˉ]dx=0.50\htmlClass{t-i}{\int_0^1} \htmlClass{t-x}{x} \big[\htmlClass{t-q}{u + d + \bar u + \bar d + s + \bar s}\big]\,\mathrm{d}x = \htmlClass{t-r}{0.50}
(6.28)

The momentum sum rule, measured. Add up the momentum carried by everything that scatters an electron or a neutrino and you get half the proton.

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.

Half. The other half of the proton’s momentum is carried by constituents with no electric charge and no weak charge — which is why nothing in the table above detects them, and why they had to be found by subtraction.

They are the gluons, and this is a genuinely strange way to discover a particle: not by observing it, but by carefully accounting for everything else and noticing that the books do not balance by a factor of two.

the sum rule, built to respect the counting rules

import numpy as np
xs = np.linspace(1e-5, 1 - 1e-9, 400000)
shape = lambda a, b: xs**a * (1 - xs)**b

# The valence normalisations are NOT free: a proton has exactly two u and one d.
Au = 2 / np.trapezoid(shape(-0.5, 3.0), xs)
Ad = 1 / np.trapezoid(shape(-0.5, 4.0), xs)
uv, dv = Au * shape(-0.5, 3.0), Ad * shape(-0.5, 4.0)
print(f"counting rules: int u_v dx = {np.trapezoid(uv, xs):.3f} (must be 2), "
      f"int d_v dx = {np.trapezoid(dv, xs):.3f} (must be 1)")

Muv, Mdv = np.trapezoid(xs*uv, xs), np.trapezoid(xs*dv, xs)
print(f"valence u carries {Muv:.3f} of the momentum, valence d {Mdv:.3f}"
      f"  ->  {Muv+Mdv:.3f} together")

# The sea normalisation is then fixed by the MEASURED total, not chosen.
sea = 0.50 - (Muv + Mdv)
print(f"the measurement says quarks + antiquarks carry 0.50, so the sea supplies {sea:.3f}")
print()
print(f"  everything the electron and the neutrino can see: {Muv+Mdv+sea:.2f}")
print(f"  everything they cannot:                          {1-0.50:.2f}")
print()
print("Half the proton's momentum belongs to partons with neither an electric")
print("nor a weak charge.  That is the gluon, found by subtraction.")
prints
counting rules: int u_v dx = 2.000 (must be 2), int d_v dx = 1.000 (must be 1)
valence u carries 0.224 of the momentum, valence d 0.092  ->  0.315 together
the measurement says quarks + antiquarks carry 0.50, so the sea supplies 0.185

everything the electron and the neutrino can see: 0.50
everything they cannot:                          0.50

Half the proton's momentum belongs to partons with neither an electric
nor a weak charge.  That is the gluon, found by subtraction.

Scaling, broken — and why that is good news

At HERA (30 GeV electrons on 800 GeV protons, so s310\sqrt s \approx 310 GeV) the range opened to 2.7<Q2<300002.7 < Q^2 < 30\,000 GeV². Over most of xx, F2F_2 is flat and scaling holds. But at small xx it rises with Q2Q^2 the scaling law fails , and it fails exactly as predicted.

The mechanism is Fig. 6.14 and it is almost embarrassingly simple. A quark with fraction xx emits a gluon and drops to xx'. Look with poor resolution and you see one object at xx; look with better resolution and you see two, at xx' and xxx-x'. Better resolution finds more partons, each softer. So the small-xx population grows with Q2Q^2 and the large-xx population shrinks.

110100100010⁴0246Q² (GeV²)F₂ + offset (curves displaced for clarity)
  • x = 0.00013 — rises steeply
  • x = 0.005
  • x = 0.13 — essentially flat
  • x = 0.40 — falls slightly
Fig. 6.13, schematically. Scaling would make every line horizontal. At large x they nearly are; at small x they climb, and the climb is steeper the smaller x gets. The DGLAP equations predicted this in 1972–1977, before HERA existed to test it — and the amount of the violation is one of the best measurements of α_s there is, precisely because α_s enters at leading order here rather than as the 4 % correction §6.1 had to fight.

⚙️ Engineer’s bridge — the parton density depends on your sampling rate

The uncomfortable thing about scaling violation is that the answer to “how many quarks are in a proton, and with what momenta?” depends on how hard you look. There is no resolution-independent answer.

An engineer has met this. Ask how many components a signal has and the honest reply is “at what window length?” — a short window resolves two close tones as one, a longer window splits them. The signal did not change. Neither did your instrument’s honesty. The question was underspecified.

PDFs are the same object: a description of contents that is only defined relative to a resolving power, with a known rule — the DGLAP equations — for translating between scales. That is why every PDF set is published at a scale, why a cross-section prediction has to evolve them to the scale of the process, and why §9.14 can predict LHC rates at all: the measurement was made at HERA at one Q2Q^2 and evolved to another.

The invariance underneath is worth holding onto: the momentum fractions barely move while the curves do. Resolution redistributes; it does not create.

Where it breaks: “resolution redistributes, it does not create” is true of momentum and false of number. The momentum sum rule is fixed — the quarks hold half at any scale — but the count of partons grows without bound as Q2Q^2 rises, because every gluon resolves into more gluons and the small-xx density diverges. So the invariant to hold onto is the sum rule, not the population. An engineer used to a lossless change of representation should note which quantity is conserved here: it is the integral, never the histogram.

🔑 If you remember only three things

  • Half the proton’s momentum belongs to something with no charge. The gluons were found by subtraction, years before anything hit one.

  • The proton stops being a thing and becomes a distribution. After this section it is described by densities rather than by a shape, which is a change of type and not of detail.

  • Everything reduces to two measured numbers. One angle and one energy give both invariants, which is why an entire section’s results fit on a single plot.

Where this goes next

  • §6.3 finally writes the theory these two sections have been demanding: SU(3), and gluons that carry the charge they mediate.
  • §6.5 justifies the impulse approximation that Feynman had to assume — asymptotic freedom is what makes a struck parton behave as free.
  • §9.14 is where PDFs stop being a result and become a tool: every LHC cross-section is a convolution of two of them.
  • §1.8 is the elastic form factor whose steep fall is the control curve in Fig. 6.10.

Check yourself — deep inelastic scattering and the parton model

0/5 answered · 0 correct

  1. 1.DIS deliberately throws away all information about the hadronic final state X. Why does discarding data make the measurement better?

  2. 2.In Fig. 6.10 the elastic curve falls four decades while the deep inelastic points stay flat. What does the flatness mean?

  3. 3.Why is x = Q²/2m_pν restricted to x ≤ 1, and what happens at exactly x = 1?

  4. 4.The measured quarks and antiquarks carry 0.50 of the proton's momentum. What makes that number a discovery rather than a disappointment?

  5. 5.Move the widget's Q² slider. The curves shift towards small x but the momentum percentages barely change. Why both?

Study aid derived from A. Bettini, Introduction to Elementary Particle Physics, 3rd ed., Cambridge University Press 2024 — published Open Access under CC-BY-NC 4.0, DOI 10.1017/9781009440745. Not the book: an independently written interactive companion, figures redrawn.