The quark-parton model

By the late 1960s the SLAC-MIT experiments had done something the theory of the day could not explain: fire a multi-GeV electron at a proton, let it lose a large fraction of its energy to a violent collision, and it scattered through large angles far more often than a soft, spread-out charge cloud should allow. A diffuse proton should have looked fuzzy and stopped deflecting hard electrons; instead it behaved as if it contained tiny, hard, point-like lumps. Richard Feynman's response was audaciously simple. Forget, for a moment, the tangle of virtual gluons and sea pairs binding the proton together. In the instant of a hard collision, the electron does not see the proton at all — it sees one of its point-like constituents, which Feynman called a parton, and it scatters off that single parton elastically, as cleanly as it would off a free muon. That one assumption converts an intractable bound-state problem into textbook QED. The messy hadron becomes a beam of free, spin-half, point-charge quarks sharing the proton's momentum, and deep inelastic scattering becomes incoherent elastic scattering, $e^-q\to e^-q$ , summed over the constituents. Two predictions fall out immediately and both were confirmed: Bjorken scaling, that the structure functions depend only on the ratio $x$ and not on $Q^2$ ; and the Callan-Gross relation $F_2=2xF_1$ , the fingerprint of spin-half constituents. The bold "free during the collision" hypothesis had no justification in 1969 — it was vindicated four years later when asymptotic freedom in QCD explained why quarks stop interacting at short distance.

Key takeaways

Feynman modelled the proton as a beam of free, point-like, spin-half partons; deep inelastic scattering is incoherent elastic scattering $e^-q\to e^-q$ off one of them.
Work in the infinite momentum frame ( $E\gg m_p$ ), where a struck quark carries four-momentum $p_q=\xi p_2$ and transverse motion and masses are negligible.
On-shell kinematics of the struck quark force $\xi=Q^2/(2p_2\cdot q)\equiv x$ : the measured Bjorken $x$ is literally the quark's share of the proton momentum.
Point spin-half constituents predict Bjorken scaling ( $F$ depends on $x$ alone) and the Callan-Gross relation $F_2(x)=2xF_1(x)$ , both confirmed at SLAC.
Structure functions become charge-weighted sums over parton distribution functions: $F_2^{ep}(x)=x\sum_i Q_i^2\,q_i(x)$ .
Momentum sum rule from $ep$ and $en$ data gives $f_u\approx0.36$ , $f_d\approx0.18$ — the quarks carry only $\approx54\%$ of the proton's momentum; the missing $\approx46\%$ is gluons.
Key idea: deep inelastic scattering at high $Q^2$ is just elastic scattering off individual free, point-like, spin-half quarks; Bjorken $x$ measures the fraction of the proton's momentum each quark carries, so structure functions become charge-weighted sums over quark momentum distributions.

Where this lives today

The parton model is the operating language of every hadron collider. At the LHC, computing the rate of any process — from Higgs production to searches for new particles — begins by convolving the hard partonic cross section with parton distribution functions $q_i(x,Q^2)$ extracted from global fits (by groups such as CT, MSHT, and NNPDF) to decades of deep inelastic and collider data. The one place the naive model needed repair was Bjorken scaling: at the electron-proton collider HERA, $F_2(x,Q^2)$ was measured over five orders of magnitude in $Q^2$ and found to scale-violate — it drifts logarithmically with $Q^2$ , rising at small $x$ — exactly as QCD's DGLAP evolution equations predict once gluon radiation is switched back on. That controlled violation, far from breaking the model, turned deep inelastic scattering into a precision measurement of $\alpha_s$ and the gluon density. The next generation is the Electron-Ion Collider under construction at Brookhaven, which will image the sea and gluon distributions and how the proton's spin is built from its partons.

The physical picture

Picture the proton not at rest but boosted to enormous energy, its internal clocks frozen by time dilation and its shape Lorentz-contracted into a thin pancake of partons streaming side by side. To a virtual photon of high $Q^2$ — short wavelength, fast interaction — the binding forces between partons are effectively switched off: they act over times far longer than the collision lasts, so during the strike each parton is a free particle. The photon picks out exactly one of them, the one carrying momentum fraction $x$ , and knocks it out; the other "spectator" partons sail on undisturbed and later dress themselves into hadrons. The first figure fixes this geometry and the single kinematic relation that turns the measured energy loss into a momentum fraction.

The second figure is the model's most quotable payoff. If the proton were just "three quarks," each would carry a third of its momentum and nothing else would matter. Instead, adding up how much momentum each quark flavour carries — read directly off the structure-function data — leaves a startling deficit: the charged, photon-visible quarks account for only about half of the proton's momentum. The rest is carried by electrically neutral constituents the photon cannot see at all. Those are the gluons, and their invisible half was the parton model's first quantitative hint that the proton is glued together by a force-carrier of its own.

Figure

The proton momentum budget measured from deep inelastic scattering: up quarks carry f_u ≈ 0.36 and down quarks f_d ≈ 0.18 of the proton momentum (each including its sea), so all charged quarks together carry only ≈ 54%. The remaining ≈ 46% is carried by electrically neutral gluons, invisible to the photon — the model's first evidence for glue

The core equations

The single conceptual leap of the model is that the Bjorken variable, defined purely from measured lab quantities, is a physical momentum fraction. Impose that the struck parton was on its mass shell before the hit and remains on shell after absorbing the photon, and the identification is forced.

x \equiv \frac{Q^2}{2\,p_2\cdot q} = \frac{Q^2}{2M\nu} \qquad (1)

where: $Q^2=-q^2$ — space-like four-momentum transfer; $p_2$ — proton four-momentum, mass $M$ ; $q$ — virtual-photon four-momentum; $\nu=E_1-E_3$ — the electron's lab energy loss. Reading it: $x$ ranges over $0<x\le1$ , and in the parton model it is the fraction $\xi$ of the proton momentum carried by the struck quark; $x=1$ is elastic scattering off the whole proton. The dynamical content of the model is a statement about the two structure functions. Because the constituents are point-like spin-half Dirac particles, the electric and magnetic responses are locked together (Callan-Gross), and the single surviving structure function is a charge-weighted census of the quarks.

F_2^{ep}(x) = 2x\,F_1^{ep}(x) = x\sum_i Q_i^2\,q_i(x) \qquad (2)

where: $q_i(x)$ — the parton distribution function, the number density of quarks of flavour $i$ near momentum fraction $x$ ; $Q_i$ — that quark's charge in units of $e$ ( $+\tfrac23$ for $u$ , $-\tfrac13$ for $d,s$ ). The charges are squared because the cross section goes as the coupling squared. That $F_2$ depends on $x$ only — not $Q^2$ — is Bjorken scaling. Folding the elementary cross section against the parton densities gives the full inelastic double-differential cross section, the quantity an experiment actually bins in $x$ and $Q^2$ .

\frac{d^2\sigma}{dx\,dQ^2} = \frac{4\pi\alpha^2}{Q^4}\left[(1-y)\,\frac{F_2(x)}{x} + y^2\,F_1(x)\right] \qquad (3)

where: $\alpha=e^2/4\pi\approx1/137$ ; $y=\dfrac{p_2\cdot q}{p_2\cdot p_1}=\dfrac{\nu}{E_1}$ — the inelasticity, the fractional energy the electron gives up in the proton rest frame. The $1/Q^4$ is the photon propagator squared, the Rutherford-like signature of one-photon exchange off a point charge. The elementary process underneath (3) is the elastic scattering of the electron off one free quark of charge $Q_i$ , which has the same form as $e^-\mu^-$ scattering — this is where the spin-half structure enters.

\frac{d\hat\sigma}{dQ^2} = \frac{2\pi\alpha^2\,Q_i^2}{Q^4}\big[\,1+(1-y)^2\,\big],\qquad \tfrac12\big[1+(1-y)^2\big]=1-y+\tfrac12 y^2 \qquad (4)

where: $\hat\sigma$ — cross section on a single quark; the bracket $1+(1-y)^2$ is the angular signature of a spin-half target. A spin-zero parton would give a flat $y$ dependence and fail Callan-Gross ( $2xF_1=0$ ); the observed shape is a direct spin measurement. Splitting equation (2) by flavour, and using isospin to relate the neutron to the proton ( $u^n=d^p$ , $d^n=u^p$ ), gives the two observables that separate the flavours.

\frac{F_2^{ep}}{x}=\tfrac49(u+\bar u)+\tfrac19(d+\bar d)+\tfrac19(s+\bar s),\quad \frac{F_2^{en}}{x}=\tfrac49(d+\bar d)+\tfrac19(u+\bar u)+\tfrac19(s+\bar s) \qquad (5)

where: $u(x)\equiv u^p(x)$ etc. are the proton's parton densities. The factor $\tfrac49=(\tfrac23)^2$ weights up quarks, $\tfrac19=(\tfrac13)^2$ the down and strange. Measuring both $ep$ and $en$ and inverting this pair isolates each flavour. Integrating the densities against $x$ gives the momentum sum rule — the total momentum fraction carried by charged quarks — whose measured value is the model's most famous number.

\sum_i\int_0^1 x\,q_i(x)\,dx \;=\; f_u+f_d+f_s+\cdots \;\approx\; 0.5 \qquad (6)

where: $f_i=\int_0^1 x\,q_i(x)\,dx$ — the momentum fraction of flavour $i$ . Experiment gives $f_u\approx0.36$ , $f_d\approx0.18$ ; the sum falling short of $1$ is the direct evidence that neutral gluons carry the balance.

Derivation ABjorken x is the struck quark's momentum fraction

The goal is to show that the Bjorken variable $x=Q^2/(2p_2\cdot q)$ , which is defined entirely from measured lab quantities and carries no obvious meaning, is forced by simple on-shell kinematics to equal the fraction $\xi$ of the proton's four-momentum carried by the quark that absorbed the virtual photon. The entire argument is the single physical demand that the parton is a real particle both before and after the collision, worked in the infinite-momentum frame where the proton mass and the quark's transverse momentum can be dropped.

Step 1. Boost to the infinite-momentum frame, where the proton energy $E_2\gg M$ so that $p_2=(E_2,0,0,E_2)$ with $p_2^2=M^2\approx0$ , and assign the struck quark a fraction $\xi$ of this momentum, $p_q=\xi p_2$ . Transverse momenta and quark masses are $O(M)$ and negligible against $E_2$ . Why: in this frame the proton is a bundle of collinear, nearly-free partons, so "the quark carries fraction $\xi$ " is a clean statement about a single four-vector rather than a fuzzy bound-state wavefunction.

Step 2. Demand that the quark, after absorbing the photon of momentum $q$ , is again a real on-shell particle: $(p_q+q)^2=m_q^2\approx0$ . Expand the square: $(\xi p_2+q)^2=\xi^2 p_2^2+2\xi\,p_2\cdot q+q^2$ . Why: elastic scattering off a constituent means the constituent survives as itself — the same on-shell logic that fixed the recoil in elastic $ep$ scattering, now applied to the quark instead of the proton.

Step 3. Drop $\xi^2 p_2^2=\xi^2M^2\approx0$ and insert $q^2=-Q^2$ : the condition collapses to $2\xi\,p_2\cdot q-Q^2=0$ , i.e. $\xi=\dfrac{Q^2}{2\,p_2\cdot q}$ . But the right-hand side is exactly the definition of Bjorken $x$ , so $\xi=x$ . Why: this is the heart of the model — an experimentally accessible number ( $x$ , from angle and energy loss) is identified with an internal structural quantity ( $\xi$ , the momentum fraction), which is only possible because the parton is treated as free.

Step 4. Translate to the lab. With $p_2=(M,\mathbf0)$ at rest, $p_2\cdot q=M\nu$ where $\nu=E_1-E_3$ , so $x=Q^2/(2M\nu)$ , and combined with $Q^2=4E_1E_3\sin^2(\theta/2)$ every event is placed at a definite $x$ . Elastic scattering off the whole proton is the endpoint $x=1$ ; deep inelastic events populate $0<x<1$ . Why: this makes the abstract fraction measurable event by event, turning a histogram in $x$ directly into a picture of the proton's momentum sharing.

Go deeper · Derivation B · 2 minStructure functions as charge-weighted quark distributions

Derivation BStructure functions as charge-weighted quark distributions

The aim is to derive equation (2), $F_2(x)=x\sum_i Q_i^2 q_i(x)$ together with the Callan-Gross relation $F_2=2xF_1$ , by treating deep inelastic scattering as incoherent elastic scattering off free spin-half quarks and folding the elementary cross section against the probability of finding each quark. The strategy is to write the DIS cross section twice — once in the model-independent structure-function language, once as a sum over point quarks — and match the two, which simultaneously fixes the $F$ 's and reveals why spin-half locks them together.

Step 1. Write the general inelastic cross section in structure-function form, equation (3), which is fixed by Lorentz invariance and one-photon exchange and makes no assumption about what is inside the proton. The two unknown functions $F_1,F_2$ encode all the target structure. Why: this is the honest parametrisation — anything the proton could be is hidden in $F_1(x,Q^2)$ and $F_2(x,Q^2)$ , so matching a model to data means predicting these two functions.

Step 2. Now compute the same cross section in the parton model. Scattering off a single free quark of charge $Q_i$ and momentum fraction $\xi$ is elastic point scattering, so its contribution to $F_2$ is a delta function $Q_i^2\,\xi\,\delta(x-\xi)$ : a point quark contributes only at the $x$ equal to its own momentum fraction. Why: "elastic off a free constituent" means the event lands at exactly $x=\xi$ , so a single quark is a spike and the smooth measured $F_2$ must come from a distribution of quarks.

Step 3. Sum incoherently over all quarks by weighting each fraction $\xi$ with the density $q_i(\xi)$ and integrating: $F_2(x)=\sum_i\int_0^1 d\xi\,q_i(\xi)\,Q_i^2\,\xi\,\delta(x-\xi)=x\sum_i Q_i^2\,q_i(x)$ . Incoherence — adding probabilities, not amplitudes — is justified because the struck quark is resolved sharply in space and time at high $Q^2$ . Why: this is equation (2); the delta function collapses the integral and leaves the charge-weighted census, and the explicit $x$ out front is why $F_2$ (not $F_2/x$ ) is the naturally scaling quantity.

Step 4. Track the spin. The same point-quark calculation gives $2x F_1(x)=x\sum_i Q_i^2 q_i(x)$ as well, hence $F_2=2xF_1$ — the Callan-Gross relation. This follows because a massless spin-half quark cannot absorb a longitudinal photon, so the longitudinal structure function $F_L=F_2-2xF_1$ vanishes. Why: the relation is a spin thermometer — spin-half gives $F_2=2xF_1$ , spin-zero would give $2xF_1=0$ , and the SLAC data sit squarely on the spin-half prediction, proving the partons are the spin-half quarks of Gell-Mann and Zweig.

Go deeper · Derivation C · 1 minThe momentum sum rule and the missing half

Derivation CThe momentum sum rule and the missing half

The purpose is to turn the two measured integrals $\int F_2^{ep}\,dx$ and $\int F_2^{en}\,dx$ into the momentum fractions $f_u$ and $f_d$ carried by up and down quarks, and to expose the deficit that the charged quarks together carry only about half the proton momentum. The method is to integrate the flavour-resolved structure functions of equation (5) over $x$ , define the momentum fractions as the $x$ -weighted integrals of the densities, and solve the resulting two-by-two linear system, neglecting the small strange contribution.

Step 1. Define the momentum fractions $f_u=\int_0^1 x(u+\bar u)\,dx$ and $f_d=\int_0^1 x(d+\bar d)\,dx$ , and integrate equation (5) over $x$ . Neglecting strangeness, $\int_0^1 F_2^{ep}\,dx=\tfrac49 f_u+\tfrac19 f_d$ and $\int_0^1 F_2^{en}\,dx=\tfrac19 f_u+\tfrac49 f_d$ . Why: the $x$ -weighting inside $F_2=x\sum Q_i^2 q_i$ is exactly what converts the number densities into momentum fractions, so the area under $F_2$ is a charge-weighted momentum fraction.

Step 2. Insert the measured areas $\int F_2^{ep}\,dx\approx0.18$ and $\int F_2^{en}\,dx\approx0.12$ : two equations, two unknowns, $\tfrac49 f_u+\tfrac19 f_d=0.18$ and $\tfrac19 f_u+\tfrac49 f_d=0.12$ . Why: these two numbers are the raw experimental content — everything about the flavour momentum split is encoded in the areas under the proton and neutron structure functions.

Step 3. Solve. Multiply the first by four and subtract the second: $\tfrac{15}{9}f_u=0.72-0.12=0.60$ , so $f_u=0.36$ . Back-substitute: $\tfrac19(0.36)+\tfrac49 f_d=0.12\Rightarrow \tfrac49 f_d=0.08\Rightarrow f_d=0.18$ . Why: the algebra cleanly separates the flavours because up and down enter the proton and neutron with swapped charge weights — isospin plus two measurements is enough to disentangle them.

Step 4. Add them up: charged quarks carry $f_u+f_d\approx0.54$ of the proton's momentum, leaving $\approx0.46$ unaccounted. Photons couple only to charge, so this missing momentum belongs to electrically neutral constituents — the gluons. Why: the sum rule's failure to reach $1$ is not an error but a discovery; it is quantitative evidence, from electron scattering alone, that half the proton is made of glue the photon cannot see.

Limiting cases

Case	Result	Meaning
$x\to1$	one quark carries all the momentum	elastic scattering off the whole proton; the deep-inelastic region is $0<x<1$
$x\to0$	$q_i(x)$ diverges, sea dominates	soft gluon-generated $q\bar q$ pairs pile up; $F_2^{en}/F_2^{ep}\to1$
Point spin-half quarks	$F_2=2xF_1$ (Callan-Gross), $F_L=0$	confirmed at SLAC — the partons are spin-½
Hypothetical spin-0 partons	$2xF_1=0$ , flat in $y$	ruled out by the data; not what is seen
Fixed $x$ , vary $Q^2$	$F_2(x)$ constant	Bjorken scaling — the signature of point-like constituents
High- $x$ valence limit	$F_2^{en}/F_2^{ep}\to \tfrac14$	$u$ -quark dominance ( $d/u\to0$ ); a clean parton-model prediction

Common trapThinking the proton is just three quarks each carrying one third of the momentum

The tidy " $uud$ , a third each" picture is a cartoon of the quantum numbers, not of the momentum. Three static valence quarks would put all the momentum in three sharp spikes at $x=\tfrac13$ and give $\int x\sum q_i\,dx=1$ . Reality is dynamic: gluon radiation smears every distribution, and virtual $q\bar q$ sea pairs — favoured at small $x$ by the $1/Q^2$ gluon propagator — add an infinite number of low- $x$ partons, which is why $q_i(x)$ rises steeply as $x\to0$ . Most importantly, the charged quarks together carry only $\approx54\%$ of the momentum; the other $\approx46\%$ is in gluons the photon never touches. So $x=\tfrac13$ is at best the average valence position, the distributions are broad, and half of the proton is invisible to electron scattering.

Worked examples

1 · The proton momentum budget from structure-function areas (fully worked)

The measured areas under the proton and neutron structure functions are $\int_0^1 F_2^{ep}\,dx\approx0.18$ and $\int_0^1 F_2^{en}\,dx\approx0.12$ . Neglecting strangeness, find the momentum fractions $f_u$ and $f_d$ , the total carried by charged quarks, and the momentum left for gluons.

Step 1. Write the integrated flavour relations from equation (5): $\int F_2^{ep}\,dx=\tfrac49 f_u+\tfrac19 f_d=0.18$ and $\int F_2^{en}\,dx=\tfrac19 f_u+\tfrac49 f_d=0.12$ , where $f_u,f_d$ are the up and down momentum fractions.

Step 2. Eliminate $f_d$ : multiply the first equation by $4$ to get $\tfrac{16}{9}f_u+\tfrac49 f_d=0.72$ , then subtract the second, $\tfrac19 f_u+\tfrac49 f_d=0.12$ . The $f_d$ terms cancel: $\tfrac{15}{9}f_u=0.60$ , so $f_u=0.60\times\tfrac{9}{15}=\mathbf{0.36}$ .

Step 3. Back-substitute into the second equation: $\tfrac19(0.36)+\tfrac49 f_d=0.12$ , i.e. $0.040+\tfrac49 f_d=0.12$ , giving $\tfrac49 f_d=0.080$ and $f_d=\mathbf{0.18}$ . Check: the first equation returns $\tfrac49(0.36)+\tfrac19(0.18)=0.160+0.020=0.180$ . Consistent.

Step 4. Sum the charged quarks: $f_u+f_d=0.36+0.18=\mathbf{0.54}$ . The momentum sum rule requires the total over all constituents to be $1$ , so the gluons carry $1-0.54=\mathbf{0.46}$ . Charged quarks carry ≈ 54% of the proton momentum and neutral gluons ≈ 46% — the electron-scattering discovery of glue.

2 · The neutron-to-proton structure-function ratio

Using equation (5) and neglecting the strange sea, find the bounds on $F_2^{en}/F_2^{ep}$ , and evaluate it in the valence-dominated high- $x$ limit where the down density falls faster than the up.

Step 1. Neglecting $s$ and writing $u\equiv u+\bar u$ , $d\equiv d+\bar d$ , the ratio is $\dfrac{F_2^{en}}{F_2^{ep}}=\dfrac{4d+u}{4u+d}$ . This is a ratio of positive quantities and is monotonic in $d/u$ , so its extremes are the two limits of $d/u$ .

Step 2. Sea-dominated small- $x$ limit: $q\bar q$ pairs are produced flavour-blind, so $u\approx d$ and the ratio $\to\dfrac{4+1}{4+1}=\mathbf{1}$ . Every flavour is equally likely, so proton and neutron look identical to the photon.

Step 3. Valence-dominated high- $x$ limit: as $x\to1$ the down density falls faster than the up, $d/u\to0$ , so the ratio $\to\dfrac{u}{4u}=\mathbf{\tfrac14}$ . The general bounds are therefore $\tfrac14\le F_2^{en}/F_2^{ep}\le4$ , with the upper end (pure $d$ ) never realised. The measured ratio runs from ≈ 1 at small x down toward ¼ at large x — a direct readout of the two-thirds/one-third quark charges and up-quark dominance in the proton.

3 · Placing a DIS event in x

An $E_1=20\ \mathrm{GeV}$ electron scatters at $\theta=10^\circ$ and emerges with $E_3=10\ \mathrm{GeV}$ . Find $Q^2$ , $\nu$ , and the Bjorken $x$ ( $M=0.938\ \mathrm{GeV}$ ).

Step 1. Momentum transfer: $Q^2=4E_1E_3\sin^2(\theta/2)=4(20)(10)\sin^2(5^\circ)$ . With $\sin^2(5^\circ)=0.007596$ , $Q^2=800\times0.007596=\mathbf{6.08\ GeV^2}$ .

Step 2. Energy loss: $\nu=E_1-E_3=20-10=\mathbf{10\ GeV}$ , and the inelasticity $y=\nu/E_1=10/20=0.50$ .

Step 3. Bjorken $x$ from equation (1): $x=\dfrac{Q^2}{2M\nu}=\dfrac{6.08}{2(0.938)(10)}=\dfrac{6.08}{18.76}=\mathbf{0.324}$ . This event struck a quark carrying about a third of the proton's momentum, at Q² ≈ 6 GeV² — deep in the scaling region where the parton picture holds.

Physics of the portrait

The quark-parton model is inseparable from the physicist who dared to ignore everything hard about the proton and keep only what was simple. Richard Feynman took the fresh SLAC data, refused to wait for a theory of the strong force, and modelled the proton as a swarm of free "partons" — a word deliberately agnostic about whether they were Gell-Mann's quarks.

Origins · 1 minRichard Feynman and how this was discovered

Richard Feynman

1918–1988 · the parton picture of deep inelastic scattering

Richard Feynman had already reshaped physics twice — with the path-integral formulation of quantum mechanics and with the diagrams and renormalised QED that won him a share of the 1965 Nobel Prize — when, visiting SLAC in 1968, he turned to deep inelastic electron scattering. James Bjorken had predicted, from current algebra, that the structure functions should "scale" — depend only on $x$ , not $Q^2$ — and the data were bearing him out. Feynman supplied the picture that made scaling intuitive: the proton, viewed in a fast-moving frame, as a beam of nearly free point-like constituents he called partons, each carrying a fraction $x$ of the momentum, with the electron scattering elastically off one of them. Scaling then follows because a point particle has no length scale to set a $Q^2$ dependence, and $x$ is just the struck parton's momentum share. Feynman kept the word "parton" deliberately noncommittal, but the identification with the fractionally charged quarks of Gell-Mann and Zweig was quickly forced by the charge-weighted sum rules and by the Callan-Gross relation, which pinned the partons as spin-½. The most audacious assumption — that tightly bound quarks behave as free particles during a hard collision — had no foundation until 1973, when Gross, Wilczek, and Politzer discovered asymptotic freedom in QCD, retroactively justifying Feynman's leap. The 1990 Nobel Prize to Friedman, Kendall, and Taylor for the SLAC experiments was, in effect, the prize for confirming the parton.

History

1964

— Gell-Mann and Zweig independently propose fractionally charged quarks as hadron constituents

1968

— the SLAC-MIT experiments observe copious large-angle deep inelastic electron scattering

1969

— Bjorken predicts scaling; Feynman proposes the parton model to explain it

1969

— Callan and Gross derive F₂ = 2xF₁ as the signature of spin-½ partons

1972

— momentum sum rule from ep and en data shows quarks carry only about half the momentum

1973

— asymptotic freedom in QCD justifies treating the bound quarks as free during the collision

Going further

Texts. Thomson, Modern Particle Physics, §8.4 (the quark-parton model), §8.4.1–8.4.2 (parton distribution functions and their determination) — the primary source for the notation and results here · Halzen & Martin, Quarks and Leptons, §8–9 (the parton model, scaling, and structure functions) · Griffiths, Introduction to Elementary Particles, 2nd ed., §8.3 (deep inelastic scattering and partons) · Peskin & Schroeder, An Introduction to Quantum Field Theory, §17.3 (the parton distributions and deep inelastic scattering) · Schwartz, Quantum Field Theory and the Standard Model, §32 (deep inelastic scattering and parton evolution). The key results can be checked directly from the equations above: $\xi=x$ follows from $(\xi p_2+q)^2\approx0$ with $p_2^2\approx0$ ; the delta-function fold gives $F_2(x)=x\sum_i Q_i^2 q_i(x)$ and the Callan-Gross relation; the linear solve $\tfrac49 f_u+\tfrac19 f_d=0.18$ , $\tfrac19 f_u+\tfrac49 f_d=0.12$ yields $f_u=0.36$ , $f_d=0.18$ , $f_u+f_d=0.54$ ; and the DIS event $E_1=20$ , $E_3=10\ \mathrm{GeV}$ , $\theta=10^\circ$ gives $Q^2=6.08\ \mathrm{GeV}^2$ , $x=0.324$ .

Prerequisites. Deep Inelastic Scattering · Bjorken Scaling · Electron-Quark Scattering · Mandelstam Variables · Form Factors and Structure Functions. Leads to. Parton Distribution Functions · Asymptotic Freedom · Colour and QCD · Hadron Collisions · Neutrino-Nucleon Scattering.

Ask the tutor: Ask me to place any deep-inelastic event at its Bjorken $x$ , invert the $ep$ and $en$ structure functions for the individual quark distributions, or show how gluon radiation turns exact scaling into the DGLAP scaling violations measured at HERA.

Why can Feynman treat a tightly bound quark as a free particle?

Because the collision is fast and hard while the binding is slow and soft. A high- $Q^2$ virtual photon interacts over a time far shorter than the time the strong force needs to communicate between partons, so during the strike the struck quark has no chance to "feel" its neighbours — it responds as if free. In 1969 this was pure hypothesis, justified only by its success. The deep reason came in 1973 with asymptotic freedom: QCD's coupling $\alpha_s$ decreases at short distance (large $Q^2$ ), so at the scales probed in deep inelastic scattering the quarks really are weakly interacting. Feynman's leap was correct before anyone could prove it.

What is the difference between valence quarks and the sea?

The valence quarks are the three ( $uud$ for the proton) that carry the proton's net quantum numbers — charge, baryon number, and flavour. The sea is the cloud of virtual quark-antiquark pairs continually produced and reabsorbed by gluons. Sea quarks carry no net quantum numbers (each $q$ is matched by a $\bar q$ ), but they are very real to a high- $Q^2$ photon, which can strike any of them. The sea is concentrated at small $x$ because the gluons that make it radiate preferentially at low momentum, so as $x\to0$ the densities are sea-dominated and proton and neutron look identical. At large $x$ the valence quarks dominate.

How does the model know the partons are spin-½ and not spin-0?

Through the Callan-Gross relation $F_2=2xF_1$ . The two structure functions measure the target's response to transversely and longitudinally polarised virtual photons. A massless spin-½ quark cannot absorb a longitudinal photon — helicity conservation forbids it — so the longitudinal structure function $F_L=F_2-2xF_1$ vanishes, giving $F_2=2xF_1$ . A spin-0 parton would do the opposite: it couples only longitudinally, forcing $2xF_1=0$ . The SLAC data sit cleanly on the spin-½ prediction, which is why the partons could be identified with the spin-½ quarks rather than some scalar constituent.

Why are the quark charges squared in the structure function?

Because the structure function is proportional to a cross section, and a cross section is an amplitude squared. The virtual photon couples to a quark with a strength proportional to that quark's electric charge $Q_i$ , so the scattering amplitude carries one factor of $Q_i$ ; squaring the amplitude to get a probability gives $Q_i^2$ . That is why up quarks (charge $+\tfrac23$ ) enter with weight $\tfrac49$ and down or strange quarks (charge $-\tfrac13$ ) with weight $\tfrac19$ — a factor of four difference that is precisely what lets the proton and neutron data be inverted for the individual flavour distributions.

Does Bjorken scaling hold exactly?

No, and its controlled failure is one of the triumphs of QCD. In the strict parton model $F_2$ depends only on $x$ , because a point-like constituent has no scale to set a $Q^2$ dependence. But quarks radiate gluons, and gluons split into $q\bar q$ pairs, so as $Q^2$ rises the photon resolves ever more low- $x$ partons. The result is a slow, logarithmic drift of $F_2$ with $Q^2$ — scaling violation — that grows the small- $x$ densities and shrinks the large- $x$ ones. This drift is predicted by the DGLAP evolution equations and was confirmed in detail at HERA, turning the "failure" of scaling into a precision probe of $\alpha_s$ and the gluon.

Why does electron scattering miss half the proton momentum?

Because the photon couples only to electric charge, and gluons are electrically neutral. Adding up the momentum of everything the photon can see — the up, down, and strange quarks and their antiquarks — gives only about $54\%$ of the proton's total. The missing $\approx46\%$ must be carried by something with no electric charge, and the only candidate is the gluon, the carrier of the strong force. So the momentum sum rule is not a discrepancy to be fixed; it is the earliest quantitative evidence, from electron scattering alone, that the proton is held together by gluons sharing half its momentum.