Parton distribution functions

Fire a high-energy electron at a proton and, if you hit hard enough, you are no longer scattering from "the proton" at all — you are scattering from one of the point-like quarks rattling around inside it. The quark-parton model turns deep inelastic scattering into elastic scattering off a quasi-free, spin-half constituent, and the single kinematic variable $x=Q^2/(2p\cdot q)$ acquires a startling meaning: it is the fraction of the proton's momentum carried by the struck quark. Ask how that momentum is shared, and the answer is a set of functions — the parton distribution functions $f(x)$ — that give the probability of finding a quark, antiquark or gluon carrying fraction $x$ . They are the proton's internal census. The picture they paint is nothing like a tidy bound state of three quarks. A PDF measurement reveals the proton as a seething, dynamic object: three valence quarks ( $uud$ ) that carry its net quantum numbers, immersed in a sea of virtual quark–antiquark pairs, all bound by gluons that themselves carry nearly half the total momentum — a fact no probe that couples only to charge could ever guess directly. And the distributions are not even fixed shapes: sharpen your probe by raising $Q^2$ and the partons proliferate at small $x$ , a scale dependence governed precisely by the DGLAP evolution equations of QCD. Two features make PDFs the load-bearing beams of collider physics. First, they cannot yet be computed from first principles — they are measured, extracted from a global fit to electron, neutrino and hadron-collider data. Second, they are universal: the same $f(x,Q^2)$ measured at HERA predicts Higgs production at the LHC. This article assembles that machinery from the structure functions up.

Key takeaways

A parton distribution function $f(x,Q^2)$ gives the probability density for finding a parton (quark, antiquark or gluon) carrying momentum fraction $x$ of the proton at resolution scale $Q^2$ .
Measured structure functions decompose into PDFs: $F_2^{ep}(x)=x\sum_i e_i^2[q_i(x)+\bar q_i(x)]$ , and the Callan–Gross relation $F_2=2xF_1$ fixes the partons as spin-½.
The proton is three valence quarks plus a sea of $q\bar q$ pairs and gluons; the momentum sum rule shows gluons carry roughly half the total momentum.
PDFs cannot be computed perturbatively — they are extracted from a global fit to many experiments, then used everywhere because they are universal (factorisation).
The scale dependence $f(x,Q^2)$ is not free but obeys the DGLAP evolution equations built from QCD splitting functions $P_{ij}(z)$ .
Counting rules pin the valence content: $\int_0^1[u-\bar u]\,dx=2$ and $\int_0^1[d-\bar d]\,dx=1$ .
Key idea: the proton's momentum is shared dynamically among quarks, antiquarks and gluons, and a PDF f(x) is the probability of finding a parton of a given type carrying fraction x — a universal input, measured not calculated, that feeds every proton cross section.

The core insight

The proton's momentum is shared dynamically among quarks, antiquarks and gluons, and a PDF $f(x)$ tells you the probability of finding a parton of type $f$ carrying fraction $x$ — a universal input, measured not calculated, that feeds every high-energy proton cross section.

Where this lives today

Parton distributions are the working currency — and the dominant systematic uncertainty — of the entire LHC physics programme. Every predicted rate, from Higgs production through gluon fusion to $W$ , $Z$ and top-pair cross sections, is a PDF convolved with a partonic calculation, so the precision of a Higgs coupling or a search for new physics is often limited not by the theory of the hard process but by how well the proton's gluon and sea are known. Modern PDF sets — NNPDF, CT, MSHT — are global fits to thousands of data points from HERA, the Tevatron and the LHC, and the frontier is statistical: NNPDF pioneered replacing rigid functional forms with neural networks whose uncertainties are propagated by Monte Carlo, so a PDF today comes as an ensemble of equally likely proton "images." At the same time, lattice QCD has begun to compute $x$ -moments and even the $x$ -shape of PDFs directly from the QCD path integral, promising a first-principles cross-check of the fitted distributions. And the next dedicated machine is already approved: the Electron-Ion Collider at Brookhaven will map the gluon and sea at small $x$ , resolve the proton's spin among quark spin, gluon spin and orbital motion, and image the transverse spatial distribution of partons — the three-dimensional successor to the one-dimensional $f(x)$ this article has built.

The physical picture

Seen from a frame where the proton moves fast, time dilation freezes its internal motion and length contraction flattens it into a pancake of nearly free partons, each carrying a slice $x$ of the total longitudinal momentum. A virtual photon of virtuality $Q^2$ resolves transverse distances of order $\hbar c/Q$ ; if that is small compared with the proton, the photon strikes a single quark and knocks it out cleanly, and the inclusive rate simply counts how many quarks of each charge sit at each $x$ . The first figure fixes this anatomy — valence quarks, sea pairs and gluons — and shows how the total momentum is shared.

Figure

The proton as a bag of partons: three valence quarks uud carry its net charge and flavour, immersed in a sea of virtual q q̄ pairs and gluons. A virtual photon of virtuality Q² strikes one quark carrying momentum fraction x. The momentum sum rule ∫₀¹ x(Σq + g)dx = 1 splits as valence ≈ 0.36, sea ≈ 0.18, gluons ≈ 0.46 — nearly half the proton momentum is carried by gluons that the photon never sees

The census the fit returns is a family of curves $xf(x,Q^2)$ , and every particle physicist carries their shape in their head. The valence distributions $xu_V$ and $xd_V$ are broad humps peaking near $x\approx0.15$ — the "typical" quark carries far less than the naive $1/3$ . Below $x\approx0.1$ the sea antiquarks and, above all, the gluon distribution take over and rise steeply toward small $x$ , so that at the LHC almost every hard collision begins with two low- $x$ gluons. The second figure is that canonical plot, drawn at $Q^2=10\ \mathrm{GeV}^2$ .

Figure

The measured proton PDFs xf(x) at Q² = 10 GeV² on a log-x axis: the valence humps xu_V and xd_V peak near x ≈ 0.15 (the typical valence quark carries far less than 1/3), while the gluon xg(x) shown divided by 5 and the quark sea rise steeply as x → 0. The valence area is fixed by the counting rule ∫₀¹(u − ū)dx = 2

The core equations

The bridge from experiment to PDFs is the structure function $F_2$ . Summing the point-like elastic cross section $e^-q\to e^-q$ over all quark types, each weighted by its squared charge $e_i^2$ and its distribution $q_i(x)$ , reproduces the inclusive DIS cross section. The result identifies $F_2$ directly with a charge-weighted sum of PDFs.

F_2^{ep}(x)=x\sum_i e_i^2\big[q_i(x)+\bar q_i(x)\big]=x\left[\tfrac{4}{9}(u+\bar u)+\tfrac{1}{9}(d+\bar d)+\tfrac{1}{9}(s+\bar s)\right] \qquad (1)

where: $q_i(x)$ — number density of quark flavour $i$ ; $e_i$ — its charge ( $+\tfrac23$ for $u$ , $-\tfrac13$ for $d,s$ ); the explicit $x$ makes $F_2$ the momentum-weighted sum. Reading it: measuring $F_2$ on protons and neutrons, and with neutrinos, gives enough independent charge weightings to separate the individual flavour distributions. Because the partons are spin-½, the two structure functions are not independent. The transverse and longitudinal responses lock together into the Callan–Gross relation — a direct experimental fingerprint of the quark's spin.

F_2(x)=2x\,F_1(x) \qquad (2)

where: $F_1$ — the magnetic (transverse) structure function; $F_2$ — the total (transverse plus longitudinal). Reading it: spin-0 partons would give $F_1=0$ ; the measured $2xF_1/F_2\approx1$ is how we know quarks carry spin ½, independent of any charge assignment. The distributions are constrained by exact sum rules that encode the proton's fixed identity. The momentum sum rule states that the fractions carried by all partons — quarks, antiquarks and gluons — must add to one.

\sum_i\int_0^1 x\big[q_i(x)+\bar q_i(x)\big]\,dx+\int_0^1 x\,g(x)\,dx=1 \qquad (3)

where: $g(x)$ — the gluon distribution. Reading it: the charged-lepton and neutrino data fix the quark integral at only $\approx0.5$ , so the gluon term must supply the missing half — the cleanest indirect proof that gluons exist and carry momentum. The counting (valence) sum rules fix the net flavour content: the sea contributes equal quarks and antiquarks, so only the valence quarks survive the difference $q-\bar q$ .

\int_0^1\big[u(x)-\bar u(x)\big]\,dx=2,\qquad \int_0^1\big[d(x)-\bar d(x)\big]\,dx=1,\qquad \int_0^1\big[s-\bar s\big]\,dx=0 \qquad (4)

where: the right-hand numbers are the valence quark counts of the $uud$ proton. Reading it: no matter how violent the sea, the proton stays a proton — these integrals are protected quantum numbers, powerful anchors on any PDF fit. The distributions run with the scale $Q^2$ , and the running is fixed by QCD through the DGLAP evolution equations. A quark at $x$ can have come from a parent at higher momentum fraction $x/z$ that radiated, so the change per logarithmic step in $Q^2$ is a convolution with splitting functions.

\frac{\partial q_i(x,Q^2)}{\partial\ln Q^2}=\frac{\alpha_s(Q^2)}{2\pi}\int_x^1\frac{dz}{z}\Big[P_{qq}(z)\,q_i\!\Big(\tfrac{x}{z},Q^2\Big)+P_{qg}(z)\,g\!\Big(\tfrac{x}{z},Q^2\Big)\Big] \qquad (5)

where: $P_{qq}(z)=C_F\frac{1+z^2}{1-z}$ ( $C_F=\tfrac43$ ) is the probability of a quark radiating to fraction $z$ ; $P_{qg}(z)=T_R[z^2+(1-z)^2]$ is a gluon splitting to a quark. Reading it: PDFs are not free functions of $Q^2$ — measure them at one scale and QCD predicts them at every other, the sharp test that turned partons into quarks-of-QCD. Finally the payoff — factorisation. Because PDFs are universal, any hadronic cross section is the PDFs convolved with a short-distance partonic cross section $\hat\sigma$ computable in perturbative QCD.

\sigma_{AB\to X}=\sum_{a,b}\int_0^1\!dx_a\int_0^1\!dx_b\;f_a(x_a,\mu_F^2)\,f_b(x_b,\mu_F^2)\;\hat\sigma_{ab\to X}\big(x_a,x_b,\mu_F^2\big) \qquad (6)

where: $f_a,f_b$ — PDFs of the two beam hadrons; $\hat\sigma$ — partonic hard cross section; $\mu_F$ — factorisation scale. Reading it: this one formula is why a proton's PDF measured in electron scattering predicts Higgs, $W$ , $Z$ and top production at a hadron collider — the same $f(x,Q^2)$ everywhere.

Derivation AFrom point-like quark scattering to the structure function F2

The goal is to show that treating deep inelastic scattering as incoherent elastic scattering off point-like, spin-½ quarks — each carrying a definite momentum fraction and each described by a probability distribution $q_i(x)$ — reproduces the measured structure function as the charge-weighted sum $F_2=x\sum e_i^2 q_i$ , and simultaneously forces the Callan–Gross relation $F_2=2xF_1$ . We build the DIS cross section from the elementary $eq$ cross section, insert the momentum-fraction bookkeeping, and read off the structure functions by comparison.

Step 1. Start from elastic electron–quark scattering. A quark of charge $e_i$ carrying four-momentum $\xi p$ (a fraction $\xi$ of the proton momentum $p$ ) scatters elastically, which forces $\xi=x=Q^2/(2p\cdot q)$ by the on-shell condition $(\xi p+q)^2\approx0$ . Why: the massless-quark elastic constraint collapses the continuous fraction $\xi$ onto the single measurable variable $x$ , which is exactly why $x$ acquires its meaning as "the struck quark's momentum fraction."

Step 2. Weight by the probability of finding such a quark. The DIS cross section is the elastic $eq$ cross section summed incoherently over flavours and integrated against the distributions: $d\sigma^{ep}=\sum_i\int dx\,q_i(x)\,d\sigma^{eq_i}$ . Why: at large $Q^2$ the struck quark is resolved as free on the collision timescale, so the proton acts as an incoherent bag of quarks and probabilities simply add — no interference between partons.

Step 3. Compare with the general DIS cross section written in terms of $F_1,F_2$ . The point-like spin-½ $eq$ cross section carries a fixed ratio of its transverse and longitudinal pieces; matching term by term gives $F_2^{ep}(x)=x\sum_i e_i^2[q_i(x)+\bar q_i(x)]$ , equation (1). Why: the squared charge $e_i^2$ is the only quark property entering the electromagnetic coupling, so the flavour sum is automatically charge-weighted — the origin of the $\tfrac49$ and $\tfrac19$ factors.

Step 4. Extract Callan–Gross. Because a spin-½ target has no longitudinal absorption at leading order, the same matching yields $2xF_1=F_2$ . Why: a spin-0 parton would instead give $F_1=0$ ; the experimental confirmation $2xF_1/F_2\to1$ is a spin measurement in disguise, independent of the charge assignments, and one of the pillars establishing quarks as spin-½ Dirac particles.

Go deeper · Derivation B · 1 minThe momentum sum rule and the missing half carried by gluons

Derivation BThe momentum sum rule and the missing half carried by gluons

This derivation turns the abstract statement "partons share the proton's momentum" into a hard number and shows how it revealed the gluon indirectly. The plan is to define the total momentum fraction carried by quarks as an integral over the measured distributions, insert the numbers that the charged-lepton and neutrino fits return, and observe that they fall dramatically short of unity — the deficit being the gluon momentum, a component the photon and the $W$ can never strike directly.

Step 1. Write the fraction of the proton momentum carried by all quarks and antiquarks as $\varepsilon_q=\sum_i\int_0^1 x[q_i(x)+\bar q_i(x)]\,dx$ . If quarks carried everything, $\varepsilon_q=1$ . Why: $xf(x)\,dx$ is the momentum fraction in the interval $dx$ , so integrating $x$ against the distribution literally adds up momentum — the natural bookkeeping for the sum rule.

Step 2. Insert the measured integrals. Combining $ep$ , $en$ and neutrino data gives $\int_0^1 x[u+\bar u]\,dx\approx0.36$ and $\int_0^1 x[d+\bar d]\,dx\approx0.18$ (strange contributes only a few percent), so $\varepsilon_q\approx0.54$ . Why: neutrinos and charged leptons weight the flavours differently, so together they separate the distributions well enough to pin the total quark momentum — an experimental input, not a theoretical guess.

Step 3. Confront the sum rule (3). The quarks account for only about half the momentum: $1-\varepsilon_q\approx0.46$ is unaccounted for by anything that carries electric or weak charge. Why: the electromagnetic and weak probes couple only to charged partons, so a large momentum deficit signals a neutral carrier hiding in the proton — the gluon.

Step 4. Identify the carrier. The only candidate is the electrically neutral, colour-charged gluon, so $\int_0^1 x\,g(x)\,dx\approx0.46$ . Why: this closes the sum rule and shows that nearly half the proton's momentum resides in glue — a striking, quantitative prediction of QCD confirmed the moment the quark integral came in near one-half rather than one.

Go deeper · Derivation C · 2 minDGLAP evolution: why the distributions run with Q2

Derivation CDGLAP evolution: why the distributions run with Q2

The aim is to explain why PDFs depend on the scale $Q^2$ at all — if $x$ is a momentum fraction, why should sharpening the probe change the census? — and to sketch how QCD turns that scale dependence into a predictive evolution equation. We picture a quark seen at higher resolution revealing that it was radiating gluons, count the logarithmically enhanced radiation, and assemble the leading-order DGLAP equation with its splitting function.

Step 1. Recognise that resolution is finite. A probe of virtuality $Q^2$ resolves structure down to $\sim1/Q$ ; raise $Q^2$ and a quark that looked isolated is seen to be surrounded by the gluons it was about to emit, plus the $q\bar q$ pairs those gluons make. Why: "a quark" is a scale-dependent notion — what counts as one parton at low resolution splits into several at high resolution, so the distributions must change with $Q^2$ .

Step 2. Count the radiation. A quark emitting a gluon and ending with fraction $z$ of its momentum has an emission probability, per unit $\ln Q^2$ , proportional to $\frac{\alpha_s}{2\pi}P_{qq}(z)$ with the splitting function $P_{qq}(z)=C_F\frac{1+z^2}{1-z}$ . The collinear region generates a factor $\ln Q^2$ , the origin of the logarithmic running. Why: soft and collinear gluon emission is enhanced by a large logarithm, so even though $\alpha_s$ is small the effect accumulates and must be resummed — the physical content of the evolution equation.

Step 3. Assemble the balance equation. A quark at $x$ is fed by parents at higher fraction $x/z$ that radiated down to $x$ , giving $\frac{\partial q(x,Q^2)}{\partial\ln Q^2}=\frac{\alpha_s}{2\pi}\int_x^1\frac{dz}{z}\,P_{qq}(z)\,q(x/z,Q^2)$ , plus a $P_{qg}\otimes g$ term because gluons also split into quarks. This is equation (5). Why: the convolution is just momentum conservation over all ways to arrive at fraction $x$ — every parent at $x/z$ radiating fraction $1-z$ contributes, weighted by the splitting probability.

Step 4. Read the consequence. Integrating upward in $Q^2$ softens the valence quarks (they shed momentum to radiation, migrating to lower $x$ ) and piles up gluons and sea at small $x$ — exactly the scaling violations seen in the data, and a parameter-free QCD prediction once the PDFs are known at any one scale. Why: DGLAP converts a measurement at a single $Q^2$ into a prediction at all others, which is what makes PDFs genuinely predictive rather than an infinite table of unrelated numbers.

Limiting cases

Case	Result	Meaning
Large $x$ ( $x\to1$ )	valence dominates, sea and glue $\to0$ ; $F_2^{en}/F_2^{ep}\to\tfrac14$	a single fast valence quark carries almost all the momentum; the ratio tests isospin + valence
Small $x$ ( $x\to0$ )	$xg(x)$ and the sea rise steeply	the proton is dominated by soft gluons; LHC hard scatters are mostly low- $x$ glue
Naive parton model	$f(x)$ independent of $Q^2$ (Bjorken scaling)	exact only if quarks were truly free; broken logarithmically by QCD
Increasing $Q^2$ (DGLAP)	valence softens, small- $x$ sea and glue grow	scaling violations — the running predicted by the evolution equations
Spin-½ partons	$F_2=2xF_1$ (Callan–Gross)	non-zero $F_1$ is the fingerprint of spin ½; scalar partons give $F_1=0$
Summed over partons	$\int x(\Sigma q+g)\,dx=1$ ; $\int(u-\bar u)dx=2$	momentum and counting sum rules — exact constraints protecting the fit

Common trapThe proton is three quarks, each carrying x ≈ 1/3

This is the single most persistent misconception, and every PDF plot refutes it. If the proton were a static bound state of three quarks sharing momentum equally, $F_2$ would peak sharply at $x=1/3$ ; instead the valence humps peak near $x\approx0.15$ and there is substantial structure everywhere from $x\approx10^{-4}$ up to $x\approx0.8$ . Three things are wrong with the naive picture. First, the valence quarks are relativistic and continually exchange momentum, so their distribution is a broad smear, not a spike. Second, there is a large sea of virtual $q\bar q$ pairs — the proton momentarily contains extra quarks and antiquarks, which is why the counting rules use the difference $q-\bar q$ to isolate the "real" three. Third, and most striking, nearly half the momentum is carried by gluons, which are not quarks at all and are invisible to the electromagnetic probe; you infer them from the momentum they are missing. The proton is not a tidy triangle of quarks — it is a dynamic, scale-dependent plasma of partons whose net quantum numbers happen to be those of $uud$ .

Worked examples

1 · Gluons carry half the proton — reading the momentum sum rule (fully worked)

Global fits give the quark momentum integrals $\int_0^1 x[u+\bar u]\,dx=0.36$ and $\int_0^1 x[d+\bar d]\,dx=0.18$ , with the strange sea contributing about $0.03$ . What fraction of the proton's momentum is carried by gluons?

Step 1. Add up the quark and antiquark momentum. Summing the flavours, $\varepsilon_q=0.36+0.18+0.03=\mathbf{0.57}$ — but taking the round textbook figures without strange, $\varepsilon_q\approx0.36+0.18=\mathbf{0.54}$ . Either way the quarks carry only a little over half.

Step 2. Apply the momentum sum rule, equation (3): the total must be one, so the gluon fraction is $\varepsilon_g=1-\varepsilon_q$ . Using $\varepsilon_q\approx0.54$ gives $\varepsilon_g=1-0.54=\mathbf{0.46}$ .

Step 3. Interpret. Nearly half the proton's momentum, $\approx46\%$ , is carried by gluons — objects with zero electric charge that the scattering electron never strikes. Their presence is deduced purely from the momentum the charged partons fail to account for.

Step 4. Appreciate the significance. Before QCD there was no reason for the quark integral to fall short of one; finding $\varepsilon_q\approx0.5$ was among the earliest quantitative evidence that a neutral momentum-carrying field — the gluon — binds the proton. Gluons carry roughly half the proton's momentum, ε_g ≈ 0.46, inferred entirely from the sum-rule deficit.

2 · The large-x ratio F2en/F2ep and the number 1/4

Show that for electron scattering at large $x$ , where valence quarks dominate and the sea is negligible, the ratio of neutron to proton structure functions tends to $1/4$ .

Step 1. Use isospin to relate neutron and proton PDFs. The neutron is the proton with $u\leftrightarrow d$ , so $u^n(x)=d^p(x)\equiv d(x)$ and $d^n(x)=u^p(x)\equiv u(x)$ . At large $x$ keep only valence quarks and drop $\bar u,\bar d,s$ .

Step 2. Write both structure functions from equation (1). For the proton, $F_2^{ep}=x\big[\tfrac49 u+\tfrac19 d\big]$ ; for the neutron, swapping flavours, $F_2^{en}=x\big[\tfrac49 d+\tfrac19 u\big]$ .

Step 3. Form the ratio: $\dfrac{F_2^{en}}{F_2^{ep}}=\dfrac{4d+u}{4u+d}$ , dividing through by $u$ to get $\dfrac{4(d/u)+1}{4+(d/u)}$ .

Step 4. Take the large- $x$ limit. Empirically $d(x)/u(x)\to0$ as $x\to1$ (the fastest quark is almost always a $u$ ), so $\dfrac{F_2^{en}}{F_2^{ep}}\to\dfrac{0+1}{4+0}=\mathbf{\tfrac14}$ . The neutron-to-proton ratio falls to 1/4 at large x — a clean consequence of two valence u quarks against one d, and a textbook confirmation of the fractional charges. (At small $x$ , where the flavour-symmetric sea dominates both, the ratio instead tends to $1$ .)

3 · A toy valence distribution and the counting rule

Model the valence up-quark density as $u_V(x)=A\,x^{-1/2}(1-x)^3$ . Fix $A$ from the counting rule $\int_0^1 u_V\,dx=2$ , then find the momentum fraction $\int_0^1 x\,u_V\,dx$ it carries.

Step 1. Recall the Beta-function integral $\int_0^1 x^{a-1}(1-x)^{b-1}\,dx=B(a,b)=\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ . The normalisation integral is $\int_0^1 x^{-1/2}(1-x)^3\,dx=B(\tfrac12,4)=\dfrac{\Gamma(1/2)\Gamma(4)}{\Gamma(9/2)}=\dfrac{\sqrt\pi\cdot6}{(105/16)\sqrt\pi}=\dfrac{96}{105}=\mathbf{0.9143}$ .

Step 2. Impose the counting rule: $A\cdot0.9143=2$ , so $A=2/0.9143=\mathbf{2.188}$ . The valence normalisation is fixed by the quantum number, not by a fit — exactly the role sum rules play.

Step 3. Compute the momentum integral with an extra power of $x$ : $\int_0^1 x\,u_V\,dx=A\,B(\tfrac32,4)=A\cdot\dfrac{\Gamma(3/2)\Gamma(4)}{\Gamma(11/2)}=2.188\cdot\dfrac{96}{945}=2.188\cdot0.1016=\mathbf{0.222}$ .

Step 4. Interpret. This toy valence $u$ carries about $22\%$ of the proton's momentum — comfortably consistent with the measured $\int x(u+\bar u)\approx0.36$ once the sea and the $d$ valence are added. The counting rule fixes the area to 2; the same distribution then carries ≈ 0.22 of the momentum, showing how number and momentum are independent moments of one PDF.

Physics of the portrait

Parton distributions became a predictive science the moment their scale dependence was understood — the step that turned Feynman's static partons into the quarks and gluons of QCD, and the achievement most associated with the Rome theorist who wrote the evolution equations in their working form.

Origins · 1 minGuido Altarelli and how this was discovered

Guido Altarelli

1941–2015 · QCD evolution of parton densities

Guido Altarelli was the theorist who made parton distributions calculable. Feynman had proposed partons in 1969 and Bjorken had predicted the approximate scaling that SLAC then saw, but the partons of that era were frozen snapshots — useful, yet with no dynamics and no way to relate a measurement at one energy to another. The breakthrough came in 1977, when Altarelli and Giorgio Parisi recast the deep results of QCD radiation into a compact, physically transparent set of integro-differential equations: the parton densities evolve with $\ln Q^2$ through convolution with universal splitting functions $P_{ij}(z)$ giving the probability that parton $j$ radiates parton $i$ carrying fraction $z$ . The same content had been reached from other directions by Gribov and Lipatov and by Dokshitzer — hence the modern name DGLAP — but the Altarelli–Parisi formulation, with its intuitive probabilistic reading, is the one every practitioner learned. Its impact was immediate and lasting: scaling violations, once a puzzle, became a precision confirmation of QCD; the gluon distribution, invisible to the photon, could be pinned down through the $Q^2$ dependence it drives; and factorisation plus evolution together made it possible to measure PDFs at one machine and predict cross sections at another. Altarelli spent his career at Rome and CERN, wrote the community's standard reviews of the Standard Model, and remained, until his death in 2015, one of the clearest voices in the field. Every PDF set fitted today — for the LHC, for the Electron-Ion Collider — is evolved with the equations that carry his name.

History

1969

— Feynman's parton model and Bjorken–Paschos give x the meaning of a momentum fraction

1969

— SLAC–MIT deep inelastic scattering reveals scaling and point-like constituents

1973

— Gross, Wilczek and Politzer discover asymptotic freedom, grounding partons in QCD

1974

— the momentum sum-rule deficit shows quarks carry only about half the proton momentum

1977

— Altarelli and Parisi write the evolution equations for the parton densities

1992

— HERA begins mapping the proton's PDFs down to x ≈ 10⁻⁴

Going further

Texts. Thomson, Modern Particle Physics, §8.5–8.6 (parton distributions and their measurement), the primary source for this article · Halzen & Martin, Quarks and Leptons, §9 (the parton model and scaling violations) · Griffiths, Introduction to Elementary Particles, 2nd ed., §8.4 · Peskin & Schroeder, An Introduction to QFT, §17.5 (DGLAP evolution and the parton distributions) · Ellis, Stirling & Webber, QCD and Collider Physics, Ch. 4 (the definitive treatment of PDFs and evolution). The key results to keep in mind across this article are: (1) $F_2^{ep}=x[\tfrac49(u+\bar u)+\tfrac19(d+\bar d+s+\bar s)]$ and Callan–Gross $F_2=2xF_1$ ; (2) the momentum sum rule with $\varepsilon_q\approx0.54\Rightarrow\varepsilon_g\approx0.46$ ; (3) the large- $x$ ratio $F_2^{en}/F_2^{ep}\to1/4$ for $d/u\to0$ ; (4) the toy-model results $B(\tfrac12,4)=96/105$ , $A=2.188$ , $\int x\,u_V\,dx=0.222$ ; (5) the DGLAP splitting function $P_{qq}(z)=C_F(1+z^2)/(1-z)$ with $C_F=4/3$ .

Prerequisites. Deep Inelastic Scattering · The Quark–Parton Model · Bjorken Scaling. Leads to. Hadron Collisions · Asymptotic Freedom · The Running of α_s.

Ask the tutor: Ask me to separate the flavour PDFs using proton, neutron and neutrino data, walk through a DGLAP evolution step by hand, or show how factorisation turns a measured gluon PDF into an LHC Higgs cross section.

Is x really "the fraction of the proton's momentum" carried by the quark?

Yes, but with one precise qualifier: it is the fraction of the proton's longitudinal momentum in a frame where the proton moves very fast — the infinite-momentum (or light-cone) frame. In that frame time dilation freezes the partons' internal motion and each carries a well-defined share $x$ of the big longitudinal momentum, with transverse and mass effects suppressed by $1/Q$ . The elastic-scattering condition for a massless quark, $(xp+q)^2=0$ , then pins the kinematic variable $x=Q^2/(2p\cdot q)$ to that momentum fraction exactly. In the proton rest frame the interpretation is murkier, which is why the parton model is always stated in the fast frame. So "momentum fraction" is correct and rigorous, provided you remember which frame makes it clean.

Why can't we just calculate PDFs from QCD instead of measuring them?

Because PDFs are intrinsically non-perturbative: they encode the low-energy, strongly-coupled structure of the bound proton, a regime where $\alpha_s$ is large and the usual Feynman-diagram expansion fails. Perturbative QCD can tell you how a PDF changes with scale — that is exactly what DGLAP evolution does — but it cannot give you the distribution at some starting scale from first principles; that boundary condition is genuinely non-perturbative. So we measure the PDFs at one scale by global fitting and evolve them everywhere else. The one route that bypasses experiment is lattice QCD, which computes the proton structure directly from the path integral; it now delivers moments and rough shapes of PDFs and is improving quickly, but the fits still set the standard for precision.

What makes PDFs "universal", and why does that matter?

Universality means the same parton distributions describe the proton no matter what process you use them in — electron scattering, neutrino scattering, or proton–proton collisions. It follows from the QCD factorisation theorem: at high scale a cross section separates cleanly into a short-distance, process-dependent partonic piece $\hat\sigma$ that you compute perturbatively, times long-distance, process-independent PDFs that belong to the hadron alone. This is what turns PDFs from a curiosity into infrastructure: you measure $f(x,Q^2)$ once, in clean deep inelastic scattering at HERA, and then predict Higgs, $W$ , $Z$ and top production at the LHC with the identical functions. Without universality every new process would need its own non-perturbative input and the predictive power would collapse.

If x is a momentum fraction, how can the PDF change with Q²?

Because "how many partons of fraction $x$ " depends on how finely you look. At low $Q^2$ a quark and the gluons it is about to radiate look like a single object; raise $Q^2$ and your probe resolves the radiation, so that one quark now appears as a quark of slightly lower $x$ plus a gluon plus possibly a $q\bar q$ pair. The total momentum is conserved — the sum rule always gives one — but it is redistributed toward smaller $x$ and toward more partons as the resolution sharpens. DGLAP evolution is precisely the QCD bookkeeping for this redistribution, and the resulting "scaling violations" (the slow drift of $F_2$ with $Q^2$ at fixed $x$ ) are one of the most beautifully confirmed predictions of the theory.

How do we know gluons carry half the momentum if the photon can't see them?

By subtraction, and it is one of the most elegant arguments in the subject. The electromagnetic and weak probes couple only to electrically- or weakly-charged partons — the quarks and antiquarks — so from deep inelastic scattering you can measure the total momentum fraction the quarks carry. That number comes out near $0.5$ , not $1$ . But the momentum sum rule is exact: everything inside the proton must add to unity. The missing $\approx0.46$ has to be carried by something neutral to the photon and the $W$ , and the only candidate in QCD is the gluon. So the gluon's momentum is inferred from a deficit, and later confirmed independently through the scaling violations it drives via DGLAP — two roads to the same conclusion.

What is the difference between valence and sea quarks?

Valence quarks are the three that give the proton its identity — two $u$ and one $d$ — and they carry its net charge, isospin and baryon number. Sea quarks are the transient $q\bar q$ pairs that gluons continually create and annihilate inside the proton; at any instant there are extra quarks and antiquarks present, but they come in particle–antiparticle pairs and carry no net quantum numbers. The distinction is made experimentally by the difference $q(x)-\bar q(x)$ : the antiquark part cancels the sea, leaving only the valence, which is exactly why the counting rules are written with that difference and integrate to $2$ , $1$ and $0$ for $u$ , $d$ and $s$ . The sea dominates at small $x$ ; the valence dominates at large $x$ .

Why does the gluon distribution blow up as x → 0?

Because soft gluon emission is copious. Every parton in the proton radiates gluons, and the probability to emit a gluon carrying a small fraction of the momentum is enhanced — the splitting function $P_{gg}(z)$ has a $1/z$ pole at small $z$ , so low- $x$ gluons are produced in profusion. DGLAP evolution then amplifies this: as $Q^2$ increases, the small- $x$ gluon density grows rapidly, and $xg(x)$ climbs steeply toward $x\to0$ . This is why the LHC, which collides at very high energy and so reaches very small $x$ , is effectively a gluon–gluon collider for its most important processes. The growth cannot continue forever — at extremely small $x$ the gluons overlap and recombination is expected to tame it (saturation), a regime the Electron-Ion Collider is designed to probe.