Deep inelastic scattering

Fire an electron hard enough at a proton and the proton does not survive. Below a few hundred MeV of momentum transfer the electron sees a soft, extended blob of charge and bounces off elastically; the cross section falls steeply with $Q^2$ exactly as the dipole form factors predict, the fingerprint of a diffuse object. Turn the beam up and something startling happens. The proton shatters into a spray of hadrons $X$ , and the cross section — measured relative to scattering off a point charge — almost stops caring about $Q^2$ . A cross section that refuses to fall with increasing resolving power is the signature of scattering off something point-like. Deep inelastic scattering, $e^-p \to e^-X$ , is the process that turned that observation into the discovery of quarks. The theoretical machinery is a controlled generalisation of elastic $e^-p$ scattering. Elastic kinematics live on a single curve — fix the angle and the outgoing energy is determined — but once the proton breaks up, the electron's energy and angle become independent, and the cross section needs two variables. The most general single-photon, parity-conserving inelastic cross section is written in terms of two structure functions $F_1(x,Q^2)$ and $F_2(x,Q^2)$ of the momentum transfer $Q^2$ and the Bjorken scaling variable $x$ . Two experimental facts about them changed physics: Bjorken scaling, that at large $Q^2$ they depend on $x$ alone, and the Callan–Gross relation $F_2(x)=2xF_1(x)$ . The first says the proton contains point-like constituents; the second says those constituents carry spin one-half. Together they are the electron-scattering proof of the quark–parton model.

Key takeaways

Unlike elastic scattering, the deep-inelastic cross section depends only weakly on $Q^2$ — the tell-tale signature of scattering off point-like objects inside the proton.
The proton is broken up, so the electron's energy and angle become independent; the cross section requires two variables and is parametrised by two structure functions $F_1(x,Q^2)$ and $F_2(x,Q^2)$ .
Bjorken scaling: at large $Q^2$ the structure functions become essentially $Q^2$ -independent, $F_i(x,Q^2)\to F_i(x)$ — direct evidence for point-like constituents.
The Callan–Gross relation $F_2(x)=2xF_1(x)$ shows those constituents are spin-½; spin-0 partons would give $F_1=0$ instead.
In the parton model the Bjorken variable $x$ is the fraction of the proton's momentum carried by the struck quark: $F_2(x)=\sum_q e_q^2\,x\,[q(x)+\bar q(x)]$ .
Key idea: a cross section that stops falling with resolving power $Q^2$ betrays point-like substructure, so the observed Bjorken scaling directly reveals hard, spin-½ quarks inside the proton, with $x$ measuring each quark's momentum share.

The core insight

When a probe scatters off point-like substructure the cross section stops depending on the resolving power $Q^2$ — so the observed Bjorken scaling of the structure functions directly reveals tiny, hard, spin-½ constituents (quarks) inside the proton, and the Bjorken variable $x$ measures the momentum fraction each one carries.

Where this lives today

Deep inelastic scattering matured from a discovery into a precision industry. Between 1992 and 2007 the HERA collider at DESY smashed $27.5\ \mathrm{GeV}$ leptons into $920\ \mathrm{GeV}$ protons, extending the reach in $x$ and $Q^2$ by orders of magnitude. HERA saw two things at once: exact Bjorken scaling breaks logarithmically with $Q^2$ — the scaling violations predicted by QCD through the DGLAP evolution equations, from which the strong coupling $\alpha_s$ and the gluon distribution are extracted — and $F_2$ rises steeply at small $x$ , the signature of a sea of gluons and quark-antiquark pairs dominating the low- $x$ proton. The parton distributions $q(x,Q^2)$ measured this way are now indispensable input to every LHC prediction: a Higgs cross section is computed by convolving the hard process with these PDFs. Open frontiers remain sharp — the proton spin puzzle (quark spins supply only $\sim30\%$ of the proton's spin), gluon saturation as $x\to0$ , and three-dimensional proton tomography — and the Electron-Ion Collider now under construction at Brookhaven is built to answer them. Deep inelastic scattering, sixty years on, is still writing the map of the proton.

The physical picture

At high momentum transfer the virtual photon's wavelength shrinks below the size of the proton, so it no longer sees a smooth charge cloud but resolves individual charged constituents and knocks one of them clean out. The proton, now missing a quark and unable to stay colour-neutral, fragments into a jet of hadrons — the system $X$ whose invariant mass $W$ is one of the two kinematic handles. The first figure fixes what the detector actually records (an electron's angle and energy) and how those two numbers translate into the deep-inelastic variables.

Figure

Deep inelastic scattering e⁻p → e⁻X in the lab frame: the electron enters with energy E₁, scatters through angle θ to energy E₃, and a single virtual photon of spacelike momentum transfer Q² = 4E₁E₃sin²(θ/2) strikes one quark inside the proton. The proton fragments into a hadronic system X of invariant mass W, with energy transfer ν = E₁ − E₃ and Bjorken x = Q²/(2m_pν)

The decisive measurement is a comparison, and the second figure is its whole content. Plot the scattering rate, divided out by the point-charge (Mott) rate, against $Q^2$ . The elastic channel plunges — the proton's dipole form factor squared falls like $(1+Q^2/0.71)^{-4}$ , dropping more than two hundred-fold between $Q^2=1$ and $6\ \mathrm{GeV}^2$ , because a spread-out charge cannot recoil coherently at short wavelength. The deep-inelastic structure function $F_2$ at fixed $x$ does almost nothing: it stays flat. A form factor that does not fall means the photon is bouncing off something with no measurable size — a point. That flatness is Bjorken scaling, and it is the experimental birth certificate of the quark.

Figure

The punchline of SLAC 1968: the deep-inelastic structure function F₂ at fixed x ≈ 0.25 stays nearly flat with Q² (Bjorken scaling, blue), whereas the elastic dipole form factor squared (1 + Q²/0.71)⁻⁴ plunges more than 200-fold from Q² = 1 to 6 GeV² (amber, log scale). A cross section that will not fall betrays point-like constituents

The core equations

The single exchanged photon carries a spacelike four-momentum $q=p_1-p_3$ , and the same invariant that ruled elastic scattering rules here — neglecting the electron mass against its energy, it reduces in the lab to a product of the two measured quantities.

Q^2 \equiv -q^2 = 4E_1 E_3 \sin^2\!\frac{\theta}{2} \qquad (1)

where: $Q^2$ — positive four-momentum transfer squared (energy $^2$ ); $E_1,E_3$ — incoming and outgoing electron energies; $\theta$ — electron scattering angle. Reading it: $Q^2$ is the resolving power of the probe — large $Q^2$ means short wavelength and fine spatial detail. Nothing so far distinguishes elastic from inelastic; the difference is what the target does next. Because the proton breaks up, a second invariant is now independent. The energy the photon dumps into the target, $\nu$ , and the invariant mass $W$ of the debris are set by the same two numbers the detector reads, and elasticity ( $W=m_p$ ) is only the special edge of a two-dimensional region.

\nu = E_1 - E_3 = \frac{p\cdot q}{m_p}, \qquad W^2 = (p+q)^2 = m_p^2 + 2m_p\nu - Q^2 \qquad (2)

where: $p$ — proton four-momentum ( $p=(m_p,\mathbf 0)$ in the lab); $\nu$ — lab energy transfer; $W$ — invariant mass of the hadronic final state $X$ . Elastic scattering is $W=m_p$ ; the nucleon resonances ( $\Delta$ , etc.) sit at $W\lesssim 2\ \mathrm{GeV}$ ; the deep-inelastic continuum is $W\gtrsim 2\ \mathrm{GeV}$ . Two dimensionless variables organise the whole region. The Bjorken variable $x$ measures how elastic the collision is relative to a single constituent, and the inelasticity $y$ is the fractional energy the electron surrenders.

x = \frac{Q^2}{2\,p\cdot q} = \frac{Q^2}{2m_p\nu}, \qquad y = \frac{p\cdot q}{p\cdot p_1} = \frac{\nu}{E_1} \qquad (3)

where: $0\le x\le 1$ with $x=1$ for elastic scattering (then $W=m_p$ ); $0\le y\le 1$ . A useful identity ties them to the CM energy $s=(p_1+p)^2$ : $Q^2 = (s-m_p^2)\,x\,y \approx 2m_pE_1\,x\,y$ . So $x$ and $y$ replace $\theta$ and $E_3$ as the natural pair. The dynamics enter through the cross section. The most general single-photon, parity-conserving inelastic cross section factorises the leptonic tensor (known exactly) against an unknown hadronic tensor whose two independent invariants are the structure functions $W_1,W_2$ .

\frac{d^2\sigma}{d\Omega\,dE_3} = \frac{\alpha^2}{4E_1^2\sin^4(\theta/2)}\left[\,W_2\cos^2\!\frac{\theta}{2} + 2W_1\sin^2\!\frac{\theta}{2}\,\right] \qquad (4)

where: $\alpha=e^2/4\pi\approx1/137$ ; the prefactor is the Mott envelope; $W_1(\nu,Q^2)$ multiplies the magnetic/transverse $\sin^2$ term and $W_2(\nu,Q^2)$ the $\cos^2$ term. These two functions carry everything the photon can learn about the proton's structure — they are not assumed, they are measured. It is cleaner to switch to dimensionless structure functions $F_1,F_2$ and to the invariant variables $x,Q^2$ . In this form the spin-½ nature of the target's parts is laid bare by the angular ( $y$ ) dependence.

F_1 = m_p W_1,\quad F_2 = \nu W_2;\qquad \frac{d^2\sigma}{dx\,dQ^2} = \frac{4\pi\alpha^2}{Q^4}\left[\,(1-y)\frac{F_2}{x} + y^2 F_1\,\right] \qquad (5)

where: $F_1(x,Q^2)$ , $F_2(x,Q^2)$ are dimensionless. Using the Callan–Gross relation $F_2=2xF_1$ the bracket collapses to $\tfrac{F_2}{2x}\,[1+(1-y)^2]$ — the same $[1+(1-y)^2]$ signature that betrays spin-½ scattering in $e^+e^-\to\mu^+\mu^-$ . The two landmark experimental facts are statements about these functions. At large $Q^2$ they forget their $Q^2$ dependence, and at every $x$ they obey a fixed ratio.

F_i(x,Q^2)\;\xrightarrow[\;x\text{ fixed}\;]{\;Q^2\to\infty\;}\;F_i(x), \qquad F_2(x)=2x\,F_1(x) \qquad (6)

where: the left equation is Bjorken scaling (point-like constituents); the right is the Callan–Gross relation (spin-½ constituents). Spin-0 partons would instead give $F_1=0$ , i.e. $2xF_1/F_2\to0$ — decisively excluded by the data. Both facts are explained at a stroke by the parton model: the photon scatters elastically off one free, point-like, spin-½ quark carrying a fraction $x$ of the proton's momentum, and $F_2$ is a charge-weighted census of those quarks.

F_2(x) = \sum_q e_q^2\,x\,\big[q(x)+\bar q(x)\big] \qquad (7)

where: $e_q$ — quark charge in units of $e$ ( $+\tfrac23$ for $u$ , $-\tfrac13$ for $d,s$ ); $q(x)$ — the parton distribution function, the number density of quarks of flavour $q$ carrying momentum fraction $x$ ; $\bar q(x)$ — the antiquark (sea) density. Equation (7) is why $x$ is called a distribution variable: measuring $F_2(x)$ maps the proton's momentum among its quarks.

Derivation AFrom scattering angle and energy to the deep-inelastic variables

The goal is to build the kinematic dictionary of DIS: to show that the two laboratory observables — the electron's scattering angle $\theta$ and its outgoing energy $E_3$ — determine the momentum transfer $Q^2$ , the energy transfer $\nu$ , the invariant hadronic mass $W$ , and the dimensionless variables $x$ and $y$ , and to prove that the elastic limit is exactly $x=1$ . Everything rests on four-momentum conservation and the single decision to keep the proton mass exact while neglecting the electron mass, valid because every beam of interest has $E_1\gg m_e=0.511\ \mathrm{MeV}$ .

Step 1. Define the photon momentum $q=p_1-p_3$ and form the invariant $q^2=(p_1-p_3)^2=m_e^2+m_e^2-2p_1\cdot p_3\approx-2E_1E_3(1-\cos\theta)$ , using $|\mathbf p_1|\approx E_1$ , $|\mathbf p_3|\approx E_3$ . The half-angle identity $1-\cos\theta=2\sin^2(\theta/2)$ gives $Q^2=-q^2=4E_1E_3\sin^2(\theta/2)$ . Why: $Q^2$ is a Lorentz invariant, so it may be evaluated in the lab where it collapses to a product of the two numbers the spectrometer records.

Step 2. Contract $q$ with the target. In the lab $p=(m_p,\mathbf 0)$ , so $p\cdot q=m_p(E_1-E_3)\equiv m_p\nu$ , which defines the energy transfer $\nu=E_1-E_3$ as a frame-dependent shadow of the invariant $p\cdot q$ . Why: $p\cdot q$ is the second independent invariant DIS needs — elastic scattering had only one ( $Q^2$ ), because there the recoil pinned $\nu$ to $Q^2$ .

Step 3. Compute the debris mass. The final hadronic system carries $p+q$ , so $W^2=(p+q)^2=m_p^2+2\,p\cdot q+q^2=m_p^2+2m_p\nu-Q^2$ . Elastic scattering means the target stays a single proton, $W=m_p$ , which forces $2m_p\nu=Q^2$ , i.e. $x=Q^2/2m_p\nu=1$ . Why: this exhibits $x=1$ as the elastic edge — deep inelastic scattering is everything with $x<1$ , where the extra invariant mass $W^2-m_p^2=2m_p\nu(1-x)$ goes into breaking the proton.

Step 4. Assemble the dimensionless pair. Define $x=Q^2/2m_p\nu$ and $y=\nu/E_1$ . With $s-m_p^2=2m_pE_1$ (lab), eliminate $\nu$ : $Q^2=2m_p\nu\,x=2m_pE_1\,y\,x=(s-m_p^2)xy$ . Why: this is the change of variables $\{\theta,E_3\}\to\{x,y\}$ that makes the cross section (5) transparent; $x$ says which constituent was hit and $y$ says how hard.

Go deeper · Derivation B · 2 minBjorken x is the struck quark's momentum fraction

Derivation BBjorken x is the struck quark's momentum fraction

The aim is the conceptual heart of the subject: to prove that in the parton model the abstract kinematic ratio $x=Q^2/2m_p\nu$ equals the fraction $\xi$ of the proton's momentum carried by the quark the photon actually strikes, and hence that Bjorken scaling and the parton census (7) follow from treating the quarks as free, point-like, on-shell objects during the brief instant of the collision. The physical licence is the impulse approximation: at large $Q^2$ the interaction time is far shorter than the time for quarks to interact with one another, so the struck quark behaves as if free.

Step 1. Work in a frame where the proton's momentum is huge (the "infinite-momentum frame"), so transverse momenta and masses are negligible and each quark carries a definite longitudinal fraction $\xi$ of the proton momentum $p$ , i.e. its four-momentum is $\xi p$ . Why: in this frame the proton is a beam of collinear, quasi-free partons, and "momentum fraction" is a clean, boost-invariant label.

Step 2. Demand that the struck quark stays on shell (massless) after absorbing the photon: $(\xi p+q)^2=0$ . Expanding, $\xi^2 p^2+2\xi\,p\cdot q+q^2=0$ ; drop $p^2=m_p^2$ against the large invariants and use $p\cdot q=m_p\nu$ , $q^2=-Q^2$ to get $2\xi\,m_p\nu-Q^2=0$ . Why: an elastic hit on a free constituent has exactly one kinematic constraint — the outgoing quark must be a real particle — and that single equation fixes which quark could have been hit.

Step 3. Solve: $\xi=Q^2/2m_p\nu=x$ . The photon can only scatter elastically off the quark whose momentum fraction is precisely the Bjorken $x$ reconstructed from the electron. Averaging over the proton's quarks weights this by the probability $q(\xi)\,d\xi$ of finding such a quark, inserting a factor $\delta(\xi-x)$ . Why: this is the punchline — $x$ is not a bookkeeping abstraction but a directly physical quantity, the momentum share of the parton that was struck.

Step 4. Sum the elastic quark cross sections weighted by charge and density. Elastic scattering off a point spin-½ charge $e_q$ contributes $e_q^2$ ; summing over flavours gives $F_2(x)=\sum_q e_q^2\,x\,[q(x)+\bar q(x)]$ and, crucially, functions of $x$ alone — there is no leftover $Q^2$ because a point has no form factor. Why: the absence of any intrinsic scale in a point constituent is exactly Bjorken scaling; the $Q^2$ -independence the SLAC data showed is the statement that quarks have no measurable size.

Go deeper · Derivation C · 1 minThe Callan-Gross relation and the spin of the quarks

Derivation CThe Callan-Gross relation and the spin of the quarks

The purpose is to show that the fixed ratio $F_2(x)=2xF_1(x)$ is not an accident of fitting but a direct consequence of the constituents being spin-½, and that the alternative — spin-0 partons — gives a qualitatively different and experimentally excluded prediction, $F_1=0$ . The tool is the separation of the virtual photon's polarisations: a spacelike photon has both transverse and longitudinal (scalar) modes, and the two structure functions weight them differently.

Step 1. Decompose the cross section into the response to transversely and longitudinally polarised virtual photons, with cross sections $\sigma_T$ and $\sigma_L$ . One finds $2xF_1\propto\sigma_T$ while $F_2-2xF_1\propto\sigma_L$ ; define the ratio $R=\sigma_L/\sigma_T=(F_2-2xF_1)/(2xF_1)$ (at large $Q^2$ , neglecting target-mass terms). Why: casting the two structure functions as two photon-absorption cross sections turns "what is the ratio $F_2/2xF_1$ " into "can the constituent absorb a longitudinal photon", a question answered by its spin.

Step 2. A massless spin-½ quark cannot absorb a longitudinal photon: helicity conservation at the point $\gamma^* q\to q$ vertex forbids it, so $\sigma_L\to0$ and $R\to0$ . Therefore $F_2-2xF_1=0$ , i.e. $F_2=2xF_1$ — the Callan–Gross relation. Why: helicity conservation for massless spin-½ fermions is the same mechanism that fixes the $[1+(1-y)^2]$ angular law; the relation is a spin fingerprint, not a dynamical detail.

Step 3. Contrast spin-0. A scalar constituent couples to the photon through its charge current with no magnetic moment, so it absorbs only longitudinal photons in this kinematics: $\sigma_T\to0$ , hence $2xF_1\to0$ and $F_1=0$ , giving $R\to\infty$ . Why: the two spin hypotheses make opposite predictions for which structure function survives, so a single measurement decides between them.

Step 4. Confront the data. SLAC and later experiments found $R\approx0.18$ , small and roughly constant — close to the spin-½ value $R=0$ (the small nonzero part is the gluon-radiation correction of QCD), and nowhere near the spin-0 divergence. The quarks are spin-½. Why: the Callan–Gross relation, verified, is the electron-scattering proof that the proton's charged constituents carry the same spin as the electron probing them.

Limiting cases

Case	Result	Meaning
Elastic edge, $x=1$	$W=m_p$ , structure functions peak at $x=1$	recovers elastic $e^-p$ scattering and the Rosenbluth formula
Resonance region, $W\lesssim2\ \mathrm{GeV}$	$F_i$ show $\Delta$ /N* bumps, no scaling	excited-baryon states, not yet the deep-inelastic continuum
Bjorken limit, $Q^2,\nu\to\infty$ , $x$ fixed	$F_i(x,Q^2)\to F_i(x)$	scaling — point-like constituents with no form factor
Spin-½ partons	$F_2=2xF_1$ , $R=\sigma_L/\sigma_T\to0$	Callan–Gross; confirmed, quarks are spin-½
Spin-0 partons	$F_1=0$ , $R\to\infty$	excluded by the data
$x\to0$	$F_2$ rises steeply	sea quarks and gluons dominate at small momentum fraction
$x\to1$	$F_2\to0$	one valence quark carries nearly all the momentum; rare

Common trapReading the structure functions as Fourier transforms of a static charge cloud

The elastic form factors $G_E(Q^2)$ and $G_M(Q^2)$ genuinely are Fourier transforms of the proton's static charge and magnetisation distributions — one number per $Q^2$ , decaying because the charge is spread out. It is tempting to carry that picture over and imagine $F_1(x,Q^2)$ and $F_2(x,Q^2)$ as Fourier transforms of some richer static distribution. They are not. Two facts break the analogy. First, they depend on two variables, not one: the extra variable $x$ encodes the dynamics of scattering off a moving constituent, not a snapshot of where charge sits. Second — and this is the whole point — they do not fall with $Q^2$ ; a Fourier transform of any finite-size distribution must decay, so the observed scaling positively forbids the static interpretation. The right picture is dynamical: the photon strikes one point-like quark carrying momentum fraction $x$ , and $F_2(x)$ is the charge-weighted probability of finding such a quark, not the density of charge at a radius.

Worked examples

1 · Reconstructing a deep-inelastic event (fully worked)

A $10.0\ \mathrm{GeV}$ electron scatters off a proton ( $m_p=0.938\ \mathrm{GeV}$ ) through $\theta=10.0^\circ$ and emerges with $E_3=5.0\ \mathrm{GeV}$ . Find $\nu$ , $y$ , $Q^2$ , $x$ , and the hadronic mass $W$ , and confirm the event is genuinely deep-inelastic.

Step 1. Energy transfer and inelasticity. $\nu=E_1-E_3=10.0-5.0=\mathbf{5.0\ GeV}$ , and $y=\nu/E_1=5.0/10.0=\mathbf{0.50}$ — the electron gives up half its energy.

Step 2. Momentum transfer from equation (1). With $\sin(5^\circ)=0.08716$ , $\sin^2(5^\circ)=0.007596$ : $Q^2=4(10.0)(5.0)(0.007596)=\mathbf{1.52\ GeV^2}$ . The wavelength probed is $\hbar/\sqrt{Q^2}\approx0.16\ \mathrm{fm}$ , well inside the $\sim0.84\ \mathrm{fm}$ proton.

Step 3. Bjorken $x$ from equation (3): $x=Q^2/(2m_p\nu)=1.52/(2\times0.938\times5.0)=1.52/9.38=\mathbf{0.162}$ . The struck quark carries about $16\%$ of the proton's momentum.

Step 4. Hadronic mass from equation (2): $W^2=m_p^2+2m_p\nu-Q^2=0.880+9.38-1.52=8.74\ \mathrm{GeV}^2$ , so $W=\mathbf{2.96\ GeV}$ . Since $W\approx3\ \mathrm{GeV}$ sits well above the resonance region ( $W\gtrsim2\ \mathrm{GeV}$ ) and $Q^2>1\ \mathrm{GeV}^2$ , this is a bona-fide deep-inelastic event.

Step 5. Consistency check with the invariant identity. $s-m_p^2=2m_pE_1=18.76\ \mathrm{GeV}^2$ , and $(s-m_p^2)\,x\,y=18.76\times0.162\times0.50=1.52\ \mathrm{GeV}^2=Q^2$ . The kinematics close exactly. A 10 GeV electron at 10° losing half its energy probes Q² = 1.52 GeV², strikes a quark carrying x = 0.16, and blows the proton into a W = 2.96 GeV hadronic shower.

2 · What the parton model predicts for the neutron and the momentum budget

Using the parton model with isospin symmetry (the proton's up-quark density equals the neutron's down-quark density, and vice versa), find the bounds on the ratio $F_2^{en}/F_2^{ep}$ and interpret the measured momentum sum rule.

Step 1. Write the structure functions from equation (7), keeping the valence quarks. For the proton (uud), $F_2^{ep}=x\big[\tfrac49 u(x)+\tfrac19 d(x)\big]$ ; for the neutron (udd), isospin swaps $u\leftrightarrow d$ , so $F_2^{en}=x\big[\tfrac49 d(x)+\tfrac19 u(x)\big]$ , where $u,d$ are the proton's densities.

Step 2. Form the ratio: $\dfrac{F_2^{en}}{F_2^{ep}}=\dfrac{4d+u}{4u+d}$ . This is bounded: if $d\to0$ (only up-quarks) it gives $\tfrac14$ ; if $u\to0$ it gives $4$ ; if $u=d$ it gives $1$ . Hence $\mathbf{\tfrac14\le F_2^{en}/F_2^{ep}\le 4}$ — a parameter-free prediction of the quark charges, spectacularly confirmed.

Step 3. Read the data at large $x$ . Measurements show the ratio falling toward $\tfrac14$ as $x\to1$ , meaning a single up-quark carries the proton's momentum there. At $x=0.5$ the ratio is about $0.5$ ; solving $(4d+u)/(4u+d)=0.5$ gives $d/u=\mathbf{0.29}$ — the down-quark density is roughly a third of the up at that momentum fraction.

Step 4. The momentum sum rule. Integrating equation (7)'s parton content, $\int_0^1 x\sum_q\big[q(x)+\bar q(x)\big]\,dx\approx\mathbf{0.5}$ — the charged quarks carry only about half the proton's momentum. The missing half is carried by electrically neutral constituents invisible to the photon: the gluons. This shortfall was the first quantitative hint that quarks are bound by glue. The quark charges alone bound F₂ᵉⁿ/F₂ᵉᵖ between ¼ and 4, and the momentum sum rule shows quarks carry ~50%, leaving the rest for gluons.

Physics of the portrait

Deep inelastic scattering is remembered as an experimental triumph — the SLAC-MIT spectrometers, the movable magnets, the plots that would not fall. But the person who told the experimenters what to look for, and why a flat cross section would matter, was a theorist working from the abstract machinery of current algebra.

Origins · 1 minJames Bjorken and how this was discovered

James Bjorken

1934–2024 · current algebra, scaling, and the birth of the parton picture

James "BJ" Bjorken predicted the central result of this article before it was measured, and from a direction almost no one expected. In the mid-1960s he was applying current algebra — a set of exact operator relations and sum rules for the currents — to inelastic electron scattering, and by 1969 he derived a scaling prediction: in the limit of large $Q^2$ and $\nu$ with $x=Q^2/2m_p\nu$ fixed, the structure functions should collapse to functions of $x$ alone. To many the argument looked like formal sleight of hand, and Bjorken presented it cautiously. When the SLAC-MIT group ran the first inelastic electron-proton experiments in 1968 and found the cross section falling only weakly with $Q^2$ , it was Bjorken's scaling the data obeyed. Richard Feynman, visiting SLAC, recognised the same behaviour as elastic scattering off free point-like constituents he called partons, and identified Bjorken's $x$ as their momentum fraction — the interpretation this article follows. Callan and Gross then tied $F_2=2xF_1$ to spin-½, and the partons were revealed as Gell-Mann and Zweig's quarks made real. Bjorken's variable, his scaling, and his sum rules run through every line of deep-inelastic physics; Friedman, Kendall, and Taylor won the 1990 Nobel Prize for confirming what his algebra had foreseen.

History

1964

— Gell-Mann and Zweig independently propose the quark model as a classification scheme

1968

— the SLAC-MIT experiment finds the inelastic cross section nearly independent of Q²

1969

— Bjorken predicts scaling and Feynman interprets it as scattering off point-like partons

1969

— Callan and Gross relate F₂ = 2xF₁ to spin-½ constituents

1990

— Friedman, Kendall, and Taylor share the Nobel Prize for the discovery

Going further

Texts. Thomson, Modern Particle Physics, §8.2–8.3 (deep inelastic scattering, structure functions, Bjorken scaling and the Callan–Gross relation) and §7.5 (Rosenbluth formula, eq. 7.33; Mott cross section eq. 7.36) — the primary source for the notation used here · Halzen & Martin, Quarks and Leptons, §8–9 (inelastic lepton–nucleon scattering and the parton model) · Griffiths, Introduction to Elementary Particles, 2nd ed., §8.3 · Peskin & Schroeder, An Introduction to Quantum Field Theory, §17.3–17.5 (DIS and the parton model) · Halzen–Martin's parton-model chapter for the sum rules. Key checkpoints in the derivations: $x=1$ is the elastic edge, recovered from $W=m_p$ ; the on-shell condition $(\xi p+q)^2=0$ gives $\xi=x$ when $m_p^2$ is neglected against the large invariants; the worked event $E_1=10\ \mathrm{GeV}$ , $\theta=10^\circ$ , $E_3=5\ \mathrm{GeV}$ yields $Q^2=1.52\ \mathrm{GeV}^2$ , $x=0.162$ , $W=2.96\ \mathrm{GeV}$ , and satisfies the identity $(s-m_p^2)xy=Q^2$ ; the quark charges alone bound the neutron-to-proton ratio $\tfrac14\le F_2^{en}/F_2^{ep}\le4$ ; the elastic dipole $(1+Q^2/0.71)^{-4}$ and the momentum sum rule $\approx0.5$ are standard results.

Prerequisites. The Rosenbluth Formula · Form Factors and Charge Distributions · Elastic versus Inelastic Scattering · Mandelstam Variables. Leads to. Bjorken Scaling · The Quark–Parton Model · Parton Distribution Functions · Electron–Quark Scattering.

Ask the tutor: Ask me to reconstruct $Q^2$ , $x$ , $y$ , and $W$ for any beam energy and angle, or to walk the parton-model prediction for $F_2$ from a set of quark charges and densities.

What is the difference between the variables x and y?

They answer two different questions about the same collision. The Bjorken variable $x=Q^2/2m_p\nu$ is a statement about what was hit: in the parton model it is the fraction of the proton's momentum carried by the struck quark, and it ranges from $0$ (a soft sea quark or gluon) to $1$ (a single quark carrying everything, which is elastic scattering). The inelasticity $y=\nu/E_1$ is a statement about how hard: the fraction of the electron's energy handed to the proton, from $0$ (a glancing forward scatter) to $1$ (the electron nearly stops). They are independent — you need both to specify a deep-inelastic event — and together with the beam energy they fix $Q^2=(s-m_p^2)xy$ . The angular dependence of the cross section is carried by $y$ ; the constituent structure is carried by $x$ .

Why does a weak Q² dependence prove the proton contains point-like objects?

Because a form factor is the Fourier transform of a size, and a point has no size. When you scatter off a spread-out charge distribution, increasing $Q^2$ means probing with a shorter wavelength, and the distribution can no longer respond coherently — the cross section falls, exactly as the proton's elastic dipole form factor falls by orders of magnitude. If instead the cross section stays flat as $Q^2$ climbs, the target is responding as strongly to a fine probe as to a coarse one, which is only possible if it has no internal scale to resolve — it is point-like. Bjorken scaling is precisely this flatness for the inelastic structure functions, and it is why the SLAC result was read immediately as evidence for hard, small constituents rather than a soft continuum.

Is the Bjorken x really the quark's momentum fraction, or just a definition?

It starts as a pure definition — $x=Q^2/2m_p\nu$ is built entirely from the electron's measured angle and energy, with no reference to quarks. The remarkable thing, proved in the second derivation, is that in the parton model this defined quantity equals the momentum fraction $\xi$ of the struck quark. The logic is one line: if the quark is free and massless, then after absorbing the photon it must still be a real particle, $(\xi p+q)^2=0$ , and that single on-shell condition forces $\xi=x$ . So the photon can only scatter elastically off the quark whose momentum fraction is exactly the $x$ you reconstructed. The definition and the physical fraction coincide — that coincidence is the content of the parton model, and it is why measuring $F_2(x)$ maps the proton's momentum distribution.

What exactly does the Callan-Gross relation tell us?

That the proton's charged constituents are spin-½ fermions, like the electron. The two structure functions measure the proton'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 response vanishes and the two structure functions lock together as $F_2=2xF_1$ . Had the constituents been spin-0, the opposite would hold: only the longitudinal response survives, $F_1=0$ , and the ratio $R=\sigma_L/\sigma_T$ would diverge. The measured $R\approx0.18$ is small and close to the spin-½ prediction of zero (the little that remains is QCD gluon radiation), decisively picking spin-½. It is the electron-scattering proof that quarks carry the same spin as the probe.

If quarks carry only half the proton's momentum, what carries the rest?

Gluons. The momentum sum rule adds up the momentum fractions of everything the photon can see — the charged quarks and antiquarks — and the integral comes out near $0.5$ , not $1$ . Half the proton's momentum is missing from the electromagnetic census. The resolution is that the proton also contains the gauge bosons of the strong force, the gluons, which are electrically neutral and therefore invisible to the photon; they carry the other half. This was one of the earliest quantitative hints that quarks are bound by a gauge theory (QCD) with its own force carriers, later confirmed directly by the observation of three-jet events and by the way the parton distributions evolve with $Q^2$ .

How is deep inelastic scattering different from elastic ep scattering?

Elastic scattering leaves the proton intact ( $W=m_p$ , i.e. $x=1$ ), so its kinematics live on a single curve — fix the angle and the outgoing energy is determined — and its cross section is described by the two form factors $G_E(Q^2)$ and $G_M(Q^2)$ , both falling steeply with $Q^2$ because the proton is extended. Deep inelastic scattering shatters the proton into a hadronic system of mass $W>m_p$ ( $x<1$ ), so the electron's angle and energy become independent and the cross section needs two variables and two structure functions $F_1(x,Q^2)$ , $F_2(x,Q^2)$ . The signature difference is behaviour with $Q^2$ : elastic falls, inelastic scales. That contrast — the same plot showing one channel plunging while the other stays flat — is the whole discovery.