Electron-quark scattering

When the SLAC-MIT experiments fired 20 GeV electrons into liquid hydrogen in 1968 and saw them recoil at large angles far more often than anyone expected, the data screamed that the proton was not a soft ball of charge but a bag of hard, pointlike scatterers. Feynman's parton model made the interpretation sharp: deep inelastic electron-proton scattering is nothing more than the incoherent sum of elastic collisions between the electron and individual charged constituents — quarks. Strip away the proton and what remains is one of the cleanest calculations in the Standard Model: the pure QED process $e^-q \to e^-q$ , a single virtual photon exchanged between two Dirac currents. This article computes that subprocess from first principles. The QED Feynman rules give an electron current $\bar u(p_3)\,\gamma^\mu\,u(p_1)$ and a quark current $\bar u(p_4)\,\gamma_\mu\,u(p_2)$ , stitched together by the photon propagator $-ig_{\mu\nu}/q^2$ carrying momentum transfer $q=p_1-p_3$ . Squaring the amplitude, averaging over initial spins and summing over final ones with trace techniques, collapses everything to a strikingly compact Lorentz-invariant expression in the Mandelstam variables: $\langle|M_{fi}|^2\rangle = 2Q_q^2 e^4\,(s^2+u^2)/t^2$ . Two pieces of physics live inside it. The $1/t^2 = 1/q^4$ factor is the Rutherford forward peak — the long-range Coulomb tail of the massless photon. The numerator $s^2+u^2$ encodes helicity conservation: in the ultra-relativistic limit chirality is exact, only four of the sixteen spin combinations survive, and their interference paints the angular distribution. Recast in the inelasticity $y$ , the same formula carries the factor $1+(1-y)^2$ that reappears — unchanged — as the backbone of every deep inelastic structure-function measurement.

Key takeaways

Deep inelastic $e^-p$ scattering is built, parton by parton, from the elementary QED subprocess $e^-q \to e^-q$ : one photon exchanged in the $t$ -channel.
The spin-averaged matrix element is $\langle|M_{fi}|^2\rangle = 2Q_q^2 e^4\,(s^2+u^2)/t^2$ — quark charge enters squared, so the rate probes flavour content.
In the centre-of-mass frame this becomes $\dfrac{d\sigma}{d\Omega^*}=\dfrac{Q_q^2\alpha^2}{2s}\dfrac{1+\tfrac14(1+\cos\theta^*)^2}{\sin^4(\theta^*/2)}$ .
The $1/\sin^4(\theta^*/2)$ factor is the Rutherford/Coulomb forward peak of the massless photon propagator; the numerator $1+\tfrac14(1+\cos\theta^*)^2$ is an independent consequence of helicity conservation.
Recast with the inelasticity $y$ , the cross section carries the factor $1+(1-y)^2$ that survives verbatim into the deep inelastic structure functions.
Key idea: deep inelastic scattering at the parton level is nothing more than QED electron-quark elastic scattering — a single photon exchange whose angular structure is fixed by helicity conservation and whose magnitude scales with the square of the quark charge.

The core insight

Deep inelastic scattering at the parton level is nothing more than QED electron-quark elastic scattering: a single photon exchange whose angular structure is fixed by helicity conservation and whose magnitude scales with the square of the quark charge $Q_q^2$ .

Where this lives today

The elementary $e^-q$ cross section is still the beating heart of every lepton-hadron scattering programme. At HERA — the electron-proton collider that ran at DESY in Hamburg from 1992 to 2007 — the H1 and ZEUS detectors measured neutral-current DIS out to momentum transfers $Q^2\sim10^4\ \mathrm{GeV^2}$ , and the data are fit with exactly the $[1+(1-y)^2]$ kernel derived here, convolved with parton distributions; those fits are how we know the proton's quark and gluon content to percent precision. At high $Q^2$ the $Z^0$ boson also couples the electron to the quark, and its interference with the photon produces a small parity-violating asymmetry — measured at SLAC (E158) and Jefferson Lab (Qweak), and now targeted by MOLLER — that tests the electroweak mixing angle at low energy. Looking forward, the Electron-Ion Collider under construction at Brookhaven will fire polarised electrons at polarised protons and nuclei to map how quark and gluon spins and momenta build the proton in three dimensions. Every one of these measurements is, at its core, the QED process $e^-q\to e^-q$ computed on this page — a single photon, a quark charge squared, and a helicity factor.

The physical picture

At lowest order there is exactly one diagram. The electron comes in, radiates a virtual photon of spacelike virtuality $t=q^2<0$ , and the quark absorbs it; both particles are deflected but neither is destroyed. Everything about the process — its strength, its angular shape, its spin selectivity — is carried by that single exchanged photon. The first figure fixes the kinematics and the currents; read it as the literal translation of the QED Feynman rules into a picture.

Figure

The lowest-order Feynman diagram for e⁻q → e⁻q a single photon of virtuality t = q² = (p₁ − p₃)² is exchanged between the electron current and the quark current, giving amplitude M ∝ Q_q e²/q² and squared amplitude 2Q_q²e⁴(s²+u²)/t²

The strength of the interaction is not the whole story — its shape is. Because the photon is massless the propagator $1/q^2$ has no scale, and the cross section diverges as the electron is barely deflected: this is the same $1/\sin^4(\theta^*/2)$ forward peak Rutherford found scattering alpha particles off a $1/r$ Coulomb potential. On top of that geometric peak sits a spin-dependent modulation. In the ultra-relativistic limit helicity is conserved at each vertex, so only same-handed and opposite-handed combinations that respect chirality survive, and their interference gives the numerator $1+\tfrac14(1+\cos\theta^*)^2$ . The second figure plots the full angular factor on a log scale and pins it to real numbers.

Figure

Angular distribution of e⁻q → e⁻q the factor g = (1 + ¼(1+cosθ)²)/sin⁴(θ/2) on a log scale rises steeply toward the forward direction from g = 1.0 at θ = 180° to g = 5.0 at 90° and g = 417 at 30°, the 1/sin⁴(θ/2) photon-propagator peak of Rutherford

The core equations

The calculation begins by writing the amplitude straight off the diagram. Each vertex contributes a current, the internal photon contributes its propagator, and the electron and quark couple with strengths $e$ and $Q_q e$ respectively (the quark charge $Q_q$ is $+\tfrac23$ for up-type and $-\tfrac13$ for down-type quarks). The overall factor of $-i$ is irrelevant once we square.

M_{fi} = \frac{-iQ_q e^2}{q^2}\,\big[\bar u(p_3)\gamma^\mu u(p_1)\big]\,\big[\bar u(p_4)\gamma_\mu u(p_2)\big] \qquad (1)

where: $e$ — the electron charge magnitude, $\alpha=e^2/4\pi\approx1/137$ ; $Q_q$ — the quark charge in units of $e$ ; $q=p_1-p_3$ — the four-momentum transfer, so $q^2=t$ ; $\bar u\gamma u$ — the two Dirac currents. Reading it: the amplitude is a product of two currents divided by the photon virtuality, the momentum-space echo of two charges talking through a $1/r$ potential. Squaring, averaging over the four initial spin states and summing over final spins — done here with the trace technology of aq-pp-trace-techniques — collapses the whole spinor structure into Mandelstam variables. This is the central result of the article.

\langle |M_{fi}|^2\rangle = 2\,Q_q^2\, e^4\,\frac{s^2+u^2}{t^2} \qquad (2)

where: $s=(p_1+p_2)^2$ is the squared centre-of-mass energy; $t=(p_1-p_3)^2=q^2$ is the squared momentum transfer; $u=(p_1-p_4)^2$ ; and for massless external particles $s+t+u=0$ . Reading it: the $1/t^2$ is the propagator squared (the forward peak), and $s^2+u^2$ is the helicity-conservation numerator — everything else is a constant. Transforming to the centre-of-mass frame and writing $t$ and $u$ in terms of the scattering angle $\theta^*$ turns (2) into the differential cross section a detector actually measures. This is the electron-quark analogue of the Mott formula.

\frac{d\sigma}{d\Omega^*} = \frac{Q_q^2\,\alpha^2}{2s}\,\frac{1+\tfrac14(1+\cos\theta^*)^2}{\sin^4(\theta^*/2)} \qquad (3)

where: $\alpha=e^2/4\pi$ ; $\theta^*$ — the CM scattering angle; $\sin^4(\theta^*/2)$ — the Rutherford forward-peak denominator; $1+\tfrac14(1+\cos\theta^*)^2$ — the helicity numerator, running from $2$ (forward) down to $1$ (backward). Reading it: the rate falls as $1/s$ with energy and diverges as $\theta^*\to0$ . The kinematic dictionary that converts (2) into (3) is worth stating on its own, because it makes the identity $s^2/t^2=1/\sin^4(\theta^*/2)$ manifest.

s+t+u=0,\qquad t=-\frac{s}{2}\,(1-\cos\theta^*)=-s\,\sin^2\!\tfrac{\theta^*}{2},\qquad u=-\frac{s}{2}\,(1+\cos\theta^*) \qquad (4)

where: the massless on-shell conditions fix all three invariants from $s$ and one angle. Reading it: $t\le0$ always (spacelike exchange), $t\to0$ forward and $t\to-s$ backward, and $u$ is the forward-backward mirror of $t$ . Finally, define the inelasticity $y=-t/s=\tfrac12(1-\cos\theta^*)$ , the fraction of the electron's energy transferred in the target rest frame. In this variable the cross section wears the form that survives, untouched, into deep inelastic scattering.

\frac{d\sigma}{dy} = \frac{2\pi\,Q_q^2\,\alpha^2}{s}\,\frac{1+(1-y)^2}{y^2} \qquad (5)

where: $y\in[0,1]$ ; $1+(1-y)^2$ — the invariant helicity factor; $1/y^2$ — the propagator peak, since $y\propto -t$ . Reading it: this is the parton-level kernel; convolving it with quark momentum distributions $q(x)$ builds the measured DIS cross section.

Derivation AFrom the QED Feynman rules to the spin-averaged matrix element

The goal is to turn the single tree diagram into the compact invariant $\langle|M_{fi}|^2\rangle=2Q_q^2e^4(s^2+u^2)/t^2$ without ever writing an explicit spinor. We do it by contracting two Dirac traces — one for the electron line, one for the quark line — using the completeness relation for massless spinors and the standard four-gamma trace identity. The pay-off is that all spin information reduces to dot products of momenta, which are just Mandelstam variables, so the answer is manifestly Lorentz invariant and depends on the quark only through its charge squared.

Step 1. Form $|M_{fi}|^2$ from equation (1) and factor it into an electron tensor times a quark tensor: $\langle|M|^2\rangle=\tfrac14\,(Q_q^2e^4/q^4)\,L_e^{\mu\nu}L^{q}_{\mu\nu}$ , where the $\tfrac14$ averages over the two electron and two quark initial spins. Why: the amplitude is a product of two independent currents, so its square factorises into two separately Lorentz-covariant pieces joined by the propagator.

Step 2. Evaluate the electron tensor with the massless spin sum $\sum u\bar u=\slashed p$ : $L_e^{\mu\nu}=\mathrm{Tr}[\slashed p_3\gamma^\mu\slashed p_1\gamma^\nu]=4\big(p_3^\mu p_1^\nu+p_3^\nu p_1^\mu-g^{\mu\nu}(p_1\!\cdot\! p_3)\big)$ , and identically for the quark tensor $L^q_{\mu\nu}$ with $p_2,p_4$ . Why: summing over final and averaging over initial spins converts the outer product of spinors into a trace, and the four-gamma trace theorem does the rest — masses drop because we neglect $m_e,m_q\ll\sqrt s$ .

Step 3. Contract the two tensors. The cross terms collapse using $g^{\mu\nu}g_{\mu\nu}=4$ to give $L_e^{\mu\nu}L^q_{\mu\nu}=32\big[(p_1\!\cdot\! p_2)(p_3\!\cdot\! p_4)+(p_1\!\cdot\! p_4)(p_3\!\cdot\! p_2)\big]$ . Why: only the symmetric momentum products survive the contraction; this is the same lepton-tensor algebra that governs e⁻μ⁻ scattering, with the muon replaced by a quark.

Step 4. Insert massless Mandelstam identities $2p_1\!\cdot\! p_2=2p_3\!\cdot\! p_4=s$ and $2p_1\!\cdot\! p_4=2p_2\!\cdot\! p_3=-u$ : the bracket becomes $s^2/4+u^2/4$ , so $L_eL_q=8(s^2+u^2)$ . Assemble: $\langle|M|^2\rangle=\tfrac14(Q_q^2e^4/t^2)\cdot8(s^2+u^2)=2Q_q^2e^4(s^2+u^2)/t^2$ . Why: the dot products are literally the invariants, so the answer is frame-independent and the quark enters only through $Q_q^2$ — the flavour-blindness of QED at fixed charge.

Go deeper · Derivation B · 1 minThe centre-of-mass differential cross section and its forward peak

Derivation BThe centre-of-mass differential cross section and its forward peak

The aim is to convert the invariant $\langle|M|^2\rangle$ of equation (2) into the measurable $d\sigma/d\Omega^*$ of equation (3), and in doing so expose the two competing structures: the Rutherford $1/\sin^4(\theta^*/2)$ divergence supplied by the massless propagator, and the helicity numerator $1+\tfrac14(1+\cos\theta^*)^2$ supplied by spin. We use the standard two-body phase-space formula for massless particles, where the incoming and outgoing momenta have equal magnitude so the flux and phase-space factors combine into a clean prefactor.

Step 1. For a $2\to2$ process with all masses negligible the CM differential cross section is $d\sigma/d\Omega^*=\langle|M|^2\rangle/(64\pi^2 s)$ , because $|\mathbf p_f|=|\mathbf p_i|$ makes the phase-space and flux factors reduce to $1/(64\pi^2 s)$ . Why: with no mass scales the only dimensionful quantity is $s$ , and the kinematic prefactor is fixed entirely by it.

Step 2. Substitute the Mandelstam-to-angle dictionary (4): $t=-s\sin^2(\theta^*/2)$ gives $t^2=s^2\sin^4(\theta^*/2)$ , so $s^2/t^2=1/\sin^4(\theta^*/2)$ exactly, while $u=-\tfrac{s}{2}(1+\cos\theta^*)$ gives $u^2/s^2=\tfrac14(1+\cos\theta^*)^2$ . Why: these are the frame-specific parametrisations of the invariants; the first is the geometric heart of Rutherford scattering, the second the angular fingerprint of the spins.

Step 3. Write $s^2+u^2=s^2\big[1+\tfrac14(1+\cos\theta^*)^2\big]$ so that $(s^2+u^2)/t^2=[1+\tfrac14(1+\cos\theta^*)^2]/\sin^4(\theta^*/2)$ , and combine with the prefactor and $e^4=16\pi^2\alpha^2$ . The result is $d\sigma/d\Omega^*=(Q_q^2\alpha^2/2s)\,[1+\tfrac14(1+\cos\theta^*)^2]/\sin^4(\theta^*/2)$ . Why: converting $e^4$ to $\alpha^2$ is what turns a field-theory amplitude into a number an experimentalist can evaluate; the $1/2s$ carries the overall energy dependence.

Step 4. Read the limits directly. Forward, $\theta^*\to0$ : $\sin^4\to0$ so the rate diverges — the Coulomb/Rutherford peak, tamed in practice only by the finite acceptance or the target's screening. Backward, $\theta^*\to\pi$ : the numerator tends to $1$ and the distribution flattens to its isotropic floor. Why: the divergence is physical and famous — it is why total Coulomb cross sections are infinite and why experiments always quote differential rates at fixed angle.

Go deeper · Derivation C · 2 minThe invariant form and the bridge to deep inelastic scattering

Derivation CThe invariant form and the bridge to deep inelastic scattering

Here the objective is to recast the cross section in the boost-invariant variable $y$ and show that the resulting factor $1+(1-y)^2$ is exactly the angular structure that organises deep inelastic scattering, so that the elastic $e^-q$ kernel becomes the integrand of the DIS cross section. This is the step that justifies the whole enterprise: it shows the parton-model claim — DIS is incoherent elastic $e^-q$ scattering — is not a slogan but an algebraic identity between two cross sections.

Step 1. Define the inelasticity $y\equiv -t/s = \tfrac12(1-\cos\theta^*)$ , ranging from $0$ (forward, no energy transfer) to $1$ (backward). Then $t=-sy$ and, from $s+t+u=0$ , $u=-s(1-y)$ , so $u^2=s^2(1-y)^2$ . Why: $y$ is the Lorentz-invariant fraction of the electron's energy given up in the target rest frame — the natural variable once we embed the quark in a moving proton.

Step 2. Substitute into (2): $s^2+u^2=s^2[1+(1-y)^2]$ and $t^2=s^2y^2$ , so $\langle|M|^2\rangle=2Q_q^2e^4[1+(1-y)^2]/y^2$ . Convert the invariant $d\sigma/dt=\langle|M|^2\rangle/(16\pi s^2)$ and change variables with $dt=-s\,dy$ to obtain $d\sigma/dy=(2\pi Q_q^2\alpha^2/s)\,[1+(1-y)^2]/y^2$ , equation (5). Why: writing the rate against $y$ removes all reference to a particular frame's angle, which is essential once the quark carries only a fraction $x$ of the proton momentum.

Step 3. Promote to a proton target. A quark of flavour $i$ carrying momentum fraction $x$ has partonic $\hat s=xs$ ; summing incoherently over flavours with parton densities $q_i(x)$ and integrating gives the DIS double-differential cross section, whose electron-quark core carries precisely the factor $[1+(1-y)^2]$ — equivalently the combination $xy^2F_1+(1-y)F_2$ of structure functions. Why: the parton model is exactly this convolution, and the fact that the same helicity factor survives is what lets DIS measure $Q_q^2$ -weighted quark distributions.

Step 4. Note the fingerprint. The coefficient of the $y$ -dependence is fixed by the quark's spin-½ nature; a spin-0 constituent would give a different, $y$ -independent shape. Measuring the $1+(1-y)^2$ behaviour in the SLAC and HERA data was direct evidence that the proton's charged partons are spin-½ Dirac particles. Why: the shape of the cross section, not just its size, is a quantum-number measurement — the Callan-Gross relation $F_2=2xF_1$ seen from the parton side.

Limiting cases

Case	Result	Meaning
Forward, $\theta^*\to0$ ( $t\to0$ )	$d\sigma/d\Omega^\to\infty$ as $1/\sin^4(\theta^/2)$	Rutherford/Coulomb divergence from the massless photon propagator; the total cross section is finite only with a cutoff or screening
Backward, $\theta^*\to\pi$ ( $u\to0$ )	numerator $\to1$ , isotropic floor	only same-helicity ( $S_z=0$ ) amplitudes survive; the $90^\circ$ rate is $5\times$ this floor
Ultra-relativistic, $s\gg m^2$	masses negligible, helicity conserved	only $4$ of $16$ spin combinations contribute; chirality is a good quantum number
$y\to0$ (soft transfer)	$d\sigma/dy\propto1/y^2$	Rutherford limit — long-range photon exchange dominates the rate
$y\to1$ (full transfer)	$1+(1-y)^2\to1$	backscattering suppressed for the $S_z=\pm1$ channels by angular-momentum conservation
Spin-0 quark (counterfactual)	numerator loses its $y$ -dependence	the measured $1+(1-y)^2$ shape is direct proof the quark is spin-½

Common trapAssuming the angular distribution is just the Rutherford propagator

It is tempting to read equation (3) as pure Rutherford — the whole angular shape coming from the $1/\sin^4(\theta^*/2)$ photon propagator, with the numerator a boring kinematic correction. That is wrong, and the mistake hides the physics. The factor $1+\tfrac14(1+\cos\theta^*)^2$ is an independent consequence of QED helicity conservation: the sum over the four surviving chiral spin configurations. The two $S_z=0$ combinations (right-right and left-left) contribute an isotropic constant; the two $S_z=\pm1$ combinations (right-left and left-right) contribute the $\tfrac14(1+\cos\theta^*)^2$ term, which vanishes backward because a spin-1 state cannot reverse its momentum without violating angular-momentum conservation. Strip the numerator away and you lose the very signature — the $y$ -dependence — that told experimenters the proton's partons were spin-½ Dirac quarks and not spinless blobs. The propagator sets where the rate is large; helicity sets how it is distributed on top.

Worked examples

1 · Mandelstam variables and the helicity factor (confidence builder)

An electron and an up quark collide at centre-of-mass energy $\sqrt s = 10$ GeV and scatter through $\theta^*=60^\circ$ . Find $s,t,u$ , verify $s+t+u=0$ , and evaluate the angular factor two independent ways.

Step 1. Square the CM energy: $s=(10\ \mathrm{GeV})^2=\mathbf{100\ GeV^2}$ . With $\cos60^\circ=\tfrac12$ , equation (4) gives $t=-\tfrac{s}{2}(1-\tfrac12)=-50\times0.5=\mathbf{-25\ GeV^2}$ and $u=-\tfrac{s}{2}(1+\tfrac12)=-50\times1.5=\mathbf{-75\ GeV^2}$ .

Step 2. Check the massless constraint: $s+t+u=100-25-75=\mathbf{0}$ exactly, as it must for four massless legs. This is the fastest sanity check on any tree-level kinematics.

Step 3. Evaluate the invariant ratio in (2): $(s^2+u^2)/t^2=(100^2+75^2)/25^2=(10000+5625)/625=15625/625=\mathbf{25.0}$ .

Step 4. Evaluate the frame form in (3) independently: $[1+\tfrac14(1+\cos60^\circ)^2]/\sin^4(30^\circ)=[1+0.25(1.5)^2]/(0.5)^4=1.5625/0.0625=\mathbf{25.0}$ . The two agree because $s^2/t^2=1/\sin^4(\theta^*/2)$ identically — the invariant and angular pictures are one calculation. The kinematics close: s+t+u = 0 and both routes give the angular factor 25.0.

2 · The differential cross section, up versus down quark (fully worked)

At the same $\sqrt s=10$ GeV, evaluate $d\sigma/d\Omega^*$ at $\theta^*=90^\circ$ for scattering off an up quark ( $Q_u=+\tfrac23$ ) and a down quark ( $Q_d=-\tfrac13$ ). Convert to nb/sr and take the ratio.

Step 1. The angular factor at $90^\circ$ : $\cos90^\circ=0$ , so $[1+\tfrac14(1)^2]/\sin^4(45^\circ)=1.25/0.25=\mathbf{5.0}$ , the amber point in the second figure.

Step 2. Assemble equation (3) for the up quark with $Q_u^2=\tfrac49$ , $\alpha=1/137.036$ so $\alpha^2=5.325\times10^{-5}$ , and $s=100\ \mathrm{GeV^2}$ :

\frac{d\sigma}{d\Omega^*}=\frac{(4/9)(5.325\times10^{-5})}{2\times100}\times5.0=5.92\times10^{-7}\ \mathrm{GeV^{-2}}.

Step 3. Convert with $1\ \mathrm{GeV^{-2}}=0.3894\ \mathrm{mb}=3.894\times10^{5}\ \mathrm{nb}$ : $5.92\times10^{-7}\times3.894\times10^{5}=\mathbf{0.23\ nb/sr}$ for the up quark. The down quark carries $Q_d^2=\tfrac19$ , one quarter as large, giving $\mathbf{0.058\ nb/sr}$ .

Step 4. The ratio is fixed purely by charge: $\sigma_u/\sigma_d=Q_u^2/Q_d^2=(4/9)/(1/9)=\mathbf{4}$ . This is why electron scattering weights each quark by $Q_q^2$ : for an isoscalar target (equal $u$ and $d$ ) the mean squared charge is $\langle Q^2\rangle=\tfrac12(\tfrac49+\tfrac19)=\tfrac{5}{18}\approx0.278$ , the famous factor relating the electromagnetic and weak DIS cross sections. Up-quark rate 0.23 nb/sr, down-quark rate 0.058 nb/sr, ratio exactly 4 — the cross section reads out quark charge squared.

Physics of the portrait

Electron-quark scattering has no single discoverer, but it has a decisive interpreter. The bare QED amplitude is textbook Feynman-diagram technology; the idea that it is the microscopic content of a proton-smashing experiment is Richard Feynman's parton model, invented in 1968-69 to explain what SLAC had just seen.

Origins · 1 minRichard Feynman and how this was discovered

Richard Feynman

1918–1988 · the parton model and QED electron-quark scattering

Richard Feynman gave twentieth-century physics both the diagrams that make electron-quark scattering calculable and the picture that makes it meaningful. His 1948-49 formulation of quantum electrodynamics — the diagrams, the propagators, the rules for turning a process into an integral — is exactly the machinery that produces $M\propto Q_q e^2/q^2$ from a single glance at the vertices. Two decades later, visiting SLAC in 1968, he confronted the puzzle of deep inelastic electron-proton data: electrons were bouncing off at large angles far more often than a soft, extended proton could explain. Feynman's answer, worked out over 1968-69, was the parton model: treat the fast-moving proton as a swarm of nearly-free pointlike constituents — "partons" — and let the electron scatter elastically and incoherently off one at a time. Deep inelastic scattering was then nothing but a sum of the elementary QED processes computed here, each weighted by the parton's charge squared and its share of the proton's momentum. The scaling this predicted was precisely James Bjorken's independent result, and its confirmation identified the charged partons with the fractionally-charged quarks of Gell-Mann and Zweig. Feynman trusted the picture over the formalism: he insisted the partons were real, momentum-carrying objects, and the $Q_q^2$ -weighting and $1+(1-y)^2$ spin factor derived here are the quantitative fossils of that insistence.

History

1911

— Rutherford's α-scattering reveals the pointlike nucleus through a 1/sin⁴(θ/2) law

1929

— Mott extends Rutherford scattering to relativistic spin-½ electrons

1948

— Feynman formulates the diagrammatic rules of quantum electrodynamics

1968

— SLAC-MIT deep inelastic experiments see abundant large-angle electron scattering

1969

— Feynman's parton model recasts the data as elastic electron-quark scattering

1990

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

Going further

Texts. Thomson, Modern Particle Physics, §8.3 (electron-quark scattering) and §6.5.5 (trace techniques), Eqs. 6.68 and 3.50 · Halzen & Martin, Quarks and Leptons, Ch. 6 (electron-muon and electron-quark scattering) · Peskin & Schroeder, An Introduction to QFT, §5.1 and §5.4 · Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 7-8 · Schwartz, QFT and the Standard Model, Ch. 13. The key results to keep in view are: $\langle|M|^2\rangle=2Q_q^2e^4(s^2+u^2)/t^2$ from the tensor contraction; $d\sigma/d\Omega^*=(Q_q^2\alpha^2/2s)[1+\tfrac14(1+\cos\theta^*)^2]/\sin^4(\theta^*/2)$ ; the identity $s^2/t^2=1/\sin^4(\theta^*/2)$ ; the worked numbers $s=100,\,t=-25,\,u=-75\ \mathrm{GeV^2}$ at $\theta^*=60^\circ$ and $d\sigma/d\Omega^*=0.23\ \mathrm{nb/sr}$ (up quark) versus $0.058\ \mathrm{nb/sr}$ (down quark) at $90^\circ$ ; and $\langle Q^2\rangle=5/18$ for an isoscalar target. The lepton-tensor algebra is identical to that of $e^-\mu^-$ scattering with the muon replaced by a charged quark.

Prerequisites. Feynman Rules of QED · Mandelstam Variables · Trace Techniques · e⁺e⁻ → μ⁺μ⁻. Leads to. Deep Inelastic Scattering · The Quark-Parton Model · Bjorken Scaling.

Ask the tutor: Ask me to walk through the trace contraction line by line, or to show how the $1+(1-y)^2$ kernel builds the deep inelastic structure function $F_2(x)$ .

Why is deep inelastic scattering "the same as" elastic electron-quark scattering?

Because at high momentum transfer the photon's wavelength is far smaller than the proton, so the electron resolves and strikes a single quark rather than the proton as a whole. During the brief collision that quark behaves as a free, pointlike Dirac particle — this is asymptotic freedom in action — so the interaction is exactly the elastic QED process $e^-q\to e^-q$ computed here. The proton's role is only to supply the quark with some fraction $x$ of its momentum and to catch the debris afterward. Summing these elastic sub-collisions incoherently over all quarks and all momentum fractions, weighted by the parton distributions $q(x)$ , reproduces the full inelastic cross section. "Deep" means high $Q^2$ (small probe), "inelastic" means the proton shatters, and "parton-level elastic" is what each individual quark experiences.

Where does the factor

1+\tfrac14(1+\cos\theta^*)^2

actually come from?

From adding the four surviving helicity amplitudes in the massless limit. QED conserves helicity at each vertex, so of the sixteen possible initial-and-final spin assignments only four are non-zero: right-right→right-right, left-left→left-left, right-left→right-left, and left-right→left-right. The two "aligned" combinations have total spin projection $S_z=0$ along the beam and scatter isotropically, contributing the constant $1$ . The two "anti-aligned" combinations have $S_z=\pm1$ and cannot backscatter — a spin-1 state reversing its momentum would violate angular-momentum conservation — so they carry the factor $\tfrac14(1+\cos\theta^*)^2$ , which is $1$ forward and $0$ backward. Their sum is the numerator. It is genuine spin physics, not kinematics, and it is the same $1+(1-y)^2$ that fingerprints spin-½ quarks in DIS.

Why does the cross section blow up in the forward direction?

Because the exchanged photon is massless, so its propagator $1/q^2=1/t$ has no minimum scale. As the electron is deflected less and less, $t\to0$ , the propagator diverges, and the cross section grows without bound as $1/\sin^4(\theta^*/2)$ . This is the identical divergence Rutherford found scattering alpha particles off the $1/r$ Coulomb potential — the long range of the electromagnetic force means very small deflections are overwhelmingly likely. In practice the divergence is cut off: a detector has a minimum angle, a real target screens the charge at large distances, and in DIS the requirement of high $Q^2$ (to resolve a quark at all) keeps you away from $t=0$ . The total cross section is formally infinite, which is why one always quotes a differential rate at a specified angle or $Q^2$ .

Why

Q_q^2

and not

Q_q

— can this process tell a quark from an antiquark?

The rate scales as $Q_q^2$ because the amplitude is proportional to the charge and the cross section is the amplitude squared. An immediate consequence is that single-photon exchange is blind to the sign of the charge: an up quark ( $+\tfrac23$ ) and an up antiquark ( $-\tfrac23$ ) scatter electrons at identical rates. Electromagnetic DIS therefore measures the charge-squared-weighted sum $\sum_q Q_q^2[q(x)+\bar q(x)]$ and cannot, by itself, separate quarks from antiquarks — for that you need the parity-violating weak interaction (charged-current neutrino scattering) or the $\gamma$ - $Z$ interference asymmetry. The flavour information electromagnetism does give is powerful, though: the factor-of-four difference between up and down is exactly what pins down the proton's $u$ versus $d$ content.

What happened to the electron and quark masses?

They were neglected, and legitimately so, because deep inelastic scattering operates at energies vastly larger than either mass. The electron mass is $0.5$ MeV and the light-quark masses are a few MeV, while the momentum transfers of interest are GeV to hundreds of GeV — ratios of $m^2/s$ below $10^{-6}$ . Dropping the masses is what let the spin sums collapse to $\sum u\bar u=\slashed p$ and made helicity a conserved quantum number, which is the origin of the clean four-amplitude structure. Masses reappear only near threshold — for heavy-quark production ( $c$ , $b$ ) or at low $Q^2$ — where they cut off the forward divergence and mix the helicity states. For light quarks in ordinary DIS the massless approximation is excellent, and it is what makes the compact form $2Q_q^2e^4(s^2+u^2)/t^2$ possible.

How does this connect to the structure functions

F_1

and

F_2

Directly. When you embed the elastic $e^-q$ kernel in a proton and sum over quarks, the double-differential DIS cross section takes the form $d^2\sigma/dx\,dy\propto xy^2F_1+(1-y)F_2$ , where the $y$ -dependence is inherited entirely from the parton-level $1+(1-y)^2$ . In the parton model $F_2(x)=\sum_q Q_q^2\,x\,q(x)$ — the charge-squared-weighted momentum distribution — and the spin-½ nature of quarks forces the Callan-Gross relation $F_2=2xF_1$ . So the two structure functions are not independent: their ratio is fixed by the same helicity physics that gave us the numerator in equation (3), and measuring $F_2/2xF_1=1$ was among the earliest confirmations that the proton's charged partons carry spin ½.

When does the single-photon picture break down?

In three regimes. First, at momentum transfers approaching the weak scale ( $Q^2\sim M_Z^2\approx8300\ \mathrm{GeV^2}$ ), the $Z^0$ boson competes with the photon; their interference adds a parity-violating piece, as HERA saw at its highest $Q^2$ . Second, QCD corrections dress the bare quark: gluon radiation and quark-antiquark splitting make the parton distributions evolve with $Q^2$ (the DGLAP equations) and add order- $\alpha_s$ corrections, so "the quark charge squared" is only the leading term. Third, at very low $Q^2$ the photon no longer resolves individual quarks and the picture dissolves back into coherent scattering off the whole proton. Between these limits — high $Q^2$ but below the weak scale — the single-photon QED result computed here is the clean, dominant description.