Bjorken scaling and the parton model
Scaling from point constituents, the infinite-momentum frame, the master formula.
In 1968 the SLAC-MIT spectrometers measured $\nu W_2$ at scattering angles and energies nobody had reached before, and found something a generation of nuclear physics said should not happen: at fixed $x$, the structure function did not care about $Q^2$. This section derives why that single fact — Bjorken scaling — means the proton contains pointlike constituents; builds the frame in which the parton picture is actually a theorem (the infinite-momentum frame); and then derives the master formula of this whole field, the one that turns $F_2$ into a weighted census of partons.
What scaling says, precisely
$F_2$ is a function of two invariants. Bjorken's 1968 prediction (from current-algebra arguments that predate QCD) concerns the limit $Q^2 \to \infty$, $\nu \to \infty$ with $x = Q^2/2M\nu$ fixed — the Bjorken limit:
A dimensionless function of a dimensionless ratio, with the probe's resolution dropping out entirely. The SLAC data showed exactly this: at $\omega \equiv 1/x = 4$, $\nu W_2$ sat flat near $0.32$ while $Q^2$ ranged over almost an order of magnitude. Nothing with internal structure on a fixed length scale behaves like that — as the elastic channel itself demonstrates.
Scatter elastically off a static charge distribution $\rho(\vec r\,)$: the amplitude picks up the Fourier transform $F(Q^2) = \int d^3r\, \rho(\vec r\,) e^{i\vec q\cdot\vec r}$. A distribution of size $R$ has $F \approx 1$ for $QR \ll 1$ and $F \to 0$ rapidly for $QR \gg 1$: once the probe's wavelength fits inside the charge cloud, contributions from different regions dephase and cancel. Only a point charge keeps $F = 1$ at all $Q^2$.
The proton's measured elastic form factors follow the dipole shape
$$ G_D(Q^2) = \Big(1 + \frac{Q^2}{0.71\ \text{GeV}^2}\Big)^{\!-2}, \qquad \langle r^2\rangle = -6\,\frac{dG_D}{dQ^2}\Big|_{0} = \frac{12}{0.71}\ \text{GeV}^{-2} \times 0.0389\ \tfrac{\text{fm}^2}{\text{GeV}^{-2}} = 0.66\ \text{fm}^2 $$
i.e. an object of radius $\approx 0.81$ fm. The numbers are brutal: at $Q^2 = 20\ \text{GeV}^2$, $G_D = (29.2)^{-2} = 1.2\times10^{-3}$ and the elastic rate $\propto G_D^2 \approx 1.4\times10^{-6}$ — six orders of magnitude of suppression — while inelastic $F_2(x = 0.25)$ has moved by tens of percent. The proton as a whole is soft; whatever the photon strikes inelastically at large $Q^2$ is not.
The infinite-momentum frame: where the impulse approximation is honest
To convert "pointlike" into a calculable model we need scattering off constituents to be incoherent and elastic — the photon hits one parton, instantaneously free, and interference with the rest of the proton is negligible. That is the impulse approximation, and it is not obviously legal: quarks are confined, permanently interacting. The trick is to view the collision from a frame where the proton moves fast.
Boost the proton to momentum $P \to \infty$ along $\hat z$ and compare timescales via energy denominators. The proton's energy is $E_p = \sqrt{P^2 + M^2} \approx P + M^2/2P$. A virtual configuration of partons carrying momentum fractions $\xi_i$ (with $\sum\xi_i = 1$) and transverse momenta $k_{Ti}$ has
$$ E_{\rm config} = \sum_i \sqrt{\xi_i^2P^2 + k_{Ti}^2 + m_i^2} \;\approx\; P + \sum_i \frac{k_{Ti}^2 + m_i^2}{2\xi_i P} $$
so the energy defect of any fluctuation is $\Delta E \sim \mu^2/2P$ with $\mu$ a hadronic scale ($k_T, m \sim$ few hundred MeV). Its lifetime is
$$ \tau_{\rm fluct} \;\sim\; \frac{1}{\Delta E} \;\sim\; \frac{2P}{\mu^2} \;\longrightarrow\; \infty $$
— internal proton dynamics is time-dilated without bound. The photon, meanwhile, is absorbed over a time $\sim 1/Q$ (its own virtuality clock), which stays fixed. In the Bjorken limit the hierarchy $1/Q \ll 2P/\mu^2$ becomes arbitrarily strong: the photon takes a snapshot of a frozen configuration of partons, each effectively free for the duration of the interaction. Hadronization — the strings and showers that reassemble the debris — happens on the slow clock, long after the count is taken, and by unitarity (we sum over all $X$) it cannot change the inclusive rate.
The master formula, derived
Now assemble the model: the proton is a beam of partons, parton $i$ carrying longitudinal fraction $\xi$ of the proton's momentum with probability density $f_i(\xi)$; the photon scatters elastically off one of them; cross-sections add incoherently. Everything from §1.2 applies at the parton level.
The struck parton has $p = \xi P$ (transverse momentum and virtuality $\sim \mu^2 \ll Q^2$ neglected — that is the impulse approximation doing its work). The parton-level tensor from §1.2 carries the elastic phase-space delta putting the final quark on shell (quark masses dropped):
$$ \delta\big((p + q)^2\big) = \delta\big(\xi^2M^2 + 2\xi\,P\!\cdot\!q - Q^2\big) $$
Solve the argument exactly — it is a quadratic in $\xi$ with one positive root:
$$ \xi = \frac{P\!\cdot\!q}{M^2}\Big[\sqrt{1 + \tfrac{M^2Q^2}{(P\cdot q)^2}} - 1\Big] = \frac{Q^2}{2P\!\cdot\!q}\Big[1 - \frac{x^2M^2}{Q^2} + \mathcal{O}\Big(\tfrac{x^4M^4}{Q^4}\Big)\Big] \;\xrightarrow{\;Q^2 \gg x^2M^2\;}\; x $$
using $(P\!\cdot\!q)^2 = Q^4/4x^2$ in the expansion. In the Bjorken limit, then,
$$ \delta\big(2\xi P\!\cdot\!q - Q^2\big) = \frac{1}{2P\!\cdot\!q}\,\delta(\xi - x) $$
This is the theorem the whole field stands on: Bjorken's $x$, defined in §1.1 as pure kinematics measured from the lepton alone, equals the momentum fraction $\xi$ of the parton that was struck. The delta function is on-shell-ness; the correction terms $\sim x^2M^2/Q^2$ are the target-mass effects, and their size at low $Q^2$ is one reason global fits impose $Q^2_{\min}$ and $W^2$ cuts on the data they trust.
Write the hadronic tensor as an incoherent sum over parton flavours and momentum fractions:
$$ W^{\mu\nu}(P, q) \;=\; \sum_i e_i^2 \int_0^1 d\xi\; f_i(\xi)\;\hat w^{\mu\nu}(\xi P,\, q) $$
with $\hat w^{\mu\nu}$ the free-fermion tensor of §1.2's Derivation 6 and $e_i$ the parton's charge in units of $e$. The delta function eats the $\xi$-integral, evaluating every $f_i$ at $\xi = x$. Reading off the coefficient of the $W_2$ projector and converting with $F_2 = \nu W_2$, $F_1 = MW_1$ (the algebra is the point-fermion matching already done in §1.2, now weighted by $f_i(x)\,dx$):
Read it slowly, because everything is in there. Scaling is derived, not assumed: the right-hand side knows nothing about $Q^2$, because pointlike scattering has no length scale to compare $1/Q$ against. Callan–Gross comes out automatically, because the constituents were taken spin-$\tfrac12$. And $F_2$ is revealed as a charge-weighted momentum census: at each $x$ it reports the total momentum density $x f_i(x)$ of partons at that fraction, weighted by squared charge (see momentum fraction). The functions $f_i(x)$ — the parton distribution functions — are the objects this entire lab exists to infer.
Nothing stops an experimenter from computing $x = Q^2/2M\nu$ for any event — it is just lepton bookkeeping. The parton model's non-trivial content is Derivation 3: that this kinematic number coincides with a property of the proton's interior, the struck parton's momentum share. That is why plotting data "versus $x$" organizes the physics so cleanly, and why the same $f_i(x)$ measured in DIS can be exported to predict Drell–Yan, $W/Z$, and LHC cross-sections. The exportability is factorization — proven, with corrections, in QCD; here it appears in embryo as the timescale separation of Derivation 2.
What "free" actually means
The impulse approximation used two distinct facts, worth separating because Topic 2 makes both precise. First, interactions among partons during the snapshot are negligible — in QCD this is asymptotic freedom: the effective coupling $\alpha_s(Q^2)$ shrinks logarithmically as the probe sharpens, so at large $Q^2$ the struck quark really is quasi-free at short distances. Second, what happens after — confinement, hadronization — is slow and inclusive-safe. Neither fact makes the corrections exactly zero: gluon radiation at order $\alpha_s(Q^2)$ violates scaling by terms $\sim \alpha_s(Q^2) \ln Q^2$, and this is a prediction, not a failure.
Scaling is almost true — and the "almost" is QCD
Look closely at modern data and $F_2(x, Q^2)$ does drift with $Q^2$, logarithmically: it rises with $Q^2$ at small $x$ ($\lesssim 0.1$) and falls at large $x$, pivoting near $x \approx 0.1\text{–}0.2$. The pattern has a physical reading that Topic 2 turns into the DGLAP equations: sharpen the microscope and a high-$x$ quark is resolved into a quark plus a radiated gluon (draining the large-$x$ region), while gluons are resolved into $q\bar q$ pairs (feeding the small-$x$ sea). Our own fits live on exactly this physics: the SLAC + BCDMS $F_2^{p,d}$ tables (648 points spanning $x \approx 0.07\text{–}0.9$) are compared to NNLO theory evolved from $Q_0 = 1.65$ GeV, and the $Q^2$-slopes in that window are a large part of what constrains the gluon (see the PPDF-10 line of results).
Given: $F_2^p(x = 0.25) \approx 0.35$ (SLAC/BCDMS ballpark at $Q^2 \sim 15\ \text{GeV}^2$). Estimate $x u_v(0.25)$.
Step 1 — model: at $x = 0.25$ valence dominates. Keep $u_v, d_v$ plus a small sea, and assume the valence shape ratio $d_v \approx u_v/2$ (motivated by counting: 2 up vs 1 down). Then
$$ F_2^p \approx x\Big[\tfrac{4}{9}u_v + \tfrac{1}{9}d_v\Big] + F_2^{\rm sea} = x\,u_v\Big[\tfrac{4}{9} + \tfrac{1}{18}\Big] + F_2^{\rm sea} = \tfrac12\, x\,u_v + F_2^{\rm sea} $$
Step 2 — crude pass (no sea): $x u_v \approx 2 \times 0.35 = 0.70$.
Step 3 — sea correction: at $x = 0.25$ typical sea values are $x\bar u \approx x\bar d \approx 0.016$, $x\bar s \approx 0.006$, contributing $F_2^{\rm sea} = \tfrac{2x}{9}(4\bar u + \bar d + \bar s) \approx \tfrac{2}{9}(0.064 + 0.016 + 0.006) \approx 0.02$. So $x u_v \approx 2(0.35 - 0.02) = 0.66$.
Step 4 — compare: global fits give $x u_v(0.25, Q^2\!\sim\!15) \approx 0.60\text{–}0.65$. A two-line valence model lands within $\sim\!10\%$ — and the residual is exactly the $d_v/u_v$ and sea detail that full fits (§1.4, Topic 3) exist to determine.
The observable: integrate the master formula. $\int_0^1 F_2\,dx = \sum_i e_i^2\,\varepsilon_i$ where $\varepsilon_i \equiv \int_0^1 x f_i(x)\,dx$ is the total momentum fraction carried by species $i$ (quark plus antiquark folded in). Keep $u$-type and $d$-type:
$$ \int F_2^p\,dx = \tfrac{4}{9}\varepsilon_u + \tfrac{1}{9}\varepsilon_d, \qquad \int F_2^n\,dx = \tfrac{4}{9}\varepsilon_d + \tfrac{1}{9}\varepsilon_u \;\;(\text{isospin, §1.4}) $$
The data (early SLAC era): $\int F_2^p\,dx \approx 0.18$, $\int F_2^n\,dx \approx 0.12$.
Solve the 2×2 system. Subtract: $\tfrac{3}{9}(\varepsilon_u - \varepsilon_d) = 0.06 \Rightarrow \varepsilon_u - \varepsilon_d = 0.18$. Add: $\tfrac{5}{9}(\varepsilon_u + \varepsilon_d) = 0.30 \Rightarrow \varepsilon_u + \varepsilon_d = 0.54$. Hence
$$ \varepsilon_u = 0.36, \qquad \varepsilon_d = 0.18 $$
(Reassuringly $\varepsilon_u = 2\varepsilon_d$, echoing the 2:1 valence count.)
The punchline: charged partons carry $\varepsilon_u + \varepsilon_d \approx 0.54$ — barely half the proton's momentum. Strange quarks add a few percent, no more. Since the photon audit weights by charge$^2$, the missing $\approx\!46\%$ must be carried by partons the photon cannot see at all: electrically neutral partons — the gluons, inferred from a momentum deficit years before any direct observation. The momentum sum rule that formalizes this audit is derived in §1.4.