The cross-section and structure functions
Leptonic and hadronic tensors, the F₁/F₂ decomposition derived, Callan–Gross.
Section 1.1 built the coordinates; this section builds the thing that is measured at each coordinate. The result has a two-part structure of great beauty: the cross-section factorizes into a leptonic tensor that QED lets us compute exactly, line by line, and a hadronic tensor that hides everything we do not know about the proton. Symmetries then squeeze all of that ignorance into just two functions, $F_1(x,Q^2)$ and $F_2(x,Q^2)$ — the objects every table in our DIS datasets reports — and one relation between them, Callan–Gross, turns out to be the experiment that proved quarks carry spin $\tfrac12$.
One photon, two tensors
Work to leading order in $\alpha$: a single virtual photon exchanged between the lepton line and the hadronic blob (metric $(+,-,-,-)$, lepton mass dropped throughout). The amplitude for $\ell(k) + p(P) \to \ell'(k') + X$ is
with $J_\mu$ the electromagnetic current of the quarks in units of $e$. Square, average over the lepton spin, sum over $X$ and over final hadronic phase space. Because the amplitude is a product of a lepton factor and a hadron factor, its square is a product of two rank-2 tensors, contracted through the two photon propagators:
Everything computable sits in $L^{\mu\nu}$; everything unknown sits in $W_{\mu\nu}$; and the $1/Q^4$ from the propagator squared is why DIS cross-sections plunge so steeply with $Q^2$ that HERA needed a collider's luminosity to map the high-$Q^2$ corner. Now build each tensor honestly.
Define $L^{\mu\nu}$ as the spin-averaged square of the lepton current:
$$ L^{\mu\nu} \;=\; \frac{1}{2}\sum_{s,s'} \big[\bar u_{s'}(k')\gamma^\mu u_s(k)\big] \big[\bar u_{s'}(k')\gamma^\nu u_s(k)\big]^{\!*} $$
The conjugate factor equals $\bar u_s(k)\gamma^\nu u_{s'}(k')$, so the spin sums close into a trace via the completeness relation $\sum_s u_s(k)\bar u_s(k) = \slashed{k}$ (massless):
$$ L^{\mu\nu} = \frac{1}{2}\,\mathrm{Tr}\big[\slashed{k}'\gamma^\mu \slashed{k}\gamma^\nu\big] = \frac{1}{2}\,k'_\alpha k_\beta\,\mathrm{Tr}\big[\gamma^\alpha\gamma^\mu\gamma^\beta\gamma^\nu\big] $$
The four-gamma trace is $\mathrm{Tr}[\gamma^\alpha\gamma^\mu\gamma^\beta\gamma^\nu] = 4\big(g^{\alpha\mu}g^{\beta\nu} - g^{\alpha\beta}g^{\mu\nu} + g^{\alpha\nu}g^{\mu\beta}\big)$. Contract:
$$ L^{\mu\nu} \;=\; 2\big(k^\mu k'^\nu + k^\nu k'^\mu - g^{\mu\nu}\, k\!\cdot\!k'\big) $$
Symmetric, real, and — check it — conserved: $q_\mu L^{\mu\nu} = 2[(q\!\cdot\!k)k'^\nu + (q\!\cdot\!k')k^\nu - q^\nu k\!\cdot\!k']$ with $q\!\cdot\!k = -k\!\cdot\!k'$ and $q\!\cdot\!k' = +k\!\cdot\!k'$ gives $2k\!\cdot\!k'\,[k^\nu - k'^\nu - q^\nu] = 0$. That conservation will do real work in Derivation 4.
Everything hadronic is the squared current matrix element, summed over all final states with four-momentum conservation. Package it (spin-averaged over the proton, conventional $1/4\pi$):
$$ W^{\mu\nu}(P,q) \;=\; \frac{1}{4\pi}\sum_X \langle P|J^\mu(0)|X\rangle\langle X|J^\nu(0)|P\rangle\, (2\pi)^4\,\delta^4(P + q - P_X) $$
Use translation invariance, $J^\mu(z) = e^{i\hat P\cdot z}J^\mu(0)e^{-i\hat P\cdot z}$, to trade the delta function for a Fourier integral and the completeness of $|X\rangle$ to remove the sum:
$$ W^{\mu\nu} \;=\; \frac{1}{4\pi}\int d^4z\; e^{iq\cdot z}\,\langle P|\,J^\mu(z)\,J^\nu(0)\,|P\rangle $$
(For $q^0 > 0$ the opposite ordering has no states to land on, so this equals the matrix element of the commutator $[J^\mu(z), J^\nu(0)]$ — the form in which the operator-product analysis of DIS is usually run.) The point to hold: $W^{\mu\nu}$ is a well-defined matrix element of QCD operators in a proton state. We cannot compute it from first principles — but we can constrain its form completely.
For an unpolarized proton and the parity-conserving electromagnetic current, $W^{\mu\nu}$ must be a symmetric tensor built from the only vectors available, $P^\mu$ and $q^\mu$, and the metric (an antisymmetric $\epsilon^{\mu\nu\rho\sigma}$ piece is parity-odd for photons — it returns for neutrinos below). The most general ansatz has four scalar coefficients:
$$ W^{\mu\nu} = -W_1\, g^{\mu\nu} + \frac{W_2}{M^2}P^\mu P^\nu + \frac{W_4}{M^2}q^\mu q^\nu + \frac{W_5}{M^2}\big(P^\mu q^\nu + q^\mu P^\nu\big) $$
Current conservation, $\partial_\mu J^\mu = 0$, forces $q_\mu W^{\mu\nu} = 0$. Contract the ansatz with $q_\mu$ and sort by the independent vectors $P^\nu$ and $q^\nu$:
$$ \Big[\frac{W_2}{M^2}(P\!\cdot\!q) - \frac{W_5}{M^2}Q^2\Big]P^\nu + \Big[-W_1 - \frac{W_4}{M^2}Q^2 + \frac{W_5}{M^2}(P\!\cdot\!q)\Big]q^\nu \;=\; 0 $$
Both brackets must vanish separately, which solves $W_5$ and $W_4$ in terms of $W_1, W_2$:
$$ W_5 = \frac{P\!\cdot\!q}{Q^2}\,W_2, \qquad W_4 = \frac{(P\!\cdot\!q)^2}{Q^4}\,W_2 - \frac{M^2}{Q^2}\,W_1 $$
Substituting back reassembles the tensor into two manifestly gauge-invariant projectors:
$$ W^{\mu\nu} = W_1\Big(\!-g^{\mu\nu} + \frac{q^\mu q^\nu}{q^2}\Big) + \frac{W_2}{M^2}\Big(P^\mu - \frac{P\!\cdot\!q}{q^2}q^\mu\Big)\Big(P^\nu - \frac{P\!\cdot\!q}{q^2}q^\nu\Big) $$
Two functions of two invariants — $W_{1,2}(\nu, Q^2)$ — carry the entire inclusive electromagnetic structure of the proton. Not an approximation: a theorem of Lorentz, parity, and gauge invariance.
Because $q_\mu L^{\mu\nu} = 0$ (Derivation 1), every $q^\mu$ in $W^{\mu\nu}$ dies in the contraction; only $-W_1 g^{\mu\nu}$ and $(W_2/M^2)P^\mu P^\nu$ survive. Using $g_{\mu\nu}L^{\mu\nu} = -4k\!\cdot\!k'$ and $L^{\mu\nu}P_\mu P_\nu = 2\big[2(P\!\cdot\!k)(P\!\cdot\!k') - M^2 k\!\cdot\!k'\big]$, then the lab-frame values $P\!\cdot\!k = ME$, $P\!\cdot\!k' = ME'$, $k\!\cdot\!k' = 2EE'\sin^2(\theta/2)$:
$$ L^{\mu\nu}W_{\mu\nu} \;=\; 4EE'\Big[\,W_2\cos^2\tfrac{\theta}{2} + 2W_1\sin^2\tfrac{\theta}{2}\,\Big] $$
Restoring flux $4ME$ and phase space $d^3k' = E'^2\,dE'\,d\Omega$ with the $1/4\pi$ of Derivation 2:
$$ \frac{d^2\sigma}{dE'\,d\Omega} = \frac{\alpha^2}{Q^4}\frac{E'}{E}\,L^{\mu\nu}W_{\mu\nu} = \frac{\alpha^2}{4E^2\sin^4(\theta/2)}\Big[W_2\cos^2\tfrac{\theta}{2} + 2W_1\sin^2\tfrac{\theta}{2}\Big] $$
— the Mott cross-section dressed by the proton's two unknowns. Elastic scattering is the special case where $W_{1,2}$ collapse onto delta functions of form factors; DIS is the general case.
Define the dimensionless structure functions $F_1 \equiv M W_1$ and $F_2 \equiv \nu W_2$ (this scaling choice is exactly what makes §1.3's statement of Bjorken scaling clean). Two short Jacobians move the lab formula onto the $(x, Q^2)$ plane: $\big|\partial(Q^2,\nu)/\partial(\cos\theta, E')\big| = 2EE'$ and, at fixed $Q^2$, $d\nu = (Q^2/2Mx^2)\,dx$. The algebra is mechanical; the result is the formula every fit code implements:
The target-mass term $x^2y^2M^2/Q^2$ matters only at low $Q^2$ (it is kept in our theory pipeline); drop it and the anatomy is stark: $F_2$ enters with weight $(1-y)$, $F_1$ with weight $xy^2$. Two unknown functions, two independent $y$-weights — that is a solvable system.
The $y$-anatomy: how experiments pry $F_1$ from $F_2$
At fixed $(x, Q^2)$, changing $\sqrt s$ changes $y = Q^2/x(s - M^2)$ and nothing else. The cross-section is linear in the two weights, so measurements at two or more beam energies separate $F_1$ and $F_2$ by a straight-line fit — the DIS incarnation of the Rosenbluth separation. This is exactly why BCDMS ran its muon beam at 100, 120, 200 and 280 GeV, and why HERA took dedicated low-proton-energy runs in 2007 to measure the longitudinal structure function directly.
Fit codes usually repackage the same content as a reduced cross-section, with $Y_+ \equiv 1 + (1-y)^2$:
The longitudinal sector: $F_L$ and $R$
The combination appearing above,
has a sharp physical meaning. A spacelike photon has three polarization states: two transverse, one longitudinal (scalar). Writing the DIS rate as a flux of virtual photons times absorption cross-sections, one finds $\sigma_T \propto W_1$ while $\sigma_L \propto (1 + \nu^2/Q^2)W_2 - W_1$, so the ratio
measures how well the proton absorbs longitudinal photons relative to transverse ones. $F_1$ is the transverse response, $F_L$ the longitudinal one, $F_2$ their weighted sum. And here sits one of the great spin-detector arguments of twentieth-century physics.
Model the proton as containing pointlike Dirac constituents (§1.3 makes this precise). Scattering off a free fermion of mass $m$ and unit charge is the same QED calculation as Derivation 1, run on the quark line: $w^{\mu\nu} \propto 2\big(p^\mu p'^\nu + p^\nu p'^\mu - g^{\mu\nu} p\!\cdot\!q\big)$ with $p' = p + q$. Expand $p^\mu p'^\nu + p^\nu p'^\mu = 2p^\mu p^\nu + (p^\mu q^\nu + p^\nu q^\mu)$: the $q$-pieces are the gauge completions, and matching the two physical structures onto Derivation 3's decomposition (target mass $m$) gives the strengths
$$ W_1^{\rm pt} : W_2^{\rm pt} \;=\; p\!\cdot\!q \,:\, 2m^2 \qquad\Longrightarrow\qquad \frac{W_1^{\rm pt}}{W_2^{\rm pt}} = \frac{m\nu}{2m^2} = \frac{Q^2}{4m^2} $$
using the elastic condition $\nu = Q^2/2m$ in the last step. A constituent that scatters elastically at Bjorken $x = Q^2/2M\nu$ has $m = xM$ (same algebra, §1.3). Now compute the Callan–Gross ratio — the normalization cancels:
$$ \frac{2xF_1}{F_2} = \frac{2x\,M\,W_1}{\nu\,W_2} = \frac{2xM}{\nu}\cdot\frac{Q^2}{4x^2M^2} = \frac{Q^2}{2xM\nu} = 1 \;\;\blacksquare $$
So spin-$\tfrac12$ constituents force $F_2 = 2xF_1$, i.e. $F_L = 0$ at leading order (QCD corrections make $F_L \sim \alpha_s$, small but nonzero — Topic 2).
Repeat for a charged scalar parton. Scalar QED's current is $j^\mu = (p + p')^\mu = (2p + q)^\mu$, so the parton tensor is
$$ w^{\mu\nu} \;\propto\; (2p+q)^\mu (2p+q)^\nu\,\delta\big((p+q)^2 - m^2\big) $$
— gauge invariant on shell, since $q\!\cdot\!(2p+q) = 2p\!\cdot\!q + q^2 = 0$ at the elastic point, but with no $-g^{\mu\nu}$ structure at all. Matching to Derivation 3: $W_1 = 0$, hence $F_1 = 0$, $F_2 = F_L$, and $R \to \infty$. The Breit-frame picture says why. There the quark reverses direction, $-\tfrac{Q}{2}\hat z \to +\tfrac{Q}{2}\hat z$. For a massless spin-$\tfrac12$ quark, chirality conservation keeps helicity fixed, so $S_z$ flips by one unit — which only a transverse photon ($J_z = \pm1$) can supply: $\sigma_T$ survives, $\sigma_L$ dies. A scalar has no spin to flip: only $J_z = 0$, longitudinal absorption is allowed. Measuring $R$ is literally measuring the spin of whatever is inside. SLAC found $R \approx 0.1\text{–}0.2$, small — the proton's charged constituents are spin-$\tfrac12$.
Nothing about quarks was assumed in Derivations 1–5. Lorentz invariance, parity, and current conservation alone force any strongly-interacting target — proton, pion, or plasma — to present exactly two functions to an unpolarized electromagnetic probe. That is why experiments publish $F_2$ and $R$ (or $\sigma_r$), why our fit's data files are tables of $F_2^{p,d}(x, Q^2)$, and why every model of proton structure is testable the same way: it must predict two functions of two variables. The parton model's prediction — scaling plus Callan–Gross — is the subject of §1.3.
Neutrinos add a third function
Replace the photon by a $W^\pm$ and two things change: the current becomes $V\!-\!A$, and parity violation unlocks the antisymmetric tensor that parity banned in Derivation 3, $W^{\mu\nu} \supset -\tfrac{i}{2M^2}\,\epsilon^{\mu\nu\rho\sigma}P_\rho q_\sigma\, W_3$. Contracted against the axial part of the lepton trace it yields a third structure function $F_3 \equiv \nu W_3$, odd under $\ell \leftrightarrow \bar\ell$ (see parity):
In the parton model $xF_3$ counts quarks minus antiquarks — pure valence. Our current fits use only electromagnetic $F_2$ (SLAC + BCDMS, 648 points), which is precisely why some flavour combinations stay degenerate (§1.4); charged-current data is one of the levers that breaks them.
Setup. Using Callan–Gross ($2xF_1 = F_2$), the $F_1$ term's share of the cross-section is
$$ \frac{xy^2F_1}{xy^2F_1 + (1-y)F_2} = \frac{y^2/2}{y^2/2 + (1-y)} = \frac{y^2}{1 + (1-y)^2} = \frac{y^2}{Y_+} $$
— identically the weight of $F_L$ in the reduced cross-section. One number controls both.
BCDMS-type point: $x = 0.25$, $y = 0.4$ (with $E = 200$ GeV this is $Q^2 = xy(s - M^2) = 0.25 \times 0.4 \times 375.2 \approx 37.5\ \text{GeV}^2$). Then $y^2 = 0.16$, $Y_+ = 1.36$: share $= 0.16/1.36 = \mathbf{11.8\%}$. The $F_2$ term carries the remaining $88.2\%$ — the ratio of the two contributions is $(1-y)F_2 : xy^2F_1 = 7.5 : 1$.
HERA-type point: $y = 0.8$ (e.g. $x = 0.001$, $Q^2 = xys \approx 0.8 \times 0.001 \times 101{,}200 \approx 81\ \text{GeV}^2$). Then $y^2 = 0.64$, $Y_+ = 1.04$: share $= 0.64/1.04 = \mathbf{61.5\%}$ — the $F_1$/longitudinal sector now dominates.
Reading. Fixed-target data at moderate $y$ is an $F_2$ measurement to percent accuracy with only $\sim\!12\%$ sensitivity to $F_L$ — which is why our fits take $F_2$ tables as data and compute the small $F_L$ correction from NNLO theory, and why high-$y$ HERA running was needed to measure $F_L$ itself.
Setup: a free quark that shows up at $x = 0.25$ acts like a point Dirac particle of mass $m = xM = 0.2345$ GeV (Derivation 6). Take $Q^2 = 20\ \text{GeV}^2$.
Step 1 — kinematics: the elastic condition gives $\nu = Q^2/2m = 20/0.4690 = 42.64$ GeV. Check consistency: $x = Q^2/2M\nu = 20/(2 \times 0.938 \times 42.64) = 20/80.0 = 0.25$ ✓.
Step 2 — the tensor ratio: $W_1/W_2 = Q^2/4m^2 = 20/(4 \times 0.05500) = 90.9$.
Step 3 — Callan–Gross: $\dfrac{2xF_1}{F_2} = \dfrac{2xM W_1}{\nu W_2} = \dfrac{2 \times 0.25 \times 0.938 \times 90.9}{42.64} = \dfrac{42.64}{42.64} = 1.000$ ✓.
Step 4 — the scalar counterfactual: had the constituent been spin-0, Step 2 would return $W_1/W_2 = 0$ and the measured $R = \sigma_L/\sigma_T$ would blow up instead of sitting at $R \approx 0.18$ (SLAC, moderate $x$). One measured ratio, one spin assignment.