The experiment and its kinematics
Setup, invariants x, Q², y, W² — derived, with event reconstruction worked in full.
Deep-inelastic scattering is the microscope of this whole field, and this chapter builds it completely: the process, every kinematic variable with its derivation, the physical meaning of each, the phase-space map on which all our datasets live, and — worked at the end — the full reconstruction chain from raw detector readings to the $(x, Q^2)$ coordinates our fits consume.
The process and its players
A lepton $\ell$ (electron, muon, or neutrino) scatters off a proton $p$ through the exchange of a single electroweak boson, and the proton breaks up:
Four-momenta in brackets. $X$ is anything — the smashed proton's debris. That inclusiveness is a feature: summing over all final hadronic states is what will let the cross-section be written in terms of just two structure functions (§1.2). The exchanged boson carries four-momentum
For charged leptons the exchange is dominantly a virtual photon $\gamma^*$ (with $Z$ interference at high energies); for neutrinos it is a $W^\pm$ — charged-current DIS, which is what makes CHORUS and NuTeV data flavour-selective in a way photons cannot be.
The invariants, one by one
Experiments measure lab-frame quantities; theory wants Lorentz invariants. There are exactly five independent scalars from $k, k', P$ (with $k^2 = m_\ell^2 \approx 0$, $P^2 = M^2$), and the standard set is:
With massless leptons, $q^2 = (k-k')^2 = -2k\!\cdot\!k'$. In the lab frame $k\!\cdot\!k' = E E' (1 - \cos\theta) = 2 E E' \sin^2(\theta/2) \ge 0$, so
$$ Q^2 = -q^2 = 4 E E' \sin^2\!\frac{\theta}{2} \;\ge\; 0 $$
The exchanged boson is spacelike: it exists only as a probe, never as a free particle. Its virtuality sets the probe's transverse resolution $\Delta b \sim \hbar/Q$ — at $Q = 1$ GeV, $\Delta b \approx 0.2$ fm (a quarter of a proton); at HERA's $Q \sim 100$ GeV, three orders finer.
The unobserved system must be at least a proton's worth of invariant mass: $W^2 \ge M^2$. Expand:
$$ W^2 = (P + q)^2 = M^2 + 2 P\!\cdot\!q - Q^2 = M^2 + \frac{Q^2}{x}(1 - x) $$
Since $Q^2 > 0$ and $P\!\cdot\!q > 0$ (energy flows into the proton), $W^2 \ge M^2$ forces $0 < x \le 1$, with equality $x = 1$ exactly when $W = M$ — elastic scattering: the proton stays intact. Inelastic events populate $x < 1$, and resonance production ($W \approx$ 1.2–2 GeV) sits just below the elastic point. This is why fits impose a $W^2$ cut: to stay in the deep-inelastic continuum, away from resonances where the parton language fails.
In the proton's rest frame, $P = (M, \vec 0\,)$, so $P\!\cdot\!q = M(E - E') \equiv M\nu$ and $P\!\cdot\!k = M E$. Hence
$$ y = \frac{\nu}{E} = \frac{E - E'}{E} \in [0, 1] $$
— the fraction of the lepton's energy handed to the proton. $y$ also controls the helicity structure of the cross-section (§1.2): the $F_1$ and $F_2$ pieces carry different $y$-weights, which is precisely how experiments separate them by running at several beam energies.
The invariants obey one exact relation (neglecting $M^2/s$):
$$ Q^2 \;=\; x\, y\, (s - M^2) \;\approx\; x\, y\, s $$
Proof: $s - M^2 = 2 P\!\cdot\!k$, so $x y (s - M^2) = \dfrac{Q^2}{2P\!\cdot\!q}\cdot \dfrac{P\!\cdot\!q}{P\!\cdot\!k} \cdot 2 P\!\cdot\!k = Q^2$. $\blacksquare$
So at fixed beam energy the whole event is pinned by two numbers — $(x, Q^2)$ — and every DIS dataset is a table of cross-sections on that plane. The relation also bounds the reach: since $y \le 1$, $Q^2_{\max} = x s$. No experiment escapes its diagonal.
Because each answers a physical question invariantly. $Q^2$: how sharp is the microscope? $x$: which momentum slice did we hit? (proved in §1.3 — at this stage $x$ is pure kinematics, defined long before partons). $y$: how violent was the collision in the target's frame? $W$: what got produced? The genius of Bjorken's variable $x$ is that it was defined without any model — and then turned out to be the parton momentum fraction.
The kinematic plane — where our data lives
Every dataset in our fits is a set of points on the $(x, Q^2)$ plane, bounded by its accelerator's $s$:
Fixed target vs collider — the $s$ arithmetic
The reach difference is pure kinematics. For a beam of energy $E$ on a stationary proton, $s = M^2 + 2ME \approx 2ME$ — linear in beam energy. For a collider with beam energies $E_\ell, E_p$, $s \approx 4 E_\ell E_p$ — the product. Concretely: BCDMS's 280 GeV muon beam gives $s \approx 2 \times 0.938 \times 280 \approx 525\,\text{GeV}^2$; HERA's $27.5 \times 920$ GeV beams give $s \approx 101{,}200\,\text{GeV}^2$ — a factor ~200 in $s$, hence in $Q^2$ reach, hence (at fixed $Q^2$) a reach in $x$ two-plus decades deeper. That is why "then scale up to HERA" multiplies our dataset's kinematic coverage, not just its point count.
Worked example — from detector readings to $(x, Q^2)$, in full
Given: muon beam $E = 200$ GeV on a fixed hydrogen target; scattered muon measured at $E' = 119$ GeV, $\theta = 1.5^\circ$ ($= 0.02618$ rad). Reconstruct everything.
Step 1 — $Q^2$: $\;Q^2 = 4 E E' \sin^2(\theta/2) = 4 \times 200 \times 119 \times \sin^2(0.01309) = 95{,}200 \times 1.7135\times10^{-4} \approx 16.3\ \text{GeV}^2.$
Step 2 — energy transfer: $\;\nu = E - E' = 81$ GeV, so $y = \nu/E = 0.405$.
Step 3 — $x$: in the target rest frame $x = \dfrac{Q^2}{2 M \nu} = \dfrac{16.3}{2 \times 0.938 \times 81} = \dfrac{16.3}{151.9} \approx 0.107.$
Step 4 — consistency by the master relation: $s \approx 2ME + M^2 = 376.1\ \text{GeV}^2$; $x y (s - M^2) = 0.107 \times 0.405 \times 375.2 \approx 16.3\ \text{GeV}^2 = Q^2$ ✓.
Step 5 — what was produced: $W^2 = M^2 + Q^2(1-x)/x = 0.88 + 16.3 \times 8.35 \approx 137\ \text{GeV}^2 \Rightarrow W \approx 11.7$ GeV — far above the resonance region; safely deep-inelastic.
Step 6 — physics reading: this event probed the proton at resolution $\hbar/Q \approx 0.05$ fm and struck a parton carrying $\approx 10.7\%$ of the proton's momentum — the heart of the valence region. Bin thousands of such events in $(x, Q^2)$, correct for QED radiation, and you have one row of the $F_2$ tables our fits consume.
Same beam, but demand the proton stay intact ($W = M$, i.e. $x = 1$): elastic $\mu p$ scattering at $\theta = 1.5^\circ$. From $x = Q^2 / 2M\nu = 1$: $\nu = Q^2/2M$, and with $Q^2 = 4EE'\sin^2(\theta/2)$, $E' = E - \nu$, solve:
$$ Q^2 = \frac{4 E^2 \sin^2(\theta/2)}{1 + \tfrac{2E}{M}\sin^2(\theta/2)} = \frac{4\times200^2\times 1.7135\times10^{-4}}{1 + 0.0731} \approx 25.6\ \text{GeV}^2 $$
— a unique $Q^2$ for that angle. Elastic scattering has one free kinematic variable, DIS has two: seeing a continuum in $(x, Q^2)$ rather than a line is, by itself, evidence of substructure. The 1968 SLAC data showed exactly that continuum — the door §1.3 walks through.