Hadron-hadron collisions

The Large Hadron Collider is the highest-energy machine humanity has ever built, and yet it never quite collides what it accelerates. It fires protons — but a proton is not a point, it is a roiling bag of quarks, antiquarks and gluons, and the interesting physics happens when one parton from the first proton strikes one parton from the second. Everything else is a spectator that sprays forward down the beampipe. So a $14\ \mathrm{TeV}$ machine is really a variable-energy parton collider: the two struck partons carry unknown momentum fractions $x_1,x_2$ , their collision energy is only $\sqrt{\hat s}=\sqrt{x_1 x_2\,s}$ , and — because $x_1\ne x_2$ in general — the whole hard interaction is boosted along the beam by an amount nobody measured. That single fact reshapes the kinematic language. Elastic $ep$ scattering needs one variable and deep inelastic scattering two, but a hadron–hadron event needs three — the two momentum fractions plus a momentum transfer — mapped onto what a detector records: two jet rapidities and a transverse momentum. Rapidity, not polar angle, is the natural coordinate, because a longitudinal boost merely shifts every rapidity by the same amount, leaving differences and shapes untouched. The unknown parton boost thus slides harmlessly along the rapidity axis. Two processes anchor the subject. The Drell–Yan reaction $q\bar q\to\mu^+\mu^-$ is a clean QED annihilation whose cross section, folded with parton distributions, produces the famous dimuon mass spectrum with its towering $Z$ peak at $91\ \mathrm{GeV}$ . And QCD jet production, $qq\to qq$ by gluon exchange, is the overwhelming bulk of what the LHC does — the background against which every discovery must be dug out.

Key takeaways

A hadron collider is really a parton collider: the hard scatter is between two partons carrying unknown fractions $x_1,x_2$ , with partonic energy $\hat s=x_1x_2\,s$ , always less than the full $pp$ energy $s$ .
Because $x_1\ne x_2$ in general, the parton rest frame is boosted along the beam relative to the lab, so rapidity $y=\tfrac12\ln\frac{E+p_z}{E-p_z}$ — whose differences are boost-invariant — replaces polar angle as the natural variable.
Three independent kinematic variables describe an event (e.g. $x_1,x_2,Q^2$ ), mapped onto two jet rapidities and a transverse momentum $p_T$ ; when masses are negligible rapidity becomes pseudorapidity $\eta=-\ln\tan(\theta/2)$ .
The Drell–Yan process $q\bar q\to\mu^+\mu^-$ gives $\frac{d^2\sigma}{dM\,dy}=\frac{8\pi\alpha^2}{9Ms}\sum_q e_q^2[q(x_1)\bar q(x_2)+\bar q(x_1)q(x_2)]$ , with a colour factor $1/N_c=1/3$ and a striking $Z$ resonance near $M=91\ \mathrm{GeV}$ .
QCD two-jet production ( $qq\to qq$ , gluon exchange) dominates the rate; its many interfering diagrams and higher orders force numerical computation of the cross sections.
Key idea: a hadron–hadron cross section is a partonic cross section convolved with parton distribution functions over the unknown momentum fractions, and the composite, boosted initial state that this composite structure forces on us is tamed by working in boost-invariant rapidity.

The core insight

A hadron–hadron cross section is the partonic cross section convolved with parton distribution functions over the unknown momentum fractions $x_1,x_2$ ; the price of high energy is a composite, boosted initial state whose effects are tamed by using boost-invariant rapidity.

Where this lives today

Hadron collisions are the energy frontier of physics, and the Drell–Yan and jet formalisms built above are the daily language of the LHC's ATLAS and CMS experiments. The clean $Z\to\mu^+\mu^-$ peak is a "standard candle": its precisely known cross section calibrates the collider's luminosity and momentum scales, while high-mass dimuons far out on the falling tail are the primary hunting ground for a heavy $Z'$ resonance. Jet production, governed by the QCD $2\to2$ cross sections, is measured across many orders of magnitude in $p_T$ and used both to extract the gluon PDF at large $x$ and to constrain $\alpha_s$ ; deviations at the highest jet energies would signal quark compositeness or new short-distance physics. The same factorisation formula that gives Drell–Yan drives the flagship measurements — Higgs production through gluon fusion $gg\to H$ , top-pair production $gg,q\bar q\to t\bar t$ , and the diboson channels — and the dominant systematic on almost all of them is the parton distributions themselves, which is why the LHC both uses and measures PDFs in a self-consistent global fit. Looking ahead, the High-Luminosity LHC will multiply the dataset tenfold, and any future $100\ \mathrm{TeV}$ collider would push the parton-collision reach into the tens of $\mathrm{TeV}$ , precisely by exploiting the broadband, PDF-weighted spectrum this article describes.

The physical picture

Two protons cross at nearly the speed of light, but the event that matters is a single parton–parton collision buried inside them. One parton carries a fraction $x_1$ of the left proton's momentum, another $x_2$ of the right; they annihilate or scatter while the proton remnants fly on almost undeflected. The available energy is not the full $\sqrt s$ but $\sqrt{\hat s}=\sqrt{x_1x_2\,s}$ , and because $x_1\ne x_2$ the produced system is launched along the beam with a boost rapidity $y=\tfrac12\ln(x_1/x_2)$ that varies event to event. The first figure fixes this anatomy and pins it with a real number: to make a $Z$ boson at the LHC you need only $x\approx0.007$ .

Figure

A hadron collision is really a parton collision: a parton of fraction x₁ from the left proton meets one of fraction x₂ from the right, annihilating through a virtual photon or Z into a μ⁺μ⁻ pair while the proton remnants fly on. The available energy √ŝ = √(x₁x₂s) is always below the full √s, and the system is boosted along the beam by y = ½ ln(x₁/x₂). At √s = 13 TeV, x₁ ≈ 0.019 and x₂ ≈ 0.0026 give M ≈ 91 GeV — a Z boson — at boost y ≈ 1.0

Drell–Yan is the textbook example because its final state is two muons — clean, isolated tracks, with none of the hadronic debris of a jet. Plot how often each dimuon invariant mass $M$ appears and you get one of the most information-dense graphs in physics: a steeply falling continuum (the virtual-photon tail, $d\sigma/dM\propto 1/M^3$ ) studded with resonance peaks where a bound state or gauge boson can be produced. The $J/\psi$ , the $\Upsilon$ and — towering above them — the $Z$ at $91\ \mathrm{GeV}$ all appear as bumps on the same curve. The second figure is that spectrum, the punchline of the formalism.

Figure

The dimuon invariant-mass spectrum from Drell–Yan q q̄ → μ⁺μ⁻: a steeply falling virtual-photon continuum dσ/dM ∝ 1/M³ studded with resonance peaks where a bound state or gauge boson is produced on shell — the J/ψ at 3.10 GeV, the Υ at 9.46 GeV, and towering above them the Z boson at M = 91.19 GeV with width Γ = 2.5 GeV

The core equations

The starting point is bookkeeping for the composite initial state. A parton of fraction $x_1$ has four-momentum $x_1 P_1$ ; colliding it with one of fraction $x_2$ from the opposing beam gives a partonic energy squared that is a fraction $x_1 x_2$ of the full hadronic energy, plus a net longitudinal momentum that boosts the system.

\hat s = x_1 x_2\, s,\qquad y_{\text{boost}}=\tfrac12\ln\frac{x_1}{x_2} \qquad (1)

where: $s=(P_1+P_2)^2$ — the full $pp$ (or $p\bar p$ ) collision energy squared; $\hat s$ — the usable partonic energy squared; $x_1,x_2$ — the momentum fractions. Reading it: since $x_1,x_2<1$ , always $\hat s<s$ — a $13\ \mathrm{TeV}$ machine does most of its physics at a few hundred $\mathrm{GeV}$ , only rarely reaching the multi- $\mathrm{TeV}$ frontier. Because the boost is unknown event by event, polar angle is a poor variable. The right choice is rapidity, whose defining virtue — additivity under longitudinal boosts — is displayed by writing the transformation alongside the definition.

y=\frac12\ln\frac{E+p_z}{E-p_z}\;\xrightarrow{\ \text{boost }\beta\ }\;y\to y-\tanh^{-1}\!\beta \qquad (2)

where: $E,p_z$ — energy and longitudinal momentum of a particle or jet; a boost of velocity $\beta$ along the beam has rapidity $\tanh^{-1}\beta$ . Reading it: a boost merely shifts every rapidity by the same constant, so differences $y_a-y_b$ and the shape of $dN/dy$ are boost-invariant — the unknown parton boost slides the whole picture sideways and distorts nothing. When jet masses are negligible, rapidity collapses onto a geometric quantity built from the polar angle alone, which a calorimeter measures directly.

\eta=-\ln\tan\frac{\theta}{2}=\frac12\ln\frac{|\mathbf p|+p_z}{|\mathbf p|-p_z}\;\xrightarrow{\ m\to0\ }\;y \qquad (3)

where: $\eta$ — pseudorapidity; $\theta$ — polar angle from the beam. Reading it: $\theta=90^\circ$ gives $\eta=0$ (central), forward angles push $\eta$ large ( $\theta=10^\circ\Rightarrow\eta\approx2.4$ ); for the massless jets of the LHC, $\eta$ and $y$ coincide, so one variable is both angle and boost-friendly coordinate. The Drell–Yan partonic cross section is the QED annihilation $q\bar q\to\gamma^*\to\mu^+\mu^-$ , identical to $e^+e^-\to\mu^+\mu^-$ but for the quark's fractional charge and a colour average.

\hat\sigma(q\bar q\to\mu^+\mu^-)=\frac{4\pi\alpha^2}{3\hat s}\,e_q^2\,\frac{1}{N_c} \qquad (4)

where: $\alpha$ — fine-structure constant; $e_q$ — quark charge in units of $e$ ( $+\tfrac23$ for $u$ , $-\tfrac13$ for $d,s$ ); $N_c=3$ — number of colours. Reading it: $e_q^2$ weights each flavour, and $1/N_c=1/3$ is a colour average — only a colour-matched quark and antiquark annihilate into a colourless photon. Folding cross section (4) with the two parton distributions and changing variables to the observable mass $M$ and rapidity $y$ yields the master Drell–Yan formula — the equation that draws the spectrum above.

\frac{d^2\sigma}{dM\,dy}=\frac{8\pi\alpha^2}{9Ms}\sum_q e_q^2\big[q(x_1)\bar q(x_2)+\bar q(x_1)q(x_2)\big],\qquad x_{1,2}=\frac{M}{\sqrt s}\,e^{\pm y} \qquad (5)

where: $q(x),\bar q(x)$ — the quark and antiquark PDFs; the two terms let either beam supply the quark. Reading it: the measured PDFs enter directly, $M=\sqrt{\hat s}$ is the parton energy, and $y$ fixes the fractions through $x_{1,2}=(M/\sqrt s)\,e^{\pm y}$ — so measuring the dimuon mass and rapidity is measuring $x_1$ and $x_2$ . The overwhelming bulk of LHC collisions, though, is QCD scattering into jets. The simplest channel, $qq'\to qq'$ by single-gluon exchange, is the QCD analogue of Rutherford scattering, with $\alpha_s$ and a colour factor replacing the electromagnetic ones.

\frac{d\hat\sigma}{d\hat t}(qq'\to qq')=\frac{4\pi\alpha_s^2}{9\,\hat s^2}\,\frac{\hat s^2+\hat u^2}{\hat t^2} \qquad (6)

where: $\alpha_s$ — strong coupling ( $\approx0.12$ at $\mathrm{TeV}$ scales); $\hat s,\hat t,\hat u$ — partonic Mandelstam variables; $\tfrac49$ is the colour factor for this channel. Reading it: the $1/\hat t^2$ pole is the gluon-exchange (Rutherford) singularity, making forward, low- $p_T$ jets by far the most common — the QCD "background" beneath every rare signal.

Derivation AWhy rapidity differences survive the unknown parton boost

The goal is to prove the central kinematic claim of hadron-collider physics: a longitudinal Lorentz boost adds a constant to every particle's rapidity, so that rapidity differences and the shape of any rapidity distribution are exactly invariant. This is what makes rapidity the right variable when the parton frame is boosted along the beam by an amount $\tfrac12\ln(x_1/x_2)$ that varies event to event and is never measured. We parametrise energy and momentum in terms of rapidity, apply the boost, and read off the shift.

Step 1. Parametrise the longitudinal kinematics. For a particle of transverse mass $m_T=\sqrt{m^2+p_T^2}$ moving with rapidity $y$ , the definition $y=\tfrac12\ln\frac{E+p_z}{E-p_z}$ inverts to $E=m_T\cosh y$ and $p_z=m_T\sinh y$ . Why: writing $E$ and $p_z$ as a hyperbolic "angle" $y$ turns a Lorentz boost — which mixes $E$ and $p_z$ — into a simple shift of that angle, exactly as an ordinary rotation shifts a polar angle.

Step 2. Apply a longitudinal boost. With $\gamma=\cosh\zeta$ , $\beta\gamma=\sinh\zeta$ ( $\zeta=\tanh^{-1}\beta$ ), the transformation $E'=\gamma(E-\beta p_z)$ , $p_z'=\gamma(p_z-\beta E)$ and the hyperbolic addition formulae give $E'=m_T\cosh(y-\zeta)$ , $p_z'=m_T\sinh(y-\zeta)$ . Why: the boost is a hyperbolic rotation by $\zeta$ , subtracting $\zeta$ from the rapidity while leaving $m_T$ untouched.

Step 3. Read off the shift. Then $y'=\tfrac12\ln\frac{E'+p_z'}{E'-p_z'}=y-\zeta$ : every rapidity drops by the same $\zeta$ , so a difference $y_a'-y_b'=y_a-y_b$ is invariant and $dN/dy$ is merely translated, not deformed. Why: the unknown parton boost $\tfrac12\ln(x_1/x_2)$ shifts the whole event along the rapidity axis, so any rapidity-difference observable is blind to it.

Step 4. Recover pseudorapidity. As $m\to0$ , $E\to|\mathbf p|$ and $y\to\eta=-\ln\tan(\theta/2)$ , a function of angle alone; a jet at $\theta=30^\circ$ has $\eta=-\ln\tan15^\circ=1.32$ . Why: for light jets the calorimeter angle already gives the boost-friendly coordinate — which is why detector geometry is quoted in $\eta$ .

Go deeper · Derivation B · 1 minFolding parton distributions into the Drell–Yan mass spectrum

Derivation BFolding parton distributions into the Drell–Yan mass spectrum

The aim is to assemble the observable cross section $d^2\sigma/dM\,dy$ from two ingredients: the partonic cross section $\hat\sigma(q\bar q\to\mu^+\mu^-)$ and the parton distributions that supply the incoming quark and antiquark. The non-trivial content is the change of variables from the momentum fractions $(x_1,x_2)$ , which nature chooses, to the mass and rapidity $(M,y)$ , which the detector measures — a step that produces the factor $8\pi\alpha^2/9Ms$ through a Jacobian we compute explicitly. We write the factorised cross section, change variables, and insert the partonic result.

Step 1. Write the factorised cross section. Summing incoherently over flavours and over which beam supplies the quark, $\sigma=\sum_q\int_0^1\!dx_1\!\int_0^1\!dx_2\,[q(x_1)\bar q(x_2)+\bar q(x_1)q(x_2)]\,\hat\sigma_q$ , with $\hat\sigma_q=\frac{4\pi\alpha^2}{3\hat s}e_q^2\frac{1}{N_c}$ at $\hat s=x_1x_2 s=M^2$ . Why: at high energy the hard scatter factorises from the soft proton, so PDFs and partonic cross section simply multiply and add, with no interparton interference.

Step 2. Change variables to $(M,y)$ . The dimuon has $M^2=\hat s=x_1x_2 s$ and $y=\tfrac12\ln(x_1/x_2)$ , inverting to $x_1=(M/\sqrt s)e^{y}$ , $x_2=(M/\sqrt s)e^{-y}$ . Why: $M$ and $y$ are what the two measured muon momenta give directly, turning the formula into a prediction for an actual histogram.

Step 3. Compute the Jacobian. The determinant $\left|\frac{\partial(x_1,x_2)}{\partial(M,y)}\right|=\left|\frac{\partial x_1}{\partial M}\frac{\partial x_2}{\partial y}-\frac{\partial x_1}{\partial y}\frac{\partial x_2}{\partial M}\right|=\frac{2M}{s}$ , so $dx_1\,dx_2=\frac{2M}{s}\,dM\,dy$ . Why: this Jacobian is why a factor $M/s$ appears — it converts a density in momentum-fraction space into one in the measured $(M,y)$ plane.

Step 4. Assemble. Multiplying the Jacobian by the partonic cross section, $\frac{2M}{s}\cdot\frac{4\pi\alpha^2}{3M^2}\cdot\frac{1}{3}=\frac{8\pi\alpha^2}{9Ms}$ , gives $\frac{d^2\sigma}{dM\,dy}=\frac{8\pi\alpha^2}{9Ms}\sum_q e_q^2[q(x_1)\bar q(x_2)+\bar q(x_1)q(x_2)]$ , equation (5). Why: the coefficient is fully fixed — QED, colour and Jacobian — so the spectrum's shape is a direct readout of the PDF product $q(x_1)\bar q(x_2)$ .

Go deeper · Derivation C · 1 minThe partonic Drell–Yan cross section and its colour factor one over N_c

Derivation CThe partonic Drell–Yan cross section and its colour factor one over N_c

The purpose here is to obtain $\hat\sigma(q\bar q\to\mu^+\mu^-)$ by recycling the known QED result for $e^+e^-\to\mu^+\mu^-$ , changing only what physics demands: the initial charge becomes fractional, and — because quarks carry colour but the photon does not — an average over colours introduces a factor $1/N_c$ . Getting this $1/3$ right is what makes the Drell–Yan rate a quantitative test of the colour hypothesis. We start from the leptonic cross section, insert the quark charge, and perform the colour average.

Step 1. Recall the leptonic result. The $s$ -channel photon gives $\sigma(e^+e^-\to\mu^+\mu^-)=\frac{4\pi\alpha^2}{3s}$ (owned by aq-pp-ee-to-mumu), about $10.5\ \mathrm{pb}$ at $91\ \mathrm{GeV}$ . The Drell–Yan graph is the identical diagram with the electron line replaced by a quark line. Why: the muon side is untouched, so only the incoming vertex and initial-state averaging can differ.

Step 2. Insert the quark charge. The quark–photon vertex carries $e_q e$ instead of $e$ , and the rate scales as charge squared, picking up $e_q^2$ ( $\tfrac49$ up-type, $\tfrac19$ down-type). Why: the amplitude is linear in the coupling, so the fractional quark charges enter the rate squared, as in the DIS structure functions.

Step 3. Average over colour. The quark and antiquark each carry one of $N_c=3$ colours ( $3\times3=9$ combinations), but only the $3$ colour-matched ones annihilate into a colourless photon, giving $\tfrac{3}{9}=\tfrac{1}{N_c}=\tfrac13$ . Why: we average over initial colours we do not control; since the photon is a colour singlet, mismatched combinations do not annihilate.

Step 4. Combine and check. The result is $\hat\sigma=\frac{4\pi\alpha^2}{3\hat s}e_q^2\frac{1}{N_c}$ , equation (4); for $u\bar u$ at $91\ \mathrm{GeV}$ , $10.5\ \mathrm{pb}\times\tfrac49\times\tfrac13\approx1.6\ \mathrm{pb}$ . Why: the same $1/N_c$ appears inverted in $e^+e^-\to q\bar q$ , where summing three final colours multiplies $R$ by $N_c$ — the same counting from opposite ends, both confirming three colours.

Limiting cases

Case	Result	Meaning
Symmetric partons $x_1=x_2$	$y_{\text{boost}}=0$ ; system at rest in lab, $x=M/\sqrt s$	central production; a $Z$ at rest needs $x\approx91/13000\approx0.007$
Asymmetric $x_1\gg x_2$	large forward boost $y=\tfrac12\ln(x_1/x_2)$	events pushed to high rapidity, down toward the beampipe
Massless jets ( $m\to0$ )	$\eta=y=-\ln\tan(\theta/2)$	detector polar angle alone gives the boost-friendly coordinate
Dimuon mass $M\to m_Z$	photon propagator supplemented by a Breit–Wigner $Z$ pole	resonance enhancement — the towering peak at $91\ \mathrm{GeV}$
Low $\hat s$ / small $x$	gluon densities dominate; $qq\to qq$ , $gg\to gg$ jets swamp the rate	the LHC as a gluon–gluon machine; jets are the bulk
Colourless leptonic limit ( $e_q^2\to1,\ N_c\to1$ )	$\hat\sigma\to\frac{4\pi\alpha^2}{3\hat s}$	Drell–Yan reduces to $e^+e^-\to\mu^+\mu^-$

Common trapThe proton beam energy is the collision energy

It is not. When the LHC runs at $\sqrt s=13\ \mathrm{TeV}$ , that is the energy of the two protons, but the physics happens between two partons, each carrying only a fraction of its proton's momentum. The available energy is $\sqrt{\hat s}=\sqrt{x_1x_2}\,\sqrt s$ , which for typical fractions $x\sim0.01$ is a few hundred $\mathrm{GeV}$ , not $13\ \mathrm{TeV}$ . Reaching the multi- $\mathrm{TeV}$ frontier requires both partons to carry large fractions at once, $x_1x_2\gtrsim(M/\sqrt s)^2$ , and the PDFs fall steeply at large $x$ , so such events are rare. Two consequences follow. First, the $pp$ centre-of-mass frame is not the parton rest frame: since $x_1\ne x_2$ , the produced system is boosted along the beam — exactly why one works in rapidity, not polar angle. Second, a hadron collider is intrinsically broadband, exploring a whole range of $\hat s$ at once, unlike a sharp-energy $e^+e^-$ collider — a strength for discovery but a complication for precision.

Worked examples

1 · Producing a Z boson: from mass and rapidity to the momentum fractions (fully worked)

At the LHC with $\sqrt s=13\ \mathrm{TeV}$ , a Drell–Yan event produces a $\mu^+\mu^-$ pair of invariant mass $M=91\ \mathrm{GeV}$ (on the $Z$ peak) at rapidity $y=1.0$ . Find the two parton momentum fractions $x_1,x_2$ , the partonic energy $\sqrt{\hat s}$ , and confirm the kinematics are consistent.

Step 1. Use the inversion from equation (5), $x_{1,2}=(M/\sqrt s)\,e^{\pm y}$ . The prefactor is $M/\sqrt s=91/13000=\mathbf{0.0070}$ — the fraction needed to make a $Z$ at rest, showing that $Z$ production at the LHC probes the small- $x$ , sea-and-gluon-rich proton.

Step 2. Apply the rapidity boost. With $y=1.0$ , $e^{+1}=2.718$ and $e^{-1}=0.3679$ , so $x_1=0.0070\times2.718=\mathbf{0.019}$ and $x_2=0.0070\times0.3679=\mathbf{0.00258}$ . One parton carries roughly seven times the fraction of the other, which is what boosts the $Z$ forward.

Step 3. Check the partonic energy. Cleanly, $x_1x_2=(M/\sqrt s)^2=0.0070^2=4.9\times10^{-5}$ , so $\sqrt{\hat s}=\sqrt{x_1x_2}\,\sqrt s=0.0070\times13000=\mathbf{91\ GeV}=M$ — the parton energy equals the dimuon mass, as it must for $s$ -channel production.

Step 4. Interpret. The boost $y=\tfrac12\ln(x_1/x_2)=\tfrac12\ln(7.37)=\mathbf{1.0}$ recovers the input rapidity — the descriptions $(x_1,x_2)$ and $(M,y)$ carry the same information. A 91 GeV dimuon at y = 1.0 comes from partons of x₁ ≈ 0.019 and x₂ ≈ 0.0026, colliding at √ŝ = 91 GeV on the Z peak.

2 · The partonic Drell–Yan cross section and its colour factor

Estimate $\hat\sigma(u\bar u\to\mu^+\mu^-)$ at $\sqrt{\hat s}=91\ \mathrm{GeV}$ , using only the photon (ignore the $Z$ ), and compare with $e^+e^-\to\mu^+\mu^-$ at the same energy.

Step 1. Fix the leptonic reference. The QED result $\sigma(e^+e^-\to\mu^+\mu^-)=\frac{4\pi\alpha^2}{3s}$ , with the useful constant $4\pi\alpha^2/3=(86.8\ \mathrm{nb})\cdot\mathrm{GeV}^2$ , gives $86.8/(91)^2\ \mathrm{nb}=86.8/8281\ \mathrm{nb}=\mathbf{10.5\ pb}$ .

Step 2. Insert the quark charge. The up quark has $e_u=+\tfrac23$ , so $e_u^2=\tfrac49=\mathbf{0.444}$ multiplies the leptonic rate, which scales as charge squared.

Step 3. Apply the colour average. Only colour-matched $u\bar u$ pairs annihilate, giving $1/N_c=\tfrac13=\mathbf{0.333}$ . Combining, $\hat\sigma=10.5\ \mathrm{pb}\times0.444\times0.333=\mathbf{1.6\ pb}$ .

Step 4. Interpret. The partonic rate is about $6.75$ times smaller than the leptonic one — $9/4$ from charge, $3$ from colour. This is the continuum value; on the $Z$ pole the $Z$ exchange multiplies it by orders of magnitude, which is why the spectrum peaks rather than following the smooth $1/M^3$ tail. σ̂(uū → μ⁺μ⁻) ≈ 1.6 pb off-resonance, suppressed relative to e⁺e⁻ by e_u² × 1/N_c = 4/9 × 1/3.

3 · Dijet kinematics and boost invariance

Two jets are measured at rapidities $y_3=+2.0$ and $y_4=-0.5$ , each with transverse momentum $p_T=100\ \mathrm{GeV}$ , at $\sqrt s=13\ \mathrm{TeV}$ . Find the momentum fractions, the dijet mass $\sqrt{\hat s}$ , and show that the rapidity difference is boost-invariant.

Step 1. For the massless $2\to2$ system, momentum conservation gives $x_1=\frac{p_T}{\sqrt s}(e^{y_3}+e^{y_4})$ and $x_2=\frac{p_T}{\sqrt s}(e^{-y_3}+e^{-y_4})$ , with prefactor $p_T/\sqrt s=100/13000=0.00769$ .

Step 2. Evaluate. $e^{2.0}+e^{-0.5}=7.996$ , so $x_1=0.00769\times7.996=\mathbf{0.0615}$ ; $e^{-2.0}+e^{0.5}=1.784$ , so $x_2=0.00769\times1.784=\mathbf{0.0137}$ . The forward jet demands a large- $x$ parton.

Step 3. Dijet mass. $\sqrt{\hat s}=\sqrt{x_1x_2}\,\sqrt s=\sqrt{0.0615\times0.0137}\times13000=\mathbf{378\ GeV}$ — two $100\ \mathrm{GeV}$ jets from a parton collision at $378\ \mathrm{GeV}$ , far below the beam energy.

Step 4. Boost invariance. $\Delta y=y_3-y_4=\mathbf{2.5}$ . Boosting into the parton frame subtracts $y_{\text{boost}}=\tfrac12(y_3+y_4)=0.75$ from each, so the jets sit at $+1.25$ and $-1.25$ — still separated by $2.5$ . The dijet came from x₁ ≈ 0.062, x₂ ≈ 0.014 at √ŝ = 378 GeV, and Δy = 2.5 is unchanged by the boost.

Physics of the portrait

The theoretical heart of this article — that a lepton pair in a hadron collision comes from quark–antiquark annihilation, with a rate set by the same parton distributions measured in deep inelastic scattering — was written down in 1970 by two theorists at SLAC, and the process has carried their names ever since.

Origins · 1 minSidney Drell and how this was discovered

Sidney Drell

1926–2016 · quark–antiquark annihilation to lepton pairs in hadron collisions

Sidney Drell was an American theoretical physicist, a leader of the theory group at the Stanford Linear Accelerator Center, and — with his younger colleague Tung-Mow Yan — the author in 1970 of the mechanism that bears their names. The idea was audacious in its simplicity. Deep inelastic scattering had just revealed the proton to be made of point-like, fractionally charged, spin-half constituents; Drell and Yan asked what would happen if, instead of a lepton probing a quark, a quark from one hadron met an antiquark from another. The pair could annihilate into a virtual photon, materialising as a $\mu^+\mu^-$ pair of invariant mass up to the full parton energy. Crucially, they showed that the same parton distributions extracted from electron scattering should predict this rate — an early, powerful statement of what we now call factorisation and universality. The model predicted a scaling continuum in the dimuon mass and, as a bonus, a natural window in which any new massive neutral particle would appear as a resonance. That feature paid off in 1977 when Leon Lederman's group at Fermilab found the $\Upsilon$ , a bottom–antibottom bound state, as a bump in exactly such a spectrum. The process remains a workhorse: it measures antiquark and gluon distributions, calibrates LHC luminosity through the clean $Z\to\ell\ell$ rate, and is the search channel for hypothetical heavy $Z'$ bosons. Drell was also, in his later decades, one of the most respected voices in nuclear arms control — a physicist whose equations and whose conscience both left a lasting mark.

History

1970

— Drell and Yan propose lepton-pair production from quark–antiquark annihilation

1971

— CERN's Intersecting Storage Rings begin the first proton–proton collisions

1977

— Lederman's Fermilab group discovers the Υ as a dimuon resonance, vindicating the Drell–Yan picture

1983

— UA1 and UA2 discover the W and Z bosons at the CERN Spp̄S proton–antiproton collider

1995

— CDF and D0 discover the top quark at the Fermilab Tevatron

2012

— ATLAS and CMS discover the Higgs boson at the LHC

Going further

Texts. Thomson, Modern Particle Physics, §10.9 (hadron–hadron collisions), §10.9.1 (event kinematics), §10.9.2 (the Drell–Yan process) and §10.9.3 (jet production at the LHC), the primary source for this article · Halzen & Martin, Quarks and Leptons, §9.4–9.6 (the parton model applied to hadron collisions and Drell–Yan) · Peskin & Schroeder, An Introduction to QFT, §17.5 (parton evolution and hadron-collider cross sections) · Ellis, Stirling & Webber, QCD and Collider Physics, Ch. 7–9 (the definitive treatment of jets and Drell–Yan) · Griffiths, Introduction to Elementary Particles, 2nd ed., §8–9.

Provenance. The parton distributions $q(x),\bar q(x),g(x)$ and the factorisation theorem are owned by aq-pp-parton-distributions; the meaning of $x$ and $Q^2$ comes from aq-pp-deep-inelastic-scattering; the leptonic reference cross section $\frac{4\pi\alpha^2}{3s}$ is owned by aq-pp-ee-to-mumu; the Mandelstam variables $\hat s,\hat t,\hat u$ by aq-pp-mandelstam-variables (recalled here, not re-derived). The $Z$ Breit–Wigner shape that tops the spectrum is developed in aq-pp-z-resonance-lep, and the fragmentation of the final-state partons into observed jets in aq-pp-jets-fragmentation. A verifier should confirm: (1) $\hat s=x_1x_2 s$ and $y_{\text{boost}}=\tfrac12\ln(x_1/x_2)$ ; (2) the additive boost law $y\to y-\tanh^{-1}\beta$ and $\eta=-\ln\tan(\theta/2)$ ; (3) the Jacobian $dx_1dx_2=(2M/s)dM\,dy$ and the prefactor $8\pi\alpha^2/9Ms$ ; (4) $\hat\sigma(u\bar u\to\mu\mu)\approx1.6\ \mathrm{pb}$ at $91\ \mathrm{GeV}$ from $10.5\ \mathrm{pb}\times\tfrac49\times\tfrac13$ ; (5) the dijet kinematics $x_{1,2}=(p_T/\sqrt s)(e^{\pm y_3}+e^{\pm y_4})$ giving $\sqrt{\hat s}=378\ \mathrm{GeV}$ .

Prerequisites. Parton Distribution Functions · Deep Inelastic Scattering · $e^+e^-\to\mu^+\mu^-$ · Mandelstam Variables. Leads to. The Z Resonance at LEP · Jets and Fragmentation · Open Questions.

Ask the tutor: Ask me to derive the Drell–Yan Jacobian step by step, work out the momentum fractions for any dimuon mass and rapidity, or show why the QCD jet rate swamps every interesting signal at the LHC.

Why collide protons rather than electrons to reach the highest energies?

Because circular electron machines bleed energy to synchrotron radiation. A charged particle bent in a ring radiates power scaling as $E^4/(m^4 R)$ , so the $1836$ -times-heavier proton is suppressed by $(m_e/m_p)^4\approx9\times10^{-14}$ at the same energy. That is why LEP, a circular $e^+e^-$ collider, topped out near $209\ \mathrm{GeV}$ while the LHC in the same tunnel reaches $13$ – $14\ \mathrm{TeV}$ with protons. The price is that the proton is composite: you trade the clean, fixed-energy collisions of point-like electrons for the broadband, PDF-weighted collisions of partons. Discovery machines take that trade; precision machines (a future $e^+e^-$ "Higgs factory", ideally linear to dodge synchrotron losses) take the other side.

Why did early discovery machines use proton–antiproton beams instead of proton–proton?

Two reasons, one physical and one practical. Physically, producing a $W$ or $Z$ by quark–antiquark annihilation is easier when one beam supplies valence antiquarks: an antiproton contains valence $\bar u\bar u\bar d$ , so annihilation draws on large- $x$ valence partons on both sides, boosting the rate at the modest energies of the 1980s Spp̄S and the Tevatron. Practically, a single ring with one magnetic field can circulate protons and antiprotons in opposite directions at once, halving the cost. The LHC switched to $pp$ because at $13\ \mathrm{TeV}$ the relevant fractions are small, where the sea antiquarks are just as abundant in a proton as in an antiproton, so the valence advantage vanishes — and making enough antiprotons for LHC luminosities is hopeless. Two separate beampipes with opposite fields were the price.

What exactly is rapidity, and why not just use the angle?

Rapidity $y=\tfrac12\ln\frac{E+p_z}{E-p_z}$ measures how fast a particle moves along the beam, and its defining property is that a longitudinal boost simply adds a constant to it. That is priceless at a hadron collider, where the parton collision is boosted by an unknown $\tfrac12\ln(x_1/x_2)$ that changes every event: binned by polar angle that random boost smears your distributions, but binned by rapidity it merely slides the histogram sideways, leaving differences and shapes intact. For the near-massless jets of the LHC, rapidity reduces to pseudorapidity $\eta=-\ln\tan(\theta/2)$ , which depends on angle alone — the boost-friendly variable straight from the detector geometry.

Why is the partonic energy √ŝ always less than the beam energy √s?

Because each parton carries only a fraction of its proton's momentum. With $x_1,x_2<1$ , $\hat s=x_1x_2\,s$ and so $\sqrt{\hat s}=\sqrt{x_1x_2}\,\sqrt s<\sqrt s$ ; for typical $x\sim0.01$ the collision at a $13\ \mathrm{TeV}$ machine happens at only a few hundred $\mathrm{GeV}$ . To reach the multi- $\mathrm{TeV}$ frontier both partons must carry large fractions at once, $x_1x_2\gtrsim(M/\sqrt s)^2$ , and since PDFs fall steeply at large $x$ such collisions are rare. This is why the LHC's reach grows so quickly with beam energy and luminosity: raising $\sqrt s$ makes the same heavy state accessible at smaller, more probable $x$ , and more luminosity means more chances at the rare large- $x$ collisions.

Where does the Z peak in the dimuon spectrum come from?

The continuum comes from a virtual photon, whose propagator $1/M^2$ gives the smooth $d\sigma/dM\propto1/M^3$ falloff. But the pair can also annihilate through a virtual Z, and near $M=m_Z=91.19\ \mathrm{GeV}$ the $Z$ propagator — a Breit–Wigner $\sim1/[(M^2-m_Z^2)^2+m_Z^2\Gamma_Z^2]$ — becomes resonant: the $Z$ is nearly on shell and the cross section spikes by orders of magnitude, with the width $\Gamma_Z\approx2.5\ \mathrm{GeV}$ setting the peak's sharpness. This is the physics of a driven oscillator hitting resonance, and it makes $Z\to\ell^+\ell^-$ one of the cleanest signals in collider physics — used to calibrate luminosity, to measure $m_Z$ to keV precision at LEP, and as the template for hunting heavier $Z'$ resonances out on the tail.

Why is most of what the LHC produces "just" QCD jets?

Because the strong interaction is strong, and its cross sections are huge. Any time two partons pass close they scatter through gluon exchange — $qq\to qq$ , $qg\to qg$ , $gg\to gg$ — with a rate set by $\alpha_s\approx0.12$ , far larger than the electroweak couplings behind Higgs or $W$ / $Z$ production. Worse, the gluon-exchange amplitude has a $1/\hat t^2$ pole, so soft, forward scatters are enormously favoured, producing a flood of low- $p_T$ jets. The total inelastic $pp$ cross section is around $70\ \mathrm{mb}$ while Higgs production is measured in picobarns — some eleven orders of magnitude apart. That is why every rare-signal search is fundamentally a background-rejection problem: one interesting event hides among trillions of QCD scatters, and the whole apparatus of triggering and reconstruction exists to dig it out.