Interaction cross sections

A decay rate answers "how fast does this particle fall apart?"; a cross section answers the harder question "how often do these two particles, hurled at each other, actually do something?" The extra ingredient is flux — how many projectiles cross unit area per unit time — and it is what makes scattering subtler than decay. Divide the quantum transition rate by the incident flux and by the number of targets and you are left with a quantity that has the dimensions of an area: the cross section $\sigma$ . Sometimes that area is almost literal — a slow neutron sees a nucleus as a disc of a couple of barns — but in general $\sigma$ is nothing so tangible. It is the probability of interaction, dressed up as an effective target size, and it is the single number every collider ultimately measures. Two demands shape the formula. First, the flux must be built from the relative motion of beam and target, not the speed of either alone, or colliding beams would be mis-counted. Second, the whole construction must be Lorentz invariant: the rate an experiment records cannot depend on whether you describe the collision in the laboratory or the centre-of-mass frame. Meeting both demands turns the intuitive "rate = flux × targets × area" into a compact master formula in which a Lorentz-invariant matrix element is integrated over invariant phase space and divided by an invariant flux factor $F=4E_aE_b(v_a+v_b)$ . Every factor of the normalisation volume cancels, and what survives is frame-independent, ready to be compared with a number on a detector.

Key takeaways

A cross section $\sigma$ has the dimensions of area and encodes the quantum probability of a reaction through rate = flux × number of targets × cross section.
The incident flux uses the relative velocity of the colliding particles, $v=v_a+v_b$ for a head-on approach — never the speed of one beam alone.
The relativistic master formula is $\sigma=\frac{1}{F}\int|M_{fi}|^2\,\mathrm{d}\Pi_{\text{LIPS}}$ , an invariant amplitude over invariant phase space divided by the invariant flux factor.
The flux factor $F=4E_aE_b(v_a+v_b)=4\sqrt{(p_a\cdot p_b)^2-m_a^2m_b^2}$ is manifestly Lorentz invariant, so $\sigma$ is the same in every frame.
For a two-body final state in the centre-of-mass frame everything collapses to $\dfrac{\mathrm{d}\sigma}{\mathrm{d}\Omega}=\dfrac{1}{64\pi^2s}\dfrac{p_f^*}{p_i^*}|M_{fi}|^2$ .
Key idea: a cross section converts a quantum transition rate into an effective target area by dividing out the incident flux — and the flux must be defined relativistically so the answer is the same in any frame.

Where this lives today

Every number a collider reports is a cross section in disguise. At the LHC the total proton–proton cross section is about $110$ mb at $\sqrt s=13$ TeV, but the interesting physics lives in the tails: Higgs production runs near $50$ pb, some twelve orders of magnitude rarer, and it is only accessible because the master formula's clean separation of amplitude from flux lets theorists predict those tiny numbers and experimenters normalise to them through the measured luminosity via $\text{rate}=\mathcal{L}\,\sigma$ . The same framework sets the reach of every frontier search — dark-matter direct detection quotes a nucleon cross section, neutrino experiments a cross section per nucleon that rises linearly with energy, and precision electroweak fits at LEP measured $\sigma(e^+e^-\to\text{hadrons})$ across the $Z$ resonance to count exactly three light neutrino species. The invariant flux factor is what lets a cross section measured in one frame be compared with a prediction or another experiment; without it, "how often does this happen?" would have as many answers as there are laboratories.

The physical picture

Send a beam of particles at a slab of targets and count the collisions. In a short time $\delta t$ each projectile, closing on the targets at the relative speed $v=v_a+v_b$ , sweeps out a tube of length $v\,\delta t$ and cross-sectional area $A$ . Every target whose centre falls in that tube is a potential collision, and if each target presents an effective interaction area $\sigma$ , the chance that the projectile hits one is the total blocked area $\delta N\,\sigma$ divided by $A$ . The whole of scattering theory is the project of computing that one number $\sigma$ from quantum mechanics; the geometry just tells you what to do with it. The first figure makes the bookkeeping concrete and reads off the rate.

Figure

A beam of area A sweeps through targets of number density n_b at the relative closing speed v = v_a + v_b. In time δt it clears a tube containing δN = n_b·v·A·δt targets, each blocking an effective area σ, so the interaction rate is rate = φ·N·σ with flux φ = n_a·v

The geometric picture is honest but incomplete: it never says how large $\sigma$ actually is, and it hides a relativistic trap — the flux must be assembled from both particles' motion or the answer will change when you switch frames. The cure is to compute everything from a Lorentz-invariant matrix element and divide by a Lorentz-invariant flux factor. The payoff is a number an experiment can check. The second figure plots one such prediction: the textbook point cross section for $e^+e^-\to\mu^+\mu^-$ , which the master formula reduces to $\sigma=4\pi\alpha^2/3s$ — a pure $1/s$ fall that is one of the most-tested lines in particle physics.

Figure

The point QED cross section σ(e⁺e⁻ → μ⁺μ⁻) = 4πα²/3s = 86.8 nb / (s in GeV²), a pure 1/s fall-off. At the centre-of-mass energy √s = 10 GeV the prediction is σ = 0.87 nb, marked on the curve

The core equations

The starting point is the operational definition read straight off the collision picture: the interaction rate per target particle is the incident flux times the cross section. This is the equation that fixes the dimensions of $\sigma$ and that every later formula must reproduce.

r_b=\sigma\,\phi_a,\qquad \text{rate}=\sigma\,\phi_a\,N_b \qquad (1)

where: $r_b$ — reaction rate per target, $\mathrm{s}^{-1}$ ; $\phi_a$ — incident flux of beam particles, $\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ ; $N_b$ — number of targets exposed; $\sigma$ — cross section, dimensions of area. Reading it: because $\phi$ carries units of (area·time) $^{-1}$ and a rate is (time) $^{-1}$ , the cross section is forced to be an area — that is not an analogy but a dimensional necessity. To make the rate relativistically well-defined the flux is rebuilt from the relative motion of the two particles, giving the flux factor $F$ . In the collinear frame where both move along one axis, it is the product of the two energies and the closing speed.

F=4E_aE_b(v_a+v_b) \qquad (2)

where: $E_a,E_b$ — total energies of the two particles; $v_a,v_b$ — their speeds along the collision axis; the factor $4$ comes from the relativistic $2E$ normalisation of each incoming state. Reading it: for two beams approaching head-on the speeds add, so $F$ counts how vigorously the two currents overlap, not how fast either one moves in isolation. That collinear expression is secretly a Lorentz scalar, and rewriting it in terms of four-vector dot products makes the invariance manifest — the same $F$ in every frame.

F=4\sqrt{(p_a\cdot p_b)^2-m_a^2m_b^2}\quad\Longleftrightarrow\quad F^2=16\big[(p_a\cdot p_b)^2-m_a^2m_b^2\big] \qquad (3)

where: $p_a,p_b$ — four-momenta; $m_a,m_b$ — rest masses; $p_a\cdot p_b=E_aE_b-\vec p_a\cdot\vec p_b$ . Reading it: only invariants ( $p_a\cdot p_b$ , the masses) appear, so an observer in any frame computes the same flux factor — the guarantee that a measured cross section is frame-independent. Assembling the pieces gives the master formula: an invariant amplitude squared, integrated over the Lorentz-invariant phase space of the final state, divided by the invariant flux. Every factor of the normalisation volume $V$ has cancelled.

\sigma=\frac{1}{F}\int|M_{fi}|^2\,(2\pi)^4\,\delta^4\!\Big(p_a+p_b-\sum_f p_f\Big)\prod_f\frac{\mathrm{d}^3p_f}{(2\pi)^3\,2E_f} \qquad (4)

where: $M_{fi}$ — Lorentz-invariant matrix element (from the Feynman rules); the $\delta^4$ enforces energy–momentum conservation; the product is the invariant phase-space measure over all final particles. Reading it: physics ( $M_{fi}$ ) and kinematics (the phase-space integral and $F$ ) are cleanly separated — compute the amplitude once and the same integral machinery delivers every cross section. For the workhorse case of a two-body final state, doing the phase-space integral in the centre-of-mass frame collapses the master formula to a single compact expression for the angular distribution.

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}\bigg|_{\text{CM}}=\frac{1}{64\pi^2 s}\,\frac{p_f^*}{p_i^*}\,|M_{fi}|^2 \qquad (5)

where: $s=(p_a+p_b)^2$ — the squared centre-of-mass energy; $p_i^*,p_f^*$ — the magnitudes of the initial and final momenta in the CM frame; $\mathrm{d}\Omega$ — solid angle of one outgoing particle. Reading it: for elastic scattering $p_f^*=p_i^*$ and the ratio is $1$ ; all the frame-dependence has been packaged into the single invariant $s$ .

Derivation AFrom counting collisions to rate equals flux times targets times cross section

The goal of this first derivation is to earn the defining relation $\text{rate}=\sigma\,\phi\,N$ from nothing but geometry and to expose the one feature that will later demand a relativistic upgrade: the flux must be built from the relative velocity of the two particles, so that a projectile and a target rushing toward each other are counted by their closing speed rather than by either speed alone. We follow a single projectile through a field of moving targets, count how many it can strike, and then sum over a beam.

Step 1. Place targets at number density $n_b$ and let a projectile move at speed $v_a$ while the targets drift the other way at $v_b$ . In a time $\delta t$ the projectile and any given target close their separation at $v=v_a+v_b$ , so relative to the projectile the targets stream past filling a tube of length $v\,\delta t$ and area $A$ ; that tube contains $\delta N=n_b\,v\,A\,\delta t$ targets. Why: interactions depend only on how fast the two approach each other, which is why the relative speed — not $v_a$ alone — sets the count.

Step 2. Give each target an effective interaction area $\sigma$ drawn broadside to the beam. The projectile strikes one if its trajectory pierces any target's disc, and since the discs occupy a total area $\delta N\,\sigma$ out of the tube's face $A$ , the interaction probability in $\delta t$ is $\delta P=\delta N\,\sigma/A=n_b\,v\,\sigma\,\delta t$ . Why: this is the step that gives $\sigma$ its meaning — it is the area each target must present to reproduce the observed hit probability, whatever the underlying physics.

Step 3. Divide by $\delta t$ to get the reaction rate per projectile, $\mathrm{d}P/\mathrm{d}t=n_b\,v\,\sigma$ . This is already $\text{rate}=\sigma\times(\text{flux seen by one projectile})$ , because $n_b v$ is the number of targets crossing unit area per unit time. Why: it shows the cross section is exactly the rate stripped of its kinematic flux, isolating the quantum content.

Step 4. Sum over a beam of $N_a$ projectiles at density $n_a$ . Writing the incident flux as $\phi=n_a v$ and the exposed target number as $N_b$ , the total rate is $\text{rate}=\sigma\,\phi\,N_b$ , and inverting gives the working definition $\sigma=\text{rate}/(\phi\,N_b)$ . Why: this is the equation an experiment inverts — measure a rate, know your flux and target count, and read off $\sigma$ — and it is the relation the relativistic formula must reproduce.

Go deeper · Derivation B · 2 minThe Lorentz-invariant cross-section master formula

Derivation BThe Lorentz-invariant cross-section master formula

The aim here is to turn the geometric definition into an expression that is manifestly the same in every inertial frame, by feeding it the relativistic golden rule with states normalised to $2E$ particles per volume $V$ , and to demonstrate the crucial bookkeeping miracle: every factor of the arbitrary box volume $V$ cancels, leaving a flux factor and a phase-space integral that are both Lorentz scalars. This is the formula from which every two-to-anything cross section in the Standard Model is computed.

Step 1. Adopt the covariant normalisation of one relativistic wavefunction to $2E$ particles in volume $V$ , so each incoming particle contributes a density $2E/V$ and each state carries a factor $1/\sqrt{2EV}$ . The golden-rule transition rate for $a+b\to$ final state is then $\Gamma=|T_{fi}|^2\times(\text{density of final states})$ , with $T_{fi}$ related to the invariant amplitude by stripping one $1/\sqrt{2E_iV}$ per external leg. Why: the $2E$ convention is exactly what makes $M_{fi}$ a Lorentz scalar, so the physics factor and the kinematic factors can be separated cleanly.

Step 2. Write the final-state phase space with the same convention: each outgoing particle brings $V\,\mathrm{d}^3p_f/(2\pi)^3$ available states and a $1/2E_f$ from its normalisation, and overall energy–momentum conservation contributes $(2\pi)^4\delta^4(p_a+p_b-\sum p_f)$ . Collecting the outgoing factors gives the Lorentz-invariant phase-space measure $\mathrm{d}\Pi=\prod_f \mathrm{d}^3p_f/[(2\pi)^3 2E_f]$ times the delta function. Why: this measure is a known Lorentz scalar, so once the incoming side is also made invariant the whole cross section will be frame-independent.

Step 3. Divide the rate by the incident flux and the target number to get $\sigma$ , exactly as Derivation 1 prescribes. The flux for two colliding particles in the box is $(\text{density}_a)\times(\text{relative velocity})=(2E_a/V)(2E_b/V)\,v_{\text{rel}}\times$ the geometric factors, and target number is one particle per $V$ . Assemble everything and the powers of $V$ from the two incoming $1/2E_iV$ normalisations, the outgoing $V$ 's, and the $1/V$ flux all cancel identically. Why: the box was a fiction introduced only to make states countable, so a physical cross section cannot remember it — the cancellation is the proof that the calculation was set up consistently.

Step 4. What remains multiplying $|M_{fi}|^2\,\mathrm{d}\Pi$ is precisely $1/F$ with $F=4E_aE_b\,v_{\text{rel}}=4E_aE_b(v_a+v_b)$ , the collinear flux factor. Recognising that $E_aE_b(v_a+v_b)=E_bp_a+E_ap_b=\sqrt{(p_a\cdot p_b)^2-m_a^2m_b^2}$ shows $F$ is a Lorentz invariant, completing equation (4). Why: this closes the loop — an invariant amplitude over invariant phase space divided by an invariant flux is invariant, so the number the formula predicts is the number any detector in any frame will record.

Go deeper · Derivation C · 1 minThe two-body cross section in the centre-of-mass frame

Derivation CThe two-body cross section in the centre-of-mass frame

This derivation carries out the phase-space integral of the master formula for the single most important case — two particles in, two particles out — working in the centre-of-mass frame where the kinematics is simplest, and lands on the compact result $\mathrm{d}\sigma/\mathrm{d}\Omega=|M_{fi}|^2\,p_f^*/(64\pi^2 s\,p_i^*)$ . Along the way it evaluates both the two-body phase space and the flux factor in terms of the single invariant $s$ , so the final expression is frame-independent despite the frame-specific route.

Step 1. In the CM frame the two incoming momenta are equal and opposite with magnitude $p_i^*$ , so the flux factor becomes $F=4E_aE_b(v_a+v_b)=4(E_bp_a+E_ap_b)=4p_i^*(E_a+E_b)=4p_i^*\sqrt{s}$ , using $E_a+E_b=\sqrt{s}$ . Why: reducing $F$ to $p_i^*\sqrt s$ removes every reference to the individual energies, so only the invariant $s$ and the CM momentum survive.

Step 2. Evaluate the two-body invariant phase space. Integrating one particle's three-momentum against the momentum delta and the remaining energy delta gives $\int\mathrm{d}\Pi_2=\frac{1}{16\pi^2}\frac{p_f^*}{\sqrt s}\int\mathrm{d}\Omega$ , where $p_f^*$ is the common magnitude of the outgoing CM momenta and $\mathrm{d}\Omega$ is the solid angle of one final particle. Why: this is the standard two-body result — the density of final states for a fixed CM energy is proportional to the outgoing momentum $p_f^*$ , which is what makes thresholds turn on as $\sqrt{s}$ rises past a mass sum.

Step 3. Insert both results into the master formula (4). The differential form is $\mathrm{d}\sigma=\frac{1}{F}|M_{fi}|^2\,\mathrm{d}\Pi_2=\frac{1}{4p_i^*\sqrt s}\,|M_{fi}|^2\,\frac{1}{16\pi^2}\frac{p_f^*}{\sqrt s}\,\mathrm{d}\Omega$ , and combining the $\sqrt s$ factors into $s$ gives $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}=\frac{1}{64\pi^2 s}\frac{p_f^*}{p_i^*}|M_{fi}|^2$ . Why: the flux momentum $p_i^*$ ends up downstairs and the phase-space momentum $p_f^*$ upstairs, so their ratio measures how much the final-state kinematics opens up relative to the initial state.

Step 4. Specialise and integrate. For elastic scattering $p_f^*=p_i^*$ so the ratio is $1$ ; a total cross section follows from integrating over $\mathrm{d}\Omega$ , and if $|M_{fi}|^2$ is isotropic the integral simply multiplies by $4\pi$ . Why: this is the launch pad for the whole differential-cross-section programme — feed in the QED, QCD or electroweak amplitude and the angular distribution and total rate follow immediately.

Limiting cases

Case	Result	Meaning
Fixed target ( $b$ at rest)	$F=4m_b\,p_a$	flux is set by beam momentum × target mass; $E_b=m_b$ , $v_b=0$
Collider, CM frame	$F=4p_i^*\sqrt{s}$	symmetric form; feeds $\mathrm{d}\sigma/\mathrm{d}\Omega=\|M\|^2p_f^/(64\pi^2 s\,p_i^)$
Ultra-relativistic ( $m\to0$ , equal beams)	$F=2s$	since $p_i^*=\sqrt s/2$ , $F=4(\sqrt s/2)\sqrt s=2s$
Elastic two-body	$p_f^=p_i^$ , so $\mathrm{d}\sigma/\mathrm{d}\Omega=\|M\|^2/64\pi^2 s$	phase-space ratio is exactly one
Slow-neutron absorption	$\sigma\to$ geometric-ish, a few barns	the one regime where $\sigma$ really is an effective disc area
High energy, point coupling	$\sigma\propto 1/s$	e.g. $\sigma(e^+e^-\to\mu^+\mu^-)=4\pi\alpha^2/3s$ , the plotted curve

Common trapThe flux uses the relative velocity, and its "speed" can exceed c without anything being wrong

The single most common error is to build the flux from the beam speed alone. The flux factor must use the relative motion of both particles, $v=v_a+v_b$ in the collinear frame; drop the target's motion and the cross section stops being Lorentz invariant and disagrees between frames. A subtler trap hides in the same expression: for two beams each approaching the speed of light, $v_a+v_b$ tends toward $2c$ . That is not a physical velocity and it does not violate relativity — it is a closing speed, the rate at which the coordinate separation of the two particles shrinks in one particular frame, and closing speeds up to $2c$ are perfectly ordinary. The genuinely invariant object is the flux factor $F=4\sqrt{(p_a\cdot p_b)^2-m_a^2m_b^2}$ , which never misbehaves; the collinear $4E_aE_b(v_a+v_b)$ is just its value written out in one frame. So "relative velocity" here is shorthand for the invariant $F$ , not a claim that something travels faster than light.

Worked examples

1 · The point cross section for e⁺e⁻ → μ⁺μ⁻ (fully worked)

Use the centre-of-mass master formula to evaluate the total cross section for $e^+e^-\to\mu^+\mu^-$ in the high-energy limit where all masses are negligible, taking the spin-averaged amplitude $\overline{|M_{fi}|^2}=e^4(1+\cos^2\theta)$ from the sibling article. Then put in numbers at $\sqrt s=10$ GeV.

Step 1. Because the electron and muon masses are dropped, the reaction is effectively elastic in the kinematic sense $p_f^*=p_i^*$ , so equation (5) becomes $\dfrac{\mathrm{d}\sigma}{\mathrm{d}\Omega}=\dfrac{1}{64\pi^2 s}\,\overline{|M_{fi}|^2}=\dfrac{e^4}{64\pi^2 s}(1+\cos^2\theta)$ .

Step 2. Trade the coupling for the fine-structure constant with $\alpha=e^2/4\pi$ , so $e^4=(4\pi\alpha)^2=16\pi^2\alpha^2$ . Substituting cancels the $16\pi^2$ and gives the clean differential form $\dfrac{\mathrm{d}\sigma}{\mathrm{d}\Omega}=\dfrac{\alpha^2}{4s}(1+\cos^2\theta)$ .

Step 3. Integrate over solid angle using $\displaystyle\int(1+\cos^2\theta)\,\mathrm{d}\Omega=4\pi+\frac{4\pi}{3}=\frac{16\pi}{3}$ . The total cross section is therefore $\boxed{\sigma=\dfrac{4\pi\alpha^2}{3s}}$ , the famous point cross section that falls as $1/s$ .

Step 4. Insert numbers at $\sqrt s=10$ GeV, so $s=100\ \mathrm{GeV}^2$ , with $\alpha=1/137.04$ . In natural units $\sigma=\dfrac{4\pi}{3(137.04)^2(100)}=2.23\times10^{-6}\ \mathrm{GeV}^{-2}$ .

Step 5. Convert to a laboratory area with $(\hbar c)^2=0.3894\ \mathrm{mb}\cdot\mathrm{GeV}^2$ : $\sigma=2.23\times10^{-6}\times0.3894\ \mathrm{mb}=8.7\times10^{-7}\ \mathrm{mb}$ . Since $1\ \mathrm{mb}=10^{6}\ \mathrm{nb}$ , this is $\boldsymbol{\sigma=0.87\ \mathrm{nb}}$ — equivalently the handy rule $\sigma=86.8\ \mathrm{nb}/(s/\mathrm{GeV}^2)$ , and exactly the point plotted in the second figure.

2 · The flux factor is the same in every frame

Two electrons collide head-on at $\sqrt s=10$ GeV (masses negligible). Compute the flux factor three independent ways — from the collinear energies-and-speeds form, from the CM form $4p_i^*\sqrt s$ , and from the invariant four-vector form — and check they agree.

Step 1. In the CM frame each electron has energy $E=\sqrt s/2=5$ GeV and, being effectively massless, momentum $p_i^*=5$ GeV with speed $v\approx1$ . The collinear form gives $F=4E_aE_b(v_a+v_b)=4(5)(5)(1+1)=\boldsymbol{200\ \mathrm{GeV}^2}$ .

Step 2. The centre-of-mass reduction gives $F=4p_i^*\sqrt s=4(5)(10)=\boldsymbol{200\ \mathrm{GeV}^2}$ , and since $p_i^*=\sqrt s/2$ this equals $2s=2(100)=200\ \mathrm{GeV}^2$ — the ultra-relativistic limiting case.

Step 3. The invariant form needs $p_a\cdot p_b$ . For head-on massless particles $p_a\cdot p_b=E_aE_b-\vec p_a\cdot\vec p_b=25-(-25)=50\ \mathrm{GeV}^2$ , so $F=4\sqrt{(p_a\cdot p_b)^2-0}=4(50)=\boldsymbol{200\ \mathrm{GeV}^2}$ .

Step 4. All three routes give 200 GeV². The agreement is guaranteed because $F$ is a Lorentz scalar, but computing it three ways is the concrete check that the "relative velocity" combination and the covariant square root are the same object — the reason a cross section measured in a fixed-target machine can be quoted for a collider without ambiguity.

Physics of the portrait

The cross section is one idea passed down a chain of physicists, each of whom made it more abstract: a literal area for Rutherford, a probability for Born, and finally an invariant for the architects of quantum field theory.

Origins · 1 mincross section and how this was discovered

cross section

the area σ from rate = flux × N × σ · from Rutherford's nucleus to the invariant flux formula

The cross section began as something you could almost see. When Hans Geiger and Ernest Marsden fired alpha particles at gold foil in 1909 and found a few bouncing straight back, Ernest Rutherford in 1911 explained the deflections by concentrating the atom's positive charge in a tiny nucleus and derived the first scattering cross section, $\mathrm{d}\sigma/\mathrm{d}\Omega\propto 1/\sin^4(\theta/2)$ — an angular distribution with the honest units of area per solid angle. For Rutherford the cross section was geometry: the effective target the nucleus presented to an incoming alpha. Max Born dissolved that literalism in 1926, when his probabilistic reading of the wavefunction turned the scattering amplitude into a probability amplitude and the cross section into $|f(\theta)|^2$ — no longer a disc but a quantum likelihood that merely carries the dimensions of area. The unit caught up in 1942, when Manhattan-Project physicists measuring neutron absorption needed a word for $10^{-24}\,\mathrm{cm}^2$ and, finding a uranium nucleus "as big as a barn," coined the barn. The final abstraction came with covariant perturbation theory around 1949, when Feynman and Dyson gave the invariant matrix element $M_{fi}$ that plugs into the modern master formula; the cross section became a Lorentz scalar, the same in every frame, computed from Feynman diagrams and divided by an invariant flux. Rutherford's tangible area survives only as the dimensions of the answer — the ghost of geometry in a quantity that is really pure probability.

History

1909

— Geiger and Marsden record large-angle alpha-particle deflections from gold foil

1911

— Rutherford derives the nuclear scattering cross section and infers the nucleus

1926

— Born reads the squared amplitude as a probability, making the cross section quantum

1942

— the barn (10⁻²⁴ cm²) is coined for neutron-absorption cross sections

1949

— Feynman and Dyson supply the covariant amplitude M that feeds the modern σ

Going further

Texts. Thomson, Modern Particle Physics, §3.4 (interaction cross sections) and §3.4.1 (Lorentz-invariant flux), eqs. (3.23)–(3.29) · Halzen & Martin, Quarks and Leptons, §4.3 (the cross section and phase space) · Peskin & Schroeder, An Introduction to Quantum Field Theory, §4.5 (cross sections and the S-matrix) · Griffiths, Introduction to Elementary Particles, 2nd ed., §6.2 (Fermi's golden rule and cross sections) · Schwartz, Quantum Field Theory and the Standard Model, §5.1. The point amplitude $\overline{|M|^2}=e^4(1+\cos^2\theta)$ used in Worked Example 1 is quoted from the $e^+e^-\to\mu^+\mu^-$ article, not re-derived here. Key results to keep in mind: (1) $F=4E_aE_b(v_a+v_b)=4\sqrt{(p_a\cdot p_b)^2-m_a^2m_b^2}$ , which evaluates to $200\ \mathrm{GeV}^2$ for a head-on massless collision at $\sqrt s=10$ GeV by all three routes; (2) the two-body phase space $\int\mathrm{d}\Pi_2=\frac{1}{16\pi^2}\frac{p_f^*}{\sqrt s}\int\mathrm{d}\Omega$ ; (3) $\mathrm{d}\sigma/\mathrm{d}\Omega=|M|^2p_f^*/(64\pi^2 s\,p_i^*)$ ; (4) $\sigma(e^+e^-\to\mu^+\mu^-)=4\pi\alpha^2/3s=86.8\ \mathrm{nb}/(s/\mathrm{GeV}^2)=0.87$ nb at $\sqrt s=10$ GeV using $(\hbar c)^2=0.3894\ \mathrm{mb\,GeV}^2$ .

Prerequisites. Fermi's Golden Rule · Lorentz-Invariant Phase Space · Wavefunction Normalisation · Four-Vector Notation. Leads to. Differential Cross Sections · $e^+e^-\to\mu^+\mu^-$ · Luminosity and Event Rates · Resonances and the Breit–Wigner Formula.

Ask the tutor: Ask me to walk you through why every factor of the volume $V$ cancels in the master formula, or to compute $\sigma$ for any two-to-two process you name.

Why is a cross section an area if nothing is physically that size?

Because the dimensions are forced, not chosen. The defining relation is rate = flux × targets × cross section, and flux carries units of (area·time) $^{-1}$ while a rate is (time) $^{-1}$ ; the only way the equation balances is for $\sigma$ to have the dimensions of area. For a slow neutron hitting a nucleus the cross section really is close to the geometric disc area, which is where the picture comes from. But for something like Higgs production the "area" is fantastically smaller than any particle — it simply encodes how improbable the reaction is. The area is a unit, not a size.

What are the barn, millibarn, and femtobarn?

The barn is the working unit of cross section, $1\ \mathrm{b}=10^{-28}\ \mathrm{m}^2=10^{-24}\ \mathrm{cm}^2$ , chosen because it is roughly a nuclear geometric area — a uranium nucleus has $\pi R^2\approx1.7$ barn. Its submultiples track the rarity of the process: strong-interaction cross sections sit near the millibarn ( $10^{-3}$ b), electroweak processes near the picobarn ( $10^{-12}$ b), and the rarest measured signals near the femtobarn ( $10^{-15}$ b). Because event yield is $\mathcal{L}\,\sigma$ , integrated luminosity is quoted in inverse femtobarns, so that " $\sigma$ in fb × luminosity in fb⁻¹" gives a plain event count.

Why must the flux use the relative velocity and not just the beam speed?

Because interactions depend on how fast the two particles approach each other, not on how fast either moves in some frame. If the target is also moving, using only the beam speed miscounts how many targets the beam sweeps through per second, and — more seriously — it makes the cross section frame-dependent, which is physically impossible since a rate is a rate in any frame. Building the flux from $v_a+v_b$ (and, invariantly, from $\sqrt{(p_a\cdot p_b)^2-m_a^2m_b^2}$ ) is exactly what guarantees every observer computes the same $\sigma$ .

How does the cross section connect to what a detector actually counts?

Through the luminosity. A collider delivers an instantaneous luminosity $\mathcal{L}$ (units of area $^{-1}$ time $^{-1}$ ) that packages beam intensities, focusing and crossing geometry, and the observed event rate for any process is simply $\text{rate}=\mathcal{L}\,\sigma$ . Integrating over a run, the number of events is $N=\sigma\int\mathcal{L}\,\mathrm{d}t$ , with $\int\mathcal{L}\,\mathrm{d}t$ the integrated luminosity in inverse barns. So the cross section is the theory-side number and the luminosity is the machine-side number; their product is the experiment. This is developed fully in the luminosity article.

What is the difference between the total and the differential cross section?

The differential cross section $\mathrm{d}\sigma/\mathrm{d}\Omega$ tells you how the scattered particles are distributed in angle — it is what equation (5) gives directly, and its shape encodes the dynamics (spin, the propagator, interference). The total cross section $\sigma$ is its integral over all angles (and, more generally, over all final-state variables), and it tells you the overall probability that anything of the specified kind happens. You almost always compute the differential form first from the amplitude, then integrate; the total is the single headline number, the differential is where the physics is visible.

Can the "relative velocity" v_a + v_b really be larger than the speed of light?

Yes, and it breaks nothing. In a frame where two beams approach head-on, each at nearly $c$ , the rate at which their coordinate separation shrinks is nearly $2c$ . That closing speed is a bookkeeping quantity, not the speed of any object or signal, so it is not bounded by $c$ — no material thing moves at $2c$ . The physically meaningful and invariant object is the flux factor $F=4\sqrt{(p_a\cdot p_b)^2-m_a^2m_b^2}$ , which is well-behaved in every frame; $4E_aE_b(v_a+v_b)$ is just its value in the collinear frame. Relativity limits the velocity of a single body, not the difference of two.