Differential cross sections
The total cross section answers "how often?"; the differential cross section answers "which way, and with how much energy?" — and it is the second question that carries almost all of the physics. When a beam strikes a target, the scattered particles do not spray out uniformly: they crowd forward or backward, they peak at special angles, they trade energy against recoil in a way fixed by the underlying interaction. The differential cross section dσ/dΩ is the count of particles scattered into a small cone of solid angle dΩ per unit time, per target, divided by the incident flux — a directly measurable, positive quantity with units of area per steradian. Integrate it over the whole sphere and the total cross section drops out, σ = ∫ (dσ/dΩ) dΩ; but throw that angular information away and you have discarded the thing that told Rutherford there was a nucleus and told the SLAC experimenters there were quarks. For any two-body reaction 1 + 2 → 3 + 4 the calculation runs through one master formula: dσ/dΩ = |ℳ|²·p_f/(64π²s·p_i*) in the centre-of-mass frame, where the Lorentz-invariant matrix element ℳ carries the dynamics and everything else is kinematics and phase space. Because s, t and |ℳ|² are frame-independent, the same content repackages into a manifestly invariant form dσ/dt that holds in every frame, and from it the laboratory-frame angular distribution follows by a single Jacobian. This article is the hinge between the abstract transition rate of Fermi's golden rule and the concrete angular plots that experiments actually publish.
- The differential cross section dσ/dΩ is the scattering rate into a solid-angle element dΩ per unit incident flux; its integral over 4π steradians is the total σ.
- For any 2 → 2 process in the centre-of-mass frame, dσ/dΩ = |ℳ|²·p_f/(64π²s·p_i*), with the invariant matrix element ℳ carrying all the dynamics.
- Rewriting the solid angle through the Mandelstam variable t gives the manifestly Lorentz-invariant form dσ/dt = |ℳ|²/(64π·s·p_i*²), valid in every frame.
- In a fixed-target lab frame the angular distribution follows by the Jacobian dt/d(cosθ) = 2E₃², giving dσ/dΩ = |ℳ|²·(E₃/m_p E₁)²/(64π²) for elastic e⁻p.
- The angular shape is the fingerprint of the interaction: spin-1 photon exchange in e⁺e⁻ → μ⁺μ⁻ produces dσ/dΩ ∝ (1 + cos²θ).
- Key idea: the differential cross section is the shape of scattering — where particles go and with what energy — and that angular dependence is the direct fingerprint of the matrix element; the total cross section is only its integral.
The differential cross section tells you the shape of scattering — where the particles go and with what energy — and its angular dependence is the direct fingerprint of the interaction's matrix element; the total cross section is just its integral over all directions.
Where this lives today
Differential cross sections are the primary data product of every modern collider. At the LHC, ATLAS and CMS publish dσ/dp_T, dσ/dy and dσ/dm distributions that are compared to QCD predictions to search for deviations; the shape, not just the total rate, is where new physics would first appear as an excess at high transverse momentum or a resonant bump in an invariant-mass spectrum. At LEP, the forward-backward asymmetry A_FB in e⁺e⁻ → μ⁺μ⁻ and e⁺e⁻ → bb̄ — a directly measured asymmetry of the differential cross section — pinned down the vector and axial couplings of the Z boson and, through them, the electroweak mixing angle sin²θ_W to four significant figures. In fixed-target physics the lineage is even more direct: elastic e⁻p scattering at Jefferson Lab measures dσ/dΩ versus q² to extract the proton's electric and magnetic form factors (the Rosenbluth separation), while the deep-inelastic double-differential d²σ/dE′dΩ measured at SLAC and HERA revealed the quark structure functions of the nucleon. Even neutrino oscillation experiments and dark-matter direct-detection searches quote their signals as differential recoil-energy spectra dσ/dE_R. The century-old logic is unchanged: measure where the particles go, and read the interaction off the shape.
The physical picture
Send a steady, uniform beam at a single target particle. Far away, ring the target with detectors, each subtending a tiny solid angle dΩ = sinθ dθ dφ about a direction (θ, φ). Every detector logs a steady count rate, and the pattern of rates across the sphere is the physical content of the differential cross section. Normalise each rate by the incident flux Φ (particles per unit area per unit time) and by the solid angle it covers, and you have dσ/dΩ — an area per steradian that no longer depends on how bright your beam was or how big your detector is, only on the physics of the collision. The first figure fixes this geometry; it is the definition made visible.
The reason the pattern matters is that different interactions paint different pictures on that sphere. A featureless contact interaction scatters isotropically; a long-range Coulomb field piles particles into the forward direction; the exchange of a spin-1 photon in e⁺e⁻ → μ⁺μ⁻ produces a gentle (1 + cos²θ) rise toward both poles. The angular distribution is a read-out of the spin and range of whatever was exchanged — which is exactly why measuring dσ/dΩ, not just σ, is how the structure of an interaction is diagnosed. The second figure is that diagnosis made quantitative: the pure-QED angular distribution of muon pair production, a real prediction with real numbers attached.
The core equations
The operational definition comes first: the differential cross section is a rate normalised to flux and solid angle. It is what a detector array measures directly, before any theory is invoked, and everything that follows is a way of computing it.
where: Φ — incident flux (particles per unit area per unit time); dR/dΩ — scattered rate per unit solid angle per target; dΩ = sinθ dθ dφ — solid-angle element. The integral over the full sphere (4π steradians) returns the total cross section, so dσ/dΩ contains strictly more information than σ. The master result for a two-body reaction 1 + 2 → 3 + 4 in the centre-of-mass frame packages the golden rule, the flux, and the two-body phase space into one line. It is derived below; here we read it.
where: s — Mandelstam invariant, the square of the total CoM energy (s = E_CM²); p_i — magnitude of each initial-state momentum in the CoM frame; p_f — magnitude of each final-state momentum in the CoM frame; ℳ_fi — the Lorentz-invariant matrix element (invariant amplitude) for the process. The starred quantities are CoM-frame values; the dynamics live entirely in |ℳ|², while the ratio p_f/p_i is pure phase space. When the collision is elastic, or the final masses equal the initial masses, the outgoing and incoming CoM momenta are equal, p_f = p_i, and the phase-space ratio collapses to one.
where: all symbols as in equation (2). This is the form used for e⁺e⁻ → μ⁺μ⁻ above the muon threshold (where the muon mass is negligible) and for any elastic 2 → 2 collision viewed in the CoM frame. Because σ, s, t and |ℳ|² are each Lorentz-invariant, one can trade the solid angle for the invariant momentum-transfer t and obtain a differential cross section that reads the same in every frame. This is the workhorse expression for fixed-target physics.
where: t = (p₁ − p₃)² — the Mandelstam momentum-transfer invariant (t ≤ 0 for physical scattering); s p_i*² = (p₁·p₂)² − m₁²m₂² — the invariant flux combination. Nothing here is starred: equation (4) is frame-independent, which is precisely why it is the right stepping-stone to a laboratory-frame result. Applying equation (4) to elastic electron–proton scattering in the laboratory frame, with the proton initially at rest, the Jacobian dt/d(cosθ) = 2E₃² (derived below) converts it into a lab angular distribution. The recoil of the finite-mass target enters through the scattered-electron energy E₃.
where: E₁ — incident electron energy; E₃ — scattered electron energy (fixed by θ through recoil kinematics); m_p — proton mass; θ — lab scattering angle. The factor (E₃/E₁)² is the recoil correction that would disappear for an infinitely heavy, static target. Finally, the concrete instance the second figure plots: inserting the spin-averaged QED matrix element for e⁺e⁻ → μ⁺μ⁻ into equation (3) gives an angular distribution whose (1 + cos²θ) shape is the signature of one-photon exchange.
where: α = e²/4π ≈ 1/137 — the fine-structure constant; θ — CoM scattering angle of the μ⁻ relative to the e⁻. The total follows by integrating (1 + cos²θ) over the sphere to give 16π/3; numerically σ = 86.8 nb divided by s in GeV².
The goal of this derivation is to show where equation (2) comes from — to assemble dσ/dΩ = |ℳ|²·p_f/(64π²s·p_i*) out of three separately meaningful pieces: the transition probability per unit time supplied by Fermi's golden rule (which carries |ℳ|²), the incident flux that normalises a rate into a cross section, and the Lorentz-invariant two-body phase space that counts the available final states. Keeping the three factors visibly separate is what makes the final formula memorable rather than magical, and it is what lets the same skeleton be reused for decays and for many-body final states.
Step 1. Write the cross section as rate over flux times phase space: dσ = |ℳ|²·(1/F)·dΠ₂, where F is the invariant flux factor and dΠ₂ is the two-body Lorentz-invariant phase-space element. Why: a cross section is a transition rate (from the golden rule, carrying |ℳ|²) divided by the incident flux, with the density of final states supplied by phase space — this factorisation is the definition dressed for relativistic use.
Step 2. Evaluate the flux factor. The invariant form is F = 4√[(p₁·p₂)² − m₁²m₂²], and in the CoM frame this reduces cleanly to F = 4 p_i √s. Why: the flux must be Lorentz-invariant so that σ is frame-independent, and (p₁·p₂)² − m₁²m₂² is the unique invariant that reduces to the intuitive "relative velocity times densities" in any frame.*
Step 3. Evaluate the two-body phase space. Integrating the four-momentum-conserving delta functions over the final momenta leaves dΠ₂ = (1/(4π)²)·(p_f/√s)·dΩ, a standard result: the final-state momentum magnitude p_f times the solid angle into which particle 3 emerges. Why: after conservation is imposed, a 2 → 2 final state has just two free angles, so the phase space is genuinely an angular distribution weighted by the outgoing momentum p_f.
Step 4. Multiply the pieces: dσ = |ℳ|²·[1/(4 p_i√s)]·[(1/16π²)(p_f/√s)] dΩ. Collecting the constants, 1/(4·16π²) = 1/(64π²) and √s·√s = s, gives dσ/dΩ = |ℳ|²·p_f/(64π²s·p_i). Why: this is equation (2) — the flux put p_i in the denominator, the phase space put p_f in the numerator, and their ratio is exactly the phase-space factor that vanishes for elastic scattering. *Note: the two factors F = 4p_i√s and dΠ₂ = (p_f/(16π²√s))dΩ combine to give exactly the coefficient 1/(64π²s) in equation (2), and the p_f/p_i ratio reduces to 1 for elastic scattering.
Go deeper · Derivation B · 1 minRepackaging the cross section into the invariant form dσ/dt
The aim here is to convert the frame-dependent centre-of-mass formula, equation (2), into the manifestly Lorentz-invariant equation (4) by trading the solid angle dΩ* for the Mandelstam invariant t. The payoff is large: because dσ/dt is built entirely from invariants (s, t, and |ℳ|²), a single calculation done in the convenient CoM frame becomes immediately usable in the laboratory frame or any other, which is indispensable for fixed-target experiments where the lab and CoM frames differ.
Step 1. Express t in CoM variables. Writing t = (p₁ − p₃)² = m₁² + m₃² − 2E₁E₃ + 2 p_i p_f cosθ, only the last term depends on the scattering angle. Why: t is Lorentz-invariant, so evaluating it in the CoM frame gives an invariant handle on the angle θ without committing to that frame.
Step 2. Differentiate to get the angle-to-invariant Jacobian: dt = 2 p_i p_f d(cosθ). Since the process is azimuthally symmetric, the solid angle integrates to dΩ = 2π d(cosθ), so dΩ = (π/(p_i p_f)) dt. Why: with no φ dependence in |ℳ|², the only nontrivial variable is cosθ, and the chain rule turns "per solid angle" into "per t" through this factor.*
Step 3. Substitute into equation (2): dσ/dt = [dσ/dΩ]·[dΩ/dt] = [|ℳ|²p_f/(64π²s p_i)]·[π/(p_i p_f)]. The p_f cancels and one power of π reduces 64π² to 64π. Why: the phase-space momentum p_f that appeared through the final-state counting is exactly cancelled by the Jacobian, leaving a result that depends only on the initial momentum p_i — a hint that the answer is now frame-neutral.*
Step 4. Collect: dσ/dt = |ℳ|²/(64π s p_i²). Every surviving symbol — s, p_i² (equivalently the invariant (p₁·p₂)² − m₁²m₂²), and |ℳ|² — is Lorentz-invariant, so the equation holds in all frames. Why: this is equation (4); its invariance is what makes it the correct launch point for the laboratory-frame calculation that follows. *Note: the relations dt = 2p_ip_f d(cosθ) and dΩ = 2π d(cosθ) combine to give dσ/dt = |ℳ|²/(64π s p_i²), confirming equation (4).*
Go deeper · Derivation C · 2 minThe laboratory-frame angular distribution and its Jacobian
This derivation takes the invariant equation (4) into the laboratory frame for elastic electron–proton scattering with the proton at rest, producing equation (5). The single nontrivial ingredient is the Jacobian dt/d(cosθ) relating the invariant t to the lab scattering angle θ, and the surprising simplicity of its result — exactly 2E₃² — is what collapses the whole transformation to one clean line. Getting this right is what separates a correct fixed-target prediction from the classic mistake of applying the CoM formula in the lab.
Step 1. Fix the lab kinematics. With the electron mass neglected and the proton at rest, energy-momentum conservation gives the scattered energy E₃ = E₁ / [1 + (E₁/m_p)(1 − cosθ)], and the momentum transfer t = −2E₁E₃(1 − cosθ). Why: the target recoils, so E₃ is not free but locked to θ; this recoil is the entire difference between a finite-mass target and a static potential.
Step 2. Compute the flux combination in the lab. Since s p_i² = (p₁·p₂)² − m₁²m₂², and in the lab p₁·p₂ = E₁ m_p with the electron mass dropped, s p_i² = (E₁ m_p)². Why: the invariant flux factor from Derivation A, evaluated with a stationary target, becomes simply the product of beam energy and target mass squared — a clean lab-frame number.
Step 3. Evaluate the Jacobian. Writing t = −2E₁²(1 − cosθ)/D with D = 1 + (E₁/m_p)(1 − cosθ), careful differentiation gives d/d(cosθ)[(1 − cosθ)/D] = −1/D², hence dt/d(cosθ) = 2E₁²/D² = 2E₃². Why: the recoil dependence of E₃ on angle conspires so that the algebra collapses to 2E₃² exactly — a small miracle that makes the lab formula as compact as the CoM one.
Step 4. Assemble. With dσ/dΩ = (1/2π)(dt/d(cosθ)) (dσ/dt) = (1/2π)(2E₃²)·|ℳ|²/(64π(E₁m_p)²), the factors combine to dσ/dΩ = |ℳ|²(E₃/m_p E₁)²/(64π²). Why: this is equation (5); the (E₃/E₁)² factor is the recoil correction, and it tends to 1 as m_p → ∞, recovering scattering off a fixed potential. *Note: explicit differentiation confirms dt/d(cosθ) = 2E₃², which evaluates numerically to 0.3117 GeV² at E₁ = 0.5 GeV, θ = 60°.*
Limiting cases
| Case | Result | Meaning |
|---|---|---|
| Lab frame = CoM frame (collider) | dσ/dΩ = dσ/dΩ* directly | no Jacobian needed; the measured angle is the CoM angle, as at LEP or the LHC |
| Elastic, or equal masses | p_f = p_i, so dσ/dΩ* = ∣ℳ∣²/(64π²s) | the phase-space ratio drops out; only the amplitude shapes the distribution |
| Isotropic ∣ℳ∣² (pure s-wave) | dσ/dΩ = constant, σ = 4π·(dσ/dΩ) | a featureless contact interaction; the angular plot is flat |
| Heavy static target (m_target → ∞) | E₃ → E₁, recoil factor → 1 | scattering off a fixed potential; the Rutherford / first-Born limit |
| e⁺e⁻ → μ⁺μ⁻, √s ≫ m_μ | dσ/dΩ ∝ (1 + cos²θ) | spin-1 photon exchange fingerprint; symmetric front-to-back, A_FB = 0 |
Worked examples
1 · The angular distribution and total rate of e⁺e⁻ → μ⁺μ⁻ (fully worked)
At a collider with √s = 10 GeV (well above the muon threshold), the spin-averaged QED matrix element for e⁺e⁻ → μ⁺μ⁻ is ⟨|ℳ|²⟩ = e⁴(1 + cos²θ). Find the differential cross section, integrate it to the total, and evaluate both numerically.
Step 1. Apply the elastic CoM formula, equation (3), which is exact here because both leptons are effectively massless at 10 GeV so p_f = p_i: dσ/dΩ = ⟨|ℳ|²⟩/(64π²s) = e⁴(1 + cos²θ)/(64π²s). Substitute e⁴ = (4πα)² = 16π²α² to clear the constants, giving dσ/dΩ = (α²/4s)(1 + cos²θ) — equation (6).
Step 2. Integrate over the sphere. Because the distribution is azimuthally symmetric, ∫(1 + cos²θ) dΩ = 2π ∫₋₁¹ (1 + c²) dc = 2π[2 + 2/3] = 16π/3. Hence σ = (α²/4s)(16π/3) = 4πα²/3s — the celebrated point-cross-section formula.
Step 3. Put in numbers. With s = 100 GeV² and the conversion (ħc)² = 3.894 × 10⁵ nb·GeV², the total is σ = 4π(1/137.036)²/(3·100) × 3.894 × 10⁵ nb = 0.87 nb (equivalently σ = 86.8 nb ÷ s[GeV²]). The differential value at θ = 90° is (α²/4s) = 0.052 nb/sr, rising to 0.104 nb/sr at the forward and backward poles — a factor of two, exactly the 1 → 2 swing of (1 + cos²θ).
Step 4. Read the physics. The distribution is symmetric under θ → π − θ, so the forward-backward asymmetry A_FB vanishes in pure QED; the (1 + cos²θ) shape is the unmistakable signature that a single spin-1 photon was exchanged. σ = 0.87 nb, dσ/dΩ ranges from 0.052 to 0.104 nb/sr, and A_FB = 0.
2 · Fixed-target electron–proton kinematics and the Jacobian
A 0.5 GeV electron scatters elastically off a proton at rest (m_p = 0.938 GeV) into a lab angle θ = 60°. Find the scattered energy E₃, the momentum transfer q² = −t, the Jacobian dt/d(cosθ), and check the flux combination.
Step 1. Scattered energy from recoil kinematics: E₃ = E₁/[1 + (E₁/m_p)(1 − cosθ)] = 0.5/[1 + (0.5/0.938)(1 − 0.5)] = 0.5/1.2665 = 0.395 GeV. The proton therefore carries off E₁ − E₃ = 0.105 GeV of recoil kinetic energy — the target does not stand still, which is the whole reason the recoil factor exists.
Step 2. Momentum transfer: q² = −t = 2E₁E₃(1 − cosθ) = 2(0.5)(0.395)(0.5) = 0.197 GeV². This spacelike invariant is what a form-factor measurement would plot the elastic cross section against.
Step 3. Jacobian from Derivation C: dt/d(cosθ) = 2E₃² = 2(0.395)² = 0.312 GeV². This is the factor that converts the invariant dσ/dt into the lab dσ/dΩ; note it uses the scattered energy E₃, not E₁, because the recoil is baked in.
Step 4. Flux check: s p_i*² = (E₁ m_p)² = (0.5 × 0.938)² = 0.220 GeV⁴, confirming the lab-frame flux combination used in equation (5). Assembling equation (5) then needs only the invariant amplitude |ℳ|² for one-photon exchange off a proton, which is the content of the Rosenbluth formula that this article leads into. E₃ = 0.395 GeV, q² = 0.197 GeV², dt/d(cosθ) = 0.312 GeV².
Physics of the portrait
The differential cross section is not an abstraction that arrived with quantum field theory — it is the quantity that founded nuclear and particle physics, in a Manchester laboratory, from the angular pattern of alpha particles bouncing off gold.
Origins · 1 minErnest Rutherford and how this was discovered
Ernest Rutherford, a New Zealand farmer's son who became the founding figure of nuclear physics, gave the differential cross section its first and most famous use. In 1909 his assistants Hans Geiger and Ernest Marsden, at his suggestion, fired alpha particles at thin metal foils and found — to everyone's astonishment — that a small fraction bounced almost straight back. Rutherford later called it "as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you." The plum-pudding atom then in vogue could not produce such large deflections; a diffuse positive charge simply cannot deliver the needed impulse. In 1911 Rutherford resolved the paradox by concentrating the atom's positive charge into a tiny central nucleus and deriving the angular distribution it would produce: dσ/dΩ ∝ 1/sin⁴(θ/2), the Rutherford scattering law. The crucial point is that he read the existence and size of the nucleus off the shape of the distribution — the steep 1/sin⁴(θ/2) rise toward small angles and the non-zero rate at large angles were the fingerprint of a point-like Coulomb scatterer. Geiger and Marsden confirmed the predicted dependence on angle, energy and nuclear charge in 1913. The method Rutherford invented — infer the interaction from the angular pattern of the scattered particles — is exactly the logic of every differential-cross-section measurement since, from Mott's relativistic electrons to the deep-inelastic experiments that later resolved the proton's quarks. The differential cross section is his enduring instrument.
Going further
Texts. Thomson, Modern Particle Physics, §3.4–3.5 (the 2 → 2 differential cross section, CoM and lab frames), Ch. 3 Summary Eqs. 3.49–3.51 · Peskin & Schroeder, An Introduction to QFT, §4.5 (cross sections and the S-matrix), Eq. 4.84 · Halzen & Martin, Quarks and Leptons, §4.3–4.4 · Griffiths, Introduction to Elementary Particles, 2nd ed., §6.2 (the golden rule for scattering) · Schwartz, QFT and the Standard Model, Ch. 5. The key results derived in this article are: (1) the coefficient 1/(64π²s) in the master formula, arising from F = 4p_i√s and dΠ₂ = (p_f/16π²√s)dΩ; (2) the invariant form dσ/dt = |ℳ|²/(64π s p_i²), obtained via dt = 2p_ip_f d(cosθ*); (3) the lab Jacobian dt/d(cosθ) = 2E₃² from explicit differentiation; (4) the e⁺e⁻ → μ⁺μ⁻ results dσ/dΩ = (α²/4s)(1 + cos²θ) and σ = 4πα²/3s = 86.8 nb/s[GeV²]; and (5) the fixed-target numbers E₃ = 0.395 GeV, q² = 0.197 GeV² at E₁ = 0.5 GeV, θ = 60°.
Prerequisites. Interaction Cross Sections · Mandelstam Variables · Centre-of-Mass Frame · Lorentz-Invariant Phase Space. Leads to. Rutherford and Mott Scattering · The Rosenbluth Formula · Deep Inelastic Scattering · e⁺e⁻ → μ⁺μ⁻.
Ask the tutor: Ask me to derive the Rutherford 1/sin⁴(θ/2) law from this master formula, or to work out the forward-backward asymmetry A_FB when the Z boson is added to e⁺e⁻ → μ⁺μ⁻.
What is the difference between the total and the differential cross section?
The total cross section σ is a single number with units of area — it tells you how likely an interaction is to happen at all, and it is what you use to compute an event rate from a beam luminosity. The differential cross section dσ/dΩ (or dσ/dt, dσ/dE) is a distribution: it tells you how that total probability is spread over scattering angles or final-state energies. They are tied by integration, σ = ∫(dσ/dΩ)dΩ, so the total is just the differential summed over every direction. The differential form always contains more information; the moment you integrate you have thrown away the angular pattern that carries the physics of the interaction's spin and range.
Why does the matrix element |ℳ|² carry all the dynamics while the rest is kinematics?
Because the master formula cleanly factorises into "how strongly the fields couple" and "how many final states are available." The invariant amplitude ℳ is what the interaction Lagrangian (the vertices and propagators of the theory) produces — it encodes the coupling strength, the spin structure, and the mass of any exchanged particle. Everything multiplying it — the 1/(64π²s), the p_f/p_i ratio, the phase-space and flux factors — is pure kinematics that would be identical for any interaction with the same external particles. So to predict a new process you compute one thing, |ℳ|²; the kinematic machinery is reusable boilerplate. This is exactly why physicists quote "the matrix element" as the deliverable of a Feynman-diagram calculation.
When can I use the simple centre-of-mass formula, and when must I transform frames?
Use the compact CoM formula dσ/dΩ = |ℳ|²p_f/(64π²s·p_i*) directly whenever your detector's angle is the CoM angle — that is, at a symmetric collider like LEP or the LHC, where the lab frame and the centre-of-mass frame coincide. You must transform when the lab and CoM frames differ, which is the case in any fixed-target experiment (beam on a stationary target). Then the correct procedure is to compute the invariant |ℳ|², form dσ/dt, and apply the frame's Jacobian dt/d(cosθ) plus the recoil factor. Skipping this and reading a lab angle into the CoM formula is the classic mistake flagged in the warning above.
Why is the invariant form dσ/dt so convenient?
Because it is built entirely from Lorentz invariants — s, t, and |ℳ|² — so it reads the same in every reference frame. That means you can do the hard part of the calculation, evaluating |ℳ|², in whatever frame is easiest (usually the CoM frame, where the kinematics are symmetric), and the resulting dσ/dt is immediately valid in the laboratory frame without any boost. You only introduce frame-dependent quantities at the very last step, through the Jacobian that converts t into whatever variable your detector actually measures (a lab angle, an energy, a rapidity). It is the cleanest possible separation between the invariant physics and the frame-dependent bookkeeping.
What does the (1 + cos²θ) shape in e⁺e⁻ → μ⁺μ⁻ actually tell us?
It is a direct read-out of the spin of the exchanged particle. One-photon exchange is mediated by a spin-1 field coupling to conserved currents, and the interference of the possible helicity configurations produces exactly the 1 + cos²θ angular dependence — more muons emitted along the beam axis than transverse to it, by a factor of two. If the exchange had been via a spin-0 scalar, the distribution would be isotropic; a different chiral structure would tilt it forward or backward. So measuring the angular shape confirms that the electromagnetic interaction proceeds through a spin-1 photon. When the Z boson contributes (near its resonance), its parity-violating coupling adds a term that breaks the front-back symmetry, producing the forward-backward asymmetry A_FB that LEP used to measure the weak mixing angle.
Why does the scattered-electron energy E₃ appear in the lab cross section but not the CoM one?
Because in the laboratory the target proton recoils, and how much it recoils depends on the scattering angle. In the CoM frame the collision is symmetric and elastic scattering leaves both momenta unchanged in magnitude (p_f = p_i), so no recoil-energy factor is needed. In the lab, by contrast, a proton of finite mass takes up momentum and energy, so the scattered electron emerges with a reduced energy E₃ = E₁/[1 + (E₁/m_p)(1 − cosθ)] that decreases toward backward angles. That energy loss enters the cross section through the recoil factor (E₃/E₁)². In the limit of an infinitely heavy target the recoil vanishes, E₃ → E₁, and the factor becomes 1 — recovering scattering off a static potential.