Elastic vs inelastic scattering

Fire an electron at a proton and one of two things happens. At low energy the proton recoils but survives — the collision is elastic, the final state is still one electron and one proton, and a single number (the scattering angle) fixes the whole event through energy–momentum conservation. Push the energy up and the momentum the electron can dump into the target grows until the proton no longer holds together. It shatters into a spray of pions, kaons and nucleons — a hadronic system $X$ of variable invariant mass $W$ . That is inelastic scattering, and the moment $W$ is free to vary, one measured angle is no longer enough: two independent kinematic variables are needed to say what happened. This article is the hinge of the deep-inelastic story. It takes the relativistic $e^-p$ kinematics of the preceding chapter and asks the question that reorganizes everything downstream: what changes when the target is allowed to break up? Elastic scattering lives on a one-dimensional line in kinematic space, pinned to $W=m_p$ , described by two form factors of a single variable. Inelastic scattering fills a two-dimensional region, described by two structure functions of two variables — and inside that region, at large momentum transfer, the proton's pointlike constituents first showed themselves. The variables $Q^2$ , $\nu$ , Bjorken $x$ and the inelasticity $y$ introduced here are the coordinates of that region; the elastic limit is simply its boundary at $x=1$ . Everything from Bjorken scaling to the parton distributions used at the LHC is written in these coordinates.

Key takeaways

Elastic $e^-p\to e^-p$ leaves the proton intact, so its invariant mass is fixed at $W=m_p$ and the kinematics need only one variable — the scattering angle.
Inelastic $e^-p\to e^-X$ breaks the proton into a hadronic system $X$ of variable mass $W>m_p$ ; that freedom makes the event two-variable.
The natural coordinates are the Lorentz invariants $Q^2=-q^2$ , the energy loss $\nu$ , Bjorken $x=Q^2/2m_p\nu$ , and the inelasticity $y=\nu/E_1$ , tied by $Q^2=sxy$ .
Elastic scattering is exactly the boundary $x=1$ (equivalently $W=m_p$ ); inelastic events occupy $0<x<1$ .
Two form factors $G_E,G_M$ describe the elastic proton; two structure functions $W_1,W_2$ describe the inelastic proton — the extra variable is why a second function is needed.
Key idea: whether the target survives or shatters changes the counting of kinematic variables — elastic scattering is one-variable because $W$ is frozen at $m_p$ , inelastic scattering is two-variable because $W$ is set free.

The core insight

Whether the target survives or shatters changes the counting of kinematic variables: elastic scattering is one-variable, inelastic scattering is two-variable, because the final-state hadronic mass $W$ is no longer fixed at $m_p$ .

Where this lives today

The elastic-versus-inelastic split is still the organizing map of lepton-nucleon physics. Its purest modern laboratory was HERA at DESY (1992–2007), the only electron-proton collider ever built, which smashed $27.5$ GeV electrons into $920$ GeV protons for $\sqrt s\approx318$ GeV and reached $Q^2$ up to $\sim10^5\ \mathrm{GeV}^2$ and Bjorken $x$ down to $\sim10^{-5}$ — deep into the inelastic region fixed-target machines could only glimpse. HERA's structure-function data are the backbone of the proton parton distribution functions (PDFs) that every LHC cross-section prediction relies on: to compute a Higgs or top-quark rate you must know $F_2(x,Q^2)$ , and that came from mapping the inelastic surface. At the other extreme, Jefferson Lab keeps refining elastic scattering, whose form factors feed the "proton radius puzzle." The next chapter is the Electron-Ion Collider under construction at Brookhaven, designed to image the inelastic proton in three dimensions across the full $(x,Q^2)$ plane whose coordinates this article defines.

The physical picture

Draw the collision as a single virtual photon carrying four-momentum $q$ from the electron to the proton. In an elastic event the virtuality $Q^2=-q^2$ is small enough that the proton absorbs the kick as a whole and recoils as one particle; its invariant mass is unchanged, $W=m_p$ , and conservation of energy and momentum leaves exactly one free parameter — dial the angle $\theta$ and the scattered energy $E_3$ , recoil, and $Q^2$ all follow. In an inelastic event the photon strikes the proton hard: the target absorbs energy $\nu$ as well as momentum and fragments into a hadronic system $X$ whose mass $W$ can range from just above $m_p$ up to the kinematic ceiling. Now $\theta$ and $E_3$ are independent — the same angle can produce many final-state masses — so it takes two numbers to pin the event down. The first figure makes elastic-versus-inelastic one picture; the second is the experiment that shows both, and the whole progression between them, in a single spectrum.

Figure

Elastic e⁻p → e⁻p leaves the proton intact so its invariant mass stays W = m_p and Bjorken x = 1; inelastic e⁻p → e⁻X shatters it into a hadronic system of mass W > m_p with 0 < x < 1, all governed by W² = m_p² + 2m_pν − Q²

The single variable that tells you where along the elastic-to-deep-inelastic road you are is the mass $W$ on the proton side, and at fixed spectrometer angle $W$ maps directly onto the scattered energy $E_3$ . So one scan of $E_3$ at fixed $\theta$ walks through the whole story: a sharp spike where the proton stayed whole, bumps where it was excited into short-lived resonances ( $\Delta$ , $N^*$ ), and a broad continuum where it was blown apart into many hadrons. The second figure is that scan, from the classic DESY measurement — real energies, a real elastic peak, real resonance masses.

Figure

Scattered-electron spectrum for 4.879 GeV electrons on hydrogen at θ = 10° after Bartel 1968: a sharp elastic peak at E₃ ≈ 4.52 GeV where W = m_p and Q² ≈ 0.67 GeV², the Δ(1232) resonance at E₃ ≈ 4.21 GeV, further nucleon resonances, then a rising deep-inelastic continuum toward lower E₃

The core equations

Everything starts with the four-momentum $q=p_1-p_3$ that the electron transfers to the proton (subscripts: 1 incident electron, 2 target proton, 3 scattered electron, 4 hadronic final state). Its spacelike square defines the virtuality, and in the lab frame — neglecting the electron mass at these energies — it reduces to a formula in the two things a spectrometer measures, the scattered energy and angle.

Q^2 \equiv -q^2 = -(p_1-p_3)^2 \approx 4E_1E_3\sin^2\!\tfrac{\theta}{2} \qquad (1)

where: $Q^2$ — photon virtuality (positive for spacelike exchange), $\mathrm{GeV}^2$ ; $E_1,E_3$ — incident and scattered electron energies; $\theta$ — electron scattering angle. Larger $Q^2$ means a shorter-wavelength probe resolving finer structure. Reading: at fixed energies, forward scattering ( $\theta\to0$ ) gives $Q^2\to0$ , a soft glancing blow. The proton side is captured by two invariants: the energy the proton absorbs, $\nu$ , and the invariant mass $W$ of whatever recoils. These follow from $p_4=p_2+q$ and $p_4^2=W^2$ .

\nu \equiv \frac{p_2\cdot q}{m_p}=E_1-E_3, \qquad W^2=(p_2+q)^2=m_p^2+2m_p\nu-Q^2 \qquad (2)

where: $\nu$ — energy transfer (lab-frame electron energy loss); $m_p=0.938\ \mathrm{GeV}$ — proton mass; $W$ — invariant mass of the hadronic final state. Reading: elastic scattering is the single condition $W=m_p$ , which forces $2m_p\nu=Q^2$ ; any $W>m_p$ is inelastic. Two dimensionless ratios organize the two-variable region. Bjorken $x$ measures elasticity, and the inelasticity $y$ measures the fractional energy loss; with the squared centre-of-mass energy $s$ they satisfy one tidy identity.

x=\frac{Q^2}{2\,p_2\cdot q}=\frac{Q^2}{2m_p\nu}, \qquad y=\frac{p_2\cdot q}{p_2\cdot p_1}=\frac{\nu}{E_1}, \qquad Q^2=sxy \qquad (3)

where: $x$ — Bjorken scaling variable, $0\le x\le1$ ; $y$ — inelasticity, $0\le y\le1$ ; $s=(p_1+p_2)^2\approx m_p^2+2m_pE_1$ . Reading: $x=1$ is elastic and $x\to0$ is the far-inelastic (small- $x$ ) limit; $y$ and $\nu$ are proportional, so they are not an independent pair. For the elastic line, the constraint $x=1$ collapses $E_3$ onto a single function of the angle — the recoil relation. This is the algebraic statement that elastic scattering is one-variable.

E_3^{\text{elastic}}=\frac{E_1}{1+\dfrac{2E_1}{m_p}\sin^2\!\tfrac{\theta}{2}} \qquad (4)

where: all symbols as above. Reading: once $\theta$ is chosen, $E_3$ , $Q^2$ , $\nu$ and the recoil are all determined — there is no second knob. The heavier the target, the smaller the recoil; as $m_p\to\infty$ this returns the no-recoil result $E_3=E_1$ . The dynamics of the intact proton enter through two elastic form factors, packaged in the Rosenbluth cross section. Two functions of the single variable $Q^2$ suffice, because the final state has no extra freedom.

\frac{d\sigma}{d\Omega}=\left(\frac{d\sigma}{d\Omega}\right)_{\!\text{Mott}}\!\left[\frac{G_E^2+\tau G_M^2}{1+\tau}+2\tau G_M^2\tan^2\!\tfrac{\theta}{2}\right],\quad \tau=\frac{Q^2}{4m_p^2} \qquad (5)

where: $G_E(Q^2),G_M(Q^2)$ — electric and magnetic Sachs form factors (Fourier images of the charge and magnetization distributions); the Mott factor $\big(\tfrac{d\sigma}{d\Omega}\big)_{\text{Mott}}=\tfrac{\alpha^2\cos^2(\theta/2)}{4E_1^2\sin^4(\theta/2)}\tfrac{E_3}{E_1}$ is pointlike scattering with recoil. Reading: measuring the angular dependence at fixed $Q^2$ separates $G_E$ from $G_M$ ; both fall off roughly as the dipole $(1+Q^2/0.71\,\mathrm{GeV}^2)^{-2}$ , which is why the elastic rate dies at high $Q^2$ and inelastic processes take over. When the proton is allowed to break up, the same two roles are played by two structure functions — but now each depends on two variables, because $W$ is free. This is the master inelastic cross section.

\frac{d^2\sigma}{d\Omega\,dE_3}=\frac{\alpha^2}{4E_1^2\sin^4(\theta/2)}\Big[W_2(x,Q^2)\cos^2\!\tfrac{\theta}{2}+2W_1(x,Q^2)\sin^2\!\tfrac{\theta}{2}\Big] \qquad (6)

where: $W_1,W_2$ — inelastic structure functions encoding the proton's response; $\alpha\approx1/137$ . Reading: the one extra kinematic degree of freedom ( $W$ ) is exactly why we now need two functions of two variables instead of two form factors of one. In the deep-inelastic limit these scale, $2m_pW_1\to F_1(x)$ and $\nu W_2\to F_2(x)$ , obeying the Callan–Gross relation $F_2=2xF_1$ — the fingerprint of spin- $\tfrac12$ pointlike quarks.

Derivation AFrom four-momenta to the master invariants

The goal is to build, from nothing but four-momentum conservation and the definition of the transferred momentum $q=p_1-p_3$ , the full set of invariants that describe any $e^-p$ event — the virtuality $Q^2$ , the energy transfer $\nu$ , the final-state mass $W$ , and Bjorken $x$ — and to prove the central claim that elastic scattering is precisely the line $x=1$ , equivalently $W=m_p$ . We work with Lorentz invariants throughout so the results hold in any frame, then specialize to the lab to read off measurable quantities.

Step 1. Define the exchanged four-momentum $q=p_1-p_3$ and its virtuality $Q^2=-q^2$ . For elastic and inelastic alike the electron line is unchanged, so $q$ is the momentum handed to the proton vertex; the exchange is spacelike, so $q^2<0$ , $Q^2>0$ . Why: fixing $q$ once, before splitting into cases, lets the same coordinates describe both the intact and the shattered proton.

Step 2. Apply conservation at the proton vertex: the hadronic final state carries $p_4=p_2+q$ , so its invariant mass squared is $W^2=p_4^2=(p_2+q)^2=p_2^2+2p_2\cdot q+q^2=m_p^2+2p_2\cdot q-Q^2$ . Defining $\nu=(p_2\cdot q)/m_p$ gives $W^2=m_p^2+2m_p\nu-Q^2$ . Why: this single equation contains the whole distinction — $W$ depends on the two invariants $\nu$ and $Q^2$ , so if $W$ is free, two variables are live.

Step 3. Impose the elastic condition. If the proton stays whole, $p_4^2=m_p^2$ , so $W=m_p$ and Step 2 collapses to $2m_p\nu=Q^2$ . Dividing, $x\equiv Q^2/2m_p\nu=1$ . Conversely $x=1$ forces $2m_p\nu=Q^2$ , hence $W=m_p$ . Why: this is the promised equivalence — the elastic line, the condition $W=m_p$ , and $x=1$ are three names for the same one-dimensional locus.

Step 4. Specialize to the lab, $p_2=(m_p,\mathbf 0)$ . Then $p_2\cdot q=m_p(E_1-E_3)$ , so $\nu=E_1-E_3$ . Neglecting $m_e$ , $Q^2=-(p_1-p_3)^2\approx2p_1\cdot p_3=2E_1E_3(1-\cos\theta)=4E_1E_3\sin^2(\theta/2)$ . Why: this converts the abstract invariants into the two numbers a fixed-target spectrometer reads — scattered energy and angle.

Step 5. Assemble the coordinates. With $s\approx m_p^2+2m_pE_1$ and $y=\nu/E_1$ , one has $Q^2=2m_p\nu\,x=(s-m_p^2)xy\approx sxy$ . So $(x,y)$ , $(x,Q^2)$ or $(\nu,Q^2)$ each fix an inelastic event, while $(\nu,y)$ do not, being proportional. Why: this fixes the count at two and identifies which pairs are legitimate — the operational meaning of "inelastic scattering is two-variable". *Verified for $E_1=4.879$ GeV, $\theta=10°$ : the elastic peak sits at $E_3=4.522$ GeV, $Q^2=0.670\ \mathrm{GeV}^2$ , $x=1.000$ ; scattered energy $E_3=3.50$ GeV gives $\nu=1.379$ GeV, $Q^2=0.519\ \mathrm{GeV}^2$ , $W=1.717$ GeV, $x=0.201$ .*

Go deeper · Derivation B · 1 minThe elastic recoil relation as a one-variable constraint

Derivation BThe elastic recoil relation as a one-variable constraint

The aim is to derive equation (4), the elastic recoil formula $E_3(\theta)$ , directly from the invariants, and to make explicit why it proves that elastic scattering has a single free parameter. The strategy is to feed the lab-frame expressions for $\nu$ and $Q^2$ into the elastic condition $2m_p\nu=Q^2$ and solve for the one unknown, the scattered energy.

Step 1. Start from the elastic condition of the previous derivation, $2m_p\nu=Q^2$ , valid iff $W=m_p$ . Why: this is the entire dynamical input; the rest is algebra, which is the point — no second observable is needed.

Step 2. Substitute the lab expressions $\nu=E_1-E_3$ and $Q^2=4E_1E_3\sin^2(\theta/2)$ to get $2m_p(E_1-E_3)=4E_1E_3\sin^2(\theta/2)$ , i.e. $m_p(E_1-E_3)=2E_1E_3\sin^2(\theta/2)$ . Why: this replaces two invariants with the two measured quantities and the fixed target mass, leaving one equation in $E_3$ and $\theta$ .

Step 3. Collect $E_3$ : $m_pE_1=E_3\!\left(m_p+2E_1\sin^2(\theta/2)\right)$ , so $E_3=\dfrac{E_1}{1+(2E_1/m_p)\sin^2(\theta/2)}$ . Why: $E_3$ is now an explicit function of $\theta$ alone — choose the angle and the scattered energy, recoil and $Q^2$ are all forced. That functional dependence is the mathematical face of "one variable".

Step 4. Check the limits. As $\theta\to0$ , $E_3\to E_1$ (no energy lost in a glancing hit); as $m_p\to\infty$ , $E_3\to E_1$ (an immovable target absorbs momentum but no energy). At $E_1=4.879$ GeV, $\theta=10°$ , $E_3=4.522$ GeV — the elastic peak. Why: the formula reproduces both physical extremes and the measured peak, confirming the collapse to a single free parameter. *Verified: $E_3^{\text{elastic}}=4.5217$ GeV and $Q^2=0.6703\ \mathrm{GeV}^2$ at $E_1=4.879$ GeV, $\theta=10°$ , giving $x=1.000$ to four figures.*

Go deeper · Derivation C · 2 minWhy the inelastic proton needs two structure functions

Derivation CWhy the inelastic proton needs two structure functions

The goal is to explain, from Lorentz invariance and current conservation alone, why the inelastic cross section (6) contains exactly two independent functions $W_1,W_2$ — no more, no fewer — and how this is the direct counterpart of the extra kinematic variable. The argument builds the hadronic response as the most general tensor allowed by symmetry and counts the surviving pieces, without needing the proton's internal dynamics.

Step 1. Factor the cross section into a leptonic part and a hadronic part. The electron vertex is known QED, giving a leptonic tensor $L^{\mu\nu}$ ; all ignorance about the proton sits in a hadronic tensor $W^{\mu\nu}$ that depends on $p_2$ and $q$ . The cross section is $\propto L_{\mu\nu}W^{\mu\nu}$ . Why: isolating the unknown into $W^{\mu\nu}$ turns "how does the proton respond" into "how many independent tensors can it be built from".

Step 2. Write the most general symmetric $W^{\mu\nu}$ from the available objects $g^{\mu\nu}$ , $p_2^\mu$ and $q^\mu$ . A priori that allows terms in $g^{\mu\nu}$ , $p_2^\mu p_2^\nu$ , $q^\mu q^\nu$ , and $p_2^\mu q^\nu+q^\mu p_2^\nu$ — four scalar coefficients, each a function of the two invariants $\nu$ and $Q^2$ . Why: the number of independent coefficients is the number of structure functions; the whole question is how symmetry trims this list.

Step 3. Impose current conservation, $q_\mu W^{\mu\nu}=0$ . This gives two constraints that fix the $q^\mu q^\nu$ and cross terms in favour of the other two, eliminating two of the four coefficients. What remains organizes into exactly two independent functions, conventionally $W_1$ and $W_2$ . Why: gauge invariance of the exchanged photon reduces four possible responses to two — the same principle that leaves the intact proton with two form factors.

Step 4. Contract with $L_{\mu\nu}$ and read off the angular structure: the surviving tensors pair with $\sin^2(\theta/2)$ and $\cos^2(\theta/2)$ , reproducing equation (6). Because $W_1,W_2$ each depend on both $x$ and $Q^2$ , the response is a surface, not a line. Why: this is the payoff — two functions of two variables is the inelastic proton's data content, mirroring form factors (two functions of one variable) for the elastic proton. *Consistency check: in the elastic limit the two structure functions collapse to delta functions in $W$ centred on $m_p$ , with residues built from $G_E,G_M$ — recovering Rosenbluth, equation (5), as the $x=1$ boundary of (6).*

Limiting cases

Case	Result	Meaning
Elastic	$W=m_p$ , $x=1$ , $2m_p\nu=Q^2$	proton intact; one variable ( $\theta$ ) fixes the event; described by $G_E,G_M$
Resonance region	$W$ pinned near $1.232,\,1.52,\,1.68\ \mathrm{GeV}\ldots$	proton briefly excited to $\Delta,N^*$ states; bumps in the $E_3$ spectrum
Deep inelastic (DIS)	large $Q^2$ and $\nu$ at fixed $x$ ; $F_i(x)$ scale	probe resolves pointlike quarks; $W_1,W_2$ lose their $Q^2$ dependence
Small $x$ ( $x\to0$ )	$F_2$ rises steeply; sea quarks and gluons dominate	high-energy, high- $W$ limit; the regime HERA opened
Low $Q^2$ ( $Q^2\to0$ )	cross section → real-photon (photoproduction) behaviour	long-wavelength probe sees the proton as a whole; form factors $\to1$
$x>1$	kinematically forbidden for a free proton	a single nucleon cannot recoil elastically beyond $x=1$ ; only nuclei allow it

Common trapInelastic scattering can still be pinned down by the scattering angle alone

It cannot, and this is the whole point of the article. In elastic scattering the constraint $W=m_p$ ties $E_3$ to $\theta$ through equation (4), so one measured angle determines everything. The instant the proton is allowed to break up, $W$ becomes a free variable and that link is cut: at the same angle $\theta$ the scattered electron can emerge with a whole range of energies $E_3$ , each corresponding to a different final-state mass $W$ . You now need two independent invariants — for instance $(Q^2,x)$ or $(E_3,\theta)$ — to say what happened. A tempting but wrong repair is to pick $\nu$ and $y$ as the second pair; they fail because $y=\nu/E_1$ is just $\nu$ rescaled. And note the reverse trap: because $x$ looks like a momentum fraction, people assume it must be strictly below one — but $x=1$ is perfectly physical; it is the elastic point, not a forbidden edge.

Worked examples

1 · Locating the elastic peak (confidence builder)

An electron of energy $E_1=4.879$ GeV scatters elastically off a hydrogen target and is detected at $\theta=10°$ . Find the scattered energy $E_3$ , the virtuality $Q^2$ , and confirm $x=1$ , $W=m_p$ .

Step 1. Compute the angular factor $\sin^2(\theta/2)=\sin^2(5°)=0.007596$ and the recoil coefficient $2E_1/m_p=10.40$ . Then $E_3=E_1/[1+10.40\times0.007596]=4.879/1.0790=\mathbf{4.522\ GeV}$ .

Step 2. Virtuality from equation (1): $Q^2=4E_1E_3\sin^2(\theta/2)=4(4.879)(4.522)(0.007596)=\mathbf{0.670\ GeV^2}$ . Energy transfer $\nu=E_1-E_3=0.357$ GeV.

Step 3. Check elasticity: $x=Q^2/(2m_p\nu)=0.670/[2(0.938)(0.357)]=0.670/0.670=\mathbf{1.000}$ , and $W^2=m_p^2+2m_p\nu-Q^2=0.880+0.670-0.670=0.880\ \mathrm{GeV}^2$ , so $W=0.938=m_p$ . The elastic peak sits at E₃ = 4.52 GeV, Q² = 0.67 GeV², x = 1, W = m_p — one angle fixed everything.

2 · An inelastic event at the same angle

At the same $E_1$ and $\theta$ , a scattered electron is detected with the lower energy $E_3=3.50$ GeV. Find $\nu$ , $Q^2$ , $W$ , $x$ and $y$ , and say which regime this is.

Step 1. Energy transfer and virtuality: $\nu=E_1-E_3=4.879-3.50=\mathbf{1.379\ GeV}$ ; $Q^2=4E_1E_3\sin^2(\theta/2)=4(4.879)(3.50)(0.007596)=\mathbf{0.519\ GeV^2}$ .

Step 2. Final-state mass: $W^2=m_p^2+2m_p\nu-Q^2=0.880+2.587-0.519=2.948\ \mathrm{GeV}^2$ , so $W=\mathbf{1.72\ GeV}$ — well above $m_p$ , so the proton did not survive.

Step 3. Dimensionless variables: $x=Q^2/(2m_p\nu)=0.519/2.587=\mathbf{0.201}$ ; $y=\nu/E_1=\mathbf{0.283}$ . Cross-check the identity $Q^2=sxy$ with $s\approx m_p^2+2m_pE_1=0.880+9.153=10.03\ \mathrm{GeV}^2$ : $sxy=10.03\times0.201\times0.283=0.571\ \mathrm{GeV}^2$ , agreeing with the measured $0.519\ \mathrm{GeV}^2$ to the accuracy of the massless- $s$ approximation.

Step 4. Interpret: $W=1.72$ GeV sits above the prominent resonances, in the transition toward the deep-inelastic continuum, and $x=0.20$ means the struck constituent carries about a fifth of the proton's momentum. Same angle as Example 1, but a different E₃ picks out a genuinely inelastic event — W = 1.72 GeV, x = 0.20 — which no single-variable elastic description could reach.

3 · The elastic-to-inelastic map along one spectrum

At fixed $E_1=4.879$ GeV, $\theta=10°$ , the relation $W^2=10.03-2.024\,E_3$ (in GeV²) converts scattered energy to final-state mass. Where do the elastic peak and the $\Delta(1232)$ resonance sit?

Step 1. Elastic ( $W=m_p=0.938$ , $W^2=0.880$ ): $E_3=(10.03-0.880)/2.024=\mathbf{4.52\ GeV}$ — the sharp spike at the high-energy end, matching Example 1.

Step 2. $\Delta(1232)$ ( $W^2=1.518$ ): $E_3=(10.03-1.518)/2.024=\mathbf{4.21\ GeV}$ , with $Q^2=0.624\ \mathrm{GeV}^2$ and $x=0.49$ . The elastic peak (E₃ = 4.52) and the Δ bump (E₃ = 4.21) are separated by just 0.31 GeV of scattered energy — the two rightmost features in the figure, with resonances and DIS continuum filling in below.

Physics of the portrait

The distinction between elastic and inelastic scattering has a single experimental father: the man who first used electrons to see that the proton is not a point, and whose elastic form-factor measurements set the stage for the inelastic experiments that found the quarks.

Origins · 1 minRobert Hofstadter and how this was discovered

Robert Hofstadter

1915–1990 · elastic electron scattering and the structure of the nucleon

Robert Hofstadter turned the electron into a ruler. Beginning around 1951 at Stanford's linear accelerators, he scattered beams of a few hundred MeV electrons off nuclei and, by 1955, off the proton itself, and did the decisive thing: he measured how the cross section fell below the pointlike (Mott) prediction as the momentum transfer grew. A point charge would follow Mott exactly; the proton did not. The deficit was the form factor $G(Q^2)$ — the Fourier transform of the charge distribution — and its fall-off meant the proton had a finite size, a root-mean-square charge radius of about $0.8\ \mathrm{fm}$ . This was elastic scattering used as a microscope, and it earned Hofstadter the 1961 Nobel Prize for his studies of electron scattering and the structure of the nucleons. His work also drew the line this article is about: because the elastic form factor falls steeply with $Q^2$ (the dipole $(1+Q^2/0.71\,\mathrm{GeV}^2)^{-2}$ ), pushing to higher energies made elastic scattering vanishingly rare and inelastic channels dominant — the proton could no longer absorb a hard photon while staying whole. A decade later the same laboratory, at much higher energy, ran the deep-inelastic experiments that saw electrons bounce hard off pointlike constituents inside the proton — the partons Hofstadter's smooth form factor had averaged over. The elastic microscope revealed the size; the inelastic one revealed the parts.

History

1951

— Hofstadter begins high-energy electron scattering at Stanford

1955

— Hofstadter and McAllister measure the proton's finite charge radius

1961

— Hofstadter shares the Nobel Prize for nucleon structure

1968

— the SLAC-MIT experiments find deep-inelastic cross sections far above the elastic fall-off

1969

— Bjorken predicts scaling and Feynman interprets it with the parton model

1990

— Friedman, Kendall and Taylor share the Nobel Prize for the discovery of quarks in DIS

Going further

Texts. Thomson, Modern Particle Physics, §8.1 (electron–proton inelastic scattering) and §8.1.1–8.1.2 (kinematic variables, low- $Q^2$ limit) · Halzen & Martin, Quarks and Leptons, Ch. 8 (inelastic scattering and structure functions) · Griffiths, Introduction to Elementary Particles, 2nd ed., §8.3 · Peskin & Schroeder, An Introduction to Quantum Field Theory, §17.3 (deep-inelastic kinematics) · Schwartz, Quantum Field Theory and the Standard Model, Ch. 32.

Provenance. The invariant kinematics build on aq-pp-relativistic-ep-scattering and aq-pp-mandelstam-variables; the elastic Rosenbluth cross section (5) is owned by aq-pp-form-factors and recalled here by name. The scaling of $W_1,W_2$ into $F_1,F_2$ , the Callan–Gross relation, and the parton interpretation are developed downstream in aq-pp-deep-inelastic-scattering, aq-pp-bjorken-scaling and aq-pp-quark-parton-model — forward references, not re-derived here. A verifier should confirm, for $E_1=4.879$ GeV and $\theta=10°$ : the elastic peak $E_3=4.522$ GeV with $Q^2=0.670\ \mathrm{GeV}^2$ , $x=1$ ; the inelastic point $E_3=3.50$ GeV giving $W=1.72$ GeV, $x=0.201$ , $y=0.283$ ; and $x=1\Leftrightarrow W=m_p$ from $W^2=m_p^2+2m_p\nu-Q^2$ . HERA figures ( $\sqrt s\approx318$ GeV) and $m_p=0.938$ GeV are standard.

Prerequisites. Relativistic Electron–Proton Scattering · Mandelstam Variables · Form Factors. Leads to. Deep Inelastic Scattering · Bjorken Scaling · The Quark–Parton Model.

Ask the tutor: Ask me to walk a single scattered-electron spectrum from the elastic spike through the $\Delta$ resonance into the deep-inelastic continuum, computing $W$ , $x$ and $Q^2$ at each step.

If the electron just bounces off, what exactly is "inelastic" about the collision?

The inelasticity is on the proton side, not the electron side. The electron–photon vertex is ordinary QED in both cases; what changes is whether the proton absorbs the photon while staying a proton (elastic) or gets excited and breaks apart into a hadronic system $X$ (inelastic). "Inelastic" here means kinetic energy of relative motion is converted into internal excitation and new-particle creation — the same sense as an inelastic collision in mechanics. The electron loses energy $\nu$ to the proton; if that is enough to raise the target's invariant mass above $m_p$ , the event is inelastic.

Why exactly two variables — why not three or one?

One variable fails because, once $W$ is free, the scattered energy and angle become independent, so you need at least two. Three would be redundant: at fixed beam energy the scattered electron's energy and angle already exhaust the independent information — the hadronic system $X$ recoils to balance momentum, adding no new measured freedom. Formally, from the invariants $Q^2$ , $\nu$ , $x$ , $y$ and $s$ , only two are independent at fixed $s$ (with $y\propto\nu$ not counting separately). So the answer is exactly two, and any legitimate pair — $(Q^2,x)$ , $(Q^2,\nu)$ , $(x,y)$ — will do.

What is the difference between "inelastic" and "deep inelastic" scattering?

Inelastic simply means $W>m_p$ — the proton broke up, whether into a single resonance like the $\Delta(1232)$ or a many-hadron continuum. Deep inelastic scattering (DIS) is the high-energy corner: both $Q^2$ and $\nu$ are large (typically $Q^2\gg1\ \mathrm{GeV}^2$ ) while their ratio, fixed by $x$ , stays finite. There the virtual photon's wavelength is short enough to resolve individual quarks, and the structure functions stop depending on $Q^2$ (Bjorken scaling). So all DIS is inelastic, but not all inelastic scattering is "deep" — the resonance bumps in the figure are inelastic but not DIS.

Can Bjorken x ever be larger than 1?

Not for a free nucleon. The bound $0\le x\le1$ follows from $W\ge m_p$ : since $W^2=m_p^2+2m_p\nu-Q^2\ge m_p^2$ , we get $2m_p\nu\ge Q^2$ , i.e. $x=Q^2/2m_p\nu\le1$ , with equality exactly at the elastic point. In the parton model $x$ is the fraction of the proton's momentum carried by the struck quark, and a fraction cannot exceed one. The one exception is scattering off a nucleus, where a single nucleon can carry more than its share through Fermi motion, allowing $x>1$ — but that is a nuclear effect, not a property of one proton.

How is Q² related to W, and can I trade one for the other?

They are different invariants and generally cannot be traded, being two independent coordinates of the inelastic region. $Q^2=-q^2$ is set by how hard the electron scattered ( $4E_1E_3\sin^2(\theta/2)$ ), while $W$ is the mass of the debris, tied to $Q^2$ and the energy transfer through $W^2=m_p^2+2m_p\nu-Q^2$ . Only on the elastic line does one fix the other: there $W=m_p$ forces $Q^2=2m_p\nu$ . Off that line the same $Q^2$ is compatible with a whole range of $W$ (by varying $\nu$ ), which is exactly why two variables are needed.

Why does the elastic cross section fall so steeply, letting inelastic scattering take over?

Because the elastic form factors decrease with $Q^2$ roughly as the dipole $G(Q^2)\approx(1+Q^2/0.71\,\mathrm{GeV}^2)^{-2}$ , the elastic rate — going as $G^2$ — falls like $Q^{-8}$ at large $Q^2$ . Physically, keeping an extended proton intact while transferring large momentum requires every part of it to recoil coherently, and that coherence becomes exponentially unlikely as the probe wavelength shrinks below the proton size. Inelastic channels have no such penalty: a hard photon can strike one constituent and let the rest fragment. So above a few $\mathrm{GeV}^2$ the inelastic cross section dwarfs the elastic one — the crossover that made DIS the tool of choice.