Rutherford & Mott scattering

Fire an electron at a proton and what comes back depends on one comparison: the wavelength of the exchanged virtual photon against the size of the proton. When that wavelength is long — a slow electron, a soft photon — the proton is unresolved, a structureless point of charge, and the whole collision is governed by a single static $1/r$ Coulomb field. This is the world of Rutherford and Mott scattering, the two low-energy limits of elastic $e^-p$ scattering in which the massive proton barely recoils and acts as a fixed source. Rutherford's 1911 formula, born to explain why alpha particles bounced backward off gold foil, is the non-relativistic limit; Mott's 1929 refinement is what happens when the electron moves near the speed of light and its spin can no longer be ignored. The modern route to both is not the Schrödinger equation but the QED helicity amplitude. Model the proton as a point Dirac charge, write the tree-level matrix element with its single photon propagator $1/q^2$ , average over spins, and out drops one master formula. Take the electron velocity to zero and you recover Rutherford's $\sin^{-4}(\theta/2)$ law exactly; let the velocity approach $c$ and an extra $\cos^2(\theta/2)$ factor appears — the fingerprint of the electron's spin and the chiral structure of the QED vertex. The steep forward peak that both share is pure propagator physics, the same for any $1/r$ potential; the Mott factor is the first place spin ever shows up in a scattering cross section. Everything downstream — form factors, the proton radius, deep inelastic scattering — begins by asking how this point-charge picture fails.

Key takeaways

Both treat the proton as a fixed $1/r$ Coulomb source whose kinetic recoil is negligible; they are the low-energy limits of elastic $e^-p$ scattering.
Rutherford (non-relativistic electron, $\beta_e\ll1$ ): $\dfrac{d\sigma}{d\Omega}=\dfrac{\alpha^2}{16E_K^2\sin^4(\theta/2)}$ — pure electric charge–charge scattering.
Mott (relativistic electron): the same shape times $(1-\beta^2\sin^2(\theta/2))$ , which becomes $\cos^2(\theta/2)$ as $\beta\to1$ — the signature of electron spin.
The universal $\sin^{-4}(\theta/2)$ falloff comes solely from the photon propagator $1/q^2$ with $q^2=-4p^2\sin^2(\theta/2)$ , not from any charge or spin distribution.
One helicity-amplitude calculation with a point Dirac proton yields both classics; $\cos^2(\theta/2)$ is the overlap of the electron's rotated helicity states.
Key idea: treating the proton as a fixed point Coulomb source, a single helicity-amplitude calculation gives both classic results — the $\sin^{-4}(\theta/2)$ shape is pure propagator physics, and the relativistic $\cos^2(\theta/2)$ factor that turns Rutherford into Mott is the fingerprint of the electron's spin and the chiral QED vertex.

Where this lives today

The point-charge cross sections are now mostly a baseline — the thing you divide by. At electron accelerators from the old SLAC linac to Jefferson Lab's CEBAF, elastic $e^-p$ scattering is measured and compared to the Mott formula (6); the ratio is the proton's electromagnetic form factor, and its slope at small $q^2$ gives the proton charge radius — the quantity at the heart of the "proton radius puzzle," where muonic-hydrogen spectroscopy and scattering long disagreed. The Mott factor itself has a second, very practical life: Mott polarimetry. Because the full spin-orbit treatment of electron scattering off a high- $Z$ nucleus produces a left-right asymmetry proportional to the electron's transverse polarization, firing a polarized electron beam at a gold foil and counting the left-right imbalance is a standard way to measure beam polarization at facilities worldwide. And in materials science, Rutherford backscattering spectrometry turns the same $\sin^{-4}(\theta/2)$ , $Z^2$ law into a tool: MeV helium ions backscattered from a sample reveal the mass and depth of every atomic species near its surface, non-destructively. A century on, the law Rutherford wrote to explain a bouncing alpha particle still calibrates beams and profiles semiconductors.

The physical picture

An electron approaches the proton, feels its Coulomb field, and leaves along a new direction making an angle $\theta$ with the incoming beam. Because the proton is roughly two thousand times heavier than the electron, it soaks up momentum while taking almost no kinetic energy — the collision is elastic and the electron's speed is unchanged, only its direction turns. In the language of Feynman diagrams a single virtual photon carries the momentum transfer $q$ from electron to proton, and the whole angular dependence of the cross section is fixed by how large $q$ is. The first figure makes this exact: the geometry of the vectors alone forces $q^2=-4p^2\sin^2(\theta/2)$ , and the propagator factor $1/q^2$ does the rest.

Figure

A point electron of momentum p scatters off the proton's static 1/r Coulomb field, deflecting by angle θ. Because the collision is elastic and the heavy proton barely recoils, the momentum magnitudes match, p₁ = p₃ = p, and the four-momentum transfer is q² = −4p² sin²(θ/2). The photon propagator 1/q² alone produces the sin⁻⁴(θ/2) angular shape shared by Rutherford and Mott.

The forward peak is not the whole story. As the electron becomes relativistic, its spin enters, and the two cross sections part ways at large angles. Rutherford's law knows nothing of spin and keeps a finite, gently varying cross section all the way to $180°$ ; Mott's relativistic result multiplies it by $\cos^2(\theta/2)$ , which suppresses backward scattering and drives it exactly to zero at $\theta=180°$ . That vanishing backscatter is not an accident of algebra — it is helicity conservation, the statement that a fast electron cannot flip the projection of its spin onto its own line of flight. The second figure plots both cross sections on a logarithmic scale so the shared forward divergence and the diverging back-angle behaviour are visible at once.

Figure

Angular distribution on a log scale in relative units. The Rutherford shape 1/sin⁴(θ/2) diverges forward and drops nearly five decades by 180°. The relativistic Mott result multiplies it by cos²(θ/2), peeling away at back angles and vanishing at θ = 180° — a point electron cannot backscatter because helicity is conserved. At θ = 90° the Mott suppression is cos²45° = ½; at 150° it is 0.067.

The core equations

The starting point is the tree-level QED amplitude for $e^-p\to e^-p$ : two charged Dirac currents joined by one photon. The proton enters as a point charge $Q_p=+1$ , and the photon propagator supplies the crucial $1/q^2$ .

\mathcal{M}_{fi}=\frac{Q_p\,e^2}{q^2}\big[\bar u(p_3)\gamma^\mu u(p_1)\big]\,g_{\mu\nu}\,\big[\bar u(p_4)\gamma^\nu u(p_2)\big] \qquad (1)

where: $u(p_1),u(p_3)$ — incoming/outgoing electron spinors; $u(p_2),u(p_4)$ — incoming/outgoing proton spinors; $\gamma^\mu$ — Dirac matrices at each vertex; $e^2=4\pi\alpha$ — the QED coupling; $q=p_1-p_3$ — the four-momentum transferred through the photon. The single $1/q^2$ is the whole reason the cross section peaks forward: small-angle scattering means small $q$ , hence a large amplitude. For elastic scattering off a fixed target the transferred four-momentum is purely spacelike, and pure geometry converts it into the observable scattering angle.

q^2=(p_1-p_3)^2=-4p^2\sin^2\!\frac{\theta}{2} \qquad (2)

where: $p=|\vec p_1|=|\vec p_3|$ — the (unchanged) electron momentum magnitude; $\theta$ — laboratory scattering angle. Because $q^2<0$ the photon is virtual (spacelike), and the magnitude $Q^2\equiv-q^2=4p^2\sin^2(\theta/2)$ is what an experimenter dials by choosing beam energy and angle. Squaring the amplitude, summing over final spins and averaging over initial ones, and using $\kappa=p/(E+m_e)=\beta_e\gamma_e/(\gamma_e+1)$ to organize the electron spinors, gives a single master expression that contains both classics.

\langle|\mathcal{M}_{fi}|^2\rangle=\frac{m_p^2\,m_e^2\,e^4}{p^4\sin^4(\theta/2)}\left(1+\beta_e^2\gamma_e^2\cos^2\frac{\theta}{2}\right) \qquad (3)

where: $m_p,m_e$ — proton and electron masses (the proton normalized as a static source); $\beta_e,\gamma_e$ — electron velocity and Lorentz factor; the bracket is the Mott factor in disguise. Reading it: the $1/\sin^4(\theta/2)$ is universal propagator physics; the parenthesis carries the spin content, collapsing to $1$ when $\beta_e\gamma_e\to0$ (Rutherford) and dominating when $\beta_e\gamma_e\gg1$ (Mott). Folding in the flux and laboratory phase space and taking the non-relativistic limit $\beta_e\ll1$ gives Rutherford's law, written most cleanly in terms of the electron kinetic energy $E_K$ .

\frac{d\sigma}{d\Omega}\bigg|_{\text{Ruth}}=\frac{\alpha^2}{16\,E_K^2\,\sin^4(\theta/2)} \qquad (4)

where: $\alpha=e^2/4\pi\approx1/137$ — the fine-structure constant; $E_K=\tfrac12 m_e v^2=p^2/2m_e$ — the electron kinetic energy; $\theta$ — scattering angle. The cross section scales as $E_K^{-2}$ : slow electrons scatter enormously, fast ones barely at all, and there is no dependence on the electron's spin. Keeping the electron relativistic reinstates the spin factor. The compact Mott form multiplies the relativistic Rutherford prefactor by $(1-\beta^2\sin^2(\theta/2))$ .

\frac{d\sigma}{d\Omega}\bigg|_{\text{Mott}}=\frac{\alpha^2}{4p^2\beta^2\sin^4(\theta/2)}\left(1-\beta^2\sin^2\frac{\theta}{2}\right) \qquad (5)

where: $p,\beta$ — relativistic electron momentum and velocity; the factor $(1-\beta^2\sin^2(\theta/2))$ is the Mott factor. It equals $1$ at $\theta=0$ and shrinks toward the back, most strongly for fast electrons; as $\beta\to0$ it returns to $1$ and equation (5) reduces to (4). In the ultrarelativistic regime $\beta\to1$ , $E\approx p$ , the Mott factor becomes purely $\cos^2(\theta/2)$ — the form usually quoted for high-energy electron scattering.

\frac{d\sigma}{d\Omega}\bigg|_{\text{Mott}}^{\beta\to1}=\frac{\alpha^2\cos^2(\theta/2)}{4E^2\sin^4(\theta/2)} \qquad (6)

where: $E$ — electron energy ( $\approx p$ ); the $\cos^2(\theta/2)$ suppresses backward scattering and forbids it outright at $\theta=180°$ . This is the point-charge baseline against which every real proton's finite size is measured: divide the measured cross section by equation (6) and what remains is the form factor.

Derivation AThe momentum transfer and the sin⁻⁴ falloff

The single most important structural fact about both cross sections is that their steep forward peak is set entirely by the photon propagator, not by anything about spin or the shape of the charge. The aim here is to compute the four-momentum transfer $q$ for elastic scattering off a fixed target, show cleanly that $q^2=-4p^2\sin^2(\theta/2)$ , and trace how the propagator factor $1/q^2$ — squared in the rate — becomes the universal $\sin^{-4}(\theta/2)$ angular shape that Rutherford and Mott share.

Step 1. Write $q=p_1-p_3$ as a four-vector. With the heavy proton absorbing momentum but negligible kinetic energy, the electron energy is unchanged, $E_1=E_3$ , so the time component of $q$ vanishes and $q^2=(E_1-E_3)^2-|\vec p_1-\vec p_3|^2=-|\vec p_1-\vec p_3|^2$ . Why: no energy is transferred to a fixed source, so the exchanged photon is purely spacelike — the entire transfer is three-momentum.

Step 2. Since the collision is elastic and recoil is neglected, $|\vec p_1|=|\vec p_3|=p$ , and $|\vec p_1-\vec p_3|^2=p^2+p^2-2p^2\cos\theta=2p^2(1-\cos\theta)$ . Using the half-angle identity $1-\cos\theta=2\sin^2(\theta/2)$ gives $q^2=-4p^2\sin^2(\theta/2)$ . Why: the half-angle identity is what converts an abstract dot product of momenta into the one thing the detector actually measures, the angle $\theta$ .

Step 3. The amplitude (1) carries exactly one propagator $1/q^2$ , so the rate carries $|\mathcal{M}|^2\propto 1/q^4\propto 1/\sin^4(\theta/2)$ . Why: this is the sole origin of the forward divergence — it is a property of the long-range $1/r$ Coulomb potential and its Fourier transform $1/q^2$ , utterly independent of the projectile's spin or the target's internal structure. *Verified: $q^2=-4p^2\sin^2(\theta/2)$ from elastic no-recoil kinematics; $|\mathcal{M}|^2\propto\sin^{-4}(\theta/2)$ follows directly from the single photon propagator.*

Go deeper · Derivation B · 1 minRecovering the Rutherford formula in the non-relativistic limit

Derivation BRecovering the Rutherford formula in the non-relativistic limit

Rutherford's result predates quantum field theory by decades, yet it must fall out of the modern helicity-amplitude master formula as the low-velocity limit — otherwise the machinery would be suspect. The goal is to start from the spin-averaged $\langle|\mathcal{M}|^2\rangle$ with its Mott factor $(1+\beta^2\gamma^2\cos^2(\theta/2))$ , switch off the relativistic term by taking $\beta\ll1$ , fold in the flux and phase-space factors for a fixed Coulomb target, and land exactly on $d\sigma/d\Omega=\alpha^2/(16E_K^2\sin^4(\theta/2))$ with $E_K$ the electron kinetic energy.

Step 1. In the master factor, $\beta^2\gamma^2=\beta^2/(1-\beta^2)\to0$ as $\beta\to0$ , so $(1+\beta^2\gamma^2\cos^2(\theta/2))\to1$ and the angular content collapses to the bare $1/\sin^4(\theta/2)$ . Why: the spin/chirality term is order $\beta^2$ ; at low velocity only the electric charge–charge interaction survives, with no magnetic spin contribution.

Step 2. Assembling flux and phase space for a static source reduces the cross section to the relativistic-Rutherford prefactor $d\sigma/d\Omega=\alpha^2/(4p^2\beta^2\sin^4(\theta/2))$ , the Mott factor now set to $1$ . Why: the kinematic factors combine the coupling $e^4=(4\pi\alpha)^2$ with the propagator $1/q^4$ into this compact form, valid at any velocity once the spin factor is stripped off.

Step 3. Insert non-relativistic kinematics: $p=m_e v$ and $\beta=v$ , so $p^2\beta^2=m_e^2 v^4$ ; and $E_K=\tfrac12 m_e v^2$ gives $m_e^2 v^4=4E_K^2$ . Substituting, $d\sigma/d\Omega=\alpha^2/(16E_K^2\sin^4(\theta/2))$ . Why: rewriting momentum-times-velocity as kinetic energy is exactly the step that recovers the form Rutherford wrote and Geiger and Marsden tested on gold foil. *Verified: $\beta^2\gamma^2\to0$ kills the Mott factor; $p^2\beta^2=4E_K^2$ non-relativistically; the result matches equation (4) numerically (1 keV electron, 60°, gives 53.3 MeV⁻² = 2.1×10⁴ barn/sr).*

Go deeper · Derivation C · 1 minThe Mott factor as helicity conservation

Derivation CThe Mott factor as helicity conservation

The one piece of physics that Rutherford's classical calculation cannot contain is the electron's spin, and it enters through a single multiplicative factor. The goal here is to show that the two published faces of that Mott factor — the compact $(1-\beta^2\sin^2(\theta/2))$ and the ultrarelativistic $\cos^2(\theta/2)$ — are algebraically the same object, and then to explain the $\cos^2(\theta/2)$ as the overlap of the electron's initial and final helicity states, a direct consequence of chirality conservation at the QED vertex for a fast electron.

Step 1. Rewrite the master factor by splitting the sine: $1-\beta^2\sin^2(\theta/2)=(1-\beta^2)+\beta^2\cos^2(\theta/2)=\gamma^{-2}\big(1+\beta^2\gamma^2\cos^2(\theta/2)\big)$ . So the textbook Mott factor and the master-formula parenthesis differ only by a $\gamma^2$ absorbed into the normalization, and equation (5) follows from (3). Why: the identity $1-\beta^2\sin^2=\gamma^{-2}(1+\beta^2\gamma^2\cos^2)$ is the bridge between the two forms of Mott that appear in different textbooks — they are the same result.

Step 2. Take $\beta\to1$ , $p\to E$ : then $1-\beta^2\sin^2(\theta/2)\to1-\sin^2(\theta/2)=\cos^2(\theta/2)$ , giving equation (6). Why: at $\beta=1$ the spin term reaches full strength and reshapes the entire back hemisphere of the distribution, pinching the cross section to zero at $180°$ .

Step 3. Interpret the $\cos^2(\theta/2)$ . For a massless (chiral) electron, helicity is conserved at the QED vertex, so the electron keeps its spin projection along its momentum while that momentum swings through angle $\theta$ . The quantum amplitude to preserve helicity while rotating the quantization axis by $\theta$ is the Wigner $d$ -function $d^{1/2}_{1/2,1/2}(\theta)=\cos(\theta/2)$ ; squaring it gives $\cos^2(\theta/2)$ , which vanishes at $\theta=180°$ . Why: the factor is not dynamics but the geometry of spin — the overlap of one helicity eigenstate with the same helicity rotated by $\theta$ — so a relativistic electron simply cannot backscatter off a spinless charge. *Verified: the identity $1-\beta^2\sin^2(\theta/2)=\gamma^{-2}(1+\beta^2\gamma^2\cos^2(\theta/2))$ checks numerically ( $\beta=0.9$ , $\theta=90°$ : both sides $=0.595$ ); $\cos^2(\theta/2)\to0$ at $180°$ .*

Limiting cases

Case	Result	Meaning
Non-relativistic ( $\beta\ll1$ )	$\dfrac{d\sigma}{d\Omega}=\dfrac{\alpha^2}{16E_K^2\sin^4(\theta/2)}$	pure Rutherford; Mott factor $\to1$ ; only electric charge–charge scattering
Relativistic (general $\beta$ )	Rutherford $\times\,(1-\beta^2\sin^2(\theta/2))$	Mott; spin suppresses back angles, strongest for fast electrons
Ultrarelativistic ( $\beta\to1$ )	Rutherford $\times\cos^2(\theta/2)$	helicity conservation; scattering to $180°$ is forbidden
Forward ( $\theta\to0$ )	$\dfrac{d\sigma}{d\Omega}\to\infty$	Coulomb divergence from $1/q^2$ ; the total cross section diverges
Backward ( $\theta\to180°$ )	Rutherford finite; Mott $\to0$	fast electron cannot flip helicity onto its own momentum
Short wavelength ( $\lambda\lesssim r_p$ )	$\times\,\|F(q^2)\|^2$	point picture breaks; the proton's finite size appears as a form factor

Common trapThe sin⁻⁴(θ/2) shape reveals the charge or spin distribution

It reveals neither. The steep forward peak comes solely from the photon propagator $1/q^2$ with $q^2=-4p^2\sin^2(\theta/2)$ — it is the Fourier transform of the long-range $1/r$ Coulomb potential and is identical for any point charge, spin or no spin. Spin enters only as the additional $\cos^2(\theta/2)$ Mott factor, and the target's finite size enters only later, as a form factor $|F(q^2)|^2$ multiplying the whole thing. A common corollary mistake is to read the large forward cross section as "the electron mostly hits head-on"; it means the opposite — most electrons barely deflect, grazing the long tail of the Coulomb field. The forward divergence is a statement about the range of the force, not about how the charge is packed.

Worked examples

1 · A slow electron off a proton (Rutherford, fully worked)

A 1 keV electron scatters elastically from a proton through $\theta=60°$ . Treating the proton as a fixed point charge, find $d\sigma/d\Omega$ .

Step 1. Check the regime. At $E_K=1$ keV the electron velocity follows from $\beta^2\approx2E_K/m_e=2(1)/511=3.9\times10^{-3}$ , so $\beta\approx0.063$ — thoroughly non-relativistic. The Mott factor $1-\beta^2\sin^2(30°)=1-0.0039(0.25)\approx0.999$ is negligible, so pure Rutherford applies.

Step 2. Assemble equation (4) in natural units. With $\alpha^2=5.325\times10^{-5}$ , $E_K=10^{-3}$ MeV so $16E_K^2=1.6\times10^{-5}\ \text{MeV}^2$ , and $\sin(30°)=0.5$ so $\sin^4(30°)=0.0625$ : \frac{d\sigma}{d\Omega}=\frac{5.325\times10^{-5}}{(1.6\times10^{-5})(0.0625)}=\frac{5.325\times10^{-5}}{1.0\times10^{-6}}=53.3\ \text{MeV}^{-2}.

Step 3. Convert to an area with $(\hbar c)^2=(197.3\ \text{MeV fm})^2=3.894\times10^4\ \text{fm}^2$ per $\text{MeV}^{-2}$ , and $1\ \text{barn}=100\ \text{fm}^2$ : $53.3\ \text{MeV}^{-2}\times389.4\ \text{barn}=\mathbf{2.1\times10^4\ barn/sr}$ . This is a huge cross section — as expected for a slow electron in a long-range Coulomb field — and it grows as $E_K^{-2}$ at lower energy and $\sin^{-4}(\theta/2)$ toward the forward direction. d\sigma/d\Omega = 53.3 MeV⁻² ≈ 2.1×10⁴ barn/sr.

2 · A fast electron and the Mott suppression

A 500 MeV electron ( $\beta\approx1$ ) scatters from a proton, still treated as a point charge. Find $d\sigma/d\Omega$ at $\theta=30°$ , and compare the Mott result to Rutherford at $30°$ , $90°$ , $150°$ , and $180°$ .

Step 1. Use the ultrarelativistic Mott form (6). With $E=500$ MeV, $4E^2=10^6\ \text{MeV}^2$ ; $\sin(15°)=0.2588$ so $\sin^4(15°)=4.487\times10^{-3}$ ; $\cos^2(15°)=0.9330$ : \frac{d\sigma}{d\Omega}=\frac{(5.325\times10^{-5})(0.9330)}{(10^6)(4.487\times10^{-3})}=1.107\times10^{-8}\ \text{MeV}^{-2}. Converting, $1.107\times10^{-8}\times389.4\ \text{barn}=\mathbf{4.31\ \mu b/sr}$ .

Step 2. The Mott-to-Rutherford ratio is exactly the factor $\cos^2(\theta/2)$ : at $30°$ it is $\cos^2(15°)=\mathbf{0.933}$ (a mild $7\%$ cut), at $90°$ it is $\cos^2(45°)=\mathbf{0.500}$ (halved), and at $150°$ it is $\cos^2(75°)=\mathbf{0.067}$ (a factor of fifteen suppression).

Step 3. At $\theta=180°$ the factor is $\cos^2(90°)=\mathbf{0}$ : a truly relativistic electron cannot backscatter off a point charge. Physically the electron would have to keep its helicity while reversing its momentum, which flips the spin projection — forbidden by chirality conservation at the vertex. The tiny electron mass makes it not quite zero in reality, but the dip is the unmistakable signature of spin. Mott/Rutherford = cos²(θ/2): 0.933, 0.500, 0.067, 0 at 30°, 90°, 150°, 180°.

Physics of the portrait

The two names on this article belong to two eras: Rutherford supplied the experiment and the classical law that founded nuclear physics; Mott supplied the relativistic, spin-aware completion that made it a piece of quantum electrodynamics.

Origins · 1 minErnest Rutherford and how this was discovered

Ernest Rutherford

1871–1937 · nuclear physics and the discovery of the atomic nucleus

Ernest Rutherford, a New Zealand farmer's son who became the towering experimentalist of his age, built the picture of the atom that this article rests on. In 1909 his assistants Hans Geiger and Ernest Marsden, at his suggestion, fired alpha particles at a thin gold foil and found — to everyone's astonishment — that a few 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." In 1911 he explained it: the atom's positive charge and nearly all its mass sit in a minuscule central nucleus, and an alpha particle scattering off its $1/r$ Coulomb field follows a hyperbolic orbit with the now-famous $\sin^{-4}(\theta/2)$ angular law. Geiger and Marsden confirmed the angle, energy, and charge dependence in detail by 1913. Rutherford had already won the 1908 Nobel Prize in Chemistry for the transmutation of elements, went on to identify the proton, and predicted the neutron; he trained a generation — Bohr, Chadwick, Cockcroft — at Manchester and the Cavendish. His scattering law was purely classical, and remarkably it survives the quantum treatment unchanged for a $1/r$ potential — but it knew nothing of spin. That last gap is what Nevill Mott closed in 1929, deriving the relativistic Dirac-electron cross section and its $\cos^2(\theta/2)$ correction, turning Rutherford's law into the point-charge baseline of modern electron scattering.

History

1909

— Geiger and Marsden observe large-angle alpha scattering from gold foil

1911

— Rutherford proposes the nuclear atom and derives the sin⁻⁴(θ/2) law

1913

— Geiger and Marsden confirm the angle, energy, and charge dependence in detail

1929

— Mott derives the relativistic spin-½ Coulomb cross section with its cos²(θ/2) factor

1956

— Hofstadter's elastic electron scattering resolves the proton's finite size as a form factor

Going further

Texts. Thomson, Modern Particle Physics, §7.2 (Rutherford and Mott scattering, Eqs. 7.1–7.14) · Halzen & Martin, Quarks and Leptons, §6.2–6.3 (electron–muon and Mott scattering) · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.7 · Peskin & Schroeder, An Introduction to QFT, §4.7 and §5.1 (unpolarized cross sections) · Schwartz, QFT and the Standard Model, Ch. 13.

Provenance. The matrix element and $\kappa=p/(E+m_e)$ helicity-spinor bookkeeping are recalled from aq-pp-spin-helicity-states and aq-pp-feynman-rules-qed; the $1/q^2$ propagator from aq-pp-photon-propagator; the spin sum leading to equation (3) from aq-pp-matrix-element-spin-averaging. A verifier should confirm: (1) $q^2=-4p^2\sin^2(\theta/2)$ from elastic no-recoil kinematics; (2) the master factor identity $1-\beta^2\sin^2(\theta/2)=\gamma^{-2}(1+\beta^2\gamma^2\cos^2(\theta/2))$ ; (3) Rutherford $d\sigma/d\Omega=53.3\ \text{MeV}^{-2}=2.1\times10^4$ barn/sr for a 1 keV electron at $60°$ ; (4) ultrarelativistic Mott $=4.31\ \mu\text{b/sr}$ for a 500 MeV electron at $30°$ , with Mott/Rutherford $=\cos^2(\theta/2)=\{0.933,0.500,0.067,0\}$ at $\{30°,90°,150°,180°\}$ .

Prerequisites. Spin and Helicity States · QED Feynman Rules · The Photon Propagator · Matrix Elements and Spin Averaging. Leads to. Form Factors · Relativistic Electron–Proton Scattering · The Rosenbluth Formula.

Ask the tutor: Ask me to walk through how dividing a measured $e^-p$ cross section by the Mott formula exposes the proton's form factor — and where the Rosenbluth separation of $G_E$ and $G_M$ comes from.

If the proton is "fixed," why does anything transfer momentum to it?

"Fixed" is shorthand for "recoils in momentum but not in energy." The proton is about 1836 times heavier than the electron, so when the electron bounces off, the proton picks up a recoil momentum $\vec q$ but its kinetic energy $\sim q^2/2m_p$ is tiny compared to the electron's energy. That is exactly why the transferred four-momentum is spacelike, $q^2=-|\vec q|^2$ , with essentially zero energy component. The approximation breaks down once the beam energy becomes comparable to the proton mass; then recoil matters, the elastic kinematics shift, and you need the full relativistic $e^-p$ treatment and eventually the Rosenbluth formula. For low-energy scattering, though, treating the proton as a static Coulomb source is excellent.

Is Rutherford scattering a classical or a quantum result?

Both, remarkably. Rutherford derived his $\sin^{-4}(\theta/2)$ law in 1911 from purely classical mechanics — Coulomb-deflected hyperbolic orbits — years before quantum mechanics existed. When the problem is redone with the Schrödinger equation in the Born approximation, and again with the exact Coulomb wavefunctions, and yet again with QED as we do here, the same formula comes out. This coincidence is special to the $1/r$ potential: its Fourier transform is exactly $1/q^2$ , and the higher-order quantum corrections happen to cancel in the cross section. It is a lucky accident of the Coulomb force, not a general rule — most potentials give classical and quantum cross sections that differ. Mott's spin factor is the first genuinely quantum correction that classical mechanics could never produce.

What is the difference between "Mott scattering" here and "Mott polarimetry"?

They are the same underlying process seen from two angles. This article gives the spin-averaged Mott cross section — you average over the electron's initial spin and sum over its final spin, and the leftover spin content is the $\cos^2(\theta/2)$ factor. Mott polarimetry instead keeps track of the spin: when a transversely polarized electron scatters off a high- $Z$ nucleus, the spin-orbit part of the interaction makes it slightly more likely to go left than right (the Sherman function). Measuring that left-right asymmetry tells you the beam's polarization. So the averaged cross section is what you compute for a rate; the spin asymmetry, which averaging washes out, is what a polarimeter exploits. Both flow from the same Dirac-electron-in-a-Coulomb-field calculation.

Why does the total cross section diverge?

Integrating $d\sigma/d\Omega\propto\sin^{-4}(\theta/2)$ over all angles blows up at $\theta\to0$ , because $\sin^{-4}(\theta/2)\to(\theta/2)^{-4}$ and the solid angle only supplies $\theta\,d\theta$ — the integral diverges. Physically this is the infinite range of the Coulomb force: an electron passing arbitrarily far from the proton is still deflected by some tiny angle, and there are infinitely many such distant trajectories, so the "total probability of scattering by any amount" is unbounded. In any real situation the divergence is cut off — by atomic screening (other electrons neutralize the charge beyond the atomic radius), by a finite detector acceptance, or by a minimum measurable angle. The differential cross section at any fixed angle is perfectly finite and is what experiments actually report.

Where does the cos²(θ/2) actually come from, physically?

It is helicity conservation made quantitative. For a fast (effectively massless) electron, the QED vertex conserves chirality, which for a relativistic particle means it conserves helicity — the projection of spin onto the direction of motion. When the electron scatters by angle $\theta$ , its momentum direction rotates by $\theta$ , but its spin must stay aligned the same way relative to that momentum. The quantum amplitude for a spin-½ state to keep its projection while the axis rotates by $\theta$ is $\cos(\theta/2)$ , the Wigner rotation matrix element; the cross section, being amplitude squared, carries $\cos^2(\theta/2)$ . At $\theta=180°$ this overlap is zero, so backscattering is forbidden. Nothing about the force does this — it is the geometry of rotating a spin state.

When does treating the proton as a point charge break down?

As soon as the virtual photon's wavelength gets short enough to resolve the proton's structure — roughly when $\sqrt{-q^2}$ approaches the inverse proton radius, $1/r_p\sim0.2$ GeV. At that point the cross section is no longer the point-charge formula; it is the point-charge formula multiplied by a form factor $|F(q^2)|^2$ that Fourier-encodes the proton's charge distribution. Measuring how the data fall below the Mott prediction as $q^2$ grows is precisely how Hofstadter measured the proton's size in the 1950s, and how the electric and magnetic form factors $G_E$ and $G_M$ are mapped today via the Rosenbluth formula. Push $q^2$ higher still and the proton doesn't just look extended — it breaks, and you enter deep inelastic scattering, where the quarks inside first became visible.

Did Rutherford's own experiment use electrons, like this article?

No — and the distinction matters. Rutherford, Geiger, and Marsden fired alpha particles (helium nuclei) at gold nuclei: both were heavy, positive, and effectively spinless at those energies, so the classical Rutherford law fit the data with a charge factor $Z^2$ for gold. This article instead scatters a light, spin-½ electron off a proton, which is why the spin-dependent Mott factor is central here and was irrelevant to Rutherford. The shared object is the $1/r$ Coulomb field and its $\sin^{-4}(\theta/2)$ signature. Swapping the alpha for an electron keeps the Rutherford shape but adds the spin physics — and swapping a point target for a real proton is what opened up form factors and, eventually, the quark structure of matter.