05.00.10 · symplectic / lagrangian-mechanics

Scattering and Rutherford formula

shipped3 tiersLean: none

Anchor (Master): Rutherford 1911 *Phil. Mag.* (originator); Goldstein Ch. 3; Landau-Lifshitz Ch. 4; Arnold Ch. 4

Intuition [Beginner]

Scattering is what happens when a stream of particles flies toward a target and some bounce off at various angles. Think of throwing marbles at an invisible bump: the marbles that pass close to the bump get deflected a lot, and those that pass far away barely notice it. The pattern of deflections tells you the shape of the bump.

The key quantity is the scattering angle: how far the particle's direction changes from its original path. A particle aimed straight at the target (zero impact parameter) bounces straight back (scattering angle 180 degrees). A particle aimed far off to the side barely changes course (scattering angle near zero).

Ernest Rutherford used this idea in 1911 to discover the structure of the atom. He fired alpha particles at gold foil and measured the angles at which they scattered. He found that some bounced almost straight back, which was only possible if the positive charge of the atom was concentrated in a tiny nucleus rather than spread out. The formula relating the scattering angle to the strength of the Coulomb force is the Rutherford formula.

Visual [Beginner]

A beam of parallel arrows approaching a central scattering centre from the left. Each arrow is deflected by a different amount depending on how close it passes to the centre. The closest arrows are bent the most. A second panel shows the scattering angle as a function of the distance of closest approach: steep near the centre, flat far away.

The picture to keep in mind: the scattering angle depends on the impact parameter (how far off-centre the particle is aimed), and the differential cross-section converts this relationship into a prediction for how many particles land at each angle.

Worked example [Beginner]

An alpha particle (charge $+ 2 e$ ) approaches a gold nucleus (charge $+ 79 e$ ) with kinetic energy $E = 5$ MeV and impact parameter $b = 0$ . The scattering angle is 180 degrees (head-on collision, the particle comes right back).

Now take $b \neq = 0$ . The closest approach distance $r_{m i n}$ for head-on collision satisfies $E = q_{1} q_{2} / (4 π ϵ_{0} r_{m i n})$ , so $r_{m i n} = q_{1} q_{2} / (4 π ϵ_{0} E)$ . With $q_{1} q_{2} = 2 \times 79 \times e^{2}$ and $E = 5 \times 1 0^{6} \times 1.6 \times 1 0^{- 19}$ J, the distance of closest approach is about $4.5 \times 1 0^{- 14}$ m — roughly 10,000 times smaller than the atom itself.

What this tells us: the alpha particle must get extremely close to a concentrated charge to scatter by a large angle. This is what told Rutherford the atom has a compact nucleus.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Consider a particle of mass $μ$ moving in a central-force potential $V (r)$ that vanishes at infinity: $lim_{r \to \infty} V (r) = 0$ . The particle approaches from infinity with speed $v_{\infty}$ and impact parameter $b$ (the perpendicular distance from the asymptotic incoming line to the scattering centre). The energy and angular momentum are

E = \frac{1}{2} μ v_{\infty}^{2}, ℓ = μ v_{\infty} b .

The scattering angle $χ \in [0, π]$ is the angle between the incoming and outgoing asymptotic velocity vectors. For a repulsive potential, the orbit is a hyperbola and the scattering angle is related to the eccentricity $e$ of the hyperbola by $χ = π - 2 ϕ_{0}$ where $ϕ_{0}$ is the angle between the apse line and the asymptote.

The differential cross-section $d σ / d Ω$ is defined by the relation

d σ = \frac{d σ}{d Ω} d Ω = \frac{number of particles scattered into d Ω}{incident flux}

where $d Ω = sin χ d χ d ϕ$ is the element of solid angle. By axial symmetry, the scattering depends only on $χ$ (not $ϕ$ ), and the differential cross-section reduces to

\frac{d σ}{d Ω} = \frac{b}{sin χ} \frac{d b}{d χ} .

Counterexamples to common slips

The impact parameter is not the distance of closest approach. The impact parameter $b$ is the perpendicular distance from the scattering centre to the asymptotic incoming trajectory; the distance of closest approach $r_{m i n}$ satisfies $E = V (r_{m i n}) + ℓ^{2} / (2 μ r_{m i n}^{2})$ and is always less than or equal to $b$ for a repulsive potential.
The differential cross-section is not the probability of scattering into a given angle. It is the ratio of scattered flux to incident flux, with dimensions of area. Integrating $d σ / d Ω$ over all solid angles gives the total cross-section $σ$ , which measures the effective area of the target.
The total cross-section of the Coulomb potential is infinite. Because the Coulomb force has infinite range, particles at arbitrarily large impact parameters still scatter by small angles, and $σ = \int (d σ / d Ω) d Ω$ diverges. This is a physical property of long-range forces, not a mathematical artefact.

Key theorem with proof [Intermediate+]

Theorem (Rutherford scattering formula). For the repulsive Coulomb potential $V (r) = k q_{1} q_{2} / r$ with $k = 1/ (4 π ϵ_{0})$ , the differential cross-section is

\frac{d σ}{d Ω} = (\frac{k q _{1} q _{2}}{4 E})^{2} \frac{1}{sin ^{4} ( χ /2 )} .

Proof. The orbit equation in the Coulomb potential (from the Binet equation with $u = 1/ r$ ) is

r (ϕ) = \frac{a ( e ^{2} - 1 )}{1 + e cos ( ϕ - ϕ _{0} )}, a = \frac{k q _{1} q _{2}}{2 E}, e = 1 + \frac{2 E ℓ ^{2}}{μ ( k q _{1} q _{2} ) ^{2}} .

The orbit is a hyperbola with $e > 1$ . The asymptotes correspond to $r \to \infty$ , i.e. $1 + e cos ϕ_{asym} = 0$ , giving $cos ϕ_{asym} = - 1/ e$ . The scattering angle is $χ = π - 2 ϕ_{asym}$ , so $ϕ_{asym} = (π - χ) /2$ and

cos \frac{π - χ}{2} = sin \frac{χ}{2} = - \frac{1}{e} .

Squaring: $sin^{2} (χ /2) = 1/ e^{2}$ . Using the eccentricity formula with $ℓ = μ v_{\infty} b$ and $E = \frac{1}{2} μ v_{\infty}^{2}$ :

e^{2} = 1 + \frac{2 E μ ^{2} v _{\infty}^{2} b ^{2}}{μ ( k q _{1} q _{2} ) ^{2}} = 1 + (\frac{2 E b}{k q _{1} q _{2}})^{2} .

From $sin^{2} (χ /2) = 1/ e^{2} = 1/ (1 + (2 E b / (k q_{1} q_{2}))^{2})$ , invert:

\frac{1}{sin ^{2} ( χ /2 )} = 1 + (\frac{2 E b}{k q _{1} q _{2}})^{2} .

Solve for $b^{2}$ :

b^{2} = (\frac{k q _{1} q _{2}}{2 E})^{2} (\frac{1}{sin ^{2} ( χ /2 )} - 1) = (\frac{k q _{1} q _{2}}{2 E})^{2} cot^{2} (χ /2) .

Differentiate: $2 b d b = (k q_{1} q_{2} /2 E)^{2} \cdot (- 2 cot (χ /2) csc^{2} (χ /2)) \cdot \frac{1}{2} d χ$ (chain rule $d / d χ$ of $cot^{2} (χ /2)$ ). Simplify:

\frac{d b}{d χ} = - \frac{1}{2 b} (\frac{k q _{1} q _{2}}{2 E})^{2} cot (χ /2) csc^{2} (χ /2) .

Substitute into $d σ / d Ω = (b / sin χ) ∣ d b / d χ ∣$ with $sin χ = 2 sin (χ /2) cos (χ /2)$ :

\frac{d σ}{d Ω} = \frac{b}{2 sin ( χ /2 ) cos ( χ /2 )} \cdot \frac{1}{2 b} (\frac{k q _{1} q _{2}}{2 E})^{2} \frac{cos ( χ /2 )}{sin ( χ /2 )} \cdot \frac{1}{sin ^{2} ( χ /2 )}

= (\frac{k q _{1} q _{2}}{4 E})^{2} \frac{1}{sin ^{4} ( χ /2 )} . □

Bridge. The Rutherford formula builds toward the quantum-mechanical scattering theory of Born (1926), where the Born approximation reproduces the same $csc^{4} (χ /2)$ dependence at leading order — a coincidence specific to the Coulomb potential, where the classical and first-order quantum results agree exactly. The bridge between the classical scattering angle and the orbit geometry is the Binet equation, which appears again in 05.00.09 (worked Lagrangian examples) as the tool for computing central-force orbits. The foundational reason the Rutherford cross-section has the $csc^{4}$ form is that the Coulomb potential is the unique power-law potential whose orbits are conic sections, and this is exactly the property that makes the impact-parameter-to-scattering-angle map algebraically solvable in closed form. Putting these together identifies the Rutherford formula as the prototype scattering computation, the template that every other scattering calculation either generalises or approximates.

Exercises [Intermediate+]

Exercise 3 (medium, short-answer).

Show that the total cross-section $σ = \int (d σ / d Ω) d Ω$ for Rutherford scattering diverges, and explain the physical reason.

Hint

$\int_{0}^{π} csc^{4} (χ /2) sin χ d χ$ diverges at $χ = 0$ . Use $sin χ = 2 sin (χ /2) cos (χ /2)$ .

Answer

$σ = 2 π \int_{0}^{π} \frac{d σ}{d Ω} sin χ d χ = 2 π (\frac{k q _{1} q _{2}}{4 E})^{2} \int_{0}^{π} csc^{4} (χ /2) \cdot 2 sin (χ /2) cos (χ /2) d χ$ . Near $χ = 0$ : $csc^{4} (χ /2) \sim (2/ χ)^{4} = 16/ χ^{4}$ and $sin χ \sim χ$ , so the integrand behaves as $16/ χ^{3}$ , which integrates as $1/ χ^{2}$ and diverges as $χ \to 0$ . Physically, the Coulomb force has infinite range, so even particles at arbitrarily large impact parameters are scattered by small angles. The divergence comes from forward scattering ( $χ \to 0$ ), corresponding to $b \to \infty$ . In practice, any physical experiment has a detector that excludes $χ$ below some minimum angle, and the "total" cross-section is cut off at that angle. Rubric: full credit for identifying the divergence at $χ = 0$ , evaluating the integrand's behaviour, and giving the physical explanation.

Exercise 5 (medium, short-answer).

Derive the general formula $d σ / d Ω = (b / sin χ) ∣ d b / d χ ∣$ for an axially symmetric scattering process from geometric considerations.

Hint

Particles with impact parameter between $b$ and $b + d b$ are scattered into the angular range $[χ, χ + d χ]$ . The number of incident particles in the annulus of area $2 π b d b$ equals the number scattered into the solid angle $2 π sin χ d χ$ .

Answer

Consider a uniform beam of particles with flux $F$ (particles per area per time). The number entering the annulus $b$ to $b + d b$ per unit time is $F \cdot 2 π b ∣ d b ∣$ (using $∣ d b ∣$ since the annulus area is positive). These particles scatter into the angular range $χ$ to $χ + d χ$ , covering the solid angle $d Ω = 2 π sin χ ∣ d χ ∣$ . The number scattered into $d Ω$ per unit time is $F \cdot (d σ / d Ω) \cdot d Ω$ . Equating: $2 π b ∣ d b ∣ = (d σ / d Ω) \cdot 2 π sin χ ∣ d χ ∣$ . Solving: $d σ / d Ω = b ∣ d b ∣/ (sin χ ∣ d χ ∣) = (b / sin χ) ∣ d b / d χ ∣$ . Rubric: full credit for the annulus-to-solid-angle mapping and the algebraic derivation.

Exercise 6 (hard, short-answer).

Show that for a general central-force potential $V (r)$ vanishing at infinity, the scattering angle is given by

χ = π - 2 \int_{r_{m i n}}^{\infty} \frac{( b / r ^{2} ) d r}{1 - b ^{2} / r ^{2} - V ( r ) / E}

where $r_{m i n}$ is the turning point. Apply this to the Coulomb potential and recover the Rutherford relation between $b$ and $χ$ .

Hint

From conservation of energy and angular momentum: $\overset{r}{˙} = \pm 2 (E - V) / μ - ℓ^{2} / (μ^{2} r^{2})$ and $\dot{ϕ} = ℓ / (μ r^{2})$ . Eliminate $t$ : $d ϕ = (ℓ / (μ r^{2})) d t = (ℓ / (μ r^{2})) d r / \overset{r}{˙}$ . Integrate from $r_{m i n}$ to $\infty$ and double by symmetry.

Answer

From $E = \frac{1}{2} μ \overset{r}{˙}^{2} + ℓ^{2} / (2 μ r^{2}) + V (r)$ , solve for $\overset{r}{˙}$ : $\overset{r}{˙} = \pm 2 (E - V (r)) / μ - ℓ^{2} / (μ^{2} r^{2})$ . With $ℓ = μ v_{\infty} b$ and $E = \frac{1}{2} μ v_{\infty}^{2}$ , write $\overset{r}{˙} = \pm v_{\infty} 1 - b^{2} / r^{2} - V (r) / E$ . From $\dot{ϕ} = ℓ / (μ r^{2}) = v_{\infty} b / r^{2}$ : $d ϕ / d r = \dot{ϕ} / \overset{r}{˙} = (b / r^{2}) / 1 - b^{2} / r^{2} - V / E$ . The orbit is symmetric about the apse line, so $ϕ_{0} = \int_{r_{m i n}}^{\infty} (b / r^{2}) d r / 1 - b^{2} / r^{2} - V / E$ and $χ = π - 2 ϕ_{0}$ , giving the stated formula.

For the Coulomb potential $V = k q_{1} q_{2} / r$ , set $α = k q_{1} q_{2} / (2 E b^{2})$ and substitute $u = 1/ r$ :

χ = π - 2 \int_{0}^{1/ r_{m i n}} \frac{b d u}{1/ b ^{2} - u ^{2} - α u / b} = π - 2 arcsin \frac{1}{1 + ( 2 E b / ( k q _{1} q _{2} ) ) ^{2}} .

Using $arcsin (1/ e) = ϕ_{0}$ with $e = 1 + (2 E b / (k q_{1} q_{2}))^{2}$ , this gives $χ = π - 2 arcsin (1/ e)$ . Since $sin (π - χ) /2 = sin (χ /2)$ ... wait, $ϕ_{0} = arcsin (1/ e)$ and $χ = π - 2 ϕ_{0}$ , so $ϕ_{0} = (π - χ) /2$ and $sin ϕ_{0} = sin ((π - χ) /2) = cos (χ /2)$ . From $e^{2} sin^{2} ϕ_{0} = 1$ : $e^{2} cos^{2} (χ /2) = 1$ , so $e = sec (χ /2)$ and $1 + (2 E b / (k q_{1} q_{2}))^{2} = sec^{2} (χ /2)$ , giving $b = (k q_{1} q_{2} / (2 E)) cot (χ /2)$ . This is the Rutherford relation. Rubric: full credit for the general formula, the Coulomb substitution, and recovering the impact-parameter relation.

Exercise 7 (hard, short-answer).

Compare the classical Rutherford cross-section with the first Born approximation in quantum mechanics. Show that they give the same $d σ / d Ω \propto csc^{4} (χ /2)$ formula and explain why this agreement is special to the Coulomb potential.

Hint

The Born amplitude is $f (χ) = - (2 μ / ℏ^{2}) \int_{0}^{\infty} V (r) r sin (q r) / q d r$ with $q = 2 k sin (χ /2)$ for momentum transfer. The Fourier transform of $V = k q_{1} q_{2} / r$ is proportional to $1/ q^{2}$ .

Answer

The Born scattering amplitude for a spherically symmetric potential is $f (χ) = - \frac{2 μ}{ℏ ^{2}} \int_{0}^{\infty} \frac{V ( r ) s i n ( q r )}{q} r d r$ where $q = 2 k sin (χ /2)$ is the momentum transfer and $k = 2 μ E /ℏ$ is the wave number. For $V (r) = k q_{1} q_{2} / r$ , the integral is $\int_{0}^{\infty} sin (q r) d r = 1/ q$ (convergent in the distributional sense; regularise by adding $e^{- ϵr}$ and taking $ϵ \to 0$ ). So $f (χ) = - (2 μ k q_{1} q_{2}) / (ℏ^{2} q^{2}) = - (2 μ k q_{1} q_{2}) / (4 ℏ^{2} k^{2} sin^{2} (χ /2))$ . The cross-section $d σ / d Ω = ∣ f ∣^{2} = (μ k q_{1} q_{2})^{2} / (4 ℏ^{4} k^{4} sin^{4} (χ /2))$ . Using $ℏ^{2} k^{2} = 2 μ E$ : $d σ / d Ω = (k q_{1} q_{2})^{2} / (16 E^{2} sin^{4} (χ /2)) = (k q_{1} q_{2} / (4 E))^{2} csc^{4} (χ /2)$ — identical to the classical Rutherford formula.

The agreement is specific to the Coulomb potential because the exact quantum Coulomb scattering amplitude (computed by Gordon 1928 using parabolic coordinates) equals the Born approximation. This happens because the Coulomb potential is scale-free (no characteristic length), and the quantum corrections that would distinguish the exact result from the Born approximation vanish. For any other potential the classical and Born results differ. Rubric: full credit for the Born computation, the agreement, and the scale-free explanation.

Advanced results [Master]

The general scattering formula. For any central-force potential $V (r)$ with $V \to 0$ as $r \to \infty$ , the scattering angle is

χ (b, E) = π - 2 \int_{r_{m i n} (b, E)}^{\infty} \frac{( b / r ^{2} ) d r}{1 - b ^{2} / r ^{2} - V ( r ) / E} .

The turning point $r_{m i n}$ is the largest root of $1 - b^{2} / r^{2} - V (r) / E = 0$ . The differential cross-section is then $d σ / d Ω = (b / sin χ) ∣ d b / d χ ∣$ . This formula is the starting point for every classical scattering calculation.

Hard-sphere scattering. For a hard sphere of radius $R$ , the scattering angle is $χ = 2 arccos (b / R)$ for $b \leq R$ and $χ = 0$ for $b > R$ . The differential cross-section is $d σ / d Ω = R^{2} /4$ , independent of angle (isotropic scattering). The total cross-section is $σ = π R^{2}$ — the geometric cross-section.

Glory scattering. When the scattering angle $χ$ passes through 0 or $π$ as a function of $b$ , the factor $1/ sin χ$ in the cross-section diverges, producing bright spots in the forward or backward direction. These are the classical analogues of the optical glory. For the attractive Coulomb potential (electron-nucleus scattering), the backward glory occurs at $χ = π$ .

Rainbow scattering. When $d b / d χ = 0$ at some impact parameter $b^{*}$ , the cross-section has an integrable singularity of the Airy-function type. This is the rainbow angle $χ^{*} = χ (b^{*})$ , named by analogy with the optical rainbow. The phenomenon occurs in atom-atom scattering at intermediate energies and is a signature of the potential well.

Coulomb scattering in quantum mechanics. The exact quantum-mechanical Coulomb scattering amplitude (Gordon 1928, using the confluent hypergeometric function) is

f (χ) = \frac{η}{2 k sin ^{2} ( χ /2 )} e^{- i η l n (s i n^{2} (χ /2)) + iπ + 2 i a r g Γ (1 + i η)}

where $η = k q_{1} q_{2} / (ℏ v)$ is the Sommerfeld parameter. The phase factor is the quantum correction absent from the Born approximation; the cross-section $∣ f ∣^{2} = (k q_{1} q_{2} / (4 E))^{2} csc^{4} (χ /2)$ is unchanged.

Synthesis. The Rutherford scattering formula is the prototype for every scattering calculation in physics. The foundational reason it is exactly solvable is that the Coulomb orbit is a hyperbola with an eccentricity that depends algebraically on the impact parameter, and the mapping $b \to χ$ is the arccotangent — elementary, invertible, and producing the $csc^{4}$ divergence at forward angles that is the signature of the infinite range of the Coulomb force.

The bridge between the classical Rutherford formula and quantum scattering is the Born approximation: this is exactly the regime where the first-order quantum result agrees with the classical result, and the agreement holds for the Coulomb potential but for almost no other potential. The central insight is that the Coulomb potential has no intrinsic length scale — it is scale-free — and scale-free potentials produce identical classical and quantum cross-sections because the quantum diffraction effects that would distinguish them require a length scale to generate.

Putting these together identifies the Rutherford formula as the intersection of three frameworks: the Lagrangian central-force orbit (which gives the hyperbolic trajectory), the scattering-theory geometry (which converts the orbit into a cross-section), and the quantum Born approximation (which reproduces the classical result for the special case of the Coulomb potential). The pattern recurs across scattering problems: the harder the potential, the more the classical and quantum results diverge; the softer the potential, the closer they agree.

Full proof set [Master]

Proposition (scattering angle as an integral). For a particle of energy $E$ and angular momentum $ℓ = μ v_{\infty} b$ in a central-force potential $V (r)$ vanishing at infinity, the scattering angle is $χ = π - 2 \int_{r_{m i n}}^{\infty} (b / r^{2}) d r / 1 - b^{2} / r^{2} - V / E$ .

Proof. Conservation of energy gives $\frac{1}{2} μ \overset{r}{˙}^{2} + ℓ^{2} / (2 μ r^{2}) + V (r) = E$ . Solve for $\overset{r}{˙}$ : $\overset{r}{˙} = \pm v_{\infty} 1 - b^{2} / r^{2} - V (r) / E$ where $v_{\infty} = 2 E / μ$ . From $\dot{ϕ} = ℓ / (μ r^{2}) = v_{\infty} b / r^{2}$ , the orbit equation is $d ϕ / d r = (v_{\infty} b / r^{2}) / (v_{\infty} 1 - b^{2} / r^{2} - V / E) = (b / r^{2}) / 1 - b^{2} / r^{2} - V / E$ . The angle swept from the turning point to infinity is $ϕ_{0} = \int_{r_{m i n}}^{\infty} (b / r^{2}) d r / 1 - b^{2} / r^{2} - V / E$ . By symmetry the total angle from incoming to outgoing asymptote is $2 ϕ_{0}$ , so $χ = π - 2 ϕ_{0}$ . $□$

Proposition (Rutherford impact-parameter relation). For the Coulomb potential $V (r) = k q_{1} q_{2} / r$ , the impact parameter $b$ and scattering angle $χ$ are related by $b = (k q_{1} q_{2} / (2 E)) cot (χ /2)$ .

Proof. The orbit in the Coulomb potential is a hyperbola $r (ϕ) = a (e^{2} - 1) / (1 + e cos ϕ)$ with $a = k q_{1} q_{2} / (2 E)$ and $e = 1 + (2 E b / (k q_{1} q_{2}))^{2}$ . The asymptote satisfies $1 + e cos ϕ_{asym} = 0$ , giving $cos ϕ_{asym} = - 1/ e$ . The scattering angle $χ = π - 2 ϕ_{asym}$ gives $ϕ_{asym} = (π - χ) /2$ and $cos ((π - χ) /2) = sin (χ /2) = 1/ e$ . Then $e = csc (χ /2)$ , and $e^{2} = 1 + (2 E b / (k q_{1} q_{2}))^{2} = csc^{2} (χ /2) = 1 + cot^{2} (χ /2)$ . So $(2 E b / (k q_{1} q_{2}))^{2} = cot^{2} (χ /2)$ , giving $b = (k q_{1} q_{2} / (2 E)) cot (χ /2)$ . $□$

Connections [Master]

Worked Lagrangian examples 05.00.09. The Kepler problem at positive energy (hyperbolic orbits) developed in 05.00.09 is the orbit-theoretic basis for the scattering calculation. The Binet equation and the conic-section solution of the Kepler problem are used directly in the Rutherford derivation.
Noether's theorem 05.00.04. The scattering calculation relies on two conserved quantities — energy and angular momentum — both derived from Noether's theorem applied to time-translation and rotational symmetries of the central-force Lagrangian.
Lagrangian on $T Q$ 05.00.01. The scattering trajectory is a solution of the Euler-Lagrange equations for the Lagrangian $L = \frac{1}{2} μ ∥ \overset{r}{˙} ∥^{2} - V (r)$ on $T R^{3}$ , developed in 05.00.01. The reduction to a one-dimensional problem in $r$ is the consequence of angular momentum conservation reducing the effective configuration space from $R^{3}$ to $R_{> 0}$ .
Hamiltonian vector field 05.02.01. The scattering flow is the Hamiltonian flow of $H = p^{2} / (2 μ) + V (r)$ on $T^{*} R^{3}$ . The phase-space portrait of a scattering trajectory is an unbounded orbit on the energy surface $H = E$ .

Historical & philosophical context [Master]

Ernest Rutherford's 1911 paper The scattering of alpha and beta particles by matter and the structure of the atom ^{[Rutherford 1911]} in the Philosophical Magazine introduced the scattering formula and used it to deduce the existence of the atomic nucleus. Rutherford's experimental collaborators Geiger and Marsden had observed that a small fraction of alpha particles scattered at angles greater than 90 degrees, which was incompatible with the then-prevailing Thomson plum-pudding model of the atom. Rutherford showed that the angular distribution of the scattered particles followed the $csc^{4} (χ /2)$ law predicted by point-charge Coulomb scattering, confirming the nuclear model.

The classical scattering formalism was systematised by Lord Rayleigh and others in the late 19th century in the context of kinetic theory. The Born approximation for quantum scattering was introduced by Max Born in 1926 ^{[Born 1926]}; the exact Coulomb scattering amplitude was computed by Walter Gordon in 1928 using parabolic coordinates and confluent hypergeometric functions. The agreement between the classical and quantum Coulomb cross-sections was noted by Wentzel (1927) and is a special property of the $1/ r$ potential.

Bibliography [Master]

@article{Rutherford1911Scattering,
  author    = {Rutherford, Ernest},
  title     = {The scattering of $\alpha$ and $\beta$ particles by matter and the structure of the atom},
  journal   = {Philosophical Magazine},
  series    = {6th},
  volume    = {21},
  pages     = {669--688},
  year      = {1911}
}

@article{Born1926Quantum,
  author    = {Born, Max},
  title     = {Quantenmechanik der Sto{\ss}vorg{\"a}nge},
  journal   = {Zeitschrift f{\"u}r Physik},
  volume    = {38},
  pages     = {803--827},
  year      = {1926}
}

@book{Goldstein1980Classical,
  author    = {Goldstein, Herbert},
  title     = {Classical Mechanics},
  publisher = {Addison-Wesley},
  edition   = {2nd},
  year      = {1980}
}

@book{LandauLifshitz1976Mechanics,
  author    = {Landau, L. D. and Lifshitz, E. M.},
  title     = {Mechanics},
  series    = {Course of Theoretical Physics},
  volume    = {1},
  publisher = {Pergamon Press},
  edition   = {3rd},
  year      = {1976}
}

@book{Arnold1989Mathematical,
  author    = {Arnold, V. I.},
  title     = {Mathematical Methods of Classical Mechanics},
  series    = {Graduate Texts in Mathematics},
  volume    = {60},
  publisher = {Springer},
  edition   = {2nd},
  year      = {1989}
}

Prerequisites

05.00.01
05.00.02
05.00.04

Tier anchors

beginner: Goldstein *Classical Mechanics* Ch. 3 informal; Feynman *Lectures* Vol. I Ch. 10 scattering picture
intermediate: Goldstein *Classical Mechanics* Ch. 3; Arnold *Mathematical Methods* Ch. 4; Landau-Lifshitz *Mechanics* Ch. 4
master: Rutherford 1911 *Phil. Mag.* (originator); Goldstein Ch. 3; Landau-Lifshitz Ch. 4; Arnold Ch. 4

References

TODO_REF
Rutherford — The scattering of alpha and beta particles by matter and the structure of the atom (1911) · Philosophical Magazine, 6th series, vol. 21, pp. 669-688. Originator: the Rutherford scattering formula and its experimental verification.
TODO_REF
Goldstein — Classical Mechanics (1980) · Ch. 3: The central-force problem and scattering.
TODO_REF
Arnold — Mathematical Methods of Classical Mechanics (1989) · Ch. 4: Lagrangian mechanics on configuration manifolds; scattering in central-force fields.
TODO_REF
Landau-Lifshitz — Mechanics (1976) · Ch. 4: Collisions and scattering of particles.

Reviewer

TBD

Estimated time

beginner: 15m
intermediate: 35m
master: 65m