San José State University

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 The Derivation of the Rutherford Scattering Formula and Its Generalization

In 1911 Ernest Rutherford published a formula which indicated that the number of particles that would be deflected by an angle θ due to scattering from fixed nuclei is inversely proportional to the fourth power of the sine function of one half the angle of deflection; i.e.,

#### n(θ)Δθ = [κ/sin4(θ/2)] Δθ

where κ is a constant.

It was brilliant analysis in support of brilliant empirical work. Rutherford allowed a beam of alpha particles (helium nuclei) to impinge upon very thin gold foil. The distribution of the deflected alpha particles corresponded to his formula. This result established that the the structure of atoms involved a small dense positively charged nucleus surrounded by the negatively charged electrons. Prior to Rutherford's work the prevalent concept of the structure of atoms was J.J. Thomson's plum pudding model in which the positive and negative charges of atoms were uniformly intermixed.

The following is a derivation of Rutherford's formula with a small degree of generalization.

Consider a body at rest at point O. Another body is coming from the left horizontally initially along the upper red line. It is deflected and travels along a trajectory that asymptotically approaches the line which makes an angle of θ with the horizontal. The problem is to establish the angle of deflection θ. Ernest Rutherford first solved this problem for the electrostatic force. The purpose of this page is present a generalization Rutherford's derivation.

The force between two bodies which is due to a field carried by particles is necessarily of the form f(r)/r² where r is the distance separating the two bodies. For a single field the form of f(r) is Ae−λr, where A are constants for the field. If the field-carrying particles do not decay, as is the case for the photons and gravitons of the electromagnetic and gravitation fields, then λ=0. If the field-carrying particles do decay, as is the case for the π mesons of the nuclear force field, then λ>0.

For a single field the constant A has the form of the product of the field magnitudes of the two bodies, such as masses, charges or number of nucleons, multiplied by a constant characteristic of the field. For example, the force between two bodies of mass m and M is GmM/r², where G is the universal gravitational constant, and thus for this case f(r)=GmM.

For the general case two bodies may be affected by more than one field so f(r) might, for example, be f(r)=JqQ+HnNexp(−λr), where q and Q are the charges of the two bodies and n and N are their nucleon numbers.

For some simple cases the deflection of incoming bodies due to scattering by the target bodies can be analyzed without resort to analyzing the detailed dynamics of the interaction. For simplicity it is assumed that the mass and other characteristics of the incoming bodies, hereafter called particles, are small compared to the target bodies, hereafter called nuclei. This allows the simplification that the target nuclei remain fixed; i.e., do not recoil as a result of the interaction with the incoming particles.

The target nuclei is assumed to be fixed at point O. The incoming particle is following a horizontal trajectory that is located a distance b above the target nuclei. The distance b is commonly termed, the impact parameter for the interaction. Although the particle starts on the horizontal trajectory the interaction with the fields of the nuclei causes it to be deflected and asymptotically approach a line making an angle of θ with the horizontal.

## Conservation of Angular Momentum

Let v0 be the velocity of the particle at the initial time and m be its mass. The angular momentum of the particle with respect to the point O is mr0. At a point X on its trajectory the location of the particle can be described by an angle φ and its distance r from the nuclei. At point X the particle has a tangential velocity with respect to O of r(dφ/dt). Thus its angular momentum with respect to O is mr²(dφ/dt). Angular momentum will be conserved therefore

## Change in Linear Momentum

A fundamentatl relation of mechanics is that the change in linear momentum p in any direction over a time interval is equal to the the force F in that direction integrated over the time interval; i.e.,

#### Δp = p(t1) − p(t0) = ∫t0t1Fdt

Consider the direction shown as OD, which splits the angle (π−θ) in half. The component of the initial linear momentum of mv0 is equal to −mv0cos(β). The angle β works out to be (π/2−θ/2) so the initial component of momentum in the direction OD is, since cos(π/2−θ/2)=sin(θ/2), −mv0sin(θ/2).

By symmetry the terminal velocity will also be v0 and its component of linear momentum in the direction OD will be +mv0sin(θ/2).

Therefore the net change in linear momentum in the direction OD is 2mv0sin(θ/2).

The component of the central force of f(r)/r² in the direction OD is [f(r)/r²]cos(φ).

Thus

#### ∫−∞∞(f(r)/r²)cos(φ)dt = 2mv0sin(θ/2);

The variable of integration in the above may be changed to φ and from the relation derived from the conservation of angular momentum dφ/dt = v0b/r² and hence dt/dφ = r²/(v0b).

At t=−∞ φ=−α=−(π−θ)/2 and at t=+∞ φ=+α=+(π−θ)/2. This means that with the change of variable the integral equation above reduces to

#### ∫−α+α(f(r)/(bv0))cos(φ)dφ = 2mv0sin(θ/2);

When f(r) is independent of r, such as JqQ where q and Q are the charges of the bodies and J is a constant, the integral can be evaluated analytically as

Therefore

#### 2(JqQ/bv0))cos(θ/2) = 2mv0sin(θ/2) and hence tan(θ/2) = JqQ/(bmv0²) which is equivalent to b = (JqQ/(mv0))cot(θ/2)

This establishes the inverse relationship between the angle of deflection θ and the impact parameter b.

There is no problem in generalizing this relation to two forces which are carried by particles which do not decay, such as the electrostatic force and graviation. If the combined force is given by (JqQ-GmM)/r²) then the numerator in the above fraction becomes (JqQ-GmM).

## The Distribution of the Angle of Deflection Resulting from the Distibution of the Impact Parameter

A variable x has a cumulative probability function P(X), which is the probability that x will have a value less than or equal to X, and a probability density function p(x). The two are related by

#### P(X) = ∫−∞Xp(x)dx and p(X) = dP(X)/dX

If a variable y is a monotonic function of x, say y=f(x) then there is an inverse function x=g(y) and dx=g'(y)dy. The probability density function for y is given by

#### q(y) = p(g(y))|g'(y)|

where |g'(y)| is the absolute value of the derivative of the function g(y) with respect to its argument.

Suppose the incoming particles are evenly distributed over a beam of radius B and that there is only one target nucleus and it is at the center of the beam. The probability that the impact parameter is less than or equal to a value b is the area of a circle of radius b around the target nucleus relative to the area of the beam, πB²; i.e.

#### P(b) = πb²/πB² = b²/B² and therefore the probability density function for b is p(b) = 2b/B²

The relationship between the impact parameter b and the deflection angle θ was found to be of the form

#### b = γcos(θ/2)/sin(θ/2) = γcot(θ/2)

Therefore db/dθ = [−γ/sin²(θ/2)](1/2) and |db/dθ| = γ/(2sin²(θ/2)).

Since the probability density function for b is p(b)=2b/B², the probability density function of θ, q(θ), is

#### q(θ) = 2[(γcot(θ/2)/B²](γ/(2sin²(θ/2)); which reduces to q(θ) = γ²[cos(θ/2)/sin³(θ/2)]/B²

The minimu deflection angle θmin corresponds to the maximum impact parameter B.

The relations found above are not quite Rutherford scattering formla. His formula is in terms of the cross section for an interaction. The cross section σ for an impact parameter b is the area of a circle with radius b. Let Θ be the deflection angle corresponding to an impact parameter of b. The probability that the impact parameter is less than or equal to b is the same as the probability that the deflection angle will be greater than or equal to Θ. Given the relations previously found then

#### σ(θ≥Θ) = πb² = πγ²cot²(Θ/2))

Rutherford's formula is in terms of dσ/dΩ where dΩ is the solid angle between Θ and Θ+dΘ. This solid angle is equal to the area of a band of width dΘ and length equal to the perimeter of a circle of radius sin(Θ); i.e., dΩ- 2πsin(Θ)dΘ.

Note that the argument of the sine function is Θ rather than Θ/2. Since sin(2x) is equal to 2sin(x)cos(x) the above solid angle dΩ is equal to 4πsin(Θ/2)cos(Θ/2). Thus

#### dσ/dΩ = (1/(4πsin(Θ/2)cos(Θ/2))(dσ/dΘ)

The derivative dσ/dΘ is given by

#### πγ²(2cot(Θ/2)(-1/sin²)(Θ/2)(1/2) which reduces to −πγ²cos(Θ/2)/sin³(Θ/2)

Therefore, taking the absolute value and cancelling the cos(Θ/2) terms which appear in the numerator and denominator,

#### dσ/dΩ = (1/4)γ²/sin4(Θ/2) or equivalently dσ/dΩ = (1/4)γ²cosec4(Θ/2)

This is the exact form of the Rutherford formula.

There are a couple of interesting generalizations. One allows for the recoil of the target nuclei as a result of their interaction with the incoming particles. The other generalization that would be of interest is when the force is of the form Aexp(−λr)/r².

For more on the scattering and diffraction of atomic particles see SCATTERING.

(To be continued.)

Sources:
W.S.C. Williams, Nuclear and Particle Physics, Clarendon Press, Oxford, 1991.