Differential Cross Sections and Rutherford Scattering

Differential Cross Sections and Rutherford Scattering

M04-10 introduced the classical scattering map

(1)

where b is the impact parameter and χ is the scattering angle in the relative or center of mass description.

For an axially symmetric central force, the geometry of an incident annulus gives

(2)

while the corresponding outgoing solid-angle band is

(3)

Therefore

(4)

This formula converts a trajectory calculation into a measurable angular distribution.

For the repulsive inverse-radius potential

(5)

the orbit can be solved exactly. The result is

(6)

where E is the asymptotic relative kinetic energy.

Substituting into the differential cross-section formula gives the classical Rutherford result:

(7)

For two point charges,

(8)

The striking angular dependence

(9)

is the hallmark of classical Coulomb scattering.

1 From trajectories to measured distributions

A single classical trajectory is specified by quantities such as

(10)

An experiment, however, sends many projectiles toward many targets.

The useful observable is not one trajectory but the number of particles scattered into a range of directions.

The differential cross section answers the question:

(11)

Its customary units are

(12)

Since the steradian is dimensionless in SI, the fundamental dimensions remain those of area.

2 Impact-parameter annulus

Consider incident trajectories whose impact parameters lie between

(13)

and

(14)

Because the central interaction is axially symmetric around the beam direction, these trajectories form an annulus in impact-parameter space.

Its area is

(15)

If the map bχ is one-to-one, that annulus is scattered into a band of polar angles from χ to χ + dχ.

The solid angle of that band is

(16)

3 Derivation of the classical differential cross section

The same set of trajectories occupies

(17)

in the incoming impact-parameter plane and

(18)

in outgoing angle space.

Therefore

	=	(19)
	= .	(20)

The absolute value is required because a common scattering map has

(21)

larger impact parameters give smaller deflections.

4 Scattering function

The relation

(22)

is often called the scattering function.

If it can be inverted,

(23)

then the differential cross section follows directly.

If several impact parameters lead to the same scattering angle,

(24)

the contributions add:

(25)

The Rutherford problem is simpler because the relevant repulsive Coulomb scattering map is monotonic.

5 Rutherford interaction

For two point charges q₁ and q₂, the electrostatic potential energy is

(26)

Define

(27)

For like charges,

(28)

and the interaction is repulsive:

(29)

The force on the relative particle is

(30)

The reduced mass is

(31)

Far from the interaction,

E	= μv∞2,	(32)
ℓ	= μv∞b.	(33)

6 Why the orbit is a hyperbola

The potential tends to zero at infinity:

(34)

A scattering projectile arrives from infinity with

(35)

For an inverse-radius potential, positive-energy trajectories are unbound conic sections.

For the repulsive case, the physical branch is a hyperbola that bends away from the force center.

7 Binet equation

To solve the orbit, define

(36)

For a central force, Binet’s equation is

(37)

For repulsive Coulomb force,

(38)

Thus

(39)

A convenient solution is

(40)

where e is a constant determined by the energy.

Therefore the orbit is

(41)

with

(42)

The angle origin has been chosen so that closest approach occurs at

(43)

8 Eccentricity from the energy

At closest approach,

(44)

From the orbit equation,

(45)

The energy at the turning point is

(46)

Using

(47)

we obtain

E	= (e − 1)2 + (e − 1)	(48)
	= .	(49)

Therefore

(50)

For a scattering orbit,

(51)

so

(52)

This confirms the hyperbolic character.

9 Asymptotes of the Rutherford orbit

As

(53)

we have

(54)

From

(55)

the asymptotes satisfy

(56)

Thus

(57)

The hyperbola is symmetric about the closest-approach axis.

The angle between the incoming and outgoing velocity directions is

(58)

Therefore

(59)

Hence

(60)

Thus

(61)

10 Deriving the impact-parameter relation

From

(62)

we have

(63)

But

(64)

Using

(65)

and

(66)

we find

e2 − 1	=	(67)
	=	(68)
	= 2.	(69)

Therefore

(70)

Solving for b,

(71)

Equivalently,

(72)

This is the Rutherford scattering function.

11 Coulomb scattering length

Define

(73)

Then

(74)

The quantity a has dimensions of length and sets the characteristic scattering scale.

It is not the distance of closest approach.

For a head-on encounter,

(75)

12 Derivation of the Rutherford differential cross section

Start with

(76)

Differentiate:

(77)

Therefore

(78)

Use

(79)

Substitute:

= csc 2.

(80)

Using

(81)

the trigonometric factors reduce to

(82)

Since

(83)

we obtain

(84)

This is the Rutherford differential cross section.

13 Rutherford formula for point charges

For charges

(85)

the Coulomb constant is

(86)

Therefore

(87)

Here E is the relative or center of mass kinetic energy at infinity.

For a very heavy target,

(88)

the center of mass energy approaches the projectile laboratory kinetic energy.

14 Angular dependence

The Rutherford formula contains

(89)

As

(90)

(91)

Therefore

(92)

for sufficiently small angles.

Small deflections are extremely common in a pure long-range Coulomb interaction.

Large-angle deflections require small impact parameters and are much less common.

15 Energy dependence

At fixed scattering angle,

(93)

Therefore doubling the relative kinetic energy reduces the differential cross section at the same angle by a factor of four.

Higher-energy particles are more difficult to deflect through a specified angle.

This agrees with the impulse picture from M04-10: a larger incoming momentum requires a larger transverse impulse to produce the same directional change.

16 Charge dependence

At fixed E and χ,

(94)

Doubling either charge magnitude increases the classical Coulomb differential cross section by a factor of four.

The sign of Z₁Z₂ determines whether the orbit bends away from or toward the force center, while the ideal Coulomb cross-section magnitude depends on k².

The Rutherford nuclear-scattering case uses repulsion between positively charged projectiles and nuclei.

17 Closest approach and scattering angle

M04-10 derived the repulsive inverse-radius turning point

(95)

With

(96)

this becomes

(97)

Using

(98)

we have

(99)

Thus

(100)

For head-on backscattering,

(101)

so

(102)

18 Integrated cross section above an angle

Suppose we count all particles scattered through angles

(103)

Because Rutherford scattering is monotonic, these events correspond to

(104)

where

(105)

Therefore the integrated cross section is simply the impact-parameter disk:

(106)

Hence

(107)

19 Why the total Coulomb cross section diverges

If one attempts to count all nonzero scattering angles, then

(108)

Since

(109)

the integrated Rutherford cross section diverges:

(110)

This is not a statement that an infinite number of particles are physically detected.

It reflects the infinite range of the ideal unscreened Coulomb force: arbitrarily distant trajectories receive arbitrarily small deflections.

Real experiments always have finite angular resolution, finite beam geometry, and screening or environmental effects.

20 Barns

Nuclear and particle cross sections are often reported in barns:

(111)

Since

(112)

we have

(113)

so

(114)

21 Example 1: alpha particle scattering from gold

Consider a 5.0 MeV alpha particle scattering from a very heavy gold nucleus.

Take

(115)

Using

(116)

the Coulomb parameter is

k	= (2)(79)(1.44) MeV fm	(117)
	= 227.5 MeV fm.	(118)

Because gold is much heavier than the alpha particle, use

(119)

Then

(120)

For

(121)

(122)

Thus

(123)

22 Example 2: differential cross section at 90^∘

For the same alpha-gold example,

(124)

At

(125)

(126)

Also,

(127)

Therefore

	= (11.38)2(4) fm2∕sr	(128)
	≈ 518 fm2∕sr.	(129)

Since

(130)

(131)

23 Example 3: head-on closest approach

For the same 5.0 MeV alpha-gold encounter,

(132)

Thus

(133)

This is much larger than the characteristic nuclear radius of a heavy nucleus.

In this energy range, a head-on alpha particle turns around because of Coulomb repulsion before reaching the nuclear surface.

24 Example 4: energy scaling

Suppose the differential cross section at a fixed angle is

(134)

at energy

(135)

At energy

(136)

the Rutherford formula gives

(137)

Therefore

(138)

25 Example 5: angle scaling

Compare

(139)

and

(140)

at the same E and charges.

The ratio is

(141)

Numerically,

(142)

Thus

(143)

Small-angle scattering is dramatically more probable.

26 From cross section to count rate

Suppose a thin target has areal number density

(144)

and the incident beam rate is

(145)

For a detector subtending small solid angle

(146)

around angle χ, the ideal single-scattering count rate is approximately

(147)

A real experiment may also require detector efficiency, dead-time, energy-loss, and multiple-scattering corrections.

This equation explains why the differential cross section is experimentally useful: it connects a microscopic trajectory law to an observable rate.

27 Center of mass energy versus laboratory energy

The Rutherford formula derived here uses the relative energy

(148)

Suppose particle 1 is the projectile and particle 2 is initially at rest in the laboratory.

Then

(149)

The projectile laboratory kinetic energy is

(150)

Therefore

E	= Elab	(151)
	= Elab.	(152)

If

(153)

then

(154)

28 Center of mass angle versus laboratory angle

The reduced-coordinate calculation gives the center of mass scattering angle χ.

For a target initially at rest, the projectile laboratory velocity after scattering is

(155)

Using the two-body velocity decomposition,

v1x,lab,f	= ,	(156)
v1y,lab,f	= m2 sin χ.	(157)

Therefore the projectile laboratory angle 𝜃_lab satisfies

(158)

For a very heavy target,

(159)

this reduces approximately to

(160)

29 Assumptions behind Rutherford scattering

The ideal Rutherford result assumes:

• Classical Mechanics,
• nonrelativistic motion,
• pointlike interacting charges for the Coulomb force,
• a pure 1∕r potential over the relevant distance range,
• elastic two-body scattering,
• negligible radiation loss,
• isolated single scattering,
• no important electron screening at the relevant impact parameters.

The formula is powerful partly because deviations from it reveal additional physics.

30 Screening at very small angles

Small scattering angles correspond to large impact parameters:

(161)

At sufficiently large distances, atomic electrons can screen the nuclear charge.

Then the interaction is no longer the unscreened nuclear Coulomb potential

(162)

over the entire trajectory.

Therefore the ideal Rutherford divergence as

(163)

is not physically realized without modification.

31 Short-distance deviations

Large scattering angles correspond to small impact parameters and smaller closest-approach distances.

If the projectile approaches distances where:

• finite nuclear size matters,
• the strong interaction contributes,
• relativity becomes important,
• quantum effects cannot be neglected,

then the simple classical Rutherford model can fail.

Historically, large-angle alpha scattering was important because it showed that positive charge and most atomic mass are concentrated in a very small region.

32 Classical and quantum significance

The Rutherford angular form is remarkable because the same

(164)

dependence also appears in the nonrelativistic quantum treatment of unscreened Coulomb scattering.

That agreement is special to the Coulomb potential.

The interpretation differs:

• classical mechanics assigns trajectories labeled by b,
• quantum mechanics assigns a scattering amplitude and probability distribution.

The present article remains entirely within classical mechanics.

33 Common mistakes

1. Using dσ∕dΩ without first identifying the scattering map b(χ).
2. Dropping the absolute value in |db∕dχ|.
3. Confusing total cross section with differential cross section.
4. Forgetting the sin χ factor from the solid-angle element.
5. Using sin χ where the Rutherford formula requires sin(χ∕2).
6. Missing the fourth Power in csc ⁴(χ∕2).
7. Using laboratory projectile energy in the exact finite-mass formula without converting to center of mass relative energy.
8. Using laboratory scattering angle in place of the center of mass angle without a frame transformation.
9. Confusing the Coulomb scattering length a = k∕(2E) with the head-on closest approach k∕E = 2a.
10. Assuming the ideal Coulomb total cross section is finite when arbitrarily small scattering angles are included.
11. Forgetting that large-angle scattering corresponds to small impact parameter.
12. Forgetting that increasing energy decreases the cross section at fixed angle as E⁻².
13. Applying the point-charge Rutherford formula in regimes where screening, nuclear structure, relativity, or quantum corrections dominate.

34 Practice exercises

1. Starting from the impact-parameter annulus and solid-angle band, derive

   (165)

2. Derive Binet’s equation for a particle of mass μ moving under a central force.
3. Solve Binet’s equation for the repulsive Coulomb force F = k∕r².
4. Show that the repulsive Coulomb orbit can be written

   (166)

5. Derive

   (167)

6. Use the asymptote condition to prove

   (168)

7. Derive the Rutherford scattering function

   (169)

8. Derive the Rutherford differential cross section.
9. Show that dσ∕dΩ ∝ E⁻² at fixed angle.
10. Show that the small-angle Rutherford cross section scales approximately as χ⁻⁴.
11. Derive

    (170)

12. Derive the integrated cross section for χ ≥ χ₀.
13. Explain why the ideal unscreened Coulomb total cross section diverges as χ₀ → 0.
14. Derive the relationship between center of mass energy and projectile laboratory energy for a target initially at rest.
15. Derive

    (171)

35 Summary

For an axially symmetric classical scattering problem,

(172)

For repulsive Coulomb scattering,

(173)

The relative orbit is hyperbolic and the scattering function is

(174)

Therefore

(175)

For point charges,

(176)

The integrated cross section for scattering through at least χ₀ is

(177)

The clean Rutherford law completes the M04 progression from momentum and particle systems through central forces, two-body reduction, and classical scattering.

References

References