PHY6200 W07

Chapter 14: Collision Theory


In Taylor, read sections 14.4 - 14.6 for today, and 15.1 to 15.3 for Monday.

The Differential Scattering Cross Section

In the definition of the scattering cross section, we counted any scattering. It is possible to be more specific, and ask how many scattering occur in a particular direction.

Solid Angle

ΔOmega; = A/r²
dΩ = sinθ dθ dφ

The Differential Cross Section

Nsc(into dΩ) = Ninc ntar dσ(into dΩ)
dσ(into dΩ) = (dσ/dΩ) dΩ
Nsc(into dΩ) = Ninc ntar (dσ/dΩ) dΩ
σ = ∫ (dσ/dΩ) dΩ = ∫ 0π sinθ dθ ∫ 0 dφ (dσ/dΩ)(θ, φ)

Example: Angular Distribution of Scattered Neutrons

The differential cross section for scattering neutrons off a heavy nucleus might have the form dσ/dΩ = σ0(1 + 3cosθ + 3cos²θ) where σ0 is a constant.

Notice that the differential cross section is independent of the azimuthal angle φ. This is quite common -- one usually needs to go to some effort to break azimuthal symmetry, for instance by aligning spins in a magnetic field. To plot this function, look at a few special angles. At θ = 0, dσ/dΩ = 7σ0. At θ = π, dσ/dΩ = σ0. At θ = π/2, dσ/dΩ = σ0. And find any minima and maxima from the zeroes of the derivative, d²σ/dΩdθ = -3σ0sinθ(1 + 2cosθ). This has zeroes at θ = 0 and π, and when cosθ = -1/2, or θ = 2π/3. The first two are maxima, and the third is a minimum. We already have the value of the function at the maxima, when cosθ = -1/2 we find dσ/dΩ = σ0/4. Note that the minimum is positive, as it must be.

Calculating the Differential Cross Section

To calculate the differential cross section, we need to know the part of the cross section particles must hit

dσ = 2πb db

to scatter into the direction of interest

dΩ = 2π sinθ dθ.
The differential cross section is then the ratio of these two quantities (more precisely, we find &Delta&sigma, and &Delta&Omega, and we calculate the limit of the ratio as ΔΩ goes to zero, but this works equally well here)
(dσ/dΩ) = (b/sinθ) |db/dθ|

where the absolute value sign is inserted because normally the scattering angle θ decreases as the impact parameter b increases. The problem now is to determine the relation between impact parameter and scattering angle and evaluate the derivative db/dθ.

Example: Hard Sphere Scattering

Find the differential cross section for scattering of a point projectile off a fixed rigid sphere of radius R. Integrate the differential cross section to find the total cross section.

Use the knowledge that for hard sphere scattering, the angle of reflection equals the angle of incidence. The angle of incidence is α = sin-1(b/R) where b is the impact parameter. The scattering angle is θ = π - 2α = π - 2sin-1(b/R). Invert this to find b as a function of θ. This yields b = Rsin(π/2 - θ/2) = Rcos(θ/2). We need the derivative db/dθ = -(R/2)sin(θ/2). Put the pieces together to find (for 0 < θ < π, check sign for -π < θ < 0)

(dσ/dΩ) = (b/sinθ)|db/dθ| = [Rcos(θ/2)/sinθ](R/2)sin(θ/2) = R²/4
The differential cross section is isotropic, meaning there is equal probability to scatter into any element of solid angle. The total cross section is
σ = ∫(dσ/dΩ)dΩ = R²/4 ∫ dΩ = πR².
That is, the total cross section is the circular cross section of the sphere, as expected.

Rutherford Scattering

For scattering of particles by a conservative force, we can begin by choosing a coordinate system so that the orbit lies in the x-y plane, and writing down the Lagrangian in polar coordinates

L = ½mrdot² + ½mr²φdot² - U(r)

where U(r) is the potential for the conservative force. The Lagrange equation for φ yields mr²φdot = constant = L, where L is the total angular momentum. The total angular momentum is related to the impact parameter by L = mvb where mv is the momentum of the projectile when far from the target. When far from the target, the projectile's energy is purely kinetic, so E = ½mv², so we can also write L = b√(2mE). Instead of proceeding to the r equation, it is easier to start from the knowledge that the total energy is constant.

E = ½mrdot² + ½mr²φdot² + U(r)

Use the angular momentum relation to eliminate φdot, separate variables (r and t), and integrate

E = ½mrdot² + ½(L²/mr²) + U(r)

Find rmin by setting rdot to zero and solving for r.

dr/dt = √{(2/m)(E - U(r)) - (L/mr)²}

Since we are interested in solving for the scattering angle, not the orbit. The angular change of the orbit can be found by again using the angular momentum expression to replace dt by (mr²/L)dφ and separating differentials to integrate

α = ∫ dφ = ∫rmin (L/r²) dr /√{(2m)(E - U(r)) - (L/r)²}

And finally, the scattering angle is θ = π - 2α.

Scattering by Electrostatic Force

Now we'll find the differential cross section for the case that U = c/r. (c = kqQ = qQ/4πε0 where q and Q are the charges of the projectile and target, respectively.) The potential is attractive if k is negative (opposite sign charges between projectile and target) and repulsive if k is positive (same sign charges). We will see that the scattering is the same in either case.

We solved this problem in chapter 8, and found

r(φ) = (L²/cm)/[εcosφ - 1]

where E = (c²m/2L²)(&epsilon² - 1). (This differs from 8.49 by a minus sign. This arises because the derivation in chapter 8 contains an explicit minus sign on the force. If you propagate a change in sign through the derivation, you will arrive at the result above.)

The energy is greater than zero for scattering, meaning that ε>1. The distance of closest approach occurs when φ = 0 yielding rmin = L²/[cm(ε - 1)]. The radius of the orbit goes to infinity for φ = ±φ0 where φ0 = cos-1(1/ε). The scattering angle is θ = π - 2φ0. Since 1/ε = cosφ0, &epsilon² - 1 = cot²φ0, and using L² = 2mEb² we find that &epsilon² - 1 = (2Eb/c)². Therefore cotφ0 = 2Eb/c, or

b = (c/2E)cot(θ/2).

The derivative is

db/dθ = -(c/4E)csc²(θ/2)

Use this to find the differential cross section

dσ/dΩ = (c/2E)[cot(θ/2)/sinθ](c/4E)csc²(θ/2)

Use the relation sinθ = 2sin(θ/2)cos(θ/2) and simplify to get

dσ/dΩ = (c/4E)²/sin4(θ/2) = [c/4Esin²(θ/2)]²

This is the Rutherford Scattering formula. It predicts a strong angular dependence, but yet does allow for a measurable amount of scattering into the backward hemisphere. Were the atom constructed with its charges more or less mixed, like in the so-called "plum pudding" model, then the force law would be considerably different, much more like the 1/r³ force in your homework problem.

Cross Sections in Various Frames*

Optional, won't be covered in lecture.

Relation of the CM and Lab Scattering Angles*

Optional, won't be covered in lecture.

© 2007 Robert Harr