1. Hour 38 Scattering Cross Sections
Physics 321
Hour 38
Scattering Cross Sections

2. The Scattering Problem
In the cm frame, two particles approach each other and scatter. Given the interaction, what is the distribution of scattered particles? 𝑃
𝑃 0 Θ Θ
𝑃 0 𝑃

3. The Inverse Scattering Problem
Given the distribution of scattered particles, what is the interaction?
This is a very hard but very interesting problem. 𝑃
𝑃 0 Θ Θ
𝑃 0 𝑃

4. The Scattering Problem
The impact need not be head-on. – The distance between the trajectories is the impact parameter, b. Θ 𝑃
𝑃 0 b

5. Relating θ to b
The first thing we need to do is find the relationship between θ and b for a given type of interaction. Θ b Θ

6. Hard Sphere Scattering – b(θ)
cos 𝜗 2 = sin 𝜋 2 − 𝜗 2 = 𝑏 𝑎
𝑏=𝑎 sin 𝜋−𝜗 2

7. Coulomb Scattering – b(θ)
First we note that the total angular momentum is ℓ=𝑝𝛼𝑏+𝑝𝛽𝑏
=𝑝𝑏
𝐸=𝑇 ∞ = 𝑝 2 2𝜇 = ℓ 2 2𝜇𝑏 2

8. Coulomb Scattering – b(θ)
Next we use a result from the central force problem.
𝐸= 𝛾 2 𝜇 2 ℓ 2 𝜀 2 −1
𝛾= 𝑘 𝑒 𝑞 1 𝑞 2

9. Coulomb Scattering – b(θ)φ
For hyperbolic orbits
𝑟 𝜑 = 𝐶 1+𝜀 cos 𝜑
cos 𝜑 𝑚𝑎𝑥 =− 1 𝜀
𝜑 𝑚𝑎𝑥 = 𝜋 2 + 𝜗 2
cos 𝜋 2 + 𝜗 2 = −sin 𝜗 2
= − 1 𝜀

10. Coulomb Scattering – b(θ)
Rotate the figure so that 𝜑 is measured with respect to the x-axis, as in Chapter 10. φ
𝑟 𝜑 = 𝐶 1+𝜀 cos 𝜑
As 𝑟→∞,
1+𝜀 cos 𝜑 →0
𝜑 𝑚𝑎𝑥 = 𝜋 2 + 𝜗 2

11. Coulomb Scattering – b(θ)φ
For hyperbolic orbits
𝑟 𝜑 = 𝐶 1+𝜀 cos 𝜑
cos 𝜑 𝑚𝑎𝑥 =− 1 𝜀
𝜑 𝑚𝑎𝑥 = 𝜋 2 + 𝜗 2
cos 𝜋 2 + 𝜗 2 = −sin 𝜗 2
= − 1 𝜀

12. Coulomb Scattering – b(θ)
Using
sin 𝜗 2 = 1 𝜀 φ
𝜗 2 𝜀
𝜀 2 −1
𝐸= 𝛾 2 𝜇 2 ℓ 2 𝜀 2 −1
= 𝛾 2 𝜇 2 ℓ 2 cot 2 𝜗 2 1

13. Coulomb Scattering – b(θ)
𝐸= 𝛾 2 𝜇 2 ℓ 2 cot 2 𝜗 2
ℓ 2 =2𝜇 𝑏 2 𝐸
cot 2 𝜗 2 = 2 2𝜇 𝑏 2 𝐸 𝐸 𝛾 2 𝜇 = 4 𝑏 2 𝐸 2 𝛾 2
cot 𝜗 2 = 2𝑏𝑇 0 𝛾

14. Differential Cross Section - Definition
Consider a small beam striking a large target. The detector has solid angle ΔΩ and detects all particles scattered. θ
detector
beam
target
𝑑𝜎 𝑑Ω = # 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑑𝑒𝑡𝑒𝑐𝑡𝑒𝑑/𝑠𝑒𝑐 # 𝑜𝑓 𝑏𝑒𝑎𝑚 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠/ sec × # 𝑡𝑎𝑟𝑔𝑒𝑡 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑎𝑟𝑒𝑎 ×∆Ω

15. Differential Cross Section - Definition
Consider a large beam striking a small target. A detector has solid angle ΔΩ. θ
detector
beam
target
𝑑𝜎 𝑑Ω = # 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑑𝑒𝑡𝑒𝑐𝑡𝑒𝑑/𝑠𝑒𝑐 # 𝑜𝑓 𝑏𝑒𝑎𝑚 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑎𝑟𝑒𝑎 ∙ 𝑠𝑒𝑐 ×# 𝑡𝑎𝑟𝑔𝑒𝑡 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠×∆Ω

16. Differential Cross Section - Theoretical
We usually take 1 target particle… θ
detector
beam
target
𝑑𝜎 𝑑Ω = # 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑑𝑒𝑡𝑒𝑐𝑡𝑒𝑑/𝑠𝑒𝑐 # 𝑜𝑓 𝑏𝑒𝑎𝑚 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑎𝑟𝑒𝑎 ∙ 𝑠𝑒𝑐 ×1×∆Ω

17. Differential Cross Section - Theoretical
𝑑𝜎 𝑑Ω = # 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑑𝑒𝑡𝑒𝑐𝑡𝑒𝑑/𝑠𝑒𝑐 # 𝑜𝑓 𝑏𝑒𝑎𝑚 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑎𝑟𝑒𝑎 ∙ 𝑠𝑒𝑐 ×1×∆Ω
Find the total number of beam particles per second between 𝑏→𝑏+∆𝑏. This is the number in a ring of radius 𝑏.
𝑁≡ # 𝑜𝑓 𝑏𝑒𝑎𝑚 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑎𝑟𝑒𝑎 ∙ 𝑠𝑒𝑐
This is: 𝑁 2𝜋𝑏 𝑑𝑏

18. Differential Cross Section - Theoretical
𝑑𝜎 𝑑Ω = # 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑑𝑒𝑡𝑒𝑐𝑡𝑒𝑑/𝑠𝑒𝑐 𝑁 ∆Ω
2) Find the total number of scattered particles between 𝜗→𝜗+∆𝜗.
𝑏=𝑏 𝜗
𝑑𝑏= 𝑏 ′ 𝜗 𝑑𝜗
This is: 𝑁 2𝜋𝑏 𝑑𝑏=𝑁 2𝜋𝑏 𝑏 ′ 𝜗 𝑑𝜗

19. Differential Cross Section - Theoretical
𝑑𝜎 𝑑Ω = # 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑑𝑒𝑡𝑒𝑐𝑡𝑒𝑑/𝑠𝑒𝑐 𝑁 ∆Ω
3) Think of the detector as subtending angles ∆𝜗 and ∆𝜑. The solid angle is defined by:
∆Ω= sin 𝜗 ∆𝜗 ∆𝜑
or more generally
∆Ω= sin 𝜗 𝑑𝜗 𝑑𝜑
(If the area of the detector is small and a distance R from the target, this can be expressed in terms of the detector’s area: 𝐴≈ 𝑅 2 ∆Ω.)
This is: 𝑁 2𝜋𝑏 𝑑𝑏=𝑁 2𝜋𝑏 𝑏 ′ 𝜗 𝑑𝜗

20. Differential Cross Section - Theoretical
𝑑𝜎 𝑑Ω = # 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑑𝑒𝑡𝑒𝑐𝑡𝑒𝑑/𝑠𝑒𝑐 𝑁 ∆Ω
4) The number of particles per second scattering into 𝑑𝜗 is:
𝑁 2𝜋𝑏 𝑏 ′ 𝜗 𝑑𝜗
The number of those hitting the detector is
𝑁 2𝜋𝑏 𝑏 ′ 𝜗 𝑑𝜗× ∆𝜑 2𝜋

21. Differential Cross Section - Theoretical
The number of particles detected per second is
𝑁 𝑏 𝑏 ′ 𝜗 𝑑𝜗𝑑𝜑
𝑑𝜎 𝑑Ω = 𝑁 𝑏 𝑏 ′ 𝜗 𝑑𝜗𝑑𝜑 𝑁 sin 𝜗 𝑑𝜗𝑑𝜑 = 𝑏 𝑏 ′ 𝜗 sin 𝜗
Or since we want a positive value, we usually write
𝑑𝜎 𝑑Ω = 1 sin 𝜗 𝑏 𝑑𝑏 𝑑𝜗

22. Differential Cross Section - Theoretical
Or since we want a positive value, we usually write
𝑑𝜎 𝑑Ω = 1 sin 𝜗 𝑏 𝑑𝑏 𝑑𝜗
This is the usual “book form,” but it is usually more convenient to write this as:
𝑑𝜎 𝑑Ω = 𝑏 2 sin 𝜗 2 cos 𝜗 𝑑𝑏 𝑑𝜗

23. Total Cross Section
The total cross section is the differential cross section integrated over all angles.
𝜎= 𝑑𝜎 𝑑Ω 𝑑Ω= 𝑑𝜎 𝑑Ω sin 𝜗𝑑𝜗𝑑𝜑

24. Hard Sphere Scattering (cm)
𝑏= 𝑟 1 + 𝑟 2 sin 𝜋−𝜗 2 = 𝑟 1 + 𝑟 2 cos 𝜗 2
𝑑𝑏 𝑑𝜗 =− 𝑟 1 + 𝑟 2 cos 𝜋−𝜗 2 =− 𝑟 1 + 𝑟 2 sin 𝜗 2
𝑑𝜎 𝑑Ω = 1 2cos 𝜗 2 sin 𝜗 𝑟 1 + 𝑟 cos 𝜗 2 sin 𝜗 2
𝑑𝜎 𝑑Ω = 𝑟 1 + 𝑟 2 2

25. Hard Sphere Scattering (cm)
𝑑𝜎 𝑑Ω = 𝑟 1 + 𝑟 2 2
𝜎= 𝑑𝜎 𝑑Ω 𝑑Ω=𝜋 𝑟 1 + 𝑟 2 2

26. Rutherford Scattering (cm)
𝑏= 𝛾 2𝑇 0 cot 𝜗 2
𝑑𝑏 𝑑𝜗 = − 𝛾 2𝑇 csc 2 𝜗 2
𝑑𝜎 𝑑Ω = 1 2cos 𝜗 2 sin 𝜗 𝛾 2𝑇 cot 𝜗 csc 2 𝜗 2
𝑑𝜎 𝑑Ω = 1 sin 4 𝜗 𝛾 4𝑇 = 𝛾 𝑇 sin 4 𝜗 2

27. CM to Lab Conversion The only thing that changes in lab vs cm is ∆Ω.
𝑑𝜎 𝑑Ω = # 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑑𝑒𝑡𝑒𝑐𝑡𝑒𝑑/𝑠𝑒𝑐 # 𝑜𝑓 𝑏𝑒𝑎𝑚 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠/ sec × # 𝑡𝑎𝑟𝑔𝑒𝑡 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑎𝑟𝑒𝑎 ×∆Ω
=−∆ cos 𝜗 𝑙𝑎𝑏 ∆ 𝜑 𝑙𝑎𝑏
∆ Ω 𝑐𝑚 = −∆ cos 𝜗 𝑐𝑚 ∆ 𝜑 𝑐𝑚
∆ 𝜑 𝑙𝑎𝑏 =∆ 𝜑 𝑐𝑚
→ 𝑑𝜎 𝑑Ω 𝑙𝑎𝑏 = 𝑑𝜎 𝑑Ω 𝑐𝑚 × 𝑑 cos 𝜗 𝑐𝑚 𝑑 cos 𝜗 𝑙𝑎𝑏