MechanicsPhysics 151

Lecture 7Scattering Problem

(Chapter 3)

What We Did Last Time

Discussed Central Force ProblemsProblem is reduced to one equation

Analyzed qualitative behaviorsUnbounded, bounded, and circular orbitsCondition for stable circular orbits

Defined orbit equation and solved it for the Kepler problem

Conic orbits depending on E

2

3 ( )lmr f rmr

= +

2

2( ) ( )2

lV r V rmr

′ = +

2

2 2

1 21 1 cos( )mk Elr l mk

θ θ⎛ ⎞

′= + + −⎜ ⎟⎜ ⎟⎝ ⎠

Goals for Today

Introduce the scattering problemWhat happens when a particle “collides”

Define scattering cross sectionHow often a particle gets scattered in a given directionHow to calculate it from the potential

Examples1/r2 force Rutherford scatteringRainbow scattering

Scattering Problem

Consider unbound motion under central forceParticle comes from infinity goes to infinity

Assume f(r) → 0 as r → ∞Orbit approaches straight lines at large r

How are sections A and B related?

InteractionStraight section A

Straight section B

Why Scattering Problem?

Physical “observations” are scattering phenomenaPhotons scattered by an object SeeingElectrons scattered by an object Electron microscope

Experiments on microscopic objectsElectron-nucleus scattering to probe nuclear structureNeutrino-electron scattering to measure neutrino energy

Classical description fails with such targetsStill a good approximation in many casesClassical framework of describing scattering used in QM as well – and it’s more intuitive to understand

Lousy Shooter Model

Imagine shooting bullets at a small targetSuppose you have very poor aim

Bullets spread uniformlyNumber of bullets / area / time = intensity I

Number of hits will be proportional to the target size

hitsN I σ= ⋅ Target cross section (m2)

Intensity (bullets/m2/sec)

Hit frequency (bullets/sec)

Spherical Target

Imagine the target is a solid sphereWe want to know which direction the bullets ricochetNumber of bullets ricocheting into solidangle dΩ around a direction Ω is

Concentrate on the scattering angle ΘTarget is round = has rotational symmetry

Number of bullets between Θ and Θ + dΘ is

Θ( )N I dσ= ΩΩ Differential cross

section (m2/str)

( ) ( )sind d dσ σ φΩ = Θ Θ ΘΩ

2

0( )sin ( )2 sinN d I d I d

πφ σ σ π= Θ Θ Θ = Θ Θ Θ∫

Differential Cross Section

Classical mechanics is deterministicScattering angle Θ is determined by the impact parameter sProbability of scattering betweenΘ and Θ + dΘ is proportionalto the area of this ring ( )sΘ

sds

dΘ

2 ( )2 sinsds dπ σ π= Θ Θ Θ

( )sin

s dsd

σ Θ =Θ Θ

Absolute value taken becauseds/dΘ might be negative

Total Cross Section

Let’s check if this matches our idea of the total cross section

Integrating over the entire solid angle

( )sin

s dsd

σ Θ =Θ Θ

04

2

0

( ) 2 sin ( )

2

T

a

d d

sds a

π

π

σ σ π σ

π π

= = Θ Θ Θ

= =

∫ ∫

∫

Ω Ω

Total area ofthe target

Central Force Scattering

Now consider scattering by general central forceHow does Θ relate to s?We need to know the shape of the orbitat large r

Look at the orbit equation

Angular momentum l is related to s by

If we assume V(r) → 0 as r → ∞

( )sin

s dsd

σ Θ =Θ Θ

0 0sinl rp spθ= × = =0r pΘ

s

0p

2

2 2

(1/ ) 0d u m dV uud l duθ

+ + =

20

2pE Tm

= = 0 2l sp s mE= =

Central Force Scattering

One can in principle solve this equation and get

r → ∞ at θ = Θ

Then we can calculate

2

2 2

1 (1/ ) 02

d u dV uud s E duθ

+ + =

( , , )u u s Eθ=

( , , ) 0u s EΘ = Solve ( , )s s E= Θ

( , )sin

s dsEd

σ Θ =Θ Θ

Let’s look at the orbit we already know

Inverse-square force

Orbit equation in terms ofthe impact parameter s

and the energy E

Inverse Square Force

Consider a repulsive 1/r2 force

Equation and solution same as Kepler problemJust flip the sign of k

Radius > 0

2( ) kf rr

= ( ) kV rr

= Ex: electrostatic force between two like-sign charged particles

2

2 2

1 21 1 cos( )mk Elr l mk

θ θ⎛ ⎞

′= − + + −⎜ ⎟⎜ ⎟⎝ ⎠

2

2

21 1Elmk

ε = + >

Eccentricity

Hyperbola

Hyperbolic Orbit

Solution is a hyperbolaε > 1 E > 01/r > 0

Scattering angle is

2

2 2

1 21 1 cos( )mk Elr l mk

θ θ⎛ ⎞

′= − + + −⎜ ⎟⎜ ⎟⎝ ⎠

2

2

21 1Elmk

ε = + >

1cos( )θ θε

′− < −

θ ′

θ θ ′−

2πΘ = − ΨΨ

cos 1/ εΨ =

2 2cot 12

Esk

εΘ= − =A bit of

work

2l s mE=

cot2 2ksE

Θ=

We’ve got what we need!

Differential Cross Section

Differential cross section is

Scattering of particles with charges Ze and Z’e

2

2

42

1( ) cot cotsin sin 2 2 2

1 14 2 sin

s ds k dd E d

kE

σ

Θ

Θ Θ⎛ ⎞ ⎛ ⎞Θ = = ⎜ ⎟ ⎜ ⎟Θ Θ Θ Θ⎝ ⎠ ⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

cot2 2ksE

Θ=

2k ZZ e′=22

42

1 1( )4 2 sin

ZZ eE

σΘ

′⎛ ⎞Θ = ⎜ ⎟

⎝ ⎠

Rutherford scattering:α particle (Z’ = 2) scatteredby atomic nuclei with Z

Existence of nuclei in atoms

Rutherford Scattering

Before Rutherford’s discoveryElectron was known to exist in matterPositive charge must exist in atoms, but the distribution was unknown

Measurement of σ(Θ) showed

Positive charge of +Ze is in one particlee.g. Z particles of +e each would give

Discovery of atomic nuclei

22

42

1 1( )4 2 sin

ZZ eE

σΘ

′⎛ ⎞Θ = ⎜ ⎟

⎝ ⎠22

42

14 2 sinZ Z e

E Θ

′⎛ ⎞⎜ ⎟⎝ ⎠

Total Cross Section

Integrating Rutherford cross section gives

Because electrostatic force is long-rangeNo matter how large is the impact parameter s, the particle still gets slightly deflectedReality: electrostatic field is shielded by the electrons around the nucleus Finite cross section

22

4024

22 12

302

1 1( ) 2 sin4 2 sin

(sin )22 sin

TZZ ed d

E

dZZ eE

π

π

σ σ π

π

Θ

Θ

Θ

′⎛ ⎞= = Θ Θ⎜ ⎟

⎝ ⎠

′⎛ ⎞=⎜

⎝ ⎠∞= ⎟

∫ ∫

∫

Ω Ω

Rainbow Scattering

Equation for σ(Θ) assumes that s(Θ) is single-valuedTrue for Rutherford scattering, but not always

If s(Θ) is not monotonous

At maximum Θ = Θm

Called rainbow scattering

( )sin

s dsd

σ Θ =Θ Θ

Θ

2s

1s

mΘ

( )sin

i i

i

s dsd

σ Θ =Θ Θ∑ Sum up for

possible s’s

0ddsΘ

= ( )σ Θ = ∞

( )σ Θ

Rainbow

You’ve probably heard of how rainbows are madeBut the scattering angle depends onwhere the light enters the dropIf you add up all possible positions,rainbow will be washed out

Real rainbow is made bythe light that reflectedinternally

Total deflection is

They lied

1θ

2θ2θs

2θ2θ

1θ

Θ1 22 4θ θ πΘ = − +

From Physics 15c

Rainbow

Θ has a minimum around 137.5°Illuminate a water dropletwith uniform light

What is the distribution oflight intensity in Θ?A bit difficult problem

Covered in Physics 143a and 151

The answer:

1sin sR

θ = 1 2sin sinnθ θ=1 22 4θ θ πΘ = − +

Θ

s R

π 1.33n =

min 2.40Θ =

( )sin

s dsId

Θ ∝Θ Θ

This goes to infinityat the turning point there

From Physics 15c

Rainbow

Minimum of Θ Sharp peak of intensity I(Θ)

Reflection observed only atΘmin This depends on n,which depends slightly on λThis is really how rainbowis created

Θ

s R

minΘ

ΘminΘ

( )I Θ ∞

mins r > rmin

r < rmin

minΘ

From Physics 15c

Attractive Force

Repulsive force can only scatter byAttractive force can do more

If the potential and the energy are just right,particle can make multiple turns before emergingCalled spiraling or orbiting

0 π< Θ <

r

( )V r′

E

Orbiting region:E – V’ is small

r varies slowly

Summary

Discussed scattering problemFoundation for all experimental particle physics

Defined and calculated cross sectionsDifferential cross sectionRutherford scattering

Done with central force problemsNext: Rigid Bodies

( )sin

s dsd

σ Θ =Θ Θ

hitsN I σ= ⋅