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= ⋅