Mechanics Scattering Problem, Lecture Notes - Physics, Study notes of Mechanics

Download Mechanics Scattering Problem, Lecture Notes - Physics and more Study notes Mechanics in PDF only on Docsity!

Mechanics

Physics 151

Lecture 7

Scattering Problem

(Chapter 3)

What We Did Last Time „^

Discussed Central Force Problems^ „

Problem is reduced to one equation „^

Analyzed qualitative behaviors^ „

Unbounded, bounded, and circular orbits „ Condition for stable circular orbits „^

Defined orbit equation and solved it for the Keplerproblem^ „

Conic orbits depending on

E

2 3

l

mr

f^

r

mr =^

2 2

l 2

V^

r^

V r

mr

′^

=^

2

2

2

1

2

1

1

cos(

)

mk

El

r^

l^

mk

θ^

θ

⎛^

⎞ ′

=^

+^

+^

−

⎜^

⎟

⎜^

⎟

⎝^

⎠

Scattering Problem „^

Consider unbound motion under central force^ „

Particle comes from infinity

Æ

goes to infinity

„^

Assume

f (

r )

0 as

r

„^

Orbit approaches straight lines at large

r

„^

How are sections A and B related?

Interaction

Straight section A

Straight section B

Why Scattering Problem? „^

Physical “observations” are scattering phenomena^ „

Photons scattered by an object

Æ

Seeing

„^

Electrons scattered by an object

Æ

Electron microscope

„^

Experiments on microscopic objects^ „

Electron-nucleus scattering to probe nuclear structure „ Neutrino-electron scattering to measure neutrino energy „^

Classical description fails with such targets^ „

Still a good approximation in many cases „ Classical framework of describing scattering used in QM aswell – and it’s more intuitive to understand

Spherical Target „^

Imagine the target is a solid sphere^ „

We want to know which direction the bullets ricochet „ Number of bullets ricocheting into solidangle

d

around a direction

is

„^

Concentrate on the scattering angle

„^

Target is round = has rotational symmetry „^

Number of bullets between

and

• d

is

(^

N^

I^

d

=^

Differential crosssection (m

2 /str)

(^

)^

(^

) sin

d^

d^

d

2 0

(^

) sin

(^

sin

N^

d^

I^

d^

I^

d

π^

σ^

=^

∫

Differential Cross Section „^

Classical mechanics is deterministic^ „

Scattering angle

is determined by the impact parameter

s

„^

Probability of scattering between Θ

and

d

is proportional

to the area of this ring

( ) s Θ

s ds

d Θ

(^

sin

sds

d

π^

σ^

=^

(^

)^

sin

s^

ds d

σ Θ =

Absolute value taken because

ds / d Θ

might be negative

Central Force Scattering „^

Now consider scattering by general central force^ „

How does

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^

ds d

σ Θ =

0

0

sin

l^

rp

sp θ

=^

×^

=^

0 r^

p

s p^0

2 2

2

)^

d u

m dV

u

u

d^

l^

du

θ^

+^

+^

(^20) p 2

E^

T^

m

=^

=^

0

l^

sp

s^

mE

=^

Central Force Scattering

„^

One can in principle solve this equation and get „^

r^ → ∞

at

θ^

=^

„^ 2 2 Then we can calculate

d u

dV

u

u

d^

s E

du

θ^

+^

+^

u^

u^

s E θ = (^

)^

u^

s E Θ

=^

Solve

(^

,^

s^

s^

E

=^

(^

,^

)^

sin

s^

ds

E^

d

=^

Let’s look at the orbit we alreadyknow Æ

Inverse-square force

Orbit equation in terms ofthe impact parameter

s

and the energy

E

Hyperbolic Orbit

„^

Solution is a hyperbola^ „

ε^

Æ

E

„^

r^ > 0

Æ

„^

Scattering angle is

2

2

2

cos(

mk

El

r^

l^

mk

θ^

θ

⎛^

+^

+^

⎜^

⎜^

⎝^

(^22) 2 1

El mk

ε =

+^

cos(

θ^

θ^

ε

′ −^

θ^ θ ′

θ ′−

π Θ =

−^

cos

ε

Ψ

2

cot

Es k

ε

Θ

=^

−^

A bit ofwork

l^

s^

mE

=

cot 2

k s^

E

We’ve got what we need!

Differential Cross Section „^

Differential cross section is^ „

Scattering of particles with charges

Ze

and

Z’e

2 2

4 2

(^

)^

cot

cot

sin

sin

sin

s^

ds

k^

d

d^

E^

d

k E

Θ

⎛^

⎞^

⎛^

=^

⎜^

⎟^

⎜^

⎝^

⎠^

⎝^

⎛^

=^

⎜^

⎝^

cot 2

k

s^

E

2

k^

ZZ e

2 2

4 2

(^

)^

sin

ZZ e

E

σ^

Θ

⎛^

⎜^

⎝^

Rutherford scattering: α^ particle (

Z’

= 2) scattered

by atomic nuclei with

Z

Existence of nuclei in atoms

Total Cross Section „^

Integrating Rutherford cross section gives^ „

Because electrostatic force is long-range „ No matter how large is the impact parameter

s , the particle

still gets slightly deflected „^

Reality: electrostatic field is shielded by the electrons aroundthe nucleus

Æ

Finite cross section

2 2

4

0

2

4

2 2

1

2 3

0

2

(^

)^

sin

sin

(sin

sin

T

ZZ e

d^

d

E

d

ZZ e

E

π

π

σ^

Θ

Θ Θ ′ ⎛^

=^

=^

⎜^

⎝^

⎛^

⎞^

⎜ ⎝^

=^

∫^

∫

∫

Rainbow Scattering „^

Equation for

σ

) assumes that

s

) is single-valued

„^

True for Rutherford scattering, but not always

„^

If

s

) is not monotonous

„^

At maximum

m

„^

Called rainbow scattering

(^

)^

sin

s^

ds d

s^2 s^1

m Θ

(^

)^

sin

i^

i

i

s^

ds d

σ^

∑

Sum up forpossible

s ’s

d^ Θ ds

=^

(^

σ^

(^

σ^

Rainbow

„^

has a minimum around 137.5°

„^

Illuminate a water dropletwith uniform light^ „

What is the distribution oflight intensity in

„^

A bit difficult problem^ „

Covered in Physics 143a and 151 „^

The answer:

1 sin

s R

1

2

sin

sin n

1

2

θ^

θ^

−^

s R

π^

n^

min

(^

)^

sin

s^

ds

I^

d

This goes to infinity at the turning point there

From Physics 15c

Rainbow „^

Minimum of

Æ

Sharp peak of intensity

I

„^

Reflection observed only at Θ

min

Å

This depends on

n

which depends slightly on

„^ Θ This is really how rainbowis created

s R

min Θ

min Θ

(^

I^

s^^ min

r^ >

r min r^ <

r min Θmin

From Physics 15c