Monday, Sep. 20, 2010PHYS 3446, Fall 2010 Andrew Brandt 1 PHYS 3446 – Lecture #4 Monday, Sep. 20 2010 Dr. Brandt 1.Differential Cross Section of Rutherford

Monday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 1

PHYS 3446 – Lecture #4Monday, Sep. 20 2010

Dr. Brandt

1. Differential Cross Section of Rutherford Scattering2. Measurement of Cross Sections3. Lab Frame and Center of Mass Frame4. Relativistic Variables

***HW now due Weds***Labs due at next lab***Everyone should go to lab Friday 24 and Fri Oct. 1 weeks

as we will have a half lecture before each lab


Assignment #2 Due Weds. Sep. 221. Plot the differential cross section of the Rutherford scattering

as a function of the scattering angle for three sensible choices of the lower limit of the angle.

(use ZAu=79, Zhe=2, E=10keV).

2. Compute the total cross section of the Rutherford scattering in unit of barns for your cut-off angles.

3. Find a plot of a cross section from a current HEP experiment, and write a few sentences about what is being measured.

4. Book problem 1.10 *New Problem*• 5 points extra credit if you are one of first 10 people to email an

electronic version of a figure showing Rutherford 1/sin^4(x/2) angular dependence


Scattering Cross Section• For a central potential, measuring the yield as a

function of , or differential cross section, is equivalent to measuring the entire effect of the scattering

• So what is the physical meaning of the differential cross section?

This is equivalent to the probability of certain process in a specific kinematic phase space

• Cross sections are measured in the unit of barns:-24 21 barn = 10 cm


Total X-Section of Rutherford Scattering• To obtain the total cross section of Rutherford scattering, one

integrates the differential cross section over all :

• What is the result of this integration?– Infinity!!

• Does this make sense?– Yes

• Why?– Since the Coulomb force’s range is infinite (particle with very large impact

parameter still contributes to integral through very small scattering angle)• What would be the sensible thing to do?

– Integrate to a cut-off angle since after certain distance the force is too weak to impact the scattering. (0>0); note this is sensible since alpha particles far away don’t even see charge of nucleus due to screening effects.

Total 0

2 sind

dd

22 1

0 3

' 18 sin

4 2 sin2

ZZ ed

E


Measuring Cross Sections• With the flux of N0 per unit area per second• Any particles in range b to b+db will be

scattered into to -d• The telescope aperture limits the measurable

area to

• How could they have increased the rate of measurement?– By constructing an annular telescope– By how much would it increase?

TeleA

2/d

sinRd R d 2R d


Measuring Cross Sections

• Fraction of incident particles (N0) approaching the target in the small area =bddb at impact parameter b is dn/N0. – so dn particles scatter into R2d, the aperture of the telescope

• This fraction is the same as – The sum of over all N nuclear centers throughout the foil divided by the total

area (S) of the foil.– Or, in other words, the probability for incident particles to enter within the N

areas divided by the probability of hitting the foil. This ratio can be expressed as

0

dn

N Nbd d

S

,N

S Eq. 1.39


Measuring Cross Sections• For a foil with thickness t, mass density , atomic weight A:

• Since from what we have learned previously

• The number of scattered into the detector angle () is

N A0: Avogadro’s number of atoms per mole

dn

0

tSA

A

0 0N tA

A

0

,dNN dS d

,d

dd

0

dn

N

Nbd d

S

Eq. 1.40


Measuring Cross Sections

dn

• This is a general expression for any scattering process, independent of the theory

• This gives an observed counts per second

0 0N tA

A

0

,dNN dS d

,d

dd

Number of detected particles/sec

Projectile particle flux

Density of the target particles

Scattering cross section

Detector acceptance


Lab Frame and Center of Mass Frame• So far, we have neglected the motion of target nuclei in

Rutherford Scattering• In reality, they recoil as a result of scattering • This complication can best be handled using the Center

of Mass frame under a central potential• This description is also useful for scattering experiments

with two beams of particles (moving target)


Lab Frame and CM Frame• The equations of motion

can be written

1 1m r

2 2m r

i

where

Since the potential depends only on relative separation of the particles, we redefine new variables, relative coordinates & coordinate of CM

1 2r r r

CMR

and 1 1 2 2

1 2

m r m r

m m

1 1 2V r r

2 1 2V r r

ii

rr

i

i ir

ˆ

sini

i i ir

1,2i

2r


Now some simple arithmetic• From the equations of motion, we obtain

1 1 2 2m r m r

1 1 2 2

1 2CM

m r m rR

m m

1 2 1 2 m m r m r

2

1 2 21 2

mm m r m r

m m

• Since the momentum of the system is conserved:

• Rearranging the terms, we obtain

1 1 2 1m r m r r 1 2 1 2m m r m r

1 21 2

ˆ ˆ ˆ2r r V r r V rr r r

0 1 1 m r

2 2m r 2 1m r r

2

11 2

m

r rm m

1 2

1 2

mmr

m m

r V r

r

1 2

1 2

2m mr

m m


Lab Frame and CM Frame• From the equations in previous slides

1 2 CMm m R and

• What do we learn from this exercise?• For a central potential, the motion of the two particles can be decoupled

when re-written in terms of – a relative coordinate– The coordinate of center of mass

1 2

1 2

m mr

m m

Thus ˆconstant CMR R

Reduced Mass

r V r

ˆ

V rr

r

CMMR 0

quiz!


Lab Frame and CM Frame• The CM is moving at a constant velocity in the lab

frame independent of the form of the central potential• The motion is that of a fictitious particle with mass

(the reduced mass) and coordinate r.• Frequently we define the Center of Mass frame as the

frame where the sum of the momenta of all the interacting particles is 0.


Relationship of variables in Lab and CM

• The speed of CM in lab frame is

• Speeds of the particles in CM frame are

• The momenta of the two particles are equal and opposite!!

CM CMv R

1v

CM

2v and1 CMv v 2 1

1 2

m v

m m CMv 1 1

1 2

m v

m m

1 1

1 2

m v

m m

1v1v

CMVCMLab


• CM represents the change in the direction of the relative position vector r as a result of the collision in CM frame

• Thus, it must be identical to the scattering angle for a particle with reduced mass, .

• Z components of the velocities of scattered particle with m1 in lab and CM are:

• The perpendicular components of the velocities are: (boost is only in the z direction)• Thus, the angles are related (for elastic scattering only) as:

Scattering angles in Lab and CM

cos Lab CMv v

sin Labv

1

sintan

cosCM

LabCM CMv v

1 2

sin

cosCM

CM m m

sin

cosCM

CM

1 cos CMv

1 sin CMv

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Monday, Sep. 20, 2010PHYS 3446, Fall 2010 Andrew Brandt 1 PHYS 3446 – Lecture #4 Monday, Sep. 20 2010 Dr. Brandt 1.Differential Cross Section of Rutherford