PHYS 3446 – Lecture #4
Monday, Sep. 20 2010
Dr. Brandt
1.
2.
3.
4.
Differential Cross Section of Rutherford Scattering
Measurement of Cross Sections
Lab Frame and Center of Mass Frame
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
Monday, Sep. 20, 2010
PHYS 3446, Fall 2010 Andrew Brandt
1
Assignment #2 Due Weds. Sep. 22
1.
Plot the differential cross section of the Rutherford scattering
as a function of the scattering angle q for three sensible
choices of the lower limit of the angle.
(use ZAu=79, Zhe=2, E=10keV).
2.
3.
4.
•
Compute the total cross section of the Rutherford scattering in
unit of barns for your cut-off angles.
Find a plot of a cross section from a current HEP experiment,
and write a few sentences about what is being measured.
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
Monday, Sep. 20, 2010
PHYS 3446, Fall 2010 Andrew Brandt
2
Scattering Cross Section
• For a central potential, measuring the yield as a
function of q, 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:
1 barn = 10
Monday, Sep. 20, 2010
-24
cm
PHYS 3446, Fall 2010 Andrew Brandt
2
3
Total X-Section of Rutherford Scattering
• To obtain the total cross section of Rutherford scattering, one
integrates the differential cross section over all q:
 T o ta l = 2 


0
 ZZ 'e 
 q  sin q d q = 8 

4
E
d


d
2
2
• What is the result of this integration?

q  1

d  sin 
0
2
3 q

sin
2
1
– 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. (q=q0>0); note this is sensible since alpha particles far
away don’t even see charge of nucleus due to screening effects.
Monday, Sep. 20, 2010
PHYS 3446, Fall 2010 Andrew Brandt
4
Measuring Cross Sections
• With the flux of N0 per unit area per second
• Any a particles in range b to b+db will be
scattered into q to q-dq
• The telescope aperture limits the measurable
area to
AT ele = Rd q  R sin q d f = R 2 d 
• How could they have increased the rate of
measurement?
– By constructing an annular telescope
– By how much would it increase? 2/df
Monday, Sep. 20, 2010
PHYS 3446, Fall 2010 Andrew Brandt
5
Measuring Cross Sections
• Fraction of incident particles (N0) approaching the target in the small
area D=bdfdb 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 D 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

dn
N0
Monday, Sep. 20, 2010
=
N
S
D  q , f  =
PHYS 3446, Fall 2010 Andrew Brandt
N bdf dq
Eq. 1.39
S
6
Measuring Cross Sections
• For a foil with thickness t, mass density r, atomic weight A:
N =
r tS
A0
A0: Avogadro’s number
of atoms per mole
A
• Since from what we have learned previously
N bdf dq
dn

=
S
N0
• The number of a scattered into the detector angle (q,f) is
d  q , f 
N d  q , f 
N
r
tA
0
0
dn =
d = N0
d
d
S
d
A
Eq. 1.40
Monday, Sep. 20, 2010
PHYS 3446, Fall 2010 Andrew Brandt
7
Measuring Cross Sections
dn =
N 0 r tA0 d   q , f 
A
Number of
detected
particles/sec
d
d = N0
Projectile
particle
flux
Detector
acceptance
N d  q , f 
S
d
Density of
the target
particles
d
Scattering
cross
section
• This is a general expression for any
scattering process, independent of the
theory
• This gives an observed counts per second
Monday, Sep. 20, 2010
PHYS 3446, Fall 2010 Andrew Brandt
8
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)
Monday, Sep. 20, 2010
PHYS 3446, Fall 2010 Andrew Brandt
9
Lab Frame and CM Frame
• The equations of
motion can be written
m1 r1 =   V  r  r 
m 2 r2 =   V  r  r 
r2
1
1
2
where
 i = rˆi

 ri

qˆi 
ri  q i

fˆi
2
1

ri sin q i  f i
2
i = 1, 2
Since the potential depends only on relative separation of the particles, we
redefine new variables, relative coordinates & coordinate of CM
r = r1  r2
Monday, Sep. 20, 2010
and
RC M =
m 1 r1  m 2 r2
m1  m 2
PHYS 3446, Fall 2010 Andrew Brandt
10
Now some simple arithmetic
• From the equations of motion, we obtain
m r  m r = m r  m r  r  = m  m r  m r
1 1
1 1
2 2



=   rˆ1
 rˆ2
 r2
  r1
2

V

1
1
 
r
=  2 rˆ

r
V
2
1
2
r 
• Since the momentum of the system is conserved:
m1 r1  m 2 r2
m1  m 2

= R C M = 0  m1 r1 =  m 2 r2 =  m 2 r1  r

• Rearranging the terms, we obtain
 m1  m 2  r1 = m 2 r
 m1  m 2 
m2
 m1  m 2 
Monday, Sep. 20, 2010
 r1 =
r  m2r =
m2
 m1  m 2 
2 m1 m 2
 m1  m 2 
r
r
m1 m 2
 m1  m 2 
PHYS 3446, Fall 2010 Andrew Brandt
r =  rˆ

r
V
r 
11
Lab Frame and CM Frame
• From the equations in previous slides
m1 m 2
m1  m 2
r   r =  V
r =
V
r 
r
rˆ
Reduced
Mass
and  m1  m 2  RC M = M R C M = 0
Thus
R C M = co n stan t Rˆ
• 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
Monday, Sep. 20, 2010
PHYS 3446, Fall 2010 Andrew Brandt
quiz!
12
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.
Monday, Sep. 20, 2010
PHYS 3446, Fall 2010 Andrew Brandt
13
Relationship of variables in Lab and CM
CM
v1
q Lab
v1
q CM
VC M
• The speed of CM in lab frame is
vC M = RC M =
m 1 v1
m1  m 2
• Speeds of the particles in CM frame are
m 2 v1
m 1 v1
v1 = v1  v C M =
and v 2 = v C M =
m1  m 2
m1  m 2
• The momenta of the two particles are equal and opposite!!
Monday, Sep. 20, 2010
PHYS 3446, Fall 2010 Andrew Brandt
14
Scattering angles in Lab and CM
• qCM 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:
v cos q L ab  v C M = v1 cos q C M
• The perpendicular components of the velocities are:
v sin q L ab = v1 sin q C M (boost is only in the z direction)
• Thus, the angles are related (for elastic scattering only) as:
tan q L a b =
sin q C M
cos q C M  v C M v1
Monday, Sep. 20, 2010
=
sin q C M
cos q C M  m1 m 2
PHYS 3446, Fall 2010 Andrew Brandt
=
sin q C M
co s q C M  
15