Monday, Sep. 13, 2010PHYS 3446, Fall 2010 Andrew Brandt 1 PHYS 3446 – Lecture #3 Monday, Sep. 13...
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Transcript of Monday, Sep. 13, 2010PHYS 3446, Fall 2010 Andrew Brandt 1 PHYS 3446 – Lecture #3 Monday, Sep. 13...
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 1
PHYS 3446 – Lecture #3Monday, Sep. 13 2010
Dr. Brandt
1. Rutherford Scattering with Coulomb force2. Scattering Cross Section3. Differential Cross Section of Rutherford Scattering4. Measurement of Cross Sections
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 2
Elastic Scattering
• From momentum and kinetic energy conservation
mt
m
m
mt
After Collision0v
v
tv
20v
2 1 tt
mv
m
2 2tt
mv v
m
2 tv v
***Eq. 1.3
***Eq. 1.2
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 3
Analysis Case 1• If mt<<m,
– change of momentum of alpha particle is negligible
2 1 2ttt
mv v v
m
• If mt>>m, – alpha particle deflected backwards
Analysis Case 2
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 4
Rutherford Scattering with EM Force 1• Need to take into account the EM force between the and
the atom • The Coulomb force is a central force->conservative force• The Coulomb potential energy between particles with Ze and
Z’e electrical charge, separated by a distance r is
• Since the total energy is conserved,
2'ZZ eV r
r
20
1constant>0
2E mv 0
2 v
E
m
PHYS 3446, Fall 2010 Andrew Brandt 5
• From the energy relation, we obtain
• From the definition of angular momentum, we obtain an equation of motion
• From energy conservation, we obtain another equation of motion
Rutherford Scattering with EM Force 2• The distance vector r is always the
same direction as the force throughout the entire motion, so the net torque (rxF) is 0.
• Since there is no net torque, the angular momentum (l=rxp) is conserved. The magnitude of the angular momentum is l=mvb.
2 2l m E m b b mE
2
2
2
2
dr lE V r
dt m mr
2d dt l mr
Centrifugal barrier
Effective potential
Impact parameter=distance of closestapproach if there were no Coulomb force
2 2 2b l mE
2
22
1
2
1
2
drE m
dt
dmr V r
dt
Eq. 1.17
Monday, Sep. 13, 2010
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 6
Rutherford Scattering with EM Force 3• Rearranging the terms, we obtain
• and
2 21V rdr l
r bdt mrb E
1 2
2 21
bdrd
V rr r b
E
Distance of closest approach
-sign since r decreases as alpha particle approaches target
Eq. 1.18
Asymptotic scattering angle:
02
Eq. 1.19
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 7
Rutherford Scattering with EM Force 4• What happens at the DCA?
– Kinetic energy goes to 0.
– Consider the case where the alpha particle is incident on the z-axis, it would reach the DCA, stop, and reverse direction!
– From Eq. 1.18, we can obtain
– This allows us to determine DCA for a given potential and 0.• Substituting 1.19 into the scattering angle equation gives:
0
0r r
dr
dt
02 20 1 0
V rr b
E
00 1 2
2 2
2 2
1r
drb
V rr r b
E
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 8
Rutherford Scattering with EM Force 5• For a Coulomb potential
• DCA can be obtained for a given impact parameter b,
• And the angular distribution becomes
2 22 2 2
0
'1 1 4 '
2
ZZ e Er b E ZZ e
01 2
22 2
2'
1r
drb
ZZ er r b
rE
2'ZZ e
V rr
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 9
Rutherford Scattering with EM Force 6• Replace the variable 1/r=x, and performing the
integration, we obtain
• This can be rewritten
• Solving this for b, we obtain
1
22 2 2
12 cos
1 4 'b
b E ZZ e
22 2 2
1cos
21 4 'b E ZZ e
2'cot
2 2
ZZ eb
E
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 10
Rutherford Scattering with EM Force 7
• From the solution for b, we can learn the following 1. For fixed b and E
– The scattering is larger for a larger value of Z or Z’ (large charge in projectile or target)– Makes perfect sense since Coulomb potential is stronger with larger Z.– Results in larger deflection.
2. For fixed b, Z and Z’– The scattering angle is larger when E is smaller.
– If particle has low energy, its velocity is smaller– Spends more time in the potential, suffering greater deflection
3. For fixed Z, Z’, and E– The scattering angle is larger for smaller impact parameter b
– Makes perfect sense also, since as the incident particle is closer to the nucleus, it feels stronger Coulomb force.
2'cot
2 2
ZZ eb
E
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 11
What do we learn from scattering?• Scattering of a particle in a potential is completely determined
when we know both – The impact parameter, b, and – The energy of the incident particle, E
• For a fixed energy, the deflection is defined by – The impact parameter, b.
• What do we need to perform a scattering experiment?– Incident flux of beam particles with known E– Device that can measure number of scattered particles at various
angle, .– Measurements of the number of scattered particles reflect
• Impact parameters of the incident particles • The effective size of the scattering center
• By measuring the scattering angle , we can learn about the potential or the forces between the target and the projectile
2'cot
2 2
ZZ eb
E
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 12
All these land here
• N0: The number of particles incident on the target foil per unit area per unit time.
• Any incident particles entering with impact parameter b and b+db will scatter to the angle and -d
• In other words, they scatter into the solid angle d2sind• So the number of particles scattered into the solid angle d per unit time
is 2N0bdb.• Note:have assumed thin foil, and large separation between nuclei—why?
Scattering Cross Section
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 13
Scattering Cross Section• For a central potential
– Such as Coulomb potential– Which has spherical symmetry
• The scattering center presents an effective transverse cross-sectional area of
• For the particles to scatter into and +d
2 bdb
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 14
Scattering Cross Section• In more generalized cases, depends on both & .
• With a spherical symmetry, can be integrated out:
,
d
d
Differential Cross Section
What is the dimension of the differential cross section?
Area!!
reorganize
bdbd ,d
dd
, sind
d dd
2 sind
dd
2 bdb
Why negative? Since the deflection and change of b are in opposite direction!!
sin
b db
d
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 15
Scattering Cross Section• For a central potential, measuring the yield as a function of the differential cross section) is equivalent to measuring the entire effect of the scattering
• So what is the physical meaning of the differential cross section?
Measurement of yield as a function of specific experimental variables
This is equivalent to measuring the probability of occurrence of a physical process in a specific kinematic phase space
• Cross sections are measured in the unit of barns:
1 barn Where does this come from?
-24 210 cmCross sectional area of a typical nucleus! nanobarn
picobarn
femptobarn
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 16
Total Cross Section• Total cross section is the integration of the differential
cross section over the entire solid angle, :
• Total cross section represents the effective size of the scattering center integrated over all possible impact parameters (and consequently all possible scattering angles)
Total 4
0,
dd
d
02 sin
dd
d
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 17
Cross Section of Rutherford Scattering• The impact parameter in Rutherford scattering is
• Thus,
• Differential cross section of Rutherford scattering is
2'cot
2 2
ZZ eb
E
db
d
d
d
224'
cosec4 2
ZZ e
E
22
4
' 1
4 sin2
ZZ e
E
221 '
cosec2 2 2
ZZ e
E
sin
b db
d
what happened? can you say “trig identity?” sin(2x)=2sin(x) cos(x)plot/applets
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 18
Measuring Cross Sections
• Rutherford scattering experiment– Used a collimated beam of particles emitted from Radon– A thin Au foil target– A scintillating glass screen with ZnS phosphor deposit– Telescope to view limited area of solid angle– Telescope only needs to move along not . Why?
• Due to the spherical symmetry, scattering only depends on not .
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 19
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
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 20
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
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 21
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
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 22
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
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 23
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
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 24
• Inclusive jet production cross section as a function of transverse energy
Example Cross Section: Jet +X
triumph ofQCD
what if there werean excess at highenergy?
Monday, Sep. 13, 2010 PHYS 3446, Fall 2010 Andrew Brandt 25
Assignment #2 Due Mon. Sep. 201. 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.
• 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