monday, sept. 15, 2008phys 3446, fall 2008 andrew brandt 1 phys 3446 – lecture #6 monday, sept....

Monday, Sept. 15, 2008 PHYS 3446, Fall 2008Andrew Brandt

1

PHYS 3446 – Lecture #6Monday, Sept. 15, 2008

Dr. Andrew Brandt

1. Relativistic Variables

2. Invariant Scalars3. Feynman Diagram


2

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

sin(2x)=2sin(x) cos(x) gives 4th factor so 1/16 is correct

3 factorsof 2


3

Rutherford Scattering Cross Section • Let’s plug in the numbers

– ZAu=79

– ZHe=2– For E=10keV

• Differential cross section of Rutherford scattering

d

d

19

4

4.0 10

sin2

barns

22

4

' 1

4 sin2

ZZ e

E

2219

3 194

79 2 1.6 10 1

4 10 10 1.6 10 sin2

432

4

4.0 10

sin2

cm


4

Relativistic Variables• Velocity of CM in the scattering of two particles

with rest mass m1 and m2 is:

• If m1 is the mass of the projectile and m2 is that of the target, for a fixed target we obtain

CMv

c

CM

12

1 2

Pc

E m c

CM 1 2

1 2

P P c

E E

1

2 2 2 4 21 1 2

Pc

P c m c m c


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Relativistic Variables-Special Cases• At very low energies where m1c2>>P1c, the velocity

reduces to:

• At very high energies where m1c2<<P1c and m2c2<<P1c , the velocity can be written as:

CM

CM CM

22 21 2

1 1

1

1m c m c

Pc Pc

2

2 1

1 1

11

2

m c m c

P P

1 12 2

1 2

m v c

m c m c

1 1

1 2

m v

m m c

Expansion


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Relativistic Variables• For high energies, if m1~m2,

• CM becomes:

• In general, for fixed target

• Thus CM becomes

2

1

1CM

m c

P

CM

21 CM

21/ 22 1 2

2 4 2 2 41 1 2 2

12

CMCM

E m c

m c E m c m c

Invariant Scalar: s

1/ 221CM

1/ 21 1CM CM

1 2

2

1

2m c

P

1

22

P

m c

2 4 2 2 41 1 2 2

221 2

2m c E m c m c

E m c

CM

12

1 2

Pc

E m c


7

Relativistic Variables• The invariant scalar, s, is defined as:

• So what is this in the CM frame?

• Thus, represents the total available energy in the CM; At the LHC

21 2s P P 2 4 2 4 21 2 1 22m c m c E m c

2 4 2 4 21 2 1 22s m c m c E m c

22 21 2 1 2CM CM CM CME E P P c

21 2CM CME E 2CMToTE

s

0

22 21 2 1 2E E P P c

see thisby evaluatingin lab frame

14TeVs


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Useful Invariant Scalar Variables• Another invariant scalar, t, the momentum transfer

(difference in four momenta), is useful for scattering:

• Since momentum and total energy are conserved in all collisions, t can be expressed in terms of target variables

• In CM frame for an elastic scattering, where PiCM=Pf

CM=PCM and Ei

CM=EfCM:

2f it P P

2 2 22 2 2 2f i f it E E P P c

2 2 22f i f iCM CM CM CMt P P P P c

2 22 1 cos .CM CMP c

2 2 21 1 1 1f i f iE E P P c


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Feynman Diagram• The variable t is always negative for elastic scattering• The variable t could be viewed as the square of the invariant

mass of a particle with and exchanged in the scattering

• While the virtual particle cannot be detected in the scattering, the consequence of its exchange can be calculated and observed!!!– A virtual particle is a particle whose mass is less than the rest mass

of an equivalent free particle

2 2f iE E

2 2f iP P

Time

t-channel diagram

Momentum of the carrier is the difference between the two particles.


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Useful Invariant Scalar Variables• For convenience we define a variable q2,• In the lab frame, , thus we obtain:

• In the non-relativistic limit:

• q2 represents “hardness of the collision”. Small CM corresponds to small q2.

2 2q c t

2 2q c

2 0iLabP

2 22 2 22 f

Labm c E m c 22 22 f

Labm c T

2fLabT

2 222 2 2f fLab LabE m c P c

22 2

1

2m v


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Relativistic Scattering Angles in Lab and CM • For a relativistic scattering, the relationship between the

scattering angles in Lab and CM is:

• For Rutherford scattering (m=m1<<m2, v~v0<<c):

• Divergence at q2~0, a characteristics of a Coulomb field

tan Lab

2dq

2

d

dq

Resulting in a cross section

sin

sinCM

CM CM CM

22 cosP d 2P d

22

2 4

4 ' 1kZZ e

v q


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Assignment 31. Derive Eq. 1.55 starting from 1.48 and 1.492. Derive the formulae for the available CM energy for

• Fixed target experiment with masses m1 and m2 with incoming energy E1.

• Collider experiment with masses m1 and m2 with incoming energies E1 and E2.

3. Given m1= m2 =proton, what beam energy would be needed for the fixed target experiment to have the same CM energy ( )as the LHC

4. End of chapter problem 1.7 Due 9/22/08

s

monday, sept. 15, 2008phys 3446, fall 2008 andrew brandt 1 phys 3446 – lecture #6 monday, sept....

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