PHYS 3446 – Lecture #4

Monday, Sept. 11, 2006 PHYS 3446, Fall 2006Jae Yu

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PHYS 3446 – Lecture #4Monday, Sept. 11, 2006

Dr. Jae Yu

1. Lab Frame and Center of Mass Frame2. Relativistic Treatment3. Feynman Diagram4. Invariant kinematic variables


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Announcements• I now have 9 on the distribution list.

– Only three more to go!! Whoopie!! – Please come and check to see if your request has been

successful• I will send a test message this week• No lecture this Wednesday but • The class time should be devoted for your project

group meetings in preparation for the workshop


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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?

Measurement of yield as a function of specific experimental variable

This is equivalent to measuring 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


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Lab Frame and Center of Mass Frame• We assumed that the target nuclei do not move through

the collision in Rutherford Scattering• In reality, they recoil as a result of scattering • Sometimes we use two beams of particles for scattering

experiments (target is moving)• This situation could be complicated but..• If the motion can be described in the Center of Mass

frame under a central potential, it can be simplified


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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


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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


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Now with some simple arithmatics• 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 ˆr 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


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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 as if that of a fictitious particle with mass

(the reduced mass) and coordinate r.• In the frame where CM is stationary, the dynamics

becomes equivalent to that of a single particle of mass scattering off of a fixed scattering center.

• Frequently we define the Center of Mass frame as the frame where the sum of the momenta of all the interacting particles is 0.


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Relationship of variables in Lab and CMS

• The speed of CM is

• Speeds of the particles in CMS are

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

CM CMv R

1v

CMS

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


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• CM represents the changes in the direction of the relative position vector r as a result of the collision in CMS

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

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

• The perpendicular components of the velocities are:

• Thus, the angles are related, for elastic scattering only, as:

Scattering angles in Lab and CMS

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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Differential cross sections in Lab and CMS• The particles that scatter in lab at an angle Lab into solid

angle dLab scatter at CM into solid angle dCM in CM.

• Since is invariant, dLab = dCM.– Why?– is perpendicular to the direction of boost, thus is invariant.

• Thus, the differential cross section becomes: sinLab Lab Lab

Lab

dd

d

LabLab

d

d

reorganize

Using Eq. 1.53 LabLab

d

d

sinCM CM CMCM

dd

d

CMCM

d

d

cos

cosCM

Lab

d

d

CMCM

d

d

3/ 221 2 cos

1 cos

CM

CM


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Some Quantities in Special Relativity• Fractional velocity

• Lorentz factor

• Relative momentum and the total energy of the particle moving at a velocity is

• Square of four momentum P=(E,pc), rest mass E

P

E

v c

v c

2

1

1

M v

M c

2 T Re

2st

E 2 2 2 4P c M c 2M c

2 2 2 2 2P Mc E p c


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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

1

1 2

Pc

E E

1

2 2 2 4 21 1 2

Pc

P c m c m c

CM 1 2

1 2

P P c

E E


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Relativistic Variables• 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:

• And

• Thus CM becomes

2

2

1

1CM

m c

P

CM

21 CM

21/ 22 1 2

2 4 2 2 41 1 2 2

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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


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

• So what is this the CMS frame?

• Thus, represents the total available energy in the CMS

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 2CM

ToTE

s

0

22 21 2 1 2E E P P c


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

(differences 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 CMS frame for an elastic scattering, where PiCM=Pf

CM=PCM and Ei

CM=EfCM:

21 2t P P

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

22 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 an 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 different than the rest

mass

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 CMS • For a relativistic scattering, the relationship between the

scattering angles in Lab and CMS 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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Assignments1. Extension of previous w/ the following (due 9/18):

• Plot the differential cross section of the Rutherford scattering as a function of the scattering angle with some sensible lower limit of the angle + express your opinion on the sensibility of the cross section, along with good physical reasons

2. Derive Eq. 1.55 starting from 1.48 and 1.493. Derive the formulae for the available CMS 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.

4. Reading assignment: Section 1.7 5. End of chapter problem 1.7• These assignments are due next Wednesday, Sept. 20.

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PHYS 3446 – Lecture #4