Physics 214 UCSD/225a UCSB

Lecture 13• Finish H&M Chapter 6• Start H&M Chapter 8

Result worthy of discussion1. σ ∝ 1/s must be so on dimensional grounds2. σ ∝ α2 two vertices!3. At higher energies, Z-propagator also

contributes:

More discussion4. Calculation of e+e- -> q qbar is identical as

long as sqrt(s) >> Mass of quark.

!(e+e"# qq ) = 3 $ eq

2!(e

+e"# µ +µ"

)q" flavor

%

# of flavorsCharge of quark

Sum over quark flavors

Measurement of this cross section was very important !!!

Measurement of R

R =!(e

+e"# qq )

!(e+e"# µ +µ"

)= 3 $ eq

2

q" flavor

%

Below charm threshold: R = 3 [ (2/3)2 + (1/3)2 + (1/3)2 ] =2

Between charm and bottom: R = 2 + 3(4/9) = 10/3

Above bottom: R = 10/3 + 3(1/9) = 11/3

Measurement of R was crucial for:a. Confirm that quarks have 3 colorsb. Search for additional quarksc. Search for additional leptons

Experimental Result

Ever more discussion5. dσ/dΩ ∝ (1 + cos2θ)

5.1 θ is defined as the angle between e+ andmu+ in com. cos2θ means that the outgoingmuons have no memory of the direction ofincoming particle vs antiparticle.Probably as expected as the e+e- annihilatebefore the mu+mu- is created.

5.2 Recall, phase space is flat in cosθ. cos2θdependence thus implies that the initial stateaxis matters to the outgoing particles. Why?

Helicity Conservation in relativistic limit

• You showed as homework that uL and uR arehelicity eigenstates in the relativistic limit, andthus:

• We’ll now show that the cross terms are zero,and helicity is thus conserved at each vertex.

• We then show how angular momentumconservation leads to the cross section wecalculated.

u ! µu = u

L+ u

R( )! µu

L+ u

R( )

u ! µu = u

L+ u

R( )! µu

L+ u

R( )

u L

= uL

T *! 0 = uT * 1

21" ! 5( )! 0 = u

1

21+ ! 5( )

uR

=1

21+ ! 5( )u

u L! µ

uR

= u 1

41+ ! 5( )! µ

1+ ! 5( )u = u ! µ 1

41" ! 5( ) 1+ ! 5( )u = 0

Let’ do one cross product explicitly:

! 5! µ= "! µ! 5

! 5 = ! 5T *

! 5! 5 =1

Here we have used:

Helicity conservation holds for allvector and axialvector currents as E>>m.

• eL- eR

+ -> muL- muR

+ Jz +1 -> +1• eL

- eR+ -> muR

- muL+ Jz +1 -> -1

• eR- eL

+ -> muL- muR

+ Jz -1 -> +1• eR

- eL+ -> muR

- muL+ Jz -1 -> -1

• Next look at the rotation matrices:

d11

1(!) =

1

21+ cos!( ) "

#u

s

d#1#1

1(!) =

1

21+ cos!( ) "

#u

s

d#11

1(!) =

1

21# cos!( ) "

#t

s

d1#1

1(!) =

1

21# cos!( ) "

#t

s

Cross products cancel inSpin average:

M2

! 1+ cos2"( )

dJ1-1

Initial Jz final Jz

Conclusion on relativistic limit• Dependence on scattering angle is given

entirely by angular momentum conservation !!!• This is a generic feature for any vector or

axialvector current.• We will thus see the exact same thing also for

V-A coupling of Electroweak interactions.

Propagators

i

p2!m

2

i u us

!

p2 " m

2=

i pµ#µ + m( )

p2 " m

2

!ig µ"

q2

i !gµ" + p

µp"M

2( )p2!M

2

Spinless:

Photon:

Spin 1/2, e.g. electron:

Massive Vector Bosons:

See H&M Ch.6.10fffor more details.

Compton Scattering: e- gamma -> e- gamma

i (p + k)µ!µ + m( )(p + k)2 "m2

u

u

p p’

k k’

!µ

! " #

*

ie! µ

ie!"

Compton Scattering: e- gamma -> e- gamma

i (p ! " k )µ#µ + m( )

(p ! " k )2 ! m

2

u

u

p p’

k k’

!µ

! " #

*

ie! µ

ie!"

M2

= M1

+ M2

2

= 2e4 !

u

s!s

u

"

# $

%

& '

Where we neglect the electron mass, and refer to H&MChapter 6.14 for details.

Pair annihilation via crossing• Like we’ve done before: s <-> t

M2

= M1+ M

2

2

= 2e4u

t+t

u

!"#

$%&

t = -2 kk’ = -2 pp’u = -2 kp’ = -2 k’p Ignoring the electron mass.

First step towards chapter 8• In chapter 8 we investigate the structure of

hadrons by scattering electrons of chargedistributions that are at rest in the lab.

• As an initial start to formalism review e- mu-scattering with the initial muon at rest.

• Let’s start with what we got last time,neglecting only terms with electron mass:

M2

=8e

4

t2

! k ! p ( ) kp( ) + ! k p( ) k ! p ( ) " M2k ! k [ ]

M2

=8e

4

t2

! k ! p ( ) kp( ) + ! k p( ) k ! p ( ) " M2k ! k [ ]

As we will want frame for which p = (M,0), it’s worth rewritingthis using q = k - k’ .As we won’t care for the muon recoil p’, we eliminate p’ via: p’ = k - k’ + p.As we ignore the electron mass, we’ll drop terms with k2, or k’2, and simplify q2 = -2kk’ .We then get:

M2

=8e

4

q4

!1

2q2

kp! " k p( ) + 2 " k p( ) kp( ) +1

2M

2q2

#

$ % &

' (

I’ll let you confirm this for yourself.

M2

=8e

4

q4

!1

2q2

kp! " k p( ) + 2 " k p( ) kp( ) +1

2M

2q2

#

$ % &

' (

Now got to muon restframe: p = (M,0)This means kp = EM and k’p = E’M .

M2

=8e

4

q4

!1

2q2M E ! "E( ) + 2E "E M 2

+1

2M

2q2#

$%&

'(

=8e

4

q42E "E M 2 !

q2

2M2

M E ! "E( )

2E "E+1+

q2

4EE '

#

$%

&

'(

=8e

4

q42E "E M 2 !

q2

2M2

M E ! "E( )

2E "E+1! sin2

)

2

#

$%

&

'(

=8e

4

q42E "E M 2 !

q2

2M2sin

2)

2+ cos

2)

2

#

$%

&

'(

q2 = -2kk’ = -4EE’sin2θ/2

q2 = -2pq = -2M(E-E’)

Aside on scattering angle

θk

k’

-2kk’ = -2 (EE’ - kk’ cosθ) = -2EE’ (1-cosθ) = -4EE’ sin2θ/2

Recall: We eliminated any need to know p’ in favor of a measurement of the labframe angle between k and k’

Aside

q2 = (k-k’)2 = (p-p’)2 q2 = -2kk’ = -2pp’

However, we already know that p’ = q + p, and thus:

q2 = -2p(q+p) = -2pq + 2m2 = -2pq

Always neglecting terms proportional to the electron mass.

M2

=8e

4

q42E ! E M

2cos

2 "

2#

q2

2M2sin

2 "

2

$

% &

'

( )

Now recall how to turn this into dσ in the labframe:

d! =1

64"2

#(4 )(pD + pC $ pA $ pB )

vAEAEB

M2 dpc

3dpD

3

EcED

(slide 15 lecture 10)

d! =1

64"2

#(4 )(p + k $ % p $ % k )

EMM

2 d3% k d3% p

% E % p 0

=M

2

4"2

#(4 )(p + k $ % p $ % k )

4EM

% E d % E d&

2

d3% p

2 % p 0

=1

d! =M

2

4"2

#(4 )(p + k $ % p $ % k )

4EM

% E d3

% E d&

2

d3% p

2 % p 0

What do we do about this ?

Recall, we are heading towards collissions between electron and hadron. The hadronic mess in the final state is not something we care to integrate over !!!

Exercise 6.7 in H&M:

!(4 )(p + q " # p )d3 # p

2 # p 0

$ =1

2M! E " # E +

q2

2M

%

& '

(

) *

k

k’

θ

Putting it all together:

d!

d " E d#=

M2

4$ 2

" E

8EM%(4 )(p + q & " p )

d3 " p

2 " p 0

'

M2

=8e

4

q42E ! E M

2cos

2 "

2#

q2

2M2sin

2 "

2

$

% &

'

( )

d!

d " E d#= 4$ 2 2M " E

2

q4

...[ ] %(4 )(p + q & " p )d3 " p

2 " p 0

'

d!d " E d#

=4 " E

2$ 2

q4

cos2 %2&

q2

2M2sin

2 %2

'

( )

*

+ , - E & " E +

q2

2M

.

/ 0

1

2 3

d!

d" lab

=4# 2

4E2sin

4 $

2

% E

Ecos

2 $

2&

q2

2M2sin

2 $

2

'

( )

*

+ ,

Now we perform the E’ integration by noticing:

1

2M! E " # E +

q2

2M

$

% &

'

( ) =

! # E " EA( )

2MA

A *1+2E

Msin

2 +2

Exercise 6.7 H&M

We then finally get:

This is completely independent of the target’s final state !!!

Experimental importance• You measure only initial and final state electron.• You have a prediction for electron scattering off a

spin 1/2 point particle with charge = -1.• Any deviation between measurement and prediction

indicates substructure of your supposed pointparticle !!!

• Next: How does one describe scattering off a chargedistribution, rather than a point particle?

=> Beginning of chapter 8 !

Probing the Structure of Hadronswith electron scattering

• All you measure is the incoming and outgoingelectron 3 momentum.

• If you had a static target then you can show that thisgives you directly the fourier transform of the chargedistribution of your target:

k

k’

θ

d!d"

=d!d" po int

• F(q)2

F(q) = #(x)eiqxd3x$Once the target is not static,we’re best of using e-muscattering as our starting point.

e-proton vs e-muon scattering• What’s different?• If proton was a spin 1/2 point particle with

magnetic moment e/2M then all one needs todo is plug in the proton mass instead of muonmass into:

• However, magnetic moment differs, and wedon’t have a point particle !!!

d!

d" lab

=4# 2

4E2sin

4 $

2

% E

Ecos

2 $

2&

q2

2M2sin

2 $

2

'

( )

*

+ ,

e-proton vs e-muon scattering• Let’s go back to where we started:

– What’s the transition current for the proton ?

Jµ = !eu ( " k )# µ

u(k)ei( " k !k )x

Jµ = !eu ( " p ) ???[ ]u(p)e

i( " p ! p )x

electron

proton

We need to find the most general parametrization for [???],and then measure its parameters.

What is [???]

• This is a two part problem:– What are the allowed 4-vectors in the current ?– What are the independent scalars that the dynamics can

depend on ?• Let’s answer the second first: M2 = p’2 = (p+q)2 = M2 +2pq + q2

• I can thus pick either pq or q2 as my scalar variableto express dependence on kinematics.

What are the 4-vectors allowed?• Most general form of the current:

• Gordon Decomposition of the current: any term with(p+p’) can be expressed as linear sum ofcomponents with γµ and σµν (p’-p).

• K4 must be zero because of current conservation.

Jµ = !eu ( " p ) ???[ ]u(p)e

i( " p ! p )x

???[ ] = # µK1 + i$ µ% " p ! p( )

%K2 + i$ µ% " p + p( )

%K3 +

+ " p ! p( )µK4 + " p + p( )

µK5

Gordon Decomposition• Exercise 6.1 in H&M:

• I leave it as a future homework to show this.

u ( ! p )" µu(p) = u ( ! p ) ! p + p( ) + i# µ$ ! p % p( )

$[ ]u(p)

Current Conservation

qµ Jµ = 0

0 = qµ u ( ! p ) " µK1 + i# µ$ ! p % p( )

$K2 + ! p + p( )

µK5[ ]u(p)

qµ"µ& = m& ' 0

qµ#µ$

q$ = 0 because sigma is anti-symmetric.

because of relativistic limit.

As a result, K5 must be zero.

Jµ = !eu ( " p ) # µ

F1

q2( ) + i$ µ%

q%

&2M

F2

q2( )

'

( ) *

+ , u(p)ei( " p ! p )x

Proton Current

Jµ = !eu ( " p ) # µ

F1

q2( ) + i$ µ%

q%

&2M

F2

q2( )

'

( ) *

+ , u(p)ei( " p ! p )x

We determine F1, F2, and kappa experimentally, with the constraint that F1(0)=1=F2(0) in order for kappa to have the meaning of the anomalous magnetic moment.

The two form factors F1 and F2 parametrize our ignorance regarding the detailed structure of the proton.