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.
Top Related