Physics 214 UCSD/225a UCSB
Lecture 13
• Finish H&M Chapter 6
• Start H&M Chapter 8
Result worthy of discussion
1. σ ∝ 1/s must be so on dimensional grounds
2. σ ∝ α2 two vertices!
3. At higher energies, Z-propagator also
contributes:
More discussion
4. Calculation of e+e- -> q qbar is identical as
long as sqrt(s) >> Mass of quark.
+ "
! (e e # qq ) = 3$
# of flavors
% e ! (e e
2
q
q" flavor
+ "
+
"
#µ µ )
Charge of quark
Sum over quark flavors
Measurement of this cross section was very important !!!
Measurement of R
! (e +e" # qq )
2
R=
= 3$ % eq
+ "
+ "
! (e e # µ µ )
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 colors
b. Search for additional quarks
c. Search for additional leptons
Experimental Result
Ever more discussion
5. dσ/dΩ ∝ (1 + cos2θ)
5.1 θ is defined as the angle between e+ and
mu+ in com. cos2θ means that the outgoing
muons have no memory of the direction of
incoming particle vs antiparticle.
Probably as expected as the e+e- annihilate
before the mu+mu- is created.
5.2 Recall, phase space is flat in cosθ. cos2θ
dependence thus implies that the initial state
axis matters to the outgoing particles. Why?
Helicity Conservation in relativistic limit
• You showed as homework that uL and uR are
helicity eigenstates in the relativistic limit, and
thus:
µ
µ
u ! u = ( uL + uR )! ( uL + uR )
• We’ll now show that the cross terms are zero,
and helicity is thus conserved at each vertex.
• We then show how angular momentum
conservation leads to the cross section we
calculated.
Let’ do one cross product explicitly:
u ! µ u = ( uL + uR )! µ ( uL + uR )
1
1
5
0
5
uL = u ! = u
1"
!
!
=
u
1+
!
(
)
(
)
2
2
1
uR = (1+ ! 5 ) u
2
1
µ
5
µ
5
µ 1
5
5
uL ! uR = u (1+ ! )! (1+ ! ) u = u !
1"
!
1+
!
u=0
(
)
(
)
4
4
T* 0
L
T*
Here we have used:
Helicity conservation holds for all
vector and axialvector currents as E>>m.
! 5! µ = "! µ ! 5
! 5 = ! 5T *
! 5! 5 = 1
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:
1
#u
1
d11 (! ) = (1+ cos ! ) "
Cross products cancel in
•
•
•
•
•
2
s
1
#u
1
d#1#1 (! ) = (1+ cos! ) "
2
s
1
#t
1
d#11 (! ) = (1# cos! ) "
2
s
1
#t
1
d1#1 (! ) = (1# cos ! ) "
2
s
Spin average:
M ! (1+ cos 2 " )
2
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
Spinless:
i
2
2
p !m
Massive Vector Bosons:
i(!g
See H&M Ch.6.10ff
for more details.
µ"
µ
"
+p p
p !M
2
M
2
Spin 1/2, e.g. electron:
i! u u
s
p "m
2
2
=
i( p µ # µ + m )
p "m
2
2
Photon:
!ig µ"
q
2
2
)
Compton Scattering: e- gamma -> e- gamma
k
!µ
"#!
i(( p + k) µ ! µ + m)
k’
*
( p + k) 2 " m 2
u
p
ie!
µ
ie!
"
u
p’
Compton Scattering: e- gamma -> e- gamma
k
ie!
ie!
"
( p ! k ") 2 ! m 2
2
M = M1 + M 2
2
" u s%
= 2e $! ! '
# s u&
4
k’
µ
i(( p ! k ") µ # µ + m)
u
p
"#!
!µ
*
u
p’
Where we neglect the electron mass, and refer to H&M
Chapter 6.14 for details.
Pair annihilation via crossing
• Like we’ve done before: s <-> t
M = M1 + M 2
2
t = -2 kk’ = -2 pp’
u = -2 kp’ = -2 k’p
2
!u t$
= 2e # + &
" t u%
4
Ignoring the electron mass.
First step towards chapter 8
• In chapter 8 we investigate the structure of
hadrons by scattering electrons of charge
distributions that are at rest in the lab.
• As an initial start to formalism review e- muscattering with the initial muon at rest.
• Let’s start with what we got last time,
neglecting only terms with electron mass:
4
8e
M = 2
t
2
2
!
!
!
!
k
p
kp
+
k
p
k
p
"
M
[( )( ) ( )( ) kk !]
8e 4
M = 2 [( k !p!)( kp) + ( k !p)( kp!) " M 2 kk !]
t
2
As we will want frame for which p = (M,0), it’s worth rewriting
this 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:
8e 4 # 1 2
1 2 2&
M = 4 %! q ( kp ! k "p) + 2( k "p)( kp) + M q (
'
q $ 2
2
2
I’ll let you confirm this for yourself.
4
8e
M = 4
q
2
# 1 2
1 2 2&
%$! 2 q ( kp ! k "p) + 2( k "p)( kp) + 2 M q ('
Now got to muon restframe: p = (M,0)
This means kp = EM and k’p = E’M .
8e4
M = 4
q
2
1 2 2&
# 1 2
2
!
q
M
E
!
E
+
2E
E
M
+
M q (
"
"
(
)
%$ 2
2
'
8e4
q2 M ( E ! E")
q2 &
2 #
= 4 2EE "M % !
+1+
2
q
2EE "
4EE ' ('
$ 2M
=
8e
q M ( E ! E")
2 #
2)&
2E
E
M
!
+
1
!
sin
"
% 2M 2
q4
2EE "
2 ('
$
4
2
8e4
q2
2 #
2)
2)&
= 4 2EE "M % !
sin
+
cos
(
2
q
2M
2
2
$
'
q2 = -2kk’ = -4EE’sin2θ/2
q2 = -2pq = -2M(E-E’)
Aside on scattering angle
k’
k
θ
-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.
4
2
$
'
8e
"
q
2
2
2
2"
M = 4 2EE !M &cos #
sin )
2
q
2 2M
2(
%
Now recall how to turn this into dσ in the labframe:
1 # (4 ) ( pD + pC $ pA $ pB ) 2 dpc3 dpD3
d! =
M
2
64 "
vA E A E B
Ec ED
(slide 15 lecture 10)
=1
1 # (4 ) ( p + k $ p% $ k %) 2 d 3 k %d 3 p%
d! =
M
2
64 "
EM
E %p%0
2
M # (4 ) ( p + k $ p% $ k %) E %dE %d& d 3 p%
=
4" 2
4 EM
2
2 p%0
2
M # (4 ) ( p + k $ p% $ k %) E %d 3 E %d& d 3 p%
d! =
4" 2
4 EM
2
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:
3
2 (
%
#
d
p
1
q
(4 )
!
$ ( p + q " p#) 2 p# = 2M !'& E " E # + 2M *)
0
k
k’
θ
Putting it all together:
2
M
d!
E"
=
dE "d# 4 $ 2 8EM
3
d
p"
(4 )
' % ( p + q & p") 2 p"
0
2
$
'
8e 4
"
q
2
2
2"
M = 4 2EE !M &cos #
sin )
2
q
2 2M
2(
%
2
2
3
"
d!
2M
E
d
p"
2
(4 )
= 4$
[...] ' % ( p + q & p")
4
dE "d#
q
2 p"0
2 1
.
*
d!
4 E "2$ 2 ' 2 %
q2
%
q
2
=
cos
&
sin
3
)
,-0 E & E " +
4
2
dE "d#
q
2 2M
2+ /
2M 2
(
Now we perform the E’ integration by noticing:
(
)
E
2 '
#
!
E
"
$
1
q
A
!& E " E # +
=
)
2M %
2M (
2MA
2E 2 +
Exercise 6.7 H&M
A * 1+
sin
M
2
We then finally get:
*
d!
4# 2
E%' 2 $
q2
2$
=
cos &
sin ,
)
2
d" lab 4 E 2 sin 4 $ E (
2 2M
2+
2
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 point
particle !!!
• Next: How does one describe scattering off a charge
distribution, rather than a point particle?
=> Beginning of chapter 8 !
Probing the Structure of Hadrons
with electron scattering
k’
k
θ
• All you measure is the incoming and outgoing
electron 3 momentum.
• If you had a static target then you can show that this
gives you directly the fourier transform of the charge
distribution of your target:
d! d!
2
=
• F(q)
d" d" po int
Once the target is not static,
we’re best of using e-mu
scattering as our starting point.
F(q) =
iqx 3
#
(x)e
d x
$
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 to
do is plug in the proton mass instead of muon
mass into:
*
d!
4# 2
E%' 2 $
q2
2$
=
cos &
sin ,
)
2
d" lab 4 E 2 sin 4 $ E (
2 2M
2+
2
• However, magnetic moment differs, and we
don’t have a point particle !!!
e-proton vs e-muon scattering
• Let’s go back to where we started:
– What’s the transition current for the proton ?
µ
µ
i( k " !k )x
"
J = !eu ( k )# u(k)e
i( p " ! p )x
"
J = !eu ( p )[???] u( p)e
µ
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 variable
to express dependence on kinematics.
What are the 4-vectors allowed?
• Most general form of the current:
i( p " ! p )x
"
J = !eu ( p )[???] u( p)e
µ
µ
µ%
µ%
"
???
=
#
K
+
i
$
p
!
p
K
+
i
$
[ ]
(
)% 2
( p" + p)% K 3 +
1
+( p" ! p) K 4 + ( p" + p) K 5
• Gordon Decomposition of the current: any term with
(p+p’) can be expressed as linear sum of
components with γµ and σµν (p’-p).
• K4 must be zero because of current conservation.
µ
µ
Gordon Decomposition
• Exercise 6.1 in H&M:
u ( p!)" µ u( p) = u ( p!)[( p! + p) + i# µ$ ( p! % p)$ ] u( p)
• I leave it as a future homework to show this.
Current Conservation
µ
qµ J = 0
[
µ
0 = qµ u ( p!) " K1 + i#
µ
qµ " & = m& ' 0
µ$
( p! % p)$ K 2 + ( p! + p)
µ
]
K 5 u( p)
because of relativistic limit.
µ$
qµ# q$ = 0 because sigma is anti-symmetric.
As a result, K5 must be zero.
' µ
&
2
µ%
2 *
J = !eu ( p"))# F1 (q ) + i$ q%
F2 (q ),u( p)e i( p "! p )x
(
+
2M
µ
Proton Current
' µ
&
2
µ%
2 *
J = !eu ( p"))# F1 (q ) + i$ q%
F2 (q ),u( p)e i( p "! p )x
(
+
2M
µ
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.