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Electromagnetic Interactions
Introduction to Elementary Particle Physics
Diego Bettoni
Anno Accademico 2010-2011
Outline
•
•
Electron-nucleus scattering
Rutherford formula
– Mott cross-section
•
•
Electron-nucleon scattering
Rosenbluth formula
– Hadron form factors
•
•
•
The process e+e- +Bhabha scattering e+e-  e+eMagnetic moments of leptons
– Measurement of the muon g-2.
e-nucleus Scattering
(neglecting spin)
e
e_
E 0,p 0
A
-
_
E,p
θ
M if 
_
W,p’
A
The transition probability (per unit time)
is given by the Fermi golden rule:
2
W 
M if

For a potential V(r) the matrix element
is given by the volume integral:
2
f
*

 f V ( r )i d
i = wave function of the incoming electron
f = wave function of the scattered electron
The Born approximation assumes
the perturbation to be weak; we can
represent i and f as plane waves
M if 
e
  
i ( k 0  k )r


V ( r ) d 3r
where k0=p0/ħ e k=p/ħ are the initial and
final propagation vectors, respectively.
p d dp
f  3
h dE f
2
1
d
p 2 dp

M if
2 4
d 2   v dE f
Ef = total energy in the final state
2
2
2
p d dp
dW  M if
3
h dE f

dW
d 
v =velocity of the
v incident beam
Let us now consider the nuclear recoil. We assume that both incident and
scattered electrons are extreme relativistic (v) and we set ħ=c=1.
p0 = k0 = E0 p = k = E
From energy conservation:
Ei = p0+M = p+W = Ef
initial
energy
From momentum conservation:
nucleus
mass
final
energy
  
p0  p  p
2
2
E f  p  W  p  p  M 2
 2
2
 p   p0  p   M
 p  p02  p 2  2 p0 p cos  M 2
 dp

 dE
 f

W
W
p
 
 E  p cos  M p
f
0
0

p
1
1


p0 1  p0 (1  cos  ) 1  2 p0 sin 2 
M
M
2
  
q  p0  p
d
1 2 W p iqr  3  2
 2p
 e V (r )d r
M p0
d 4
momentum transfer
 dp

 dE
 f

1
1
 

  dE f 
2 p  2 p0 cos
 

1

2
2
2



2
2
cos

p
p
p
p
M
dp


0
0
W

W  p  p0 cos
W

E f  p0 cos
W
M

M p0  M  p0 cos

W p
M p0
p0  M  p  p02  p 2  2 p0 p cos   M 2
 p0  M  p 2  p02  p 2  2 p0 p cos   M 2
p02  M 2  p 2  2 p0 M  2 p0 p  2 Mp  p02  p 2  2 p0 p cos   M 2
: 2 p02
M p
p
p
  M 2  cos   0
p0 p0
p0 p0
p
M
1


p0 p0  M  p0 cos  1  p0 (1  cos  )
M
Let us now calculate the matrix element, taking for V(r) the coulomb
 interaction.
We represent the nucleus by a sphere of charge density Ze (R )
 3
  ( R)d R  1
 3
dq  Ze ( R)d R
  
s r R

R

r
e
edq
dV 

4 s
 3
Ze  ( R)d R

 
4 r  R
 3
2
Ze  ( R )d R

V (r )  
 
4 r  R
2

M if   d 3 r e

iq  r

2

Ze  ( R)
3
 d R 4 r  R

Ze
3
3 e

d rd R

4
2
  
 
iq ( r  R ) iq  R
e
 
r R

 ( R)
 polar angle
Ze 2 3   iqR eiqs cos
2
between s and q
(
)
2

d
R

R
e

s
dsd (cos  )


4
s
 iqR 3 

Let us define the Nuclear Form Factor F ( q )    ( R )e d R
If (R) is spherically symmetric (R) = (R)
  2




(q  R)


3
   ( R ) d R
F ( q )    1  iq  R 
2


1 2 2
 1 q R 
6
2
q2  q
F  F (q )
2
The matrix element becomes:
1

Ze 2
M if 
F (q 2 )  sds  eiqs cos d (cos  )
2
0
1

Ze 2
eiqs  e iqs
2
F (q )  s
ds

2
iqs
0
This integral diverges.
r / a
We modify V(r) by a factor e
which takes into account the screening of the
nucleus by atomic electrons.
Since a  atomic dimensions >> R  nuclear dimension, we can write:
e r / a  e s / a
R  a
And the matrix element becomes:

Ze 2
2
M if 
F (q )  e  s / a (eiqs  e iqs )ds
2iq
0
e
1
 s ( iq )
a
ds   e
1
 s (  iq )
a
ds
1
1
 iq   iq
1
1
2iq
a


a

1
1
1
1
2
2
 iq
 iq
q  2
q  2
a
a
a
a
Ze 2 F (q 2 )
M if 
2
1
q 2   
a
a  10-8 cm = 10-10 m  5105 GeV-1
1/a  0.210-5 GeV = 2 keV
q2 >> (1/a)2
Ze 2
M if  2 F (q 2 )
q
2
d
2 e  1
2 W p
2 2

F (q )
 4Z   4 p
M p0
d
 4  q
2
Let us assume
p  q  p0
W
1
M

(Non relativistic nuclear recoil)
p  p0
p
(  1)
p0
  2
2
2
2
2
q  ( p0  p )  2 p0  2 p0 cos   4 p0 sin
2
2
2
d
Z 2 2
1
2
2 2
2 2
2 e 

p0 F (q ) 
F (q )
 4Z  
d
2
4
 4  16 p 4 sin 4 
4 p0 sin
0
2
2

2
[d  2d (cos ); dq2  2 p02 d (cos ); d 
2
dq
]
2
p0
For an effectively pointlike nucleus, i.e. low values of q2, F(q2)  1
d
Z 2 2

d 4 p 2 sin 4 
0
2
d 4Z 2 2

2
dq
q4
Rutherford
cross section
Four-momentum Transfer
Initial state

 P0  ( E0 , p0 )


 P0  ( M ,0)
Final state
incident electron
nucleus at rest in the laboratory

P
(
E
,
p
)





 P  (W , p)
scattered electron
recoiling nucleus
  2
2
2
q 2   P0  P    E0  E    p0  p 
 
 E02  E 2  2 E0 E  p02  p 2  2 p0  p
m  E
 2m 2  2 E0 E  2 p0 p cos 
 2 p0 p (1  cos  )  4 p0 p sin 2
q2 < 0
q2 > 0
spacelike
timelike

2
(scattering processes)
(annihilation, e.g. e+e- +-)
Considering the four-momentum transfer to the nucleus:
2
2
2


 q  ( P0  P )  ( M  W )  p  2 M 2  2 MW  2 MK
2
where K = W-M = kinetic energy of the nucleus
Thus:
W
q2
 1
M
2M 2
The nucleus recoils coherently for q2 << 2M2
W
1
M
The nuclear form factor:
F (q 2 )    ( R )e iqR cos 2R 2 d (cos  )dR
e iqR  e iqR
   ( R)
2R 2 dR
iqR
sin qR
   ( R)
4R 2 dR
qR
Typical nuclear radius is R  a few fm. For example, if R = 4 fm, qR = 1, q=1/R
qc 
c 197 MeV  fm

 50MeV
R
4 fm
Therefore if q << 50 MeV/c, qR  0 and F(q2) 1
e
d 4 2
 4
2
dq
q
e

1
q2
p

p
Electron Spin


For a relativistic fermion the spin vector  is aligned with the momentum vector p
 z  1
x  y  0

if p defines the z axis.
 
The helicity H is defines as:
H=+1 right-handed R
p
H    1
H= -1 left-handed L
p
In electromagnetic interactions helicity is conserved.
L
L
R
e-
e-
e-
Jz = 1
R
e-
L
R
e+
e-
Transverse photon
In the relativistic limit fermion and antifermion have opposite helicities.
L
 L
e
L

allowed
forbidden


Jz =-½

Jz =-½
1
1
Jz    Jz  
2
2
1
1
Jz    Jz  
2
2
1
1
Jz    Jz  
2
2
Helicity amplitudes
d
Mott
cross section

 L
j
mm
 

Z  cos
1

d


2




2p

 d  Mott
4 p02 sin 4 1  0 sin 2
2
M
2
2
2
2
d  ,   cos
1
2
1
2
d
1
2
1
2
d
1
2
1
2
,  12
,  12
1
2
 cos
 sin

2

2

2
the factor p/p0
takes into account
the nuclear recoil.
e-N Scattering
Let us now consider also the spin of the target. In the scattering of electrons by
hypothetical pointlike protons there will be a magnetic and an electrical interaction.
 2
d 
q2

 d 
2 

sin 
 cos 



2
2 2M
2
 d  Dirac  d  Rutherford 
Z 2 2
1

2p

4 p02 sin 4 1  0 sin 2
2
M
2
electrical
magnetic
(non spin-flip)
(spin-flip)
If nucleons were pointlike this would be the cross section. However p and n are not
pointlike, as shown by their anomalous magnetic moments:
(Dirac)
(experimental)
p eħ/2mc = 1 n.m.
+2.79 n.m.
n
-1.91 n.m.
0
Nucleons have an
extended structure
Proton Structure
p0
p
j
J
k
k
 
j   eu ( p )  u ( p0 )ei ( p  p ) x
0
J   eu (k )u (k )ei ( k  k ) x
most general 4-vector which
can be constructed from
k, k, q and the Dirac matrices .
There are only two independent terms,  and iq,
and their coefficients are functions of q2.
  F1 q 2    
2M
F2 q 2 i  q
F1 and F2 are the Form Factors of the Proton
As
q2
 0 we see a particle of charge e
F1(0) =1
F1(0) =0
and magnetic moment
F2(0)=1
F2(0)=1
proton
neutron
1   e
2 Mc
 2  2 q 2 2  2 
q2
2
 d    d  
2
  F1 
F  cos 
F1  F2  sin 

 

2 2
2
4M
2 2M
2
 d   d  Rutherford 

Sachs Form Factors
GE  F1 
q 2
4M
GM  F1  F2
2
F2
electric Form Factor
magnetic Form Factor
2
2




d
d
G
G




2
2
2
E
M

cos  2GM sin 

 

2
2
 d   d  Rutherford  1  
2
2

d
d
G
G






  E
2
2
M
 2GM tan 

 


2
 d   d  Mott  1  
Rosenbluth Formula
q2

4M 2
GE  GE q 2 
GM  GM q 2 
 d 



d


 d 


 d  Mott
GEp 0   1
GEn 0  0
GMp 0   2.79 GMn 0  1.91
 Aq   B q  tan
2
2
2

2
Rosenbluth
Plot
The experimental determination of the nucleon form factors in the spacelike region
(q2 < 0) is carried out by directing e- beams of energy between 400 MeV and
16 GeV at a hydrogen target (for the proton) or deuterium (for the neutron).
e  p  e  p
e  d  e  d
For the neutron:
d
d
(en) 
ed   d (ep)  fattori di correzione
d
d
d
Scaling laws for the form factors:
2
2
n
p




G
q
G
q
2
p
M
M
GE q  

 G q 2 
GEn q 2   0
p
n
Nucleon Spacelike Form Factors
GMp
GEp
p
GMn
n
Proton Form Factors at High q2
G q  
1
2
Dipole formula:
M  (0.84 GeV )
2
V
2

q 
1  2 
 MV 
2
2
The dipole form corresponds to an exponential
charge distribution
 ( R)   0e  M
V
R
with an rms radius

M R 2 3

e
R d R
 0
V
R
2

0

M R 3

e
d R
 0
V
12
 2
MV
0
For the proton:
R
2
12

 0.80 fm
MV
Form Factor Measurements Using Polarization
Rosenbluth
polarization
Linear
deviation
from dipole
GE≠GM
• Different charge and
magnetization distributions
• Quark angular momentum
contribution?
Timelike Proton Form Factors
They can be measured via the processes e+e-  pp, or pp  e+e-.
For the latter in the CMS frame:
e+
p
p(E,p)
*
e2
4
m
 2  2 c 2 
d
2
2
p
2 *
2 *




G
1
cos

G
1
cos

 M
E
*
2 xs 
s
d cos

s

= CM Energy
One measures the
total cross section .







d
  0 d  cos *
d (cos  *)
d
2
2

m
4

2
2
p

B GE 
 A GM 
s
2 p s 

2
 cos * max
max
Proton Magnetic Form Factor
The dashed line is a
fit to the PQCD prediction
GM
p
C

 s 
s 2 ln 2  2 
 
The expected Q2 behaviour
is reached quite early,
however ...
... there is still a factor of 2
between timelike and
spacelike.
Timelike Form Factor of the 
e  e  




e+ beam with energies
100, 125, 150, 175 GeV
Spacelike Form Factor of the 
   e     e
300 GeV - beam
r2  0.439  0.008 fm 2
r  0.66 fm
e+e-  ++

e+
4 2
 e e     
3s
e-
 
-

s = 4E1E2 if s >> me2, m2
e
e





1
q2
•Each vertex gives a contribution () to the matrix element. The cross section
is therefore proportional to 2. 2.
1
1
•For a timelike process q2=s, hence the propagator 2
q
s
• has dimensions of (length)2, i.e. (energy)-2. If s >> me2, m2
1
s is the only energy scale in the process:  s
•The factor (4/3) comes from integration over solid angle and averaging over spins.
Angular Distribution for e+e-  +d
 1 cos 2  
d
J 1
J z  1
Because electromagnetic interactions conserve parity, both Jz=+1 and Jz=-1 occur
with equal probability. The amplitude for emitting the + at angle  to the e+
starting from a JZ=+1 state is given by:
d11,1  12 1  cos  
If the initial state has JZ=-1 we have to replace  by - hence the amplitude becomes:
1
2
1  cos  
Squaring and adding the amplitudes of these two orthogonal states we obtain:
d
2
2
 1  cos    1  cos    1  cos 2  
d
The process e+e-  +- (or
e+e-  +-) is not purely
electromagnetic: there is a
weak contribution, due to
Z0 exchange.
e
e




  e


e
G
G
Z0

d
d
d d

QED   ( weak )  interference
d d
d
d
2
G2s
G
s
The asymmetry arises from the interference term, the effect is of the order of
10 % for s = 1000 GeV2.


e+e-  e+eBhabha Scattering
The dimensional arguments used for e+e- +- apply equally well to BhaBha
scattering to predict a 1/s dependence for the total cross section.
The angular distribution is however more complex, because two diagrams
contribute:
The first diagram dominates at small angles. In this region the cross section
is large and is used to monitor the luminosity in e+e- colliders.
Lepton Magnetic Moments
According to the Dirac theory a pointlike fermion possesses a magnetic moment
equal to the Bohr magneton . If e and m are the lepton charge and mass:
e
B 
2m
In general the magnetic moment is related to the spin vector s by:


  g B s
Where g is called the Landé factor and gB is the gyromagnetic ratio.
For electron and muon |s|=½ and the Dirac theory predicts g=2.
The actual g-values have been measured experimentally with great precision
and have been found to differ by a small amount (0.2 %) from the value 2.
The Dirac picture of a structureless, point particle is not exact for the electron
and muon.
The magnetic moment of a charged particle depends on the spatial
distributions of charge and mass (e/m ratio). For a spin ½ a value g≠2
argues that processes are taking place which distort the relative charge
and mass distributions. For example, for the proton g=5.59, due to its
internal structure.
The electron, the , the  consist of a bare, pointlike object surrounded by
a cloud of virtual  which are continually being emitted and reabsorbed.
These  carry part of the mass energy of the lepton, and hence the
e/m ratio (and thus the magnetic moment) changes.
In terms of QED:
The Landé can be written as a perturbation series in (/).
To lowest order:
e 


g2
At the next order:
g  2



2m


e  
1  
2m   
g 2
2
2
3



 
 
 
 0.5   0.32848   1.19   
 
 
 
We can define the anomaly: a 
 g  2
ae  

 2 e
QED
 (1159652140  28) 10 12
 g  2
a  

 2 
QED
 
 
 
 0.5   0.76578   24.45   
 
 
 
2
 (1165847008  18  28) 10 12
3
For the  the measured value differs by 9 standard deviations from the
QED calculation. This arises from the fact that for the  there are further
corrections to a due to the strong and weak interactions.
Hadrons do not couple directly to the , but they can couple to the virtual photon.
We therefore expect hadronic contributions to vacuum polarization of the kind:

+
-

This contribution, which would be small due to the high  mass, gets amplified
by resonances in the  system (vector resonances).
The strong contribution to the anomaly
can be computed starting from the
measurable cross section
e+e-  hadrons
via dispersion relations the two diagrams
can be related one to the other.
m2   e  e   adroni 
ds
a ( forte) 
3 
s
12 0
Weak contributions
Measurement of the Muon (g-2)
Consider a longitudinally polarized charged particle moving in a static
magnetic field B. The particle momentum rotates at the cyclotron frequency:
eB
c 
mc
The spin precesses at the frequency:
eB
eB
s  g
 1  a 
mc
2mc
If g=2, i.e. a=0, s=c and the particle will maintain its longitudinal polarization.
If however g>2 (a>0), s>c, spin precesses faster than momentum. The
rotation frequency a of the spin with respect to the momentum is given by:
eB
 a   s  c  a 
mc
The measurement principle of a is the following: muons are kept turning in a known
magnetic field B, the angle between the spin and the direction of motion is measured
as a function of time and from this the value of a can be determined.
Since a ≈ 1/800, the muon must make roughly 800 turns in the field for the spin to
make 801 and the polarization to change gradually through 2.
y
x
The muons take roughtly 2000 turns in the field. The field gradient
displaces the orbit to the right. At the end a very large gradient is used
to eject the muons, which are then stopped in the polarization analyzer.
B  1.6 T
Bz  B0 1  ay  by 2 
p  90 MeV / c
t     2.2 s
Polarization analyzer. When a muon stops
in the liquid methylene iodide (E) a pulse of
current in coil G is used to flip the spin through
900. Backward or forward decay electrons
are detected in the counter telescopes 66 and
77. The asymmetry in counting rates as a
function of time is given by:
c  c
A
 A0 sin  s
c  c
  e 

A  A0 sin a   Bt   
  mc 

Using this method the anomaly a
was measured with an accuracy of 0.4 %
a  1162  5 10 6
Muon Storage Ring
In order to improve the measurement
accuracy it was necessary to increase
the number of (g-2) cycles, either by
increasing the field B or by lengthening
the storage time.
The usage of an electric field for the
vertical focussing allowed the use of
a uniform magnetic field.
For the precession of spin in combined
electric and magnetic fields:
e   
1   
a 
a B   a  2    E 

  1
mc 



For =29.3 (magic ) the coefficient
of the second term vanishes and the
precession is again  a.
This method allowed to considerably increase the accuracy in the measurement of a.
a  1165924  9 10 9
Present Situation for a
a SM [e+e– ] = (11 659 182.8 ± 6.3had ± 3.5LBL ± 0.3QED+EW)  10 –10
BNL E821 (2004) : aexp = (11 659 208.0  5.8) 10 10
aexp  aSM  25.2  9.2 10 10
The discrepancy between theory and
experiment is 2.7 standard deviations.