Electromagnetic Interactions

Introduction to Elementary Particle Physics

Diego BettoniAnno 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+e-

• Magnetic moments of leptons– Measurement of the muon g-2.

e-nucleus Scattering(neglecting spin)

The transition probability (per unit time) is given by the Fermi golden rule:

For a potential V(r) the matrix elementis given by the volume integral:

i = wave function of the incoming electronf = wave function of the scattered electronThe Born approximation assumesthe perturbation to be weak; we canrepresent i and f as plane waves

where k0=p0/ħ e k=p/ħ are the initial and final propagation vectors, respectively.

0 0_

,p_

W,p’

E,p_

-

E

e

θe A

A

-

fifMW 22

drVM ifif )(*

rdrVeM rkkiif

3)( )(0

ff dE

dphdp3

2

fif dE

dphdpMdW 3

222

vdWd

22

4221

iff

MdEdp

vp

dd

Ef = total energy in the final state

v =velocity of the incident beam

Let us now consider the nuclear recoil. We assume that both incident andscattered 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 nucleus finalenergy mass energy

From momentum conservation:

ppp 0

2

022

0

220

22

cos2 Mppppp

Mppp

MppWpE f

00 cos pp

MW

pEW

dEdp

ff

2sin21

1

)cos1(1

12000

Mp

Mpp

p

23

0

22 )(

41

rdrVe

pp

MWp

dd rqi

ppq 0 momentum transfer

0

00

0

0

20

220

0

cos

cos

cos

cos22cos221

11

pp

MW

pMpM

MW

pEW

ppWW

Mpppppp

dpdEdE

dp

f

ff

)cos1(1

1cos

0cos

2:

cos2222

cos2

cos2

0000

02000

20

20

22000

2220

20

220

20

20

2200

MppMp

Mpp

pp

ppM

pp

pM

p

MppppMpppMppMp

MpppppMp

MpppppMp

R

r

Rrs

RdRZedq 3)(

e

Let us now calculate the matrix element, taking for V(r) the coulomb interaction.We represent the nucleus by a sphere of charge density )(RZe

1)( 3 RdR

RrRdRZe

sedqdV

4)(

432

Rr

RdRZerV

4)()(

32

)(cos2)(4

)(4

4)(

2cos

32

)(33

2

233

dsdss

eeRRdZe

RrReeRdrdZe

RrRZeRderdM

iqsRqi

RqiRrqi

rqiif

polar anglebetween s and q

Let us define the Nuclear Form Factor RdeRqF Rqi 3)()( If (R) is spherically symmetric (R) = (R)

22

32

611

)(2

)(1)(

Rq

RdRRqRqiqF 22 qq

)( 2qFF

The matrix element becomes:

0

22

1

1

cos

0

22

)(2

)(cos)(2

dsiqs

eesqFZe

desdsqFZeM

iqsiqs

iqsif

This integral diverges.We modify V(r) by a factor which takes into account the screening of thenucleus by atomic electrons.Since a atomic dimensions >> R nuclear dimension, we can write:

are /

aRee asar //

And the matrix element becomes:

0

/22

)()(2

dseeeqFiq

ZeM iqsiqsasif

22

22

)1()1(

12

1

11

11

11

aq

iq

aq

iqa

iqa

iqa

iqa

dsedseiq

asiq

as

22

22

1)(

aq

qFZeMif

a 10-8 cm = 10-10 m 5105 GeV-1

1/a 0.210-5 GeV = 2 keVq2 >> (1/a)2

)( 22

2

qFqZeMif

22

0

24

222 )(1

44 qF

pp

MWp

qeZ

dd

Let us assume 1MW

(Non relativistic nuclear recoil)

)1(0

00 pppppqp

2sin4cos22)( 22

020

20

20

2 pppppq

22

420

22222

044

0

222 )(

2sin4

)(

2sin16

14

4 qFp

ZqFpp

eZdd

]);(cos2);(cos2[ 220

20

2 dqp

ddpdqdd

For an effectively pointlike nucleus, i.e. low values of q2, F(q2) 1

4

22

2

420

22

42

sin4

qZ

dqd

p

Zdd

Rutherford

cross section

Four-momentum Transfer

)0,(),(

0

000

MPpEP

Initial state

incident electron

nucleus at rest in the laboratory

),(),(pWP

pEP

Final state

scattered electron

recoiling nucleus

2sin4)cos1(2

cos222

22

200

002

022

0022

0

20

20

20

2

pppp

ppEEm

ppppEEEE

ppEEPPq

Em

q2 < 0 spacelike (scattering processes)q2 > 0 timelike (annihilation, e.g. e+e- +-)

Considering the four-momentum transfer to the nucleus:

MKMWMpWMPPq 222)()( 22220

2

where K = W-M = kinetic energy of the nucleus

Thus: 2

2

21

Mq

MW

The nucleus recoils coherently for q2 << 2M2

1MW

The nuclear form factor:

dRRqR

qRR

dRRiqR

eeR

dRdReRqFiqRiqR

iqR

2

2

2cos2

4sin)(

2)(

)(cos2)()(

Typical nuclear radius is R a few fm. For example, if R = 4 fm, qR = 1, q=1/R

MeVfm

fmMeVRcqc 50

4197

Therefore if q << 50 MeV/c, qR 0 and F(q2) 1

4

2

24

qdqd

21q

e e

p p

Electron Spin

For a relativistic fermion the spin vector is aligned with the momentum vector

if defines the z axis.The helicity H is defines as:

p

01 yxz p

1

ppH

H=+1 right-handed RH= -1 left-handed L

In electromagnetic interactions helicity is conserved. L L R R R L

e- e- e- e- e+ e-

Jz = 1 Transverse photon

In the relativistic limit fermion and antifermion have opposite helicities.

L

e

L

L

L

allowed forbidden

Jz =-½

Jz =-½

Helicity amplitudes

jmmd

2sin

21

21

2cos

21

21

2cos

21

21

21

21

21

21

21

21

21

21

21

,

,

,

dJJ

dJJ

dJJ

zz

zz

zz

2sin21

1

2sin4

2cos

20420

222

Mpp

Z

dd

Mott

Mottcross section

the factor p/p0takes into accountthe nuclear recoil.

e-N Scattering

Let us now consider also the spin of the target. In the scattering of electrons byhypothetical pointlike protons there will be a magnetic and an electrical interaction.

2

sin22

cos 22

22

Mq

dd

dd

RutherfordDirac

2sin21

1

2sin4 2042

0

22

Mpp

Z

electrical(non spin-flip)

magnetic(spin-flip)

If nucleons were pointlike this would be the cross section. However p and n are notpointlike, as shown by their anomalous magnetic moments:

(Dirac) (experimental)p eħ/2mc = 1 n.m. +2.79 n.m.n 0 -1.91 n.m.

Nucleons have anextended structure

Proton Structure

p0 p

k k

jJ

xkki

xppi

ekukueJ

epupuej

)(

)(0

)()(

)()( 0

most general 4-vector whichcan be constructed from k, k, q and the Dirac matrices .

There are only two independent terms, and iq,and their coefficients are functions of q2.

qiqF

MqF 2

22

1 2

F1 and F2 are the Form Factors of the Proton

As q2 0 we see a particle of charge and magnetic momente Mc

e2

1

F1(0) =1 F2(0)=1 protonF1(0) =0 F2(0)=1 neutron

2

sin22

cos4

22212

222

22

222

1 FF

MqF

MqF

dd

dd

Rutherford

Sachs Form Factors

21

22

2

1 4FFG

FMqFG

M

E

electric Form Factor

magnetic Form Factor

2

sin22

cos1

22222

MME

Rutherford

GGGdd

dd

2

tan21

2222

MME

Mott

GGGdd

dd

Rosenbluth Formula

2

2

4Mq

22 qGGqGG MMEE 91.1079.20

0010

nM

pM

nE

pE

GGGG

2

tan222

qBqA

dd

dd

Mott

RosenbluthPlot

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 and16 GeV at a hydrogen target (for the proton) or deuterium (for the neutron).

dedepepe

For the neutron:

correzionedifattoriepdded

dden

dd

)()(

Scaling laws for the form factors:

02

222

2

qG

qGqGqGqG

nE

n

nM

p

pMp

E

Nucleon Spacelike Form Factors

p

pMG

n

nMG

pEG

Proton Form Factors at High q2

Dipole formula: 2

2

2

2

1

1

VMq

qG22 )84.0( GeVMV

RMVeR 0)(

2

0

30

0

320

2 12VRM

RM

MRde

RdReR

V

V

fmM

RV

80.0122

The dipole form corresponds to an exponentialcharge distribution

with an rms radius

For the proton:

• Different charge and magnetization distributions

• Quark angular momentum contribution?

Rosenbluth

polarization

Linear deviationfrom dipoleGE≠GM

Form Factor Measurements Using Polarization

Timelike Proton Form Factors

They can be measured via the processes e+e- pp, or pp e+e-. For the latter in the CMS frame:

*22

2*22

222

* cos14

cos12cos

E

pM G

sm

Gxs

cd

d

p(E,p) pe+

e-*

22

22

*cos

*cos

2

0

42

)*(cosmax

max

Ep

Mp

GBsm

GAs

dddd

s = CM Energy

One measures thetotal cross section .

Proton Magnetic Form Factor

222 ln ss

CG

p

M

The dashed line is afit to the PQCD prediction

The expected Q2 behaviouris reached quite early, however ...... there is still a factor of 2between timelike and spacelike.

Timelike Form Factor of the

ee e+ beam with energies100, 125, 150, 175 GeV

Spacelike Form Factor of the

ee 300 GeV - beam

fmr

fmr

66.0

008.0439.0 22

e+e- +-

e+ e-

+

-

s

ee3

4 2

s = 4E1E2 if s >> me2, m

2

21q

e

e

•Each vertex gives a contribution () to the matrix element. The cross sectionis therefore proportional to 2. 2.

•For a timelike process q2=s, hence the propagator

• has dimensions of (length)2, i.e. (energy)-2. If s >> me2, m

2

s is the only energy scale in the process:•The factor (4/3) comes from integration over solid angle and averaging over spins.

sq11

2

s1

Angular Distribution for e+e- +-

2cos1d

d

11 zJJ

Because electromagnetic interactions conserve parity, both Jz=+1 and Jz=-1 occurwith equal probability. The amplitude for emitting the + at angle to the e+

starting from a JZ=+1 state is given by:

cos1211

1,1 dIf the initial state has JZ=-1 we have to replace by - hence the amplitude becomes:

cos121

Squaring and adding the amplitudes of these two orthogonal states we obtain:

222 cos1cos1cos1 d

d

The process e+e- +- (ore+e- +-) is not purelyelectromagnetic: there is aweak contribution, due toZ0 exchange.

e

e

e

e

0Z

G G

ceinterferenddweak

ddQED

dd

dd

)(

s

2 sG2 G

The asymmetry arises from the interference term, the effect is of the order of10 % for s = 1000 GeV2.

e+e- e+e-

Bhabha 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 sectionis 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 momentequal to the Bohr magneton . If e and m are the lepton charge and mass:

In general the magnetic moment is related to the spin vector s by:

Where g is called the Landé factor and gB 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 precisionand 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 electronand muon.

me

B 2

sg B

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 chargeand mass distributions. For example, for the proton g=5.59, due to itsinternal structure.The electron, the , the consist of a bare, pointlike object surrounded bya cloud of virtual which are continually being emitted and reabsorbed.These carry part of the mass energy of the lepton, and hence thee/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:

meg

22

At the next order:

1

22

meg

We can define the anomaly:2

2 ga

12

32

10)281159652140(

19.132848.05.02

2

QED

ee

ga

12

32

10)28181165847008(

45.2476578.05.02

2

QEDga

For the the measured value differs by 9 standard deviations from the QED calculation. This arises from the fact that for the there are furthercorrections 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 amplifiedby resonances in the system (vector resonances).

-+

The strong contribution to the anomalycan be computed starting from the measurable cross section

e+e- hadronsvia dispersion relations the two diagramscan be related one to the other.

dss

adronieemfortea

03

2

12)(

Weak contributions

Measurement of the Muon (g-2)

Consider a longitudinally polarized charged particle moving in a staticmagnetic field B. The particle momentum rotates at the cyclotron frequency:

The spin precesses at the frequency:

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. Therotation frequency a of the spin with respect to the momentum is given by:

mceB

c

mceBa

mceBgs 1

2

mceBacsa

The measurement principle of a is the following: muons are kept turning in a knownmagnetic field B, the angle between the spin and the direction of motion is measuredas 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 tomake 801 and the polarization to change gradually through 2.

The muons take roughtly 2000 turns in the field. The field gradientdisplaces the orbit to the right. At the end a very large gradient is usedto eject the muons, which are then stopped in the polarization analyzer.

20 1 byayBBz

stcMeVpTB 2.2/906.1

x

y

Polarization analyzer. When a muon stopsin the liquid methylene iodide (E) a pulse ofcurrent in coil G is used to flip the spin through900. Backward or forward decay electronsare detected in the counter telescopes 66 and 77. The asymmetry in counting rates as a function of time is given by:

sAccccA sin0

BtmceaAA sin0

Using this method the anomaly awas measured with an accuracy of 0.4 %

61051162 a

Muon Storage RingIn order to improve the measurementaccuracy it was necessary to increasethe number of (g-2) cycles, either byincreasing the field B or by lengtheningthe storage time. The usage of an electric field for thevertical focussing allowed the use ofa uniform magnetic field.For the precession of spin in combinedelectric and magnetic fields:

EaBa

mce

a

11

2

For =29.3 (magic ) the coefficientof the second term vanishes and theprecession is again a.

This method allowed to considerably increase the accuracy in the measurement of a.

91091165924 a

aSM [e+e– ] = (11 659 182.8 ± 6.3had ± 3.5LBL ± 0.3QED+EW) 10 –10

BNL E821 (2004) : aexp = (11 659 208.0 5.8) 10 10

10102.92.25 SMexp aa

Present Situation for a

The discrepancy between theory and experiment is 2.7 standard deviations.