Relativistic Kinematics

1

1. Costituents of Matter2. Fundamental Forces3. Particle Detectors (N. Neri)4. Experimental highlights (N. Neri)5. Symmetries and Conservation Laws6. Relativistic Kinematics7. The Static Quark Model8. The Weak Interaction9. Introduction to the Standard Model10. CP Violation in the Standard Model (N. Neri)

Recalling Relativistic Kinematics

Basic Principles

Every experiment will give the same results whenever executed in reference frames that are in uniform rectilinear motion with respect to one another.

Physical laws are the same in every inertial frame.Energy, total momentum and total angular momentum of

a physical system are constant in time.The speed of light in vacuum is the same in every inertial

frame : c=2.9979108 m/s (Time is not a relativistic invariant) (Space is not a relativistic invariant)

2

Four-vector :

For example, for a particle Minkoski pseudo-euclidean metricScalar product:

Lorentz transformations Given 2 inertial frames Oxyz, Ox’y’z’ in relative motion and assuming that the origin of the axis coincide at a common t=t’=0 and also assuming that the uniform translatory motion be along the x axis:β=vx/c with vx velocity di O’ rispetto a O e con γ=1/(1-b2)1/2

By applying a Lorentz transformation L(b) to a four-vector A in the system O, one gets A’ in the O’ system:

3

2

10

10

3

2

1

0

'3

'2

'1

'0

1000

0100

00

00

a

a

aa

aa

a

a

a

a

a

a

a

a

),(),,,( 03210 aaaaaaA

),( pEp

)()( 0033221100 babababababaAB

3

The Lorentzian four-vector :

xxyxyxyx

00

likespacex

likelightx

liketimex

xxxxxxx

0

0

0

2

2

2

002

The Lorentz Boost :

TT

LL

LL

pp

vppp

vppp

'

0'

0'0

4

The Special-Relativity spacetime :

)()()()( 2211 plkpek

21

21

21

21 kk

22

22

22

22 kk

221

21

21 mpEp

222

22

22 pEp

iii

iii

pEp

kk

,

,

*2

*2

*1

*1 pkpk

In the LAB

In the CM

**

**

,

,

iii

iii

pEp

kk

Center of Mass Energy

2*1

*1

2*1

*1

2*1

*1

211 )()()()( EpkEpks

*1

*1 Es

Maximal energy that can be transformed in mass 5

*1k

*1p

*2k

*2p

*

A prototype reaction

Dispersion Relations

In a fixed-target configuration :

)()()()( 2211 plkpek

m)( 1k

)( 2k

)( 2pl

6

),0(2

222

11122

111122

1121

21

211

mEpmm

pkEmpkpkpks

At high energies (masses neglected) : ms 12

In a collider situation :

1

21

2111 ,: kkk

2

22

2222 ,: kkk

Let’s assume 12 kk

121 2)( ks

At high energies (masses neglected) :

221

2

212

212

21 kkkks

1k

2k

7

Threshold of a Reaction i

ims Sum of masses in the final state

Example 1: production of a muon with a neutrino beam impinging on e

m)( 1k

)( 2k

)( 2pl

2

12

1222

11

2

2

mm

mmpks

Muon mass GeVm

m11

02.1

3.0130,11

2

22

1

Example 2: muon production in e+e- collisoins (collider)

mk ,

mk ,

22)( 21 ks

Two muons to conserve leptonic numbers MeVk 106

8

Unstable particle: two-body decay

),( 111 pEp

),( 222 pEp

M21 ppP

021

22

22

21

2121

pp

MpmpmEE

Mpmpm 222

221

pp

in this section only

222

221 pmpmM

22

221

21

2 2 mpmMmM 2212121

2 2)( pmMmmmmM

)(42)( 221

22121

2221

221

4 pmMmmmmMmmmmM

2

22

221

2221

221

42

4

22)(

M

mMmMmmmmMp

2

221

2221

2221

221

42

4

)()()(

M

mmMmmMmmmmMp

M

mmMmmMpp

2

)()( 221

2221

2

21

• Possible only if • Momentum uniquely defined

21 mmM

9

…and the energies of the two particles

22

22

22

21

21

21 mEppmE

MEE

mmEE

21

22

21

212

22

21

211 mmEEM

22

21

212 mmMME

2221

21 2

1mmM

ME and, similarly : 212

22

2 2

1mmM

ME

Because of momenum conservation, 1 and 2 are heading in opposite directions in the M reference frame

If 1 and 2 happen to have the same mass :

MEE2

121 22

21 42

1mMpp

21 EEM

10

Two body decays in flight

),0,0,( pEP),( 1,111 zT ppEP

vectorp

vectorP

3

4

),( 2,222 zT ppEP

2-vectorsTp

TTT ppp 21

Momentum conservation in the transverse direction :

*11

*1

*11

*1

*11

TT

zz

z

pp

vEpp

vpEE

*22

*2

*22

*2

*22

TT

zz

z

pp

vEpp

vpEE

Between the CM and the laboratory :

M

E

E

pv 22

222

2

2

2

2

2

2

2

222

)1(11

11

1

vvm

m

E

p

E

m

E

p

mpE

2cmE

222

2222

)1( mcTmccm

mcmccmcmE

in this slide

Kinetic energy and mass energy :

11

Mandelstam variables

Let’s introduce three Lorentz scalars : 1p

2p

3p

4p 223

241

224

231

243

221

ppppu

ppppt

pppps

And :

32312122

23

23

21

22

21

223

231

221 222)()()( pppppppppppppppppputs

4324

2321

22

214321 22 pppppppppppp

24

23

22

2121433

24

23

22

21

3231433344

23

22

21

32312122

23

23

2143

24

2321

)(2

2222

22222

mmmmpppppmmmm

pppppppppppp

pppppppppppppppputs

12

Physical meaning of s: energy available in the center-of-mass

2212

21 EEpps

Physical meaning of t: let us see it in the CM

In the case of an unstable particle decaying : 221 0 Mps

k,1 k

,2

',3 k

*

',4 k

31*'2

321

'2'231

23

21

2'231

231

2cos2

22

)()(

EEkkmm

kkkkEEEE

kkEEppt

Θ*< 900

31'2

3210 22 EEkkmmt

2sin41cos2

*2'

0*'

0

kktkktt

2sin4)(

2sin4

*2'

min

*2'

0

kktkktt

Momentum transfer

13

Three body decay: the Dalitz plot

),( 111 pEp

),( 222 pEp

M

),( 333 pEp

321 pppP

221

233

231

222

232

211

22

)()(

)()(

)()(

pppPs

pppPs

pppPs

MPs

Invariant mass of subsystems

The subsystem invariant masses :

23

22

21

2321 mmmMsss

Let us study the limits of the kinematics variable’s space (phase space)In the CM system:

121

221

21

21

2

121

2211

21

221

21

211

22

22)0()()(

MmmMmpMmM

MEmMpMEEMpEMpPs

211 )(max mMs

14

To find the lower limit we use the CM system of 2 ,3:

232

223

23

22

22

232

232

232

232

211

)(

)()()()(

mmmpmpEE

ppEEpppPs

So that, for every s:

233

221

222

231

211

232

)()(

)()(

)()(

mMsmm

mMsmm

mMsmm

A parallelogram !

One can actually devise a better limit by considering the correlation between the variables. To this goal, let’s use the Jackson frame, defined by )(, 123 pPpp

In this frame : 21

2321 )( EEEEs

221

21

21

22

21

21

22211 )( pmpMpmPMEEs

15

221

21

21

21 pmpMs

),,(4

1 21

21

1

21 mMs

sp xzyzxyzyxzyx 222),,( 222

Inverting, to find the momentum

In addition :

223

23

22

22

232

232

232

232 pmpmEEppEEpps

),,(4

1 23

221

1

23

22 mms

spp

At this point, let us consider the invariant 2312 pps

depends only on

16

)3,1(cos2

cos2

313123

21

313123

21

2312

ppEEmm

ppEEpppps

Let us now suppose to fix 1s

),,(4

1 21

21

1

21 mMs

sp

),,(4

1 23

221

1

23

22 mms

spp

The momenta of 1,2,3 are fixed in magnitude :

2s

)3,1(cos2 3123

23

21

21

23

212 pppmpmmms

23123

23

21

21

23

212 2max spppmpmmms

23123

23

21

21

23

212 2min spppmpmmms

It is possible to express the energies of 1 and 3 as a function of 1, ss

17

22231

1

3211

1

12

1

2

1mms

sEmss

sE

In this way, one obtains the limits of the Dalitz Plot:

),,(),,(2

1 23

221

2/1211

2/123

221

211

1

23

212 mmsmssmmsmss

smms

The Dalitz plot represents the transition between an initial state and a three-body final state. It is built up by using two independent variables.

The Dalitz Plot contours are given by kinematics

The density of dots in the Dalitz Plot is giving information on the dynamics of the final state particles :

21

2

21 ),( dsdsssM

18

19

Invariant Mass

Let us consider the decay of a particle in flight.Let us suppose it decays in three particles (with n particles would be the same)

pE,

),( 111 pEp

),( 333 pEp

),( 222 pEp

The states 1,2,3 are observed in the spectrometerMomenta get measuredA mass hypotesis is made, based on the information from the spectrometer

332211 ,,,,, pmpmpm

Ingredients :

20

Bump hunting in invariant mass distributions :

2321

223

23

22

22

21

21 pppmpmpmpA

This quantity is built up :

But this is a Lorentz scalar. Then, I can compute it (for instance), in the rest frame of the decaying particle :

which can also be written as : 23212

321 pppEEEA

222 0 MMA

???

???

The Upsilon peaks

B0 decay

21

Types of Collisions : the Elastic case

1p

2p

3p

4p

The identity of particles does not change between the initial and the final state

2121

4321

22

22

24

21

21

23

pppp

mppmpp

How many invariants can be used to characterize the collision ? There’s 16 of them… 4,3,2,1, jipp ji

…..both four of them are trivial, since 22ii mp

The remaining 12 are really only 6 six because of symmetry ijji pppp

The remaining six are just two since we have the four conditions of conservation of Energy-Momentum

4321 pppp

We can use 3 Mandelstam variables s,t,u keeping in mind 22

21 22 mmuts

22

Type of Collisions di Collisione : the Inelastic case

1p

2p

3p

4p

np

...

pp

ee

)(inclusiveXepe

And, clearly nppppp .....4321

In a fixed target laboratory frame, with 1 (projectile) impinging on 2

)0,0,0,(),0,0,( 221 mppEp lablab

2432

22

21

221 )....(2)( nlab pppEmmmpps

2432**

4*3

243 ....)....()....( nnn mmmEEEppps

Which can also be calculated in the CM using the final state

23

Threshold Energy in the Center of Mass :

nthr mmmsE .....43min*

….and in the Lab System:

labEmmms 222

21 2 222

12

432

)....(2

1mmmmm

mE n

thrlab

We can also use the Kinetic Energy in the Lab Frame :

2212

432

)(...2

1mmmmm

mT n

thrlab

1mET lablab

Homework - calculate the threshold kinetic energy for the reaction :

pp

24

Wave-Optical description of Hadron Scattering

Propagation of a wave packet: superposition of particle waves of a number of different frequencies:

The wavepacket impinges on a scattering (diffusion) center

)exp(exp)(),(),( ikzEtxpi

pcpdtxtx int

• Neglecting an exp(-iωt) term• Neglecting the structure of the wave-packet

dBk /2

mk 1510 Range of Nuclear Forces

25

kzii e

unaltoutinl

likrikrlkzi

i Peelkr

ie

)(cos)1()12(2

Beam of particles propagating along zDepicted as a time-independent inde plane wave

Spinless collision center

z

ikreikre

Expansion of the incident wave in spherical harmonic functions, in the kr>>1 approximation

entering and exiting

If we now introduce the effect of the diffusion center, we will have a phase shift and a reduction of the amplitude of the out wave

26

ll

il P

i

el

kF

l

)(cos2

1)12(

1)(

2

outinl

likri

likrl

total Peeelkr

il )(cos)1()12(

22

10

2

l

l

Asymptotic form of the global wave

The diffused wave: difference between incident and total wave :

)()(cos

2

1)12(

2

Fr

eP

i

el

kr

e ikr

ll

il

ikr

unaltoutoutitotalscatt

l

Scattering amplitude

Elastic diffusion, with k staying the same (but of general validity in the CM system)

27

Physical meaning of the scattering amplitude

r

efeAr

ikrikz )()( In a situation of the type :

We can consider an incident flux equal to the number of incident particles per cross sectional area of the collision center. This is given by the probability density times the velocity :

fluxscms

cm

cmAvv

23

2* 11

And we have a diffusion flux given by :2

22

r

fAv

Diffusion cross section defined as the number of particles scattered per unit flux in an area subtended by a solid angle dΩ:

vA

dr

r

fAvd 2

2

2

22

2)(

fd

d

28

2)(

Fd

d

el

12

4

ldPP lk

kl

l

il

el i

el

l22

2

2

1)12(4

As a general result :

ll

il P

i

el

kF

l

)(cos2

1)12(

1)(

2

Legendre polynomials orthogonality

Integrating over the solid angle :

Total elastic cross section

1l No absorption and diffusion only due to phase shifts

l

llel l 22 sin)12(4)1(

29

droutinr222

llr l 22 1)12(

l

llelrT l 2cos12)12(2

In a more general case (η<1) we can divide the cross section between a reaction part and an elastic part :

l

likrl

in Pelkr

i)(cos)1()12(

2

ll

ikrilout Peel

kr

il )(cos)12(

22

The total cross section :

Phase shift part (with or without absorption)

Non-zero absorption

Computed with the probability loss Effect on the outgoing wave

30

Optical Theorem :

Let us consider the amplitude for forward scattering :

ll

il P

i

el

kF

l

)1(2

1)12(

1)0(

2

l

lllk

F 2cos12)12(2

1)0(Im

T

kF

4)0(Im

Relation between the total cross section and the forward amplitude

31

)12(4),1( 2max llel

)12(),0( 2max llr

Unitarity Limit on the cross section due to conservation of probability

If one starts from the fully elastic case :

l

llel l 22 sin)12(4)1(

The maximum cross-section for the l wave takes place when 2

l

The maximum absorption cross section takes place when

This gives rise to a semiclassical interpretation: angular momentum and impact parameter

0l

32

b

p

lbp lb

Particles between l and l+1 are absorbed by an annular area

)12(2221 lbb lll

Role of the various angular momentum waves : a given angular momentum is related to a given impact parameter :

)12(),0( 2max llr

Semiclassical interpretation: angular momentum and impact parameter

33

l

lil

il e

ii

i

elf

22

222

1)(

Scattering amplitude for the l wave

Im f

Re f0

i/2Unitarity Circle

f(η=1)

2δ

η=1: f traces a circle with radius ½, centered in i/2, with phase shift between 0 and π/2

The maximum module is reached at π/2: resonance in the scattering amplitude

η<1 : f has a raiuds smaller than the Unitarity Circle

The vector cannot exceed the Unitarity Circle a limit to the cross section

iii

lf l )1(22

)2/,1,( i

34

Resonance and Breit –Wigner formula

i

ei

eeef i

iii

cot

1sin

2

2...........)(cot)()(cot)(cot R

EERR EEE

dE

dEEEE

R

0)(cot RE

REE

EdE

d

)(cot

2

Goal: to express the behaviour of the cross section near to a resonance, i.e. when the scattering amplitudes goes through π/2 (spinless particles case)

At resonance δ = π/2Power series expansion

Resonance energy

Assuming

We obtain :

RR EEE

2/)(

2/

cot

1)(

iEEi

EfR

Breit – Wignerformula

35

4/)(

4/)12(4)(

22

22

R

el EElE

2/exp)0()0()( 2/ R

tti iEteet R

0 0

2/exp)0()()( iEiEtdtdtetE RtiE

Using the Breit-Wigner formula, one obtains - for the case when a given l is predominant :

l

il

el i

el

l22

2

2

1)12(4

This is a quantum dependence on energy, that corresponds to a temporale dependence of the state of the type :

/* )0()( teItI Decay law of a particle

The Fourier transform of the decay law gives the E dependence :

36

2/)(

2/exp)(0

iEE

KiEiEtdtE

RR

4/)(

4/)12(4)(

22

22

R

el EElE

4/)(

4/

)12()12(

)12(4)(

22

22

Rba

el EEss

JE

In the case of an elastic resonance, the cross section is proportional to the square modulus of this amplitude :

This holds for elastic collisions of spinless particles. In general, if we form a spin J resonance by making spin Sa and Sb particles collide, one has :