Relativistic Kinematics

1

1. Costituents of Matter 2. Fundamental Forces 3. Particle Detectors 4. Symmetries and Conservation Laws 5. Relativistic Kinematics 6. The Static Quark Model 7. The Weak Interaction 8. Introduction to the Standard Model 9. CP Violation in the Standard Model (N. Neri)

Recalling Relativistic Kinematics (Special Relativity) 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=2.9979⋅108 m/s (Time is not a relativistic invariant) (Space is not a relativistic invariant)

2

Four-vector : For example, for a particle Minkowski pseudo-euclidean metric Scalar 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

100001000000

aa

aaaa

aaaa

aaaa

γβγβγγ

γβγβγγ

),(),,,( 03210 aaaaaaA ==

),( pEp =

)()( 0033221100 babababababaAB⋅−=−−−=

3

The Lorentzian four-vector :

νµµνη xxyxyxyx =−=

00

likespacexlikelightxliketimex

xxxxxxx

−<

−=

−>

−==

000

2

2

2

002

The Lorentz Boost :

( )( )

TT

LL

LL

ppvpppvppp

=

+=

+=

'

0'

0'0

γ

γ

4

The Special-Relativity spacetime :

)()()()( 2211 plkpek +→+ νν

21

21

21

21 εω =−= kk

22

22

22

22 εω =−= kk

221

21

21 mpEp =−=

222

22

22 µ=−= pEp

( )( )iii

iii

pEpkk

,,

== ω

0*2

*2

*1

*1 =+=+ pkpk

In the LAB

In the CM

( )( )**

**

,

,

iii

iii

pEpkk

=

= ω

Center of Mass Energy

2*1

*1

2*1

*1

2*1

*1

211 )()()()( EpkEpks +=+−+=+= ωω

*1

*1 Es +=ω

Maximal energy that can be transformed in mass 5

*1k

*1p

*2k

*2p

*θ

A prototype reaction

Dispersion Relations

22112211 pkpkEE +=++=+ ωω4-momentum conservation

*2

*2

*1

*1 EE +=+ ωω

3-momentum

Energy conservation

In a fixed-target configuration :

)()()()( 2211 plkpek +→+ νν•

m)( 1kν

)( 2kν

)( 2pl

6

( )),0(2

222

11122

111122

1121

21

211

mEpmm

pkEmpkpkpks

==++=

=−++=++=+=

ωε

ωε

At high energies (masses neglected) : ms 12ω≅

In a collider situation :

+= 1

21

2111 ,: kkk

εω

+= 2

22

2222 ,: kkk

εωLet’s assume 12 kk

−=

121 2)( ks

≅+= ωωAt high energies (masses neglected) :

( ) ( ) ( ) ( )221

2

212

212

21 ωωωω +=+−+=+= 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)( 1kν

)( 2kν

)( 2pl

( )2

12

1222

11

2

2

µω

ωε

≥+

≅++=+=

mm

mmpks

Muon mass GeVm

m 1102.1

3.0130,112

22

1 ≅−

≅−

≥µω

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

mk ,

mk ,

µ

µ

µωω 22)( 21 =≅+= ks

Two muons to conserve leptonic numbers MeVk 106== µ

8

Unstable particle: two-body decay

•),( 111 pEp

= ),( 222 pEp =M

21 ppP +=021

22

22

21

2121

=+

=+++=+

ppMpmpmEE

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

422)(

MmMmMmmmmMp −−+−+

=

( )2

221

2221

2221

221

42

4)()()(

MmmMmmMmmmmMp −−+−+−+

=

( )( )M

mmMmmMpp

2)()( 2

2122

212

21

+−−−==

• Possible only if • Momentum uniquely defined

21 mmM +≥

9

…and the energies of the two particles

22

22

22

21

21

21 mEppmE −===−

MEEmmEE

=+

+−=

21

22

21

212

22

21

211 mmEEM +−=−

22

21

212 mmMME +−=+−

( )22

21

21 2

1 mmMM

E −+= and, similarly : ( )21

22

22 2

1 mmMM

E −+=

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

If 1 and 2 happen to have the same mass :

MEE21

21 == 2221 4

21 mMpp −==

21 EEM =−

10

Two body decays in flight

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

=

vectorpvectorP

−=−=

34

),( 2,222 zT ppEP =

2-vectors Tp

TTT ppp 21

−=≡Momentum conservation in the transverse direction :

( )( )*11

*1

*11

*1

*11

TT

zz

z

ppvEpp

vpEE

=

+=

+=

γ

γ ( )( )*22

*2

*22

*2

*22

TT

zz

z

ppvEpp

vpEE

=

+=

+=

γ

γBetween the CM and the laboratory :

ME

Epv == γ22

222

2

2

2

2

2

2

2

222

)1(1111

1

vvm

mEp

Em

Ep

mpE

=−−=−=−=

−=

+=

γγ2cmE γ=

222

2222

)1( mcTmccmmcmccmcmE

+=+−=

=+−==

γ

γγ

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

mmmmpppppmmmmpppppppppppp

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

*θ

',4 k

−

( )

31*'2

321

'2'231

23

21

2'231

231

2cos2

22

)()(

EEkkmm

kkkkEEEE

kkEEppt

−++=

=+−−−+=

=−−−=−=

θ

Θ*< 900

022 31'2

3210 <−++= EEkkmmt

( )2

sin41cos2*

2'0

*'0

θθ kktkktt

−=−+=

2sin4)(

2sin4

*2'

min

*2'

0θθ kktkktt

+−=+−=−

Momentum transfer

13

Elements of Collisions Theory for the specific case of :

Classical Collisions • Mass is conserved

• 3-momentum is conserved

• T may or may not be conserved

DCBA +→+

DCBA mmmm +=+

DCBA pppp +=+

• T decreases (it can be transformed into heat or chemical energy) . Extreme case: A+BC (two particles sticking together).

• T increases (internal energy is released). Extreme case: AC+D (a particle breaks up)

• Classical Elastic Collision

DCBA TTTT +>+

DCBA TTTT +<+

DCBA TTTT +=+

Because of the conservation of mass, the extreme cases need to be :

CBA mmm =+

DCA mmm +=

14

Relativistic Collisions

• Energy is conserved

• 3-momentum is conserved

• T may or may not be conserved

DCBA +→+

DCBA EEEE +=+

DCBA pppp +=+

• T decreases (rest energy increases) . • Heavier particles can be produced.

• T increases (rest energy decreases). Extreme case: AC+D (a particle decays)

• Elastic Collision (rest energy conserved)

DCBA TTTT +>+

DCBA TTTT +<+

DCBA TTTT +=+

Mass is conserved only in elastic collisions

15

Reactions in the s or in the t channel (timelike or spacelike photons)

a b

q 2200

22 )()()( bababaq

baqqba

−−−=−=

−=+=

ba mm =

)(2

22

002

2200

20

20

2

babam

babababaq

+−

=+−−−+=Now, choosing a frame where

ba

−=

0)(4

)(2

)(2

20

2

220

20

2

220

22

<−

=+−−

=−−=

ammaamaamq This is a space-like photon that

enters diffusion-like graphs

a b

q The final state does not have photon quantum numbers.

16

a

b

q

ba mm = 2200

22 )()()( bababaq

baq

+−+=+=

+=

Now, choosing a frame where

ba

−=

In the case where the final state does have the photon quantum numbers :

0)( 200

2 >+= baq

This is a time-like photon that enters annihilation graphs :

a

b

q

17

Three body decay: the Dalitz plot

•

),( 111 pEp = ),( 222 pEp

=M

),( 333 pEp =

321 pppP ++=

221

233

231

222

232

211

22

)()(

)()(

)()(

pppPspppPspppPs

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 −=

18

To find the lower limit we use the CM system of 2 ,3 (Jackson frame) :

( ) ( )232

223

23

22

22

232

232

232

232

211

)(

)()()()(

mmmpmpEE

ppEEpppPs

+≥+++=+=

=+−+=+=−=

So that, for every s:

233

221

222

231

211

232

)()(

)()(

)()(

mMsmmmMsmmmMsmm

−≤≤+

−≤≤+

−≤≤+

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

+−+=+−+=−=

19

( )221

21

21

21 pmpMs

+−+=

),,(41 2

12

11

21 mMs

sp λ= xzyzxyzyxzyx 222),,( 222 −−−++=λ

Inverting, to find the momentum

In addition :

( ) ( ) ( ) ( ) ( )223

23

22

22

232

232

232

232 pmpmEEppEEpps

+++=+=+−+=+=

),,(41 2

3221

1

23

22 mms

spp λ==

At this point, let us consider the invariant ( )2312 pps +=

depends only on

20

( ) ( )( ) )3,1(cos2

cos2

313123

21

313123

21

2312

ααα

α

=−++=

=−++=+=

ppEEmm

ppEEpppps

Let us now suppose to fix 1s

),,(41 2

12

11

21 mMs

sp λ= ),,(

41 2

3221

1

23

22 mms

spp λ==

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

2s α

( ) )3,1(cos2 3123

23

21

21

23

212 ααα =−++++= 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

21

( ) ( )22

231

13

211

11 2

12

1 mmss

Emsss

E −+=−−=

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

( )( )[ ]),,(),,(21 2

3221

2/1211

2/123

221

211

1

23

212 mmsmssmmsmss

smms λλ±+−−−++=±

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 :

212

21 ),( dsdsssM≈ρ

22

23

What does really a Dalitz Plot represent ?

2* )()()( 21 πφπ αα KKAeKKAeAR ii ++≈

+−++ → πKKDs

Which admits a formal analogy with :

i

πφ

*KK

πKK

f

1ψ

2ψ

3ψ

24

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 spectrometer Momenta get measured A mass hypotesis is made, based on the information from the spectrometer

332211 ,,,,, pmpmpm Ingredients :

25

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

26

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

ppppmppmpp

+=+

====

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

…..but 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 +=++

A new definition of Elasticity !

27

Type of Collisions : the Inelastic case

1p

2p

3p

4p

np

...

++++ +++→+ ππππ pp

−+−+ → µµee

)(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

A new definition of Elasticity !

28

Threshold Energy in the Center of Mass :

nthr mmmsE +++== .....43min*

….and in the Lab System:

labEmmms 222

21 2++= [ ]2

221

243

2

)....(2

1 mmmmmm

E nthrlab −−+++=

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

( )[ ]221

243

2

)(...2

1 mmmmmm

T nthr

lab +−+++=

1mET lablab −=

Homework - calculate the threshold kinetic energy for the reaction :

++++ +++→+ ππππ pp

29

Particle propagation in Space

For a photon one has : λ

ν hphW ==

So that the phase :

−=−=−

TtxhthxhWtpx

λν

λ

According to the de Broglie hypotesis, for any particle:

−= )(2exp),( Wtxp

hiAtxu π

pdWtxph

ipctx ∫

−= )(2exp)(),( πψ

A real particle is a superposition of plane waves :

22)( mppW +=

mppW2

)(2

=

with the appropriate dispersion law :

30

Particle propagation in Space

For a photon one has : λ

ν hphW ==

So that the phase :

−=−=−

TtxhthxhWtpx

λν

λ

According to the de Broglie hypotesis, for any particle:

−= )(2exp),( Wtxp

hiAtxu π

pdWtxph

ipctx ∫

−= )(2exp)(),( πψ

A real particle is a superposition of plane waves :

22)( mppW +=

mppW2

)(2

=

with the appropriate dispersion law :

31

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)(),(),( ikzEtxpipcpdtxtx int≅

−≈→ ∫−∞→

ψψ

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

dBk λλλπ == /2

mk 1510−≈ Range of Nuclear Forces

32

kzii e=ψ

[ ] unaltoutinl

likrikrlkzi

i Peelkrie −

− +=−−+== ∑ ψψθψ )(cos)1()12(2

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

We wll consider a spinless collision center in the origin :

z θ

ikre+≈ikre−≈

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

33

( )∑ −+=

ll

il P

iel

kF

l

)(cos2

1)12(1)(2

θηθδ

[ ] outinl

likri

likrl

total Peeelkri

l ψψθηψ δ +=−−+= ∑ − )(cos)1()12(2

2

102<<

∈=∆

l

ll Rη

δδφAsymptotic form of the global wave

The diffused wave: difference between incident and total wave :

( ) )()(cos2

1)12(2

θθηψψψψψδ

Fr

ePi

elkre ikr

ll

il

ikr

unaltoutoutitotalscatt

l

=−

+=−=−= ∑−

Scattering amplitude

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

34

Physical meaning of the scattering amplitude

+=

reFeA

ikrikz

total )(θψWhat we have written is equivalent to :

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

cmcm

Avv === 232* 11ψψ

And we have a diffusion flux given by : 2

22

rF

Av

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

vAdr

rf

Avd 2

2

2

22 Ω

=σ 2)(θσ fdd

=Ω

35

2)(θσ Fdd

el

=

Ω

∫ +=Ω

124

ldPP lk

klδπ

( )∑ −+=

l

il

el iel

l22

2

21)12(4

δηπσ

As a general result :

( )∑ −+=

ll

il P

iel

kF

l

)(cos2

1)12(1)(2

θηθδ

Legendre polynomials orthogonality

Integrating over the solid angle :

Total elastic cross section

1=lη No absorption and diffusion only due to phase shifts

∑ +==l

llel l δπησ 22 sin)12(4)1(

36

( )∫ Ω−= droutinr222 ψψσ ( )∑ −+=

llr l 22 1)12( ηπσ

( )∑ −+=+=l

llelrT l δηπσσσ 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 Pelkri )(cos)1()12(

2θψ [ ]∑ +=

ll

ikrilout Peel

kri

l )(cos)12(2

2 θηψ δ

The total cross section :

Phase shift part (with or without absorption)

Non-zero absorption

Computed with the probability loss Effect on the outgoing wave

37

Optical Theorem : The scattering amplitude is a complex quantity The scattering amplitude is not well defined for θ = 0

How can we get information on F close to (or at) θ = 0 ?

Let us consider the amplitude for forward scattering :

( )∑ −+==

ll

il P

iel

kF

l

)1(2

1)12(1)0(2 δηθ

( )∑ −+=l

lllk

F δη 2cos12)12(21)0(Im T

kF σπ4

)0(Im =

Relation between the total cross section and the forward amplitude

38

)12(4),1( 2max +== llel πησ

)12(),0( 2max +== llr πησ

Unitarity Limit on the cross section due to conservation of probability

If one starts from the fully elastic case :

∑ +==l

llel l δπησ 22 sin)12(4)1(

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

The maximum absorption cross section takes place when 0=lη

And in any case, for a given non-zero ηl the value of Also gives a maximum for the total cross section 2

πδ =l

( )∑ −+=+=l

llelrT l δηπσσσ 2cos12)12(2

39

b

p

lbp = lb =

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

( ) )12(2221 +=−= + lbb lll ππσ

Role of the various angular momentum waves : a given angular momentum is related to a given impact parameter . Suppose a particle is impinging on a target, with impact parameter b. The angular momentum can be expressed in units of the Planck constant :

)12(),0( 2max +== llr πησ

Semiclassical interpretation: angular momentum and impact parameter

lbl =

A bigger impact parameter is related to a different unit of angular momentum )1(1 +=+ lbl

40

ll

ili

l eiii

elf δδ ηη 2

2

2221)( −=

−=

Scattering amplitude for the l wave

Im f

Re f 0

i/2 Unitarity 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

iiilf l =−−=== )1(22

)2/,1,( πδηi

41

Resonance and Breit –Wigner formula

( )i

ei

eeef iiii

−==

−=

−

δδδ

δδδ

cot1sin

2

( )Γ

−−≅+

−+=

=

2...........)(cot)()(cot)(cot REE

RR EEEdEdEEEE

R

δδδ

0)(cot =REδ

REE

EdEd

=

−=

Γ)(cot2 δ

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/2 Power series expansion

Resonance energy

Assuming

We obtain :

RR EEE <<Γ≅−

2/)(2/

cot1)(

Γ−−Γ

=−

=iEEi

EfRδ

Breit – Wigner formula

42

4/)(4/)12(4)( 22

22

Γ+−Γ

+=R

el EElE πσ

( )[ ]2/exp)0()0()( 2/ Γ+−== −−R

tti iEteet R ψψψ τω

( )[ ]∫ ∫∞ ∞

−+Γ−==0 0

2/exp)0()()( iEiEtdtdtetE RtiE ψψχ

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

( )∑ −+=

l

il

el iel

l22

2

21)12(4

δηπσ

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

τψψ /* )0()( teItI −== Decay law of a particle

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

43

( )[ ]2/)(

2/exp)(0 Γ−−

=−+Γ−= ∫∞

iEEKiEiEtdtE

RRχ

4/)(4/)12(4)( 22

22

Γ+−Γ

+=R

el EElE πσ

4/)(4/

)12()12()12(4)( 22

22

Γ+−Γ

+++

=Rba

el EEssJE πσ

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 :

44

22max 8)12(42)0,1( ππησ =+×=== llel

In the case of an elastic scattering in l-wave : pp ++++ →∆→ ππ )1232(

A spin factor (for the proton)