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)
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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 :
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(