Relativistic Kinematics
description
Transcript of Relativistic Kinematics
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 :