Transcript Recalling Relativistic Kinematics Basic Principles

Relativistic Kinematics
1. Costituents of Matter
2. Fundamental Forces
3. Particle Detectors (N. Neri)
4. Experimental highlights (N. Neri)
5. Symmetries and Conservation Laws
6. Relativistic Kinematics
7. The Static Quark Model
8. The Weak Interaction
9. Introduction to the Standard Model
10. CP Violation in the Standard Model (N. Neri)
1
 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=2.9979108 m/s
(Time is not a relativistic invariant)
(Space is not a relativistic invariant)
2
Four-vector :

A  (a0 , a1 , a2 , a3 )  (a0 , a)
For example, for a particle
Minkoski pseudo-euclidean metric
Scalar product:

p  ( E , p)
 
AB  (a0b0  a1b1  a2b2  a3b3 )  (a0b0  a  b )
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:
a'   
  0 0   a   a  a 
 0'  
 a1    
a'    0
 2 
 a'   0
 3 

0
0
1

  0  0
0 0   a1    a 0  a1 



a2
1 0   a2  

   





a3
0 1   a3  

3
The Lorentzian four-vector :

x y  x0 y0  x y  x x
The Special-Relativity spacetime :

x2  x x  x0 x0  x x
x2  0
tim e like
x2  0
light  like
x2  0
space like
The Lorentz Boost :

  p

 vp 
p L0 '   p 0  vpL
p L'


pT'  pT
0
L
4
 (k1 )  e ( p1 )  (k2 )  l ( p2 )
A prototype reaction
In the
LAB
In the
CM



k i  i , k i

pi  Ei , pi 


*
*
k i  i , k i
* *
p i  E i , pi

Dispersion
Relations

2 2
k   k1  1

p12  E12  p12  m2
2
1
*  * *  *
k1  p1  k2  p2
2
1
*
k2
*
k1
2 2
k   k2   2

p22  E22  p22   2
2
2
*
2
2

p1*

p2*
Center of Mass Energy
*  * 2
s  (k1  p1 )  (  E )  (k1  p1 )  (1*  E1* )2
2
*
1
* 2
1
s  1*  E1*
Maximal energy that can be transformed in mass
5
In a fixed-target configuration :
 (k 2 )
 (k1 )  e ( p1 )  (k2 )  l ( p2 )

 (k1 )
m

2
2
2
2
2
s k1  p1   k1  p1  2k1 p1   m  21E1  2k1 p1
 
2
2
  m  21m
( p1  0, E1  m)
s 
At high energies (masses neglected) :
Let’s assume

k2
2

2

2
2



k1 :  1  k1   1 , k1  k 2 :  2  k 2   2 , k 2 






k2   k1
s  k1  k2    1   2 
2
21m

k1
In a collider situation :
l ( p2 )
2

 
 k1  k2
At high energies (masses neglected) :

2
  1  2 
2

s  ( 1  2)  2 k1
6
Threshold of a Reaction
s   mi
Sum of masses in
the final state
i
Example 1: production of a muon with a neutrino beam impinging on e
s  k1  p1    2  m 2  2 1m 
 (k 2 )
2
m 2  2 1m   2
Muon mass

 (k1 )
1 
m
 2  m2
2m
l ( p2 )

Example 2: muon production in e+e- collisoins (collider)

s  ( 1  2)  2 k  2 

k   106 MeV
Two muons
to conserve
leptonic
numbers
11,130 0.3
 11 GeV
1.02


k,m

k,m

7
Unstable particle: two-body decay

E1  E2  m12  p12 
 
p1  p2  0
P  p1  p2

p p

m22  p22  M

p1  ( E1 , p1 )
M


p2  ( E2 , p2 )
in this section only
m12  p 2  m22  p 2  M
M  m12  p 2  m22  p 2
M 2  m1  m2  (m1  m2 )  2M m12  p 2
M 2  m12  2M m12  p 2  m22
M 4 m1  m2  (m1  m2 )2  2M 2 m1  m2 m1  m2   4M 2 (m12  p2 )
2
M 4  m1  m2  (m1  m2 ) 2  2M 2 m12  2M 2 m22
p 
4M 2
2
M 4  m1  m2  (m1  m2 ) 2  M 2 (m1  m2 ) 2  M 2 (m1  m2 ) 2
2
p 
4M 2
2
2


p1  p2 
M
2

 (m1  m2 ) 2 M 2  (m1  m2 ) 2
2M

• Possible only if
M  m1  m2
• Momentum uniquely defined
8
…and the energies of the two particles


E12  m12  p12  p22  E22  m22
E2  E12  m12  m22
E1  E2  M
M  E1  E2
M  E1  E12  m12  m22
 2ME1  M 2   m12  m22
E1 

1
M 2  m12  m22
2M

and, similarly :
E2 

1
M 2  m22  m12
2M

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 :
1
E1  E2  M
2


1
p1  p2 
M 2  4m 2
2
9
P  4  vector
Two body decays in flight
p  3  vector
in this slide

P1  ( E1 , p1T , p1z )
P  ( E,0,0, p)

pT
2-vectors

P2  ( E2 , p2T , p2 z )



pT  p1T   p2T
Momentum conservation in the transverse direction :
Between the CM and the
laboratory :

  p

 vE 

  p

 vE 
E1   E1*  vp1*z
E2   E2*  vp2* z
*
p1z
1z


p1T  p1*T
*
p2 z
2z


p2T  p2*T
Kinetic energy and mass energy :
E 2  p 2  m2
p2
m2
 1 2
E2
E
2
p
m2
1

1


1

1  (1  v 2 )  v 2
2
2 2
2
E
 m

*
1
*
2
E  m c 2  m c 2  m c2  m c2 
 m (  1) c 2  m c2  T  m c2
E m c
2
v
p
E

E
M
10
Mandelstam variables
Let’s introduce three Lorentz scalars :
p1
p3
p2
p4
s   p1  p2    p3  p4 
2
2
t   p1  p3    p4  p2 
2
2
u   p1  p4    p3  p2 
2
2
And :
s  t  u  ( p1  p2 )2  ( p1  p3 )2  ( p3  p2 )2  p12  p22  p12  p32  p32  p22  2 p1 p2  2 p1 p3  2 p2 p3
p1  p2  p3  p4  p12  p22   2 p1 p2  p32  p42  2 p3 p4
s  t  u   2 p1 p2  p32  p42  2 p3 p4  p12  p32  p32  p22  2 p1 p2  2 p1 p3  2 p2 p3 
 p12  p22  p32  p44  2 p3 p3  2 p3 p4  2 p1 p3  2 p2 p3 
 m12  m22  m32  m42  2 p3 ( p3  p4  p1  p2 )  m12  m22  m32  m42
11
Physical meaning of s: energy available in the center-of-mass
s   p1  p2   E1  E2 
2
2
s   p1  0  M 2
2
In the case of an unstable particle decaying :
'
3, k
Physical meaning of t: let us see it in the CM


 '
t0  m  m  2 k k  2 E1E3
2
1
2
3

2,  k
*

1, k
 ' 2
2
2
t  ( p1  p3 )  ( E1  E3 )  k  k 
 2  '2
'
2
2
 E1  E3  2 E1 E3  k  k  2k k 
 '
2
2
 m1  m3  2 k k cos *  2 E1 E3
'
4,  k
Θ*< 900
*
 '
 '
*
2
t  t0  2 k k cos  1  t0  4 k k sin
2


*
*
 '
 '
2
2
 t   t0  4 k k sin
 (t ) min  4 k k sin
2
2
Momentum
transfer
12
Three body decay: the Dalitz plot
sP M
2
P  p1  p2  p3

p1  ( E1 , p1 )
2
s1  ( P  p1 )  ( p2  p3 )
2
Invariant mass of
subsystems
2
s2  ( P  p2 ) 2  ( p1  p3 ) 2
s3  ( P  p3 )  ( p1  p2 )
2
2
M


p2  ( E2 , p2 )

p3  ( E3 , p3 )
The subsystem invariant masses :
s1  s2  s3  M 2  m12  m22  m32
Let us study the limits of the kinematics variable’s space (phase space)
In the CM system:
  2
2
2
2
s1  ( P  p1 )  ( M  E1 )  (0  p1 )  M  E1  2ME1  p1  M 2  m12  2ME1 
2
2
2
 M  m1  2M p1  m12  M 2  m12  2Mm1
2
2
max s1  (M  m1 )2
13
To find the lower limit we use the CM system of 2 ,3:
 
s1 ( P  p1 ) 2  ( p2  p3 ) 2  ( E2  E3 ) 2 ( p2  p3 ) 2 
2
2
2
2
2
2
2
 ( E2  E3 )  p2  m2  p3  m3  m2  m3 


So that, for every s:
(m2  m3 ) 2  s1  ( M  m1 ) 2
(m1  m3 ) 2  s2  ( M  m2 ) 2
(m1  m2 )  s3  ( M  m3)
2
A parallelogram !
2
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 
p3   p2 , ( P  p1 )
s1  E2  E3   (E  E1 )2
2
In this frame :
s1  ( E  E1 ) 
2

2

M  P  m12  p12
2
 
2


M  p12  m12  p12
2

2
14
s1 



M  p12  m12  p12
2

2
Inverting, to find the momentum
2
2
2
2 1
2
2

(
x
,
y
,
z
)

x

y

z
 2xy  2 yz  2xz
p1 
 ( s1 , M , m1 )
4s1
In addition :
s   p2  p3   E2  E3 
2
2

 2
2
  p2  p3   E2  E3  

2
2
2
m  p2  m3  p3
2
2

2
2 2 1
p2  p3 
 ( s1 , m22 , m32 )
4s1
At this point, let us consider the invariant
s2   p1  p3 
2
15
 
2
s2   p1  p3   p12  p32  2 E1E3  p1 p3 cos 
 
 m12  m32  2E1E3  p1 p3 cos 
   (1,3)
Let us now suppose to fix
The momenta of 1,2,3
are fixed in magnitude :
s1
2 2 1
2 1
2
2
p
p1 
 ( s1 , M , m1 ) 2  p3   ( s1 , m22 , m32 )
4s1
4s1
s2
depends only on


s2  m12  m32  2 m12  p12


 
m32  p32  p1 p3 cos 


min s  m  m  2 m  p

max s2  m12  m32  2 m12  p12
2
2
1
2
3
2
1
2
1

   (1,3)


p  s

 
m32  p32  p1 p3  s2


m32  p32  p1
3
2
It is possible to express the energies of 1 and 3 as a function of s, s1
16
E1 

1
s  s1  m12
2 s1

E3 

1
s1  m32  m22
2 s1

In this way, one obtains the limits of the Dalitz Plot:
s2  m12  m32 



1
s  s1  m12 s1  m22  m32  1/ 2 ( s1 , s, m12 )1/ 2 ( s1 , m22 , m32 )
2s1

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 :
  M ( s1 , s2 ) ds1ds 2
2
17
18
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)

p1  ( E1 , p1 )

p2  ( E2 , p2 )

E, p

p3  ( E3 , p3 )
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
Ingredients :



m1 , p1 , m2 , p2 , m3 , p3
19
This quantity is built up :
A




p12  m12  p22  m22  p32  m32
A  E1  E2  E3 
which can also be written as :
2
   p  p  p 
2
2
1
2
3
   2
  p1  p2  p3 
But this is a Lorentz scalar. Then, I can compute it (for instance), in the rest
frame of the decaying particle :
A  M 
2

2
 0 M 2
Bump hunting in invariant mass distributions :
???
The Upsilon peaks
???
B0 decay
20
Types of Collisions : the Elastic case
The identity of particles does not change
between the initial and the final state
1 2 1 2
p32  p12  m12
p42  p22  m22
p1  p2  p3  p4
p1
p3
p2
p4
How many invariants can be used to characterize the collision ?
There’s 16 of them…
p p
i, j  1,2,3,4
i
j
…..both four of them are trivial, since
pi2  mi2
The remaining 12 are really only 6 six because of symmetry
pi p j  p j pi
The remaining six are just two since we have the four
conditions of conservation of Energy-Momentum
p1  p2  p3  p4
We can use 3 Mandelstam variables s,t,u keeping in mind
s  t  u  2m12  2m22
21
Type of Collisions di Collisione : the
Inelastic case
p3
   p     p    
p1
p4
e e     
e  p  e  X
And, clearly
...
p2
(inclusive)
pn
p1  p2  p3  p4  ..... pn
In a fixed target laboratory frame, with 1 (projectile) impinging on 2
p1  ( Elab ,0,0, plab )
p2  (m2 ,0,0,0)
s  ( p1  p2 )2  m12  m22  2m2 Elab  ( p3  p4  .... pn )2
Which can also be calculated in the CM using the final state
s  ( p3  p4  .... pn )2  (E3*  E4*  .... En* )2  m3  m4  .... mn 
2
22
Threshold Energy in the Center of Mass :
*
Ethr
 smin  m3  m4  ..... mn
….and in the Lab System:
thr
Elab

s  m  m  2m2 Elab
2
1
2
2
We can also use the Kinetic Energy in
the Lab Frame :
thr
Tlab


1
(m3  m4  .... mn ) 2  m12  m22
2m2
Tlab  Elab  m1

1
m3  m4  ... mn 2  (m1  m2 ) 2
2m2

Homework - calculate the threshold kinetic energy for the reaction :
   p     p    
23

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


 
 i 

 ( x,t )   in ( x,t )   dp c( p)exp  px  Et   exp(ikz) k  2 / 
t 


  dB
• Neglecting an exp(-iωt) term
• Neglecting the structure of the wave-packet
k 1015 m
Range of Nuclear Forces
24
Beam of particles propagating along z
Depicted as a time-independent inde plane wave
 i  ei kz
 e ikr
Spinless collision center
 e ikr

z
Expansion of the incident wave in spherical harmonic functions, in the kr>>1
approximation
i  e
i kz


i
l ikr
ikr

(
2
l

1
)
(

1
)
e

e
Pl (cos )  in  out unalt

2kr l
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
25


i
2i l ikr
l ikr
 total 
(
2
l

1
)
(

1
)
e


e
e Pl (cos )  in  out

l
2kr l
  2 l
Asymptotic form of the global wave
0  l  1
The diffused wave: difference between incident and total wave :
 scatt   total  i   out  out unalt



l e2i l  1
eikr
eikr

(2l  1)
Pl (cos )  F ( )

kr l
2i
r

l e 2i l  1
1
F ( )   (2l  1)
Pl (cos )
k l
2i
Elastic diffusion, with k staying the same (but of
general validity in the CM system)
Scattering amplitude
26
Physical meaning of the scattering amplitude
In a situation of the type :
 ikz
eikr 
 (r  )  A e  f ( ) 
r 

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 :
  v v A
*
2
And we have a diffusion flux given by :
1 cm
1
 2  flux
3
cm s cm s
vA
2
f
2
r2
Diffusion cross section defined as the number of particles scattered per unit flux
in an area subtended by a solid angle dΩ:
d  v A
2
f
2
r2
r 2d 
2
A v
d
2
 f ( )
d
27
As a general result :
 d 

  F ( )
 d el

2

l e 2i l  1
1
F ( )   (2l  1)
Pl (cos )
k l
2i
4  lk
 Pl Pk d  2l  1
Legendre polynomials orthogonality
Integrating over the solid angle :
 el  4
2

l
l 1

e
(2l  1)
2 i l
l

1
2
Total elastic cross section
2i
No absorption and diffusion only due to phase shifts
 el (l  1)  4 2  (2l  1) sin 2  l
l
28
In a more general case (η<1) we can divide the cross section between a reaction
part and an elastic part :
 in 




i
i
l ikr
(
2
l

1
)
(

1
)
e
P
(cos

)


(2l  1) l e 2i l eikr Pl (cos )


l
out
2kr l
2kr l

 r    in   out
2
2
r
2
 r   2  (2l  1) 1 l2 
d
l
The total cross section :
 T  r   el   2  (2l  1) 2 1 l cos2 l 
l
Non-zero
absorption
Computed with the probability loss
Phase shift part (with or
without absorption)
Effect on the outgoing wave
29
Optical Theorem :
Let us consider the amplitude for forward scattering :


l e2i l  1
1
F (  0)   (2l  1)
Pl (1)
k l
2i
Im F (0) 
1
(2l  1) 2 1  l cos 2 l 

2k l
k
Im F (0) 
T
4
Relation between the total cross section and the forward amplitude
30
Unitarity
Limit on the cross section due to
conservation of probability
If one starts from the fully elastic case :
 el (l  1)  4 2  (2l  1) sin 2  l
l
The maximum cross-section for the l wave takes place when
l 

2
 elmax (  1, l )  4 2 (2l 1)
The maximum absorption cross section takes place when
l  0
 rmax (  0, l )   2 (2l  1)
This gives rise to a semiclassical interpretation: angular momentum and impact
parameter
31
Semiclassical interpretation: angular momentum and impact parameter
 rmax (  0, l )   2 (2l  1)
Role of the various angular momentum waves : a given angular momentum
is related to a given impact parameter :
p
b
pb  l
b  l
Particles between l and l+1 are absorbed by an annular area
 l   bl21  bl2   2 (2l  1)
32
Scattering amplitude for the l wave
f (l ) 
Im f
l
2i
il 2i l
 
e
2 2
f (l , 1, l  / 2) 
i
Unitarity
Circle
l e 2i  1 i
i i
 (1)  i
2 2
i/2
2δ
f(η=1)
0
Re f
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
33
Resonance and Breit –Wigner formula
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)


ei ei  e i
1
f
 ei sin  
2i
cot  i
At resonance δ = π/2
Power series expansion
2
d

cot ( E)  cot ( ER )  ( E  ER )  cot ( E)
 ...........   E  ER 

 dE
 E  ER
Resonance energy
Assuming
We obtain :
cot ( ER )  0
E  ER    ER
f (E) 
2
d

   cot ( E )

 dE
 E  ER
1
/2

cot  i ( E  ER )  i / 2
Breit – Wigner
formula
34
Using the Breit-Wigner formula, one obtains - for the case when a given l is
predominant :
 el  4 2 
 e
(2l  1)
2 i l
l
l

1
2 / 4
 el ( E)  4 (2l  1)
( E  ER ) 2   2 / 4
2
2
2i
This is a quantum dependence on energy, that
corresponds to a temporale dependence of the
state of the type :
 (t )  (0) ei t et / 2  (0) exp  t iER   / 2
R
I (t )   *  I (0) et /
Decay law of a particle
The Fourier transform of the decay law gives the E dependence :


0
0
 ( E )   (t ) eiE t dt  (0)  dt exp t  / 2  iER  iE 
35

K
 ( E )   dt exp t  / 2  iER  iE  
( ER  E )  i  / 2
0
In the case of an elastic resonance, the cross section is proportional to the
square modulus of this amplitude :
2

/4
2
 el ( E)  4 (2l  1)
( E  ER ) 2   2 / 4
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 :
4 2 (2 J  1)
2 / 4
 el ( E ) 
(2sa  1) (2sb  1) ( E  ER ) 2   2 / 4
36