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Subnuclear Physics in the
1970s
IFIC Valencia. 4-8 November 2013
Lecture 1
Basic concepts
25-May-17
A. Bettini LSC, Padova University and INFN
1
The summary
1. Basic concepts
2. Detectors physics principles
1. Cosmic rays
2. Accelerators
3. Neutrinos from accelerators
1. Origins and evolution in the 1960s
2. Gargamelle
3. Neutral currents
4. The weak charge and chirality
4. The weak mixing angle
1. Parity violation in atoms
2. Polarised electron-deuteron
scattering
5. New instruments
1. Electron-positron colliders
2. Integrated circuits and the wires
revolution
25-May-17
6. The second and third families
1. Tau
2. Charm (and Japan)
3. November revolution
4. Hidden beauty
7. Looking inside the nucleon
1. with different probes
2. with increasing resolution
(since the 1960s)
8. The coloured world
1. Discover of the gluon
2. Coloured charges
3. The mass of the proton
9. The SPS collider and UA1
1. The discovery of the
intermediate vector bosons
10. What next?
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A text-book
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Relativity
1. The physical processes in subnuclear physics happen at energies usually larger or much
larger than the particles rest energies
In cosmic rays and in accelerators particles have speed close to c
Constituents of atoms (electrons and nuclei) and of nuclei (nucleons) are non-relativistic,
constituents of hadrons are relativistic, quarks speed close to c (mu and md << mh)
Need special relativity
Two types of phenomena
1.Collision: initial state= two particles; final state = two or more particles
2.Decay: one particle goes in two or more “new” particles
In both cases interaction time << measurable times. In practice instantaneous
Initial state particles and final states ones are free (exception: beta decay of atomic nucleus)
Baryons and mesons (hadrons) are composite particles
Quark are never free particles
Need relativistic quantum mechanics = field theory
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Lorentz transformations

V
;
c

1
1  2
coordinates 4-vector(ict , r)
x '   x   ct 
y'  y
z'  z
ct '   ct   x 
its norm (scalar) is the interval
energy-momentum 4-vector (icE , p)
ds  c2 dt 2  dr 2
px '   px   cE 
py '  py
pz '  pz
its norm (scalar) is the mass squared
cE '   cE   px 
m2 c 4  E 2  p2 c2
N.B. The Lorentz transformations contain a constant: c2
Physical meaning (Poncaré 1905): squared velocity of all the fundamental waves;
electromagnetic, gravitational,..?
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Energy, momentum, mass
p
The momentum of a system of energy E and velocity v is
p  m v
For systems of non-zero mass, m≠0, an equivalent expression is
For a free body, or system of free bodies,
the square of the mass is the norm of the
energy-momentum 4-vector
 
mc
2 2
E
2 v
c
 E 2 pc
2
m2  E 2  p2
Natural units
Mass is an invariant quantity, in particular it does not depend on velocity
 an observable trajectory, like electron, muon,
Mass of stable or metastable particles (giving
proton,…) is determined by measuring two quantities
•Energy (e.g. calorimeter) and momentum (e.g. spectrometer)
•Velocity (e.g. time of flight) and momentum

 

Mass of unstable particle (hadron resonance) = (invariant) m 2 
 Ei    p i 
mass of the decay products
 i   i 
2
2
Quarks are never free, are inside hadrons. Mass is extracted from hadrons properties
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Mass and energy
m 2c4  E 2  p 2c2
Energy of a free body is the quadratic sum of rest energy and
motion energy

For a free body at rest only mass (rest) energy E0=mc2
For a relativistic body the kinetic contribution is small
If massless (never at rest) only motion energy E = pc
Attention!
The “famous Einstein equation” E=mc2 is wrong
The correct equation is E0=mc2
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Energy, momentum, mass of a system
Free
(non interacting) particles
n
Energy of the system E   Ei
1 n  n r
m  2  Ei    cpi 
c  i1   i1 
2
i1
r n r
Momentum of the system P   pi
2
i1
Mass of the system ≠ sum of the masses of its components
Two photons, same energy
same directions
opposite directions
E
p=E/c
E
p=E/c
E
p=E/c
E
p=E/c
any direction
Etot =2 E, ptot = 2E/c
mtot=0
mtot= 2E/c2
0 < mtot< 2E/c2
Interacting particles (atom, nucleus, proton,..)
n
Energy E   Ei
i1
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r n r
Momentum P   pi
Field contains energy and momentum
i1
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h =6.58x10–22 MeV s
c = 3 x 1023 fm/s
hc = 197 MeV fm (GeV am)
Natural units
1 MeV = 1.52  1021 s–1
1 s–1= 6.5  10-16 eV
We put c = 1, by redefinition of the unit of length
1 s = 3  1023 fm
Unit of time = second
1 m = 5.07  104 eV–1
Unit of length = distance made by light in 1 s [L] = [T]
Mass, energy, momentum have the same physical dimensions 1 fm = 5.07  10–3 MeV–1
We put h =1, by re-definition of the mass unit
1 MeV–1 = 197 fm
Energy dimensions [E]=[L–1]=[T–1]
The symbol m for a particle may mean
•The mass m
•The rest energy mc2
•The reciprocal of the Compton length h/mc
•The reciprocal of the time light takes to
make a Compton length h/mc2

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Compton length of π (m=140 MeV)
1
1fm
  MeV–1 
 1.42fm
–3
m
140  5 10
Time to go through it at c

1.42fm
–24

5
10
s
23
310 fm/s
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Angular frequency and energy
In NU symbol  may mean angular frequency or energy h
 t 
 t 
t   0 exp– cos 0t  0 exp– cos 0t,
 2 
 2 
 = 1/

Fourier transform F() (<<
F  
2

2
1

 2 2 – 2
0
 –  2 2
2
Measuring the width of a resonance
gives its lifetime
Example: r, a meson strongly decaying in 2π. Width   150MeV
1
1
1
–24
 


4
10
s
21
–1
Typical time for a strong process
 150MeV 150 1.52 10 s
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Wave vector and momentum
In NU symbol p may mean momentum or wave vector k=hp (=2π/k)
A

D/2 ik x
x
D/2
e
dx
A

f x  kxsin  kx x


T x eikx x dx
I



T r e
ikT r
2
dr
The scattered intensity is the square of the (spatial) Fourier transform of the amplitude
density of the target
The conjugate variable is kx, better the transverse component of wave-vector the scattered
particle (kT)
If D</2 the total phase change Df induced by the target on its section D is small becoming nonobservable (Fourier transform is uniform)
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A. Bettini LSC,
Padova University
INFN
Resolving
power = minimum observable
diameter
Dmin=/2and
=π/k
T
The elementary fermions
Three quark families
Q
Iz
S
C
B
T
Mass
d
–1/3
–1/2
0
0
0
0
4.8+0.7–0.3 MeV
u
+2/3 +1/2
0
0
0
0
2.3+0.7–0.5 MeV
s
–1/3
0
–1
0
0
0
95±5 MeV
c
+2/3
0
0
+1
0
0
1.275±0.0025 GeV
b
–1/3
0
0
0
–1
0
4.18±0.03 GeV
t
+2/3
0
0
0
0
+1
173.5±1.0 GeV
Quark masses are not
observable
The mass parameters are
determined from the
properties (masses,
couplings,…) of the hadrons.
Possible only within a
defined theoretical scheme
Three lepton families
m(MeV)
lifetime
e–
0.511
>5 1026 y
e
n.a.
n.a
–
105.7
2.2 µs

n.a.
n.a
–
1776.8
291 fs

n.a.
n.a.
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The Standard Model assumes zero mass neutrinos
Observation of neutrino oscillations
neutrino masses are nonzero
e, ,  are not mass eigenstates, but linear
superpositions of them
Physics beyond the MS
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Fundamental interactions
Interaction
Boson
M(GeV)
JP
Weak CC
W±
91.2
1–
Weak NC
Z0
80.4
1–
El. Mag.

0
1–
Strong
8 gluons
0
1–
Gravity does not have a microscopic theory (yet)
Forces binding the nuclei are not fundamental, but “tails” of QCD
The photon has no charge. Gauge symmetry U(1)
Ws have weak and electric charge. Z has weak charge. Symmetry SU(2)
Gluons have colour charges. Symmetry SU(3)
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The Dirac field
In Quantum Field Theory (QFT) both bosons and fermions are quanta of a field
Field and corresponding wave function obey the same equation
SM assumes spin ½ fermions obey Dirac equation (not proven for neutrinos)
i   mx  0
1 
 
2 

 x  
 3 
 
 4 
x = (x0, x1, x2, x3)
The 4 elements of the bi-spinor correspond to particle and antiparticle and,
for each the two possible polarisation (or helicity) states: Sz=+1/2 and Sz=–1/2
Dirac matrices can have different representations. We take
1 0 
 
,

 0 –1
0
 0
  i
 
i
i
0 
 0 1
 
,
 1 0
1
 0 i 
 
,
 i 0 
2
1 0 
 
 0 1
3
 0 1 
 

 1 0 
5
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Dirac covariants
For monochromatic plane wave
 x  ue
 u1 
u 
The bispinor conjugate of u   2 
 u3 

 u 
4
ip x
is
Dirac equation becomes
u  u 0 
u
*
1



p
m u0

u3* u4*
u2*



u   p  m  0
Satisfying equation
With two bispinors a e b (corresponding to equal or different particles) and the Dirac matrices 5
covariants can be built.
Nature uses only two of them in the interactions: vector and axial currnets
ab
a 5b
a  b
scalar
pseudoscalar
vector
a  5b
axial vector
1
2 2
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a        b
E.M.
Weak
QCD
V
V&A
V
tensor
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Meaning of the four components
j
  
 c
The bi-spinor can be written with two spinor components j and
c, corresponding to particle and antparticle
Three possible choices of the two elements of j and the two of c depending of the quantity we
want definie
1. Polarisation. The two projections of spin 1/2 on a physically defined (magnetic field, electric
field) z axis
 1 
 0 


j 
 and j  

0
1




are eigenstates of the 3rd
spin component
1  1 1  1 0   1
1
z    


;
2  0 2  0 1  0
2
1  0 1  1 0   0
1
z    


2  1 2  0 1  1
2
2. Helicity. the component of the spin on the direction of velocity (particle not at rest)
Photn helicity corrsponds to circular polarisation of light
r r
1p
Helicity opertor is
2 p
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Chirality and Helicity
3. Chirality states are the eigenstates of 5
Bi-spinors can be written as
1
1  5  ,
2

1
 R  1  5  ,
2
L 
  L  R
 5 L   L
 5  R   R
Left and Right bi-spinor are the states
of negative and positive chirality
Not of positive and negative helicity
5 does not commute with the mass term of the Hamiltonian

CC weak interactions couple to Left fields only
Both have positive (+) and negative (–) helicity components
For E>>m
j
m 
j L  j L
E
m 
c  c  cL
E

L
If m=0 neutrinos are pure h=– states, antineutrinos pure h=+ states
But m≠0 and E>>m, then neutrinos have a small (m/E) “wrong” helicity component
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Feynman diagrams. t channel
Example e–+µ– e–+µ–
Incoming particles are annihilated by the e free, incoming
field
Outgoing particles are created by the field
propagator,
virtual photon
Four-momentum transfer
r2
r
r 2
2
t  E 2  p  E2  E1   p2  p1  µ free, incoming
the virtual particle in the tchannel has imaginary mass=
√t
qe2
1

;
4  0 hc 137
1

e free, outgoing
µ, free, outgoing
vertex factor=charge
Amplitude proportional to the product of the
currents at the two vertices, hence to  e and to
the propagator
1
propagator

 A e  e  A   
m2  t


This is the same diagram
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The s channel

e

s is the centre of mass energy squared >0 s  Ei  Ei
  p
2
e
i
 pi
  E
2
e
f
 E f
  p
2
e
f
 p f

2
s channel
Amplitude is the sum of the two (complex) contributions
Probability = absolute square of the amplitude
+
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e++ e– hadrons
s GeV

 
s 
3
s
 e f
 sM 
2
R
25-May-17
2

4
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Virtual particles
Amplitude is an analytical function. Does not become zero in the light cone (b)
If interval AB is space like the (virtual) electron travels faster than light
In another frame (c) event B is earlier than A. Electron goes back in time. Observer says: in B
photon becomes an e–e+ pair, in A positron meets electron and annihlialtes in the final photon .
The virtual particle of an observer is a virtual antiparticle of the other one
Quantum mechanics + Lorentz invariance imply the existence of antiparticles
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Coupling constants run
s  sin 2 W
s(MZ)=0.118±0.002
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Experimental an theoretical discoveries
Known in 1945
•proton, neutron, electron and positron
•Dirac and Majorana equations, Fermi theory of weak interactions, Yukawa theory of strong
interactions
1947
1948
1952
1955
1956
1957
1961+
1962
1963
1964
1964-67
1967
Lamb shift, Muon is a lepton, pion, V˚ particles
QED
3/2 3/2 resonance
Antiproton
(Electron anti) neutrino. Hypothesis of parity violation
Discovery of parity violation
The hadronic resonances zoo
Muon neutrino
Cabibbo mixing
CP violation, quark model, Brout+Englert+Higgs mechanism
EW theory
The quark structure of the proton with electron beam
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Experimental an theoretical discoveries
1970
GIM and charm hypothesis
1971
Charm discovery in cosmic rays
1972
Renoramisability of EW theory
1973
Weak neutral currents, QCD
1974
Charm at accelerators
1975
 lepton
1976
Hidden beauty
1979
Discovery of the gluon
1983
W and Z bosons
1989-2000. Precision tests at LEP and HERA
1995
Top quark
1998
Neutrino oscillations in solar and atmospheric neutrinos
2001
nu-tau
2013
BEH boson
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Two body decay
Exercise. Consider the decay of a particle of mass M in two particles of masses m1 and m2
Give the expressions for energies and momenta of the decay products in the CM frame.
M  m1  m2
r
r
p*f  p1*f  – p2* f ;
E1*f 
2
*
p*2

m

M

E
f
1
2f

E1*f  E2* f  M
2
2
*2
*
p*2

m

M

E

2ME
f
1
2
2
2ME2* f  M 2  m22  m12
E
*
2f
M 2  m22  m12
M 2  m12  m22
*

; E1 f 
2M
2M
p*f  E1*2f  m12  E2*2f  m22
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Kinematic formulas
s  E1  m2   p  m  m  2E1m2
2
E1* 
2
1
2
1
2
2
s  m12  m22
E1 
2m2
p1*2  m12  s  E2*
p1*2  m12  s  2E2* s  E2*2


2E2* s  s  E2*2  p1*2  m12  s  m22  m12
s  m22  m12
s  m12  m22
*
E 
E1 
2 s
2 s
*
2
*2
2
p1*  p2*  E1/2
 m1/2
s  m32  m42
s  m42  m32 *
*2
2
*
p3  p4*  E3/4
 m3/4
E 
E4 
2 s
2 s
*
3
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Kinematic formulas

 p  p   m  m  2 p p cos
 p  p  p  p   m  m  2 p p cos 
µ
t  p3  p1
3
µ
2
3
1
µ
4
2 µ
4
2
1
2
4
2
1
2
2
2
3
13
4
24
 2E1E3 
 2E2 E4
t  m12  m32  2E1*E3*
cos 
2 p1* p3*
*
13
Nel CM
t  m22  m42  2E2*E4*
cos 
2 p2* p4*
*
24
m22  m42  t
E4 
2m2
Nel L (p2=0) t  m  m  2m2 E4
2
2
2
4
s  t  m12  m42
E3  m2  E1  E4 
2m2

 p  p   m  m  2 p p cos  2E E
 p  p  p  p   m  m  2 p p cos   2E E
u  p4  p1
µ
4
µ
µ
3
Nel L (p2 =0)
25-May-17
2 µ
3
2
4
1
2
2
3
2
1
2
2
u  m22  m32  2m2 E3
1
2
4
3
E3 
14
23
1
2
4

3
m m u
2m2
2
2
2
3
s  t  u  m12  m22  m32  m42
A. Bettini LSC, Padova University and INFN
27
Exercise 2
A L hyperon decays as
L p–
In the lab frame the momentum of the L is pL2 GeV.
In the CM frame the angle between the line of flight of the L and the proton is *p = 30˚
Find
1.
Energy, momentum and Lorentz factors in the L of the L
2.
Energy, momentum and Lorentz factors in the CM of the L
3.
The momenta of the L e and π in the lab frame
4.
The angle between their directions
mL = 1.15 GeV
m = 0.139 GeV
mp = 0.938 GeV
25-May-17
A. Bettini LSC, Padova University and INFN
28
Solution 2
s  m22  m12
E 
2 s
*
= 30˚
pL2 GeV
mL = 1.15 GeV
m = 0.139 GeV
mp = 0.938 GeV
*
2
p
EL 
L 
pL2  mL2  2.307 GeV
pL
 0.867
EL
L 
r
r
p*  – p*p  p*
EL
 2.00
mL
E*  E *p  mL
E *p 
p*2  m 2p  mL  E*
p*2  m 2p  mL2  E*2  2mL E*
E *p  mL  E*  0.95 GeV
25-May-17
2mL E*  mL2  m 2p  m2
A. Bettini LSC, Padova University and INFN
29
Solution 2
*p = 30˚
pL2 GeV
mL = 1.15 GeV
m = 0.139 GeV
mp = 0.938 GeV
p sin   p* sin *  0.145  sin 210Þ 0.073 GeV


p cos   L p* cos*   L E*  2.0(0.145  cos 210Þ0.867  0.201)  0.097 GeV
tan 
p 
0.072
 0.742
0.097
p sin   p cos 
2
L  0.867
  36.6Þ
2
 L  2.00
 0.121 GeV
p p sin p  p*p sin *p  0.072 GeV


p p cos p   L p*p cos *p  L E *p  2.0(0.145  cos 30Þ0.867  0.95)  1.90 GeV
tan p 
pp 
25-May-17
0.072
 0.038  p  2.2Þ
1.90
p
p
sin  p
  p
2
p
cos p
  1.9
2
GeV
   p    38.8Þ
A. Bettini LSC, Padova University and INFN
30