Particle%20Physics%20-I

1
Particle Physics -I

• Nucleon-nucleon interaction meson exchange
  Exchange giving rise to the four forces of
  nature properties of pion
• Pion-nucleon interaction Collision kinematics
  and conservation lawsPhase shift Analysis More
  conservation laws Ns and ?s Photon-nucleon
  interaction

2
Book K R Krane, Introductory Nuclear Physics,
Wiley

• Nucleon-Nucleon force -Chapter 4
• Intro/ the deuteron pp 80-81
• Spin and Parity pp 83-84
• Magnetic Dipole moment pp 84-85
• Electric Quadrupole moment p 85
• Properties of force pp 100-102
• Exchange force pp 108-112
• Isospin pp 388-389
• Properties of pions pp 656-665
• Pion-proton interaction pp 671-675
• Scattering theory - partial wave analysis pp
  408-411
• Conservation of Baryon number p715

3
NUCLEONS

• Nucleon is the generic name given to the
  proton(p) and the neutron(n0) - particles with
  essentially the same mass but different charges.
  We view the proton and the neutron as different
  CHARGE STATES of the nucleon.

4
Properties of Nucleons

• Property Unit Proton Neutron
• Charge e e 0
• Mass MeV/c2 938.3 939.6
• Spin h/2? ½ ½
• ? ?N 2.793 -1.913
• Radius fm(10-15m) 1 1
• Stability Stable ?gt1032 y ? 11.1 min
• Decay Mode ----- n? p e- ??e

5
Di- Nucleons

• Only ONE stable - the DEUTERON (D)
• Charge e
• Mass(MeV/c2) 1875.7 938.3 939.6 - 2.2
• Binding Energy 2.2 MeV
• Spin (h/2?) 1 ½ ½
• ?( ?N) 0.857 2.793 -1.913
• Radius fm(10-15m) 2 1 1

6
Deuteron Cross- section
? ? rD2 12.5 fm2 125 mb
Deformation from circle 3mb
P
N
7
What is the deuteron state?

• From ? ?p ?n we infer that the proton and
  neutron are in an orbital angular momentum L0
  state most of the time
• This would imply a spherical nucleus. From the
  fact there is a deformation this must mean for a
  small fraction of the time the proton and neutron
  are in a different orbital angular momentum state

8
Unstable Di - nucleons

• D - Unbound excited state of Deuteron. Eex
  70 KeV. Spin0. ? 10-22 sec
• Di-proton - Unbound state of two protons.
  Spin0. ? 10-22 sec
• Di-neutron - Unbound state of two neutrons.
  Spin0. ? 10-22 sec

9
Evidence for existence of di-proton (2-p)
Nn
En(MeV)
p D ? p n p
p D ? n (2-p)
10
Possible combinations of two S½ Nucleons in L0
state
11
States of two nucleons with L0

• Only the deuteron - 1 CHARGE STATE -exists
  with S1, Sz 1,0, -1
• i.e. 1 CHARGE STATE, 3 possible spin
  orientations
• D, 2-p and 2-n - 3 CHARGE STATES -exist with
  S0, Sz 0
• i.e. 3 CHARGE STATES, 1 spin orientation
• There is a certain symmetry here!

12
2-p D
2-n
S0
Sz 0
D
Sz 1,0,-1
S1
13
Introducing ISOSPIN

Formally, Isospin is a vector in a Charge
Space orthogonal to normal space, whose
orientation in charge space defines the charge
on the particle. As a vector it obeys the same
rules as the Spin vector - hence its name.Other
than that it is totally unrelated to spin. For
the nucleon, the Isospin vector t ½, tz
?½. The relation between charge and tz is Q/?e?
tz ½ For two nucleons Isospin T t1 t2, Tz
t1z t2z. There are two
possibilities T1, Tz 1, 0, -1 T0,
Tz 0. and Q/?e? Tz 1
14
2-p D
2-n
S0 T1
Sz 0
Tz 1 Tz 0
Tz -1
D
Sz 1,0,-1
S1
T0, Tz 0
15
COMPONENTS OF NUCLEON-NUCLEON FORCE

• Spin dependent - D bound, D unbound , only
  difference orientation of spins
• Charge or Isospin Independent - 2-p, D, 2-n have
  same mass
• Range of force 1- 2 fm
• Evidence RD 2fm
• RA 1.2A1/3 V 4/3?RA3 A ?
  V(Nucleon)
• Attractive - nuclei exist
• Repulsive core
• Evidence RD 2fm R(3He) 2fm R(4He) 1.7 fm

16
?D 2 ? ?( 2 Nucleons) ?(3He) 1.33 ? ?(3
Nucleons) ?(4He) 0.7 ?
?(4 nucleons)
D 3He 4He
Binding Energy per nucleon pair B/½A(A-1)
2.23 2.53 4.67
17
The ?-particle (4He nucleus) is the smallest and
most strongly bound nucleus in the periodic
table. The nucleons within it clearly overlap.
Why doesnt it become even smaller? The N-N
interaction appears to have a Repulsive Core of
range 0.7 fm. The magnitude of Binding Energies
(MeV) means that the force must be a STRONG
interaction ( 100 times stronger than the EM
interaction). Tensor Component To explain the
non-spherical deuteron the N-N force must have a
component which is non-central ie depends on the
orientation of spins relative to spatial position.
18
Deuteron
r
r
19
Force between two dipole magnets

N S
N S
d1
d2
r

N S
V(r ) -1/r33(d1.r)(d2.r) - d1.d2 where r is
the unit vector
d1
r
N S
d2
20
Deuteron
r
r
The EM potential V(r ) -1/r33(?p.r)(?n.r) -
?p. ?n leads to a deformation which is of order
?b. A strong interaction potential is
needed 2spsnVT(r) V(r ) VT(r)
3(sp.r)(sn.r) - sp.sn to give a deformation of
order mb. -spsnVT(r)
21
Meson Exchange

• What can give rise to the tensor potential part
  of the N-N interaction?
• The answer is thought to be the virtual
  exchange of mesons between the nucleons.
• What do we mean by this?

22
The 4 exchange forces
Force Range Exchange particle Strength Gravitati
onal ? Graviton,g, 0
10-39 Electromagnetic ? Photon, ?, 0
10-2 Strong lt 2fm Pion, ?, 140 MeV/c2
1 Weak 0 W-boson,W,81 GeV/c2 10-5
23
The Electromagnetic Interaction
In the interaction between two charged particles
(eg an electron and a proton) photons are
exchanged between the two particles. However we
do not observe any photons being exchanged. As
long as the exchange is described by the
Uncertainty Principle it can still occur even if
it is unobservable.
?E?t ? R c ?t ? c/?E If ?E 0 (photon)
the range R ?. Also as E2 p2c2 m02 c4 and m0
0 then the change in momentum of the emitting
particle will be ?p ?E/c and F ?p/ ?t ? c/R2
24
The relativistic relation between total energy,
momentum and mass helps us to visualise what may
be happening in this virtual ( ie real but non -
observable) exchange of photons. We see that E
?? (p2c2 m02c4) The negative solution implies
that we can view the vacuum as a sea of negative
energies
p
e-
?
?
E0
The photon interacts with the vacuum locally.
Energy is then transmitted out in all directions.
When another charged particle is encountered this
is then transferred to the particle by a photon.
25
The Strong Interaction
Meson Exchange ?E?t ? R c ?t ? c/?E

For the strong interaction the long range
part of the interaction is produced by the
exchange of the lightest meson the pion, ?.
M? 140 MeV/c2 ie ?E ? 140 MeV
R 6.6 10-22 MeV sec ? 3 1023 fm
sec-1/ 140 MeV ie R 1.4 fm
?
?
E0
26
Properties of Pions
Pions exist in three charge states ? , ?- and
?0 Therefore the isospin of the Pion T? 1 with
orientations in charge space T?z 1, -1 and 0
respectively. Q/?e? T?z
M? M?- 139.57 MeV/c2 M?0 134.97
MeV/c2. The spin of each pion is S? 0.
Another important property of the Pion is its
PARITY
27
PARITY
This is defined for wave-functions in a symmetric
potential. E.g. The N-N potential is symmetric
about the mid-point of the two nucleons
V(r) V(-r). Then for the probability density of
the two nucleons??(r)?2 ??(-r)?2
and ?(r)
P?(-r) where P ?1 A wave-function has
parity P 1 if it is even under reflection (
i.e. a swap) of both the particles co-ordinates
and parity P -1 if it is odd under reflection.
In a symmetric potential wave-functions
with Angular Momentum L0,2,4, are Even Parity,
those with L1,3,5, are Odd Parity. All Strong
and EM potentials are symmetric and each energy
state has a definite parity - either Even or Odd.
Particles of finite size with internal structure
held together by a symmetric interaction to be
discussed later - e.g. Nucleons, Pions - have
INTRINSIC PARITIES. Parity is conserved in Strong
or EM interactions between particles.
28
Intrinsic Parity of the ?-
Consider the reaction AT REST between a ?- and a
deuteron ?- D ? n n Spin (S)
0 1 ½ ½ Angular Momentumtum (L)
L?D 0 Lnn? Now Total angular
Momentum J is conserved in all interactions.
Therefore for Total Angular Momentum(J)
1 1
What is the value of Lnn? As the neutrons
are like particles with S ½ they are subject to
Pauli Principle. i.e. the overall wave-function
must be anti-symmetric
29

• The overall wave-function of two neutrons
  consists of two parts ?nn ?(L) ?(S)
• a space wave-function produced by solution of
  Schrodingers equation in a potential Vnn (r)
  i.e. containing a Legendre polynomial
  dependent on angular momentum state
  L. Wavefunctions with L 0,2, .. Are Symmetric
  - Even Parity. Those with L1,3,.. Are
  Anti-symmetric - Odd Parity.
• a spin wave-function dependent on the spin of
  the two neutrons, S. This too can be symmetric or
  anti-symmetric.

30
The Spin Wave-function, ?(S)
Two neutrons can exist with their spins parallel
This is a Symmetric state S 1, Sz
1,0 0r -1 Or anti-parallel
This is an anti-symmetric state S0, Sz 0
31

• For two neutrons J 1 cannot occur if either
• L0 S1 or L2 S1- Symmetric Space Wavefunction,
  Symmetric spin Wavefunction ie Overall Symmetry
• L1, S0 - Anti-Symmetric Space Wavefunction,
  Anti-Symmetric spin Wavefunction. ie Overall
  Symmetry The ONLY possibility is L1 S1 -
  Anti-Symmetric Space Wavefunction, Symmetric spin
  Wavefunction. ie Overall Anti-Symmetry.
  J can then have values J2
• or importantly J1

J0
L1 S1
L1 S1
S1
L1
J1
32
So now, in the reaction AT REST between a ?- and
a deuteron ?- D ? n n Spin (S) 0
1 ½ ½ Angular Momentum (L) L?D
0 Lnn? Total Angular Momentum(J) 1
1 We see that the value of Lnn has to be
Lnn 1. Thus in the reaction, an initial even
parity WF (L?D 0 ) goes to an odd parity Lnn1
WF. But parity is conserved in the Strong
Interaction.We must thus consider Intrinsic
Parities. Then Pinitial P??Pn?Pp?P(L?D 0 )
Pfinal Pn?Pn ?P (Lnn 1) The intrinsic
parities of the nucleons cancel each other.
So P??P(L?D 0 ) P (Lnn 1) i.e. -1 ?
1 -1 The INTRINSIC PARITY of the PION is
NEGATIVE. Incidentally the INTRINSIC PARITY of
the NUCLEON is POSITIVE - by convention.
33
Pion Exchange between Nucleons

• The negative intrinsic parity of the pion has
  important consequences when it is exchanged
  between nucleons.
• It gives rise to the TENSOR force.
• When the pion is emitted by a nucleon, as the
  nucleon continues to have positive parity, the
  pion must be emitted with L 1 - to cancel out
  its intrinsic negative parity. In order for this
  to happen, to conserve total angular momentum
  either
• (a) the emitting and receiving nucleons must
  reverse their initial spin directions
• (b) the emitting and receiving nucleons must
  recoil with L1.
• Depending on their initial spatial
  orientation (a) or (b) happens.

34
Lnp0
Lnp0
Pion transfer occurs by nucleons changing spin
direction
?0
L1

Lnp2
Lnp0
Pion transfer by angular momentum change
L1 L1
?0 L1
35
So in the loosely bound deuteron ONE PION
exchange is the dominant process leading to the
tensor force. In 3He, with nucleons closer
together, TWO PION exchange is more likely. Here
no nett angular momentum need be transferred
between two nucleons - i.e. a central force
?0
?0
This has a range 0.7 fm. In 4He, with the
nucleons overlapping, even heavier mesons can be
exchanged. The ?,? and ? mesons have masses
between 600 and 900 MeV/c2, negative parity and
S1. When these are transferred (with L1 to
conserve parity) nucleons are forced to L2 state
-i.e . Must move apart, are effectively repelled.
36
Pion-Proton interaction

• We can create beams of pions by bombarding a
  heavy nucleus with a high energy proton beam from
  an accelerator.

? ?0
Beams of charged pions are formed using electric
and magnetic fields.
Proton beam T 3000 MeV
n
Tungsten (W) target
37
Liquid hydrogen target
Area AH, length tH,density ?H
? beam
Pion detectors

• Number of scattered pions N(IN) N(OUT)
• N(IN) (NAV?HtH AH )(?T /AH)
• By analogy with throwing point-like darts at a
  board ?T would just be the cross-sectional area
  of a proton, but this analogy needs to be
  extended as
• there is an interaction between the pion and the
  nucleon, predominantly the STRONG force with a
  certain strength and range, but there is also a
  small contribution from the Coulomb force.
• So we extend the definition of ?T to
• THE TOTAL CROSS-SECTION FOR THE STRONG
  INTERACTION OF A PION WITH A PROTON
• ?T is measured in barns (10-24 cm2) and, when
  divided by AH, is the probability for a pion to
  interact with a proton. It is energy dependent.

38
Pion-proton total cross-sections
195 600 900 1380
T ? (MeV)
39

• Peaks in the ?p cross-section are interpreted as
  resonances in the ?p system ?? p ? (?? p) ?
  ?? p
• The lowest resonant state occurs in both the ?
  p and the ?-p cross-sections at a pion incident
  laboratory energy of T? 195 MeV.
• The width of this resonance is 100 MeV. Using
  the Uncertainty Principle ? E? t ? we find that
  the lifetime of the resonance is 10-23
  sec.
• What is its mass? This can be calculated
  by applying the laws of Conservation of Energy
  and Momentum to the formation of the resonance.

?- p (?-p) T?-, p?-,m?-
mp T,P,m,v
40
C.of E. T?- m?-c2 mpc2 mc2 T
(1) C.of M. p?-
p and hence p?-2c2 p2c2
(2) Relativistic kinematic relationships Total
Energy E m0c2/ ?(1-v2/c2) Momentum p m0/
?(1-v2/c2)v Eliminating v, E2 p2c2
m02c4. Also E T m0c2 and hence E2 T2
2m0c2T m02 c4 Thus p2c2
T2 2m0c2T In (2) we then have T?-2 2m?-c2T?-
T2 2mc2T and
substituting for T from (1) mc2 ?
(m?- mp)2c4 2T?-mpc2 Substituting pion,
proton masses and T?- 195 MeV we find m
1235 MeV/c2
41
The Total centre-of-mass Energy is E mc2 1236
MeV. To see this impose a back velocity on the
laboratory such that the resonant state is
brought to rest ( i.e. - v). Then the pion and
the proton will have equal and opposite momenta.
pCM
pCM mc2 To find the
spin state of the resonance we need scattering
theory.
42
Scattering Theory -Spin-less particles, Elastic
Scattering,CM reference frame
r
?
Incident pion wave
Proton
Unscattered pion wave
z
Scattered pion wave ?scatt

• ?inc eikz
• Solution of -?2/2m?(?2 ?inc/?z2) E ?inc. As
  EpCM2/2m? and pCM ?k this reduces to ?2
  ?inc/?z2 -k2?inc.

43

• ?scatt is a solution of -?2/2m(?2?scatt) (E
  -V(r))?scatt
• ?scatt f(?) eikr/r
• at large r where Egtgt V
• Solution independent of ? as there is cylindrical
  symmetry about the incident beam direction.
• The total wavefunction after the ?p interaction
  consists of an unscattered part (eikz) as well as
  ?scatt.
• ?final eikz ?scatt

44

• How are ?inc, ?scatt related to observables?
• We know ? ? ? ??2 is the probability of a
  particles existence at a point in space. If
  there are
• n pions/unit area in the incident beam and
• the CM beam velocity is v then the number of
  pions incident per second per unit area is
  nveikze-ikz nv. The number of pions
  elastically scattered at angle ? per unit area
  per second is nvf(?)f(?)eikre-ikr/r2

45
Experimentally we measure the number of pions
scattered into unit solid angle (unit area /r2)
at angle ? per second. This is nv f(?)
f(?). Then the differential cross-section for
elastic scattering d?el/d?(?) number of pions
scattered into unit solid angle at angle ? per
second / number of pions incident per unit area
per second i.e. d?el/d?(?)
f(?) f(?) . ?el is the cross-section for
elastic scattering. ?el/unit area is the
probability for scattering into 4? solid angle.
46
?el ?4? d ?el/d ?(?) d ? ?4? f(?)f(?) d ?
Now, we cannot specify the x- and y- positions
of the incident pions, momentum p, relative to a
target proton.
?p angular momentum state l?p? p ? d
d
2 fm
1013 fm
47
The pions can interact, in principle, with the
protons in ALL possible angular momentum states
l?p?. This allows us to rectify the asymmetry in
the scattering - initial plane wave, final
spherical scattered and plane wave - by
expressing the plane waves in terms of ingoing
and outgoing spherical waves representing a
pion-proton state of angular momentum number l
l?p. ?inc eikz eikrcos? 1/kr ?0?
(2l 1) Pl(cos?)eikr - e-i(kr-l?)/2i where
Pl(cos?) is a Legendre polynomial.
outgoing wave ingoing wave
48
When the ?p interaction occurs only outgoing
waves will be affected. Assuming only elastic
scattering occurs ( i.e. no inelastic scattering)
then these will be advanced (if ?p interaction
attractive) or retarded (if ?p interaction
repulsive) compared to the situation if there
were no ?p interaction. i.e. there will be a
phase shift in the outgoing waves. By convention
this is 2?(l). Thus the wavefunction describing
the total situation after scattering ?final
1/kr?0?(2l 1) Pl(cos?)ei(kr 2?(l)) -
e-i(kr-l?)/2i eikz f(?) eikr/r
(3)
49
As eikz 1/kr ?0? (2l 1) Pl(cos?)eikr -
e-i(kr-l?)/2i we then find that, by
rearrangement of (3) f(?) 1/2ik ?0? (e2i?(l)
-1)(2l 1) Pl(cos?) 1/2ik ?0? ei?(l) (ei?(l) -
e-i?(l) )(2l 1) Pl(cos?) 1/2ik ?0? ei?(l)
2isin ?(l) (2l 1) Pl(cos?) 1/k ?0? ei?(l)
sin ?(l) (2l 1) Pl(cos?) Now ?el ?4? d ?el/d
?(?) d ? ?4? f(?)f(?) d ?. As ?4? Pl(cos?)
Pl(cos?) d ? 4?/(2l1) when ll 0
otherwise ?el 4?/k2 ?0? sin2?(l) (2l 1)
50
Applying scattering theory to the 1235 Mev/c2
resonance

• Scattering theory applies to spinless particles
  only and predicts
• ?el 4?/k2 ?0? sin2?(l) (2l 1)
• We measure ?T ?el ?R
• When does ?T ?el ???
• i.e.When is ?R 0 ???

51
When a pion interacts with a proton many events
other than elastic scattering may occur ?- p ?
?- p Elastic Low Energy
High Energy ? ?0 n T?- lt300MeV
T?- gt300MeV ? ?- ?0 p ? ?- ?
?- p ? p ? ? p Elastic Low
Energy High Energy
T? lt300MeV T? gt300MeV ? ? ?0 p
? ? ? ?- p
52
When pions interact with nucleons as many pions
are produced as there is energy available. The
number of nucleons remains constant. Analogy
Turning on an electric light. As many photons
are produced as there is energy available. The
number of electrons remains constant. Conservation
Laws Baryon Number B 1 Nucleons B0
Pions B is conserved in the All
Interactions (Strong, EM, Weak, Grav.) Epton
Number Le 1 Electrons Le 0
Photons Le is conserved in All
Interactions We will generalise these laws to
include other particles later.
53
We can only apply our calculation of ?el
4?/k2 ?0? sin2?(l) (2l 1) to ? p scattering at
low energy where ?T ?el . How many l-values
contribute? lmax? pCM ? d? p
where d? p is the range of the ? p
interaction (1fm) and pCM is the momentum of
both the pion and the proton in the CM reference
frame. The Centre-of -Momentum ( Centre-of-Mass,
CM) reference frame is formed by imposing a back
velocity on the laboratory such that the CM
momenta of the pion and the proton are equal and
opposite. To work out lmax we need to calculate
first pCM.
Pcm
Pcm
(? p)
?
p
1 fm
54
We know m (T? 195 MeV) 1236 MeV/c2. Then
E? Ep mc2 i.e. ?
(pCM2c2 m?2c4) ?(pCM2c2 mp2c4)
mc2 Rearranging ? (pCM2c2 m?2c4) mc2
- ?(pCM2c2 mp2c4) Squaring pCM2c2
m?2c4 m2c4 pCM2c2 mp2c4 - 2mc2 ?(pCM2c2
mp2c4) Rearranging 2mc2 ?(pCM2c2 mp2c4)
m2c4 mp2c4 - m?2c4 Squaring again 4m2c4
(pCM2c2 mp2c4) (m2c4 mp2c4 -
m?2c4)2 And pCM2c2 (m2c4 mp2c4 - m?2c4)2
- mp2c4 4m2c4 Substituting the mass
values we find pCM 240 MeV/c
55
Now lmax? pCM ? d? p Thus
lmax 240 MeV/c ? 1fm/6.6 10-22 MeV sec 240
MeV?1 fm/ (3 1023 fm sec-1?6.6 10-22MeV
sec) 1.2 Therefore lmax 1
And ?el 4??01 sin2?(l) (2l 1).
k2
56
How does sin2?(l) vary with pion energy T?
? sin2?(0) At T?0, k0 and for ?el to be
finite this means that sin2?(0) ? 0 and hence
?(0) ? 0. As T? increases l? pCM ? d? p 0
then d? p 0 at all energies. I.e. l0 waves
see the same amount of the ? p potential at
all energies in ahead-on collision. ?
p
57
Therefore the phase-shift ?(0), which is caused
by the potential seen, is a constant at all
energies except T?0. 90? ?(0)
T? This is non-resonant
behaviour. I.e. the l0 phase shift cannot cause
the peak in the cross-section - sin2?(0) is a
constant at all T?.
58
sin2?(1) Again, at T?0, k0 and for ?el to be
finite this means that sin2?(1) ? 0 and hence
?(1) ? 0. As T? increases l? pCM ? d? p
1 I.e. as T? and hence pCM increases d? p must
decrease. low T? d? p high T?
I.e. as T? increases the l1 wave sees more
and more of the potential and hence ?(1)
increases.
59
180? ?(1) 90? 0?
T? sin2?(1) 1
0 195 MeV T?
?(1) 90?
60
This is resonant behaviour. Thus the l1 phase
shift behaviour causes the resonance at T? 195
MeV. Thus ?el 4? sin2?(l) ? 3
small l0 bit k2 At the peak of the
resonance ?(l) 90? and k 240 MeV/c / ?
240 MeV/ (3 1023 fm s-1 ?6.6 10-22 MeV s)
1.2 fm-1 I.e. 1/k2 6.94 mb as 1fm2
10 mb Thus ?T ?el 255 mb at the peak
whereas it is measured to be only 200 mb. Why do
we get an over prediction?
61
In the formulation of our scattering theory we
assume the pion and proton to be spin-less -
true for the pion but the proton has
spin-1/2. Thus the interaction with l1 can be in
either of two total angular momentum states j
3/2 or 1/2. Thus our formula becomes
?el 4?/2?j1/2,3/2 sin2?(j) (2j 1)
k2 Is the resonance in the j1/2 or the j3/2
state? Assume ?(j3/2) is small and ?(j1/2)
90?at T? 195 MeV then ?el
(4?/2k2)?2 85 mb
62
Clearly this is far too much of an under
prediction! However if we assume ?(j1/2) is
small and ?(j3/2) 90?at T? 195 MeV then at
the peak ?el (4?/2k2)?4 170 mb
small j1/2 and l0 bits. This is then in good
agreement with the measured cross-section of 200
mb. The resonance in the ?p cross-section at T?
195 MeV is in the j3/2 state.
What is its isospin? As
it appears in the ?p total cross-section and tz
(?) 1, tz(p) 1/2 then tz ( resonance)
3/2 and hence t ? 3/2. It also occurs in the
?-p, ?n and ?-n cross-sections so tz 3/2,
1/2, -1/2, -3/2 (i.e. only 4 states) and hence t
3/2 .
63
It is called the (3/2,3/2) - or in even shorter
shorthand - the (3,3) resonance and as it occurs
in four charge states is an example of a ?-
resonance - the ?(1236). There are many others at
higher excitation energies ( see particle
data sheet handed out). In the ?-p
cross-section the next higher peak at ECM 1.5
GeV) does NOT occur in the ?p cross-section . It
only occurs in only two charge states - in the
?-p and ?n cross-sections i.e. it has isospin
t1/2 and tz -1/2, 1/2. This , and similar
resonant states, are called Ns.The peak at 1.5
Gev is in fact due to a combination of 3 states
the N(1440), N(1520) and N(1535) - these can
only be sorted out by doing many different ?-p
scattering experiments.
64
Now we think the peaks in the ?-p and ?p
cross-sections are due to excited resonant states
of nucleons - rather than compound ?-p states.
Evidence for this is available in ?p total
cross-section measurements.Both N and ?
resonances are produced -in fact to a good
approximation ?T(?p) ? ?T(?-p ) ?T(?p
) ( Incidentally this is an example of Tz but
not T being conserved in the EM interaction). As
?s are pure energy the states can only be nucleon
excited states.
65
Photon-Proton Total Cross-sections
E? (GeV)
66
Particle Physics -II

• Particles and Anti-particles. Kaons and Hyperons.
  Associated Production and Strangeness. Properties
  of hyperons and kaons. Hyperon excited states.
  K-p and Kp cross-sections. Isospin, Strangeness
  and Charge. Hypercharge. Families of particles.
• Quarks ...

67
Particles, anti-particles and the vacuum
perceived as a sea of negative energies are
not discussed in Krane. Kaons, hyperons and
strangeness - Krane pp 686-689 The rest of
particle physics is covered in Chapter 18.
68
Particles and Anti-particles

• We are familiar from the undergraduate laboratory
  with two phenomena
• Pair production e-
  e
• Positron annihilation ?
  e e- ?
  

69
We can understand these phenomena in terms of the
vacuum viewed as a sea of negative energies.For
the first, shown below, a photon is incident on
the vacuum, raising an e-, mass energy mec2, and
producing simultaneously a hole in the vacuum- an
absence of negative energy -(-mec2) - a positron
e. In the second an electron fills
the hole and the energy released is emitted as
two back-to-back photons ( annihilation photons)
- to conserve momentum. e- E0
e

70
We arbitrarily define the electron as a particle
and the positron as an anti-particle.With
higher energy photons we can in principle produce
more massive particles and anti-particles than
the electron and positron. Anti-particles for
the particles we have encountered thus far are
shown below. The anti-particle has the same mass
as the particle but has opposite scalar
attributes e.g. charge (and hence for
strongly-interacting particles - TZ), Baryon
Number and Epton Number. Vector attributes -
spin, isospin - are the same for both.

• Particles
• Electron e-
• Proton p
• Neutron n
• Positive pion ?
• Neutral pion ?0

• Antiparticles
• Positron e (Le -1)
• Anti-proton ?p (B -1)
• Anti-neutron ?n (B-1)
• Negative pion ?-
• Neutral pion ?0

Photon ?
71
Kaons and Hyperons

• At higher pion incident energies new particles
  (Hyperons) such as the ?0 and the ?, ?- and ?0
  were discovered in the 1950s in reactions ?-
  p ? ?- K ? ?0 K0 ? p ?
  ? K
• They are always produced in association with a
  KAON - a heavy meson with mass 500 MeV/c2.
• They have mass gt mp and decay into protons and
  pions. e.g. ?0 ? ?- p

72
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73
A side remark! Knowing the kaon, pion and proton
masses and the pion momentum and being able to
measure to the exit angles of the ?0 K0 we can
calculate - by applying conservation of energy
and momentum to the collision - the mass and
momentum of the ?0. Knowing the momentum of the
?0 and by measuring its flight length before
decay we can work out how long it
lived. Repeating this for many events we can work
out the mean life of the ?0 - 2.6310-10 sec.
An odd quantum mechanics phenomenon!
We can do the same sort of
calculation for the K0 - we find for this
particular event and similar ones a time of
10-10 secs. From other events we find apparently
much longer times 10-8sec - as though the K0
has two lifetimes. You can read about this (NOT
COMPULSORY!!) in Krane pp. 692-695. K and K-
mesons have only one mean life - 1.210-8 secs.
74
At even higher pion energies heavier hyperons
were discovered - the ?- and ?0 ? p ? ?0
K K The ?s are often called the Cascade
particles because of their decay mode e.g.
?0? ?0 ?0 p ?- Even
if we use extremely high pion energies the
Hyperons are ALWAYS produced in the STRONG
interaction with protons with either one Kaon (
the ?0 and ? hyperons) or 2 Kaons ( the ?- and ?0
hyperons). This phenomenon is called ASSOCIATED
PRODUCTION. The hyperons and the kaons decay via
the WEAK interaction.
75
STRANGENESS

• To describe both the phenomenon of ASSOCIATED
  PRODUCTION and the Weak Decays of hyperons and
  kaons we need a new attribute STRANGENESS and an
  associated Quantum Number S
• S 1 K, K0, ?0 and ?,-,0
• S -1 K-, K0,?0 and ?,-,0
• S 0 ?s and Nucleons and a Conservation
  Law
• S is conserved in the Strong Interaction but NOT
  in the Weak interaction ( where in fact ?S 1).

76
Other properties of hyperons and kaons
Mass(MeV/c2) B T TZ
Spin ?0 1115
1 0 0 1/2 ?0 1115
-1 0 0 1/2 ?,0,-
1190 1 1 1,0,-1
1/2 ?,0,- 1190 -1 1 -1,0,1
1/2 ?0,- 1320
1 1/2 1/2,-1/2 1/2 ?0,- 1320
-1 1/2 -1/2,1/2 1/2 K,K0 495
0 1/2 1/2,-1/2 0 K-,K0 495
0 1/2 -1/2,1/2 0
77
? and ? Excited States
As the K- has the same strangeness as the
hyperons it should be possible to observe any
resonant states in the K- p total cross-section
plotted versus energy K- p ? ?0 or ?0 ?
decay products. As the kaon mass is 495 MeV/c2
and the proton mass is 938 MeV/c2 giving a total
mass at rest of 1433 MeV/c2 the ground state ?0
or ?0 ( masses 1118 and 1192 MeV/c2
respectively) cannot be seen but excited states
should be visible.
78
K-p Total Cross-section
1525 1670 1750 ECM
79
As we see there are resonant states around ECM
1525, 1670 and 1750 MeV. These are the ?0(1520),
?0(1670), ?0(1660), ?0(1670), ?0(1750) plus some
other states which can be sorted out by also
doing differential cross-section measurements
versus scattering angle. The rising cross-section
as zero momentum is approached is a reflection of
two sub-threshold resonances - the ?0(1385)
which has a spin of 3/2 and the ?0(1405) which
have widths of 36 and 50 MeV respectively. The
Kp Total Cross-section K plus a proton
has S 1 and hence cannot give rise to S -1
hyperon resonant states. It cannot produce S 1
anti-hyperon resonant states due to
non-conservation of Baryon Number. B 1
initially (BK 0, Bp 1). B -1 for any
produced anti-hyperon. We would expect a flat
cross-section versus plab at low momenta due to
Kp elastic scattering. At higher momenta the
cross-section should increase as there is enough
energy for more mesons to be produced.
80
Kp Total Cross-sections K p ? K p 1,2
or 3 ?s
81
Excited States of the ? particles

• As these have Strangeness Number S -2 and there
  is no S -2 meson, they cannot be seen in any
  meson-proton total cross-section measurement.
  They can be produced in collisions though
  ?- p ? ?- K0 K ?0 ?- ?0
  ?0 p ?-and observed in - for example
  - a bubble chamber.?- (1530), ?0(1530) have
  spin3/2

82
Isospin, Strangeness and Charge For pions and
nucleons and their anti-particles the charge on a
particle can be expressed as Q/?e? TZ
B/2 This formula is too simple for hyperons and
kaons . A more general formula is needed
Q/?e? TZ B/2 S/2 This is often written
as Q/?e? TZ Y/2 where

Y BS is called the HYPERCHARGE.
83
Families of Particles
Historically the proton was discovered, followed
by its charge neutral ,spin-1/2 partner the
neutron - different charge (ISOSPIN) states of
the NUCLEON. Subsequently by adding energy to the
nucleon via pion bombardment more heavier
spin-1/2 particles were found -the hyperons -
differing in Strangeness or HYPERCHARGE. We can
plot these on a graph with TZ and Y as axes
1 p n Y 0 ? ?0,?
?- -1 ?0 ?- TZ
1 1/2 0 -1/2 -1
84
An Analogy

• In Atomic Physics a similar historical
  progression led to the discovery of HYPERFINE
  STRUCTURE in the Spectral Lines of Atoms.
• For example in the hydrogen atom characteristic
  lines were observed (the Balmer, Lyman series
  etc) which were explained as transitions between
  discrete electron energy levels in a Coulomb
  Potential - the e-p interaction.
• Later, fine structure of these lines was observed
  - explained as due to an extra interaction
  between the electron magnetic moment ?e and the
  orbital magnetic field of the electron He -
  ?e.He, splitting the energy level.
• Then hyperfine structure was discovered -
  explained as due to a further interaction between
  (?e He) and the proton magnetic moment ?p
  - (?e He). ?p .

85
(?e He). ?p
?e.He Coulomb Basic energy
level Fine Structure
Hyperfine Structure
Can the spin-1/2 baryon octet be considered
similarly? E.M. Strong N
p n Super-strong ?,?
? ?0,?0
?- ? ?0
?-
Basic spin-1/2 baryon
86
What are the constituents of the basic spin-1/2
baryon? We can study the proton. It is thought
to be composed of 3 point-like objects - the
quarks. The evidence for this comes from
experiments analogous to Rutherford
back-scattering of ?-particles - which give us
our picture of the atom having a central small,
heavy, nucleus ( radius a few fm) surrounded by a
light, electron cloud( radius many Angstroms). To
penetrate the nucleon very high energy electrons
- or neutrinos or anti-neutrinos - have to be
used and backscattering corresponds to a few
degrees deflection from the incident direction.
87
Deep-inelastic electron scattering
88

• Main conclusions from deep-inelastic electron-,
  neutrino - and anti-neutrino-proton scattering
  measurements
• The nucleon contains three point-like objects -
  quarks
• They have spin -1/2
• Two have fractional electric charges consistent
  with 2/3?e?, the up-quark qu and one with
  -1/3?e?, the down quark qd

89

To explain the properties of the hyperons there
is a need also for a third quark - the strange
quark qs
The quark model T Tz S B Y Q/?e? Spin qu
(u) 1/2 1/2 0 1/3 1/3 2/3 1/2 qd
(d) 1/2 -1/2 0 1/3 1/3 -1/3 1/2 qs (s) 0
0 -1 1/3 -2/3 -1/3 1/2 There are also
corresponding anti-quarks u, d, s with
corresponding attributes u 1/2 -1/2 0 -1/3 -1/3 -
2/3 1/2 d 1/2 1/2 0 -1/3 -1/3 1/3 1/2 s 0
0 1 -1/3 2/3 1/3 1/2
90
The nucleons are composed of the non-strange
quarks p (u, u, d) in an L0
state, T1/2 n (d, d, u) The spin-1/2
hyperons include one or two strange quarks
? (u, u, s) ?0 (u, d,
s) in an L0 state, T1 ?- (d, d, s)
?0 (u, d, s) in an L0 state, T0 ?-
(d, s, s) in an L0 state, T1/2 ?0
(u, s, s)
91
An Example A proton consists of u u
d Charge 2/3 2/3 -1/3 1 Spin 1/2
1/2 1/2 1/2 B 1/3 1/3
1/3 1 T 1/2 1/2 1/2
1/2 TZ 1/2 1/2 -1/2 1/2
92
Families of ParticlesThe family of spin-1/2
baryons is thus
1 p (uud) n(ddu)
Y 0 ?(uus) ?0,?(uds)
?-(dds) -1 ?0(uss)
?-(dss) TZ 1
1/2 0 -1/2 -1 There
is also a family of spin-0 mesons composed of
quark- anti-quark pairs - whose spins must be
opposed to give spin-0. 1 K0( d s) K( u
s) Y 0 ?-( d u) ?0,?,?( u u, d d, s
s) ?(u d) -1 K-( s u) K0( s d)
-1 -1/2 0 1/2
1 TZ
93
The ? is a T0 meson with a mass of 550
MeV/c2 The ? is a T0 meson with a mass of 960
MeV/c2 The ?0 is formed from pairs of uu and dd
quarks - its wave-function is a linear
combination of these quark wave-functions The ?
and ? mesons are linear combinations of uu, dd
and ss wave-functions.
94
How are the quarks oriented inside the baryons
and what are their masses?

• Magnetic moments of protons and neutrons
• The proton and neutron magnetic moments are
  anomalous in the sense that if they were point
  particles like electrons you would expect
  that ?p e?/2mpc 1 Nuclear Magneton (NM) and
  ?n 0
• In fact ?p 2.8 NM, ?n -1.9 NM i.e. ?n /?p
  -2/3.
• They both MUST contain some charged substructure.
  How does the quark model do?
• Assume the u,d quarks ARE point-like and have
  same mass m. Then ?u 2/3 e?/2mc and ?d -1/3
  e?/2mc
• Two of the quarks must be parallel (spin 1) and
  one anti-parallel (spin 1/2) to give the
  nucleons spin-1/2.

95

• Let the two quarks forming the spin-1 bit have
  net magnetic moment ?a the remaining
  anti-parallel quark ?b. Then rules for vector
  addition say that ? 2/3?a -1/3 ?b.
• Then if a proton is ( u d ) u as we might
  expect if Pauli principle holds for spin-1/2
  quarks ?p 2/3(2/3 -1/3) -1/3(2/3)
  e?/2mc 0!! Similarly if the neutron is (
  u d) d then ?n 2/3(2/3 -1/3)
  -1/3(-1/3) e?/2mc 1/3 e?/2mc Both are
  obviously wrong.
• However if we take the proton to be ( u u) d
  then ?p 2/3(2/3 2/3) -1/3(-1/3) e?/2mc
  e?/2mc and if the neutron is ( d d)
  u then ?n 2/3(-1/3 -1/3) -1/3(2/3)
  e?/2mc -2/3 e?/2mc Thus ?n /?p -2/3
• This is an excellent result, implies quarks ARE
  point-like and have charges 2/3e and -1/3e but
  seems to require them to disobey Pauli principle!
  This may be very important.

96
Families of ParticlesThe family of spin-1/2
baryons is now
1 p ( u u d) n(d d u)
Y 0 ?( u u s) ?0( u d s)
?-( d d s) ? ( u d s) -1
?0( u s s) ?-( d s s)
TZ 1 1/2 0
-1/2 -1
97
The spin-3/2 family of baryons If we look at the
first excited states of particles we find these
too are in agreement with the nucleon model that
predicts magnetic moments i.e Pauli principle
appears to be violated for these too. Also the ?-
- hyperon is predicted. Y 1 ?- ( d d d)
?0( d d u) ?( d u u) ?( u u u) 0
?-( d d s) ?0( d u s) ?( u u s)
-1 ?-( d s s) ?0( u s s)
-2 ?- ( s s s)
-3/2 -1
-1/2 0 1/2 1 3/2
TZ
98
Quark Masses

• From ?p e?/2mc 2.79 e?/2mpc we find that the
  mass of both the up quark and down quark (assumed
  the same) is m 336 MeV/c2
• We can get an estimate of the strange quark mass
  from ?? -0.61 e?/2mpc. If we take the ?-
  hyperon to be ( u d) s then the
  spin-0 (u d) combination does not contribute to
  ?? and ?? (-1/3) e?/2msc From this we
  find ms510 MeV/c2.
• This agrees with the mean estimate of the qs -
  qu,d mass difference we get by comparing the mass
  increases e.g M(?-) -M(?-) ms - m in the
  spin-3/2 decuplet ms -m
  160 MeV/c2

99
The space wave-function ?(L)
?(L0)
?(L1)
r
r
P 1, Symmetric P-1, Antisymmetric
?(L2)
r
P 1, Symmetric
100
Table of masses of s3/2
101
Reconciliation of Pauli Principle and Quarks

• If we want to both
• retain the dictum that all spin-1/2 identical
  particles obey Pauli Principle and
• retain the static quark model
• then clearly the quarks must have some extra
  attribute
• that nucleons, mesons, hyperons dont have. If we
  
• look at the point-like particle we are most
  familiar
• with, the electron, it has the attribute of
  charge and
• interacts with the exchange of mass-less photons.

102
Color Charge and Gluons

• For quarks which have to have
• an extra attribute AND
• some mechanism for interacting with other quarks
• it seems natural to assign the extra attribute as
  one which could be associated with the
  interaction - COLOR charge - and introduce a new
  mass-less exchange particle to allow the
  interactions - the GLUON.
• To allow three otherwise identical u-quarks to
  exist with parallel spins in the ?, COLOR
  charge clearly needs to have THREE different
  forms.

103
When these three COLOR charges are added together
then the resulting COLOR charge must be zero - as
protons, neutrons etc do not have any COLOR
charge. This is how the word COLOR arose - by
analogy with red blue green mixed together
making white . The quarks are deemed to have a r,
b or g COLOR charge. Also anti -quarks must have
anti-COLOR r, b or g charges so that the mesons
dont have any COLOR charge.
white
104
The GLUONS like the photons have spin-1 but
differ by also having COLOR charge. There are
three COLOR charges and three anti-COLOR
charges. As the GLUONS are exchange particles
like mesons they are given COLOR- anti-COLOR
combinations. There are eight of these rb rg
b g b r g b g r (r r g g -2bb)/?6 (r r -
g g)/ ?2 The exchange of colored gluons gives
rise to the COLOR force.
105
The COLOR force
106
Suppose the COLOR lines of force have been pulled
together until they form a tube AND the
interaction energy is then so high the quark
masses are small by comparison. If this system is
now considered to be rotating we have a crude
model for a meson with spin. k2 is the energy
per unit length of the force tube. The ends of
the tube rotate at velocity c, the half-length of
the tube is ?. Then the mass of the system is
given by ? Mc2 2 0 ?
k2dr/ (1-v2/c2)1/2. This is true as k2dr is the
REST MASS energy of dr so that the relativistic
total energy is k2dr/ (1-v2/c2)1/2. At a distance
r from the centre v c r/ ?. ?

Thus Mc2 2 0 ? k2dr/ (1-r2/?2)1/2 ?k2?



107
The angular momentum of the infinitesimal mass at
distance r is (k2dr/c2 (1-r2/?2)1/2 ) v r
(k2dr/c2 (1-r2/?2)1/2 ) (cr/?)r (k2r2/?c
(1-r2/?2)1/2 )dr Thus the total angular momentum
of the tube, in units of ?, is ?

J 2/? 0 ? (k2r2/?c (1-r2/?2)1/2
)dr ?k2?2/2?c (Mc2)2/
(2?k2?c)
108
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109
We see that dJ/d(Mc2)2 0.9 Gev-2
(2??k2c)-1 and knowing ? 6.6 10-22 MeV sec c
3 1023 fm sec-1 we find k2 1Gev fm-1 As the
spin-1/2 baryons have masses 1GeV and radii
about 1 fm this is not a bad result! At a
distance of 1 fm it means the energy holding
quarks in the baryons is 1 GeV - 100 times
nuclear binding energies. Put another way the
force holding quarks inside a hadron is 1015
Gev/m or in traditional UK units -about 10 tons
per pointlike quark!
110
Quarks on springs We see now that the
potential between quarks is of the
form Vk2r i.e. as the distance between the
quarks increases so does the magnitude of the
potential binding the quarks in (e.g.) the
proton. A classical example of such a potential
is a spring
111
As two nucleons approach each other one can
envisage gluon transfer between quarks being
attempted. However at distances of 1 fm the
energy involved 1 GeV is sufficient to allow
pair production to occur
Thus mesons are emitted rather than gluons! Three
quarks always remain confined in the nucleon.
112
Evidence for the COLOR Force
In addition to the spin-0 mesons
1 K0( d s) K( u s) Y 0 ?-( du)
?0,?,?( u u, d d, s s) ?(u d) -1 K-( s
u) K0( s d) -1 -1/2 0
1/2 1 TZ there are the
spin-1 mesons
1 K0(d s) K(u s) Y
0 ?- ( u d) ?0,?,?( u u, d d, s s) ?( d
u) -1 K-(s u) K0(s d) -1 -1/2
0 1/2 1
TZ
113
These have masses of K s 890 MeV/c2
? s 770 MeV/c2 ? 782 MeV/c2
? 1020 MeV/c2 These spin-0 and spin-1 mesons
exhaust the possibilities for u,d,s and u,d,s
combinations. The ? in particular is an ss meson
- which we might have expected given a static
model s-quark mass of 510 MeV/c2. Its lifetime is
very short - from its energy width 4.4 MeV -
using ?E?t ? - it is 1.510-22 secs. It decays
into KK- predominantly rather than ??-?0 . The
only known explanation for this is in terms of
gluons!
114
Zweigs Rule
In a strong decay completely disconnected q?q
decays - q?q? gluons ? q?q - are strongly
suppressed relative to direct q?q production -
q?q ? q?q q?q.
115
The Discovery of the J/?
It was quite a surprise then when in 1975 a new
spin-0 meson was discovered simultaneously in
both e-e- collisions at high energies (shown
below) and in the reaction p Be
?J/? anything
? e-e- with an energy of
3.1 GeV/c2.
116
As all spin-1 mesons predicted by the quark model
had been found the only way the J/ ? could be
understood was if another heavier quark existed
called qC - the charm quark. Associated with this
is another quantum number C1. The J/ ? is a cc
state ie C0. It is said to have hidden charm.
It must have J1 as e-e- annihilate to produce a
virtual photon (spin 1) which forms the J/ ? in a
3S1 state. By analogy with the ? -meson - an ss
state- we infer the mass of the c-quark to be
1550 MeV/c2. Its other properties
are T Tz S B Y C Q/?e? Spin qC
(c) 0 0 0 1/3 1/3 1 2/3 1/2
117
Other states involving the charm quark
Subsequently many other meson states made from cu
(the D0), uc (D0), cd (D ),dc (D- ) - all with
masses around1870 MeV/c2, and cs (DS) ,sc (DS-)
pairs - with masses around 1970 MeV/c2. There are
also charmed baryons ?C (udc), ?C(uuc) ,
?C(udc), ?C0(ddc), ?C(usc), ?C0(dsc),
?C0(ssc). Also other hidden charm cc states
like the 3S1 J/? except in configurations 1So ,
3P0,1,2 were found - the ? and ? mesons
respectively. As these are made of very heavy
quarks they are likely to be moving very slowly
and hence can be compared with states formed by
another particle-anti-particle pair - the e e-
pair- which form a set of very loosely bound
states called positronium. The new hidden charm
set of meson states is called charmonium.
118
Lifetime ( energy width) of the J/? The
?-meson (ss) has sufficient mass to decay
directly into a pair of K(us) and K- (su)
mesons and hence, as we have seen this is not
suppressed by Zweigs rule, has a broad width -
when produced in e- - e collisions - of 4.4
MeV. The J/? (cc ) however has insufficient mass
to decay directly into D(cd) and D- (dc) mesons.
Also - see examples below- other possible decay
modes are inhibited by Zweigs rule. Hence it has
a very narrow width - 88 keV.
119
Charge of Charmed Particles The much-modified
expression for charge is now further generalised
to Q/?e? TZ (B S C)/2
120
Charmonium
109 difference in energies!
Positronium
121
Potential deduced from charmonium states
122
QUARKS

• Evidence for point-like objects in nucleons from
  deep inelastic lepton-nucleon scattering
• 3 flavor (u,d,s) static quark model of baryons-3q
  mesons- qq
• qq orientation in mesons - anti-parallel
  (spin-0), parallel (spin-1)
• Magnetic moments of baryons ? q orientation in
  baryons
• Parallel orientation of like flavor quarks in L0
  state ? Apparent violation of Pauli Principle
• Magnetic moments and J3/2 baryon masses ? u,d,s
  quark masses in static model
• Reconciliation of Pauli Principle and Quarks ?
  Need for COLOR charge ? COLOR force through
  exchange of massless gluons
• Constant Energy / unit distance of COLOR force
  from simple interaction model ? quarks bound by
  spring-like force
• Need for fourth flavor quark qC ? charmonium ?
  evidence for shape of COLOR force potential.

123
As a confirmation that the interaction between
quarks is due to the special property of COLOR
charge (that ALL quarks have regardless of
flavor) AND the exchange of COLORED gluons, at
even higher e- e- collision energies another
meson was found the ? (Upsilon) with a mass of
9460 MeV/c2. This could not be explained as a
combination of u,d,s,c quarks and
anti-quarks. Another quark attribute was needed
beauty and a quantum number b. For qb, b-1and
hence, with an even further revised expression
for charge Q/?e? TZ (B S C b)/2 the
charge of the b-quark is -1/3 ?e? - as it is a
singlet with Tz 0. Its mass is I/2 the upsilon
(bb) mass- 4730 MeV/c2. Otherwise T Tz S B Y
C b Q/?e? Spin qb (b) 0 0 0 1/3 1/3 0
-1 -1/3 1/2
124
Another set of states was then discovered -
bottomonium.
It can be fitted with exactly the same potential
as for charmonium!
125
Leptons
Just as more and more massive hadrons - strongly
interacting particles - were discovered as higher
and higher energy accelerators became available,
a heavier charged lepton was discovered in 1975
- the ?-lepton or (tauon) with a mass of 1784
MeV/c2. There are thus three known charged
leptons Lifetime Mass (sec)
(MeV/c2) Electron e- Stable
0.511 Muon ?- 2.210-6 105.66 Tauon
?- 3.410-13 1784 They of course all have
antiparticles. Both the ?s and the ?s decay via
the WEAK INTERACTION.
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The muon decays ?- ? e- ?? ?e The
e-anti-neutrino, ?e ,conserves epton number in
the decay the ?-neutrino, ?? conserves mepton
number Le 0 ? 1 0 ( -1) L ?
1 ? 0 1 0 The masses of ?? and ?e are
near zero. The tauon can decay in many ways
(three shown here) ?- ? e- ?? ?e
? ?- ?? ?? ? ?- ?? The
tau-neutrino, ??, conserves tepton number in the
decays - L? 1 (?-, ??), L? -1 (?, ??)
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Families of Quarks and Leptons
Normal matter thus consists of the up and down
quarks, the electron and ?e. As more energy is
available the strange and charm quarks can be
created, together with the muon and ?? u d
s c e- ?e ?- ?? At even higher
energies we find the b-quark, the tauon and ?? Is
there a further quark - the top quark (t-quark)
with the extra attribute truth? b t ?-
??
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Such a quark has been found in extremely high
energy proton- antiproton colliding beam
experiments (Tp Tp 900 GeV) . In these
experiments quarks in the proton annihilate with
antiquarks in the antiproton
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The tt pair then decay leading to lighter
quarks which initiate jets of hadrons by a
complicated process called quark
fragmentation. Tracking back the jets to their
origin the top quark mass can be determined
as Mt 175 GeV/c2 The other properties of the
top quark are T Tz S B Y C b t
Q/?e? Spin qt (t) 0 0 0 1/3 1/3 0 0 1
2/3 1/2
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Quark counting!
Production of hadrons or muons in e-e-
collisions can be viewed as Rutherford Scattering.
e- q e- ?-
e- q e- ?-
q ? q ?- ? ?
e- e e- e

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As there are different quark species ?(e-e- ?
hadrons) ?i ?(e-e- ? qiqi) The ratio of the
cross-sections for hadron production and muon
production at a particular energy will take out
any kinematic factors AND the cross-sections will
be proportional to the square of the charges of
the scatterer (quarks or muons) - Rutherford
scattering formula. Therefore we have R
?(e-e- ? hadrons) ?i Qqi2
?(e-e- ? ? ?-) Q?2 i.e. R counts the
number of quarks produced at a given energy!

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We see that if there are only 3 flavors of quark
-u, d and s - and COLOR is not a real attribute
then R (2/3)2 (1/3)2 (1/3)2 2
12 3 If COLOR is a real attribute the we
have , ur,ub, ug,dr,db,dg,sr,sb,sg quarks
and R 3(2/3)2 (1/3)2 (1/3)2 2
What do we get experimentally?
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When we have five flavors of quarks we get R
3(2/3)2 (1/3)2 (1/3)2 (2/3)2 (1/3)2
11/3 in excellent agreement with the
data. Clearly with the t quark R 15/3
5 but we do not have data at sufficiently high
energies yet. We do though have specific evidence
of the reality of both quarks of different
flavors and color charge - evidence as strong as
for the existence of nuclei.
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The Weak Interaction
Examples of this we have met already are neutron
decay and muon decay n?p e- ?e
?- ? e- ?? ?e These are both very slow in
particle physics terms - mean lives 887 s and
2.10-6 s respectively. Another example is
neutrino scattering e.g. ?e e- ? ?e e- The
cross-sections for such reactions are incredibly
small indicating near-zero range for the
interaction. These two facts indicate that the
reaction must be mediated by the production of a
very heavy particle .
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These have been discovered in PP collisions at
very high energy and are the INTERMEDIATE VECTOR
BOSONS W? MW 80 GeV/c2 Z0 MZ 91
GeV/c2 The Z0 mediates such reactions as ??
e- ? ?? e-
e- ??
e- Z0 ??
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When both leptons are of the same generation
(e.g. eptons) both the Z0 and the W- can
mediate ?e e-? ?e e-
e- Z0 ?e
e- ?e
e- W- ?e
e- ?e
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In weak decays of particles only the W and W-
are involved
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Strong decays occur by the excited state energy
being converted into quark-antiquark pairs.
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Strong interaction reactions annihilation is
followed by pair production
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Electromagnetic Decay Example
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And finally Decay by Annihilation!
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