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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 laws;Phase shift Analysis; More conservation laws; N*s and s; Photon-nucleon interaction
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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
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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.
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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 >1032 y ~ 11.1 min
Decay Mode ----- n p + e- +e
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Di- NucleonsOnly 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
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Deuteron Cross- section
~ rD2 ~ 12.5 fm2 = 125 mb
Deformation from circle ~ 3mb
P N
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What is the deuteron state?
• From ~ p + n we infer that the proton and neutron are in an orbital angular momentum L=0 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
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Unstable Di - nucleons
D* - Unbound excited state of Deuteron. Eex ~ + 70 KeV. Spin=0. ~ 10-22 sec
Di-proton - Unbound state of two protons. Spin=0. ~ 10-22 sec
Di-neutron - Unbound state of two neutrons. Spin=0. ~ 10-22 sec
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p + D p + n + p
p + D n + (2-p)
Nn
En(MeV)
Evidence for existence of di-proton (2-p)
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Possible combinations of two S=½ Nucleons in L=0 state
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States of two nucleons with L=0
Only the deuteron - 1 CHARGE STATE -exists with S=1, Sz = +1,0, -1
i.e. 1 CHARGE STATE, 3 possible spin orientations
D*, 2-p and 2-n - 3 CHARGE STATES -exist with S=0, Sz =0
i.e. 3 CHARGE STATES, 1 spin orientation
There is a certain symmetry here!
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2-p D* 2-n
D
S=0
S=1
Sz = 0
Sz = +1,0,-1
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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:
T=1, Tz = +1, 0, -1 T=0, Tz = 0.
and Q/e = Tz + 1
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2-p D* 2-n
D
S=0
T=1 Tz = +1 Tz = 0 Tz = -1
S=1
Sz = 0
Sz = +1,0,-1
T=0, Tz = 0
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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= 4/3RA3 ~ A V(Nucleon)
•Attractive - nuclei exist
Repulsive core
Evidence: RD= 2fm R(3He) = 2fm R(4He) = 1.7 fm
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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
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The -particle (4He nucleus) is the smallest and most strongly bound nucleus in the periodic table. The nucleons within it clearly overlap. Why doesn’t 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.
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r
r
Deuteron
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N
S
N
S
N
S
N
S
Force between two dipole magnets
d1 d2r
d1
r
d2
V(r ) = -1/r3{3(d1.r)(d2.r) - d1.d2}
where r is the unit vector
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r
Deuteron
The EM potential V(r ) = -1/r3{3(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)
r
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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?
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The 4 exchange forces
Force Range Exchange particle Strength
Gravitational Graviton,g, 0 10-39
Electromagnetic Photon, , 0 10-2
Strong < 2fm Pion, , 140 MeV/c2 1
Weak ~ 0 W-boson,W,81 GeV/c2 10-5
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The Electromagnetic InteractionIn 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.
Et ~
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
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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:
e- p
E=0
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.
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The Strong InteractionMeson Exchange
Et ~ 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
E=0
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Properties of PionsPions exist in three charge states + , - and 0
Therefore the isospin of the Pion T = 1 with orientations
in charge space Tz = +1, -1 and 0 respectively.
Q/e = Tz
M+ = M- = 139.57 MeV/c2 M0 = 134.97 MeV/c2.
The spin of each pion is S = 0.Another important property of the Pion is its
PARITY
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PARITYThis 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 L=0,2,4,… are Even Parity, those with L=1,3,5,… 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.
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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) LD= 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
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The overall wave-function of two neutrons consists of two parts: nn = (L) (S)
• a space wave-function produced by solution of Schrodinger’s 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 L=1,3,.. 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.
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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 S=0, Sz =0
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For two neutrons J= 1 cannot occur if either:
•L=0 S=1 or L=2 S=1- Symmetric Space Wavefunction, Symmetric spin Wavefunction ie Overall Symmetry
•L=1, S=0 - Anti-Symmetric Space Wavefunction, Anti-Symmetric spin Wavefunction. ie Overall Symmetry The ONLY possibility is L=1 S=1 - Anti-Symmetric Space Wavefunction, Symmetric spin Wavefunction. ie Overall Anti-Symmetry. J can then have values J=2
or importantly J=1
L=1 S=1 L=1 S=1J=0
L=1 S=1
J=1
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So now, in the reaction AT REST between a - and a deuteron:
- + D n + n
Spin (S) 0 1 ½ ½
Angular Momentum (L) LD = 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 (LD = 0 ) goes to an odd parity Lnn=1 WF. But parity is conserved in the Strong Interaction.We must thus consider Intrinsic Parities. Then: Pinitial = PPnPpP(LD = 0 ) = Pfinal= PnPn P (Lnn =1)The intrinsic parities of the nucleons cancel each other. So:
PP(LD = 0 ) = P (Lnn =1) i.e. -1 +1 = -1The INTRINSIC PARITY of the PION is NEGATIVE. Incidentally the INTRINSIC PARITY of the NUCLEON is POSITIVE - by convention.
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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 L=1.
Depending on their initial spatial orientation (a) or (b) happens.
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0
L=1
0
L=1
L=1
L=1
Lnp=2Lnp=0
Lnp=0 Lnp=0Pion transfer occurs by nucleons changing spin direction
Pion transfer by angular momentum change
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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 S=1. When these are transferred (with L=1 to conserve parity) nucleons are forced to L=2 state -i.e . Must move apart, are effectively repelled.
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Pion-Proton interaction
We can create beams of pions by bombarding a heavy nucleus with a high energy proton beam from an accelerator.
Proton beam
T= 3000 MeV
Tungsten (W) target
+
0
n
Beams of charged pions are formed using electric and magnetic fields.
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+ beam
Pion detectors
Liquid hydrogen target
Number of scattered pions = N(IN) –N(OUT) =
N(IN)* (NAV*H*tH *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.
Area AH, length tH,density H
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Pion-proton total cross-sections 195 600 900 1380 T (MeV)
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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?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*
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C.of E. T- + m-c2 + mpc2 = m*c2 + T* (1)
C.of M. p- = p* and hence p-2c2
= p*2c2 (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- = T*2 + 2m*c2T*
and substituting for T* from (1):m*c2 = {(m-
+ mp)2c4 + 2T-mpc2}Substituting pion, proton masses and T- = 195 MeV we find:
m* = 1235 MeV/c2
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The Total centre-of-mass Energy is E =m*c2 = 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
m*c2
To find the spin state of the resonance we need scattering theory.
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Scattering Theory -Spin-less particles, Elastic Scattering,CM reference frame
inc = eikz
Solution of: -2/2m(2 inc/z2) = E inc. As E=pCM2/2m
and pCM = k this reduces to 2 inc/z2 = -k2inc.
r
z
Incident pion wave Proton Unscattered pion wave
Scattered pion wave scatt
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scatt is a solution of: -2/2m(2scatt) = (E -V(r))scatt
scatt= f() eikr/r
•at large r where E>> 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
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How are inc, scatt
related to observables?
We know * = 2 is the probability of a particle’s 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
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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 del/d() =number of pions scattered into unit solid angle at angle per second / number of pions incident per unit area per second i.e. del/d() = f() f*() .el is the cross-section for elastic scattering.el/unit area is the probability for scattering into 4 solid angle.
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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.
1013 fm 2 fm
d p angular momentum state
lp = p d
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The pions can interact, in principle, with the protons in ALL possible angular momentum states lp. 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 = lp.inc = eikz = eikrcos = 1/kr 0
(2l +1) Pl(cos){eikr - e-i(kr-l)}/2i where Pl(cos) is aLegendre polynomial.
outgoing wave ingoing wave
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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/kr0(2l +1) Pl(cos){ei(kr+ 2(l)) - e-i(kr-l)}/2i
= eikz +f() eikr/r (3)
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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/(2l+1) when l=l’’ = 0 otherwise
el = 4/k2 0 sin2(l) (2l +1)
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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 ???
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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- <300MeV T- >300MeV
- 0 p
- + - p
+ p + p Elastic Low Energy High Energy
T+ <300MeV T+ >300MeV
+ 0 p
+ + - p
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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 NucleonsB=0 Pions
B is conserved in the All Interactions (Strong, EM, Weak, Grav.)
Epton Number Le = +1 ElectronsLe = 0 Photons
Le is conserved in All Interactions
We will generalise these laws to include other particles later.
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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.
+
p
Pcm
Pcm
(+ p)*
1 fm
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We know m* (T+ = 195 MeV) = 1236 MeV/c2.
Then E+ + Ep = m*c2
i.e. (pCM2c2 + m+
2c4) + (pCM2c2 + mp
2c4) = m*c2
Rearranging: (pCM2c2 + m+
2c4) = m*c2 - (pCM2c2 + mp
2c4)
Squaring:pCM
2c2 + m+2c4 = m*2c4 + pCM
2c2 + mp2c4 - 2m*c2 (pCM
2c2 + mp2c4)
Rearranging: 2m*c2 (pCM2c2 + mp
2c4) = m*2c4 + mp2c4 - m+
2c4
Squaring again:
4m*2c4 (pCM2c2 + mp
2c4) = (m*2c4 + mp2c4 - m+
2c4)2
And: pCM2c2 =(m*2c4 + mp
2c4 - m+2c4)2 - mp
2c4
4m*2c4 Substituting the mass values we find: pCM = 240 MeV/c
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Now lmax= pCM d+ p
Thus lmax = 240 MeV/c 1fm/6.6 10-22 MeV sec
=240 MeV1 fm/ (3 1023 fm sec-16.6 10-22MeV sec)
~ 1.2
Therefore lmax = 1
And el = 401 sin2(l) (2l +1).
k2
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How does sin2(l) vary with pion energy T ?
sin2(0)
At T=0, k=0 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. l=0 waves ‘see’ the same amount of the + p potential at all energies in ahead-on collision.
+ p
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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 l=0 phase shift cannot cause the peak in the cross-section - sin2(0) is a constant at all T.
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sin2(1)
Again, at T=0, k=0 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 l=1 wave ‘sees’ more and more of the potential and hence (1) increases.
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180
(1) 90
0 T
sin2(1)
1
0 195 MeV T
(1) = 90
60
This is resonant behaviour. Thus the l=1 phase shift behaviour causes the resonance at T = 195 MeV. Thus el = 4 sin2(l) 3 + small l=0 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 l=1 can be in either of two total angular momentum states j = 3/2 or 1/2. Thus our formula becomes:
el = 4/2j=1/2,3/2 sin2(j) (2j +1)
k2
Is the resonance in the j=1/2 or the j=3/2 state?
Assume (j=3/2) is small and (j=1/2)= 90at 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 (j=1/2) is small and (j=3/2)= 90at T = 195 MeV then at the peak:
el = (4/2k2)4 ~ 170 mb + small j=1/2 and l=0 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 j=3/2 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 t=1/2 and tz = -1/2, +1/2. This , and similar resonant states, are called N*s.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 K+p 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-
E=0
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 1950’s 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 > mp and decay into protons and pions.e.g. 0 - + p
72
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
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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 K+p 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 K+p elastic scattering. At higher momenta the cross-section should increase as there is enough energy for more mesons to be produced.
80
K+p 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= B+S
is called the HYPERCHARGE.
83
Families of ParticlesHistorically 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 AnalogyIn 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
nSuper-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 +2/3e, the up-quark qu and one with -1/3e, the down quark qd
89
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
To explain the properties of the hyperons there is a need also for a third quark - the strange quark qs
90
The nucleons are composed of the non-strange quarks:
p (u, u, d) in an L=0 state, T=1/2
n (d, d, u)
The spin-1/2 hyperons include one or two strange quarks: + (u, u, s)
0 (u, d, s) in an L=0 state, T=1
- (d, d, s)
0 (u, d, s) in an L=0 state, T=0
- (d, s, s)
in an L=0 state, T=1/2
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 T=0 meson with a mass of ~550 MeV/c2
The ’ is a T=0 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: = 2/3a -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/2mcThus 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 ms=510 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)
(L=0)
r
(L=1)
r
(L=2)
r
P= +1, Symmetric P=-1, Antisymmetric
P= +1, Symmetric
100
Table of masses of s=3/2
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 don’t 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 GluonsFor 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 don’t 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
= k22/2c
= (Mc2)2/ (2k2c)
108
109
We see that :
dJ/d(Mc2)2 = 0.9 Gev-2 = (2k2c)-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:
V=k2r
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 ForceIn 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 K*0(d s) K*+(u s)
Y 0 - ( u d) 0,,( u u, d d, s s) +( d u)
-1 K*-(s u) K*0(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 Et ~ - it is ~ 1.510-22 secs. It decays into K+K- predominantly rather than +-0 .
The only known explanation for this is in terms of gluons!
114
Zweig’s Rule
In a strong decay completely ‘disconnected’ qq decays - qq gluons q’q’ - are strongly suppressed relative to direct qq production - qq qq 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 C=+1. The J/ is a cc state ie C=0. It is said to have ‘hidden charm’. It must have J=1 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 quarkSubsequently 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), C
0(ddc), C+(usc), C
0(dsc), C
0(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 Zweig’s 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 Zweig’s 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
Positronium
109
difference in energies!
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 L=0 state Apparent violation of Pauli
Principle• Magnetic moments and J=3/2 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/eSpin
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
LeptonsJust 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.
126
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 (+, )
127
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
-
128
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:
129
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
130
Quark counting!
e-
e+
e-
e+
q
q
-
+
e-
q
e-
-
e-
q
e-
-
Production of hadrons or muons in e+-e- collisions can be viewed as Rutherford Scattering.
131
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- + -) Q2
i.e. R counts the number of quarks produced at a given energy!
132
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?
133
134
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.
135
The Weak Interaction
Examples of this we have met already are neutron decay and muon decay :
np + 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 .
136
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-
Z0
e-
137
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
138
In weak decays of particles only the W+ and W- are involved:
139
140
Strong decays occur by the excited state energy being converted into quark-antiquark pairs.
141
Strong interaction reactions: annihilation is followed by pair production:
142
Electromagnetic Decay Example
143
And finally: Decay by Annihilation!
144
145
146
147
