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UNIT 2: OUTLINE SYLLABUS:
1st Lecture Introduction
Hadrons and Leptons
Spin & Anti-Particles
The conservation laws: Lepton Number
Baryon number
Strangeness
2nd Lecture Problem solving
Check a decay for violation of conservation laws
Quarks
Properties of a particle given quark combination
3rd Lecture Follow-up
Fundamental forces and field particles
The standard model
Checking Baryon Numbers
a) p+ + n
b) p+ + n
p+ + p+ + n + p
_
+
p + p + p+
Method: B=+1 for baryons, -1 for anti-baryons, 0 for everything else
Answer: a) B = 1+1 on left hand side
B = 2 on right hand side too!
Allowed reaction!
b) B = 2 on left hand side
B = 1 on right hand side
Forbidden reaction
Checking Lepton Numbers
µ-
a)
b) π+
_
e- + ne + nm
µ+ + nm + ne
Method: L=+1 for Leptons, -1 for anti-Leptons, 0 for non-Leptons BUT separate
Lepton number for e-, m, t and their neutrinos
Answer: a) Before decay Le = 0 and Lm = +1
After decay Le = 0 and Lm = +1
Allowed reaction!
b) Before decay Lm = 0 and Le = 0
After decay Lm = 0 and Le = 1
Forbidden reaction!
Is Strangeness Conserved?
a) π+ + n
b) π- + p
K+ + 
 -+ 
Method: Strangeness is a property of some hadrons – see Tipler Fig 41-2 P.1322
Answer: a) Initial state has S = 0
Final state has S = +1 - 1 = 0
Allowed reaction!
b) Initial state has S = 0
Final state has S = -1
Forbidden reaction!
Conservation Laws
(Tipler Chap 41 Q14)
• Test the following decays for violation of the conservation
of energy, electric charge, baryon number and lepton
number. Assume that linear and angular momentum are
conserved.
• (a) n ->     m  m
• (b) 0 > e+ + e- + g
Conservation Laws
(Tipler Chap 41 Q14)
Solution
• Method: Use Table 41-1 and the conservation laws for Baryon number
and Lepton number
•
•
(a) n ->     m  m
– mn > 2m  2mm
– Total charge on both sides = 0 : conserved
– Baryon number changes from +1 to 0: violated
– Lm = 0 on both sides : conserved
– Process not allowed
(b) 0 > e+ + e- + g
– m > 2me
– Total charge on both sides = 0 : conserved
– Baryon number on both sides = 0 : conserved
– Le = 0 on both sides: conserved
– Process is allowed
Quarks - The Smallest Building Blocks of Matter
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Gell–Mann & Zweig 1963: see Tipler section 41-4
Three Different Types of QUARKS
There are three elementary quarks (flavors)
That make up the fundamental particles:
Up
Down
Strange
Name
Up
Down
Strange
u
d
s
u
d
s
Spin Charge (e)
1/2
+2/3
1/2
-1/3
1/2
-1/3
π+
u
d
p
Baryon
Baryon Strangeness
1/3
0
1/3
0
1/3
-1
Anti-quarks maintain spin, but change sign of S and B!
Meson
u
u
d
Different types of quarks contd.
• Mesons – quark + anti-quark ( q q )
• Baryons – three quarks ( q q q )
• Anti-baryons – three anti-quarks ( q q q)
By 1967 it was realised that new kinds of quarks were required
to explain discrepancies between the model and experiment
Charm (c)
Bottom (b) – discovered 1977
Top (t) – discovered 1995
Quark combinations
(Tipler Chap 41 problem 17)
• Find the baryon number, charge & strangeness of the
following quark combinations and identify the hadron:
• (a) uud
• (b) udd
• (c) uus
• (d) dds
Quark combinations
(Tipler Chap 41 problem 17)
Solution
Method: for each quark combination determine the baryon number B, the
charge q and the strangeness S; then use Tipler Table 41-1 and Fig. 412 to find a match.
• (a) uud
–
–
–
–
B = 1/3 + 1/3 + 1/3 = 1
q = 2/3 + 2/3 – 1/3 = 1
S=0
It is a proton
• (b) udd
–
–
–
–
B = 1/3 + 1/3 + 1/3 = 1
q = 2/3 – 1 /3 – 1/ 3 = 0
S=0
It is a neutron
• (c) uus
– Ditto, B=1, q=1, S= -1 and it is a +
• (d) dds
– Ditto, B=1, q=-1, S= -1 and it is a -
Quark spin
(orange booklet)
• The angular momentum vector of a spin ½ quark
can have one of two settings up or down
• So a meson can have its two quark spins parallel
with each other or anti-parallel:
Spin 1
Spin 0
Quark spin contd.
• Baryons e.g. uud:
Spin 3/2
Spin 1/2
The spin ½ particle is a proton, spin 3/2 particle is a D
Note that
is also spin ½ (parallel, parallel, anti-parallel)
EIGHT FOLD WAY PATTERNS
(see also Orange booklet)
(ddu)
(uud)
n






The Baryon
Octet Eight Spin 1/2
Baryons
S=0
p
S = -1
S = -2
Q = +1
Q = -1
Q=0