# 1. Relativistic kinematics

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13 Νοε 2013 (πριν από 4 χρόνια και 7 μήνες)

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U. Uwer
1
II.Pre-requisites
1.Relativistic kinematics
2.Wave description of free particles
3.Scattering matrix and transition amplitudes
4.Cross section and phase space
5.Decay width, lifetimes and Dalitz plots
Literature:
F. Halzen, A.D. Martin, “Quarks and Leptons”
O. Nachtmann, “Elementarteilchenphysik”
1. Relativistic kinematics
1.1 Notations
 4-vector
• contra-variant form
• covariant form
(
)
(
)
Φ,(,Φ,(,
00
pEpppxtxxx
r
r
r
r
====
µµ
(
)
(
)
),(,),(,
00
pEpppxtxxx
r
r
r
r
−=−=−=−=
µµ

==
1000
0100
0010
0001
µν
µν
gg
ν
µνµ
ν
µνµ
xgx
xgx
=
=

=

=∂

∇−

=

=∂
rr
,,
t
x
tx
µ
µ
µ
µ
)(
00
bababagbaab
r
r
⋅−===
µν
µν
µ
µ
• Metric tensor
• Derivative operator
• Scalar product
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2
1.2 Lorentz invariants
Lorentz transformation:

=

=

=

tt
ll
pp
p
E
p
E
p
rr
rr
γβγ
βγγ
),( withparticle moving pEp
r
=
Scalar products are invariant under Lorentz transformations:
abba =

Example 1:
invariant mass
2222
mpEppp =−==
r
P
P
Example 2:
CMS energy of 2 particle
collision calculated in any frame
2
21
2
21
2
21
)()()( ppEEpps
r
r
+−+=+=
x
t
E
p
r

2/12
)1(

−= βγ
c
v

m
E

w/r to rest frame:
t
x’
t’
Remark: Lorentz invariants (e.g. cross sections) can only depend on scalars
1.3 Mandelstam variables
C
D
A
B
DCBA +→+
(unpolarized particles)
What are the Lorentz scalars the
cross section can depend on ?
DCBAikiki
ppppppp,,, with
,
=

4 constraints
4-mom. conservation: 4 constraints
2 independent products
22
ii
mp =
2
2
2
)(
)(
)(
DA
cA
BA
ppu
ppt
pps
−=
−=
+=
2222
DCBA
mmmm
uts
+++
=
+
+
i
p
k
use 2 out
of the 3 Mandelstam
variables
10 combinations
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3
ψψ
2
2
1
∇−=

mt
i
[ ]
)(exp
1
),( Etxpi
V
tr −=
r
rr
ψ
m
p
E
2
2
=
(
)
ψψψψ
)()(
2
1
∗∗
∇−∇=
im
j
r
2
ψρ=
2.1 Schrödinger Equation for non-relativistic free particles
Solution for energy
0
=∇+

j
t
r
ρ
Continuity equation:
Schrödinger Eq uses classical E-p
relation E
2
=p
2
/2m and the replacement
∇−=

r
r
ip
t
iE and
2. Wave description of free particles
2.2 Klein-Gordon Equation
Starts from relativistic energy relation
E
2
=p
2
+m
2
:
Describes relativistic Spin 0 particles
0
22
2
2
=+∇−

φφφ m
t
0
22 >

+±= mpE
[ ]
)(exp),( tExpiNtr
±
−=
r
r
r
I
Solutions for energy values:
negative E values cannot be
ignored as otherwise
solutions are incomplete
0=∇+

j
t
r
ρ
Continuity equation:

=
∗∗
φφφφρ
t
i
t
i
(
)
)()(
∗∗
∇−∇−= φφφφ iij
r
with
and
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4

=
∗∗
φφφφρ
t
i
t
i
(
)
)()(
∗∗
∇−∇−= φφφφ iij
r
[
]
)(exp),( tExpiNtr
±
−=
r
r
r
I
2
2 NE=ρ
2
2 Npj
r
r
=
For the solution:
What are negative probabilities
for the E < 0 solutions ?
Normalization schemes:
N = 1/√(2EV) ⇒ 1 particle per unit volume V
N = 1/√V ⇒ 2E particles per unit volume V
2.3 Anti-particles
Dirac interpretation for fermions:
2m
e
E
0
Vacuum = sea of occupied neg. E levels
For fermions the negative energy levels
are w/o influence as long as they are fully
occupied
e
+
e

annihilation:
Free energy level in the sea. e

drops into the hole and releases
energy by photon emission: E
γ
> 2m
e
Missing e

w/ negative energy
corresponds to to a positron w/ E>0
Photon conversion for E
γ
> 2m
e
Excitation of e

from neg. energy
level to pos. level: γ →e
+
e

Model predicts anti-particles (Discovery of positron by Anderson in 1933)
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5
Discovery of the positron
Absorber:
Energy loss

B field
small curvature
strong curvature
BvqF
L
r
r
r
×=
eq +=
Anderson, 1933
Feynman Stückelberg interpretation
)exp(:
)exp(:
iEtxpiNEE
iEtxpiNEE
+=−=
−==
−−
++
r
r
r
r
φ
φ
Solutions with neg. energy
propagate backwards in time:
Solutions describe anti-particles propagating forward in time:
x
t
1
2
p
E
r
0<

x
t
2
1
p
E
r

>

0
2
2 NE=ρ
2
2 Npj
r
r
=
2
0
2 NEqJ ⋅=
2
2 NpqJ
r
r
⋅=
q
×
Neg.
probability
density
Charge
density /
currents
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6
Example
Particle T

with q= −e and energy E

= −E < 0
)()(2)(2)()(
)()(2)()(2)()(
22
0
22
0
+−
+−
=−⋅+=⋅−=
=+⋅+=−⋅−=
TJNpeNpeTJ
TJNEeNEeTJ
r
rr
r
−+
−=>
++
TT
ppTET
r
r
,0)( with
Description of creation and annihilation:
• Emission of anti-particle ⎯T with p
µ
= (E, p) ⇔
absorption of particle T with p
µ
= (-E, -p)
• Absorption of anti-particle⎯T with p
µ
= (E, p) ⇔
emission of T with p
µ
= (-E, -p)
3. Scattering matrix and transition amplitude
Scattering process:
p
p
π
π

p
c
p
D
p
A
p
B
Described through quantum
numbers of initial and final state:
i
i

Scattering operator (S matrix):
ii S=

Measurement selects a specific state f.
Probability to find f:
fi
ifif SS ==

As there is the probability that it is useful to introduce the
transition operator T
ii =

if TT1S =+=
fi
T with
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7
fi
, conventionally one uses the transition or scattering amplitude M
fi
MppppNNNNiT ⋅−−+⋅−= )()2(
44
δπ
normalization:
N
k
=1/√V →2E particles/V
4-momentum conservation
Feynman rules for
calculation
Transition probability:
In this convention the transition probability is given for a single „possible final
state“. It turns out that the final state particles C(p
C
) and D(p
D
) can be in more
than one state. The number of possible final states is described by the phase
space factor and will be considered when calculating observable quantites.
2
2428
2
)]([)()2(
MppppNNNNTw −−+== δπ
Transition rate per unit volume:
2
2428
2
)]([)()2(
1
fi
fi
MppppNNNN
VTVT
T
W −−+

=

= δπ
)(
)2(
)]([
4
4
24
pppp
VT
pppp −−+=−−+ δ
π
δ
2
4
4
4
)(
1
)2(
fiDCBA
Mpppp
V
−−+= δπ
Fermi‘s Trick:
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8
4. Cross section and phase space
p
c
p
D
p
A
p
B
DCBA +→+
2
4
4
4
)(
)2(
fiDCBAfi
Mpppp
V
W ⋅−−+= δ
π
)states final of number(
flux) (initial
W
section Cross
fi
=
Cross section:
),( DC
F
W
f
fi
ρσ=
ρ
f
number of final states for given configuration
F incident particle flux of A and B
Transition rate:
4.1 Number of final states (phase space)
[
]
pdpp
r
r
r
+∈,
3
3
33
3
3
3
)2(2)2(22 ππ
ρ
E
pVd
E
pVd
hE
pVd
d
f
===
h
3
3
3
3
)2(2)2(2
),(
ππ
ρ
D
D
C
C
f
E
pVd
E
pVd
DCd =
1=h
Quantum mechanics restricts the number
of final states dρ
f
of a single particle in a
volume V with momentum
For particle C and D scattered into
momentum elements d
3
p
C
and d
3
p
D
Factor 2E is the result of
normalization of the wave
function: 2E particles / V
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4.2 Incident particle flux F
Choose rest frame of particle B to calculate F (simplest)
F = (flux density A) × (density B)
V
E
V
E
vF
BA
A
22
⋅=
r
A
B
A
v
r
A
A
p
r
B
B
p
r
iBA
ppp
r
r
r
=−=
A
A
A
E
p
v
r
r
=with
sp
V
EEp
V
F
iBAi
rr
22
4
)(
4
=+⋅=
CMS frame:
General form:
2
2
2
2
1
),,(2
V
mmsw
F =
(
)
2
1
2
21
2
2121
))()()((),,( mmsmmsmmsw −−+−=
with
(see Nachtmann)
4.3 Lorentz invariant phase space factor
3
3
3
3
44
2
3
3
3
3
2
2
4
4
4
)2(2)2(2
)()2(
22
)2(2)2(2
22
)(
)2(
ππ
δπ
ππ
δ
π
σ
D
D
C
c
DCBA
BAA
fi
D
D
C
c
BAA
fiDCBA
E
pd
E
pd
pppp
EEv
M
E
pVd
E
pVd
EEv
V
Mpppp
V
d
⋅⋅−−+⋅=
⋅⋅⋅−−+=
r
r
Lorentz invariant 2-particle phase space factor dΦ
2
Particle flux F
Remark: volume V drops out !
f
fi
d
F
W
d ρσ =
Putting everything together
sp
i
r
4=
CMS
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+++−=Φ
final
f
f
nnn
E
pd
pppPpppPd
2)2(
))(()2(),,,,(
3
3
21
44
21
π
δπ KK
Final state
http://pdg.lbl.gov/2007/reviews/kinemarpp.pdf
Phase space factor for n particles in the final state:
Phase space integration for two-particles final-state (CMS)
D
D
C
C
E
pd
E
pd
EEEEppd
d
22
)()(
4
1
33
3
2
2
2
∫∫
−−++=Φ
⎯→⎯Φ

δδ
π
rr
2
)(
:System CM
BA
DCfBAi
EEs
pppppp
+=
−==−==
rrrrrr
f
f
d
s
p
d Ω=Φ
∫∫
r
2
2
16
1
π
A
B
C
D
CCf
ddd
θ
ϕ
cos
=

f
p
r
i
p
r
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11
4.4 Differential cross section …putting everything together
ffi
i
f
fi
dM
p
p
s
d
F
M
d Ω⋅⋅⋅=Φ=
2
2
2
2
1
64
1
r
r
π
σ
2
2
1
64
1
fi
i
f
f
M
p
p
sd
d
⋅⋅⋅=

r
r
π
σ
p
c
p
D
p
A
p
B
DCBA
+
→+
CMS
• The dynamics of the scattering process is contained in the matrix element M
fi
which can be calculated using Feynman rules
• 1/s dependence of the cross section because of initial/final state kinematics