1. Elementary Particle Physics
Concepts
Lectures 2 & 3
Mark Thomson:
Chapter 2
Chapter 3
David Griffiths:
Frank Linde, Nikhef, H044,

2. Concepts (2&3):  Units  Relativistic kinematics  cross section, lifetime,   Feynman calculus toy model 2 3

3. Units: 0=0= h=c=1

4. Electromagnetism: Heaviside-Lorentz unitsq1 q2
r12
units:
0 allows to decouple unit of q from Time, Mass and Length units
Coulomb:
jullie!
 nu
ikke …
Thomson
Griffiths
unit of charge: qSI = 0 qHL = 04 qG
Ampère:

5. Electromagnetism: Maxwell equations
And for our applications:
We experiment in vacuum so none of the complications of e.m. behavior in matter.
And: often  = 0 and j = 0 as well!

6. ‘Natural’ units: h=c=1 Units: Length [L]
Time [T]
Mass [M]
Fundamental quantities: h   1034 Js (1 Js=1 kg m2/s)
c  m/s (definition of meter)
Simplify life by setting: c1
h1
[L]=[T]
[M]=[L]=[T]
Remaining freedom: pick as main unit Energy [E] =eV, keV, MeV, GeV, TeV, PeV, …
Energy [E]  GeV
Momentum  GeV/c
Mass [M]  GeV/c2
Time [T]  h/GeV
Lenght [L]  hc/GeV +e e

7. Common units in GeV x Relations between SI & natural units quantity
SI units
[kg,m,s]
[ h,c,GeV ]
natural units
h=c=1
Energy
kg m2 s2
GeV
Momentum
kg m s1
GeV/c
Mass kg
GeV/c2
Time s
h/GeV
GeV1
Length m
hc/GeV
Area m2
(hc/GeV)2
GeV2
h = [Energy][Time]
1/h gets you from [s] to [GeV1]
1/(hc)2 gets you from [m2] to [GeV2]
c = [Length] / [Time]

8. Units: lifetime – decay width
Lifetime  : [] = time s  GeV1
Decay width  1/ : [ ] = 1/time s1 GeV
conversion factor:
 = 106 s 
 3.31018 GeV1
   31010 eV

9. Units: cross section L   Ldt  L = # of events
1 barn  U-nucleus
Cross section  : [] = area m2  GeV2
ATLAS & CMS
LHCb
ALICE
1 nb1s1
conversion factor:
L=fn 𝑁 1 𝑁 2 𝐴
Ni: number of particles/bunch
n: number of bunches/beam
f: revolution frequency
A: cross sectional area of beams
Luminosity:
LEP: cm2s1= 10 b1s1  100 pb1year1
“B-factory”: cm2s1= 1 nb1s1  10 fb1year1
LHC: cm2s1= 10 nb1s1  100 fb1year1
typical
“luminosities”
L   Ldt
Technology!
(other course)
1 b = 1024 cm2
1 nb = 1033 cm2
1 pb = 1036 cm2
1 fb = 1039 cm2
typical
cross sections 
Physics!
(this course)
L = # of events

10. Relativistic kinematics

11. Lorentz transformation
And: sorry for the many c’s still floating around here … v S S’ x x’
Co-moving coordinate systems:
Einstein’s postulate:
speed of light in vacuum always c y y’
Time dilatation: moving clocks tick slower!
Time  between events in rest frame x= (0,0) & (,0)
What measures S’ as time difference? Transform: (0,0) & (,…)
Example: Cosmic-ray muons (210-6 s, m106 MeV):
E=p=10 GeV  vc, 100  travel typically 100c60 km

12. 4-vector notation: conventions
Invariant: same in frames S & S’ connected by Lorentz transformation:
Introduce metric:
Note: ‘g= g’
Contra-variant 4-vector:
Co-variant 4-vector:
Generally for 4-vectors a and b:

13. Elegance/power of notation!
𝒕´ 𝒙´ 𝒚´ 𝒛´ = 𝜸 𝟎 𝟎 𝟏 𝟎 −𝜷𝜸 𝟎 𝟎 𝟎 𝟎 −𝜷𝜸 𝟎 𝟏 𝟎 𝟎 𝜸 𝒕 𝒙 𝒚 𝒛
S S’
   
For boosts
𝒂 𝝁  = 𝝏 𝒙′ 𝝁 𝝏 𝒙 
𝒙′ 𝝁 = 𝒂 𝝁  𝒙 
contra-variant 4-vector:
𝒙′ 𝝁 = 𝒂 𝝁  𝒙 
𝒕´ −𝒙´ −𝒚´ −𝒛´ = 𝜸 𝟎 𝟎 𝟏 𝟎 +𝜷𝜸 𝟎 𝟎 𝟎 𝟎 +𝜷𝜸 𝟎 𝟏 𝟎 𝟎 𝜸 𝒕 −𝒙 −𝒚 −𝒛
𝒙′ 𝝁 = 𝒂 −𝟏 𝝁  𝒙 
= 𝒂 𝝁  𝒙 
co-variant 4-vector:
𝒙′ 𝝁 = 𝒂 𝝁  𝒙 
𝒕 𝒙 𝒚 𝒛 = 𝜸 𝟎 𝟎 𝟏 𝟎 +𝜷𝜸 𝟎 𝟎 𝟎 𝟎 +𝜷𝜸 𝟎 𝟏 𝟎 𝟎 𝜸 𝒕′ 𝒙′ 𝒚′ 𝒛′
S’ S
𝒙 𝝁 = 𝒂 −𝟏 𝝁  𝒙′ 
For rotations upper/lower indices story remains the same i.e. same red expressions, numerically x & x transform identically - only spatial coordinates k=1, 2 & 3 enter the game

14. Derivatives:   &  4-vectors
and this really makes
a very elegant notation!
𝒂 𝝁  = 𝝏 𝒙′ 𝝁 𝝏 𝒙 
𝑡 𝑥 𝑦 𝑧 = 𝛾 𝛽𝛾 𝛽𝛾 𝛾 𝑡′ 𝑥′ 𝑦′ 𝑧′
Using:
𝜕 𝜕𝑧′ = 𝜕𝑧 𝜕𝑧′ 𝜕 𝜕𝑧 + 𝜕𝑡 𝜕𝑧′ 𝜕 𝜕𝑡 = 𝛾 𝜕 𝜕𝑧 +𝛽𝛾 𝜕 𝜕𝑡
And chain rule:
𝜕 𝜕𝑡′ = 𝜕𝑧 𝜕𝑡′ 𝜕 𝜕𝑧 + 𝜕𝑡 𝜕𝑡′ 𝜕 𝜕𝑡 = 𝛽𝛾 𝜕 𝜕𝑧 +𝛾 𝜕 𝜕𝑡
𝜕/𝜕𝑡′ 𝜕/𝜕𝑥′ 𝜕/𝜕𝑦′ 𝜕/𝜕𝑧′ = 𝛾 𝛽𝛾 𝛽𝛾 𝛾 𝜕/𝜕𝑡 𝜕/𝜕𝑥 𝜕/𝜕𝑦 𝜕/𝜕𝑧
Yields:
′ 𝝁 = 𝒂 𝝁   
Hence:
𝜕 𝜕𝑡 , 𝜕 𝜕𝑥 , 𝜕 𝜕𝑦 , 𝜕 𝜕𝑧
transforms as a co-variant 4-vector!
𝝏 𝝏𝒕 , 𝝏 𝝏𝒙 , 𝝏 𝝏𝒚 , 𝝏 𝝏𝒛 = 𝝏 𝝏 𝒙 𝝁 = 𝝏 𝝁
corresponding contra-variant 4-vector:
𝝏 𝝏𝒕 ,− 𝝏 𝝏𝒙 ,− 𝝏 𝝏𝒚 ,− 𝝏 𝝏𝒛 = 𝝏 𝒕 ,−  = 𝝏 𝝏 𝒙 𝝁 = 𝝏 𝝁

15. Lorentz transformations*
Lorentz transformations
Lorentz-transformations:
These matrices must obey:
with
invariant
hence
For infinitesimal transformations you find:
Because:
with or
# of independent transformations: +
Hence: 4x4–10=6 independent transformations: 3 rotations: around X-, Y- & Z-axis
3 ‘boosts’: along X-, Y- & Z-axis

16. infinitesimal  macroscopic*
infinitesimal  macroscopic
from expression for ex
… via Taylor expansion
to cos & sin

17. Lorentz transformations: rotations*
Lorentz transformations: rotations S S’ 
Lorentz
Rotations around Z-axis (indices 1 & 2):  1 2 3 
pick d as infinitesimal parameter:
finite rotation  around Z-axis yields:
Similarly you find for the indices: 1 & 3 the rotations around the Y-axis
2 & 3 the rotations around the X-axis

18. *
cos, sin & cosh, sinh
𝑐𝑜𝑠 2 𝑥+ 𝑠𝑖𝑛 2 𝑥=1
𝑐𝑜𝑠ℎ 2 𝑥− 𝑠𝑖𝑛ℎ 2 𝑥=1

19. Lorentz transformations: boosts*
Lorentz transformations: boosts
Lorentz v
Boosts along Z-axis (indices 0 & 3): S’ S  1 2 3 
pick d as infinitesimal parameter:
finite boost  along Z-axis yields: =
Relation between x0=ct, x3=z & x’0=ct’, x’3=z’:
origin of S’ (z’=0) is seen by S at any time as: (t, vt)  - - 
and with cosh2sinh2=1, you get: -

20. Velocity & Momentum 4-vectorsx u S’ x’ u’
Standard velocity definition: u x/t:
For observer S: u = x /t
For observer S’: u’= x’/t’
u (and u’) not 4-vectors!
(not surprising: division of vector components …)
Lorentz transformation links u and u’:
2nd attempt to define a velocity: use proper time  =t/ or: d =dt/
(hard to imagine nicer invariant)
4-velocity:
4-momentum:
why p0E/c?
For particles with mass=0:

21. Transformation to center-of-mass
N particles in the Lab-frame
Useful expressions:
For transformation to the center-of-mass frame, use: m1 m2 m3 m5 m4
N particles in the c.m. frame
You get to the center-of-mass frame with as boost :
Verification:

22. 2-body decay: A B+C mB mA mC Example: -decay E Before decay:
2-body decay in c.m. frame
Before decay:
After decay:
With:
m  140 MeV
m  106 MeV
Example: -decay E
You get:

23. Example particle decay: -decay
m  106 MeV
me  MeV
Ee 53 MeV >2 particles
½m53 MeV

24. Example particle decay:  0-decay
 0   +
c.m. frame 1 2  0
12
Lab-frame 1 2 0
m  140 MeV
𝜸≫𝟏 → 𝛃≈𝟏 + 
c.m.
Lab.
typical opening angle in Lab. frame:

25. Example particle decay: -beam (theory)
-beams typically from K,   (or  ) decays
c.m. frame  
 cm K
Boost c.m. kinematics to find Lab. Kinematics:
Need to express cos cm in terms of  lab
boost along Z: , 1
 lab
Lab-frame  K   

26. Example particle decay: -beam (experiment)
Plug in a real example:
Neutrinos from a 200 GeV beam with K (mK = 494 MeV) &  (m = 139 MeV)
GeV
 
𝐊: 𝐄  𝐜𝐦 = 𝐦 𝐊 𝟐 − 𝐦 𝛍 𝟐 𝟐𝐦 𝐊 ≈𝟐𝟑𝟓 𝐌𝐞𝐕
: 𝐄  𝐜𝐦 = 𝐦 𝛑 𝟐 − 𝐦 𝛍 𝟐 𝟐𝐦 𝐊 ≈𝟐𝟗 𝐌𝐞𝐕 K 
lab
CDHS
detector 

27. Threshold energy: anti-proton discovery
Lab.: 
Lab. EA B
Threshold energy in Lab. 1 2
c.m.: 
c.m. 3 4
(assumed mA=mB …)
Threshold energy in c.m. frame: all particles produced at rest!
 energy in c.m. frame:
𝑝 𝑡𝑜𝑡 = 𝑀, 0
c.m. 
Example: anti-proton
p+p  p+p+p+p
 6.4 GeV
Use this minimum energy (M) in the c.m to calculate threshold energy in Lab.:

28. Bevatron: 6.5 billion electron volts
anti-proton
annihilation

29. Compton scattering ,E ’,E’ Pe Pe’ k k’
𝒑 𝒆 = 𝒎 𝒆 , 𝟎
𝒌 𝜸 = 𝑬,𝟎,𝟎,𝑬
Scattering of light on electrons: 
Lab-frame
,E
’,E’ Pe
Pe’ k
k’
𝒌+𝒑= 𝒌 ′ + 𝒑 ′  𝒑 ′ =𝒌− 𝒌 ′ +𝒑
 𝒑′ 𝟐 = 𝒌− 𝒌 ′ +𝒑 𝟐
=−𝟐𝒌. 𝒌 ′ + 𝒑 𝟐 +𝟐𝒑.(𝒌− 𝒌 ′ )
 𝒌. 𝒌 ′ =𝒑.(𝒌− 𝒌 ′ )
𝒌′ 𝜸 = 𝑬 ′ ,…,…, 𝑬 ′ 𝒄𝒐𝒔𝜽
 𝑬𝑬′ 𝟏−𝒄𝒐𝒔𝜽 = 𝒎 𝒆 (𝑬− 𝑬 ′ )
 𝟏−𝒄𝒐𝒔𝜽 = 𝒎 𝒆 𝟏 𝑬′ − 𝟏 𝑬  𝟏 𝑬′ = 𝟏 𝑬 + 𝟏 𝒎 𝒆 𝟏−𝒄𝒐𝒔𝜽
𝒑′ 𝒆 = …,…,…,…
and with:
you get Compton’s result, apart from units:
using ‘proper’ units, you get really Compton’s result:

30. Mandelstam variables I: s, t & u4 4 3 3 16
masses 12
momentum conservation 8
rotations 5
boosts 2
A + B  C + D process characterized by 2 parameters
cm
c.m. frame A B C D
E.g. in the c.m. frame: EAcm &  cm EA cm
E.g. in the lab. frame: EAlab &  lab
Lab-frame A B C D
lab EA
lab
Useful Lorentz-invariant variables:
For which you can easily prove:

31. Mandelstam variables II: s, t & u
Useful expressions in terms of s, t & u:
Lab. frame:
beam energy, EA: 
&
total energy in c.m.:
c.m. frame:
&
beam energy, EA: 
For A + A  A + A, in the c.m. frame:
= 4E2
 2E2
(also if all m<< E)

32. Mandelstam variables III: s, t & u
s (p1+p2)2=q2
‘s-channel’
t (p1-p3)2=q2
‘t-channel’
u (p1-p4)2=q2
‘u-channel’
Remark:
p1, p2, p3 & p4 in these figures are the physical 4-momenta,
arrows just flag particles (forward in time) and anti-particles (backward in time)
And:
direction of q you can pick, expressions for s, t & u do not depend on q-direction

33. Lecture 2
Study Thomson Chapter 2!

34. Elementary Particle Physics
Concepts
Lectures 2 & 3
Mark Thomson:
Chapter 2
Chapter 3
David Griffiths:
Frank Linde, Nikhef, H044,

35. Lifetime, Decay width, Cross section, …

36. Lifetime & Decay width mA mB mC
c.m. frame
Particle decay mathematics ( is decay width):
# particles
start
decay:
particles
Lifetime ( ) decay-width ( ) connection:
With multiple decay channels:
branching fraction:
and  = 1/tot
partial decay-widths: i
decays
Branching fractions: i /tot

37. Lifetime & Decay width W+ l l+

38. A+B C+D cross section
Several factors playing a role: physics & ‘administration’:
1. Physics: transition probability Wfi (most of this course)
probability for initial state ‘i’ to become final state ‘f’
units: [Wfi] = 1/(time volume)
2a. Administration: flux i.e. beam- & target densities
beam intensity: # of particles / (area time)
target intensity: # of particles / (volume)
units: [Flux] = 1/(area time volume)
beam
target
2b. Administration: ’density’ of states ‘f’: phase space N=2 (or N for N particles)
what really counts is the number density of final states with more or less same properties i.e. 2 is just a weight/number (dimensionless)
Cross section for A+B  C+D becomes: [] = area
Caveat: dimensions referred to don't account for wave function normalisations (later)

39. Example: cross section solid sphere
Scattering upon a solid sphere
(very simple, stupid geometry …)  b R 
Geometry: 
Differential cross section: 
Total cross section: as it should be of course!

40. Intermezzo: Dirac  - ‘function’
𝒇 𝒙 ′ = −∞ +∞ 𝜹 𝒙− 𝒙 ′ 𝒇 𝒙 𝒅𝒙
Definition of the Dirac -function: 
Fourier transforms:
𝑭 𝒕 = −∞ +∞ 𝒇(𝒙) 𝒆 −𝟐𝝅𝒊𝒕𝒙 𝒅𝒙
𝒇 𝒙 = −∞ +∞ 𝑭(𝒕) 𝒆 +𝟐𝝅𝒊𝒕𝒙 𝒅𝒕
Take 𝒇(𝒙) = 𝜹(𝒙):
𝑭 𝒕 = −∞ +∞ 𝜹 𝒙 𝒆 −𝟐𝝅𝒊𝒕𝒙 𝒅𝒙= 𝒆 𝟎 =𝟏
𝜹 𝒙 = −∞ +∞ 𝟏× 𝒆 +𝟐𝝅𝒊𝒕𝒙 𝒅𝒕
And therefore:
= −∞ +∞ 𝒆 +𝟐𝝅𝒊𝒕𝒙 𝒅𝒕
𝟐𝝅𝒕=𝒑
= 𝟏 𝟐𝝅 −∞ +∞ 𝒆 +𝒊𝒑𝒙 𝒅𝒑
𝜹 𝒂𝒙 = 𝟏 𝒂 𝜹 𝒙
𝛿 𝑥 = lim 𝜎→0 1 𝜎 𝜋 𝑒 − 𝑥 2 / 𝜎 2
𝛿 𝑥 = 1 𝜋 lim 𝜎→0 𝜎 𝑥 2 + 𝜎 2
𝛿 𝑥 = 1 𝜋 lim 𝑁→∞ sin 𝑁𝑥 𝑥
𝜹(𝒙) examples:

41. Fermi’s ‘golden rule’  classical
‘Standard’ non-relativistic QM:
complete set of states n for non-perturbed system
for perturbed system express  using n:
solve iteratively for coefficients a(t) i.e. solve af (t) while assuming all
coefficients ak, except the ith, to be 0:
hence (f i):
transition amplitude Tfi (f i):
and for a time independent perturbation (f i):
with:

42. Fermi’s ‘golden rule’  classical
Does |Tfi|2 represent the probability of an i f transition?
no! Better ansatz: use |Tfi|2/T as i f transition probability per unit time:
note:
T-divergence
no surprise:
V(x) lasts forever!
in real life:  you control specific state with Ei  nature picks any state with Ef  sum over states with Ef
 introduce: # states with energy E:
Fermi’s
golden rule:
𝑾 𝒇𝒊 =𝟐𝝅 𝒅𝑬 𝝆 𝑬 𝑽 𝒇𝒊 𝟐 𝜹 𝑬− 𝑬 𝒊 =𝟐𝝅 𝑽 𝒇𝒊 𝟐 𝝆 𝑬 𝒊

43. Relativistic expressions A+B C+D
Relativistic free particle states: plane waves
transition amplitude also becomes simple since the physics part only involves the 4-momenta of in- & out-going particle pA, pB, pC & pD. All hidden in M for now
gives for the transition probability:
 transition amplitude scaled per unit time & unit volume:
note:
T,V-divergence
no surprise:
plane waves:
 last forever &
 are everywhere!
4-momentum conservation!

44. Relativistic expressions A+B C+D
N-particle phase-space:
We will later show the particle density of to be:
To calculate we put everything in a cube with periodic boundary conditions : L L L = V L
particles/box 2E
2EV 1 L
L/ v
Incidently: With # particles  E, you nicely maintain consistent picture when boosting between frames!
E  E

45. Relativistic expressions A+B C+D
Periodic boundary conditions:
and similarly for y & z
𝒆 𝒊 𝒑 𝒙 𝒙 = 𝒆 𝒊 𝒑 𝒙 (𝒙+𝑳)
N-particle phase-space: L
 spacing allowed momentum states is 2/L for px, py & pz
 density of states:
𝒅 𝟑 𝒑 𝟐𝝅 𝑳 𝟑 = 𝑽 𝒅 𝟑 𝒑 𝟐𝝅 𝟑

46. Relativistic expressions A+B C+D
N-particle phase-space:
We will later show the particle density of to be:
To calculate we put everything in a cube with periodic boundary conditions : L L L = V L
particles/box 2E
2EV 1
1 particle/box:
density of states in momentum space (periodic boundaray conditions):
2E particle/box:

47. Relativistic expressions A+B C+D
Flux factor: v
beam
target
2EA/V
2EB/V
remark on ‘volumes’:
picture suggests different V for beam & target.
not needed, but if you do:
cancels against the N-dependent V terms in Wfi

48. Relativistic expressions A+B C+D
Transition amplitude scaled by TV:
N-particle phase-space:
Flux factor:

49. Relativistic expressions A+B C+D
Invariant form for A+B flux factor
Lab. frame C D 1 2
p1=(E, p) & p2=(m2,0)
4-momenta:
𝒗 = 𝒑 𝑬
c.m. frame D C 1 2
p1=(E1,+p) & p2=(E2,p)
4-momenta:

50. Relativistic expressions A+B C+D
Explicitly Lorentz invariant forms
Lorentz invariant (later)
matrix element
Lorentz invariant
(E,p) conservation
Lorentz invariant
flux
phase space
Lorentz invariant
𝒅𝑬𝜹 𝒑 𝟐 − 𝒎 𝟐 = 𝒅𝑬𝜹 𝑬 𝟐 − 𝒑 𝟐 − 𝒎 𝟐 = 𝟏 𝟐 𝑬= 𝒑 𝟐 + 𝒎 𝟐
because:

51. Relativistic expressions A+B C+D
Useful expressions for 2-particle phase-space
Simplify 2-particle phase from 6 2 variables using the -function
1. Integrate over the 3-momentum of p4:
E4 not an independent variable! &
2. With spherical coordinates d3p3=|p3|2d dp3 for p3:
the dp3 integration is less trivial than it appears, but can be done using the -function paying attention to the p3 dependence hidden in its argument

52. Continue integration c.m. frame 1 2 4 3 3. In c.m. frame: remark: 𝑝≡ 𝒑
𝑝≡ 𝒑
3. In c.m. frame:
𝒑 𝟑 =( 𝑬 𝟑 ,+ 𝒑 )
𝑬 𝟑 𝒅 𝑬 𝟑 =𝒑𝒅𝒑
𝑬 𝟒 𝒅 𝑬 𝟒 =𝒑𝒅𝒑
𝒑 𝟒 =( 𝑬 𝟒 ,− 𝒑 )
define E=E1+E2 & E’=E3+E4 to find:
to perform the integration over dp3=dp:
remark:
you can also do 𝒅 𝒑 𝟑 integration
using: 𝒅𝒙𝜹 𝒇 𝒙 = 𝟏 𝒇′ 𝒙= 𝒙 𝟎

53. Generic expressions for A+B C+D & A C+D
scattering A B D C
A + B  C + D cross section
decay D C A
A  C + D decay
these expressions we will use repeatedly throughout this class

54. Example: A+B C+D scattering
c.m. frame A B D C
phase space: 2 = =
flux:
cross section:
question: units of 𝑴 𝟐 ?

55. Example: A C+D decay =2s decay C A D phase space: 2 = = flux:
decay width:
question: units of 𝑴 𝟐 ?

56. Guestimate  0  + decay width0
m=140 MeV
two identical particles: reduction 1/2! = ½
decay width:
need to guess matrix element: u  
 10-17 s, PDG: = s

58. Feynman calculus in a toy model
Griffiths: sections

59. Feynman ‘rules’  ABC theory
Soon we will discover Feynman rules for real physics (electromagnetism, …)
Now I give rules for a toy model to illustrate how you calculate the matric element
You get –iM with following rules:
Label all external lines: p1, p2, … Label all internal lines: q1, q2, …
Each vertex gets: ig
Each vertex gets: (2)4( ki)
Each propagator gets: i/(q2m2)
Integrate the qi:  d4qi
Drop overall (2)4  ( pi=0) p1 p2 p3
ig B C A
only one vertex: ‘ABC’ p3 p4 p1 p2 q
ig B A C
(2)4 1

60. Lifetime of A ig  ½ mA B p2 p1 A p3 C 8 8
Very simple matrix element: p1 p2 p3
ig B C A 8
Decay width using the master expression: 8
And the lifetime of particle A:
And if you like the momentum of particles B & C in c.m. frame:
 ½ mA
(mB, mC << mA)

61. A+A  B+B scattering (A) (B) (A): (B): (A+B): p1 p2 p4 p3 q A B C p1

62. A+A  B+B scattering c.m. frame p3 p1 A p2 p4 B
For simplicity, assume: mA=mB=m and mC=0:
Gives for matrix element:
And for cross section:
identical particles in final state, factor ½

63. A+B  A+B scattering (B) (A) (A): (B): (A+B): p1 p2 p4 p3 q A B C p4

64. A+B  A+B scattering c.m. frame B A p1 p2 p3 p4
For simplicity, assume again: mA=mB=m and mC=0:
Gives for matrix element:
And for cross section:

65. Lecture 3
Study Thomson Chapter 3
Griffiths (ABC theory)