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Physics Formulary
By ir
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C
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Wevers
Contents
Contents
I
Physical Constants
1
1 Mechanics
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1
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1 Definitions
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1
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1
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1
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1
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1 Force, (angular)momentum and energy
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3
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1
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3 Gravitation
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3
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1
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5 The virial theorem
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4 Point dynamics in a moving coordinate system
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4
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1
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2 Tensor notation
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5 Dynamics of masspoint collections
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5
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1
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2 Collisions
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6 Dynamics of rigid bodies
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6
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1
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2 Principal axes
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6
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1
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7
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1
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2 Hamilton mechanics
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7
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1
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4 Phase space, Liouville’s equation
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7
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2
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6
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7
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8
2 Electricity & Magnetism
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2
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2
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2
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2
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2
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1 Electromagnetic waves in vacuum
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2 Electromagnetic waves in matter
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6 Multipoles
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7 Electric currents
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8 Depolarizing field
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9 Mixtures of materials
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J
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A
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1 Special relativity
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1
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3
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2 Red and blue shift
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1
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2 General relativity
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2
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3
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2 The line element
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2
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3
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4 The trajectory of a photon
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2
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3
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6 Cosmology
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13
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17
4 Oscillations
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4
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4
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4
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4
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6 Pendulums
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1 The wave equation
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2 Solutions of the wave equation
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2
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5
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2 Spherical waves
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2
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5
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4 The general solution in one dimension
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20
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23
6 Optics
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6
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6
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1 Lenses
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2
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6
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3 Principal planes
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2
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6
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1 Degrees of freedom
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2 The energy distribution function
7
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J
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A
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6
III
Interaction between molecules
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1 Mathematical introduction
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2 Definitions
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3 Thermal heat capacity
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4 The laws of thermodynamics
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5 State functions and Maxwell relations
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32
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33
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38
9 Transport phenomena
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9
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1 Flow boundary layers
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39
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44
10 Quantum physics
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10
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1 Black body radiation
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1
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10
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3 Electron diffraction
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2 Wave functions
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3 Operators in quantum physics
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4 The uncertainty principle
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5 The Schr¨ dinger equation
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6 Parity
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7 The tunnel effect
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8 The harmonic oscillator
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9 Angular momentum
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10 Spin
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11 The Dirac formalism
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12 Atomic physics
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12
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10
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2 Eigenvalue equations
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12
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10
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4 Selection rules
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13 Interaction with electromagnetic fields
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14 Perturbation theory
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J
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A
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15 N-particle systems
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1 General
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15
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16 Quantum statistics
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51
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11 Plasma physics
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11
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1 General
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3
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11
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3 The induced dipole interaction
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3
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11
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5 Scattering of light
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4 Thermodynamic equilibrium and reversibility
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11
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1 Types of collisions
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1 Crystal structure
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2 Crystal binding
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3 Crystal vibrations
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3
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12
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2 A lattice with two types of atoms
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3
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12
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4 Thermal heat capacity
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4 Magnetic field in the solid state
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4
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12
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2 Paramagnetism
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1 Thermal heat capacity
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3 The Hall-effect
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1 Description
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3 Flux quantisation in a superconducting ring
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4 Macroscopic quantum interference
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6 The BCS model
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J
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A
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1 Introduction
...
1
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13
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2 The Cayley table
...
1
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13
...
4 Isomorfism and homomorfism; representations
...
1
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13
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13
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1 Schur’s lemma
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2
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13
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3 Character
...
3 The relation with quantum mechanics
...
3
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13
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2 Breaking of degeneracy by a perturbation
...
3
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13
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4 The direct product of representations
...
3
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13
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6 Symmetric transformations of operators, irreducible tensor operators
...
3
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13
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13
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1 The 3-dimensional translation group
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4
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13
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3 Properties of continuous groups
...
5 The group SO(3)
...
6 Applications to quantum mechanics
...
6
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13
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2 Irreducible tensor operators, matrixelements and selection rules
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7 Applications to particle physics
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71
71
71
71
71
72
72
72
72
72
72
73
73
73
73
74
74
74
75
75
75
75
76
77
77
77
78
79
14 Nuclear physics
14
...
14
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14
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14
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14
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1 Kinetic model
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4
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14
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3 Conservation of energy and momentum in nuclear reactions
14
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81
81
82
82
83
83
83
84
84
15 Quantum field theory & Particle physics
15
...
15
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15
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15
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15
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1
15
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1
15
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15
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15
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15
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15
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15
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15
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15
...
1 The electroweak theory
...
13
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J
...
A
...
13
...
15
...
15
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16 Astrophysics
16
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16
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16
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4 Composition and evolution of stars
16
...
The
-operator
The SI units
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94
95
95
96
96
96
97
97
98
99
100
Physical Constants
Name
Number π
Number e
Symbol
π
e
Euler’s constant
Value
3
...
71828182845904523536
γ = lim
n
n→∞
1/k − ln(n)
Unit
= 0
...
60217733 · 10−19
6
...
99792458 · 108
8
...
9876 · 109
C
m3 kg−1 s−2
Planck’s constant
Dirac’s constant
Bohr magneton
Bohr radius
Rydberg’s constant
Electron Compton wavelength
Proton Compton wavelength
Reduced mass of the H-atom
h
h
¯ = h/2π
µB = e¯ /2me
h
a0
Ry
λCe = h/me c
λCp = h/mp c
µH
6
...
0545727 · 10−34
9
...
52918
13
...
2463 · 10−12
1
...
1045755 · 10−31
Js
Js
Am2
˚
A
eV
m
m
kg
Stefan-Boltzmann’s constant
Wien’s constant
Molar gasconstant
Avogadro’s constant
Boltzmann’s constant
σ
kW
R
NA
k = R/NA
5
...
8978 · 10−3
8
...
0221367 · 1023
1
...
1093897 · 10−31
1
...
674954 · 10−27
1
...
0508 · 10−27
kg
kg
kg
kg
J/T
Diameter of the Sun
Mass of the Sun
Rotational period of the Sun
Radius of Earth
Mass of Earth
Rotational period of Earth
Earth orbital period
Astronomical unit
Light year
Parsec
Hubble constant
D
M
T
RA
MA
TA
Tropical year
AU
lj
pc
H
1392 · 106
1
...
38
6
...
976 · 1024
23
...
24219879
1
...
4605 · 1015
3
...
1 Point-kinetics in a fixed coordinate system
1
...
1 Definitions
The position r, the velocity v and the acceleration a are defined by: r = (x, y, z), v = (x, y, z), a = (¨, y , z )
...
2
For the unit vectors in a direction ⊥ to the orbit et and parallel to it en holds:
dr ˙
v
v
e˙t
=
et = en ; en =
|v|
ds
ρ
|e˙t |
et =
For the curvature k and the radius of curvature ρ holds:
k=
d2 r
1
dϕ
det
= 2 =
; ρ=
ds
ds
ds
|k|
1
...
2 Polar coordinates
Polar coordinates are defined by: x = r cos(θ), y = r sin(θ)
...
˙
r
˙˙
1
...
r
...
a point Q holds: rD = rQ +
ω × vQ
˙
with QD = rD − rQ and ω = θ
...
means that the quantity is defined in a moving system of coordinates
...
3 Point-dynamics in a fixed coordinate system
1
...
1 Force, (angular)momentum and energy
Newton’s 2nd law connects the force on an object and the resulting acceleration of the object where the momentum is given by p = mv:
F (r, v, t) =
d(mv )
dv
dm m=const
dp
=
=m
+v
= ma
dt
dt
dt
dt
2
Chapter 1: Mechanics
3
Newton’s 3rd law is given by: Faction = −Freaction
...
For the total energy W , the kinetic energy T and the potential energy
˙
˙
U holds: W = T + U ; T = −U with T = 1 mv 2
...
The following equation is valid:
τ =−
∂U
∂θ
Fi = 0 and
Hence, the conditions for a mechanical equilibrium are:
τi = 0
...
1
...
2 Conservative force fields
A conservative force can be written as the gradient of a potential: Fcons = − U
...
For such a force field also holds:
r1
F · ds
F · ds = 0 ⇒ U = U0 −
r0
So the work delivered by a conservative force field depends not on the trajectory covered but only on the
starting and ending points of the motion
...
3
...
From Gauss law it then follows:
2
V = 4πG
...
3
...
Physics Formulary by ir
...
C
...
Wevers
4
Kepler’s orbital equations
In a force field F = kr−2 , the orbits are conic sections with the origin of the force in one of the foci (Kepler’s
1st law)
...
Half the length of the short
axis is b = a
...
Orbits with an equal ε are of equal shape
...
k < 0 and ε = 0: a circle
...
k < 0 and 0 < ε < 1: an ellipse
...
k < 0 and ε = 1: a parabole
...
k < 0 and ε > 1: a hyperbole, curved towards the centre of force
...
k > 0 and ε > 1: a hyperbole, curved away from the centre of force
...
If the surface between the orbit covered between t1 and t2 and the focus C around which the planet moves is
A(t1 , t2 ), Kepler’s 2nd law is
LC
(t2 − t1 )
A(t1 , t2 ) =
2m
Kepler’s 3rd law is, with T the period and Mtot the total mass of the system:
4π 2
T2
=
3
a
GMtot
1
...
5 The virial theorem
The virial theorem for one particle is:
mv · r = 0 ⇒ T = − 1 F · r =
2
1
2
r
dU
dr
= 1 n U if U = −
2
k
rn
The virial theorem for a collection of particles is:
Fi · ri +
T = −1
2
Fij · rij
pairs
particles
These propositions can also be written as: 2Ekin + Epot = 0
...
4 Point dynamics in a moving coordinate system
1
...
1 Apparent forces
The total force in a moving coordinate system can be found by subtracting the apparent forces from the forces
working in the reference frame: F = F − Fapp
...
Transformation of the origin: For = −maa
2
...
Coriolis force: Fcor = −2mω × v
4
...
4
...
dt dt
1
...
5
...
r
...
the centre of mass R is given by v − R
...
5
...
The changes in the relative velocities can be derived from: S = ∆p =
2
S =constant and L w
...
t
...
µ(vaft − vbefore )
...
J
...
A
...
6 Dynamics of rigid bodies
1
...
1 Moment of Inertia
The angular momentum in a moving coordinate system is given by:
L = Iω + Ln
where I is the moment of inertia with respect to a central axis, which is given by:
mi ri
I=
2
; T = Wrot = 1 ωIij ei ej = 1 Iω 2
2
2
i
or, in the continuous case:
I=
m
V
2
r n dV =
2
r n dm
Further holds:
Li = I ij ωj ; Iii = Ii ; Iij = Iji = −
mk xi xj
k
Steiner’s theorem is: Iw
...
t
...
r
...
C + m(DM )2 if axis C axis D
...
o
...
I=
1
2
12 ml
Bar, axis ⊥ through end
I = 1 ml2
3
Rectangle, axis ⊥ plane thr
...
o
...
I=
1
2
12 m(a
+ b2 )
Rectangle, axis
b thr
...
6
...
For a principal axis holds:
∂I
∂I
∂I
=
=
= 0 so Ln = 0
∂ωx
∂ωy
∂ωz
The following holds: ωk = −aijk ωi ωj with aijk =
˙
Ii − Ij
if I1 ≤ I2 ≤ I3
...
6
...
1
...
7
...
1
...
2 Hamilton mechanics
The Lagrangian is given by: L =
T (qi ) − V (qi )
...
2
qi pi − L
...
With new coordinates
2
√
(θ, I), obtained by the canonical transformation x = 2I/mω cos(θ) and p = − 2Imω sin(θ), with inverse
θ = arctan(−p/mωx) and I = p2 /2mω + 1 mωx2 it follows: H(θ, I) = ωI
...
This is elegant from a relativistic point of view: this is equivalent to the
transformation of the momentum 4-vector pα → pα − qAα
...
1
...
3 Motion around an equilibrium, linearization
For natural systems around equilibrium the following equations are valid:
∂V
∂qi
= 0 ; V (q) = V (0) + Vik qi qk with Vik =
0
∂2V
∂qi ∂qk
0
With T = 1 (Mik qi qk ) one receives the set of equations M q + V q = 0
...
This leads to the eigenfrequencies of the problem:
aT V ak
2
k
2
...
The general solution is a superposition if
ωk = T
ak M ak
eigenvibrations
...
7
...
J
...
A
...
7
...
If pi qi − H = Pi Qi − K(Pi , Qi , t) −
pi =
dF1 (qi , Qi , t)
, the coordinates follow from:
dt
∂F1
∂F1
dF1
; Pi =
; K=H+
∂qi
∂Qi
dt
dF2 (qi , Pi , t)
˙
, the coordinates follow from:
˙
2
...
If −pi qi − H = Pi Qi − K(Pi , Qi , t) +
˙
qi = −
dF3 (pi , Qi , t)
, the coordinates follow from:
dt
∂F3
∂F3
∂F3
; Pi = −
; K =H+
∂pi
∂Qi
∂t
4
...
Chapter 2
Electricity & Magnetism
2
...
Those can be written both as
differential and integral equations:
(D · n )d2 A = Qfree,included
· D = ρfree
(B · n )d2 A = 0
·B =0
E · ds = −
dΦ
dt
×E =−
H · ds = Ifree,included +
For the fluxes holds: Ψ =
dΨ
dt
(D · n )d2 A, Φ =
∂B
∂t
× H = Jfree +
∂D
∂t
(B · n )d2 A
...
2 Force and potential
The force and the electric field between 2 point charges are given by:
F12 =
Q1 Q2
F
er ; E =
4πε0 εr r2
Q
The Lorentzforce is the force which is felt by a charged particle that moves through a magnetic field
...
The magnetic field in point P which results from an electric current is given by the law of Biot-Savart, also
known als the law of Laplace
...
2
E · ds and A = 1 B × r
...
J
...
A
...
The fields can be derived from the potentials as
follows:
∂A
, B = ×A
E=− V −
∂t
Further holds the relation: c2 B = v × E
...
3 Gauge transformations
The potentials of the electromagnetic fields transform as follows when a gauge transformation is applied:
A =A− f
∂f
V =V +
∂t
so the fields E and B do not change
...
Further,
the freedom remains to apply a limiting condition
...
This separates the differential equations for A and V : 2V = − ,
1
...
· A = 0
...
2
...
4 Energy of the electromagnetic field
The energy density of the electromagnetic field is:
dW
= w = HdB + EdD
dVol
The energy density can be expressed in the potentials and currents as follows:
wmag =
1
2
J · A d3 x , wel =
ρV d3 x
1
2
2
...
5
...
The irradiance is the
time-averaged of the Poynting vector: I = |S | t
...
Chapter 2: Electricity & Magnetism
11
2
...
2 Electromagnetic waves in matter
The wave equations in matter, with cmat = (εµ)−1/2 the lightspeed in matter, are:
2
− εµ
∂2
µ ∂
−
∂t2
ρ ∂t
E =0,
2
− εµ
∂2
µ ∂
−
∂t2
ρ ∂t
B=0
give, after substitution of monochromatic plane waves: E = E exp(i(k ·r −ωt)) and B = B exp(i(k ·r −ωt))
the dispersion relation:
iµω
k 2 = εµω 2 +
ρ
The first term arises from the displacement current, the second from the conductance current
...
If the material is a good conductor,
µω
...
6 Multipoles
Because
1
1
=
|r − r |
r
∞
0
l
r
r
Pl (cos θ) the potential can be written as: V =
Q
4πε
n
kn
rn
For the lowest-order terms this results in:
• Monopole: l = 0, k0 =
• Dipole: l = 1, k1 =
ρdV
r cos(θ)ρdV
• Quadrupole: l = 2, k2 =
1
2
i
2
2
(3zi − ri )
1
...
Q
3p · r
− p
...
The magnetic dipole: dipole moment: if r
A: µ = I × (Ae⊥ ), F = (µ · )Bout
2
mv⊥
, W = −µ × Bout
|µ| =
2B
−µ 3µ · r
Magnetic field: B =
− µ
...
7 Electric currents
The continuity equation for charge is:
∂ρ
+
∂t
I=
· J = 0
...
)Eext , and
Physics Formulary by ir
...
C
...
Wevers
12
dΦ
...
If a conductor encloses a flux Φ holds: Φ = LI
...
The energy contained within a coil is given by W = 1 LI 2 and L = µN 2 A/l
...
For a capacitor holds: C = ε0 εr A/d where d is the distance between
the plates and A the surface of one plate
...
The accumulated energy is given by W = 1 CV 2
...
capacity is given by I = −C
dt
For most PTC resistors holds approximately: R = R0 (1 + αT ), where R0 = ρl/A
...
If a current flows through two different, connecting conductors x and y, the contact area will heat up or cool
down, depending on the direction of the current: the Peltier effect
...
This effect can be amplified with semiconductors
...
For a Cu-Konstantane connection holds:
γ ≈ 0
...
7 mV/K
...
In = 0,
2
...
If the medium has an ellipsoidal
shape and one of the principal axes is parallel with the external field E0 or B0 then the depolarizing is field
homogeneous
...
For a few
limiting cases of an ellipsoid holds: a thin plane: N = 1, a long, thin bar: N = 0, a sphere: N = 1
...
9 Mixtures of materials
The average electric displacement in a material which is inhomogenious on a mesoscopic scale is given by:
−1
φ2 (1 − x)
where x = ε1 /ε2
...
Further holds:
i
φi
εi
−1
≤ ε∗ ≤
φi εi
i
Chapter 3
Relativity
3
...
1
...
The general form of the Lorentz
transformation is given by:
x =x+
(γ − 1)(x · v )v
x·v
− γvt , t = γ t − 2
|v|2
c
where
γ=
1
2
1 − v2
c
The velocity difference v between two observers transforms according to:
v =
γ 1−
v1 · v2
c2
−1
v2 + (γ − 1)
v1 · v2
2 v1 − γv1
v1
If the velocity is parallel to the x-axis, this becomes y = y, z = z and:
x = γ(x − vt) , x = γ(x + vt )
xv
xv
t =γ t− 2 , t=γ t + 2
c
c
, v =
v2 − v1
v1 v2
1− 2
c
If v = vex holds:
px = γ px −
βW
c
, W = γ(W − vpx )
With β = v/c the electric field of a moving charge is given by:
E=
(1 − β 2 )er
Q
4πε0 r2 (1 − β 2 sin2 (θ))3/2
The electromagnetic field transforms according to:
E = γ(E + v × B ) , B = γ
B−
v×E
c2
Length, mass and time transform according to: ∆tr = γ∆t0 , mr = γm0 , lr = l0 /γ, with 0 the quantities
in a co-moving reference frame and r the quantities in a frame moving with velocity v w
...
t
...
The proper
time τ is defined as: dτ 2 = ds2 /c2 , so ∆τ = ∆t/γ
...
J
...
A
...
p = mr v = γm0 v = W v/c2 , and pc = W β where β = v/c
...
4-vectors have the property that their modulus is independent of the observer: their components can change
after a coordinate transformation but not their modulus
...
The relation with the “common” velocity
a 4-vector
...
For particles with nonzero restmass holds: U α Uα = −c2 , for particles
with zero restmass (so with v = c) holds: U α Uα = 0
...
So: pα pα = −m2 c2 = p2 − W 2 /c2
...
1
...
f
c
This can give both red- and blueshift, also ⊥ to the direction of motion
...
Motion: with ev · er = cos(ϕ) follows:
2
...
f
rc
3
...
g
...
λ1
R1
3
...
3 The stress-energy tensor and the field tensor
The stress-energy tensor is given by:
Tµν = ( c2 + p)uµ uν + pgµν +
The conservation laws can than be written as:
νT
Fαβ =
µν
1
α
Fµα Fν + 1 gµν F αβ Fαβ
4
c2
= 0
...
The Maxwell equations can than be written as:
∂ν F µν = µ0 J µ , ∂λ Fµν + ∂µ Fνλ + ∂ν Fλµ = 0
The equations of motion for a charged particle in an EM field become with the field tensor:
dpα
= qFαβ uβ
dτ
3
...
2
...
The geodesic postulate: free falling particles move along geodesics of space-time with the proper time
τ or arc length s as parameter
...
From δ ds = 0 the equations of motion can be derived:
dxβ dxγ
d2 xα
=0
+ Γα
βγ
ds2
ds ds
Chapter 3: Relativity
15
2
...
3
...
µ
The Riemann tensor is defined as: Rναβ T ν := α β T µ −
by j ai = ∂j ai + Γi ak and j ai = ∂j ai − Γk ak
...
For a second-order tensor holds: [ α , β ]Tν = Rσαβ Tν + Rναβ Tσ , k ai =
j
j il
i
l
i
i l
l
l
ij
ij
i lj
∂k aj − Γkj al + Γkl aj , k aij = ∂k aij − Γki alj − Γkj ajl and k a = ∂k a + Γkl a + Γkl a
...
σµ βν
σν βµ
βν
βµ
µ
The Ricci tensor is a contraction of the Riemann tensor: Rαβ := Rαµβ , which is symmetric: Rαβ = Rβα
...
α
The Einstein tensor is given by: Gαβ := Rαβ − 1 g αβ R, where R := Rα is the Ricci scalar, for which
2
holds: β Gαβ = 0
...
The equation Rαβµν = 0 has as only solution a flat space
...
From this, the Laplace
1
...
8πκ
Tαβ
c2
where Λ is the cosmological constant
...
The most general form of the field equations is: Rαβ − 1 gαβ R + Λgαβ =
2
3
...
2 The line element
The metric tensor in an Euclidean space is given by: gij =
k
∂ xk ∂ xk
¯ ¯
...
In special relativity this becomes ds2 = −c2 dt2 + dx2 + dy 2 + dz 2
...
2
µ
ν
The external Schwarzschild metric applies in vacuum outside a spherical mass distribution, and is given by:
ds2 =
−1 +
2m
r
c2 dt2 + 1 −
2m
r
−1
dr2 + r2 dΩ2
Here, m := M κ/c2 is the geometrical mass of an object with mass M , and dΩ2 = dθ2 + sin2 θdϕ2
...
If an object is smaller than its event horizon 2m, that implies that
its escape velocity is > c, it is called a black hole
...
In general relativity, the components of gµν are
associated with the potentials and the derivatives of gµν with the field strength
...
They are defined by:
Physics Formulary by ir
...
C
...
Wevers
16
• r > 2m:
u
v
• r < 2m:
u
v
=
r
r
− 1 exp
cosh
2m
4m
t
4m
=
r
r
− 1 exp
sinh
2m
4m
t
4m
=
1−
r
r
exp
sinh
2m
4m
t
4m
=
1−
r
r
exp
cosh
2m
4m
t
4m
• r = 2m: here, the Kruskal coordinates are singular, which is necessary to eliminate the coordinate
singularity there
...
The
Kruskal coordinates are only singular on the hyperbole v 2 − u2 = 1, this corresponds with r = 0
...
For the metric outside a rotating, charged spherical mass the Newman metric applies:
ds2
=
1−
2mr − e2
r2 + a2 cos2 θ
r 2 + a2 +
c2 dt2 −
r2 + a2 cos2 θ
r2 − 2mr + a2 − e2
(2mr − e2 )a2 sin2 θ
r2 + a2 cos2 θ
sin2 θdϕ2 +
dr2 − (r2 + a2 cos2 θ)dθ2 −
2a(2mr − e2 )
r2 + a2 cos2 θ
sin2 θ(dϕ)(cdt)
where m = κM/c2 , a = L/M c and e = κQ/ε0 c2
...
Near rotating black holes frame dragging occurs because gtϕ = 0
...
3
...
3 Planetary orbits and the perihelion shift
To find a planetary orbit, the variational problem δ ds = 0 has to be solved
...
Substituting the external Schwarzschild metric yields for a planetary orbit:
du
dϕ
d2 u
+u
dϕ2
=
m
du
3mu + 2
dϕ
h
˙
where u := 1/r and h = r2 ϕ =constant
...
This term can
h2
κM
1+ 2
...
In zeroth order, this results in an elliptical orbit: u0 (ϕ) = A + B cos(ϕ) with A = m/h2 and B an
arbitrary constant
...
The perihelion of a planet is the point for which r is minimal, or u maximal
...
For the perihelion shift then follows: ∆ϕ = 2πε =
6πm2 /h2 per orbit
...
2
...
Substituting the
external Schwarzschild metric results in the following orbital equation:
du
dϕ
d2 u
+ u − 3mu
dϕ2
=0
3
...
5 Gravitational waves
Starting with the approximation gµν = ηµν + hµν for weak gravitational fields and the definition hµν =
hµν − 1 ηµν hα it follows that 2hµν = 0 if the gauge condition ∂hµν /∂xν = 0 is satisfied
...
3
3
...
6 Cosmology
If for the universe as a whole is assumed:
1
...
The 3-dimensional spaces are isotrope for a certain value of x0 ,
3
...
then the Robertson-Walker metric can be derived for the line element:
ds2 = −c2 dt2 +
2
r0
R2 (t)
kr2
1− 2
4r0
(dr2 + r2 dΩ2 )
For the scalefactor R(t) the following equations can be derived:
˙
˙
¨
R2 + kc2
Λ
8πκp
8πκ
2R R2 + kc2
+
+
= − 2 + Λ and
=
2
2
R
R
c
R
3
3
where p is the pressure and
parameter q:
the density of the universe
...
This is a measure of the velocity with which galaxies far away are
moving away from each other, and has the value ≈ (75 ± 25) km·s−1 ·Mpc−1
...
Parabolical universe: k = 0, W = 0, q = 1
...
2
The hereto related critical density is c = 3H 2 /8πκ
...
Hyperbolical universe: k = −1, W < 0, q <
positive forever
...
The expansion velocity of the universe remains
3
...
The expansion velocity of the universe becomes negative
2
after some time: the universe starts collapsing
...
1 Harmonic oscillations
ˆ
ˆ
The general form of a harmonic oscillation is: Ψ(t) = Ψei(ωt±ϕ) ≡ Ψ cos(ωt ± ϕ),
ˆ
where Ψ is the amplitude
...
iω
dtn
4
...
x
˙
2
With complex amplitudes, this becomes −mω 2 x = F − Cx − ikωx
...
The quantity Z = F/x is called the impedance of the system
...
is given by Q =
k
The frequency with minimal |Z| is called velocity resonance frequency
...
In the resonance
√
curve |Z|/ Cm is plotted against ω/ω0
...
In these points holds: R = X and δ = ±Q−1 , and the width is 2∆ωB = ω0 /Q
...
The amplitude resonance frequency ωA is the frequency
where iωZ is minimal
...
2
The damping frequency ωD is a measure for the time in which an oscillating system comes to rest
...
A weak damped oscillation (k 2 < 4mC) dies out after TD = 2π/ωD
...
A strong damped oscillation (k 2 > 4mC) drops like (if
4mC) x(t) ≈ x0 exp(−t/τ )
...
3 Electric oscillations
The impedance is given by: Z = R + iX
...
The impedance of a
resistor is R, of a capacitor 1/iωC and of a self inductor iωL
...
The total
impedance in case several elements are positioned is given by:
18
Chapter 4: Oscillations
19
1
...
parallel connection: V = IZ,
1
=
Ztot
Here, Z0 =
i
1
1
,
=
Zi Ltot
i
1
, Ctot =
Li
Ci , Q =
i
R
R
, Z=
Z0
1 + iQδ
1
L
...
4
...
g
...
For them holds: Z0 =
The transmission velocity is given by v =
dL dx
...
dL dC
4
...
For the coefficients of mutual induction
Mij holds:
N 1 Φ1
N 2 Φ2
=
∼ N1 N2
M12 = M21 := M = k L1 L2 =
I2
I1
where 0 ≤ k ≤ 1 is the coupling factor
...
At full load holds:
I2
iωM
V1
=
=−
≈−
V2
I1
iωL2 + Rload
N1
L1
=−
L2
N2
4
...
I/τ with τ the moment of force and I the moment of inertia
...
2lm
the constant of torsion and I the moment of inertia
...
1 The wave equation
The general form of the wave equation is: 2u = 0, or:
2
u−
1 ∂2u
∂2u ∂2u ∂2u
1 ∂2u
=
+ 2 + 2 − 2 2 =0
v 2 ∂t2
∂x2
∂y
∂z
v ∂t
where u is the disturbance and v the propagation velocity
...
By definition holds:
kλ = 2π and ω = 2πf
...
Longitudinal waves: for these holds k
2
...
v ⊥ u
...
The group velocity is given by:
dω
dvph
k dn
= vph + k
= vph 1 −
dk
dk
n dk
vg =
where n is the refractive index of the medium
...
In a dispersive
medium it is possible that vg > vph or vg < vph , and vg · vf = c2
...
g
...
This velocity is often almost equal to the group velocity
...
• For pressure waves in a gas also holds: v =
γp/ =
γRT /M
...
If h
• Surface waves on a liquid: v =
5
...
2
...
Chapter 5: Waves
21
The equation for a harmonic traveling plane wave is: u(x, t) = u cos(k · x ± ωt + ϕ)
ˆ
If waves reflect at the end of a spring this will result in a change in phase
...
A lose end gives no change in the phase of the
reflected wave, with boundary condition (∂u/∂x)l = 0
...
r
...
the wave with a velocity vobs , he will observe a change in frequency: the
vf − vobs
f
=
...
This is given by:
f0
vf
5
...
2 Spherical waves
When the situation is spherical symmetric, the homogeneous wave equation is given by:
1 ∂ 2 (ru) ∂ 2 (ru)
−
=0
v 2 ∂t2
∂r2
with general solution:
u(r, t) = C1
f (r − vt)
g(r + vt)
+ C2
r
r
5
...
3 Cylindrical waves
When the situation has a cylindrical symmetry, the homogeneous wave equation becomes:
1 ∂2u 1 ∂
−
v 2 ∂t2
r ∂r
r
∂u
∂r
=0
This is a Bessel equation, with solutions which can be written as Hankel functions
...
2
...
where bm ∈ I Substituting u(x, t) = Aei(kx−ωt) gives two solutions ωj = ωj (k) as dispersion relations
...
5
...
If ωj (k) ∈ I the stationary
R
phase method can be applied
...
The only areas contributing significantly to the integral are areas
d
(kx − ω(k)t) = 0
...
J
...
A
...
4 Green functions for the initial-value problem
This method is preferable if the solutions deviate much from the stationary solutions, like point-like excitations
...
They are defined by:
Q(x, x , t) =
1
2 [δ(x
P (x, x , t) =
− x − vt) + δ(x − x + vt)]
1
2v
0
Further holds the relation: Q(x, x , t) =
if |x − x | < vt
if |x − x | > vt
∂P (x, x , t)
∂t
5
...
If n is a unit
vector ⊥ the surface, pointed from 1 to 2, and K is a surface current density, than holds:
n · (D2 − D1 ) = σ
n · (B2 − B1 ) = 0
n × (E2 − E1 ) = 0
n × (H2 − H1 ) = K
In a waveguide holds because of the cylindrical symmetry: E(x, t) = E(x, y)ei(kz−ωt) and B(x, t) =
B(x, y)ei(kz−ωt)
...
Bz ≡ 0: the Transversal Magnetic modes (TM)
...
2
...
Boundary condition:
∂Bz
∂n
= 0
...
Bz with boundary conditions:
∂2
∂2
+ 2
∂x2
∂y
ψ = −γ 2 ψ with eigenvalues γ 2 := εµω 2 − k 2
This gives a discrete solution ψ with eigenvalue γ 2 : k = εµω 2 − γ 2
...
Therefore, ω is called the cut-off frequency
...
Ez and Bz are zero everywhere: the Transversal electromagnetic mode (TEM)
...
Further k ∈ I so there exists no cut-off
frequency
...
5
...
Substitution of x ∼ eiωt gives: ω =
2 1−
1 2
2 ε )
...
1
2 ω0 (iε
±
While x is growing, the 2nd term becomes larger
−1
and diminishes the growth
...
If x is expanded as x = x(0) +
(1)
2 (2)
εx + ε x + · · · and this is substituted one obtains, besides periodic, secular terms ∼ εt
...
Energy is conserved if the left-hand side is 0
...
This mechanism limits the growth
of oscillations
...
For this equation, soliton solutions
of the following form exist:
−d
u(x − ct) =
cosh2 (e(x − ct))
with c = 1 + 1 ad and e2 = ad/(12b2 )
...
1 The bending of light
For the refraction at a surface holds: ni sin(θi ) = nt sin(θt ) where n is the refractive index of the material
...
The refraction of light in a
material is caused by scattering from atoms
...
From this follows
j
that vg = c/(1 + (ne e2 /2ε0 mω 2 ))
...
More
n
ak
...
The path, followed by a light ray in material can be found from Fermat’s principle:
2
2
dt = δ
δ
1
2
n(s)
ds = 0 ⇒ δ
c
1
n(s)ds = 0
1
6
...
2
...
For the refraction at a spherical surface with radius R holds:
n2
n1 − n2
n1
−
=
v
b
R
where |v| is the distance of the object and |b| the distance of the image
...
For a double concave lens holds R1 < 0, R2 > 0, for a double convex lens holds R1 > 0 and
R2 < 0
...
For a lens with thickness d and diameter D holds to a good
approximation: 1/f = 8(n − 1)d/D2
...
2
...
Spherical
aberration can be reduced by not using spherical mirrors
...
The used signs are:
Quantity
R
f
v
b
+
Concave mirror
Concave mirror
Real object
Real image
−
Convex mirror
Convex mirror
Virtual object
Virtual image
6
...
3 Principal planes
The nodal points N of a lens are defined by the figure on the right
...
The plane ⊥ the optical axis through the principal points
is called the principal plane
...
2
...
Further holds: N · Nα = 1
...
The f-number
is defined by f /Dobjective
...
J
...
A
...
3 Matrix methods
A light ray can be described by a vector (nα, y) with α the angle with the optical axis and y the distance to
the optical axis
...
M is a product of elementary matrices
...
Transfer along length l: MR =
1 0
l/n 1
2
...
4 Aberrations
Lenses usually do not give a perfect image
...
Chromatic aberration is caused by the fact that n = n(λ)
...
Using N lenses makes it possible to
obtain the same f for N wavelengths
...
Spherical aberration is caused by second-order effects which are usually ignored; a spherical surface
does not make a perfect lens
...
3
...
Further
away of the optical axis they are curved
...
4
...
5
...
6
...
This can be corrected with a combination of
positive and negative lenses
...
5 Reflection and transmission
If an electromagnetic wave hits a transparent medium part of the wave will reflect at the same angle as the
incident angle, and a part will be refracted at an angle according to Snell’s law
...
r
...
the surface
...
Then the Fresnel equations are:
r =
t =
tan(θi − θt )
sin(θt − θi )
, r⊥ =
tan(θi + θt )
sin(θt + θi )
2 sin(θt ) cos(θi )
2 sin(θt ) cos(θi )
, t⊥ =
sin(θt + θi ) cos(θt − θi )
sin(θt + θi )
The following holds: t⊥ − r⊥ = 1 and t + r = 1
...
A special case is r = 0
...
From Snell’s law it then follows: tan(θi ) = n
...
The situation with r⊥ = 0 is not possible
...
6 Polarization
The polarization is defined as: P =
Ip
Imax − Imin
=
Ip + Iu
Imax + Imin
where the intensity of the polarized light is given by Ip and the intensity of the unpolarized light is given by
Iu
...
If polarized
light passes through a polarizer Malus law applies: I(θ) = I(0) cos2 (θ) where θ is the angle of the polarizer
...
The first is independent of the polarization, the second and third are linear polarizers with the
transmission axes horizontal and at +45◦ , while the fourth is a circular polarizer which is opaque for L-states
...
The state of a polarized light ray can also be described by the Jones vector:
E=
E0x eiϕx
E0y eiϕy
For the √
horizontal P -state holds: E = (1, 0), for the vertical P -state E = (0, 1), the R-state is given by
√
E = 1 2(1, −i) and the L-state by E = 1 2(1, i)
...
For some types of optical equipment the Jones matrix M
is given by:
Horizontal linear polarizer:
1 0
0 0
Vertical linear polarizer:
0 0
0 1
Linear polarizer at +45◦
Lineair polarizer at −45◦
1 1
1 1
1
2
1 −1
−1 1
1
2
1
0
0
−i
1
4 -λ
plate, fast axis vertical
1
4 -λ
plate, fast axis horizontal
eiπ/4
Homogene circular polarizor right
1
2
1 i
−i 1
Homogene circular polarizer left
1
2
1
i
eiπ/4
1 0
0 i
−i
1
6
...
r
...
the incident direction, where α is the apex angle, θi is the angle between the incident
angle and a line perpendicular to the surface and θi is the angle between the ray leaving the prism and a line
perpendicular to the surface
...
For the refractive
index of the prism now holds:
sin( 1 (δmin + α))
2
n=
sin( 1 α)
2
Physics Formulary by ir
...
C
...
Wevers
28
The dispersion of a prism is defined by:
dδ dn
dδ
=
dλ
dn dλ
where the first factor depends on the shape and the second on the composition of the prism
...
The refractive
index in this area can usually be approximated by Cauchy’s formula
...
8 Diffraction
Fraunhofer diffraction occurs far away from the source(s)
...
N is the number of slits, b the width of a slit and d the distance
between the slits
...
The diffraction through a spherical aperture with radius a is described by:
I(θ)
=
I0
J1 (ka sin(θ))
ka sin(θ)
2
The diffraction pattern of a rectangular aperture at distance R with length a in the x-direction and b in the
y-direction is described by:
2
2
sin(α )
sin(β )
I(x, y)
=
I0
α
β
where α = kax/2R and β = kby/2R
...
Close at the source the Fraunhofermodel is invalid because it ignores the angle-dependence of the reflected
waves
...
r
...
the optical axis
...
This is the minimum angle ∆θmin between two incident rays
coming from points far away for which their refraction patterns can be detected separately
...
22λ/D where D is the diameter of the slit
...
The minimum difference between two wavelengths that gives a separated diffraction pattern
in a multiple slit geometry is given by ∆λ/λ = nN where N is the number of lines and n the order of the
pattern
...
9 Special optical effects
• Birefringe and dichroism
...
There are at least 3 directions, the principal axes, in which they are parallel
...
In case n2 = n3 = n1 ,
which happens e
...
at trigonal, hexagonal and tetragonal crystals there is one optical axis in the direction
of n1
...
The extraordinary wave is linear polarized
Chapter 6: Optics
29
in the plane through the transmission direction and the optical axis
...
Double images occur when the
incident ray makes an angle with the optical axis: the extraordinary wave will refract, the ordinary will
not
...
Incident light will have a phase shift of ∆ϕ = 2πd(|n0 −
ne |)/λ0 if an uniaxial crystal is cut in such a way that the optical axis is parallel with the front and back
plane
...
For a quarter-wave plate holds: ∆ϕ = π/2
...
In that case, the optical axis is parallel to E
...
If the electrodes have an
effective length and are separated by a distance d, the retardation is given by: ∆ϕ = 2πK V 2 /d2 ,
where V is the applied voltage
...
These crystals are also piezoelectric: their polarization
changes when a pressure is applied and vice versa: P = pd + ε0 χE
...
0
• The Faraday effect: the polarization of light passing through material with length d and to which a
magnetic field is applied in the propagation direction is rotated by an angle β = VBd where V is the
Verdet constant
...
The radiation is emitted within
a cone with an apex angle α with sin(α) = c/cmedium = c/nvq
...
10 The Fabry-Perot interferometer
For a Fabry-Perot interferometer holds in
general: T + R + A = 1 where T is the
transmission factor, R the reflection factor
and A the absorption factor
...
q
'E
d
Screen
Focussing lens
√
√
The width of the peaks at half height is given by γ = 4/ F
...
The
2
maximum resolution is then given by ∆fmin = c/2ndF
...
1 Degrees of freedom
A molecule consisting of n atoms has s = 3n degrees of freedom
...
A
linear molecule has 2 rotational degrees of freedom and a non-linear molecule 3
...
So,
for linear molecules this results in a total of s = 6n − 5
...
The
average energy of a molecule in thermodynamic equilibrium is Etot = 1 skT
...
The rotational and vibrational energy of a molecule are:
Wrot =
h
¯2
l(l + 1) = Bl(l + 1) , Wvib = (v + 1 )¯ ω0
2 h
2I
The vibrational levels are excited if kT ≈ ¯ ω, the rotational levels of a hetronuclear molecule are excited if
h
kT ≈ 2B
...
7
...
The average velocity is given by v =
√
2α/ π, and v 2 = 3 α2
...
Even s: s = 2l: c(s) =
1
(l − 1)!
2
...
3 Pressure on a wall
The number of molecules that collides with a wall with surface A within a time τ is given by:
∞ π 2π
3
d N=
nAvτ cos(θ)P (v, θ, ϕ)dvdθdϕ
0
0
0
From this follows for the particle flux on the wall: Φ = 1 n v
...
4 The equation of state
If intermolecular forces and the volume of the molecules can be neglected then for gases from p =
and E = 3 kT can be derived:
2
1
pV = ns RT = N m v 2
3
2
3n
E
Here, ns is the number of moles particles and N is the total number of particles within volume V
...
In the Van der Waals equation this corresponds
with the critical temperature, pressure and volume of the gas
...
From dp/dV = 0 and d2 p/dV 2 = 0 follows:
Tcr =
8a
a
, pcr =
, Vcr = 3bns
27bR
27b2
For the critical point holds: pcr Vm,cr /RTcr =
general gas law
...
A virial
expansion is used for even more accurate views:
p(T, Vm ) = RT
1
B(T ) C(T )
+
+
+ ···
2
3
Vm
Vm
Vm
The Boyle temperature TB is the temperature for which the 2nd virial coefficient is 0
...
The inversion temperature Ti = 2TB
...
J
...
A
...
5 Collisions between molecules
The collision probability of a particle in a gas that is translated over a distance dx is given by nσdx, where σ is
v1
2
2
with u = v1 + v2 the relative velocity between
the cross section
...
This means
...
If m1
v1
m2
nσ
nσ 2
1
...
The average distance between two molecules is 0
...
Collisions between molecules and small particles in a solution result in the Brownian motion
...
i
3
A gas is called a Knudsen gas if
the dimensions of the gas, something that can easily occur at low
pressures
...
Together with the general gas law follows: p1 / T1 = p2 / T2
...
d
The velocity profile between the plates is in that case given by w(z) = zwx /d
...
3
If two plates move along each other at a distance d with velocity wx the viscosity η is given by: Fx = η
T2 − T1
dQ
= κA
, which results in a temperdt
d
ature profile T (z) = T1 + z(T2 − T1 )/d
...
Also holds: κ = CV η
...
The heat conductance in a non-moving gas is described by:
7
...
If the distance between two
molecules approaches the molecular diameter D a repulsing force between the electron clouds appears
...
This results in the
Lennard-Jones potential for intermolecular forces:
ULJ = 4
D
r
12
−
D
r
6
with a minimum at r = rm
...
89rm
...
275NAD3 , b = 1
...
2 and Vm,kr = 3
...
A more simple model for intermolecular forces assumes a potential U (r) = ∞ for r < D, U (r) = ULJ for
D ≤ r ≤ 3D and U (r) = 0 for r ≥ 3D
...
D
with F (r) the spatial distribution function in spherical coordinates, which for a homogeneous distribution is
given by: F (r)dr = 4nπr2 dr
...
1 Mathematical introduction
If there exists a relation f (x, y, z) = 0 between 3 variables, one can write: x = x(y, z), y = y(x, z) and
z = z(x, y)
...
A homogeneous function of degree m obeys: εm F (x, y, z) = F (εx, εy, εz)
...
2 Definitions
• The isochoric pressure coefficient: βV =
• The isothermal compressibility: κT = −
• The isobaric volume coefficient: γp =
1
p
∂p
∂T
V
1
V
∂V
∂p
T
1
V
∂V
∂T
p
1
V
∂V
∂p
S
• The adiabatic compressibility: κS = −
For an ideal gas follows: γp = 1/T , κT = 1/p and βV = −1/V
...
3 Thermal heat capacity
• The specific heat at constant X is: CX = T
• The specific heat at constant pressure: Cp =
• The specific heat at constant volume: CV =
∂S
∂T
∂H
∂T
∂U
∂T
33
X
p
V
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...
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...
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34
For an ideal gas holds: Cmp − CmV = R
...
Hence Cp = 1 (s + 2)R
...
For a lower T one needs only to consider the thermalized degrees of freedom
...
2
In general holds:
Cp − CV = T
∂p
∂T
·
V
∂V
∂T
∂V
∂T
= −T
p
2
p
∂p
∂V
≥0
T
Because (∂p/∂V )T is always < 0, the following is always valid: Cp ≥ CV
...
8
...
The first law is the conservation of
energy
...
In differential form this becomes: d Q = dU + d W , where d means
that the it is not a differential of a quantity of state
...
So for a
reversible process holds: d Q = dU + pdV
...
One can extract an amount
of work Wt from the system or add Wt = −Wi to the system
...
So, the entropy difference after a
reversible process is:
2
d Qrev
T
S2 − S1 =
1
So, for a reversible cycle holds:
d Qrev
= 0
...
T
The third law of thermodynamics is (Nernst):
lim
T →0
∂S
∂X
=0
T
From this it can be concluded that the thermal heat capacity → 0 if T → 0, so absolute zero temperature
cannot be reached by cooling through a finite number of steps
...
5 State functions and Maxwell relations
The quantities of state and their differentials are:
Internal energy:
Enthalpy:
Free energy:
Gibbs free enthalpy:
U
H = U + pV
F = U − TS
G = H − TS
dU = T dS − pdV
dH = T dS + V dp
dF = −SdT − pdV
dG = −SdT + V dp
Chapter 8: Thermodynamics
35
From this one can derive Maxwell’s relations:
∂T
∂V
∂p
∂S
=−
S
∂T
∂p
,
V
∂V
∂S
=
S
∂p
∂T
,
p
∂S
∂V
=
V
,
T
∂V
∂T
=−
p
∂S
∂p
T
From the total differential and the definitions of CV and Cp it can be derived that:
T dS = CV dT + T
∂p
∂T
dV and T dS = Cp dT − T
V
∂V
∂T
dp
p
For an ideal gas also holds:
Sm = CV ln
T
T0
+ R ln
V
V0
+ Sm0 and Sm = Cp ln
T
T0
− R ln
p
p0
+ Sm0
Helmholtz’ equations are:
∂U
∂V
=T
T
∂p
∂T
∂H
∂p
−p ,
V
=V −T
T
∂V
∂T
p
for an enlarged surface holds: d Wrev = −γdA, with γ the surface tension
...
6 Processes
The efficiency η of a process is given by: η =
Work done
Heat added
The Cold factor ξ of a cooling down process is given by: ξ =
Cold delivered
Work added
Reversible adiabatic processes
For adiabatic processes holds: W = U1 − U2
...
Also holds: T V γ−1 =constant and T γ p1−γ =constant
...
Isobaric processes
Here holds: H2 − H1 =
2
1
Cp dT
...
The throttle process
This is also called the Joule-Kelvin effect and is an adiabatic expansion of a gas through a porous material or a
small opening
...
In general this is accompanied with a change in
temperature
...
If T > Ti the gas heats up, if T < Ti the gas cools down
...
The throttle process is e
...
applied in refridgerators
...
Isothermic expansion at T1
...
2
...
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...
C
...
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36
3
...
4
...
The efficiency for Carnot’s process is:
η =1−
T2
|Q2 |
=1−
:= ηC
|Q1 |
T1
The Carnot efficiency ηC is the maximal efficiency at which a heat machine can operate
...
The efficiency in the ideal case is the same as for
Carnot’s cycle
...
7 Maximal work
Consider a system that changes from state 1 into state 2, with the temperature and pressure of the surroundings
given by T0 and p0
...
Closed system: Wmax = (U1 − U2 ) − T0 (S1 − S2 ) + p0 (V1 − V2 )
...
Open system: Wmax = (H1 − H2 ) − T0 (S1 − S2 ) − ∆Ekin − ∆Epot
...
8
...
When the phases are indicated by α, β and γ holds:
Gα = Gβ and
m
m
rβα
α
β
∆Sm = Sm − Sm =
T0
where rβα is the transition heat of phase β to phase α and T0 is the transition temperature
...
Further
Sm =
∂Gm
∂T
p
so G has a twist in the transition point
...
For those there will occur only a discontinuity in the second
derivates of Gm
...
A phase-change of the 3rd order, so with e
...
[∂ 3 Gm /∂T 3 ]p non continuous arises e
...
when ferromagnetic
iron changes to the paramagnetic state
...
9 Thermodynamic potential
When the number of particles within a system changes this number becomes a third quantity of state
...
If a system exists of more
components this becomes:
µi dni
dG = −SdT + V dp +
i
where µ =
∂G
∂ni
is called the thermodynamic potential
...
For V holds:
p,T,nj
c
∂V
∂ni
ni
V =
i=1
c
:=
nj ,p,T
ni Vi
i=1
where Vi is the partial volume of component i
...
The molar volume of a mixture of two components
can be a concave line in a V -x2 diagram: the mixing contracts the volume
...
It can be derived that
ni dµi = −SdT + V dp, this gives at constant p and T :
xi dµi = 0 (Gibbs-Duhmen)
...
The number of free parameters in a system with c
components and p different phases is given by f = c + 2 − p
...
10 Ideal mixtures
For a mixture of n components holds (the index 0 is the value for the pure component):
ni Ui0 , Hmixture =
Umixture =
i
ni Hi0 , Smixture = n
i
where for ideal gases holds: ∆Smix = −nR
0
xi Si + ∆Smix
i
xi ln(xi )
...
A mixture of two liquids is rarely ideal:
i
i
this is usually only the case for chemically related components or isotopes
...
Here is xi the fraction of the ith
i
component in liquid phase and yi the fraction of the ith component in gas phase
...
For x2
∆Tk =
2
RTk
RT 2
x2 , ∆Ts = − s x2
rβα
rγβ
with rβα the evaporation heat and rγβ < 0 the melting heat
...
8
...
In equilibrium for each
component holds: µα = µβ = µγ
...
J
...
A
...
12 Statistical basis for thermodynamics
The number of possibilities P to distribute N particles on n possible energy levels, each with a g-fold degeneracy is called the thermodynamic probability and is given by:
P = N!
i
n
gi i
ni !
The most probable distribution, that with the maximum value for P , is the equilibrium state
...
The occupation numbers in equilibrium are then given by:
ni =
N
Wi
gi exp −
Z
kT
gi exp(−Wi /kT )
...
For a system in thermodynamic equilibrium this becomes:
S=
U
+ kN ln
T
Z
N
+ kN ≈
U
+ k ln
T
For an ideal gas, with U = 3 kT then holds: S = 5 kN + kN ln
2
2
ZN
N!
V (2πmkT )3/2
N h3
8
...
To do this the term d W = pdV has
to be replaced with the correct work term, like d Wrev = −F dl for the stretching of a wire, d Wrev = −γdA
for the expansion of a soap bubble or d Wrev = −BdM for a magnetic system
...
It has an entropy S = Akc3 /4¯ κ
h
with A the area of its event horizon
...
Hawkings
area theorem states that dA/dt ≥ 0
...
Chapter 9
Transport phenomena
9
...
Xd3 V =
∂
∂t
Xd3 V +
where the volume V is surrounded by surface A
...
Some important integral theorems are:
(v · n )d2 A =
Gauss:
(divv )d3 V
Stokes for a scalar field:
(φ · et )ds =
(n × gradφ)d2 A
Stokes for a vector field:
(v · et )ds =
(rotv · n )d2 A
This results in:
(rotv · n )d2 A = 0
Ostrogradsky:
(n × v )d2 A =
(φn )d2 A =
Here, the orientable surface
d2 A is limited by the Jordan curve
(rotv )d3 A
(gradφ)d3 V
ds
...
2 Conservation laws
On a volume work two types of forces:
1
...
For gravity holds: f0 = g
...
Surface forces working only on the margins: t
...
39
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...
C
...
Wevers
40
T can be split in a part pI representing the normal tensions and a part T representing the shear stresses:
T = T + pI, where I is the unit tensor
...
When the flow velocity is v at position r holds on position r + dr:
v(dr ) =
v(r )
dr · (gradv )
+
translation
rotation, deformation, dilatation
The quantity L:=gradv can be split in a symmetric part D and an antisymmetric part W
...
ω represents the local rotation
2
velocity: dr · W = 1 ω × dr
...
Here, η is the dynamical viscosity
...
From equating the thermodynamical and
mechanical pressure it follows: 3η + 2η = 0
...
The conservation laws for mass, momentum and energy for continuous media can be written in both integral
and differential form
...
Conservation of mass:
∂
∂t
d3 V +
2
...
Conservation of energy:
∂
∂t
∂
∂t
(v · n )d2 A = 0
v(v · n )d2 A =
vd3 V +
( 1 v 2 + e) d3 V +
2
−
(q · n )d2 A +
f0 d3 V +
n · T d2 A
( 1 v 2 + e) (v · n )d2 A =
2
(v · f0 )d3 V +
(v · n T)d2 A
Differential notation:
1
...
Conservation of momentum:
3
...
q = −κ T is
the heat flow
...
Chapter 9: Transport phenomena
41
From this one can derive the Navier-Stokes equations for an incompressible, viscous and heat-conducting
medium:
divv
C
=
∂v
+ (v ·
∂t
)v
=
∂T
+ C(v ·
∂t
)T
=
0
g − gradp + η
κ
2
2
v
T + 2ηD : D
with C the thermal heat capacity
...
If a surface A surrounds the object outside the
boundary layer holds:
F =−
[pn + v(v · n )]d2 A
9
...
If also holds rotv = 0 and
2
the entropy is equal on each streamline holds 1 v 2 + gh + dp/ =constant everywhere
...
For ideal gases with constant Cp and CV holds,
2
with γ = Cp /CV :
γ p
c2
1 2
= 1 v2 +
= constant
2v + γ − 1
2
γ −1
With a velocity potential defined by v = gradφ holds for instationary flows:
∂φ 1 2
+ 2 v + gh +
∂t
dp
= constant everywhere
9
...
One can also deduce functional equalities without solving the differential equations
...
α follows from the equation for heat transport κ∂y T = α∆T and a = κ/ c is the thermal diffusion coefficient
...
J
...
A
...
For this angle holds Ma=
1/ arctan(θ)
...
Now, the dimensionless Navier-Stokes equation becomes, with x = x/L, v = v/V , grad = Lgrad,
L2 2 and t = tω:
2
g
v
∂v
+
+ (v · )v = −grad p +
Sr
∂t
Fr
Re
2
=
9
...
For an incompressible laminar flow through a straight, circular tube holds for the
velocity profile:
1 dp 2
(R − r2 )
v(r) = −
4η dx
R
v(r)2πrdr = −
For the volume flow holds: ΦV =
π dp 4
R
8η dx
0
The entrance length Le is given by:
1
...
056ReD
2
...
For the total force on a sphere with radius R in a
flow then holds: F = 6πηRv
...
2
For gas transport at low pressures (Knudsen-gas) holds: ΦV =
9
...
In the incompressible case follows
from conservation of mass 2 φ = 0
...
In general holds:
∂2ψ
∂2ψ
+
= −ωz
2
∂x
∂y 2
In polar coordinates holds:
∂φ
1 ∂φ
1 ∂ψ
∂ψ
=
, vθ = −
=
r ∂θ
∂r
∂r
r ∂θ
Q
ln(r) so that vr = Q/2πr, vθ = 0
...
For a vortex holds: φ = Γθ/2π
...
The statement that Fx = 0 is d’Alembert’s
paradox and originates from the neglection of viscous effects
...
Henxe rotating bodies also create a force perpendicular to their direction of motion: the
Magnus effect
...
7 Boundary layers
9
...
1 Flow boundary layers
√
If for the thickness of the boundary layer holds: δ
L holds: δ ≈ L/ Re
...
Blasius’ equation for the boundary layer is,
with vy /v∞ = f (y/δ): 2f + f f = 0 with boundary conditions f (0) = f (0) = 0, f (∞) = 1
...
664 Rex
...
7
...
This is equivalent with
y=0
12ηv∞
dp
=
...
If Pr ≤ 1: δ/δT ≈ √
Pr
...
If Pr
1: δ/δT ≈ 3 Pr
...
8 Heat conductance
For non-stationairy heat conductance in one dimension without flow holds:
κ ∂2T
∂T
=
+Φ
∂t
c ∂x2
where Φ is a source term
...
J
...
A
...
At x = πD the temperature variation is in anti-phase with the surface
...
The flow density J = −D n
...
9 Turbulence
√
The time scale of turbulent velocity variations τt is of the order of: τt = τ Re/Ma2 with τ the molecular
time scale
...
The Navier-Stokes
equation now becomes:
∂ v
+( v ·
∂t
) v =−
p
+ν
2
v +
divSR
where SR ij = − vi vj is the turbulent stress tensor
...
It is
stated that, analogous to Newtonian media: SR = 2 νt D
...
9
...
So if ν = 0, ω is conserved
...
In three-dimensional flows the situation is just the opposite
...
1 Introduction to quantum physics
10
...
1 Black body radiation
Planck’s law for the energy distribution for the radiation of a black body is:
w(f ) =
8πhc
1
1
8πhf 3
, w(λ) =
3
5 ehc/λkT − 1
hf /kT − 1
c
λ
e
Stefan-Boltzmann’s law for the total power density can be derived from this: P = AσT 4
...
10
...
2 The Compton effect
For the wavelength of scattered light, if light is considered to exist of particles, can be derived:
λ =λ+
h
(1 − cos θ) = λ + λC (1 − cos θ)
mc
10
...
3 Electron diffraction
Diffraction of electrons at a crystal can be explained by assuming that particles have a wave character with
wavelength λ = h/p
...
10
...
This wavefunction can be described in
normal or momentum space
...
The wavefunction can be interpreted as a measure for the probability P to find a particle somewhere (Born):
dP = |ψ|2 d3 V
...
The normalizing condition for wavefunctions follows from this:
Φ|Φ = Ψ|Ψ = 1
...
3 Operators in quantum physics
In quantum mechanics, classical quantities are translated into operators
...
J
...
A
...
If this basis is taken orthonormal, then follows for the coefficients:
a basis of eigenfunctions: Ψ =
n
cn = un |Ψ
...
The matrix element Aij is given by: Aij = ui |A|uj
...
ui |A |un un |B|uj holds:
n
n
The time-dependence of an operator is given by (Heisenberg):
∂A [A, H]
dA
=
+
dt
∂t
i¯
h
with [A, B] ≡ AB − BA the commutator of A and B
...
If [A, B] = 0, the operators A and B have a common set of eigenfunctions
...
The first order approximation F (x)
t
≈ F ( x ), with F = −dU/dx represents the classical equation
...
2
10
...
2h
2h
2h
10
...
The position operator is: xop = i¯ p
...
The Hamiltonian of a particle with mass m, potential energy U and total
o
energy E is given by: H = p2 /2m + U
...
In one dimension it is:
ψ(x, t) =
+
dE c(E)uE (x) exp −
iEt
h
¯
h
¯
(ψ ∗ ψ − ψ ψ ∗ )
2im
∂P (x, t)
= − J(x, t)
The following conservation law holds:
∂t
The current density J is given by: J =
10
...
If the wavefunction is split in even and
odd functions, it can be expanded into eigenfunctions of P:
ψ(x) = 1 (ψ(x) + ψ(−x)) + 1 (ψ(x) − ψ(−x))
2
2
even:
ψ+
odd:
ψ−
[P, H] = 0
...
Hence, parity is a conserved quantity
...
7 The tunnel effect
The wavefunction of a particle in an ∞ high potential step from x = 0 to x = a is given by ψ(x) =
a−1/2 sin(kx)
...
If the wavefunction with energy W meets a potential well of W0 > W the wavefunction will, unlike the
classical case, be non-zero within the potential well
...
Using the boundary conditions requiring continuity: ψ =
h
continuous and ∂ψ/∂x =continuous at x = 0 and x = a gives B, C and D and A expressed in A
...
If W > W0 and 2a = nλ = 2πn/k
holds: T = 1
...
8 The harmonic oscillator
2
For a harmonic oscillator holds: U = 1 bx2 and ω0 = b/m
...
[A, A† ] = h and [A, H] = hωA
...
HAuE = (E − ¯ ω)AuE
...
The energy in this ground state is 1 ¯ ω: the zero point energy
...
2
10
...
However, cyclically holds:
h
[Lx , Ly ] = i¯ Lz
...
For Lz
holds:
∂
∂
∂
= −i¯ x
h
−y
h
Lz = −i¯
∂ϕ
∂y
∂x
h
The ladder operators L± are defined by: L± = Lx ± iLy
...
Further,
z
¯
L± = he±iϕ ±
∂
∂
+ i cot(θ)
∂θ
∂ϕ
h
h
From [L+ , Lz ] = −¯ L+ follows: Lz (L+ Ylm ) = (m + 1)¯ (L+ Ylm )
...
h
From [L2 , L± ] = 0 follows: L2 (L± Ylm ) = l(l + 1)¯ 2 (L± Ylm )
...
Further follows that l has to be integral or half-integral
...
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...
C
...
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48
10
...
Because the spin operators
h
do not act in the physical space (x, y, z) the uniqueness of the wavefunction is not a criterium here: also half
odd-integer values are allowed for the spin
...
The spin operators are given by S = 1 ¯ σ, with
2h
σx =
0 1
1 0
, σy =
0 −i
i 0
1 0
0 −1
, σz =
The eigenstates of Sz are called spinors: χ = α+ χ+ + α− χ− , where χ+ = (1, 0) represents the state with
spin up (Sz = 1 ¯ ) and χ− = (0, 1) represents the state with spin down (Sz = − 1 ¯ )
...
Of
course holds |α+ |2 + |α− |2 = 1
...
In the presence of an external magnetic field this gives
a potential energy U = −M · B
...
If B = Bez there are two eigenvalues for this problem: χ± for E = ±egS ¯ B/4m =
h
h
±¯ ω
...
From this can be derived: Sx = 1 ¯ cos(2ωt)
h
2
and Sy = 1 ¯ sin(2ωt)
...
This causes the normal
2h
Zeeman splitting of spectral lines
...
10
...
From this it can be derived that the γ are
hermitian 4 × 4 matrices given by:
γk =
0
iσk
−iσk
0
, γ4 =
I
0
With this, the Dirac equation becomes:
γλ
m 2 c2
∂
+ 02
∂xλ
h
¯
where ψ(x) = (ψ1 (x), ψ2 (x), ψ3 (x), ψ4 (x)) is a spinor
...
12 Atomic physics
10
...
1 Solutions
The solutions of the Schr¨ dinger equation in spherical coordinates if the potential energy is a function of r
o
alone can be written as: ψ(r, θ, ϕ) = Rnl (r)Yl,ml (θ, ϕ)χms , with
Clm
Ylm = √ Plm (cos θ)eimϕ
2π
For an atom or ion with one electron holds: Rlm (ρ) = Clm e−ρ/2 ρl L2l+1 (ρ)
n−l−1
with ρ = 2rZ/na0 with a0 = ε0 h2 /πme e2
...
The functions are 2
(−1)m n! −x −m dn−m −x n
e x
(e x )
(n − m)!
dxn−m
n−1
(2l + 1) = 2n2 -folded degenerated
...
12
...
12
...
The total magnetic dipole moment of an electron is then
M = ML + MS = −(e/2me )(L + gS S) where gS = 2
...
Further holds: J 2 = L2 + S 2 + 2L · S = L2 + S 2 + 2Lz Sz + L+ S− + L− S+
...
, 0,
...
If the interaction
2
energy between S and L is small it can be stated that: E = En + ESL = En + aS · L
...
With gS = 2 follows for the average magnetic moment:
h
e
Mav = −(e/2me )g¯ J, where g is the Land´ -factor:
g =1+
j(j + 1) + s(s + 1) − l(l + 1)
S·J
=1+
J2
2j(j + 1)
For atoms with more than one electron the following limiting situations occur:
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...
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...
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50
1
...
J ∈ {|L − S|,
...
, J − 1, J}
...
2S + 1 is the multiplicity of a multiplet
...
j − j coupling: for larger atoms the electrostatic interaction is smaller than the Li · si interaction of
an electron
...
jn , J, mJ where only the ji of the not completely filled
subshells are to be taken into account
...
For a transition between two singlet states the line splits in 3 parts, for ∆mJ = −1, 0 + 1
...
At higher S the line splits up in more parts: the anomalous Zeeman effect
...
10
...
4 Selection rules
For the dipole transition matrix elements follows: p0 ∼ | l2 m2 |E · r |l1 m1 |
...
For an atom where L − S coupling is dominant further holds: ∆S = 0 (but not strict), ∆L = 0, ±1, ∆J =
0, ±1 except for J = 0 → J = 0 transitions, ∆mJ = 0, ±1, but ∆mJ = 0 is forbidden if ∆J = 0
...
For the total atom holds: ∆J = 0, ±1 but no
J = 0 → J = 0 transitions and ∆mJ = 0, ±1, but ∆mJ = 0 is forbidden if ∆J = 0
...
13 Interaction with electromagnetic fields
The Hamiltonian of an electron in an electromagnetic field is given by:
H=
h
¯2
1
(p + eA)2 − eV = −
2µ
2µ
2
+
e2 2
e
B·L+
A − eV
2µ
2µ
where µ is the reduced mass of the system
...
For B = Bez it is given by e2 B 2 (x2 + y 2 )/8µ
...
Because f = f (x, t), this
is called a local gauge transformation, in contrast with a global gauge transformation which can always be
applied
...
14 Perturbation theory
10
...
1 Time-independent perturbation theory
To solve the equation (H0 + λH1 )ψn = En ψn one has to find the eigenfunctions of H = H0 + λH1
...
Because
φn is a complete set holds:
cnk (λ)φk
ψn = N (λ) φn +
k=n
(1)
(2)
When cnk and En are being expanded into λ: cnk = λcnk + λ2 cnk + · · ·
(1)
(2)
0
En = En + λEn + λ2 En + · · ·
Chapter 10: Quantum physics
51
(1)
and this is put into the Schr¨ dinger equation the result is: En = φn |H1 |φn and
o
φm |H1 |φn
if m = n
...
So to first order holds: ψn = φn +
φk
...
In that case an orthonormal set eigenfunctions φni
is chosen for each level n, so that φmi |φnj = δmn δij
...
Substitution in the Schr¨ dinger equation and taking dot
(1)
αi φnj |H1 |φni = En αj
...
product with φni gives:
i
i
10
...
2 Time-dependent perturbation theory
From the Schr¨ dinger equation i¯
o
h
∂ψ(t)
= (H0 + λV (t))ψ(t)
∂t
cn (t) exp
and the expansion ψ(t) =
n
0
−iEn t
h
¯
t
follows:
c(1) (t)
n
λ
=
i¯
h
φn |V (t )|φk exp
(1)
φn with cn (t) = δnk + λcn (t) + · · ·
0
0
i(En − Ek )t
h
¯
dt
0
10
...
15
...
For the total wavefunction of a system of identical indistinguishable
particles holds:
1
...
r
...
interchange of
the coordinates (spatial and spin) of each pair of particles
...
2
...
r
...
interchange of the coordinates
(spatial and spin) of each pair of particles
...
When a and b are the
quantum numbers of electron 1 and 2 holds:
ψS (1, 2) = ψa (1)ψb (2) + ψa (2)ψb (1) , ψA (1, 2) = ψa (1)ψb (2) − ψa (2)ψb (1)
Because the particles do not approach each other closely the repulsion energy at ψA in this state is smaller
...
Physics Formulary by ir
...
C
...
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52
For N particles the symmetric spatial function is given by:
ψS (1,
...
N )
1
The antisymmetric wavefunction is given by the determinant ψA (1,
...
15
...
If the 2 atoms approach each other there are √ possibilities:
two
the total wavefunction approaches the bonding function with lower total energy ψB = 1 2(φa + φb ) or
2
√
approaches the anti-bonding function with higher energy ψAB = 1 2(φa − φb )
...
r
...
the connecting axis, like a combination of two s-orbitals it is called a σ-orbital, otherwise a
π-orbital, like the combination of two p-orbitals along two axes
...
ψ|ψ
The energy calculated with this method is always higher than the real energy if ψ is only an approximation for
the solutions of Hψ = Eψ
...
Applying this to the function ψ =
ci φi one finds: (Hij − ESij )ci = 0
...
Here, Hij = φi |H|φj and
Sij = φi |φj
...
Sii = 1 and Sij is
the overlap integral
...
This results in a large electron density between the nuclei and
therefore a repulsion
...
6
...
There are three ways of hybridization in C:
√
1
...
There are 2 hybrid orbitals which are placed on one line
2
under 180◦
...
√
√
√
2
...
The 3 SP2 orbitals lay in one plane, with symmetry axes which are at an angle of
120◦
...
SP3 hybridization: ψsp3 = 1 (ψ2s ± ψ2pz ± ψ2py ± ψ2px )
...
10
...
If the probability that the system is in state ψi is given by ai ,
ri ψi |A|ψi
...
The probability to find
where ρlk = c∗ cl
...
Further holds ρ =
k
eigenvalue an when measuring A is given by ρnn if one uses a basis of eigenvectors of A for {φk }
...
If all quantumstates with the same energy are
equally probable: Pi = P (Ei ), one can obtain the distribution:
Pn (E) = ρnn =
e−En /kT
with the state sum Z =
Z
e−En /kT
n
The thermodynamic quantities are related to these definitions as follows: F = −kT ln(Z), U = H =
∂
ln(Z), S = −k Pn ln(Pn )
...
The distribution function for the internal states for a system in thermal equilibrium is the most probable function
...
For identical,
k
k
indistinguishable particles which obey the Pauli exclusion principle the possible number of states is given by:
P =
k
gk !
nk !(gk − nk )!
This results in the Fermi-Dirac statistics
...
So the distribution functions which explain how particles are
distributed over the different one-particle states k which are each gk -fold degenerate depend on the spin of the
particles
...
Fermi-Dirac statistics: integer spin
...
N
gk
Zg exp((Ek − µ)/kT ) + 1
N
2
...
nk ∈ I , nk =
with ln(Zg ) = −
gk ln[1 − exp((Ei − µ)/kT )]
...
It is found by demanding nk = N ,
and for it holds: lim µ = EF , the Fermi-energy
...
The Maxwell-Boltzmann
T →0
distribution can be derived from this in the limit Ek − µ
nk =
Ek
N
exp −
Z
kT
kT :
gk exp −
with Z =
k
Ek
kT
With the Fermi-energy, the Fermi-Dirac and Bose-Einstein statistics can be written as:
1
...
exp((Ek − EF )/kT ) + 1
2
...
exp((Ek − EF )/kT ) − 1
Chapter 11
Plasma physics
11
...
If a plasma contains also negative charged
ions α is not well defined
...
The collision frequency νc = 1/τc = nσv
...
The rate coefficient
K is defined by K = σv
...
The potential of an electron is given by:
V (r) =
r
−e
exp −
4πε0 r
λD
with λD =
ε0 kTe Ti
≈
e2 (ne Ti + ni Te )
ε0 kTe
ne e 2
because charge is shielded in a plasma
...
For distances < λD the plasma
cannot be assumed to be quasi-neutral
...
2b0 = e2 /(4πε0 1 mv 2 )
...
11
...
Starting with σm = 4πb2 ln(ΛC ) and with 1 mv 2 = kT it can be
0
2
found that:
√
2 2 3
2√
3/2
8 2πε0 m(kT )
4πε m v
=
τm = 4 0
ne ln(ΛC )
ne4 ln(ΛC )
For momentum transfer between electrons and ions holds for a Maxwellian velocity distribution:
√ √
√ √
6π 3ε2 me (kTe )3/2
6π 3ε2 mi (kTi )3/2
0
0
≈ τei , τii =
τee =
ne e4 ln(ΛC )
ni e4 ln(ΛC )
The energy relaxation times for identical particles are equal to the momentum relaxation times
...
Approximately holds: τee : τei :
E
τie : τie = 1 : 1 : mi /me : mi /me
...
For T > 10 eV holds approximately: σeo =
−2/5
, for lower energies this can be a factor 10 lower
...
The equation
of continuity is ∂t n + (nvdiff ) = 0 ⇒ ∂t n = D 2 n
...
A rough estimate gives
3
τD = Lp /D = L2 τc /λ2
...
In electrical
p
v
fields also holds J = neµE = e(ne µe + ni µi )E with µ = e/mνc the mobility of the particles
...
The coefficient of ambipolar diffusion Damb is defined by Γ = Γi = Γe = −Damb ne,i
...
The helical orbit is perturbed by collisions
...
So the electrons are magnetized if
√
me e3 ne ln(ΛC )
ρe
√ 2
=
<1
λee
6π 3ε0 (kTe )3/2 B0
Magnetization of only the electrons is sufficient to confine the plasma reasonable because they are coupled
to the ions by charge neutrality
...
Combined with the
two stationary Maxwell equations for the B-field these form the ideal magneto-hydrodynamic equations
...
If both magnetic and electric fields are present electrons and ions will move in the same direction
...
˙
11
...
3
...
The differential
cross section is then defined as:
I(Ω) =
b
s
d
d
ra
b T
c
ϕ
χ
M
b ∂b
dσ
=
dΩ
sin(χ) ∂χ
For a potential energy W (r) = kr−n follows: I(Ω, v) ∼ v −4/n
...
It arises from the interference of matter waves behind
the object
...
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...
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...
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56
11
...
2 The Coulomb interaction
2
For the Coulomb interaction holds: 2b0 = q1 q2 /2πε0 mv0 , so W (r) = 2b0 /r
...
Because dp = d(mv) =
0
D
mv0 (1 − cos χ) a cross section related to momentum transfer σm is given by:
(1 − cos χ)I(Ω)dΩ = 4πb2 ln
0
σm =
1
= 4πb2 ln
0
sin( 1 χmin )
2
λD
b0
:= 4πb2 ln(ΛC ) ∼
0
ln(v 4 )
v4
where ln(ΛC ) is the Coulomb-logarithm
...
11
...
3 The induced dipole interaction
The induced dipole interaction, with p = αE, gives a potential V and an energy W in a dipole field given by:
V (r) =
with ba =
4
|e|p
αe2
p · er
, W (r) = −
=−
4πε0 r2
8πε0 r2
2(4πε0 )2 r4
2e2 α
holds: χ = π − 2b
2
(4πε0 )2 1 mv0
2
∞
ra
dr
r2
1−
b2
b4
+ a4
2
r
4r
If b ≥ ba the charge would hit the atom
...
If the scattering
angle is a lot times 2π it is called capture
...
11
...
4 The centre of mass system
If collisions of two particles with masses m1 and m2 which scatter in the centre of mass system by an angle χ
are compared with the scattering under an angle θ in the laboratory system holds:
tan(θ) =
m2 sin(χ)
m1 + m2 cos(χ)
The energy loss ∆E of the incoming particle is given by:
∆E
=
E
1
2
2 m2 v2
1
2
2 m1 v1
=
2m1 m2
(1 − cos(χ))
(m1 + m2 )2
11
...
5 Scattering of light
Scattering of light by free electrons is called Thomson scattering
...
The cross section σ = 6
...
If λ
0
0
scattering
...
Chapter 11: Plasma physics
57
11
...
For
the reaction
Xforward →
Xback
←
forward
back
holds in a plasma in equilibrium microscopic reversibility:
ηforward =
ˆ
forward
ηback
ˆ
back
If the velocity distribution is Maxwellian, this gives:
ηx =
ˆ
nx
h3
e−Ekin /kT
gx (2πmx kT )3/2
where g is the statistical weight of the state and n/g := η
...
With this one finds for the Boltzmann balance, Xp + e− → X1 + e− + (E1p ):
←
nB
gp
p
=
exp
n1
g1
Ep − E1
kTe
And for the Saha balance, Xp + e− + (Epi ) → X+ + 2e− :
← 1
nS
n+ ne
h3
p
1
= +
exp
gp
g1 ge (2πme kTe )3/2
Epi
kTe
Because the number of particles on the left-hand side and right-hand side of the equation is different, a factor
g/Ve remains
...
From microscopic reversibility one can derive that for the rate coefficients K(p, q, T ) := σv
K(q, p, T ) =
pq
holds:
∆Epq
gp
K(p, q, T ) exp
gq
kT
11
...
5
...
The energy in the centre of
mass system is available for reactions
...
Excitation: Ap + e− → Aq + e−
←
2
...
J
...
A
...
Ionisation and 3-particles recombination: Ap + e− → A+ + 2e−
←
4
...
Stimulated emission: Aq + hf → Ap + 2hf
6
...
Penning ionisation: b
...
Ne∗ + Ar → Ar+ + Ne + e−
←
8
...
Resonant charge transfer: A+ + A → A + A+
←
11
...
2 Cross sections
Collisions between an electron and an atom can be approximated by a collision between an electron and one
of the electrons of that atom
...
25βE
Ep
A[1 − B ln(E)]2
1 + CE 3
...
6 Radiation
In equilibrium holds for radiation processes:
np Apq + np Bpq ρ(ν, T ) = nq Bqp ρ(ν, T )
emission
stimulated emission
absorption
Here, Apq is the matrix element of the transition p → q, and is given by:
Apq =
8π 2 e2 ν 3 |rpq |2
with rpq = ψp |r |ψq
3¯ ε0 c3
h
For hydrogenic atoms holds: Ap = 1
...
5 , with Ap = 1/τp =
Apq
...
The Einstein coefficients B are given by:
Bpq =
c3 Apq
Bpq
gq
and
=
8πhν 3
Bqp
gp
A spectral line is broadened by several mechanisms:
1
...
The natural life time of a state p is given by τp = 1/
Apq
...
The Doppler broadening is caused by the thermal motion of the particles:
2
∆λ
=
λ
c
2 ln(2)kTi
mi
This broadening results in a Gaussian line profile:
√
kν = k0 exp(−[2 ln 2(ν − ν0 )/∆νD ]2 ), with k the coefficient of absorption or emission
...
The Stark broadening is caused by the electric field of the electrons:
∆λ1/2 =
2/3
ne
C(ne , Te )
˚
with for the H-β line: C(ne , Te ) ≈ 3 · 1014 A−3/2 cm−3
...
The total line shape is a convolution of the Gauss- and Lorentz profile
2
and is called a Voigt profile
...
Then follows for the cross section of absorption processes: σa = Bpq hν/cdν
...
Free-Bound radiation, originating from radiative recombination
...
63 · 10−43 Wm4 K1/2 sr−1 and ξ the Biberman factor
...
Free-free radiation, originating from the acceleration of particles in the EM-field of other particles:
εf f =
C1 zi ni ne
hc
√
exp −
λ2 kTe
λkTe
ξf f (λ, Te )
11
...
This is also true in magnetic fields because ∂ai /∂xi = 0
...
The total density is given by n = F dv and vF dv = nw
...
Mass balance:
(BTE)dv ⇒
2
...
Energy balance:
∂n
+
∂t
· (nw) =
(BTE)mvdv ⇒ mn
(BTE)mv 2 dv ⇒
dw
+
dt
3 dp 5
+ p
2 dt
2
∂n
∂t
cr
T +
·w+
p = mn a + R
·q = Q
Physics Formulary by ir
...
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...
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60
Here, a = e/m(E + w × B ) is the average acceleration, q = 1 nm vt2 vt the heat flow,
2
2
mvt ∂F
dv the source term for energy production, R is a friction term and p = nkT the
Q =
r
∂t cr
pressure
...
11
...
One assumes the Quasi-steady-state solution is
valid, where ∀p>1 [(∂np /∂t = 0) ∧ ( · (np wp ) = 0)]
...
For systems in ESP, where only collisional (de)excitation
between levels p and p ± 1 is taken into account holds x = 6
...
0
1
rp nS +rp nB
p
p
b p nS
...
excit
...
deexcit
...
recomb
...
recomb
rad
...
to
Kpq + ne np
Kpq + np
q
q>p
coll
...
coll
...
Apq + ne np Kp+
q
coll
...
rad
...
from
11
...
For electromagnetic waves with complex wave number k = ω(n + iκ)/c in one dimension one finds:
Ex = E0 e−κωx/c cos[ω(t − nx/c)]
...
The dielectric tensor E, with property:
k · (E · E) = 0
is given by E = I − σ/iε0 ω
...
er is connected with a right rotating field for which
iEx /Ey = 1 and el is connected with a left rotating field for which iEx /Ey = −1
...
From this the following solutions can be obtained:
A
...
1
...
This describes a longitudinal linear polarized wave
...
n2 = L: a left, circular polarized wave
...
n2 = R: a right, circular polarized wave
...
θ = π/2: transmission ⊥ the B-field
...
n2 = P : the ordinary mode: Ex = Ey = 0
...
2
...
Resonance frequencies are frequencies for which n2 → ∞, so vf = 0
...
For R → ∞ this gives the electron cyclotron resonance frequency ω = Ωe , for L → ∞ the ion cyclotron
resonance frequency ω = Ωi and for S = 0 holds for the extraordinary mode:
α2 1 −
m i Ω2
i
me ω 2
=
1−
m 2 Ω2
i
i
m2 ω 2
e
1−
Ω2
i
ω2
Cut-off frequencies are frequencies for which n2 = 0, so vf → ∞
...
In the case that β 2
1 one finds Alfv´ n waves propagating parallel to the field lines
...
Chapter 12
Solid state physics
12
...
A
lattice can be constructed from primitive cells
...
If G is written as G = v1 b1 + v2 b2 + v3 b3 with vi ∈ I , it follows for the
vectors bi , cyclically:
ai+1 × ai+2
bi = 2π
ai · (ai+1 × ai+2 )
The set of G-vectors determines the R¨ ntgen diffractions: a maximum in the reflected radiation occurs if:
o
∆k = G with ∆k = k − k
...
From this follows for parallel lattice planes (Bragg reflection)
that for the maxima holds: 2d sin(θ) = nλ
...
12
...
Van der Waals bond
2
...
Covalent or homopolar bond
4
...
For the ion binding of NaCl the energy per molecule is calculated by:
E = cohesive energy(NaCl) – ionization energy(Na) + electron affinity(Cl)
The interaction in a covalent bond depends on the relative spin orientations of the electrons constituing the
bond
...
Furthermore the potential energy for two parallel spins has sometimes no minimum
...
62
Chapter 12: Solid state physics
63
12
...
3
...
The force on
atom s with mass M can then be written as:
Fs = M
d2 us
= C(us+1 − us ) + C(us−1 − us )
dt2
Assuming that all solutions have the same time-dependence exp(−iωt) this results in:
−M ω 2 us = C(us+1 + us−1 − 2us )
Further it is postulated that: us±1 = u exp(isKa) exp(±iKa)
...
Substituting the later two equations in the fist results in a system of linear
equations, which has only a solution if their determinant is 0
...
This requires
that −π < Ka ≤ π
...
2
M
dω
=
dK
and is 0 on the edge of a Brillouin Zone
...
12
...
2 A lattice with two types of atoms
ω
T
Now the solutions are:
ω2 = C
1
1
+
M1
M2
±C
1
1
+
M1
M2
2
−
4 sin2 (Ka)
M1 M2
Connected with each value of K are two values of ω, as can be
seen in the graph
...
In the optical branch,
both types of ions oscillate in opposite phases, in the acoustical
branch they oscillate in the same phase
...
Furthermore each
branch has 3 polarization directions, one longitudinal and two
transversal
...
3
...
Phonons
h
can be viewed as quasi-particles: with collisions, they behave as particles with momentum ¯ K
...
When they collide, their momentum need not be conserved: for a normal process holds:
K1 + K2 = K3 , for an umklapp process holds: K1 + K2 = K3 + G
...
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...
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...
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64
12
...
4 Thermal heat capacity
The total energy of the crystal vibrations can be calculated by multiplying each mode with its energy and sum
over all branches K and polarizations P :
h
¯ ω nk,p =
U=
K
Dλ (ω)
P
λ
h
¯ω
dω
exp(¯ ω/kT ) − 1
h
for a given polarization λ
...
Because exp(2πi) = 1 this is only possible if:
Kx , Ky , Kz = 0; ±
4π
6π
2N π
2π
; ±
; ±
;
...
The total number of
states with a wave vector < K is:
3
4πK 3
L
N=
2π
3
for each polarization
...
Here, only the acoustic phonons are taken into account (3 polarizations), and one assumes that v = ωK,
independent of the polarization
...
This
gives:
ωD
U =3
D(ω) n ¯ ωdω =
h
0
3V k 2 T 4
V ω2
h
¯ω
dω =
2π 2 v 3 exp(¯ ω/kT ) − 1
h
2π 2 v 3 ¯ 3
h
xD
0
x3 dx
...
θD is the Debye temperature and is defined by:
θD =
h
¯v
k
6π 2 N
V
1/3
where N is the number of primitive cells
...
Chapter 12: Solid state physics
65
12
...
4
...
Starting
with a circular electron orbit in an atom with two electrons, there is a Coulomb force Fc and a magnetic force
on each electron
...
One electron is accelerated, the other decelerated
...
The circular current is given by I = −ZeωL /2π, and
µ = IA = Iπ ρ2 = 2 Iπ r2
...
4
...
After
distribution fm ∼ exp(−∆Um /kT ), one finds for the average magnetic moment µ =
linearization and because mJ = 0, J = 2J + 1 and m2 = 2 J(J + 1)(J + 1 ) it follows that:
J
3
2
χp =
µ0 N µ
µ0 J(J + 1)g 2 µ2 N
µ0 M
B
=
=
B
B
3kT
This is the Curie law, χp ∼ 1/T
...
4
...
To describe ferromagnetism a field
BE parallel with M is postulated: BE = λµ0 M
...
Ba
T − Tc
If BE is estimated this way it results in values of about 1000 T
...
A quantum mechanical approach from Heisenberg postulates an interaction between two neighbouring atoms: U = −2J Si · Sj ≡ −µ · BE
...
A distinction between 2 cases can now be made:
1
...
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...
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...
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66
2
...
Heisenberg’s theory predicts quantized spin waves: magnons
...
This results in a system of linear equations with solution:
h
¯ ω = 4JS(1 − cos(ka))
12
...
5
...
From this follows
h
¯ 2 nπ 2
E=
2m L
In a linear lattice the only important quantum numbers are n and ms
...
If nF is the quantum number of the Fermi level, it
¯
can be expressed as: 2nF = N so EF = h2 π 2 N 2 /8mL
...
√
E=
3N
...
01 times the classical expected value 3 N k
...
There is a fraction ≈ T /TF excited thermally
...
Together with the T 3 dependence of the
2
thermal heat capacity of the phonons the total thermal heat capacity of metals is described by: C = γT + AT 3
...
5
...
The variation of k is given by
¯
δ k = k(t) − k(0) = −eEt/¯
...
Then holds: v = µE, with µ = eτ /m the mobility of the electrons
...
Because for the collision time holds:
1/τ = 1/τL + 1/τi , where τL is the collision time with the lattice phonons and τi the collision time with the
impurities follows for the resistivity ρ = ρL + ρi , with lim ρL = 0
...
5
...
This results in a magnetic field ⊥ to the flow direction of the current
...
The Hall coefficient is defined by: RH = Ey /Jx B, and RH = −1/ne if Jx = neµEx
...
12
...
4 Thermal heat conductivity
With = vF τ the mean free path of the electrons follows from κ = 1 C v
3
From this follows for the Wiedemann-Franz ratio: κ/σ = 1 (πk/e)2 T
...
12
...
From this follows for the energy:
E = ψ|H|ψ = Eat − α − 2β cos(ka)
...
If it is assumed that the electron is nearly free one can
postulate: ψ = exp(ik · r )
...
This wave can be decomposed into two standing waves:
ψ(+) =
ψ(−) =
exp(iπx/a) + exp(−iπx/a) = 2 cos(πx/a)
exp(iπx/a) − exp(−iπx/a) = 2i sin(πx/a)
The probability density |ψ(+)|2 is high near the atoms of the lattice and low in between
...
Hence the energy of ψ(+) is also
lower than the energy of ψ)(−)
...
7 Semiconductors
The band structures and the transitions between them of direct and indirect semiconductors are shown in
the figures below
...
For an
indirect semiconductor a transition from the valence- to the conduction band is also possible if the energy of
the absorbed photon is smaller than the band gap: then, also a phonon is absorbed
...
J
...
A
...
...
...
...
...
...
...
¯
Eg + hΩ
EE
h
¯ ωg
Direct semiconductor
E E
Indirect semiconductor
So indirect semiconductors, like Si and Ge, cannot emit any light and are therefore not usable to fabricate
lasers
...
Instead of the normal electron mass one has to use the effective mass within a lattice
...
¯
¯
With the distribution function fe (E) ≈ exp((µ − E)/kT ) for the electrons and fh (E) = 1 − fe (E) for the
holes the density of states is given by:
D(E) =
1
2π 2
2m∗
h
¯2
3/2
E − Ec
with Ec the energy at the edge of the conductance band
...
For the product np follows: np = 4
An exciton is a bound electron-hole pair, rotating on each other as in positronium
...
This causes a peak in the absorption just under Eg
...
8 Superconductivity
12
...
1 Description
A superconductor is characterized by a zero resistivity if certain quantities are smaller than some critical values:
T < Tc , I < Ic and H < Hc
...
14ΘD exp
−1
U D(EF )
while experiments find for Hc approximately:
Hc (T ) ≈ Hc (Tc ) 1 −
T2
2
Tc
...
There are type I and type II superconductors
...
This holds for a type I superconductor, for
a type II superconductor this only holds to a certain value Hc1 , for higher values of H the superconductor is in
a vortex state to a value Hc2 , which can be 100 times Hc1
...
This is shown in the figures below
...
This means that
there is a twist in the T − S diagram and a discontinuity in the CX − T diagram
...
8
...
The electron waveo
function in one superconductor is ψ1 , in the other ψ2
...
The electron
√
√
wavefunctions are written as ψ1 = n1 exp(iθ1 ) and ψ2 = n2 exp(iθ2 )
...
From this follows, if n1 ≈ n2 :
∂θ2
∂n2
∂n1
∂θ1
=
and
=−
∂t
∂t
∂t
∂t
The Josephson effect results in a current density through the insulator depending on the phase difference as:
o
J = J0 sin(θ2 − θ1 ) = J0 sin(δ), where J0 ∼ T
...
h
¯
Hence there is an oscillation with ω = 2eV /¯
...
8
...
Because: Adl = (rotA, n )dσ = (B, n )dσ = Ψ follows: Ψ = 2π¯ s/q
...
0678 · 10−15 Tm2
...
J
...
A
...
8
...
h
12
...
5 The London equation
A current density in a superconductor proportional to the vector potential A is postulated:
J=
−B
−A
or rotJ =
2
µ0 λL
µ0 λ2
L
2
ε0 mc2 /nq 2
...
L
The Meissner effect is the solution of this equation: B(x) = B0 exp(−x/λL )
...
12
...
6 The BCS model
The BCS model can explain superconductivity in metals
...
A new ground state where the electrons behave like independent fermions is postulated
...
This causes two electrons with
opposite spin to combine to a Cooper pair
...
The infinite conductivity is more difficult to explain because a ring with a persisting current is not a real
equilibrium: a state with zero current has a lower energy
...
Flux quantization is related to the existence of a coherent many-particle wavefunction
...
So if the flux has to change with one flux quantum there has to occur
a transition of many electrons, which is very improbable, or the system must go through intermediary states
where the flux is not quantized so they have a higher energy
...
Some useful mathematical relations are:
∞
π2
xdx
=
,
ax + 1
e
12a2
0
−∞
π2
x2 dx
,
=
x + 1)2
(e
3
∞
∞
And, when
∞
n
(−1) =
n=0
1
2
0
∞
sin(px)dx =
follows:
0
∞
cos(px)dx =
0
1
...
1 Introduction
13
...
1 Definition of a group
G is a group for the operation • if:
1
...
2
...
3
...
4
...
If also holds:
5
...
13
...
2 The Cayley table
Each element arises only once in each row and column of the Cayley- or multiplication table: because EAi =
A−1 (Ak Ai ) = Ai each Ai appears once
...
13
...
3 Conjugated elements, subgroups and classes
B is conjugate to A if ∃X∈G such that B = XAX −1
...
If B and C are conjugate to A, B is also conjugate with C
...
r
...
the same operation
...
Each group can be split up in conjugacy
classes
...
• E is a class itself: for each other element in this class would hold: A = XEX −1 = E
...
• In an Abelian group each element is a separate class
...
Elements of one class are then the same kind of operations
...
71
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...
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72
13
...
4 Isomorfism and homomorfism; representations
Two groups are isomorphic if they have the same multiplication table
...
It need not be isomorphic
...
This is symbolized by Γ
...
A faithful representation: all matrices are different
...
The representation A → det(Γ(A))
...
The identical representation: A → 1
...
13
...
5 Reducible and irreducible representations
If the same unitary transformation can bring all matrices of a representation Γ in the same block structure the
representation is called reducible:
Γ(A) =
Γ(1) (A)
0
0
Γ(2) (A)
This is written as: Γ = Γ(1) ⊕ Γ(2)
...
The number of irreducible representations equals the number of conjugacy classes
...
2 The fundamental orthogonality theorem
13
...
1 Schur’s lemma
Lemma: Each matrix which commutes with all matrices of an irreducible representation is a constant ×I
I,
where I is the unit matrix
...
I
Lemma: If there exists a matrix M so that for two irreducible representations of group G, γ (1) (Ai ) and
γ (2) (Ai ), holds: M γ (1) (Ai ) = γ (2) (Ai )M , than the representations are equivalent, or M = 0
...
2
...
2
...
3 The relation with quantum mechanics
13
...
1 Representations, energy levels and degeneracy
Consider a set of symmetry transformations x = Rx which leave the Hamiltonian H invariant
...
An isomorfic operation on the wavefunction is given by: PR ψ(x ) = ψ(R−1 x )
...
These operators commute with H: PR H = HPR , and leave the volume
element unchanged: d(Rx ) = dx
...
It causes degeneracy: if ψn is a solution of Hψn = En ψn
than also holds: H(PR ψn ) = En (PR ψn )
...
Assume an
(n)
PR ψν
n -fold
(n)
degeneracy at En : then choose an orthonormal set ψν , ν = 1, 2,
...
The function
n
(n)
is in the same subspace: PR ψν =
(n)
ψκ Γ(n) (R)
κν
κ=1
where Γ is an irreducible, unitary representation of the symmetry group G of the system
...
One can purely mathematical derive irreducible representations of a symmetry group and label the energy levels with a quantum number this way
...
This way one can also label each separate base function with a quantum number
...
Then: G = {A, A2 , A3 ,
...
The group is Abelian so all irreducible representations are one-dimensional
...
If one defines: k = −
13
...
2 Breaking of degeneracy by a perturbation
Suppose the unperturbed system has Hamiltonian H0 and symmetry group G0
...
If Γ(n) (R) is an irreducible representation of G0 , it is also a
representation of G but not all elements of Γ(n) in G0 are also in G
...
The degeneracy is then (possibly partially) removed: see the figure
below
...
with energy En is a basis for an
n -dimensional
13
...
3 The construction of a base function
n
j
(j)
fκ
Each function F in configuration space can be decomposed into symmetry types: F =
j=1 κ=1
The following operator extracts the symmetry types:
j
h
(j)∗
Γκκ (R)PR
R∈G
(j)
F = fκ
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...
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...
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74
(j)
This is expressed as: fκ is the part of F that transforms according to the κth row of Γ(j)
...
The functions fκ
are in general not
ajκ
transformed into each other by elements of the group
...
Theorem: Two wavefunctions transforming according to non-equivalent unitary representations or according
to different rows of an unitary irreducible representation are orthogonal:
(i) (j)
(i) (i)
ϕκ |ψλ ∼ δij δκλ , and ϕκ |ψκ is independent of κ
...
3
...
The subspace D (i) of the system transforms according
(i)
(1)
(2)
to Γ(i)
...
Now form all 1 × 2 products ϕκ (x1 )ϕλ (x2 )
...
These product functions transform as:
(2)
(2)
PR (ϕ(1) (x1 )ϕλ (x2 )) = (PR ϕ(1) (x1 ))(PR ϕλ (x2 ))
κ
κ
In general the space D(1) ⊗ D(2) can be split up in a number of invariant subspaces:
Γ(1) ⊗ Γ(2) =
ni Γ(i)
i
A useful tool for this reduction is that for the characters hold:
χ(1) (R)χ(2) (R) =
ni χ(i) (R)
i
13
...
5 Clebsch-Gordan coefficients
(i)
(j)
(aκ)
With the reduction of the direct-product matrix w
...
t
...
The unitary base transformation is given by:
...
3
...
If a set of
(j)
operators Aκ with 0 ≤ κ ≤ j transform into each other under the transformations of G holds:
−1
PR A(j) PR =
κ
A(j) Γ(j) (R)
ν
νκ
ν
(j)
If Γ(j) is irreducible they are called irreducible tensor operators A(j) with components Aκ
...
Further ϕκ |H|ψκ is independent of κ
...
This is applied in quantum mechanics in perturbation theory and variational calculus
...
Solutions can be found within each category of functions ϕκ with common i and κ: H is already
diagonal in categories as a whole
...
With variational calculus the try function can be chosen within a separate category because the exact eigenfunctions transform according to a row
of an irreducible representation
...
3
...
⊕ Γ(i) ⊕
...
For more a-values relations between the intensity ratios can be
stated
...
13
...
However, not all groups with h = ∞ are continuous, e
...
the translation
group of an spatially infinite periodic potential is not continuous but does have h = ∞
...
4
...
Taylor expansion near x
gives:
dψ(x) 1 2 d2 ψ(x)
+ a
− +
...
4
...
For an infinitesimal rotation around the x-axis holds:
Pδθx ψ(x, y, z) ≈
=
=
ψ(x, y + zδθx , z − yδθx )
∂
∂
− yδθx
ψ(x, y, z) + zδθx
∂y
∂z
iδθx Lx
ψ(x, y, z)
1−
h
¯
ψ(x, y, z)
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...
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76
h
¯
i
Because the angular momentum operator is given by: Lx =
So in an arbitrary direction holds:
Rotations:
Translations:
z
∂
∂
−y
...
rotation group, px , py and pz are called the generators of
the 3-dim
...
The commutation rules for the generators can be derived from the properties of the group for multiplications:
translations are interchangeable ↔ px py − py px = 0
...
Then holds: Pα,n = P−θ,y Pα,x Pθ,y , so:
h
h
h
h
e−iα(n·J )/¯ = eiθJy /¯ e−iαJx /¯ e−iθJy /¯
If α and θ are very small and are expanded to second order, and the corresponding terms are put equal with
h
n · J = Jx cos θ + Jz sin θ, it follows from the αθ term: Jx Jy − Jy Jx = i¯ Jz
...
4
...
, pn ) depend continuously on parameters p1 ,
...
For the translation group this are
e
...
anx , any and anz
...
The statement that each element arises only once in each row and column of the Cayley table holds also for
continuous groups
...
The notion representation is fitted by demanding continuity: each matrix element depends continuously on
pi (R)
...
If f (R) is a function
defined on G, e
...
Γαβ (R), holds:
···
f (R)dR :=
G
p1
f (R(p1 ,
...
, pn ))dp1 · · · dpn
pn
Here, g(R) is the density function
...
This fixes g(R) except
for a constant factor
...
If one writes: dV := dp1 · · · dpn holds:
g(S) = g(E)
Here,
dV
dV
is the Jacobian:
= det
dV
dV
∂pi
∂pj
dV
dV
, and g(E) is constant
...
This leads to the fundamental orthogonality theorem:
(j)
Γ(i)∗ (R)Γαβ (R)dR =
µν
G
1
δij δµα δνβ
i
dR
G
and for the characters hold:
χ(i)∗ (R)χ(j) (R)dR = δij
G
Compact groups are groups with a finite group volume:
dR
G
G
dR < ∞
...
5 The group SO(3)
One can take 2 parameters for the direction of the rotational axis and one for the angle of rotation ϕ
...
The diametrical points on this sphere
are equivalent because Rn,π = Rn,−π
...
If α, β and γ are the 3 Euler angles, defined
as:
1
...
r
...
xyz are θ, ϕ := β, α
...
2
...
r
...
123 are θ, ϕ := β, π − γ
...
So PR = e−iε(n·J )/¯
...
The character of D (l) is given by:
l
χ(l) (α) =
l
eimα = 1 + 2
m=−l
cos(kα) =
k=0
sin([l + 1 ]α)
2
sin( 1 α)
2
In the performed derivation α is the rotational angle around the z-axis
...
Via the fundamental orthogonality theorem for characters one obtains the following expression for the density
function (which is normalized so that g(0) = 1):
g(α) =
sin2 ( 1 α)
2
( 1 α)2
2
With this result one can see that the given representations of SO(3) are the only ones: the character of another
representation χ would have to be ⊥ to the already found ones, so χ (α) sin2 ( 1 α) = 0∀α ⇒ χ (α) = 0∀α
...
1
Because fermions have an half-odd integer spin the states ψsms with s = 1 and ms = ± 2 constitute a 2-dim
...
A problem arises for rotations over 2π:
h
ψ 1 ms → e−2πiSz /¯ ψ 1 ms = e−2πims ψ 1 ms = −ψ 1 ms
2
2
2
2
However, in SO(3) holds: Rz,2π = E
...
written as φ|ψ or φ|A|ψ , and are bilinear in the states, they do not change sign if the states do
...
The existence of these half-odd integer representations is connected with the topological properties of SO(3):
the group is two-fold coherent through the identification R0 = R2π = E
...
6 Applications to quantum mechanics
13
...
1 Vectormodel for the addition of angular momentum
If two subsystems have angular momentum quantum numbers j1 and j2 the only possible values for the total
angular momentum are J = j1 + j2 , j1 + j2 − 1,
...
This can be derived from group theory as follows:
from χ(j1 ) (α)χ(j2 ) (α) = nj χ(J) (α) follows:
J
D(j1 ) ⊗ D(j2 ) = D(j1 +j2 ) ⊕ D(j1 +j2 −1) ⊕
...
J
...
A
...
The Clebsch-Gordan coefficients, for SO(3) called the Wigner coefficients, can be chosen real, so:
ψj1 j2 JM
=
ψj1 m1 j2 m2
=
m1 m2
JM
ψj1 m1 j2 m2 (j1 m1 j2 m2 |JM )
ψj1 j2 JM (j1 m1 j2 m2 |JM )
13
...
2 Irreducible tensor operators, matrixelements and selection rules
Some examples of the behaviour of operators under SO(3)
1
...
This state is described by a
(0) −1
(0)
scalar operator
...
g
...
Then holds: J M |H|JM ∼ δMM δJJ
...
A vector operator: A = (Ax , Ay , Az )
...
So for rotations around the z-axis holds:
cos α − sin α 0
D(Rα,z ) = sin α cos α 0
0
0
1
The transformed operator has the same matrix elements w
...
t
...
According to the equation for
characters this means one can choose base operators which transform like Y1m (θ, ϕ)
...
A cartesian tensor of rank 2: Tij is a quantity which transforms under rotations like Ui Vj , where U and
−1
Tkl Dki (R)Dlj (R), so like D(1) ⊗ D(1) =
V are vectors
...
The 9 components can be split in 3 invariant subspaces with dimension 1 (D (0) ),
3 (D(1) ) and 5 (D(2) )
...
Tr(T ) = Txx + Tyy + Tzz
...
II
...
These transform as the vector U × V ,
2
so as D(1)
...
The 5 independent components of the traceless, symmetric tensor S:
Sij = 1 (Tij + Tji ) − 1 δij Tr(T )
...
2
3
Selection rules for dipole transitions
Dipole operators transform as D(1) : for an electric dipole transfer is the operator er, for a magnetic e(L +
2S )/2m
...
This means that J ∈ {J + 1, J, |J − 1|}: J = J or J = J ± 1, except
J = J = 0
...
This can also be understood
Chapter 13: Theory of groups
79
from the Wigner-Eckart theorem: from this follows that the matrix elements from all vector operators show a
certain proportionality
...
7 Applications to particle physics
The physics of a system does not change after performing a transformation ψ = eiδ ψ where δ is a constant
...
There exists some freedom in the choice of the potentials A and φ at the same E and B: gauge transformations
o
of the potentials do not change E and B (See chapter 2 and 10)
...
This is a local gauge transformation: the phase of the wavefunction changes different at each position
...
This is now stated as a guide principle:
the “right of existence” of the electromagnetic field is to allow local gauge invariance
...
The split-off of charge
in the exponent is essential: it allows one gauge field for all charged particles, independent of their charge
...
The group elements now are
PR = exp(−iQΘ)
...
The weak interaction together
with the electromagnetic interaction can be described by a force field that transforms according to U(1)⊗SU(2),
and consists of the photon and three intermediary vector bosons
...
In general the group elements are given by PR = exp(−iT · Θ), where Θn are real constants and Tn operators
(generators), like Q
...
The cijk are the structure
k
constants of the group
...
These constants can be found with the help of group product elements: because G is closed holds:
eiΘ·T eiΘ ·T e−iΘ·T e−iΘ ·T = e−iΘ ·T
...
The group SU(2) has 3 free parameters: because it is unitary there are 4 real conditions over 4 complex
parameters, and the determinant has to be +1, remaining 3 free parameters
...
Here, H is a Hermitian matrix
...
For each matrix of SU(2) holds that Tr(H)=0
...
So these matrices are a choice for the operators of SU(2)
...
2
In abstraction, one can consider an isomorphic group where only the commutation rules are considered to be
known regarding the operators Ti : [T1 , T2 ] = iT3 , etc
...
g
...
Elementary particles can
be classified in isospin-multiplets, these are the irreducible representations of SU(2)
...
The isospin-singlet ≡ the identical representation: e−iT ·Θ = 1 ⇒ Ti = 0
2
...
80
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...
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The group SU(3) has 8 free parameters
...
The Hermitian,
traceless operators are 3 SU(2)-subgroups in the e1 e2 , e1 e3 and the e2 e3 plane
...
By taking a linear combination one gets 8 matrices
...
Chapter 14
Nuclear physics
14
...
The top is at 56 Fe,
26
the most stable nucleus
...
760
17
...
711
23
...
000
MeV
MeV
MeV
MeV
MeV
9
8
↑ 7
E 6
(MeV) 5
4
3
2
1
0
0
40
80
120 160
A→
200
240
and A = Z + N , in the droplet or collective model of the nucleus the binding energy Ebind is given by:
Z(Z − 1)
(N − Z)2
Ebind
+ a5 A−3/4
= a1 A − a2 A2/3 − a3
− a4
2
1/3
c
A
A
These terms arise from:
1
...
2
...
3
...
4
...
5
...
The following holds:
Z even, N even: = +1, Z odd, N odd: = −1
...
The Yukawa potential can be derived if the nuclear force can to first approximation, be considered as an
exchange of virtual pions:
r
W0 r0
exp −
U (r) = −
r
r0
¯
With ∆E · ∆t ≈ ¯ , Eγ = m0 c2 and r0 = c∆t follows: r0 = h/m0 c
...
h
Further, there is a contribution of the spin-orbit coupling ∼ L · S: ∆Vls = 1 (2l + 1)¯ ω
...
This is just
2
2
the opposite for electrons, which is an indication that the L − S interaction is not electromagnetical
...
N = nx + ny + nz = 2(n − 1) + l
2 h
where n ≥ 1 is the main oscillator number
...
J
...
A
...
This gives rise to the so called magical numbers:
nuclei where each state in the outermost level are filled are particulary stable
...
14
...
Here, R0 ≈ 1
...
If the nuclear radius is measured including the charge distribution one obtains R0 ≈ 1
...
The shape of oscillating nuclei can be described by spherical harmonics:
alm Ylm (θ, ϕ)
R = R0 1 +
lm
l = 0 gives rise to monopole vibrations, density vibrations, which can be applied to the theory of neutron stars
...
The multipole moment is given by µl = Zerl Ylm (θ, ϕ)
...
e
e
L and MS = gS
S
...
For protons holds gS = 5
...
8263
...
The resulting
magnetic moment is related to the nuclear spin I according to M = gI (e/2mp)I
...
14
...
This gives for the number
of nuclei N : N (t) = N0 exp(−λt)
...
The average life time
2
of a nucleus is τ = 1/λ
...
So the fraction decaying into state i is
λi / λi
...
α-decay: the nucleus emits a He2+ nucleus
...
The tunnel
probability P is
P =
1
incoming amplitude
= e−2G with G =
outgoing amplitude
h
¯
2m
[V (r) − E]dr
G is called the Gamow factor
...
β-decay
...
3
...
4
...
5
...
The decay constant is given by
λ=
Eγ
P (l)
∼
h
¯ω
(¯ c)2
h
Eγ R
h
¯c
2l
∼ 10−4l
Chapter 14: Nuclear physics
83
where l is the quantum number for the angular momentum and P the radiated power
...
2
The energy of the photon is Eγ = Ei − Ef − TR , with TR = Eγ /2mc2 the recoil energy, which
can usually be neglected
...
With I the quantum
number of angular momentum of the nucleus, L = h I(I + 1), holds the following selection rule:
¯
|Ii − If | ≤ ∆l ≤ |Ii + If |
...
4 Scattering and nuclear reactions
14
...
1 Kinetic model
If a beam with intensity I hits a target with density n and length x (Rutherford scattering) the number of
scatterings R per unit of time is equal to R = Inxσ
...
This results in I = I0 e−nσx = I0 e−µx
...
4
...
At large distances from the scattering point they have approximately a spherical
wavefunction ψscat = f (θ)eikr /r where f (θ) is the scattering amplitude
...
From this it follows that σ(θ) = |f (θ)|2
...
At very low energy
h
only particles with l = 0 are scattered, so
ψ = ψ0 +
ψl and ψ0 =
l>0
If the potential is approximately rectangular holds: ψ0 = C
The cross section is then σ(θ) =
sin2 (δ0 )
so σ =
k2
At very low energies holds: sin2 (δ0 ) =
sin(kr)
kr
sin(kr + δ0 )
kr
σ(θ)dΩ =
4π sin2 (δ0 )
k2
h
¯ 2 k 2 /2m
W0 + W
with W0 the depth of the potential well
...
J
...
A
...
4
...
r
...
the laboratory system and other particles are
created, so
Pk
P1 + P2 →
k>2
the total energy Q gained or required is given by Q = (m1 + m2 −
mk )c2
...
14
...
Dosimetric quantities are related to
the energy transfer from radiation to matter
...
The intensity of a beam of particles in matter decreases according to I(s) = I0 exp(−µs)
...
The flux is given by φ = dΦ/dt
...
The absorption coefficient is given by µ = (dN/N )/dx
...
The radiation dose X is the amount of charge produced by the radiation per unit of mass, with unit C/kg
...
58 · 10−4 C/kg
...
The absorbed dose D is given by D = dEabs /dm, with unit Gy=J/kg
...
01 Gy
...
It can be derived that
D=
µE
Ψ
The Kerma K is the amount of kinetic energy of secundary produced particles which is produced per mass
unit of the radiated object
...
These weight factors are called the
quality factors
...
H = QD
...
For some types of radiation holds:
Radiation type
R¨ ntgen, gamma radiation
o
β, electrons, mesons
Thermic neutrons
Fast neutrons
protons
α, fission products
Q
1
1
3 to 5
10 to 20
10
20
Chapter 15
Quantum field theory & Particle physics
15
...
Hence the
vacuum state is given by |000 · · ·
...
The states are orthonormal:
∞
n1 n2 n3 · · · |n1 n2 n3 · · · =
δni ni
i=1
The time-dependent state vector is given by
cn1 n2 ··· (t)|n1 n2 · · ·
Ψ(t) =
n1 n2 ···
The coefficients c can be interpreted as follows: |cn1 n2 ··· |2 is the probability to find n1 particles with momentum k1 , n2 particles with momentum k2 , etc
...
The expansion of the states
in time is described by the Schr¨ dinger equation
o
i
d
|Ψ(t) = H|Ψ(t)
dt
where H = H0 + Hint
...
All operators which can change occupation numbers can be expanded in the a and a† operators
...
So aa† is an occupation
number operator
...
2 Classical and quantum fields
Starting with a real field Φα (x) (complex fields can be split in a real and an imaginary part), the Lagrange
density L is a function of the position x = (x, ict) through the fields: L = L(Φα (x), ∂ν Φα (x))
...
Using the variational principle δI(Ω) = 0 and with the action-integral
I(Ω) = L(Φα , ∂ν Φα )d4 x the field equation can be derived:
∂
∂L
∂L
=0
−
α
∂Φ
∂xν ∂(∂ν Φα )
The conjugated field is, analogous to momentum in classical mechanics, defined as:
Πα (x) =
85
∂L
˙
∂ Φα
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86
˙
With this, the Hamilton density becomes H(x) = Πα Φα − L(x)
...
3 The interaction picture
Some equivalent formulations of quantum mechanics are possible:
1
...
o
2
...
3
...
The interaction picture can be obtained from the Schr¨ dinger picture by an unitary transformation:
o
S
S
S
|Φ(t) = eiH0 |ΦS (t) and O(t) = eiH0 OS e−iH0
The index S denotes the Schr¨ dinger picture
...
4 Real scalar field in the interaction picture
It is easy to find that, with M := m2 c2 /¯ 2 , holds:
h
0
∂
∂
Φ(x) = Π(x) and
Π(x) = (
∂t
∂t
2
− M 2 )Φ(x)
2
From this follows that Φ obeys the Klein-Gordon equation (2 − M 2 )Φ = 0
...
Usually one can take the limit
V → ∞
...
the coefficients have to be each others hermitian conjugate because Φ is hermitian
...
From this follows that the
commutation rules are given by [Φ(x), Φ(x )] = i∆(x − x ) with
∆(y) =
1
(2π)3
sin(ky) 3
d k
ωk
∆(y) is an odd function which is invariant for proper Lorentz transformations (no mirroring)
...
In general holds that ∆(y) = 0 outside the light cone
...
The Lagrange density is given by: L(Φ, ∂ν Φ) = − 1 (∂ν Φ∂ν Φ + m2 Φ2 )
...
5 Charged spin-0 particles, conservation of charge
The Lagrange density of charged spin-0 particles is given by: L = −(∂ν Φ∂ν Φ∗ + M 2 ΦΦ∗ )
...
Suppose that
L ((Φα ) , ∂ν (Φα ) ) = L (Φα , ∂ν Φα ) and there exists a continuous transformation between Φα and Φα such
as Φα = Φα + f α (Φ)
...
Which quantity is conserved depends on the symmetry
...
The
conserved quantity is the current density Jµ (x) = −ie(Φ∂µ Φ∗ − Φ∗ ∂µ Φ)
...
When this field is quantized the field operators
are given by
1
Φ(x) = √
V
k
1
√
a(k )eikx + b† (k )e−ikx
2ωk
1
, Φ† (x) = √
V
k
1
√
a† (k )eikx + b(k )e−ikx
2ωk
Hence the energy operator is given by:
h
¯ ωk a† (k )a(k ) + b† (k )b(k )
H=
k
and the charge operator is given by:
Q(t) = −i
e a† (k )a(k ) − b† (k )b(k )
J4 (x)d3 x ⇒ Q =
k
From this follows that a† a := N+ (k ) is an occupation number operator for particles with a positive charge
and b† b := N− (k ) is an occupation number operator for particles with a negative charge
...
6 Field functions for spin- 1 particles
2
Spin is defined by the behaviour of the solutions ψ of the Dirac equation
...
Λ denotes
4-dimensional rotations: the proper Lorentz transformations
...
˜
˜
A rotated field ψ obeys the Dirac equation if the following condition holds: ψ(x) = D(Λ)ψ(Λ−1 x)
...
Hence:
h
results in the condition D γλ D = Λλµ γµ
...
With the notation
v r (p ) = ur (−p ) and ur (p ) = ur (p ) one can write for the dot products of these spinors:
−
+
ur (p )ur (p ) =
+
+
E
E
δrr , ur (p )ur (p ) =
δrr , ur (p )ur (p ) = 0
−
−
−
+
M
M
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88
Because of the factor E/M this is not relativistic invariant
...
From this follows:
ur (p )ur (p ) = δrr , v r (p )v r (p ) = −δrr , ur (p )v r (p ) = 0
Combinations of the type aa give a 4 × 4 matrix:
2
u
r
(p )ur (p )
r=1
2
−iγλ pλ + M
,
=
2M
v r (p )v r (p ) =
r=1
−iγλ pλ − M
2M
The Lagrange density which results in the Dirac equation and having the correct energy normalization is:
L(x) = −ψ(x) γµ
∂
+M
∂xµ
ψ(x)
and the current density is Jµ (x) = −ieψγµ ψ
...
7 Quantization of spin- 1 fields
2
The general solution for the fieldoperators is in this case:
ψ(x) =
M
V
p
1
√
E
p
1
√
E
cr (p )ur (p )eipx + d† (p )v r (p )e−ipx
r
r
and
ψ(x) =
M
V
c† (p )ur (p )e−ipx + dr (p )v r (p )eipx
r
r
†
Here, c and c are the creation respectively annihilation operators for an electron and d† and d the creation
respectively annihilation operators for a positron
...
r
r
The field operators obey
[ψα (x), ψβ (x )] = 0 , [ψα (x), ψβ (x )] = 0 , [ψα (x), ψβ (x )]+ = −iSαβ (x − x )
∂
− M ∆(x)
∂xλ
The anti commutation rules give besides the positive-definite energy also the Pauli exclusion principle and the
Fermi-Dirac statistics: because c† (p )c† (p ) = −c† (p )c† (p ) holds: {c† (p)}2 = 0
...
This is the exclusion principle
...
with S(x) =
γλ
To avoid infinite vacuum contributions to the energy and charge the normal product is introduced
...
This product is obtained by:
• Expand all fields into creation and annihilation operators,
• Keep all terms which have no annihilation operators, or in which they are on the right of the creation
operators,
• In all other terms interchange the factors so that the annihilation operators go to the right
...
Assume
hereby that all commutators are zero
...
8 Quantization of the electromagnetic field
Starting with the Lagrange density L = − 1
2
∂Aν ∂Aν
∂xµ ∂xµ
it follows for the field operators A(x):
1
A(x) = √
V
k
1
√
2ωk
4
am (k )
m
(k )eikx + a† (k )
m
(k )∗ e−ikx
m=1
The operators obey [am (k ), a† (k )] = δmm δkk
...
m gives the polarization
m
direction of the photon: m = 1, 2 gives transversal polarized, m = 3 longitudinal polarized and m = 4
timelike polarized photons
...
By changing the definition of the inner product in
R
configuration space the expectation values for A1,2,3 (x) ∈ I and for A4 (x) become imaginary
...
However, this gives problems with the commutation rules
...
From this follows that (a3 (k ) − a4 (k ))|Φ = 0
...
However, this only applies to free EM-fields: in intermediary states in interactions there
can exist longitudinal and timelike photons
...
15
...
If the Schr¨ dinger equation is integrated:
o
t
|Φ(t) = |Φ(−∞) − i
Hint (t1 )|Φ(t1 ) dt1
−∞
and perturbation theory is applied one finds that:
∞
S=
(−i)n
n!
n=0
∞
···
T {Hint (x1 ) · · · Hint (xn )} d4 x1 · · · d4 xn ≡
S (n)
n=0
Here, the T -operator means a time-ordered product: the terms in such a product must be ordered in increasing
time order from the right to the left so that the earliest terms act first
...
The interaction Hamilton density for the interaction between the electromagnetic and the electron-positron
field is: Hint (x) = −Jµ (x)Aµ (x) = ieN (ψγµ ψAµ )
When this is expanded as: Hint = ieN (ψ + + ψ − )γµ (ψ + + ψ − )(A+ + A− )
µ
µ
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...
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90
eight terms appear
...
The term ieψ + γµ ψ + A− acting on |Φ
µ
gives transitions where A− creates a photon, ψ + annihilates an electron and ψ + annihilates a positron
...
Further the factors in Hint can create and thereafter annihilate particles: the virtual particles
...
This can be written as a sum of
normal products
...
The effects of the virtual particles are described by the (anti)commutator functions
...
Wick’s theorem gives an expression for the time-ordened product of
an arbitrary number of field operators
...
In the x-representation each diagram describes a number of processes
...
This means that energy and momentum
is conserved
...
15
...
An integration over one of them becomes ∞
...
This is
solved by discounting all divergent diagrams in a renormalization of e and M
...
In the Hamilton and Lagrange density of the free electron-positron field appears M0
...
15
...
Hadrons: these exist of quarks and can be categorized in:
I
...
Chapter 15: Quantum field theory & Particle physics
91
II
...
2
...
3
...
An overview of particles and antiparticles is given in the following table:
Particle
u
d
s
c
b
t
e−
µ−
τ−
νe
νµ
ντ
γ
gluon
W+
Z
graviton
spin (¯ ) B
h
1/2 1/3
1/2 1/3
1/2 1/3
1/2 1/3
1/2 1/3
1/2 1/3
1/2 0
1/2 0
1/2 0
1/2 0
1/2 0
1/2 0
1
0
1
0
1
0
1
0
2
0
L
0
0
0
0
0
0
1
1
1
1
1
1
0
0
0
0
0
T
1/2
1/2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
T3
1/2
−1/2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
S
0
0
−1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
C
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
B∗
0
0
0
0
−1
0
0
0
0
0
0
0
0
0
0
0
0
charge (e)
+2/3
−1/3
−1/3
+2/3
−1/3
+2/3
−1
−1
−1
0
0
0
0
0
+1
0
0
m0 (MeV)
5
9
175
1350
4500
173000
0
...
658
1777
...
u
d
s
c
b
t
e+
µ+
τ+
νe
νµ
ντ
γ
gluon
W−
Z
graviton
Here B is the baryon number and L the lepton number
...
T is the isospin, with T3 the projection of the isospin on
the third axis, C the charmness, S the strangeness and B∗ the bottomness
...
The composition of (anti)quarks of the hadrons
is given in the following table, together with their mass in MeV in their ground state:
π0
π+
π−
K0
K0
K+
K−
D+
D−
D0
D0
F+
F−
1
2
√
2(uu+dd)
ud
du
sd
ds
us
su
cd
dc
cu
uc
cs
sc
134
...
56995
139
...
672
497
...
677
493
...
4
1869
...
6
1864
...
0
1969
...
8
9460
...
27231
938
...
56563
939
...
684
1115
...
37
1189
...
55
1192
...
436
Σ+
Ξ0
0
Ξ
Ξ−
Ξ+
Ω−
Ω+
Λ+
c
∆2−
∆2+
∆+
∆0
∆−
dds
uss
uss
dss
dss
sss
sss
udc
uuu
uuu
uud
udd
ddd
1197
...
9
1314
...
32
1321
...
45
1672
...
1
1232
...
0
1232
...
0
1232
...
So mesons are bosons with spin 0 or 1 in their ground state, while
baryons are fermions with spin 1 or 3
...
Neutrino’s have a
2
2
1
1
helicity of − 2 while antineutrino’s have only + 2 as possible value
...
These can be derived from symmetries in the Lagrange density: continuous symmetries give rise to additive conservation laws, discrete symmetries result in
multiplicative conservation laws
...
These are:
1
...
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2
...
3
...
Dynamical conservation laws are invariant under the CPT-operation
...
Electrical charge because the Maxwell equations are invariant under gauge transformations
...
Colour charge is conserved
...
Isospin because QCD is invariant for rotations in T-space
...
Baryon number and lepton number are conserved but not under a possible SU(5) symmetry of the laws
of nature
...
Quarks type is only conserved under the colour interaction
...
Parity is conserved except for weak interactions
...
The color
force does not decrease with distance
...
This will result in two hadrons and not
in free quarks
...
12 P and CP-violation
It is found that the weak interaction violates P-symmetry, and even CP-symmetry is not conserved
...
µ-decay: µ− → e− + νµ + ν e
...
2
...
More electrons with a spin parallel to the Co
than with a spin antiparallel are created: (parallel−antiparallel)/(total)=20%
...
There is no connection with the neutrino: the decay of the Λ particle through: Λ → p+ + π − and
Λ → n0 + π 0 has also these properties
...
These are the lowest possible states
with a s-quark so they can decay only weakly
...
Further holds P|K0 = −|K0 because K0 and K0 have an intrinsic parity of −1
...
The linear combinations
√
√
|K0 := 1 2(|K0 + |K0 ) and |K0 := 1 2(|K0 − |K0 )
1
2
2
2
are eigenstates of CP: CP|K0 = +|K0 and CP|K0 = −|K0
...
For colour interactions a base of K0 and K0 is practical because then the number
u−number u is constant
...
2 | K1 |K
1
2|
93
K0 |K0 |2 , the probability of CP=+1 decay is
2
The relation between the mass eigenvalues of the quarks (unaccented) and the fields arising in the weak currents
(accented) is (u , c , t ) = (u, c, t), and:
sin θ1 0
1
0
0
1 0 0
cos θ1
d
s = 0 cos θ2
sin θ2 0 1 0 − sin θ1 cos θ1 0
b
0 − sin θ2 cos θ2
0
0
1
0 0 eiδ
d
1
0
0
0 cos θ3
sin θ3 s
0 − sin θ3 cos θ3
b
θ1 ≡ θC is the Cabibbo angle: sin(θC ) ≈ 0
...
01
...
13 The standard model
When one wants to make the Lagrange density which describes a field invariant for local gauge transformations
from a certain group, one has to perform the transformation
D
∂
g
∂
→
=
− i Lk Ak
µ
∂xµ
Dxµ
∂xµ
h
¯
Here the Lk are the generators of the gauge group (the “charges”) and the Ak are the gauge fields
...
The Lagrange density for a scalar field becomes:
a
µν
1
L = − 2 (Dµ Φ∗ Dµ Φ + M 2 Φ∗ Φ) − 1 Fµν Fa
4
a
and the field tensors are given by: Fµν = ∂µ Aa − ∂ν Aa + gca Al Am
...
13
...
Right- and left-handed spin states are treated different because the
weak interaction does not conserve parity
...
The hypercharge
Y , for quarks given by Y = B + S + C + B∗ + T , is defined by:
Q = 1 Y + T3
2
so [Y, Tk ] = 0
...
The multiplets are classified as follows:
e−
R
T
νeL e−
L
uL dL
uR
dR
0
1
2
1
2
0
0
0
0
4
3
−2
3
T3
0
Y
−2
1
2
−
−1
1
2
1
2
−
1
3
1
2
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...
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94
Now, 1 field Bµ (x) is connected with gauge group U(1) and 3 gauge fields Aµ (x) are connected with SU(2)
...
2
15
...
2 Spontaneous symmetry breaking: the Higgs mechanism
All leptons are massless in the equations above
...
This means that the dynamic equations which describe the system have a symmetry which the ground
state does not have
...
Their antiparticles have charges −1 and 0
...
Hence the state with the lowest energy corresponds with the state Φ∗ (x)Φ(x) = v = µ2 /2λ =constant
...
When the
gauge fields in the resulting Lagrange density are separated one obtains:
√
√
−
+
Wµ = 1 2(A1 + iA2 ) , Wµ = 1 2(A1 − iA2 )
µ
µ
µ
µ
2
2
Zµ
=
Aµ
=
gA3 − g Bµ
µ
g2 + g 2
g A3 + gBµ
µ
g2 + g 2
≡ A3 cos(θW ) − Bµ sin(θW )
µ
≡ A3 sin(θW ) + Bµ cos(θW )
µ
where θW is called the Weinberg angle
...
255 ± 0
...
Relations for the
masses of the field quanta can be obtained from the remaining terms: MW = 1 vg and MZ = 1 v g 2 + g 2 ,
2
2
gg
= g cos(θW ) = g sin(θW )
and for the elementary charge holds: e =
g2 + g 2
Experimentally it is found that MW = 80
...
26 GeV/c2 and MZ = 91
...
007 GeV/c2
...
0 ± 0
...
8 ± 2
...
15
...
3 Quantumchromodynamics
Coloured particles interact because the Lagrange density is invariant for the transformations of the group SU(3)
of the colour interaction
...
“White” particles: they have no colour charge, the generator T = 0
...
“Coloured” particles: the generators T are 8 3 × 3 matrices
...
The Lagrange density for coloured particles is given by
LQCD = i
a
µν
Ψk Mkl Ψl − 1 Fµν Fa
4
Ψ k γ µ Dµ Ψ k +
k
k,l
The gluons remain massless because this Lagrange density does not contain spinless particles
...
This term can be
brought in the form Mkl = mk δkl
...
14 Path integrals
The development in time of a quantum mechanical system can, besides with Schr¨
odingers equation, also be
described by a path integral (Feynman):
ψ(x , t ) =
F (x , t , x, t)ψ(x, t)dx
in which F (x , t , x, t) is the amplitude of probability to find a system on time t in x if it was in x on time t
...
The notation d[x] means that the integral has to be
˙
1
d[x] := lim
n→∞ N
n
∞
−∞
dx(tn )
in which N is a normalization constant
...
The
h
classical limit can be found by taking δS = 0: the average of the exponent vanishes, except where it is
stationary
...
15 Unification and quantum gravity
The strength of the forces varies with energy and the reciprocal coupling constants approach each other with
increasing energy
...
It also predicts 12 extra X bosons which couple leptons and quarks and are i
...
responsible
for proton decay, with dominant channel p+ → π 0 + e+ , with an average lifetime of the proton of 1031 year
...
Supersymmetric models assume a symmetry between bosons and fermions and predict partners for the currently known particles with a spin which differs 1
...
The dominant decay channels in this theory are
p+ → K+ + ν µ and p+ → K0 + µ+
...
Chapter 16
Astrophysics
16
...
The parallax is the angular difference
between two measurements of the position of the object from different view-points
...
The clusterparallax is used to determine the distance of a group of stars by using their motion w
...
t
...
The tangential velocity vt and the radial velocity vr of the stars along the sky are given by
vr = V cos(θ) , vt = V sin(θ) = ωR
ˆ
where θ is the angle between the star and the point of convergence and R the
distance in pc
...
The parallax is then given by
p=
4
...
A method to determine the distance of objects which are somewhat
further away, like galaxies and star clusters, uses the period-Brightness relation for Cepheids
...
16
...
Earth receives s0 = 1
...
Hence, the brightness of the Sun is given by L = 4πr2 s0 = 3
...
It is also given by:
∞
L = 4πR
2
πFν dν
0
where πFν is the monochromatic radiation flux
...
If Aν is the fraction of the flux which reaches Earth’s surface, the transmission factor
is given by Rν and the surface of the detector is given by πa2 , then the apparent brightness b is given by:
∞
b = πa
2
fν Aν Rν dν
0
The magnitude m is defined by:
1
b1
= (100) 5 (m2 −m1 ) = (2
...
From this follows that m2 − m1 = 2
...
5 ·10 log(b) + C
...
The absolute magnitude is then given by
r
r
M = −2
...
When an interstellar absorption of 10−4 /pc is taken
into account one finds:
ˆ
r
M = (m − 4 · 10−4 r ) + 5 − 5 ·10 log(ˆ)
If a detector detects all radiation emitted by a source one would measure the absolute bolometric magnitude
...
5 ·10 log
Energy flux received
Energy flux detected
= 2
...
Further holds
Mb = −2
...
72
16
...
When there is no absorption the quantity Iν is independent
of the distance to the source
...
ds is the thickness of the layer
...
The layer is optically thin if τν
1
...
Then also holds:
is optically thick if τν
Iν (s) = Iν (0)e−τν + Bν (T )(1 − e−τν )
16
...
J
...
A
...
Further holds:
κ(r) = f ( (r), T (r), composition) and ε(r) = g( (r), T (r), composition)
Convection will occur when the star meets the Schwartzschild criterium:
dT
dr
<
conv
dT
dr
rad
Otherwise the energy transfer takes place by radiation
...
For pp-chains holds µ ≈ 5 and for the CNO chains holds µ = 12 tot 18
...
Further holds: L ∼ R4 ∼ Teff
...
5 Energy production in stars
The net reaction from which most stars gain their energy is: 41 H → 4 He + 2e+ + 2νe + γ
...
72 MeV
...
The slowest, speedlimiting reaction is shown in boldface
...
1
...
I
...
There is 26
...
51) MeV released
...
pp2: 3 He + α → 7 Be + γ
i
...
25
...
80) MeV
...
7 Be + p+ → 8 B + γ, then 8 B + e+ → 24 He + ν
...
5 + (7
...
Both 7 Be chains become more important with raising T
...
The CNO cycle
...
03 + (1
...
74 + (1
...
The
reactions are shown below
...
The unit vectors are then given by:
eu =
1 ∂x
1 ∂x
1 ∂x
, ev =
, ew =
h1 ∂u
h2 ∂v
h3 ∂w
where the factors hi set the norm to 1
...
temp
...
current
Luminous intens
...
Unit
metre
kilogram
second
kelvin
ampere
candela
mol
Sym
...
Derivation
Frequency
Force
Pressure
Energy
Power
Charge
El
...
Capacitance
El
...
Conductance
Mag
...
flux density
Inductance
Luminous flux
Illuminance
Activity
Absorbed dose
Dose equivalent
Extra units
Unit
hertz
newton
pascal
joule
watt
coulomb
volt
farad
ohm
siemens
weber
tesla
henry
lumen
lux
bequerel
gray
sievert
Hz
N
Pa
J
W
C
V
F
Ω
S
Wb
T
H
lm
lx
Bq
Gy
Sv
s−1
kg · m · s−2
N · m−2
N·m
J · s−1
A·s
W · A−1
C · V−1
V · A−1
A · V−1
V·s
Wb · m−2
Wb · A−1
cd · sr
lm · m−2
s−1
J · kg−1
J · kg−1
Prefixes
yotta
zetta
exa
peta
tera
Y
Z
E
P
T
1024
1021
1018
1015
1012
giga
mega
kilo
hecto
deca
G
M
k
h
da
109
106
103
102
10
deci
centi
milli
micro
nano
d
c
m
µ
n
10−1
10−2
10−3
10−6
10−9
pico
femto
atto
zepto
yocto
p
f
a
z
y
10−12
10−15
10−18
10−21
10−24