script to - uni-due.de5 5 atomic physics 146 5.1 experiments that led to atomic models ...146 (a)...
TRANSCRIPT
Prof. Dr. H. Franke
Script to
"ISE Physics" Prof. Dr. H. Franke University Duisburg-Essen Campus Duisburg Department of physics / applied physics Lotharstr. 1, D - 47048 Duisburg Phone: ++49 (0)203 - 379 2865 Email: [email protected] Web: http://www.ofm.uni-due.de
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CONTENTS 0 INTRODUCTION………………………………………………………...6 0.1 Foreword……………………………………………………………...6 0.2 Literature……………………………………………………………..7 0.3 What is physics?...................................................................................7 0.4 Physical quantities……………………………………………………8 0.4.1 basic units / SI system 0.4.2 derived units & physical constants 0.4.3 angular units 1 TRANSLATIONS……………………………………………………….11 1.1 Coordinates, position-vector…………………………………….....11 1.2 Tracks……………………………………………………………..…13 1.3 Position, velocity & acceleration…………………………………...14 1.3.1 1D motion 1.3.2 3D motion 1.4 Special translations………………………………………………….18 1.4.1 uniform motion 1.4.2 motion with constant acceleration 1.4.3 circular motion (a) general case (b) uniform circular motion (c) circular motion with constant angular acceleration 1.4.4 summary 1.5 Assembled motion / coordinate transformations…………….23 1.5.1 systems with constant velocity 1.5.2 systems with constant acceleration 1.6 Forces………………………………………………………………..24 1.6.1 Newton's laws 1.6.2 consequences from Newton's laws: conservation of momentum 1.6.3 consequences from Newton's laws: equation of motion for a particle (a) case 1: m = const (1) constant forces (2) position dependent forces: gravitational force (3) position dependent forces: spring force (b) case 2: m ≠ const, rocket equation
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1.7 Energy & work……………………………………………………....30 1.7.1 kinetic and potential energy, energy conservation 1.7.2 work and power 1.8 Collisions……………………………………………………………..34 1.8.1 1D collisions 1.8.2 3D collisions 2 OSCILLATIONS…………………………………………………......…40 2.1 Simple harmonic motion (oscillation): spring-pendulum………...42 2.1.1 spring pendulum (a) equilibrium (b) equation of motion (c) solution of the equ. of motion (d) discussion of the solution 2.1.2 simple pendulum 2.2 Overlapping oscillations…………………………………………….49 2.2.1 ω1 = ω2 2.2.2 ω1 ≠ ω2, ω1 ≈ ω2
2.3 Energy during harmonic oscillations……………………………....51 2.4 Free, damped oscillation…………………………………………....52 (a) equation of motion (b) solution of the equation of motion (c) case: small damping, γ < ω0 (d) case: large damping: γ > ω0 (e) case: critical damping (aperiodic): γ ≈ ω0 2.5 Enforced oscillations and resonance……………………………….58 (a) equation of motion (b) solution (c) stationary state
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3 WAVES AND ACCOUSTICS……………………………………….61 3.1 Harmonic waves……………………………………………………..63 3.2 Superposition of waves (interference)……………………………...66 3.3 Standing waves………………………………………………………67 3.3.1 Standing waves is different systems: resonance 3.4 Harmonic sound waves……………………………………………..71 3.5 Energy and intensity………………………………………………..74 3.5.1 Intensity and sound level at sound waves 3.6 Plane, circular and spherical waves………………………………..77 3.7 Doppler effect………………………………………………………..79 4 OPTICS………………………………………………………………..83 4.1 Introduction to electromagnetic waves…………………………….83 4.1.1 Maxwell's equations 4.1.2 electromagnetic wave-equation, speed of light 4.1.3 electromagnetic wave-functions 4.1.4 elm. waves in different media - phase velocity and refractive index 4.1.5 wavefronts and Huygen's principle 4.1.6 law of reflection and refraction 4.2 Geometrical optics…………………………………………………93 4.2.1 ray approximation 4.2.2 image formation by reflection: mirrors (a) plane mirror (b) spherical mirror
4.2.3 image formation by refraction at a spherical surface 4.2.4 image formation by thin lenses 4.2.5 optical instruments (a) human eye (b) simple magnifier (c) optical microscope (d) telescope 4.2.6 prism, dispersion, spectral analysis 4.3 Wave optics………………………………………………………...120 4.3.1 interference fringes from thin layers 4.3.2 diffraction slit 4.3.3 diffraction grating, spectral analysis 4.4 Quantum optics…………………………………………………….138 4.4.1 wave particle duality I 4.4.2 photons 4.4.3 photoelectric effect
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5 ATOMIC PHYSICS……………………………………………………146 5.1 Experiments that led to atomic models…………………………...146 (a) photoelectric effect (b) x-ray diffraction (c) Rutherford's scattering experiment (d) atomic gases in discharges - line spectra (e) Zeeman effect 5.2 Historic atomic models…………………………………………….151 (a) J.J.Thompson's "plum pudding" model (b) Rutherford's atomic model (c) Bohr's atomic model 5.3 Wave particle duality II: electrons, neutrons, αααα-particles, ... …160 5.4 Schrödinger's equation and quantum mechanics………………..163 5.4.1 Schrödinger's axiom 5.4.2 example: particle in a box 5.5 Atomic model in quantum mechanics…………………………….168 5.5.1 the hydrogen atom 5.5.2 atoms with more electrons 5.5.3 the periodic system 6 SPECIAL THEORY Of RELATIVITY………………………………180 6.1 Lorentz transformation……………………………………………181 6.2 Time dilatation……………………………………………………..182 6.3 Length contraction………………………………………………...184 6.4 Relativistic mass, momentum & energy………………………….184 7 NUCLEAR PHYSICS……………………………………………….…187 7.1 Properties of atoms………………………………………………...187 7.2 Properties of nuclei………………………………………………...191 7.3 Radioactivity……………………………………………………….195 7.4 Decay types…………………………………………………………199 7.5 Natural radioactivity………………………………………………204 7.6 Fusion & fission……………………………………………………206 Exemplary problems for the written test……….………………………..212
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Chapter 0. Introduction 0.1. Foreword
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0.2. Literature P.A.Tipler: "Physics for scientists and engineers" A.Beiser: “Modern Technical Physics”, Addison Wesley, 1988 F.Bueche: “Principles of physics”, Mc Graw Hill, 1977 G.A. Williams: “Elementary physics”, Mc Graw Hill, 1969 Van Neie,P.Riley: “OHANIAN´s physics”, W.W.Norton, 1989 more detailed: Keller, Gettys, Skove: “physics”, Mc Graw Hill, 1993
0.3. What is physics?
physics = natural science, which deals with the basic modules of our environment and with their interactions Goal: discovery legalities in diversity of natural phenomenons, explanation of observed phenomena with few physical laws How to do? →Experiment: o systematic question to nature o meaning: gain the possibility to predict physical processes
quantitately 0.4. physical quantities
• objective description of nature demands quantitative (=numerically)
ascertainable correlations: measurement – result
needs mesure unit to measure means: comparison of two physical quantities e.g.: How long is the distance between Essen and Duisburg? possible answers: 25 km, 5 hours walk, 250 x length of soccer fields,... • measuring units:
length: e.g. see above: “km”, “velocity of walking,”,
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“soccer field” • time: e.g. “pulse frequence”, “day of sun”,...
→ every physical quantity consists of a number and a unit! note: number can be a vector or scalar! e.g.:
i. time: How long does a soccer game take? → t = 90 min → physical quantity „time“ is a scalar
ii. position: Where does my neighbour sit? → from my point of view (reference point), she / he sits 1m on the left →physical quantity „position“ is a vector
iii. other scalar quantities: mass m , current I , electric charge Q , luminous intensity I ,...
iv. other vector quantites: force F, magnetic field B , momentum p , ...
physical laws: contain fundamental constants, the physical constants e.g.: interaction between two bodies with masses 1 2,m m with a distance r to each other is called gravitational force.
The norm is: 1 212 2F
m mr
γ= with 12F = gravitational force
from object 1 on object 2, γ = gravitational constant 0.4.1. basic units
• How many?
→ 3 basic quantites: length, time, mass → functional: 3 additional quantities: amount of substance, temperature, (electric) current
• Basic units in physics: SI-system (system internatonal d’unites)
[length] = m ("meter") [time] = s ("second") [mass] = kg ("kilogramm") [current] = A ("ampere") [termperature] = K ("kelvin") [amount of substance] = mol ("mole")
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• Definitions
1) 1 m is the distance light travels through space in the
time-interval 1
299792458 s.
2) 1 s is the time-interval a cesium-clock oscillates 9 192 631 770,0 times.
3) 1 kg is the mass of the platinum-iridium stab which is stored in Paris. It is used as the mass-standard.
4) 1 mol is the amount of substance of a system that contains as many particles as atoms in 0,012 kg 12C .
5) 1 K is the 273,15th part of the thermodynamic temperature of water at its triple-point (that means: 0 0C = 273,15 K, 0 K = -273,15 0C )
6) 1 A is the electric current traveling through 2 parallel conductors in the distance of 1m to each other in vacuum, if the force between the 2 conductors is
72 10−⋅ N/m. 0.4.2. derived units & physical constants
a) derived units every unit can be derived from the basic units:
[force] = N ("newton"), 1 N = 1 kg m / s2 [work] = J ("joule") 1 J = 1 N m (=[energy]) [power] = W ("watt") 1 W = 1 J / s [frequency] = Hz ("hertz") 1 Hz = 1 s-1 [pressure] = Pa ("pascal") 1 Pa = 1 N / m2 [charge] = C ("coulomb") 1 C = 1 As [voltage] = V ("volt") 1 V = 1 J / C [resistance] = Ω ("ohm") 1 Ω = 1 V / A [capacitance] = F ("farad") 1 F = 1 C / V [magnetic field] = T ("tesla") 1 T = 1 N / (A m) [magnetic flux] = Wb ("weber") 1 Wb = 1 T / m2 [inductance] = H ("henry") 1 H = 1 J / A2 ...to be continued...
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b) physical constants
Avogadro's number: NA = 6,022 × 1023 particles/mol Boltzmann's constant: Bk = 1,38 × 1023 J/K Gravitational constant: γ = 6,67 × 10-11 Nm2/kg2 Mass of electron: me = 9,109 × 10-31 kg proton: mp = 1,673 × 10-27 kg neutron: mn = 1,675 × 10-27 kg earth: mearth = 5,974 × 1024 kg Dielectric constant (vacuum): 0ε = 8,8542 × 10-12 As/(Vm) Permeability (vacuum): 0µ = 4π × 107 N/A2 Speed of light (vacuum): c = 3 × 108 m/s Unified mass unit: u = 1,66 × 10-27 kg Earth-acceleration: g = 9,81 m/s2 ... to be continued....
0.4.3. angular units - flat angles
FIG. 0-01. Degree and radian measure, α = + 90 ° = π/2 rad, β = 360 ° = 2 π rad = 2 π.
i. degree: full circle: 360 , [ ]β β= ° = °
ii. radian: L
radr
α = (rad = physical unit of angle, in equations
unitless), L : arc length, r : radius, unit: [ ] [ ] 1L m radα= → = =
e.g.: full circle: 2
2 2 2r
L rad r rad radrππ β π π= ⋅ → = = =
conversion between degree / radian: =2 360radα π = °
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Chapter 1. Translations – motion of a mass point model of a mass point: • motion of a body in force field • neglect of body’s dimension • body: dot-shaped with mass m concentrated in one single point
(mass point) 1.1. Coordinates, position-vector a) coordinates
i.cartesian coordinates (see FIG. 1-01a) 3D-space 3
1 2 3( ) : ( , , ) ( , , ) ...x y z x x x= = 2D ( 2 ,plane): 1 2( , ) ( , )x y x x= 1D ( ,plane): ( )x
ii.spherical coordinates (see FIG. 1-01b) 3D: ( , ,r ϑ ϕ )
cos coscos sinsin
x r
y r
z r
ϑ ϕϑ ϕϑ
===
2D: ( ,r ϕ ) – polar coordinates iii.cylindrical coordinates (see FIG. 1-01c)
3D: ( , ,h zϕ )
cossin
x h
y h
z z
ϕϕ
===
b) position vector contains the coordinates: r( ) ( ( ), ( ), ( )) ( ( ), ( ), ( )),...t x t y t z t r t t tϑ ϕ= = r( )t : track in space = path of mass-point Translation = motion of a mass point while traversing its pathway r( )t
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FIG. 1-01. (a) Cartesian coordinates [1.1], (b) spherical coordinates [1.2], (c) cylindrical coordinates [1.2].
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1.2 tracks
FIG. 1-02. 1D uniform motion of a mass point along the x-axis.
FIG. 1-03. 2D uniform circular motion. (a) x(t) = R cos(ωt) in m, R = 2 m, ω = 2π/4s, (b) y(t) = R sin(ωt) in m, (c) r(t) in m with position-vector r(t ≈ 0,67 s) and the time-dependent argument ϕ(t ≈ 0,67 s) ≈ 60 °.
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1.3. Position, velocity & acceleration 1.3.1. 1D motion 1D-motion = motion along one axis
FIG. 1-04. 1D motion. (a) displacement ∆x = x2 - x1. (b) Average velocity in the time interval [t1,t2], vav = ∆x/ ∆t.
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FIG. 1-05. 1D motion, instantaneous velocity v = dx / dt.
FIG. 1-06. To ex. 1-06: Average and instantaneous velocity in 1 dimension.
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description: time t = parameter
position and displacement: 2 1 2 1
r = r( ) ( )r(t) = r r
t x t
x x
=∆ − = −
average velocity: av 2 1
r 1v x x
t t∆= = −∆ ∆
(instantaneous) velocity: 0
r( )v lim : r( ) ( ) :
t
t t dxt x t
t dt∆ →
+ ∆= = = =∆
average acceleration (vector): av
va
xt t
∆ ∆= =∆ ∆
acceleration (vector): 2 2
0
va lim : v = r(t) ( ) /
tx t d x dt
t∆ →
∆= = ∆ = =∆
1.3.2. 3D motion 3D-motion = motion along three axis (=: vector, consists of amount and direction)
position and displacement: 2 1
2 1 2 1
2 1
r = r( ) ( ( ), ( ), ( ))
r(t) = r r
t x t y t z t
x x
y y
z z
=−
∆ − = − −
average velocity: 2 1
av 2 1
2 1
r 1v
x x
y yt t
z z
− ∆ = = − ∆ ∆ −
(instantaneous) velocity: 0
( ) /r( )
v lim : r( ) ( ) /( ) /
t
x t dx dtt t
t y t dy dtt
z t dz dt∆ →
+ ∆ = = = = ∆
average acceleration: av
va
t∆=∆
acceleration:
2 2
2 2
02 2
/v ( ) x ( )v
a lim : v = r(t)= v ( ) x ( ) /
/v ( ) x ( )
x x
y yt
z z
d x dtt t
t t d y dtt
d z dtt t∆ →
∆ = = ∆ = = ∆
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FIG. 1-07. Example for displacement, velocity and acceleration in 2 dimensions. (a) Position r and displacement ∆r. (b) Average velocity vav, its components vav,x, vav,y and its direction Θ.
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1.4. special translations 1.4.1. uniform motion motion with constant velocity = uniform motion with a(t) 0=
How to get velocity and position?
0
v va(t) a(t)dt= dt= v v(t)
v(t)= a(t)dt= 0 v const
d dd
dt dt
dt
= → =
→ ≡ =
analog:
0 0 0r( ) v( ) v v rt t dt dt t= = = +
note: 0 0v , r : velocity and position at 0t = , set by boundary conditions
FIG. 1-08. example for an uniform motion. 1.4.2. motion with constant acceleration
0
0 0
20 0 0
a(t) a const 0
v( ) a(t) = a v
r( ) v(t) = 1/2a v r
t dt t
t dt t t
= = ≠
→ = +
→ = + +
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FIG. 1-09. Example for a motion with constant acceleration. 1.4.3. circular motion motion with non – constant acceleration
FIG. 1-10. Circular motion. (a) ∆S = S2 - S1: circular arc, ϕ: angle, r: position-vector, R: radius, C: center = origin. (b) r: Position, v: track velocity, a: acceleration with tangential and centripetal parts ac and at.
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a) general case:
important quantities: see FIG. 1-10a
• radius: R = constant, [R]=m • angle: ( ), [ ] radtϕ ϕ ϕ= = or °
note: 0ϕ > mathematicly positive direction → counter clockwise
• (instantaneous) angular velocity: ( ) ( )d
t tdtϕω ϕ= =
units: [ ] rad / rpmsω = = (revolutions per minute)
• average angular velocity: ( )ttϕω ∆=
∆
• (instantaneous) angular acceleration:
2
rad( ) ( ) ( ), [ ]
dt t t
dt sωα ω ϕ α= = = =
• average angular acceleration: ( )ttωα ∆=
∆
• circular arc: S R ϕ∆ = ∆ • position-vector: see FIG. 1-10b
cos ( )
ˆr( ) r( )sin ( )
tt R R t
t
ϕϕ
= = ⋅
with r( )t = unit vector in r -direction
• velocity (track velocity):
( )
cos ( ) -sin ( ) ( ) -sin ( )ˆv( )=r( )=R v( )
cos ( ) ( ) cos ( )sin ( ) t
dt t t tdtt t R R R t
d t t tt
dtϕ ω
ϕ ϕ ϕ ϕω ω
ϕ ϕ ϕϕ =
⋅ = = = ⋅
direction:
cos ( ) sin ( )ˆ ˆr( ) v( )
sin ( ) cos ( )
cos ( )( 1)sin ( ) sin ( )cos ( ) 0ˆ ˆr( ) v( )
t tt t
t t
t t t t
t t
ϕ ϕϕ ϕ
ϕ ϕ ϕ ϕ
− ⋅ = ⋅
= − + =→ ⊥
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• acceleration:
2
ˆv r
aa
( )sin ( )va( )
( )cos ( )
( )sin ( ) ( )cos ( ) ( )( )cos ( ) ( )sin ( ) ( )
sin ( ) cos ( )( ) ( )
cos ( ) sin ( )
ct
t td dt R
t tdt dt
t t t t tR
t t t t t
t tR t R t
t t
ϕ ϕϕ ϕ
ϕ ϕ ϕ ϕ ϕϕ ϕ ϕ ϕ ϕ
ϕ ϕα ω
ϕ ϕ
− = =
− − = −
− = −
decomposition in two parts: a( ) = a ( ) + a ( )t ct t t
i. tangential acceleration (= track acceleration)
t
t
t
ˆa ( ) ( )v( )
ˆ ˆdirection : a ( ) v( ) (parallel to v( ))
norm: ( ) ( )
t R t t
t t t
a t R t
α
α
=
↑↑=
ii. centripetal acceleration:
2c
t
2 2c
v
ˆa ( ) r( )
ˆ ˆdirection : a ( ) r( ) (antiparallel to r( ))
points to the center of the circle
norm: ( ) v /R
t R t
t t t
a t R Rω
ω
ω=
= −
↓↑
= =
• note: formulars for uniform motion and motion with constant acceleration can be transferred: circular arc, track velocity and track acceleration
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b) uniform circular motion: (t)=0α
0
0
0 0
0
0 0 0 0
( ) ( ) 0
( ) ( )
( ) 0, ( )
( ) ( )
t
v
t t dt dt
t t dt t
a R t v R t R t
S t R t R t R v t s
ω α ω
ϕ ω ω ϕ
α ω ω
ϕ ω ϕ
→ = = =
= = +
→ = = = =
→ = = + = +
c) circular motion with constant angular acceleration:
0
20 0
20 0 0
( ) const
( ) ( )
( ) ( ) 1/ 2
, , 1/ 2t
t
t t dt t
t t dt t t
a R v R t R S R t R t R
α α
ω α α ω
ϕ ω α ω ϕ
α α ω α ω ϕ
= =
→ = = +
→ = = + +
→ = = + = + +
1.4.4. summary a lot of similarities in formulars of straight & circular translations: uniform motion uniform circular motion a( ) 0t = ( ) 0, a 0ttα = =
0v( ) v = constt = 0
0
( ) const
v( ) const
t
t R
ω ωω
= == =
0 0r( ) v rt t= + 0 0
0 0
( )
( )
t t
S t R t R
ϕ ω ϕω ϕ
= += +
motion with constant acceleration circular motion with constant
angular acceleration 0a( ) a constt = = 0( ) consttα α= =
0 0v( ) + vt a t= 0 0
0 0
( ) +
v( ) +
t t
t R t R
ω α ωα ω
==
200 0
ar( ) t +v r
2t t= + 20
0 0
200 0
(t)= t +2
( ) t +2
t
S t R R t R
αϕ ω ϕ
α ω ϕ
+
= +
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1.5. assembled motion / coordinate transformation 1.5.1. systems with constant velocity
FIG. 1-11. Assembled motion. System ' has velocity (acceleration) u (w0) relative to system . consider: 2 coordinate systems , 'Σ Σ , 'Σ has constant velocity u relative to Σ (see FIG. 1-11)
• position vector: r = r' + S → position-vectors combine! uniform motion: 0S = u St +
at 00 : ' : S 0 S = u t t= Σ = Σ = → • velocity:
r
v (r' + S) v' + ud ddt dt
= = = → velocity combines
• acceleration:
u = const
v v 'a (v ' + u) = =a '
d d ddt dt dt
= = → acceleration is equal
1.5.2. systems with constant acceleration
consider: 2 coordinate systems , 'Σ Σ , 'Σ has constant acceleration 0ω relative to Σ (see FIG. 1-11)
• position vector: r = r' + S → position-vectors combine!
motion with constant acceleration: 200 0S( ) = + u S
2t t t
ω +
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at 200 00 : ' : S 0 S( ) = + u
2t t t t
ω= Σ = Σ = →
• velocity:
r
v (r ' + S) v ' + ud ddt dt
= = = → velocity combines
0 0u = utω + • acceleration:
0
va (v ' + u)=a '+
d ddt dt
ω= = → acceleration is different!
→ forces different (see below) 1.6. forces Why does a body move? e.g.: Why does the earth circle around the sun? Answer: Newton (1643-1727): Origin for a body to change its stable of motion is an interaction between the body and its environment physical description:
• interaction = F force (between body and environment) • stable of motion: momentum p vm=
m : mass of body, v : velocity, unit :
[p]kg m
s=
1.6.1. Newton’s laws
1) An object stays at rest or in uniform motion unless acted on by an external force. note: object, which isn’t effected by an external force = free object →different phrasing of Newton 1: The momentum of a free object remains constant
2) definition of force:
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The origin of a change in the momentum of a body is an external
force acting on the body. The force is defined by p
Fddt
= ,
unit: 2
[F] (Newton)
kg mN
s= =
note: v
p= v F vd dm
m mdt dt
→ = +
o mass is a function of velocity or time o examples: bodies with velocity = speed of light (relativistic problems), rocket
most of cases: v
constant F ad
m m mdt
= → = =
• principle of superposition: forces don’t affect each other → if abody is acted on by n different forces iF , 1,..., ,i n= the net force netF is the
superposition: net i1
F Fn
i=
=
Newton 2: net i1
p=F F
n
i
ddt =
=
• equilibrium of forces
net i1
F F 0n
i=
= = → a free body is under equilibrium of forces
3) actio = reactio Forces always occur in equal and opposite pairs. If object A exerts a force ABF on object B, an equal but opposite force BAF is exerted by object B on object A: AB BAF F= − .
1.6.2. Consequences from Newton’s laws conservation of momentum consider: system of n particles with different momentum ip Newton 2 for every particle:
i i net1 1
p p = F F F
n ni i
i i
d ddt dt= =
→ = = (= net force on the system of n particles)
Newton 1: if the system is free, the net force on the system is zero: netF 0=
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Free system of n particles: 1
p0
ni
i
ddt=
=
1
1 1 1
p0 const
pdt = p p const p
ni
i
n n ni
i i neti i i
ddt dt
dt
dd
dt
=
= = =
→ = =
→ = = =
→ the net of a free system is constant! 1.6.3. equation of motion of a particle acted by a force
case 1: m = const (=constant forces): Newton2: net
vF
dm
dt=
net net
net 0
0
net 0 0
F Fx x=
1v( ) F v
r( ) v( ) r
1 [ F ] v r
d ddt dt
dt m dt m
t dtm
t t dt
dt dt tm
→ → =
→ = +
→ = +
= + +
(1) net 0F F= = constant forces Newton 2: 0F am=
20 0 00 0 0
F F F1a , v( ) v , r( ) v r
2t t t t t
m m m= = = + = + +
(2) position – dependent – forces: gravitational force 2 objects with masses 1m and 2m in the distance gravitate toward each other with the gravitational force (see FIG: 1-12)
1 22,1 1,2 1 22
ˆF (r) r gm m
mr
γ= =
(force from object 2 acting on object 1) γ : gravitational constant
1,2r : unit-vector pointing from object 1 to object 2 r : distance between objects
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FIG. 1-12. (a) Gravitational force F12 from object 1 acting on object 2. (b) Gravitational force F21 from object 2 acting on object 1. Correlation between the two forces given by Newton's 3. law: F12 = - F21.
(3) position dependent forces: spring force
FIG. 1-13. Force exerted by a spring to a mass connected to the spring. (a) Horizontal arrangement, rest position = point of equilibrium, where FS = 0. (b) Mass is elongated at ∆x.
system: mass connected to a spring (see FIG. 1-13) a) rest position: 0x : no force is exerted → mass is free b) mass is pulled, spring is streched at he distance 0x x x∆ = − → Hooke’s law: 0F (r - r )s k= − (here 0( ) ( )sF x k x x= − − )
o Fs points to rest position 0r o k spring-constant, depends on spring’s material,
geometry,... unit: [ ] /k N m=
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o x∆ is called elongation o Hooke’s law is important also to describe interaction
of atoms in crystal case 2: m ≠ const
net
p vF v
d d dmm
dt dt dt= = +
e.g. rocket equation: consider: rocket with mass m in free space. Rocket burns fuel with
constant burn-rate 0dm
Rdt
= < , mass of rocket: 0( )m t m Rt= +
conservation of momentum: time rocket
net: p =p ( )t t time rocket
net fuel: p =p ( ) + ut dt t dt m+ + fuel , u :m Rdt= − velocity of fuel conservation: rocket rocket
netp = const p ( ) - u - p ( ) 0t dt Rdt t→ + ⋅ =
rocket rocket
rocket
netNewton 2
p ( ) - p ( ) u
p vu = F v
t dt tR
dtd d dm
R mdt dt dt
+→ =
→ = = +
netF : 2 parts: netF changes the momentum: one part changes velocity
(= thrust, th
vF
dm
dx= ), other part changes mass (= v
dmdt
)
v
= u - v = (u - v)d dm dm
m Rdt dt dt
→
coordinate system:
29
FIG. 1-14. Coordinate transformations and rocket equation. (a) Velocities in systems at rest and at motion. Σ : external system, rocket has velocity v , fuel has velocity u
'Σ : rocket-system: travels with v relative to Σ , rocket at rest, fuel has constant velocity u' (relative to rocket) velocities: v + u' = u u' = u - v→
th
vu' = F
dm R
dt→ = (rocket-equation in free space, th = thrust)
note: net th
vF u v F v
d dm dmR m
dt dt dt= = + = +
thv = u - F (u - u') :dm
R Rdt
→ = force for mass – change
note: in reality you must add an gravity term:
net
gravity-term
vF u + g v
d dmR m m
dt dt= = +
many difficult calculations later:
… 20
0 0
r( ) u' (1 + ) ln(1 + ) - u' 1/ 2gm R R
t t t t tR m m
→ = +
30
1.7. Energy & work energy & work: important concepts, closely associated work in physics: a force does work when it acts on an object energy: when work is done from one system to another, energy is transferred between two systems forms of energy:
• kinetic energy: associated with motion of an object • potential energy: associated with configuration of a system • thermal energy: associated with random motion of molecules, closely connected to system’s temperature • chemical energy: associated with chemical composition of system • rotational enrgy: ... • ...
1.7.1. kinetic and potential energy, energy – conservation consider: system of body with constant mass m acted on by net force netF Newton 2: netF am=
netv F v am→ ⋅ = ⋅
FIG. 1-15. Potential and kinetic energy.
31
system: see FIG. 1-15 0t : object (here: dangerous spider) at position 0r r( 0)t= = , acted on by
net force net 0F ( )t (reference time) t : object at position r r( )t= (here: directly above the poor golf-player), acted by force netF ( )t and has passed from 0r to r through (arbitrary) curve C( )t (time is arbitary)
0 0
t t
nett t
v F dt' v a dt '
I II
m→ ⋅ =
potential energy:
0 0 0
t t r
net net net 0substitution definitiont t r ,C
v F dt' = r F dt' = F r (r ) (r)pot potI d E E= ⋅ ⋅ ≡ −
substitution: 0 0
rv r r ' r, r( ) = r , r( ) r
ddt d t t
dt= = → = =
(r)potE : potential energy of an object acted on by a force only
determinable for a constant 0(r )potE (r)potE depends on netF acting on the body → depends on system’s configuration (body in “force -field” of the earth, charge in “Coulomb-field” of other charges, mass connected to a spring, mass in force-field of both, spring + earth, ...)
0(r )potE : reference-point, point of 0(r )potE =0, fit to the system, 0(r )potE sets potential energy (r)potE for all points r for examples see FIG. 1-16 kineticl energy:
0 0
t v2 2
0substitutiont v ,C
1 1v a dt' = v' v' v v
2 2II m m d m m= ⋅ = −
substitution: 0 0
va v v = a ', boundaries: v( ) = v , v( ) v
'd
d dt t tdt
= = → =
32
kinetic energy:
22
p v
p1/2 v
2 kinm
m Em=
= =
associated with momentum of body (momentum = measure fort he motion)
FIG. 1-16. (a) Potential energy of a body somewhere close to the earth's surface. (b) Potential energy of a body far away from the earth's surface. conservation of energy:
2 2pot 0 pot 0
2 2pot 0 pot 0
(r ) (r) 1/ 2 v 1/ 2 v
1 1v (r) v (r ) constant
2 2
I II E E m m
m E m E E
= → − = −
→ + = + = =
energy of an enclosed system remains constant!
→ If you know the energy at he reference point (= 20 pot 0
1v (r )
2m E+ ), you
know the energy at all times! →calculations can be simplified using the conservation of energy →potential and kinetic energy can be converted into each other
33
1.7.2. work and power
1) work: 2
1
r
net netr C
: F r = F rW d d=
mechanical work, done by net force netF on object along the path C( )t (displacement of object from 1r to 2r ) work and potential energy:
definition of (r)potE : 0r
0 netr
(r) (r ) F (r') r'pot potE E d− =
0 0
1 2
0 2 2
1 0 1
r r
1 2 net netr r
r r r
net net netr r r
(r ) (r ) F (r') r' - F (r') r'
= F (r') r'+ F (r') r' = F (r') r' = W
pot potE E d d
d d d
→ − =
→ work is a change in potential energy :
i. W > 0: 1 2(r ) (r ) :pot potE E> potential energy decreases, net force does work on object (usually: object gets faster)
ii. W < 0: 1 2(r ) (r ) :pot potE E< potential energy increases, object does work against the net force
unit: [W]=Nm=J (“Joule”)
2) power:
definition:
2 2
1 1
r
net net netsubstitutionr
F r F v r F vt
t
dW d dP d d
dt dt dt= = = ⋅ = ⋅
substitution : r
v=ddt
unit : [P] = J/s = W („Watt“) note: difference between power and work: consider: 2 motors lift a given load a given distance →same amount of work → the one that does it in the least time supplies more power
34
1.8 Collisions phenomenon:
• 2 (or more) objects approach and interact strongly for a short time • during brief collision time: external forces are much smaller than
interaction forces → all external forces can be neglected • interaction forces: 12 21F = -F (Newton 3)
• total momentum:
2 112 21
2 11 2
p pF = -F
p p0 p p constant
d ddt dt
d ddt dt
⇔ = −
→ + = → + =
→ total momentum remains constant! • small collision time → neglection of displacement during collision • before and after collision: interaction of objects small → only external net force • Examples: cue ball hitting object ball (billiard), baseball being hit
by bat, dart colliding with dart board, comet swinging aroung sun,..
• elastic collision: total kinetic energy remains constant (remember: during collision netF =0 = constantpotE→ )
• inelastic collision: kinE constant≠ → no conservation of energy (but conservation of momentum!)
e.g. perfectly inelastic collision: 2 objects stick together after collision
1.8.1. Collisions in one dimension (1D-collision) consider: object 1: mass: 1m , initial velocity: 1v , final velocity 1,fv (directly after collision) (v < 0: object moves to the left) object 2 : 2 2 2 f, v ,vm conservation of momentum:
1 1, i 2 2, i 1 1, f 2 2, f
momentum directly momentum directlybefore collision after collision
v v v vm m m m+ = +
→ 1 equation with 2 unknown variables 1 f 2 fv , v → 2. relation necessary
35
a) perfectly inelastic collision (see FIG. 1-17)
FIG. 1-17. Perfectly inelastic collision [1.3].
particles stick together after collision: 1 2 1 + 2, fp = ( )v m m+
conservation of momentum: 1 1i 2 2i 1 2 1 + 2, f v + v ( )vm m m m= +
1 1 i 2 2 i 1 + 2, f
1 2
v + vv
m m
m m→ =
+
conservation (loss) of energy: kinetic energy is being converted into thermal energy, energy of deformation,...
: converted (lost) energy: initial kinetic energy
: final kinetic energyi i f i f
f
E
E E E E E E E
E
∆ = + ∆ → ∆ = −
here: 2 2 21 1 2 2 1 2 1 2,1/ 2 v 1/ 2 v 1/ 2( )vi i fE m m m m +∆ = + − +
for example see FIG. 1-28:
FIG. 1-18. 1D perfectly inelastic collision [1.3].
36
b) partly inelastic collision objects don’t stick together after collision. Kinetic energy of object 1 is transferred into kinetic energy of object 2 and thermal energy or energy of deformation. conservation of momentum:
1 1 i 2 2 i 1 1, f 2 2, fv + v v vm m m m= + energy-consideration (no conservation!):
2 2 2 2
1 1 i 2 2 i 1 1, f 2 2, f
1 1 1 1v + v v v2 2 2 2m m m m E= + +∆
for example see FIG. 1-21:
FIG. 1-19.. Ballistic pendulum [1.4].
c) elastic collision
conservation of energy:2 2 2 2
1 1 i 2 2 i 1 1, f 2 2, f
1 1 1 1v + v v v2 2 2 2m m m m= + (i)
conservation of momentum: 1 1 i 2 2 i 1 1, f 2 2, fv + v v v m m m m= + (ii)
2 2 2 2
1 1 i 1 f 2 2, f 2, i
1 1 i 1 f 1 i 1 f 2 2, f 2, i 2, f 2, i
( i ) (v + v ) (v - v )
(v - v )(v + v ) (v - v )(v + v ) (i ')
m m
m m
→ =
⇔ =
1 1 i 1 f 2 2, f 2, i( ii ) (v - v ) (v - v ) (ii ')m m→ =
37
1 i 1 f 2 f 2 i
1 f 2 f 2 i 1 i
2 f 1 i 1 f 2 i
( i')v + v v + v(ii')
v v + v v (iii a)
v v + v v (iii b)
→ =
→ = −
→ = −
(iiia) in (ii’):
1 1 i 2, f 2 i 1 i 2 2 f 2 i
1 1 i 1 2, f 1 2 i 2 2 f 2 2i
1 1 i 2 1 2 i 2 f 1 2
[v - (v + v - v ) ] = (v - v )
2 v v v v - v
2 v ( )vv
m m
m m m m m
m m mm m
⇔ − − =
+ −→ = +
(iiib) in (ii’):
1 1 i 1 f 2 1 i 1 f 2 i 2 i
1 1 i 1 1 f 2 1 i 2 1 f 2 2 i
1 i 1 2 2 2 i 1 f 1 2
[v - v )] [( v + v - v ) - v ]
v v v v - 2 v
v ( ) 2 vv
m m
m m m m m
m m mm m
=
⇔ − = +
− +→ = +
38
1.8.2. Collisions in three dimensions (3D-collisions)
a) perfectly inelastic collision (for example see FIG. 1-20)
FIG. 1-20. (a) 3D perfectly inelastic collision. note: same results as in one dimension, but the velocities and momentums are now vectors and no more scalars (here: directions are given by means of sign)!
b) Elastic collisions object 1: 1 1 1f, v , v ;m object 2: 2 2 2 f, v , v ;m conservation of momentum:
1 1 i 2 2 i 1 1 f 2 2 f
1 1 i 1 f 2 2 f 2 i
v + v = v + v
(v v ) (v v ) (i')
m m m m
m m⇔ − = −
conservation of energy:
2 2 2 21 1 i 2 2 i 1 1 f 2 2 f
1 1 i 1 f 1 i 1 f 2 2 f 2 i 2 f 2 i
1/ 2 v + 1/2 v = 1/2 v + 1/2 v
(v v )(v + v ) (v v )(v + v ) (ii')
m m m m
m m⇔ − = −
big problem: (ii’)/(i’) not allowd of vectors! → solution only for special case:
2 i 1 2v 0, m m m= = = (almost the case for billiard):
39
momentum: 1 1 i 2 2 i 1 1 f 2 2 f
0
1 i 1 f 2 f
v + v = v + v
v v v (i'')
m m m m
→ = +
energy: 2 2 2 2
1 1 i 2 2 i 1 1 f 2 2 f
0
2 2 2 1 i 1 f 2 f
1/ 2 v + 1/2 v = 1/2 v + 1/2 v
v v v (ii'')
m m m m
→ = +
(i’’) → velocities from a triangle (ii’’) → Pythagoras: angle between 1 fv and 2 fv 90= °
FIG. 1-21. 3D elastic collision for the special case: 1 2m m m= = , 2 0iv = .
40
Chapter 2. Oscillations (harmonic motions) What and when? →Oscillations occurs when a system is disturbed from a position of stable equilibrium What happens? →Restering force acts on system, tries to restore the position of equilibrium. E.g.
i. spring pendulum: mass connected to a spring FIG. 2-01. Spring pendulum: body with mass m connected to a spring. (a) System in position of static equilibrium x0. (b) Disturbed system, mass moved out of equilibrium position, spring force FS serves as restoring force. (c) Oscillating system, mass swing between xmas and - xmas.
a) position of stable equilibrium: 0x b) system is disturbed: mass is moved out of equilibrium
position →spring force Fs acts as restoring force pointing to
0x c) if mass is released, oscillation starts without damping:
oscillations lasts eternally, system is never at static equilibrium, mass swings between maximum values maxx and
maxx− (turning points) around positions of equilibrium.
41
ii. simple pendulum: mass hanging vertically on a string in gravity-field of the earth (surface)
FIG. 2-02. Simple pendulum: body with mass m vertically connected to a string in gravity field of earth (close to surface). (a) System in position of static equilibrium (x0,z0). (b) Disturbed system, mass moved out of equilibrium position, force FR (part of gravity force FG) serves as restoring force. (c) Oscillating system, mass swing between (xmax,zmax) and (-xmax,zmax).
a) position of stable equilibrium ( 0 0,x z ) b) mass moved out of equilibrium → restoring force acts
(FR =part of gravity force FG ) c) mass is released, system swings between turning points
( max max,x z ); (- max max,x z ) around ( 0 0,x z ) iii. further examples:
→boat bobing up and down →clock pendulums swinging back and forth →strings of musical instruments vibrating →oscillations of air molecules are sound-waves →oscillations of electric currents are radios
Why is study of oscillations important?
• Understanding of electromagnetic oscillations (cell phones, remote control, data transfer, …)
• Understanding and elimination of undesired resonance-phenomena at engines, buildings, bridges, …
42
2.1. Simple harmonic motion easiest oscillation, e.g. simple pendulum, spring-pendulum
FIG. 2-03. Spring pendulum - equilibrium. (a) Horizontal arrangement. (b) Vertical arrangement. 2.1.1. spring pendulum
a) equilibrium example: spring-pendulum (see FIG. 2-03)
position of equilibrium: net force on object netF 0= horizontal arrangement: spring at ist natural length l, spring exerts no force on object (see FIG. 2-03a) vertical arrangement: spring elongated by l∆ ,spring force and gravity force compensate each other, netF 0= (see FIG. 2-03b) (see example 1-15: determination of spring constant k out of l∆ )
b) equation of motion Newton 2: netF am=
spring: Hooke’s law: s 0F (r-r )k= − with 0r r− : elongation of spring
43
FIG. 2-04. Spring pendulum - non-equilibrium. (a) Horizontal arrangement. (b) Vertical arrangement.
i. horizontal arrangement: (see FIG. 2-04a)
net 0 0
2
2
0
0
ˆF (r r ) ( )x a
ˆa x, =
ˆ ˆx+ / ( )x = 0
+ / ( ) = 0
k k x x m
d xa a x
dta k m x x
x k m x x
= − − = − − =
→ = =
⇔ −→ −
coordinate system: 0 0 / 0x x k mx≡ → + =
equation of motion for a harmonic motion (free oscillator)
ii. vertical arrangement: (see FIG. 2-04b) spring force:
s
0
ˆF =- (z-l)= - ( - ) z positive directionk k z l z<
−
gravity force: G ˆF = g=- z negative directionm mg z −
net force: net s G ˆ ˆF = F +F = - ( - )z - zk z l mg
equilibrium: net s G ˆ ˆF = 0 = F +F = - ( - )z - z
ˆ ˆ- ( - )z - z = 0
k z l mg
k z l mg→
→ determination of k ! (exercise) Here:
44
net
0 0trick!
0
0 0
0!
2
0 2Newton 2
0
ˆ ˆF = - ( - )z - z
ˆ ˆ = - ( - )z - z
ˆ ˆ ˆ= - ( - )z - ( )z - z
ˆ ˆ= - ( - )z = a a = z , =
/ ( - ) 0
k z l mg
k z l z z mg
k z z k z l mg
d zk z z m a a z
dtz k m z z
=
+ −
−
→ =
→ + =
Coordinate system: 0 0 / 0z z k mz≡ → + =
→equation of motion the same for vertical and horizontal arrangement!
c) Solution of the equation of motion / 0x k mx+ = : homogenous differential equation 2.order
approach: 2
( ) , const
( ) , ( )
t
t t
x t e
dxx t e x t e
dt
λ
λ λ
λ
λ λ
= =
→ = = ⋅ = ⋅
2 2 2/ 0 / 0 /t te k me k m k mλ λλ λ λ→ + = ⇔ + = → = −
with i = imaginary unit, 2 1i = − (respectively (0,1)i = is the unit vector of the complex plane) / /
1,2 1 2/ ( ) , ( )i k m t i k m ti k m x t e x t eλ + ⋅ − ⋅→ = ± → = =
general solution is complex linear-combination: 1 2( ) , ,c c c cx t u x v x u v= + : complex numbers polar rotation: ,u vi i
c cu ue v ueζ ζ−≡ ≡
/ /
( / ) ( / )
( ) u v
u v
i ii k m t i k m t
i k m t i k m t
x t ue e ve e
ue ve
ζ ζ
ζ ζ
⋅ − ⋅
⋅ + ⋅ +
→ = +
= +
one single requirement: ( )x t has to be real!
( ) cos( / ) sin( / )
cos( / ) sin( / )u u
v v
x t u k m t iu k m t
v k m t iv k m t
ζ ζ
ζ ζ
= ⋅ + + ⋅ +
+ ⋅ + − ⋅ +
(we used the Euler-notation: cos sinie iϕ ϕ ϕ= + ) imaginary part has to vanish: sin( / ) sin( / ) 0u vu k m t v k m tζ ζ⋅ + − ⋅ + =
satisfied, if / 2, u vu v A ζ ζ ε= = = =
45
( / ) ( / )( ) / 2 / 2
cos( / ) ( )
i k m t i k m tx t A e A e
A k m t x t
ε ε
ε
⋅ + − ⋅ += +
= ⋅ + =
general solution of the equation of motion for harmonic motion (we
used cos 1 / 2( )i i
e eϕ ϕ
ϕ−
= + )
d) Discussion of the general solution ( ) cos( / )x t A k m t ε= ⋅ +
46
FIG. 2-05. Simple harmonic motion. (left) Period T and amplitude A. (right) Phase-shift ε and phase difference 2π between two points of same state of motion. ( 2 , 7,4 / , 3 , / 3 60A cm k N m m kg ϕ π= = = = = ° )
• amplitude: , [ ]A A m= , meaning: maximum elongation • phase: ( ) / , [ ( )] radt k m t tψ ε ψ= ⋅ + = • phase-shift (or phase constant): , [ ] radε ε = meaning: set by boundary-condition: position and velocity at 0t =
• period: , [ ]T T s= meaning: time interval after which oscillator reaches the same state of motion: ( ) ( ), v( ) v( )x t T x t t T t+ = + =
satisfied if phases differ by 2π :
( ) 2 ( )
/ 2 / ( )
2 /
2 2
/
t t T
k m t k m t T
k m T
mT
kk m
ψ π ψε π ε
ππ π
+ = +
⇔ ⋅ + + = + +
→ = ⋅
→ = =
• frequency: , [ ] 1/f f s Hz= = („Hertz“) meaning: number of oscillations per second
/
1/2k m
f Tπ
→ = = frequency of a free oscillator
e.g. 311 10 / 1000 /
1T ms f s s
ms= → = = =
→1000 oscillations per second
• angular frequency (angular velocity): rad
, [ ]s
ω ω =
meaning: phase change done in 1 second:
2 rad 2
2 rad 2f ft T T
ψ π πω π π∆= = = = =∆
here: 0
0
/2 2 /
2
/
k mf k m
k m
ω ω π ππ
ω
= = = =
→ =
angular frequency of a free oscillator
47
→ in the following: 0( ) cos( )x t A tω ε= +
• velocity of the oscillator: 0 0v( ) ( ) sin( )t x t A tω ω ε= = − + • acceleration of the oscillator: 2
0 0( ) v( ) ( ) cos( )a t t x t A tω ω ε= = = − + 2.1.2. simple pendulum equation of motion: Newton 2: netr = Fm
FIG. 2-06. Simple pendulum.
sin
ˆr rcos
L Lϕ
ϕ = = −
, L = thread length
48
2
net G R N
cos cos ˆv = r = tsin sin
-sin ˆ ˆ ˆr t + t - rcos
F =F =F +F
L L L
L L L L
ϕ ϕ ϕϕ ϕ
ϕ ϕ ϕϕ ϕ
ϕ ϕ ϕ ϕϕ ϕ
⋅ → = = ⋅
⋅ → = = ⋅
→ restoring force: RˆF sin tmg ϕ= −
→ normal force: N ˆF cos rmg ϕ=
→ 2ˆ ˆˆ ˆ t - r sin t + cos rmL mL mg mgϕ ϕ ϕ ϕ= −
centripetal force (= thread force LF ):
L
2
F
2
2
0
ˆ ˆr cos r (Newton 3)
ˆ ˆ 0 = cos r + rˆ ˆ ˆ ˆt = - sin t (+ cos r + r)
t : = - sin
mL mg
mg mL
mL mg mg mL
L g
ϕ ϕ
ϕ ϕϕ ϕ ϕ ϕ
ϕ ϕ
−
− =
→→
→
series expansion for 3 5 7
sin ...3! 5! 7!ϕ ϕ ϕϕ ϕ= − + − +
for 22,4 : sinϕ ϕ ϕ≤ °
mistake: 3 3 21% ( /3!=0,39 /3! 0,01 with 22,4 in rad)
360πϕ ϕ≤ = ° ⋅
°
0gL
ϕ ϕ→ + = equation of motion for simple pendulum
(only for small angles) solution: analogous to spring-pendulum: 0 0 0( ) cos( / ) cos( )t g L t tϕ ϕ ε ϕ ω ε= ⋅ + = +
period: 2 /T L gπ=
frequency: 0 00
1 1/ , 2 /
2f g L T L g
Tπ
π= = =
angular frequency: 0 /g Lω =
49
2.2. Overlapping oscillations nature: not simple harmonic oscillations but complex forms of periodic of simple harmonic motions, can be described at the superposition of simple harmonic oscillations with different amplitudes, frequencies, phase-constants (=overlapping oscillations) in general: direction of oscillations different →3D overlapped oscillatins in the following: 1D superposition: direction of oscillation the same superposition of 2 simple harmonic oscillations:
1 2
1 1 1 1
2 2 2 2
( ) ( ) ( )
( ) cos( )
( ) cos( )
x t x t x t
x t A t
x t A t
ω ϕω ϕ
= += += +
general case: complicated: see FIG. 2-07a: 1 1 1 2 2 22 , / 2 , / 6; 2,5 , 3 / 4 , / 4 A cm Hz A cm Hzω π ϕ π ω π ϕ π= = = = = =
FIG. 2-07. Overlapping oscillations. (a) General case. (b) ω1 = ω2 = ω0. (c) Constructive overlap. (d) Destructive overlap.
50
2.2.1. overlapping oscillations with 1 2 0ω ω ω= =
See FIG. 2-07b:
1 2 0 1 22 , 1,5 , / 2 , / 6, /5 A cm A cm Hzω π ϕ π ϕ π= = = = = −
1 0 1 2 0 2
0
( ) cos( ) cos( )
... cos( )
x t A t A t
B t
ω ϕ ω ϕω ε
= + + += = +
2 2 21 2 1 2 1 2
1 1 2 2
1 1 2 2
2 cos( )
sin sintan
cos cos
B A A A A
A AA A
ϕ ϕϕ ϕεϕ ϕ
= + + −+=+
special cases: i. constructive overlap: see FIG. 2.07c
1 2 1 2
2 2
0
2 , / 6
4 2 4 , / 6( ) 2 cos( )
A A A cm
B A B A cm
x t A t
ϕ ϕ ϕ πε ϕ π
ω ϕ
= = = = = =
→ = → = = = =→ = +
ii. destructive overlap: see FIG. 2.07d
1 2 1 2 1
1 2
1 2
0 1 0 1
0 1 0 1cos( ) cos
2 , / 6, 7 / 6 : phase difference of
sin sin 02 4 , tan ?
cos cos 0
( ) cos( ) cos( )
cos( ) cos( ) 0
A A A cm
B A cm
x t A t A t
A t A tψ π ψ
ϕ π ϕ ϕ π π πϕ ϕεϕ ϕ
ω ϕ ω ϕ πω ϕ ω ϕ
+ =−
= = = = = + =− →→ = = =− →
→ = + + + += + − + =
2.2.2. overlapping oscillations with 1 2 1 2,ω ω ω ω≠ ≈ :beats
simplification: 1 2 1 22 , 0A A cm ϕ ϕ ϕ= = = = =
1 2 1 2
1 2 1 2
changes amplitude slowly but periodically
1 2
( ) cos( ) cos( ) [cos( ) cos( )]
... 2 cos( ) cos( )2 2
cos( )2
x t A t A t A t t
A t t
C t
ω ω ω ωω ω ω ω
ω ω
= + = +− +=
+=
x→ oscillates with mean (angular) frequency 1 2
2ω ωω +=
51
→amplitude 1 22 cos( )2
C A tω ω−= oscillates slowly with beat
frequency 1 2
2b
ω ωω −=
FIG. 2-08. Beat. Amplitude oscillates with beat frequency (ω1 - ω2)/2. Oscillation has mean frequency (ω1 + ω2)/2. 2.3. Energy during harmonic oscillations kinetic energy:
0
2 2 20 0
( ) cos( )
1/ 2 ( ) 1/ 2 sin ( )kinx t A t
E mx t mA tω ϕ
ω ω ϕ= +
= = +
potential energy:
0 0
0 0
20
0 , 0
02 2 2
00 ( ) cos( )
2 2 20 0
/
2 2 20 0 0
( ) ( ) F ( )
1/ 2 1/ 2 cos ( )
1/ 2 cos ( )
( ) 0 ( ( )) 1/ 2 cos ( )
x x
pot pot s xx x
x x t A tx
k m
pot pot
E x E x dx k x x dx
k xdx kx kA t
m A t
E x E x t m A t
ω ϕ
ω
ω ϕ
ω ω ϕ
ω ω ϕ
≡ = +
=
− = = − −
= − = = +
= +
≡ → = +
energy conservation:
2 2 2 20 0 0
1
2 20
( ( )) ( ( )) 1/ 2 [sin ( ) cos ( )]
1/ 2 const (energy of the oscillator)
kin potE x t E x t mA t t
mA E
ω ω ϕ ω ϕ
ω
+ = + + +
= = =
52
mean values over one period T :
2 2 20 0
0 0
2 220
00
1/ ( ) 1/ 1/ 2 sin ( )
sin ( )2
T T
kin kin
T
I
E T E t dt T mA t dt
mAt dt
T
ω ω ε
ω ω ε
= = +
= + =
Idea: 0 00 '
sin( )sin( ) ... / 2T
u v
I t t dt Tω ε ω ε= + + = =
You need two integration by parts, use 2 2cos 1 sinx x= − and add 2sin xdx on both sides. Then divide with 2 and simple calculations
gives T/2. 2 201/ 4kinE m Aω→ =
analogous calculation: 2 201/ 4pot kinE m A Eω= = !
2.4. free, damped oscillation
a) equation of motion general case: oscillation in surrounding medium (gas, fluid), not in space →additional force: friction Stoke’s law of friction:
rF v , const, [ ] /fr
db b b b kg s
dt= − = − = =
→Friction acts against the motion equation of motion (for spring pendulum):
net s 0F = F + F a= - (r-r ) - rfr m k b=
1D-problem: 0 0(r-r ) - , r , ax x x x→ → →
0 0 :
0
x kx bx mx
b kx x x
m m
≡ − − =
→ + + =
abbreviations: 20 / , 2 /k m b mω γ= =
53
202 0x x xγ ω→ + + =
equation of motion for a free, damped oscillation
b) solution of the equations of motion
approach: 2
2 2 2 20 1,2 0
( ) ( ) ( )
2 0
t t tx t e x t e x t eλ λ λλ λ
λ γλ ω λ γ γ ω
= → = → =
→ + + = → = − ± −
general solution: 2 2 2 2
0 0( ) ( )( ) t tc cx t u e v eγ γ ω γ γ ω− + − − − −= +
, :c cu v complex numbers
c) case: small damping: 0γ ω<
2 2 2 21,2 0 0
0 0
( ) ( )( ) ( )i t i t t i t i tc c c c
i i
x t u e v e e u e v eγ ω γ ω γ ω ω
λ γ γ ω γ ω γ γ ω< >
− + − − − −
→ = − ± − = − ± − = − ±
→ = + = +
analogous to free oscillation: 2 2
0( ) cos( ) , tx t Ae tγ ω ε ω ω γ−= + = −
discussion:
FIG. 2-09. (a) Free, undamped oscillation. (b) Free, damped oscillation with γ = 5 % ω0.
• ε : phase shift • A: amplitude of undamped oscillation
• tAe γ− : amplitude of damped oscillation, decays with damping-constant ([ ] 1/ )sγ γ = to zero
• ω : reduced (angular) frequency of damped oscillation,
54
• 02f f
ωπ
= < (frequency), 0 01/ 1/T f T f= > = (period)
• See FIG. 2-09: values: 22 , 0, 3 , 3 / 29,61 /A cm m kg k N m N mε π= = = =
(a) free, undamped oscillation: 20 0 0 00 0, / 2 / 4 , 0,25 , 1/ 4b k m Hz f Hz T f sγ ω π= → = = = = = =
(b) free, damped oscillation:
2 20 0 0 0 0
3 0,05 / 0,47 /
0,05 < = , , =
b kg s kg s
T T f f
π
γ ω ω ω ω γ
= ⋅
→ = → −
• decay time τ : time when amplitude is decreased to 1e
of
its initial value:
0 1
( ) / 1/ ln(1/ )
ln(1/ ) ln(1) ln( ) 1
x A e Ae e e e
e e
γτ γττ γτ
τ τγ γ γ
− −= = ⇔ = ⇔ = −
−⇔ = = = =− −
• half time HT : time, when amplitude is at its half initial
value: as above
ln(2)HT
γ→ =
d) case: large damping: 0γ ω>
FIG. 2-10. Free, damped oscillation, case: γ > ω0. Boundary conditions: (a):
0( 0) 0, ( 0)x t v t v= = = = . (b) 0( 0) , ( 0) 0x t x v t= = = = .
55
2 2 2 21,2 0 0,
( ) ( )t t tc cx t e u e v eγ α α
λ γ γ ω γ α α γ ω− −
= − ± − = − ± = −
→ = +
→ general solution is complex! Real solution only for special cases analytically determinable:
i. case: 0( 0) 0, ( 0) vx t x t= = = = 0 0 0( 0) ( ) 0 ( 1)c c c cx t e u e v e u v iγ α α− ⋅ ⋅ − ⋅= = + = + =
0
0( 1)
( ) ( ) ( )
( 0) ( ) ( )
v( ) v ( 2)
t t t t t tc c c c
c c c c
c c c ci
x t e u e v e e u e v e
x t u v u v
u v u v i
γ α α γ α αγ α αγ α
αα
− ⋅ ⋅ − ⋅ − ⋅ ⋅ − ⋅= − + + −= = − + + −
= − = → = +
0 0
0 0 0
0 0
0
2sinh( )
2 20
v v( 2) in ( 1) : 0 ( 3)
22v v v
( 3) in ( 1) : ( 4)2 2 2
v v( ) ( )
2 2v
( ) sinh( )
, 2
x x
c c c
c
t t t
t
x e e
i i v v v i
i i u i
x t e e e
x t e t
bm
γ α α
γ
α α
α α α
α α
αα
α γ ω γ
−
− ⋅ ⋅ − ⋅
− ⋅
= −
+ + = → = −
= − =
→ = −
→ =
= − =
ii. case: 0( 0) , ( 0) 0x t x x t= = = = 0( 0) ( 1)c cx t u v x ii= = + =
00
( 1)
( 0) ( ) ( )
( ) 0 ( 2)
c c c c
c c c cii
x t u v u v
xx u v u v ii
γ αγγ αα
= = − + + −
= − + − = → = +
56
00
0 0
00 0
( 2) in ( 1) :
2 (1 ) 1/ 2 (1 ) ( 3)
( 3) in ( 2) : 1/ 2 (1 ) =1/ 2 (1 )
c c
c c
c
xii ii v v x
v x v x ii
xii ii u x x
+ + =
⇔ = − ⇔ = −
= + − +
0 0
0 0
0 0
( ) 1/ 2 (1 ) 1/ 2 (1 )
1/ 2 ( ) 1/ 2 ( )
cosh( ) sinh( ) ( )
t t t t
t t t t t t
t t
x t e x e e x e
x e e e x e e e
x e t x e t x t
γ α γ α
γ α α γ α α
γ γα α
− − −
− − − −
− −
→ = + + −
= + + +
= + =
e) case: aperiodic special case: 0γ ω=
λ γ→ = −
→ general solution: ( ) tcx t u e γ−=
Problem: case: 0 0
0
0
( 0) ( )
( 0) 0, ( )
( 0) 0!
t
t
x t x x t x e
x t x t x e
x t x
γ
γγγ
−
−
= = → =
= = = −= = − ≠
solution: cu time-dependent: ( )c cu u t=
2
2
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) 2 ( ) ( )
tc
t tc c
t t t tc c c c
t t tc c c
x t u t e
x t u t e u t e
x t u t e u t e u t e u t e
u t e u t e u t e
γ
γ γ
γ γ γ γ
γ γ γ
γγ γ γ
γ γ
−
− −
− − − −
− − −
→ =
= −
= − − +
= − +
set into equation of motion 202 0x x xγ ω+ + = :
2 20
2 20
here:
2
1 1 2
1 2 1,2
( ) 2 ( ) ( )
2 ( ) 2 ( ) 0
( ) 0 ( ) 0
( ) , ( )
( ) ( ) , ( ) real real
t t t tc c c c
t tc c
tc c
c c
t
u t e u t e u t e u e
u t e u t e
u t e u t
u t c u t c t c
x t c t c e x t c
γ γ γ γ
ω γ
γ γ
γ
γ
γ γ ω
γ γ
− − − −
=
− −
−
−
− + +
+ − =
⇔ = ⇔ =→ = = +
→ = + →
57
(i) case: 0( 0) 0, ( 0) vx t x t= = = =
1 2
2 2
2 0 0
( 0) 0 ( ) ,
( ) ( )
( 0) v ( ) v
t
t t
t
x t c x t c te
x t c e c t e
x t c x t te
γ
γ γ
γ
γ
−
− −
−
= = = → =
= + −
= = = → =
(ii) case: 0( 0) , ( 0) 0x t x x t= = = =
2 0 1 0
1 1 0
1 0 1 0
0 0 0
( 0) ( ) ( ) ,
( ) ( )( )
( 0) 0
( ) ( ) (1 ) ( )
t
t t
t t
x t c x x t c t x e
x t c e c t x e
x t c x c x
x t x t x e x t e x t
γ
γ γ
γ γ
γγ γ
γ γ
−
− −
− −
= = = → = +
= + + −= = − = → =
→ = + = + =
FIG. 2-11. Free, damped oscillation, case: γ = ω0. Boundary conditions: (a):
0( 0) 0, ( 0)x t v t v= = = = . (b) 0( 0) , ( 0) 0x t x v t= = = = .
58
2.5. enforced oscillations and resonance System: mass connected to a spring in surrounding medium, upper end of spring is moved periodically up and down by external force ext 0 ˆF =F cos( )xtω
FIG. 2-12. Enforced oscillation [2.1].
a) equation of motion analogous to chapter 2.1:
20
00
2 cos( )
F, / ,
2
x x x t
bk m
m m
γ ω κ ω
γ ω κ
+ + =
= = =
b) solution • inhomogenous differential equation • solution = general solution of homogenous equation + one solution of inhomogenous equation homogenous equation: 2
02 0x x xγ ω+ + =
→chapter 2.2: ( ) cos( )tf fx t A e tγ ω ε−= +
2 20fω ω γ= − (small damping)
inhomogenous equation: 2
02 cos( )x x x tγ ω κ ω+ + =
approach: real functions → complex functions 202 , i tz z z e z x iyωγ ω κ+ + = = +
with , i tc cz A e Aω= complex amplitude, follows
2, i t i tc cz i A e z A eω ωω ω= = −
59
2 20
2 2 2 20
( 2 )... , real amplitude
( ) (2 )i t
c
iA Ae Aωκ ω ω γω
ω ω γω− −→ = =
− +
a ib= +
2 2
2 2 2 20
2 20
( ) (2 )
2tan
( ) cos( )
A a b
ba
x t A t
κω ω γω
γωϕω ω
ω ϕ
→ = + =− +
→ = = −−
→ = +
→general solution ( ) ( ) cos( )fx t x t A tω ϕ= + +
c) stationary state
0 : ( ) 0
( ) cos( )
tft x t e
x t A t
γ
ω ϕ
−>> ∝ →
→ = +
• oscillation with frequency of driving force! • phase shift ϕ between driving force and oscillation
2 20
2tan
γωϕω ω
= −−
o dependent of damping constant γ o dependent of frequency of free undamped oscillation 0ω
FIG. 2-13. Enforced oscillation. (a) Phase-shift ϕ for different damping constants γ. For ω → ω0 : ϕ = ± π/2. (b) Amplitude A(ω) for different damping constants γ. Resonance for γ << ω0, ω → ω0 .
60
o dependent of driving force ω o see FIG. 2-13a: 0 / 2Hzω π= o for 0 : / 2 90ω ω ϕ π= = ± = ± °
(that means: 0F=F cos( ) ( ) sin( )t x t A tω ω→ = )
• amplitude: 2 2 2 20( ) (2 )
Aκ
ω ω γω→ =
− +
o dependent on amplitude 0F / mκ = of driving force o dependent of frequencies 0,ω ω o dependent of damping constant γ o see FIG. 2-13b: 0 0F 0,1 , 3 , / 2N m kg Hzω π= = = o resonance for 0 :γ ω< maximum of ( ) : ( ) 0A Aω ω =
2 20 0 0 2 , Rω ω γ ω γ ω→ = −
o for 00, : A( ) γ ω ω ω= → → ∞ → disaster! (bridges collapse under marching soldiers, glasses crash at high voices,..)
61
Chapter 3 Waves and acoustic
wave: transport of energy and momentum without transporting matter mechanical wave: transport via disturbance in a medium e.g. string: disturbance = charge in shape of string from its equilibrium shape sound wave: disturbance = local density changes out of equilibrium density electromagnetic wave: transport without medium, disturbance = oscillating electric & magnetic fields transverse waves: disturbance perpendicular to propagation e.g. string: segments of string move perpendicular to string, wave propagates down the string
FIG. 3-01. Transverse mechanical wave [3.1]. longitudinal waves: disturbance parallel to propagation e.g. soundwaves: molecules of gas, liquid, solid move back and forth along line of propagation
FIG. 3-02. Longitudinal mechanical wave [3.2].
62
general wave function: consider: 2 coordinate systems , 'Σ Σ Σ : fixed system, at t = 0 wave has shape ( )y f x= ,
'Σ : system moves with wave’s velocity v, wave has shape ( ')f x for all times coordinate transformation: ' v ' vx x t x x t= + → = − →shape of wave for all times: ( v )f f x t= −
FIG. 3-03. General wave function. Disturbance can be described by function
( )f f x vt= − . discussion: → f depends on position x and time t (compare oscillation: ( )x t depends on time only) →string: ( ) ( , )f x vt y x t− = : elongation differs with position and time →v: phase velocity, depends on medium (see TABLE 3-01) →direction of propagation: v > 0: wave moves to right v < 0: wave moves to left medium vPh [m/s] aluminium 3040 plumb 1960 glass 5640
63
water 1402 mercury 1450 air 331,5 hydrogen 1284 oxygen 316 helium 965
TABLE 3-01. Phase velocities in different solids, liquids, gases [3-03] wave equation:
( , ) ( ) ( )y x t f x vt f u= − =
chain rule
1
( v )y df u df x t dfx du x du x du
∂ ∂ ∂ −→ = ⋅ = ⋅ =∂ ∂ ∂
2 2 2
2 2 2chain rule
y d f u d fx du x d u
∂ ∂→ = ⋅ =∂ ∂
(i)
chain rule
v
vy df u dft du t du
−
∂ ∂→ = ⋅ = −∂ ∂
2 2 22
2 2 2
v
v vy d f u d f
t du t d u−
∂ ∂→ = − ⋅ =∂ ∂
(ii)
(i) = (ii) 2 2
2 2 2
1v
y yx t
∂ ∂→ =∂ ∂
wave equation
mechanical waves: v out of Newton’s laws (difficult!) electromagnetic waves: v out of Maxwell’s equations (later) solution: wave function ( , ) ( )y x t f x vt= − Chapter 3.1. Maxwell’s equations one solution of wave equation: plane harmonic wave
( , ) sin[ ( v ) ] ( v )y x t A k x t f x tδ= − + = − illustration: “photo” of the wave: t = const
64
FIG. 3-04. "Photo" of a wave ( , )y x t . discussion: • A: amplitude, unit: sound-waves: [A] = [pressure] = N/m², string: [A] = [elongation] = m elctric field: [A] = N/C (later), ... • k : wave number • wave length λ : distance between two maxima:
after x λ= phase (argument of sin-function) is 2π higher: 0 0( v ) 2 ( v )
2
2
k x t k x t
k
k
δ π λ δπ λ
πλ
+ + + = + + +⇔ = +
⇔ =
• phase shift δ : set by boundary condition • oscillation: oscillators at different positions x fulfill oscillations with different phase – shift δ depending on position, same frequencies!
65
FIG. 3-05. Oscillation for oscillators at different positions x0. (a) "Photo" of the wave at t = 0 (black) and t = 0,25 s (grey). (b) x0 = 0, (c) x0 = 3 m, (d) x0 = 4,5 m. oscillation at 0x x= :
0 0 0
0sin( ) sin( )
( , ) sin[ ( v ) ] sin[ v ]
sin[ v ] sin[ ]
y x t A k x t A kx k t
A kx k t A tα π α
δ δπ δ ω ε
= −
= − + = − += − + − = +
(simple harmonic oscillation) 0 0( )kx xε π δ ε→ = − − = : phase shift (position dependent, independent of wave length because we are in the same medium) vkω = : (angular) frequency equal for every oscillator, depends on wave number and phase velocity
e.g.
2 11 Hz 3,14 Hz
2
0,52
ms m
f s
πω π
ωπ
= =
→ = =
see FIG. 3-05:
66
(a) “photo”: 0, 0t δ= = (wave) phase: kxψ = (b) 0 00 0 = ( 0, ) - sin( )x y x t A tψ ε π ω= → = → → = =
(c) 0
0
33 3 = - 2
( 3 , ) sin( )
x m m
y x m t A t
πψ π ε ππ
ω
= → = = →
→ = =
(d) 0
0
9/ 29 / 2 4,5 =-9/2 + -7 / 2
( 4,5 , ) sin( 7 / 2 ) sin( 1/ 2 )
x m m
y x m t A t A t
πψ π ε π π ππ
ω π ω π
= → = = → =
→ = = − = +
• usual equation: 2
( , ) sin( ) , , vy x t A kx t k kπω δ ωλ
= − + = =
• plane wave: for given 0 0 0: ( , , , )x x y x y z t= = const → same elongation for all oscillators in yz – plane • spherical wave: wave generated by point source, propagation in all direction (e.g. sound wave coming out of speakers)
0( ) ,
( , ) sin( )
AA A r x r
rA
y r t kr tr
ω δ
→ = →
→ = − +
→ y depends on distance oscillator – wave source →amplitude decreases with increasing distance from source (e.g. sound!) 3.2 Superposition of waves (interference) analogous to chapter 2.2: Overlapping oscillations
1 1 1 1 1
2 2 2 2 2
1 2
( , ) sin( )
( , ) sin( )
( , ) ( , ) ( , )
y x t A k r t
y x t A k r t
y x t y x t y x t
ω δω δ
= − += − +
= +
special cases:
i. destructive interference: 1 2 1 2 1 2 1 2 1 2 1 2, ( ), ( ), (phase shift!)
( , ) 0
A A k k f f
y x t
λ λ ω ω δ δ= = → = = → = =→ =
ii. constructive interference:
67
1 2 1 2 1 2 1 2, , , (waves are in phase)
( , ) 2 sin( )
A A k k
y x t A kx t
ω ω δ δω δ
= = = =→ = − +
iii. beats:
1 2 1 2 1 2 1 2
1 2 1 2 1 2 1 2
, , , =0
( , ) 2 cos( )sin( )2 2 2 2
A A k k
k k k ky x t A x t x t
ω ω δ δω ω ω ω
= = = =− − + +→ = − −
FIG. 3-06. Beat. (a,b) "Photo" of the wave at t = 0. (c,d) Oscillation at x0 = 0.
3.3 Standing waves special case of interference:
1 2 1 2 1 2 1 2, = , = , 0, 0 A A A k k k ω ω ω δ δ= = = = = ≠ 1 2v 0, v 0> < (wave 1 moves to right, 2 moves to left)
1 2
1 2 1 21 2
( , ) sin( ), ( , ) sin( )
( , ) ( , ) ( , ) 2 cos( )sin( )2 2
y x t A kx t y x t A kx t
y x t y x t y x t A
ω ω δψ ψ ψ ψ
→ = − = + +− +→ = + =
with phases 1 2,ψ ψ :
68
1 2
1 2
1 2
cos( ) cos( )
1 2
( ) 2
( ) 2
2 cos( ) cos( ) cos( )
2 2 2
sin( ) sin( )2 2
( , ) 2 cos( )sin( )2 2
x x
kx t kx t t
kx t kx t kx
tt
kx
y x t A t kx
ψ ψ ω ω δ ω δψ ψ ω ω δ δ
ψ ψ ω δ δω
ψ ψ δ
δ δω
− =
− = − − + + = − −+ = − + + + = −
− −→ = − = +
+→ = +
→ = + +
,x t→ in different functions! • nodes: positions nx for which ( ) 0ny x = for all times!
sin( ) 0 , 2 2
2
n n
n
kx kx n n
nx
k
δ δ π
δπ
+ = ⇔ + = ∈
−→ =
• antinodes: positions anx for which amplitude of wave is at
maximum: ( , ) 2 cos( )2any x t A tδω= + for
sin( ) 1,
2(2 1)
(2 1)2 2 2
an
an an
kx n
nkx n x
k
δ
δ π π δ
+ = ± ∈
+ +→ + = + ⇔ =
3.3.1 Standing waves in different systems – resonance most cases: standing waves only possible for particular wavelengths and frequencies ,R Rfλ mode: standing wave with special ,R Rfλ , mode is resonance of the system with ,R Rfλ : resonance frequency/wavelength
69
FIG. 3-08. Standing waves for different boundary conditions. (a) "Guitar string" as a system with two fixed ends [3-4]. (b) "Organ pipe" as a system with one loose and one fixed end [3-5]. A: position of antinodes, N: position of nodes.
a) “guitar - string”: standing waves between two fixed ends (see FIG. 3-08a) boundary – condition:
( 0, ) 0, ( , ) 0, length of string
( 0, ) 2 cos( )sin( 0 ) 02 2
sin( ) 0 2 ,2 2
y x t y x L t L
y x t A t k
n n n
δ δω
δ δ π δ π
= = = = =
→ = = + ⋅ + =
→ = → = ⇔ = ⋅ ∈
( , ) 2 cos( )sin( ) 0sin( ) 0 ,
( ) ,
2 22R
y x L t A t n k L n
k L n k L n m m
kL m n z z
k L z L z
ω π ππ π ππ π
λπ π πλ λ
→ = = + ⋅ + =→ ⋅ + = → ⋅ + = ∈→ = − = ∈
= → = → =
(z=0 senseless) →Fundamental, First harmonic: z = 1 2R Lλ→ =
→Second harmonic: z = 2 R Lλ→ =
.... higher harmonics: z = 1 11
2, 1: R R
LL z
z zλλ λ λ→ = ≠ = = =
70
wavelength of higher harmonics are integer deviders of fundamental wavelength
frequencies: v / v
v2 2
R
R
ff L z z
fλ= → = =
1
1
1
v v1:
2 2
v v1: , 2,3,4,...
2 2R RR
z L ff L
zz L f z zf f z
f L
= = → =
≠ = → = = = =
→ frequencies of harmonics are integers of frequency of fundamental wave functions: 0 0n δ= → = fundamental: 1 1 1( , ) 2 cos( )sin( )y x t A t k xω= z harmonic: ( , ) 2 cos( )sin( )z z zy x t A t k xω=
b) “pipe of an organ”: standing waves between one fixed and one loose end (see FIG. 3-08b) boundary conditions:
1
2
( 0, ) 0 2 ,
( , ) 2 cos( ) maximum amplitude2
sin( + )= 1 (2 1) , 2 2
( ) (2 1) , 2 2 2
2 2(2 1) (2 1)
2 4
n
z
R
y x t n n
y x L t A t
kL kL n m m
kL m n m n z z
k L z L z
δ π
δ πδω
δ ππ
π π ππ π π
λπ π πλ λ
= ⋅
= = → = ⋅ ∈
= = + =
→ ± ⇔ + = + ∈
⇔ = + − = − + = + ∈
= → = + ⇔ = +
11 1
fundamental
0 : 4 , 42 1zz L L
zλλ λ λ= = → = =+
(2z +1): odd integer, z = 1,2,3,...
frequencies: vv f
fλ λ= ⇔ =
71
1 1
fundamental
v v(2 1) (2 1)
4 4
v(2 1)
4
R
z
L z f zf L
f f z ff
→ = + ⇔ = +
→ = → = +
wave functions:
fundamental: 1 1 1
2 2 2
( , ) 2 cos( )sin( )
( , ) 2 cos( )sin( )
y x t A t k x
y x t A t k x
ωω
==
3.4 Harmonic sound waves Generation: Tuning fork or loudspeaker vibrating with simple harmonic motion What happens? • air molecules next to vibrating source oscillate with simple harmonic
motion around equilibrium position. • “Neighbour-molecules” of vibration source collide with their
neighbours causing them to oscillate • sound wave propagates
72
FIG. 3-09. Harmonic sound wave. (a) Displacement of air molecules from equilibrium position. (b) Some representative air molecules before sound wave arrives. (c) Air molecules near points x1, x2, x3 after sound wave arrived. (d) Density of air molecules. (e) Pressure change (proportional to density) [3.6]. Displacement of air molecules:
0
0, before wave arrived, air-molecules at equilibrium position( , ) (see FIG. 3-09b)
sin( - ), after wave arrived (see FIG. 3-09a,c)s x t
s kx tω
=
Density and pressure: minima at 2 , kx t n nω π− = ⋅ ∈ maxima at (2 1) , kx t n nω π− = − ∈ compare displacement:
maxima at (2 1) , 2
kx t n nπω− = − ∈ and n uneven
73
minima at (2 1) , 2
kx t n nπω− = − ∈ and n even
FIG. 3-10. Oscillators with masses m at position x and another oscillator with mass m' while wave passes. • pressure and density are -90° out of phase with displacement
(see FIG. 3-09d,e) 0
0
0
( , ) sin( ), / 2
sin( / 2)
cos( ) ( , )
p x t p kx t
p kx t
p kx t p x t
ω ε ε πω π
ω
= − + = −= − −
= − =
pressure of air after sound wave arrived • note: before sound wave arrived: ( , ) 0p x t = discussion: • displacement, density, pressure varies in propagation direction (x-
direction) → longitudinal wave • usually: source = point source, emits wave in all directions
→ sound wave = spherical wave with 0( , ) sin( )p
p r t kx tr
ω= −
(see chapter 3.6)
74
3.5 Energy, energy density and intensity mechanical wave: • oscillators with mass m vibrate → oscillators have kinetic and
potential energy of oscillation (see chapter 2.3) • via coupling oscillation is transferred from one oscillator to the other
in wave-propagation direction, oscillator stays at its position → transfer of energy with phase velocity in direction of wave propagation
FIG. 3-11. Sound wave as spherical wave traversing through space. energy of oscillator at position r :
2 2 20
0 0
( , ) 1/ 2 cos ( )
( ( , ) sin( ) ( , ) cos( ))kinE r t m x kr t
y r t x kr t y r t y kr t
ω ωω ω ω
= −= − → = − −
0
0
0 0
0
2 2 2 20
harmonic motion with r 0 0
( , ) ( , ) ( ') '
' ' 1/ 2 1/ 2 sin ( )
r
pot potr
r
E r t E r t F r dr
D r dy Dr m x kx tω ω
≡
=
− = −
= = = −
(in the last step we used 0( , ) sin( ), /r x t x kx t D mω ω= − = )
75
→ 2 201/ 2tot kin potE E E m x Eω= + = =
problem: definition of the “unit-oscillator”: see FIG. 3-10: Energy E is different for oscillators with masses , 'm m ! solution: energy-density: mass can be expressed by density /m Vρ =
2 201/ 2E V xρ ω→ =
consider: infinitely small unit-oscillator: 0m → or 0V → : 2 2
00lim ( ) 1/ 2V
dEE V x
dVη ρω
→→ = = =
η is the energy-density of a wave, independent of single oscillators! intensity:
definition: power of wave
intensity = area through which wave is transmited
2, [ ]P W
I IA m
= =
power: , , vdE dE
p dV A dtdt dV
η= = = (see FIG. 3-11)
v/ v =
dE dV A dtI P A I
Adt Adt Adtη η η→ = = = = = ⋅
→ intensity = energy-density x phase velocity 3.5.1. Intensity and sound-level of sound waves intensity: intensity-intervall hearable for human ear: from 12 2
0 10 /I W m−= : hearing threshold to 2 12
01 / 10I W m I= = : pain threshold
76
TABLE 3-2. Intensity and sound- or intensity-level of common sounds [3.7]. sound-level (or sound-intensity): psychological sensation of loudness varies approximately logarithmically with sound-intensity →sound-level β is measured in logarithmic dB – scale (dB = decibel):
0
10lg( ), [ ]I
dBI
β β= = (unitless “unit” - similar to rad-scale)
12 20 10 /I W m−= : hearing threshold
e.g. Niagara Falls: 3 210 /I W m−=
3 23 12 9
12 20
9
9
10 /10 10
10 /
10lg(10 ) 90
I W mI W m
dBβ
−− +
−→ = = =
→ = =
77
3.6 Plane, circular and spherical wave wave-function: 0( , ) sin( )y r t y kr tω= −
intensity: vdE
IAdt
η= = ⋅ , v = phase - velocity
total energy that passes in the time t∆ through area A perpendicular to propagation-direction:
0 0
v v
' v ' ( ) v'
t t
dE dEA
Adt dtdE
dt A dt E t A tdt
η η
η η∆ ∆
= ⇔ =
→ = → ∆ = ∆
( , , vA η = constant)
2 2 2 20 01/ 2 ( ) 1/ 2 vx E t A x tη ρω ρω= → ∆ = ∆
energy-conservation: t∆ = const → E = const! a. plane waves
FIG. 3-12. Plane wave moving in x - direction.
ˆ ˆv v x xk k k kr kxω= = ⋅ → = ⋅ → =
x
: wave propagation in x-direction 0A A= = constant 2 2
0 01/ 2 vE A txρω→ = ∆ = const (if t = const)
0
0
const
( , )= sin( )
x
r x t x kx tω→ =→ −
78
b. circular waves
FIG. 3-13. Circular wave traveling in radial direction away from a point source (e.g. stone dropped into water).
cos
ˆ ˆv v r, r (polar coordinates)sin
k r
k r kr
ϕω
ϕ = = ⋅ =
→ =
!
2 2 2 20 0
constant
0
2 , = constant
E=1/2 2 rz v z v constant
1x
A r z z
trx t rx
r
π
π ρω π ρω
= ⋅
→ ⋅ ∆ = ∆ ⋅ =
→ ∝
0( , )= sin( )x
r x t kx tr
ω→ −
79
c. spherical waves (see FIG: 3-11)
cos cosˆ ˆv v r, r sin sin (spherical coordinates)
sink r
k r kr
ϕ ϑω ϕ ϑ
ϑ
= = ⋅ =
→ =
2
!2 2
0
0
4
E= constant
1x
A r
const r x
r
π=
→ ⋅ =
→ ∝
0( , )= sin( )x
r x t kx tr
ω→ −
3.7 Doppler-effect wave source and receiver move relative to each other →observed frequency and emetted frequency different! source and receiver toward each other → frequency higher source and receiver away from each other → frequency lower e.g. passing ambulance: pitch shifts o case: moving source, receiver at rest → wavelength changes
→ find new wavelength 'λ , then v
''
fλ
= (v=phase velocity)
80
FIG. 3-14. Doppler effect. Wave-source moves with uS to the right. (a) Behind the source: wave expanded, λ' > λ0. (b) In front of source: wave compressed, λ' < λ0 [3.8].
• number of waves emitted in : t N f t∆ = ∆ • displacement of first emitted wavefront after : vwft s t∆ = ∆
• displacement of first emitted source after : s st s u t∆ = ∆ distance between source & first wavefront
new wavelength = number of emitted waves
→
0 0
0 0 0 0
s0
0
v v'
v= v/
v-u' (1 ) '
v/ v
wf s s s
s
s s t u t usN N f t f
f f
u
λ
λ λ
λ λ λλ
− ∆ − ∆ −∆= = = =∆
→ =
→ = = − =
81
→ frequency: 0
0
vv ' ' ' v/ ' ... '
(v- ) / 1v
ss
ff f f
uu fλ λ= → = = = = =
−
FIG. 3-15. Doppler effect, moving source. The "new" wavelength λ' can be determined out of the distance ∆s between source and receiver. discussion: source towards receiver: 0su > 0(1 / v) 0 ' , 'su f f λ λ→ − > → > <
e.g. passing ambulance: pitch decreases directly after passing o case: moving receiver, source at rest → number of waves passing receiver changes
FIG. 3-16. Doppler effect, moving receiver. The number of wavefronts passing changes. → determine total number of wavefronts passing receiver in time t∆ → frequency ' '/f N t= ∆
• number of emitted wavefronts 0 0N f t= ∆ • number of additionally observed wavefronts:
00 0 0v / v
R R Ru t u t u tsN f
fλ λ∆ ∆ ∆∆= = = =
• total number: 0 0 0 0' / v= (1 / v)R RN N N f t f tu f t u= + = ∆ + ∆ ∆ +
82
→ frequency: 0
'' (1 ) '
vRuN
f f ft
= = + =∆
→ wavelength:
0 0
0
v0
vv ' ' ' v/ ' '
(1 / v) 1vRfR
f f fuf u λ
λλ λ=
= → = = = =+ +
discussion: o receiver towards source: 0 0>0 (1+ /v)>1 ' , 'R Ru u f f λ λ→ → > < o receiver away from source: 0 0<0 ' , 'Ru f f λ λ→ < >
o general case: moving source: 0 0' /(1 / v), ' (1 / v)s sf f u uλ λ= − = − moving receiver: 0 0' (1 / v), ' /(1 / v)R Rf f u uλ λ= + = +
→ general case: 0
1 / v'
1 / vR
S
uf f
u+=−
, 0
1 / v'
1 / vs
R
uu
λ λ −=+
,s Ru u > 0: source and receiver moving towards each other in general: v ,s Ru u>>
83
Chapter 4 Optics
Chapter 4.1. Introduction to electromagnetic waves
Chapter 4.1.1. Maxwell’s equtions
• laws of electricity and magnetism (classical electromagnetism)
• analogous: Newton's laws of mechanics
vacuum: / t∇ × = −∂ ∂E B , 0 0 / tµ ε∇ × = ∂ ∂B E , 0∇ ⋅ =E , 0∇ ⋅ =B E: electric field, / q= CE F , [ ] N / C=E
CF : Coulomb force from charge Q on charge q, 20/(4 )Qq rπε=CF r ,
[ ] N=CF , attractive for equal signs of Q, q, repulsive for different signs
FIG. 4-01. (a) Repulsive Coulomb force of charge Q on charge q with same sign (Q, q < 0 or Q, q > 0). (b) Attractive Coulomb force of charge Q on charge q with opposite sign (Q < 0, q > 0 or Q > 0, q > 0). q: electric charge, [ ] Cq = (“Coulomb”), charge is bound to mass,
negatively charged body: excess of electrons (e-), positively charged body: lack of e-, fundamental unit charge: 191,602 10 Ce −= × , e.g. electron: eq e= − ,
proton: pq e= charge quantization: , 0, 1, 2,...Q Ze z= = ± ± , (charges appear as multipliers of fundamental unit-charge) B: magnetic field, [B] = T ("Tesla")
84
LF : Lorentz-force, ( )q= ×LF v B , v: velocity of body with charge q in magnetic field B, case: ⊥v B → FL: centripetal force (circular motion)
FIG. 4-02. Lorentz force FL on charge q moving with velocity v through constant magnetic field B. Lorentz force acts as centripetal force, q moves on circular track. 0µ : permeability (of free space), 7 2
0 4 10 N/Aµ π= × 0ε : dielectric constant (of free space), 12 2
0 8,8542 10 mA /kgε −= × ∇ : Nabla operator, ( / , / , / )x y z∇ = ∂ ∂ ∂ ∂ ∂ ∂ ∇ × E = rot E (rotation) ∇ E = div E (divergence) grad ∇ Φ = Φ (gradient) – scalar field
Chapter 4.1.2. electromagnetic wave-equation
Maxwell
Maxwell Maxwell 0 0
( ) ( )
B( ) ( B) ( / )t
t t tµ ε
∇ × ∇ × = ∇ ∇ − ∆ = −∆∂ ∂ ∂= ∇ × − = − ∇ × = − ∂ ∂∂ ∂ ∂
E E E E
E
=> 2 20 0 / tµ ε∆ = ∂ ∂E E ,
analogous: 2 20 0 / tµ ε∆ = ∂ ∂B B ,
85
2 2 2
2 2 2x y z∂ ∂ ∂∆ = + +∂ ∂ ∂
(Laplace-op.)
see chapter 3: wave-equation of mechanical waves:
2 2
2 2 2
( , ) 1 ( , )v
y x t y x tx t
∂ ∂=∂ ∂
solution: wave functions, mech. waves, travelling with phase-velocity v here: wave-equation of electromagnetic waves (in free space), phase-velocity v = speed of light c :
80 01/ 3 10 m/sc µ ε= = × (vacuum: speed of light)
wave-functions: ( , ) sin( )t tω δ= − +0E r E kr ( , ) sin( )t tω δ= − +0B r B kr Chapter 4.1.3. electromagnetic waves
FIG. 4-03. Electromagnetic wave, electric and magnetic fields swing in phase, E, B are perpendicular to each other and both to the propagation direction [3.9].
• consist of electric & magnetic field swinging in phase • directions: ⊥E B , , ⊥E B k (propagation-direction): elm. wave = transverse wave • electromagnetic spectrum: all wavelengths (frequencies) possible!
e.g. light: (400,700) nmλ ∈ (VIS), infrared (IR) (heat-radiation of bodies), ultraviolett (UV),
conversion between , fλ : c fλ=
86
TABLE. 4-1. The electromagnetic spectrum [3.10].
• production of elm. waves: (i) acceleration of free charges (e.g. radio-waves, bremsstrahlung as part of x-rays, synchrotron-radiation) (ii) electrons bound to atoms make transitions to lower energy states (e.g. x-ray line spectrum, heat as blackbody radiation, light waves)
87
Chapter 4.1.4. electromagnetic waves in different media
• electromagnetic wave forces electron sheath of atoms to oscillate,
or atomic core remains at its position
• accelerated charges send out radiation with lower phase velocity
v = c / n, n = refractive index, depend on material, wavelength, ...
e.g. vacuum n =1, air n =1,0003, glass n =1,515
• enforced oscillation; stationary case: frequency doesn’t change!
• v /
/nn
c n fn
c f
λλ λ
λ= = ⋅
== ⋅ ,
λ : wavelength in vacuum,
nλ : charged wavelength in medium with refractive index n
Chapter 4.1.5. Wavefronts and Huygen’s principle
Wavefront : All points with maximum amplitude
→wavefronts move with phase velocity v in propagation-
direction
FIG. 4-04. Wavefronts of a circular water wave generated by a point source moving up and down with simple harmonic motion [3.11].
88
shapes: (i) plane wave: plane
(ii) circular waves: circle
(iii) spherical waves: sperical
(iv)
FIG. 4-05. Part of a wavefront of a spherical electromagnetic wave before and after propagation through an unplane piece of glass. Behind the "obstacle" the wavefront is deformed [3.12].
Huygen’s principle of wave propagation:
FIG. 4-06. Huygens's principle for the propagation of waves.
89
FIG. 4-07. Huygens's principle of wave propagation for a (a) plane wave and (b) circular or spherical wave. The wavefront is the envelope of the wavelets outgoing from every point on the previous wavefront [3.13].
Chapter 4.1.6. Application of Huygen’s principle:
the law of reflection and refraction
Consider: plane wave hitting a plane surface
• part of wave is reflected, part is refracted
FIG. 4-08. Reflection and refraction. Plane wave hitting a plane surface at the incidence angle Θ1. The reflected wave is also a plane wave [3.13].
90
• plane wavefront AA’ striking mirror at point A
• angle of incidence: angle between the perpendiculars on
wavefront and mirror
• position of wavefront afte time t : construction of wavelets of
radius r c t= ⋅ on every point of AA’
→ (i) wavelets that do not hit the mirror:
wavefront BB’
(ii) wavelets that hit the mirror:
wavefront BB’’
→ after t wavefront B’B’’ isn’t plane
• analoguos: construction of wavefront C’C’’
• reflection-angle:
FIG. 4-09. Derivation of the law of reflection using Huygens's principle [3.14].(enlargement of 4-08)
91
o AP: part of wavefront AA’
o after time t : wavelet from P reached B,
wavelet from A reaches B’’
o reflection-angle: 1 'Θ
o 1 11 1
v vABP: sin ' ; AB''B: sin '
AB ABt t∆ Θ = ∆ Θ = ( 1v phase-velocity)
o → 1 1 'Θ = Θ law of reflection
• refraction-angle:
FIG. 4-10. Derivation of the Snell's law of refraction using Huygens's principle [3.14].
o AP: part of wavefront AA’
o After time t : wavelet from P reached B, wavelet from A
reached B’ → but: AB' PB<
(phase-velocity for wavelet in medium lower)
o Refraction-angle: 2Θ
o 1 21 2
v vABP : sin , ABB' : sin
AB ABt t∆ Θ = ∆ Θ =
→ 1 2 1 2
1 2 1 2
v v v vAB
sin sin sin sint t= = ⇔ =Θ Θ Θ Θ
with 1 21 2 1 1 2 2
v , v : sin sin
c c c cn n n n
= = =Θ Θ
92
1 1 2 2sin sinn n⇔ Θ = Θ Snell’s law of refraction
• special case: total internal reflection:
for
crit1 2 2 1 1 2 2
1
; 90 : sin sin ,n n n n=
> Θ = ° Θ = Θ
crit1 1Θ = Θ (critical angle of total reflection)
crit 11
2
arcsin( )nn
Θ =
for angles crit1 1Θ > Θ : no refracted beam!
93
Chapter 4.2. Geometrical optics
Chapter 4.2.1. ray approximation
Many applications: propagation-direction important,
diffraction (later) can be neglected
propagation-direction: perpendicular to wavefronts
→perpendiculars on wavefronts = (light) rays
FIG. 4-11. Rays as perpendiculars on the wavefront [3.11].
FIG. 4-12. Ray approximation. The wave hitting an object with refractive index n2 can be depicted by a ray directing in propagation direction.
94
Chapter 4.2.2. Image formation by reflection mirrors
Goal of many optical set ups: generation of images of objects
object: each point P emits rays in all directions
optical instrument: bundles those rays in point P’
image: all points P’ together
(a) plane mirror
FIG. 4-13. (a) Optical image of a plane mirror. The image of the object "arrow" is virtual: it seems as if the rays come from the points Pi'. (b) Image construction.
(i) real & virtual images (see FIG 4-13a)
object: arrow
each point iP emits rays: red
rays are reflected at mirror
if human eye looks into mirror: seems as if rays come from 1P '
→ I is virtual image of O
real image: rays emerge from image (e.g. spherical mirror)
95
(ii) mirror-equation: 2
2
sP AB : tan(90 )
ABs'
P 'AB : tan(90 )AB
∆ ° − Θ =
∆ ° − Θ =
→ 's s= mirror-equation for plane mirrors s: object-distance s’:image-distance (iii) lateral magnification
1 2 1 2
O I IP P O : tan ; P 'P 'A: tan =
s s' sI
1O
m
∆ Θ = ∆ Θ =
→ = =
(image and object have same size) image construction: 2 rays:
o perpendicular ray o non-perpendicular ray both cases: reflected beam is extended to “behind mirror” side (dashed) → intersection marks image-point
(b) spherical mirror (i) optical axes straight line through vertex P and center C of mirror (see FIG. 4-14a)
96
FIG. 4-14. Spherical mirror. Derivation of the (a) focal length and (b) mirror equation.
(ii) focal point and focal length to optical axis focal point = focus of parallel incident rays (only for rays close to axes)
BF BF
CFB : tan ; ABF : tanCB BA
α α∆ = ∆ =
97
1 1
CB=BA= CA ,2 2
r→ = r = radius of mirror
CB
CFB : cos CFCF 2cos
rαα
∆ = → =
1
f=FP CF (1 ) f2cos 2cos
rr r r
α α→ = − = − = − =
(focal length of spherical mirror) rays close to axes: α small → cos 1α
→ f2r= , f = focal length of spherical mirror
r = radius of spherical mirror
(iii) mirror equation (see FIG. 4-14b)
1
1
11
11
P CA: = +2
P 'CA: = +
ABABP : tan
P B
ABABC: tan
CBAB
ABP ': tanP 'B
δ γ αδ γ β
β δ α
γ
δ
β
∆ → = +∆
∆ =
∆ =
∆ =
approximation for rays close to optical axes: 1 1P B s; P 'B s'; CB ; tan( , , ) , ,r γ δ β γ δ β (small angles)
AB AB AB
2s s'r
→ = +
1 1 1f s s'
→ = + (we used here f = r /2)
mirror-equation for spherical mirrors (s: object-distance, s’: image-distance)
98
(iv) image construction
FIG. 4-15. Image construction for spherical mirrors. Red: the parallel ray is reflected through the focal point F. Green: the "focal point ray" propagates parallel to the optical axes after reflection. Blue: the "center ray" is moves back the same way after reflection. The focusing point of the rays is the image point.
3 rays are helpful: red: “parallel” ray: reflected through focal point green: “focal point” ray: reflected parallel blue: “center” ray: reflected back through center 2 beams necessary, 3rd beam for to make sure → image iss maller, opposite directed, real image (rays emante from image point)
FIG. 4-16. Lateral magnification for a spherical mirror.
99
FIG. 4-17. Concave spherical mirror. (a) Case s > 2 f → s' > 0, 0 > m > -1, real image. (b) Case 2 f > s > f → s' > 0, m < -1, real image. (c) Case f > s > 0 → s' < 0, m > 1, virtual image, dashed: virtual rays.
100
FIG. 4-18. Convex spherical mirror (e.g. supermarket: observation mirrors). The formulas for spherical mirror can be applied with s > 0, f < 0 → s', m < 0, only virtual images.
(v) lateral magnification (see FIG. 4-16):
O -I
sin =s s'
Θ = (I < 0 !)
I s'=-
O sm = lateral magnification for sperical mirror
(vi) real and virtual images 3 different images, depending on object’s position regarding to center, focal point (see FIG. 4-17): (a): s > 2f → s’ > 0, [ 1,0]m∈ − , real image (b): 2f > s > f → s’ > 0, 1m < − , real image (c): 0 < s < f → s’ < 0, 1m > , virtual image (vii) convex spherical mirror (see FIG. 4-18): s > 0, f < 0 → s’ > 0, m < 0, always virtual images (e.g. supermarket: observation mirror) (viii) sign-convention
FIG. 4-19. Sign convention for concave and convex spherical mirrors [3.15].
101
Chapter 4.2.3. Image formed by refraction at a spherical surface
compare: 4.2.2. Image formation by reflection mirrors
(i) focal length:
FIG. 4-20. Images formed by refraction at a spherical surface, derivation of the focal length f, C = center of curvature, F = focal point.
2 1ABC : 180°= + (180 )ε∆ Θ + ° − Θ
1 2ε→ = Θ − Θ
1ABC : ; ABF : f
l lr
ε∆ Θ = ∆ (f isn’t exactly the radius)
Snell’s law: 1 1 2 2sin sinn nΘ = Θ
→ 12 1
2
sin sinnn
Θ = Θ , small angles: 1,2 1,2sin Θ Θ
102
1 12 1
2 2
1 2
2 2 1
f = r( )f
n n ln n r
n nl l lr n r n n
→ Θ = Θ =
→ = − →−
discussion:
1 2 2 1
1 2 2 1
: 0, f 0
: 0, f 0
n n n n
n n n n
< − > >> − < <
most cases: 1 21, f ( )1
nn n n r
n= = → =
−
(ii) imaging-equation:
FIG. 4-21. Images formed by refraction at a spherical surface, derivation of the imaging equation [3.16].
2
2
ACP' : 180 (180 )γ βγ β
∆ ° = + Θ + ° −→ = − Θ
, s '
l lr
β γ= (s’ isn’t radius regarding l )
Snellius: 1 1 2 2sin sinn nΘ = Θ
small angles: 12 1
2
sinnn
Θ Θ → Θ = Θ
103
1
1
PAC : 180 (180 )α βα β
∆ ° = + + ° − Θ→ Θ = +
small 12
2
: ( )s s
nl l ln r
α α → Θ = +
1 1
2 2
1 2 2 1
1 1 1 1( ) ( )
s' s s ' s
s s'
n nl l l lr n r r n r
n n n nr
→ = − + → = − +
−→ + =
with 2 2 1 2
2 1
f ( )f
n n n nr
n n r−= ⇔ =
−
1 2 2
s s' fn n n
+ =
(iii) sign convention:
FIG. 4-22. Sign convention for the imaging by spherical surfaces. Attention: for virtual objects and virtual images the refractive indices have to be exchanged regarding the side of the object / image in the imaging equation, see also derivation of lens-maker's equation [3.16].
104
Chapter 4.2.4. Image formation by thin lenses
effect: subsequent image formation by refraction at 2 spherical
surfaces
FIG. 4-23. Derivation of the lens-maker's formula. (a) Image formed by refraction at the first spherical surface. (b) Image formed by refraction at the second spherical surface. Object = image formed by first surface (virtual object). (c) Approximation: object and image distance are measured relative to the lens axis.
105
106
(i) lens maker’s formula and focal length
(1) Image by refraction at surface 1 (see FIG. 4-23a):
imaging-equation: 1 2 2 21 1
1 1 1 2 1
, fs s ' fn n n n
rn n
+ = =−
1 2 2 1
1 1 1s s 'n n n n
r−+ = (1)
(2) Image by refraction at surface 2 (see FIG. 4-23b): object: image made by surface 1 → virtual object on transmission side → exchange of 1 2,n n on left side of equation (1), 2s 0< radius: center of curvature on incident side → exchange of 1 2,n n on right side of equation (1)
1 2 2 1
2 2 2s s 'n n n n
r−+ = (2)
(3) Approximation: object and image distance measured relative to lens axis: 1 2 2 1s =s, s '=s', s =-s '
(1) + (2) 2 11 2
1 1 1 1+ =( - )( - )
s s'n n
r r→
(4) Focal length: for s → ∞ : incident rays parallel, s'=f:
2 11 2
0
1 1 1 1( )( )
s fn n
r r→
+ = − −
usually: incident medium: air with 1 21, n n n= =
1 2
1 1 1( 1)( )
fn
r r= − − lens maker’s formular
(ii) imaging equation
see above: 1 2
1 1 1 1 1( 1)( )
s s' fn
r r+ = − − =
1 1 1s s' f
→ + =
(iii) sign convention see FIG. 4-22
107
(iv) image construction, real and virtual images converging lenses:
FIG. 4-24. Image construction for converging lenses. (a) s > f, inverse real image. (b) s < f, upright virtual image. 3 beams: red: parallel ray: through focal point blue: central ray: through vertex of lense green: focal ray: emerges parallel to optical axes Attention: ray goes through at lens axis mirrored focal point F’! real & virtual images possible, depending on object pos. regarding focal point diverging lenses (see FIG. 4-25): only virtual images converging & diverging lenses, focal length (see table 4-2)
108
FIG. 4-25. Image construction for diverging lenses. (a,b) all cases: upright virtual image.
TABLE 4-2. Converging and diverging lens.
109
FIG. 4-26. Object magnification by thin lenses.
(v) lateral magnification (see FIG. 4-26)
O -ytan = , I 0!
s s'I s'
= -O s
m m
α = <
→ = =
110
Chapter 4.2.5: Optical instruments
(a) the eye
FIG. 4-27. The human eye [3.17]
how it works:
• pupil: entry for light
• iris: determines light intensity propagating through pupil,
darkness: iris widely opened
• Cornea / lens system: focuses incoming light onto retina
• Ciliary muscle: changes radii of lens, needed for
accommodation
• accommodation:
far objects: c.m. is released, lens system has maximum
focal length fmax, image is focused on retina
( max' 2,5 cms f= ≈ )
closer objects: c.m. increases curvature of lens, focal
length of lens system is decreased in order to place
the image again on retina (always: ' 2,5 cms ≈ )
111
• retina: contains sensing structure, receives image
• optic nerve: transmits image to the brain
near point:
• closest point for which lens can focus image on retina
• varies with age (@ 10 years: ~ 7 cm ,
@ 60 years: ~200 cm )
• standard value: 25 cmnpx =
apparent object size / angle of vision:
• determined by image size on retina
• image appears larger if object is closer (see FIG. 4-28)
FIG. 4-28. Human eye, apparent object size. (a) The object is far away, the image appears small. (b) The closer the object the greater the image size on the retina. The object appears greater [3.18].
112
• measure for apparent object size: angle of vision Θ
• determination of Θ (see FIG. 4-28a,b):
1 21 2
max max
' ',
y yf f
Θ = Θ = , 1 21 2
tan( ) , tan( )y ys s
Θ = Θ =
for small angles: tan( )Θ ≈ Θ
→ 1 21 2
, y ys s
Θ = Θ =
• standard angle of vision:
object at standard near point, 0np
yx
Θ =
0Θ = angle of vision for (average) largest apparent object
size that is possible (see FIG. 4-29a)
FIG. 4-29. The simple magnifier consisting of a single converging lens with short focal length [3.19]. (b) simple magnifier
goal: object magnification
compounds: converging lens
how it works: (see FIG. 4-29)
• object is placed close to focal point of converging lens:
, s f s f< ≈
• lens produces virtual image:
113
's → −∞ , image size: I → ∞
• eye easily focuses virtual image as an infinitively far object on
retina
→ simple magnifier allows the object to be brought closer to the
eye
angular magnification or magnifying power M :
0
MΘ=Θ
Θ = angle of vision with simple magnifier
0Θ = standard value of angle of vision, maximum apparent
object size without simple magnifier
(see FIG. 4-29a:) 0 0tan( )np
yx
Θ ≈ Θ =
(see FIG. 4-29b:) tan( )yf
Θ ≈ Θ =
→ 0 npM x f= Θ Θ =
114
(c) optical microscope (compound microscope)
FIG. 4-30. Optical or compound microscope [3.20].
goal: object magnification of very small objects at short distances
compounds: 2 converging lenses (objective + eyepiece)
how it works: (see FIG. 4-30)
• objective forms real, enlarged and inverted image
• eyepiece is used as simple magnifier, image formed by
objective serves as object
magnifying power: (see FIG. 4-30)
'
tan( )o
y yf L
β −= = , ' 0!y < , L = tube length, 0f = focal length of
objective
'
tan( )e
yf
−Θ = ≈ Θ−
, 0ef < (F' on incident side!)
see FIG: 4-29a:
115
0 0tan( )np
yx
Θ ≈ Θ =
→ 0
'//
npe
np o e
L xy fM
y x f fΘ= = = −Θ
(d) telescope
FIG. 4-31. Astronomical telescope [3.21].
FIG. 4-32. Reflecting telescope [3.22].
goal: (1) view far & large objects, (see FIG. 4-31)
(2) brighten the image (view dark objects)
(see FIG. 4-32)
116
goal (1):
compounds: 2 converging lenses (objective + eyepiece)
how it works:
• objective forms a real, downsized and inverted image at
focal point
• eyepiece is used as simple magnifier image of objective
serves as object
magnifying power:
' '
tan( )e ee e
y yf f
−Θ ≈ Θ = =−
0Θ = changed reference point:
before: object at near point
here: makes no sense (eye 25 cm in front of moon, sun,
Jupiter, ...)
→ 0Θ = angle of vision of object viewed directly
with unaided eye
0
'tan( )o
o
yf
−Θ ≈ Θ =
→ 0
'/'/
e e o
o e
y f fM
y f fΘ= = = −Θ −
goal (2):
• the larger the objective, the brighter the image
• problem: large lenses difficult to produce
117
→ reflecting telescope: parabolic mirror used instead of lens as
objective
→ bodies appear that are not viewable with the unaided eye!
Chapter 4.2.6. prism, dispersion, spectral analysis
(i) prism
118
FIG. 4-33. Optical prism. The deflection angle δ of light propagating through the prism depends on the incidence angle Θ1, the prism's refractive index n and the prism angle γ.
compounds: combination of 2 optical interfaces with angle γ in
between
how it works: light propagating through it gets deflected
determination of deflection angle δ: (see FIG. 4-33)
• goal: 1( , , )nδ δ γ= Θ
• ∆ ACE: 1 2 2 1180 ( ) ( ) (180 )α α δ° = Θ − Θ + − + ° −
→ 1 2 2 1δ α α= Θ − Θ + −
• ABCD: 360 90 90γ ε° = + ° + + °
→ 180ε γ= ° −
• ∆ ABC: 2 1 2 1180 (180 )ε α γ α° = Θ + + = Θ + ° − +
→ 2 1α γΘ + =
• intermediate result: 1 2δ γ α= Θ − +
• @ point C: Snell's law: 1 2sin( ) sin( )n α α= , (air) 1n =
→ 2 1arcsin[ sin( )]nα α=
• @ point A: Snell's law: 1 2sin( ) sin( )nΘ = Θ
→ 2 1arcsin[sin( ) / ]nΘ = Θ
→
1 2 2 1 1 2 1 2
11 2 1
1
( )
sinarcsin sin[arcsin( )]
( , , )
nn
n
δ α α α α
γ α γ
δ γ
= Θ − Θ + − = Θ − Θ + + =Θ= Θ − + = Θ − +
= Θ
119
(exam: beam deflections through prisms...)
(ii) dispersion
dispersion: refractive index dependent on wavelength, ( )n n λ=
normal dispersion: usual case, n decreases with increasing
wavelength, / 0dn dλ <
anormal dispersion: / 0dn dλ >
(iii) spectral analysis
FIG. 4-34. Spectral analysis with a prism. Because of dispersion white light propagating through a glass prism can be separated into parts with different wavelengths [3.23].
( )nδ δ= , ( )n n λ= → ( )δ δ λ=
parallel white light can be separated into parts with different
wavelength (see FIG. 4-34)
120
Chapter 4.3. Wave optics
interference + diffraction: wave phenomena
interference: combination of waves by superposition, 2 or more
waves meet at one point
diffraction: bending of waves around corners, occurs when portion
of wavefront is cut off by barrier
model: Huygens's principle, interference of N → ∞ waves with
phase difference ∆Ψ :
• wave 1: 1 0 1sin( )E E= Ψ
• wave 2: 2 0 2sin( )E E= Ψ
→ 2 1∆Ψ = Ψ − Ψ
constructive interference: 2m π∆Ψ = , 0, 1, 2,...m = ± ±
destructive interference: (2 1) m π∆Ψ = + , 0, 1, 2,...m = ± ±
origin of phase difference:
(1) waves propagate through different paths:
optical path difference ∆ :
• path: only absolute value important: 3D → 1D
• wave 1 at point P: 1 nk x tωΨ = −
wave 2 at point P: 2 ( )nk x x tωΨ = + ∆ −
121
→ 2 1
2n
n
k x xπ
λ∆Ψ = Ψ − Ψ = ∆ = ∆
x∆ = path difference between waves 1 and 2
• optical path difference = corresponding path difference in
vacuum
• with :n nλλ =
2 2 2
n
x n xπ π π
λ λ λ∆Ψ = ∆ = ∆ = ∆ , n x∆ = ∆
• constructive interference: mλ∆ = , 0, 1, 2,...m = ± ±
→ 2
2m mπ λ πλ
∆Ψ = =
• destructive interference: (2 1)2
mλ∆ = + , 0, 1,...m = ± →
(2 1) m π∆Ψ = +
(2) phase change δ due to reflection at
(i) medium with lower refractive index: 0δ =
(ii) medium with higher refractive index: δ π=
Chapter 4.3.1. Interference fringes from thin layers
FIG. 4-35. Soap bubbles as interference fringes of interfering light reflected on top and bottom of the film. The varying film thickness results in coloration [3.24].
122
FIG. 4-36. Interference fringes from thin layers of interfering light reflected on top and bottom of the film.
consider: wave I incident on thin layer (thickness t , refractive index
n ),
e.g. soap bubble (see FIG. 4-35)
what happens?
if film is observed with the eye under angle 1Θ , reflected light
on top and bottom of film ( 1 2&R R ) interferes on retina
(see FIG. 4-36)
determination of phase difference ∆Ψ :
• from point A,D to retina both rays have same paths
→ ∆Ψ is determined at points A,D
123
(1) optical path difference
nAB nBC AD∆ = + −
2cos( ) /t ABΘ = → 2/ cos( )AB t= Θ
AB BC=
1sin( ) /AD ACΘ = → 1sin( )AD AC= Θ , 2AC EB= ,
2sin( ) /EB ABΘ = → 2 12 sin( )sin( )AD AB= Θ Θ
→ 2 1
2 1
2 12
2 2 sin( )sin( )
2 [ sin( )sin( )]
2 [ sin( )sin( )]cos( )
nAB AB
AB n
tn
∆ = − Θ Θ =
= − Θ Θ =
= − Θ ΘΘ
Snell's law @ point A: 1 2sin( ) sin( )nΘ = Θ
→
2 22 2
2 2
22 2
2
22
22 [ sin ( )] [1 sin ( )]
cos( ) cos( )
2cos ( ) 2 cos( )
cos( )
2 1 sin ( )
t ntn n
ntnt
nt
∆ = − Θ = − Θ =Θ Θ
= Θ = Θ =Θ
= − Θ
Snell's law @ point A: 1 2sin( ) sin( )nΘ = Θ
→
22 21
12
2 21
sin ( ) 12 1 2 sin ( )
2 sin ( )
nt nt nn n
t n
Θ∆ = − = − Θ =
= − Θ
(2) phase change δ :
(i) 1R : @ A: reflection at medium with higher refractive index:
1δ π=
(ii) 2R : @ B: 2 0δ =
→ 2 1δ δ δ π= − = −
124
→ phase difference
2 21
2 4sin ( )
2 , constructive interference(2 1) , destructive interference
tn
m
m
π πδ πλ λ
ππ
∆∆Ψ = + = − Θ − =
= +
discussion:
FIG. 4-37. Interference fringes from thin layers. (a,b) One thin layer, the interference pattern starts with a dark spot in the center. (c,d) Layer system with refractive indices nsub > n. Due to an additional phase change at the bottom of the thin layer the interference pattern is inverted (bright spot in the center).
(i) case: variable incidence angle:
constλ = , t const= , 1 1[ , ]Θ∈ Θ − ∆Θ Θ + ∆Θ
interference fringes = rings with dark spot in center
(see FIG. 4-37a,b)
dark center: minimum 0. order
125
1. bright circle: maximum 1st order
1. dark circle: minimum 1st order
2. bright circle: maximum 2nd order
etc.
(ii) case: consider thin film (refr. index n , thickness t ) deposited
on substrate with subn n> , ,t constλ = , variable incidence
angle (as above)
e.g. water film ( 1,33n = ) on glass substrate 1,515subn =
optical path difference: 2 212 sin ( )t n∆ = − Θ (as above)
phase change: 1δ π= (as above), but: 2δ π=
→ 0δ = !
→
2 21
2 4sin ( )
2 , constructive interference(2 1) , destructive interference
tn
m
m
π πδλ λ
ππ
∆∆Ψ = + = − Θ =
= +
intensity pattern is inverted! (see FIG. 3-37c,d)
bright center: maximum 0. order
1. dark circle: minimum 1. order
1. bright circle: maximum 1. order
etc.
(iii) case: variable wavelength & thickness:
1 constΘ = , [400,700] nmλ ∈ (white light), t increasing
(see FIG. 3-36)
2 21
4sin ( )
tn
π πλ
∆Ψ = − Θ −
@ top: 0t → , → π∆Ψ → − → dark area
126
below top: thickness increases → ∆Ψ increases
→ 2 21
4sin ( )
tn
π πλ
− Θ → , for small λ first)
→ 0∆Ψ =
→ first rainbow arises (constructive interference for blue →
yellow → orange → red)
→ second rainbow for 2 21
4sin ( ) 3
tn
π πλ
− Θ → , 2π∆Ψ =
127
Chapter 4.3.2. Diffraction at a single slit
FIG. 4-38. Diffraction pattern of a single slit with slit width a ≈ λ. Intensity pattern I(Φ)=I0 [sin(Φ)/Φ]2, Φ = π a sin(Θ). Most intensity is concentrated in the forward direction (central maximum). Points of zero intensity are determined by: a sin(Θ) = m λ, m = ±1 (minimum ±1. order), ±2 (minimum ±2. order), ... . Points of maximum intensity are determined by a sin(Θ) = n λ, n = 0 (maximum 0. order), ±1.4303 (maximum ±1. order), ±2.459 (maximum ±2. order),±3.4707 (maximum ±3. order), ... .
consider: straight propagating light hits a single slit with slit width
a λ≈ (see FIG. 4-38)
what happens?
• part of wave front is cut off
• propagation direction not straight → light bending (diffraction)
128
determination of intensity pattern:
• approach: Huygens's principle, waves emerging from each
point of slit interfere
• derivation: complicated
• result: 20( ) [sin( ) / ]I I φ φΘ = , ( ) sin( ) /aφ π λΘ = Θ
Θ = viewing angle, a =slit width, λ = wavelength
(see FIG. 4-38)
maxima & minima of intensity pattern:
search for points with ( ) ( ) 0I I φΘ = =
necessary condition: / 0dI dφ =
→ 0 2
0 3
sin( ) cos( ) sin( )2
sin( )[ cos( ) sin( )]2 0
I
I
φ φ φ φφ φ
φ φ φ φφ
− =
−= =
case 1:
sin( ) 0φ = → , 2 , 3 ,...φ π π π= ± ± ± ( 0φ = not allowed)
→ 20( ) [sin( ) / ] 0I Iφ φ φ= = → minima
with sin( ) /aφ π λ= Θ :
→ minimum condition: sin( )a mλΘ = , 1, 2,...m = ± ±
1m = ± : minimum ±1. order
2m = ± : minimum ±2. order
etc.
case 2:
cos( ) sin( ) 0φ φ φ− = → tan( )φ φ=
only graphical solution possible:
129
FIG. 4-39. Diffraction at a single slit, the solution of maxima of the intensity pattern is only possible graphically (numerically).
result: 1.4303 , 2.459 , 3.4707 ,...φ π π π= ± ± ±
for 0φ = : rule of de L'Hôpitale:
2
0 0
2 2
0
2sin( )cos( )lim[sin( ) / ] lim
2
cos ( ) sin ( )lim 1
1
φ φ
φ
φ φφ φφ
φ φ
→ →
→
=
−= =
→ 0( )I Iφ = (central maximum)
→ maximum condition:
sin( )a nλΘ = , 0, 1.4303, 2.459,3.4707,...n = ± ±
0n = : maximum 0. order
1.4303n = ± : maximum ±1. order
2.459n = ± : maximum ±2. order
etc.
130
discussion:
(i) determination of positions of maxima / minima on screen:
FIG. 4-40. Diffraction of a single slit, determination of the width of the central maximum [3.25].
e.g. width w of central maximum
w = distance minimum ±1. order
position minimum +1.order:
1
1tan( )yL
++Θ = , L = distance slit / screen
1sin( )a λ+Θ = ( 1m = )
→ 1 tan[arcsin( / )]y L aλ+ =
intensity symmetric regarding 0Θ = → 12w y+=
131
(ii) width of intensity pattern depends strongly on slit width a :
FIG. 4-41. Diffraction at a single slit. The intensity pattern strongly depends on the slit width. Border cases: a → 0: slit acts as point source, I = I0 = const, a → ∞: no diffraction, light propagates unimpeded through slit.
border cases:
(1) 0a → : 0I I const= = , slit acts as point source
(2) a → ∞ : ( 0) 0I Θ = ≠ , ( 0) 0I Θ ≠ = , light
propagates freely through slit
(iii) illumination with white light - spectral analysis
condition for maximum:
132
FIG. 4-42. Wavelength separation by diffraction at a single slit. The higher order maxima are spread out.
maxsin( )na nλΘ = → max arcsin( / )n n aλΘ =
max max ( )n n λΘ = Θ → wavelength separation:
smaller angles for smaller wavelengths
stronger diffraction for longer wavelengths
133
Chapter 4.3.3. Diffraction gratings
FIG. 4-43. Grating constant d and slit width a of a diffraction grating.
FIG. 4-44. Conventional fabrication of diffraction gratings: scratches on surfaces act as optical slits with M scratches per length L. Grating constant d = L/(M-1).
diffraction grating = combination of equally spaced slits
(see FIG. 4-43)
grating constant d = distance between center of slits (with slit width
a ) = period of the grating
conventional fabrication:
• equidistant scratches or coatings on flat surfaces with M
lines per length unit L (see FIG. 4-44)
• grating constant: ( 1)L M d= − → /( 1)d L M= −
• number N of slits illuminated with light ray of diameter D
(red circle in FIG. 4-44): /N D d=
134
determination of light intensity pattern:
• approach: Huygens's principle, diffraction pattern of single
slits overlap
• derivation: even more complicated
• result:
2 20( ) [sin( ) / ] [sin( ) / sin( )]
sin( ) / , sin( ) / viewing angle, slit width, grating constant, wavelength
I I N
a d
a
d
φ φ α αφ π λ α π λ
λ
Θ == Θ = Θ
Θ = == =
FIG. 4-45. Diffraction grating. (a,b) Intensity pattern consists of two functions f1 & f2. (c) Function f2 with higher resolution. f2 consists of principal maxima with f2 = N 2 & subsidiary maxima. (d) The resulting intensity pattern I/I0 = f1 × f2. (e) I/I0 with higher resolution. The intensity pattern consists of principal & subsidiary maxima. The plot includes the envelope function N 2 × f1.
135
discussion:
(i) plot 2 m, 8 m, 6, 632,8 nm (red light)a d N λ= = = =
(see FIG. 4-45)
(ii) different parts: 0 1 2/I I f f=
1f : intensity pattern of single slit, envelope function
(see FIG. 4-45a,e)
2f : intensity pattern of N slits acting as point
sources
consists of principal maxima (with 22f N= ) &
subsidiary maxima (see FIG. 4-45b,c)
0 1 2/I I f f= : resulting pattern, consists of principal &
subsidiary maxima (see FIG. 4-45d,e)
determination of position of principal maxima:
FIG. 4-47. Diffraction grating, determination of the position of principal maxima.
136
@ principal maximum: 22f N=
→ sin( ) / sin( )N Nα α = → sin( ) sin( )N Nα α=
→ 0, , 2 ,...α π π= ± ± (because N is integer)
with sin( ) /dα π λ= Θ
→ condition of principal maxima:
sin( )d nλΘ = , 0, 1, 2,...n = ± ±
d = grating constant, Θ = viewing angle, λ = wavelength
0n = : maximum 0. order (central maximum)
1n = ± : maximum ±1. order
etc.
FIG. 4-46. Diffraction at a grating. The position of principal maxima is independent of the number of slits N (a) and the slit width a (b).
discussion:
(i) position of principal maxima independent of slit number N :
(see FIG. 4-46a)
→ with increasing N the width of principal maxima
decreases, subsidiary maxima vanish
137
(ii) position of principal maxima independent of slit width a :
(see FIG. 4-46b)
→ with increasing slit width the intensity of higher order
maxima decreases
(iii) determination of maxima - positions on screen
(see FIG. 4-47)
tan( ) /n ny LΘ = , L = distance grating / screen
sin( )nd nλΘ =
→ tan[arcsin( / )]ny L n dλ=
(iv) wavelength dependency - spectral analysis
FIG. 4-48. Spectral analysis with a diffraction grating. Because of higher intensities and better resolution in higher order maxima a grating with high number of slits N is preferred compared to the single slit.
(v)
( )n ny y λ=
→ white light can be separated into different colors
→ analogous to single slit:
stronger diffraction for higher wavelengths
(e.g. exam: determination of the m. maximum...)
for N → ∞ : higher order maxima narrow, high intensity
→ great solution → grating preferable to single slit
138
Chapter 4.4. Quantum optics
Chapter 4.4.1. wave-particle duality
wave phenomena:
• interference, diffraction
• radiation = electromagnetic waves, wavelength λ , phase
velocity Phv
particle phenomena:
• photoelectric effect, Compton effect, black body radiation, etc.
• radiation = beam of particles
• particles = photons
Chapter 4.4.2. photons:
momentum: p mv=
de Broglie: /p h λ= , combination of wave & particle properties
( 2 2/ / / /m E c hf c p mc hf c h λ= = → = = = )
Planck's constant: 346,626 10 Jsh −= ×
velocity of photon: photon = particle with light velocity, Phv v c= = ,
medium: /v c n=
rest mass of photon: 0 0m = (see chapter about relativistic effects
- later)
energy: Einstein: E hf=
energy-density η : → photon density n : /( )n hfη=
Chapter 4.4.3 photoelectric effect
139
Heinrich Hertz (1897), Wilhelm Hallwachs (1895):
Exposure of a negatively charged metal plate with UV light
leads to charge reduction
→ during exposure electrons left the plate
FIG. 4-49. Experimental results of Heinrich Hertz (1897) and Wilhelm Hallwachs (1895). During UV exposure the negative charge of a metal plate decreases, electrons must have left the plate.
Lenard (1902): Quantitative examination of this phenomenon
experiment:
FIG. 4-50. Experimental set up to measure the photo current IPh as a function of the voltage U between two metal plates cathode and anode (Lenard, 1902).
• application of a voltage U between two metal plates
→ metal plates get charged oppositely
→ constant electric field /E U d= between plates,
140
d = distance between plates
• UV exposure of cathode (= negatively charged plate)
→ release of electrons out of cathode (see above)
→ force on electrons: F qE eE= = − , electrons are
accelerated towards anode (positive plate)
→ kinetic energy of electrons at anode: cathkin kinE E eU= + ,
cathkinE = kinetic energy @ cathode
→ @ anode: electrons are absorbed, flow through outer
circuit again to cathode
• measurement of the photo current ( )Ph PhI I U=
FIG. 4-51. Photo current IPh as a function of applied voltage U. For U = 0, IPh ≠ 0, IPh decays at countervoltage U0 < 0! For U > 0 the photo current saturates to saturation value IPH,S.
141
FIG. 4-52. Photoelectric effect. The saturation value of the photo current depends only on the UV intensity (a), the countervoltage depends only on the UV frequency (b).
FIG. 4-53. Photoelectric effect. Energy for inner and outer electrons in cathode material.
142
results: (see FIG. 4-51)
• for 0U = , photo current 0PhI >
• for 0 0U U≤ < , photo current decays 0PhI =
0U = counter voltage, cathode is positive!
• for 0SU U≥ > , photo current saturates, ,Ph Ph SI I=
• variation of light intensity I : 0U const= , , , ( )Ph S Ph SI I I=
(see FIG. 4-52a)
• variation of light frequency f : 0 0 ( )U U f= , ,Ph SI const=
(see FIG. 4-52b)
• no time retardation between exposure and electron emission
explanation: Einstein (1905):
• UV light consists of photons (= light particles)
• in an absorption process inside cathode material each photon
transmits its entire energy E hf= onto electron, photon
disappears
• if fE E> , electrons can leave cathode
• fE : electrons have different energies in cathode material,
f AE E E= + ∆ , for outer electrons 0E∆ = , f AE E=
(see FIG. 4-53)
• AE = work function, minimum energy necessary to free
electrons out of material (see FIG. 4-53)
• energy relation @ cathode: cathkin AE E E= +
• kinetic energy of electrons @ cathode
143
,maxcath cathkin kinE E E= − ∆ (see FIG.4-53)
,maxcathkinE = maximum kinetic energy of an electron =
kinetic energy of outer electrons
• ,maxcathkinE is determined by 0U ! (see FIG. 4-51)
case 0U > : all electrons with fE E< (released
electrons) are accelerated towards anode
and contribute to PhI
case SU U→ :
• for increasing U more & more electrons occupy outer
positions (cathode is higher charged, inner positions are
occupied)
• at SU there are so many electrons in outer positions
that every photon frees an electron
case SU U> : the number of released electrons can't
be increased because every photon
already frees one electron, PhI remains
at saturation value ,Ph SI
case 0U < :
• the number of electrons in high-energy
positions decreases (positive charge - lack
of electrons), PhI decreases
• field is opposite directed, electrons get broken on
the way towards anode, released electrons with velocities
144
other than directly towards the anode cannot overcome
the counter field
case 0U U→ : only electrons with maximum kinetic
energy ,maxcathkinE and flight direction
towards the anode can get over the
counter field.
case 0U U= :
• highest energy electrons have velocity
0v = at anode.
• photo current decays.
• kinetic energy ,maxcathkinE equals kinetic energy for a
positron at cathode starting at anode with 0v = :
,max 0cathkinE eU= −
→ energy relation for outer electrons with initial velocity towards
anode:
,max 0cathkin A AE hf E E eU E= = + = − +
0 AeU hf E− = −
discussion:
(i) determination of work function AE and Planck's constant h
measurement of 0U for different frequencies f :
→ 0 / /AU hf e E e− = − (see FIG. 4-54)
AE : extrapolation to 0f = → 0 ( 0)AE eU f= − =
h : slope tan( ) /m h eα= =
145
(ii) photoelectric effect not explainable with wave model:
• wave should need retardation time to put enough energy
onto electron to release it, retardation should depend on
wave's intensity, for all wavelengths it should be possible
to release electrons
• experimental results:
(1) no retardation time,
(2) for wavelengths with / AE hf hc Eλ= = <
no electrons can be released, minimum
frequency, maximum wavelength:
min max/ AE hf hc Eλ= = =
(determination of AE possible)
FIG. 4-54. Photoelectric effect. Determination of Planck's constant h and the work function EA of the cathode's material.
146
Chapter 5. Atomic physics
Chapter 5.1. Experiments that led to atomic models
(a) photo-effect
• see chapter 4.4.3: electrons can be freed out of metals
• matter: usually electrically neutral → atoms must be electrically
neutral
result: atoms contain electrons with negative charge + particles
with equal amount of positive charge
(b) x-ray diffraction
• x-rays: 8 12[~ 10 ,10 ] mλ − −∈
• for diffraction you need grating with grating constant d λ≈ :
atomic crystals (solids with periodically located atoms)
FIG. 5-01. Atoms in crystal serve as a grating for x-ray diffraction (see also chapter 4.3.3) → atomic crystal serves as grating
(see also chapter 4.3.3)
147
result: atoms have dimension atom 0,1 nmφ ≈
(c) atomic gases in discharges - line spectra
• experiment:
FIG. 5-02. Experimental set up to measure the line spectrum of different atoms in gases.
• free electrons are generated by applying an alternate voltage
to glowing cathode
• free electrons are accelerated towards anode
• on the way to anode electrons collide with gas atoms
• in collision process gas atoms are excited or ionized
148
• excited atoms make transition to lower energy state, ionized
atoms recombine with free electrons
both cases: emission of electromagnetic radiation
• part of emitted wave front is blocked by slit
• different wavelengths are viewable on different positions on
screen behind slit → line spectrum (see chapter 4.3.2),
determination of emitted wavelengths
result: • each atom emits characteristic set of wavelengths called
its characteristic line spectrum
• line spectrum of hydrogen and mercury (only visible
wavelengths):
FIG. 5-03. Visible part of the line spectrum of hydrogen (a) and mercury (b) measured with an experimental set up according to FIG. 5-02 [5.01].
• general expression: Rydberg-Ritz formula
1 2
2 2, 1 2
1 1 1
n n
Rn nλ
= −
, 1,2n : integers, 2 1n n> ,
149
R : Rydberg constant, varies slightly from element to
element, hydrogen: 71,097 10 1/mHR = × , expression
empirically found, no theoretical model at that time!
(d) Zeeman effect
FIG. 5-04. Applying of a magnetic field B to the gas chamber leads to splitting of single lines.
• applying of a constant magnetic field to gas chamber,
measurement of emitted wavelengths (see FIG. 5-04)
150
FIG. 5-05. (a) The blue line in the line spectrum of hydrogen. (b) After application of a magnetic field the blue line is split into 5 different lines lying symmetrically around the previous line. Different colors are not detectable with the eye because the wavelength lie close together.
result: some single lines become split into different number of
lines with close wavelengths (see FIG. 5-05)
(e) Rutherford's scattering experiment
• scattering of α-particles from radioactive radium (later) by atoms
in gold-foil, measurement of scattering angle
result: nearly all mass and all positive charge is concentrated in
nucleus, dimension -6nucleus 1 fm = 10 nmφ ≈
151
Chapter 5.2. Historic atomic models
Requirements for atomic models:
• explain different line spectra of atoms
(must hold Rydberg Ritz formula)
• explain change of line spectra in magnetic fields
• explain systematics of periodic table of elements
(a) J.J.Thomson's plum pudding mode
FIG. 5-06. "Plum pudding" model of J.J.Thomson [5.02].
• electrons embedded in fluid with dimension atom 0,1 nmφ ≈
• fluid contains most of mass and all positive charge
• explanation of experiments (a,b):
• explanation of experiments (c,d):
no appropriate particle arrangement found
• Rutherford's experiment (e): disprove of Thomson's model
152
(b) Rutherford's atomic model
FIG. 5-07. Rutherford's atomic model. Atom consists of nucleus and electron shell. Nucleus has dimension φ ≈ 1 fm, contains most of the mass and all positive charge. Electron shell has dimension φ ≈ 100000 fm = 0,1 nm, contains almost no mass and all negative charge. The arrangement of the particles is still unknown. • Atom consistss of tiny nucleus and electron shell
• nucleus has dimension 1 fmφ ≈ , contains most of the mass and
positive charge
• electron shell contains almost no mass, all negative charge
• explanation of experiments (a,b,e):
• explanation of experiments (c,d):
problem: finding the electron arrangement in the electron shell
→ Bohr
(c) Bohr's atomic model (Nils Bohr, 1913) for hydrogen
• classical approach: Rutherford's atomic model, electron under
influence of Coulomb attraction to positive nucleus, Coulomb
force acts as centripetal force, electron moves on circular orbits
around nucleus
153
FIG. 5-08. Bohr's atomic model. Attractive Coulomb force on an electron in the Coulomb field of the positively charged nucleus, k = 1 / (4πε0) = coulomb constant. The Coulomb force acts as centripetal force, the electron moves in circular orbits around nucleus.
(i) forces
Coulomb force = centripetal force: 2
2e nq q v
k mr r
= −r r ,
01/(4 )k πε= : Coulomb's constant, 0ε : dielectric constant (of
free space), ,e nq q : charge of electron shell / nucleus v :
velocity of electron, r : distance electron / nucleus, r : unit
vector pointing from nucleus to electron
with eq e= − , nq e= : 2 2
2
e vk m
r r− = −r r →
22e
k mvr
=
(ii) energy
154
potential energy of electrons: ( ) ( ) pot potE E− = 0r
0 Cr
r r F dr
with 2
2
ek
r= −CF r , dr=dr r :
002
2 22 2
1 1( ) ( )
rr
pot potrr
eE E k dr ke dr ke
r r r − = − = − =
0r
0r
r r r r
reference point: 0r → ∞ , definition: ( ) 0potE ∞ =
2( ) ( ) /pot potE E r ke r= = −r
kinetic energy: 2 / 2kinE mv=
with 2
2ek mv
r= :
2
2kin
eE k
r=
total energy: 2 2 2
2 2pot kin
ke ke keE E E
r r r= + = − + = −
FIG. 5-09. Classical model, total energy of a classical electron circling around positive nucleus.
(iii) orbits:
2
2ke
Er
= − → 2
2ke
rE
= −
155
(iv) orbit-problems:
• because of Coulomb force = centripetal force circular orbit
mechanically stable
• electron moving in circular orbit = accelerated charge!
• accelerated charge emits electromagnetic waves
• total energy decreases continuously
• FIG. 5-09 → decreasing energy leads to decreasing
distance electron / nucleus
→ atom collapses!
• solution: Bohr's postulates:
1. postulate:
The electron in the hydrogen atom can move only in
circular orbits called stationary states.
FIG. 5-10. Bohr's atomic model for hydrogen. When an electron makes a transition from a stationary state with high energy to one with a lower energy, a photon hf is emitted. The process is called photon emission. The contrary process is called photon absorption.
156
2. postulate:
When electrons move from stationary states with higher
energies iE to those of lower energies fE , during the
transition a photon of frequency f is emitted which holds
the relation i fhf E E= − .
note: process is called photon emission, contrary process
is called photon absorption
3. postulate:
FIG. 5-11. Bohr's 3. postulate. The orbital angular momentum of the electron is quantized.
The orbital angular momentum L is quantized:
mvr n= × = =L r p z z → L mvr n= = (see FIG. 5-11)
157
p = momentum of the electron, m = electron mass, v =
electron velocity, /(2 )h π= ,
1,2,3,...n = quantum number
• consequences from Bohr's postulates:
(i) orbits (see FIG. 5-10)
circular orbits: 2
2ek mv
r= →
22 ke
vmr
=
quantization: mvr n= → 2
2 nv
mr =
→ 2 2n ke
mr mr =
→
2 22
02n
nr a n
mke= =
• orbits are quantized
• 1n = : 21 0 /( ) 0,0529 nmr a mke= = = : first Bohr radius,
characteristic length in atomic physics
(ii) energy levels:
FIG. 5-12. Energy levels of hydrogen according to Bohr's atomic model.
158
2
2ke
Er
= − → 2
022n
n
EkeE
r n= − = −
• energy quantized
• 2
00
113,6 eV
2ke
Ea
= = → 1 0 13,6 eVnE E= = − = −
• free electron: n → ∞ , 0nE →
(iii) stationary states:
• ground state: 1n = , lowest energy: 1 13,6 eVE = − , lowest
radius: 1 0 0,0529 nmr a= =
• excited states: 1n > , higher energies, larger radii
(iv) transitions between stationary states (see FIG. 5-10):
• transition of 1 electron → emission of 1 photon
• photon wavelength / line spectrum:
consider: electron on state with 2n moves to state
with 1 2n n<
photon energy: 2 1n nhf E E= −
with 20 /nE E n= − → ( )2 2
0 2 11/ 1/hf E n n= −
with c fλ= → ( )2 202 11/ 1/ 1/
En n
chλ = −
0 13,6 eVE = , 83 10 m/sc = × , 346,626 10 Jsh −= ×
→ ( )2 22 11/ 1/ 1/HR n nλ = − , 71,096 10 1/mHR = ×
agrees with Rydberg Ritz formula!
159
→ line spectrum of hydrogen could be explained!
• ionization: atom in ground state, removal of outer electron,
necessary energy = ionization energy IE , for hydrogen:
13,6 eVIE = −
(see also work function for electrons in solids,
chapter 4.4.3)
• problems of Bohr's atomic model:
explanation of experiments (a,b,c,e):
explanation of experiment (d):
transfer of the model to other atoms or molecules:
(line spectra of other atoms can't be explained)
periodic systems can't be explained
→ Bohr's atomic model holds only for hydrogen!
• solution: Schrödinger's theory of quantum mechanics
160
Chapter 5.3. Wave particle duality II: electrons, neutrons, αααα-
particles, ...
photon:
• see chapter 4.4: photon has wave and particle properties
• wave-properties: Maxwell's equations, wavelength, phase velocity
• particle properties: momentum p
• combination: Einstein, de Broglie: E hf= , /p h λ=
FIG. 5-13. Diffraction for different types of particles. (a) Diffraction pattern produced by x-rays of wavelength λ = 0,071 nm on an aluminium-foil target. (b) Diffraction pattern produced by Ekin = 600 eV electrons (de Broglie: λ = 0,05 nm) on an aluminium-foil target. (c) Diffraction pattern of Ekin = 0,0568 eV neutrons (λ = 0,12 nm) incident on a copper foil. (d) A two slit electron interference pattern [5.02].
electron, neutron, α-particle and other classical particles:
• experiments: electron, neutron, hydrogen, helium, ... diffraction at
atomic crystals – (see FIG. 5-13)
• classical particles have wave properties!
• particle properties: mass m, momentum p
• wave properties: energy E hf= , phase velocity c fλ=
• combination: de Broglie: /p h λ=
When can one "see" wave properties of particles?
161
Obstacles like slits or gratings and wavelengths must have equal
dimensions
e.g.: • 600 eVkinE = electron: 2
212 2kin
pE mv
m= = , /p h λ=
→ 2
22kin
hE
mλ= →
2
... 0,05 nm2 kin
hmE
λ = = =
→ appropriate obstacle: atomic crystal
• baseball with mass 0,17 kgm = moving with
100 km/hv = : 2
34... 1,4 10 m2 kin
hmE
λ −= = = ×
→ appropriate obstacles don't exist on earth!
Wave functions for classical particles:
• string: displacement ( , )y x t
• sound: displacement ( , )s r t of air molecules or density ( , )r tρ or
pressure ( , )p r t
• light (photons): electric / magnetic field: ( , )tE r / ( , )tB r
• electrons, neutrons, ...: ( , )r tψ
Interpretation of the particles' wave function:
• energy density of light:
particle model: /dE dVη = , E Nhf= , N = number of photons →
dE dNhf= → / /dNhf dV dN dVη = ∝
wave model: 2η ∝
162
→ 2/dN dV ∝ E : square of photons wave function
proportional to number of photons per
unit volume!
• consider: 1 single photon traveling:
2 1, photon is in unit volume0, otherwise
dNdV
∝ =
E
→ 2( )P r dV= E = probability of finding photon in unit volume
(proportional to size of volume dV also)!
• here: 2( , ) ( , )P r t r t dVψ= = probability to find classical particle at
time t in volume dV !!!
2 ( , )r tψ is called probability density
→ square of wave function can be interpreted, but the function
itself?
• photon: ( , )r tE = vector of electric field, makes charges moving
(Coulomb force acts)
• particle: ( , )r tψ : no idea, would be one of my questions to god...
Normalization:
• the particle at least must be somewhere:
2 1V
dVψ→∞
= (normalization condition)
163
Chapter 5.4. Schrödinger's equation and quantum mechanics
5.4.1 Schrödinger's axiom (1928):
2 2
2
( , ) ( , )( , )
2x t x t
U x t im x t
ψ ψψ∂ ∂− + =∂ ∂
→ basic equation for theory of quantum mechanics (books: wave
mechanics, quantum theory)
solution: wave function, holds for particles with wave and particle
properties
( )0( , ) i kx tx t e ωψ ψ −=
normalization:
• boundary condition, leads to standing waves
• compare: standing waves on string (chapter 3.3):
quantization of frequencies, wavelengths, depending on
boundary condition
• here: boundary condition leads to quantization of frequency,
energy!
function U :
• potential energy in quantum mechanics
• depends which forces act on particle
5.4.2 example: particle in a box
• classical pendant: string, fixed at both ends
(see FIG. 5-14), wavelength results: 2
n
Ln
λ = , 1,2,3,...n =
164
FIG. 5-14. Standing wave on a string fixed at both ends.
FIG. 5-15. Potential energy in quantum mechanics for a particle in a b
165
• here: particle is free if he moves between walls, at and beyond
the walls infinitively high forces bring it back between the
walls (see FIG. 5-15).
potential energy:
(i) between the walls ( 0(0, )x x∈ ): 0F =
→ 0
( ) ( 0) ( ') ' 0pot pot refx
E x E x F x dx− = = =
with ( 0) 0pot refE x = = → ( ) ( ) 0potE x U x= =
(ii) left wall ( 0x ≤ ): F → ∞ → ( ) ( )potE x U x= → ∞
(iii) right wall ( 0x x≥ ): F → ∞ , ( ) ( )potE x U x= → ∞
• Schrödinger's equation:
(ii) left wall ( 0x ≤ ): ( )U x → ∞ → ( , ) 0x tψ =
(iii) right wall ( 0x x≥ ): ( )U x → ∞ → ( , ) 0x tψ =
(i) between the walls, 0(0, )x x∈ : ( ) 0U x =
2 2
2
( , ) ( , )2
x t x ti
m x tψ ψ∂ ∂− =∂ ∂
approach: ( )0( , ) i kx tx t e ωψ ψ −=
→ 2
2( ) ( , ) ( ) ( , )2
ik x t i i x tm
ψ ω ψ− = −
→ 2 2
2k
Em
ω= = → 2
2mEk = ±
→ solution: ( ) ( )0,1 0,2( , ) i k x t i k x tx t e eω ωψ ψ ψ
+ −− −= +
• boundary conditions: ( 0, ) 0x tψ = = , 0( , ) 0x x tψ = =
0,1 0,2(0, ) 0i t i tt e eω ωψ ψ ψ− −= + = → 0,1 0,2 0ψ ψ ψ= − =
166
0 0( ) ( )0 0 0( , ) 0i k x t i k x tx t e eω ωψ ψ ψ
+ −− −= − =
with k k k+ −= = −
→ 0 00 0 0 0[2sin( )] 0ikx ikxe e kxψ ψ ψ−− = =
→ 0sin( ) 0kx = → 0nk x nπ= , 0, 1, 2,...n = ± ±
with 2 /n nk π λ= → 02 /n x nλ λ= = , 1,2,3,...n =
(values 0n ≤ don't make sense)
→ same result as for wave on a string fixed at both ends!!!
energy: 2 2
2n
n
kE E
m= =
→ boundary conditions lead to quantization of
energy!!!
• normalization condition:
0
2
0
1x
dxψ= , with 0( , ) 2 sin( )x t kx tψ ψ ω= −
→ 0 0
2 2 20
0 0
1 4 sin ( )x x
dx kx t dxψ ψ ω= = −
→ 0 02/ xψ =
• solution: 0( , ) 2 / sin( )nx t x k x tψ ω= − , 0/nk n xπ= ,
2 2
2n
n
kE
m= , 1,2,3,...n = (quantum number)
(see FIG. 5-16a)
• probability density:
167
2 20( , ) 2 / sin ( )nx t x k x tψ ω= − - see FIG. 5-16b
FIG. 5-16. Wave function and probability density for a particle in a box.
168
Chapter 5.5. Atomic model in quantum mechanics
5.5.1 The hydrogen atom
(a) The Schrödinger equation
• approach: 2 2
2
( , ) ( , )( , )
2x t x t
U x t im x t
ψ ψψ∂ ∂− + =∂ ∂
• potential energy U : electron in Coulomb field of positively
charged nucleus
classical potential energy: see chapter: 5.2c (Bohr's model): 2( ) / 1/potE r ke r r= − ∝
→ ( ) 1/U r r∝ `
• mathematical problems:
before: ( )U U x= → 1 dimensional problem
know: ( )U U r= → 3 dimensional problem, spherical
symmetry
→ calculation of , Eψ difficult
(but possible only for hydrogen!)
(b) solutions: wave functions
• boundary conditions in 3 dimensions → wave function depends
on 3 quantum numbers , ,n l m : , ,n l mψ ψ=
• n : determines range of , ,n l mψ
• ,l m : determine shape of , ,n l mψ , (see FIG. 5-17)
• @ Bohr's radii: maxima of 2, ,n l mψ → electron most probably at
Bohr's positions!
169
FIG. 5-17. Probability density ψ2 for electrons in hydrogen atom [5.03a].
FIG. 5-18. Possible directions of orbital angular momentum L [5.03].
170
(c) quantum numbers
• , , , sn l m m
• n : principal quantum number,
values: 1,2,3,...n =
associated with distance electron / nucleus, large
values: electron is probably far away from nucleus
• l : associated with orbital angular momentum of electron
values: 0,1,2,..., 1l n= −
orbital angular momentum in quantum mechanics:
( 1)l l= +L , unit: [ ] 2kg m /sL =
• m associated with orbital angular momentum of electron
values: , 1, 2,..., 1,0,1,..., 1,l l l l l− − + − + − −
( 0l = → 0m = , 1l = → 1,0,1m = − , ...)
orientation of L in magnetic field in (z-direction),
z-component quantized with quantum number m :
zL m= (see FIG. 5-18)
• sm associated with eigen angular momentum of electron
values: 1/ 2sm = ±
spin-up electron: electron rotating counter clockwise,
1/ 2sm =
spin-down electron: electron rotating clockwise, 1/ 2sm = −
(see FIG. 5-19)
pendant: earth moving around the sun while spinning
around its own axes
171
eigen angular momentum: ( 1)s s= +S , 1/ 2s const= =
(same S ) for spin up/down electron
FIG. 5-19. Spin-up (ms = 1/2) and spin-down (ms = -1/2) electrons.
(d) energy levels
without magnetic field: 02n
EE E
n= = − , 0 13,6 eVE =
same as in Bohr's model!
independent of , , sl m m !
with magnetic field: ,n mE E= (m dependent!)
→ explanation of Zeeman effect!
(e) electron states:
all quantum numbers , , , sn l m m together define an electron state
ground state of hydrogen: 1, 0, 0, 1/ 2sn l m m= = = =
first excited state: 2, 0, 0, 1/ 2sn l m m= = = =
(f) transitions
transition from excited state 2 with higher energy to state 1 (can be
either excited or ground state) with lower energy: emission of
photon with energy 2 1hf E E= −
selection rules: only transitions are allowed for which holds:
0, 1m∆ = ± , 1l∆ = ± (see FIG. 5-20)
172
FIG. 5-20. Selection rules for hydrogen. Not all possible transition between different states are allowed, ∆l = ±1 [5.04].
(g) comparison of Bohr's and Schrödinger's model:
Bohr Schrödinger
electron on orbits with electron most probably at
quantized radii nr Bohr's radii - maxima of
probability density
173
energy is quantized energy is quantized
explanation of Zeeman explanation of Zeeman
effect: no effect: yes
holds for hydrogen only holds for all atoms and
molecules
5.5.2 Atoms with more electrons
general information:
• ~ 100 different elements in universe
• each element characterized by atoms with different number of
protons, electrons and neutrons
• atom contains same number Z of electrons/protons (atomic
number) and a number N of neutrons
• atomic mass: concentrated in nucleus, measured in units of
atomic mass unit: 271 1,66 10 kgu −= ×
• atomic mass number A: atomic mass in u ,
e.g. hydrogen: 1,008A = → 27(H) 1,008 1,66 10 kgm −= ∗ ×
• symbols: hydrogen: H, helium: He, lithium: Li, ...
• nucleus: charge nq Ze= , contains protons, neutrons, arrangement:
Schrödinger's equation
• electron shell: charge eq Ze= − , contains electrons, arrangement:
Schrödinger's equation
• ions: eq Ze< − (negative ion), eq Ze> − (positive ion)
174
• chemical & physical properties: determined by number Z and
arrangement 2ψ of electrons in electron shell
(a) The Schrödinger equation
• approach: 2 2
2
( , ) ( , )( , )
2x t x t
U x t im x t
ψ ψψ∂ ∂− + =∂ ∂
• potential energy U : very complicated, electron in field of
nucleus and other electrons! → only numerical solutions
(b) solutions: wave functions
• only slightly different from the hydrogen ones
→ , ,n l mψ ψ= , (see FIG. 5-17)
(c) quantum numbers
same as for hydrogen!
(d) energy levels
,n lE E= (without magnetic field), , ,n l mE E= with magnetic
field
most cases: the lower the quantum numbers the lower the
energy! - see below!
, ,n l mE independent of sm ! → Spin-up and spin-down electrons
have same energies!
(e) electron states:
all quantum numbers , , , sn l m m together define electron state
175
code: shell: 0n = : K, 1n = : L, 2n = : M, 3n = : O, 4n = : P, ...
subshell: 0l = : s, 1l = : p, 2l = : d, 3l = : f, 4l = : g, ...
(f) transitions
selection rules 0, 1m∆ = ± , 1l∆ = ± hold for all atoms!
5.5.3 The periodic system
3 laws are necessary to know the electronic configuration of any atom
in ground state:
1. Pauli's principle:
No two electrons can have the same set of values , , , sn l m m !
In other words: no two electrons can be in the same state!
2. Hund's law:
In the ground state the overall spin has its maximum value!
3. Energy minimization:
The distribution of electrons in states , , , sn l m m works so that the
total energy of all electrons is minimized for the ground state of the
atom.
successive occupation of states with electrons for atoms with low
number Z of electrons to those with larger numbers:
1Z = : hydrogen (H)
law 3: electron has to be in state: 1, 0, 0n l m= = =
176
law 2: electron has to be a spin-up electron: 1/ 2sm =
electron configuration: 1s1 (means that hydrogen has
1 electron in the 1, 0n l= = state)
FIG. 5-21. Electron configuration for hydrogen.
2Z = : helium (He)
1. electron: see hydrogen: 1, 0, 0, 1/ 2sn l m m= = = =
2. electron: law 3: electron has to be in state:
1, 0, 0n l m= = =
law 1: electron must have 1/ 2sm = −
electron configuration: 1s2 (that means: helium has
two electrons in the 1, 0n l= = state)
FIG. 5-22. Electron configuration for helium.
3Z = : lithium (Li)
first two electrons: see helium
3. electron: law 1: electron can't be in state:
1, 0, 0n l m= = = (state fully
occupied)
177
law 3: electron has to be in state
2, 0, 0n l m= = =
law 2: electron has 1/ 2sm =
electron configuration: 1s2 2s1
FIG. 5-23. Electron configuration for lithium. 6Z = : carbon
first 3 electrons: see lithium
4. electron: law 3: has to be in state: 2, 0, 0n l m= = =
law 1: must have 1/ 2sm = −
5. electron: law 1: can't be in state: 2, 0, 0n l m= = =
law 3: must be in state: 2, 1n l= = , no
magnetic field: 1,0,1m = −
possible
law 2: must have 1/ 2sm =
6. electron: law 3: must be in state: 2, 1n l= =
law 2: must have 1/ 2sm = → m must be
different to electron number 5!
electron configuration: 1s2 2s2 2p2
FIG. 5-24. Electron configuration for carbon.
178
FIG. 5-25. Electron configuration for the elements. Sometimes energy minimization doesn't mean that quantum numbers have lowest values: After filling up atom of element Ar (3p orbital is full) 4s state is being filled (K,Ca) before the 3d orbital gets filled (Sc → Zn) [5.05].
179
special cases:
sometimes law 3 doesn't mean that quantum numbers have
lowest values to minimize the energy: (see FIG. 5-25)
e.g. after filling up Ar (3p orbital is full, electron
configuration 1s2 2s2 2p6 3s2 3p6) the 4s state is being
filled (K,CA) before the 3d orbital gets filled (Sc → Zn).
see also: Kr → Rb → Sr → Y, ..., → Cd → In, ...
summary:
quantum theory can explain the electron configuration and
therefore the energy levels of all atoms and molecules
maximum numbers of electrons in shells, subshells, states:
shell
n
sub-shell
l
state
m
no. of sub-
shells
spin
ms
max. no. electrons
in subshell
max. no. electrons in shell
K (n = 1)
0 1s 0 1 ±1/2 2 2
0 2s 0 1 ±1/2 2 L (n = 2) 1 2p 0,±1 3 ±1/2 6
8
0 3s 0 1 ±1/2 2
1 3p 0,±1 3 ±1/2 6 M
(n = 3) 2 3d 0,±1,±2 5 ±1/2 10
18
0 4s 0 1 ±1/2 2
1 4p 0,±1 3 ±1/2 6
2 4d 0,±1,±2 5 ±1/2 10
N (n = 4)
3 4f 0,±1,±2,±3 7 ±1/2 14
32
X (n)
0 ...
n - 1
2n2
180
Chapter 6. Special theory of relativity
physics: data of measurements (distance, velocity, ...) depend on
measuring frame (coordinate system)
FIG. 6-01. Galilei transformation, coordinate transformation in classical physics.
classical physics: Galilei transformation between two systems moving
with constant velocity relative to each other
(see chapter 1.5) (see FIG. 6-01)
results: • distances combine: ' t= +r r v
• velocities combine: '= +u u v
• accelerations identical: '=a a
problems: very accurate measurements of light velocity:
light velocity is always 83 10 m/sc const= × = , in every system!
contradiction: light emitted in system 'Σ has velocity c in 'Σ , should
have velocity c v+ but has velocity c when measured in Σ !
solution: Einstein's postulate: the speed of light is constant!
181
Chapter 6.1 Lorentz transformation
classical Galilei transformation:
' t= +r r v , '= +u u v , '=a a
inverse Galilei transformation:
' t= −r r v , ' = −u u v , ' =a a
relativistic Lorentz transformation:
same as Galilei transformation except for factor γ and different time:
( ' ')tγ= +r r v
inverse Lorentz transformation: ' ( )tγ= −r r v
Lorentz transformation must make velocity of light being constant, no
matter in which system the light is generated and the measurement
is done!
determination of γ :
consider: v const= =v x , wave pulse starting at the origin of Σ at
0t = :
→ ( ' ')tγ= +r r v → ( ' ')x x vtγ= + , 'y y= , 'z z=
→ ' ( )tγ= −r r v → ' ( )x x vtγ= − , 'y y= , 'z z=
after t wave pulse went distance x ct= in Σ , ' 'x ct= in 'Σ
→ ( ' ') ( ) 'ct ct vt c v tγ γ= + = + , ' ( ) ( )ct ct vt c v tγ γ= − = −
182
→ ( )
( )c v t
ct c vc
γγ −= + ↔ 22 2
2 2
11
c v vc cγ− = = −
→ 2 2
1
1 /v cγ =
−
note: 0 v c< < → 1γ ≥
time relations for transition between , 'Σ Σ :
( ' ')x x vtγ= + , ' ( )x x vtγ= −
→ 2[ ( ) '] ( ) 'x x vt vt x vt vtγ γ γ γ= − + = − +
→
2 2
2
2 2
( ) (1 / )'
x x vt x x vt x v c x vtt
v v v v v v
x xv x xvt t
v c v c
γ γ γγ γ
γ γ
− − − − = − = − = −
= − − + = −
inverse: 2
''
x vt t
cγ = +
Chapter 6.2 Time dilatation
consider: two events occur at a single point 0 'x at times 1 't , 2 't in frame
'Σ
times 1t , 2t of these events in frame Σ :
01 1 2
''
x vt t
cγ = +
, 02 2 2
''
x vt t
cγ = +
note: in Σ events don't happen at the same point!
proper time pt∆ : time between events that happen at the same place in
reference frame: 2 1' 'pt t t∆ = −
183
time interval in frame Σ : 2 1 2 1( ' ') pt t t t t tγ γ∆ = − = − = ∆
in Σ time interval between events is longer!!!
explanation for constant light velocity:
consider: flash of light generated at point 'A at time 1 't in 'Σ ,
reflected at mirror and returned to point 'A at time 2 't , beam
propagated distance 2D :
FIG. 6-02. Time dilatation as an explanation of the constant light velocity [6.01].
in Σ :
• flash of light and return happen at two different points
• beam propagates a greater distance 2s D> , but:
• time interval t∆ for the beam to return is longer because of
time dilatation
→ light velocity 8/ 2 / ' 3 10 m/sdc s t D t c const= ∆ = ∆ = = × = !!!
184
Chapter 6.3 Length contraction
consider: stick in frame 'Σ , determination of both ends 1,2 'x at the same
time 0 't (that means: instantaneous length determination)
proper length: 2 1' 'pL x x= −
in Σ : length 2 1L x x= − , measured at the same time 2 1t t= :
• 2 2 2' ( )x x vtγ= − , 1 1 1' ( )x x vtγ= −
→ 2 1 2 1 2 1
0
' ' ( ) ( )pL x x x x v t t Lγ γ γ= − = − − − =
→ /pL L γ=
length appears always shorter as in reference frame!
Chapter 6.4 Relativistic mass and momentum & energy
problem of classical momentum in relativistic systems:
conservation of classical momentum m=p u doesn't hold!
solution: relativistic momentum: rm=rp u
relativistic mass: 0
2 21 /r
mm
v c=
−, 0m = rest mass
relativistic energy:
• rest energy: Einstein: 20 0E m c= , any body has rest energy!
185
• total relativistic energy: 2r rE m c=
• kinetic relativistic energy: , 0r kin rE E E= +
→ 2, 0( )kin r rE m m c= −
example: acceleration of electrons
FIG. 6-03. Acceleration of charged particles between two metal plates C (cathode) and A (anode).
consider: electrons are accelerated on their way from cathode to
anode with voltage 430 kVaccU = in between
• classical calculation:
kinetic energy of electrons: 2 / 2kin accE eU mv= =
→ 2 /accv eU m=
with: 191,602 10 Ce −= × , 430 kVaccU = , 319,109 10 kgm −= × :
→ 1,3v c c≈ > !!!
• relativistic calculation:
186
kinetic energy: 2, 0( )kin r r accE m m c eU= − =
relativistic mass: 2 20 / 1 /rm m v c= −
→ 00 22 21 /
accm eUm
cv c− =
−
↔ 2
02 22 2
0 0
11
1 /acc acceU eU m c
m c m cv c
+= + =−
↔
↔ 2
2 2 02
0
1 /acc
m cv c
eU m c− =
+ ↔
222 2 0
20
/ 1acc
m cv c
eU m c
− = − +
↔ 22
2 2 02
0
1acc
m cv c
eU m c
= − +
with 191,602 10 Ce −= × , 430 kVaccU = , 310 9,109 10 kgm −= × ,
20 81,981 J 0,512 MeVm c = = , 430 kV 0,43 MeVacceU e= × = :
→ [ ]2 2 2 2 21 [0,512 /(0,43 0,512)] 1 0,295 0,705v c c c= − + = − =
→ 0,84v c c= < !!
187
Chapter 7. Nuclear Physics
chemist: nucleus =
• positive point charge
• contains almost all mass
• lays negligible role in structure of atoms / molecules
physicist: close examinations:
• nucleus contains protons p & neutrons n
• interaction of p, n plays important roles, e.g.:
→ fusion of light nuclei (energy source of stars, power
source of future?)
→ fission of heavy nuclei (current major energy source)
Chapter 7.1 Properties of atoms
(a) atoms:
• contain nucleus, electron shell
• nucleus contains number Z of protons p (Z = atomic
number), number N of neutrons n , protons & neutrons =
nucleons
• electronic shell contains number Z of electrons
(= number of protons)
• atomic mass number A Z N= +
• relative atomic mass AM , given in periodic system,
meaning: atomic mass in units of atomic mass unit u ,
mass of an atom of element X : ( ) ( )Am X M X u= ,
188
unit: [ ] 1AM = (unit less), in general:
• atomic mass unit: 1u = 1/12 of mass of carbon nuclide
12C (contains 6 protons, 6 neutrons, 6 electrons),
271 1,6605402 10 kgu −= ×
• mass M of 1 mol particles (atoms or molecules): 1 mol
contains 236,0221367 10AN = × particles, AN = Avogadro's
constant →
→ AM = atomic mass in g/mol (molar mass)
→ in general: AM A≈
• designation of element X in periodic system:
, AMAZ ZX X (more accurate), ...
• isotopes: atoms with same Z , different N
→ isotopes differ in atomic mass
→ chemical properties remain unchanged because of
unchanged electron shell
e.g.:
isotopes of Hydrogen:
- ordinary Hydrogen: 11 H , contains 1 ,1p e
(most stable)
- Deuterium: 21 H , contains 1 ,1 ,1p n e
- Tritium: 31 H , contains 1 ,2 ,1p n e
isotopes of Helium:
23 27
4
6,022 10 1,661 10 kg
10,0025 10 kg = g
A A A A
A
A A
M N M u M N u
M
M M
−
−
= × = ×
= × × ×
= ×
189
- 32 He , contains 2 ,1 ,2p n e
- 42 He , contains 2 ,2 ,2p n e (most stable)
• periodic system: contains most stable isotope
FIG. 7-01. Periodic table of elements. The number above the element's symbol represents the atomic number Z, the number below the relative atomic mass MA [7.01].
• nuclide: nuclei of isotope
e.g. nuclides of hydrogen:
- ordinary hydrogen, contains 1p , deuterium, ...
190
(b) nucleons:
• proton p : charge, 191,602 10 Ce −= × (elementary
charge), mass 271,6726231 10 kg 1,00727647pm u−= × =
• neutron n : charge 0nq = ,
mass 271,6749286 10 kg 1,008664904nm u−= × =
• compare: electron e− : charge pq e= − ,
mass 49,1093897 10 kg 5,48579903 10em u−= × = ×-31
191
Chapter 7.2 Properties of nuclei
(a) forces: strong nuclear force / hadronic force
• fundamental force (like Coulomb or Lorentz force, gravity)
• strong attractive force of nucleons on their nearby
neighbors
• much stronger than repulsive Coulomb force between
protons
• much stronger than gravitational force → gravity always
neglected in nuclear physics
• roughly the same for all kinds of interactions ( p p↔ ,
p n↔ , n n↔ )
• decreases rapidly with distance (ranges ~ fm , only "next
neighbor" interaction)
(b) shape & size
• determined by scattering experiments of high energy
particles (e.g. Rutherford)
• spherical shape, radii 1/30R R A= , 0 1,5 fmR = , A = atomic
mass number
• volume: 1/3R A∝ → 3V R A∝ ∝
• mass: m A∝
• density: m A∝ , V A∝ → / / 1m V A Aρ = ∝ = → density of
nuclei constant for all kinds of nuclei!
192
• liquid drop model: constant density in nucleus analogous
to constant density in a drop of fluid → nucleus behaves
like drop of charged fluid
(c) stability
• light nuclei: N Z=
• heavier nuclei: instability caused by electrostatic
repulsion between protons, minimized by increased
number of neutrons: N Z>
FIG. 7-02. Plot of number of neutrons N versus number of protons Z fro the stable nuclides. The dashed line is N = Z [7.02].
193
• e.g.: AZ X N Z N Z∆ = −
168O 8 8 0
4020Ca 20 20 0
5626 Fe 30 26 4
20782 Pb 125 82 43
23892U 146 92 54
(d) mass & binding energy
• in general: mass (nucleus) < mass (its nucleons)
• e.g. Helium
- atomic mass ( ) ( ) 4,002603Am He M He u u= × =
- mass of nucleons and electrons:
( ) 2 2 2
2 1,007276 2 1,008665 2 0,0005494,03298
p n em p n e m m m
u u u
u
−+ + = + + =
= × + × + ×=
→ ( ) ( )
4,03298 4,002603 0,030377m m p m e m He
u u u
−∆ = + + − == − =
• m∆ = mass defect
• binding energy bE : Einstein: 2E mc= (equivalence of
mass & energy
→ 2bE m c= ∆ × , e.g. Helium:
194
27 16 2 2
11
1119
8
0,030377 1,661 10 kg 9 10 m /s
0,4541 10 J1 e
0,4541 10 CV 1,602 10 C
0,28346 10 eV 28,346 MeV
bE −
−
−−
= × × × ×
= ×
= × ××
= × =
• energy production by nuclear fission or fusion:
plot: binding energy per nucleon /bE A versus A
FIG. 7-03. Binding energy per nucleon Eb/A versus the atomic mass number A. For nuclei with low atomic mass number the binding energy per nucleon increases with increasing A. This can be used to produce energy via nuclear fusion of light nuclei. For A > 50, Eb/A decreases with increasing A. This is used to produce energy via nuclear fission of heavy nuclei.
final product must have less value of /bE A
→ nuclear fission of heavy nuclei ( 50A > )
→ nuclear fusion of light nuclei ( 50A << ) - see later
195
Chapter 7.3 Radioactivity
many nuclei: radioactive
meaning: heavy nuclei decay into lighter nuclei by emission of particles
(photons, electrons, neutrons, α particles)
designation:
• α - decay: emission of 42 He nuclei (Helium atom without electrons)
• β- - decay: emission of electrons e−
• β+ - decay: emission of positrons e+ ( em m= , q e= + )
• γ - decay: emission of γ - rays (high energy photons)
• K - capture: emission of characteristic X-rays
mass defect m∆ : during decay mass of initial nucleus is higher than
mass of final product (final product have less binding energy). Binding
energy difference 2E mc= ∆ appears as kinetic energy of radioactive
radiation, can be used to gain energy ( 2E mc= ∆ = decay energy).
(a) decay law
• radioactive decay is statistical process
(for all decay types!!)
• rate of decay decreases exponentially over the time
(not constant!)
• pressure and temperature effects have little or no effect
because nucleus is well shielded from others by atomic
electrons
• derivation of the decay law, consider
- N = number of radioactive nuclei at time t
196
- M = number of decayed nuclei in time interval dt
- decay of a nucleus is a random event
→ number of decayed nuclei proportional to number of
radioactive nuclei: M N∝
→ number of decayed nuclei proportional to
time interval dt : M dt∝
→ M Ndtλ= , λ = decay constant, unit [ ] 1/sλ =
- number of nuclei at time t dt+ :
( ) ( ) ( )N t dt N t M N t dN+ = − ≡ + , dN = change of
radioactive nuclei after time interval dt ( 0dN < !)
→ dN M Ndtλ= − = −
- integration: dN Ndtλ= − → /( )dN Ndt λ= −
→ dN
dt dtNdt
λ= − → ln( )N t Cλ= − + ,
C = integration constant, → ln( )N t C C te e e eλ λ− + −= =
→ 0( ) tN t N e λ−=
197
• discussion:
FIG. 7-04. Radioactive decay of 235U. The exponential decay is the result of that the radioactive decay is a statistical process. N0 = initial number of radioactive nuclei, τ = mean lifetime, t1/2 = half life time.
- 0CN e= = number of radioactive nuclei at 0t =
(initial value) (see FIG. 7-04a)
- mean lifetime τ : time at which N is decreased to
1/ e of its initial value: 0( ) /N N eτ = → 0 0 /N e N eλτ− =
↔ 1/e eλτ− = ↔ ln( ) ln(1/ )e eλτ− =
↔
0 1
ln(1) ln( ) 1eλτ− = − = − ↔ 1/τ λ= (see FIG. 7-04a)
- half life time 1/ 2t : time at which N is decreased to
1/ 2 of its initial value: 1/ 2 0( ) / 2N t N= → ...
→ 1/ 2 ln(2) /t λ= (see FIG. 7-04b)
- e.g. radioactive decay of nuclide 23592U :
230 6,022 10N = × , 8
1/ 2 7,1 10 at = × !!! (see FIG. 7-04)
• decay rate R : rate at which radioactive nuclei decay:
0 0/ t tR dN dt N e R eλ λλ − −= = − = , 0 0R Nλ= − (initial rate)
unit: [ ] 1 BqR = ("Becquerel"), 1 Bq = 1 decay/s
198
e.g.: after how many half life times is the decay rate
decreased to 1 % of its initial value?
x = number of half life times
→ 1/ 21/ 2 0 0 0( ) 1% 0,01xtR xt R e R Rλ−= = =
→ 1/ 2 1/100xte λ− = ↔ 1/ 2 ln(1/100)xtλ− =
↔ 1/ 2 1/ 2
0
[ln(1) ln(100)]/( ) ln(100) /( )x t tλ λ= − − =
with 1/ 2 ln(2) /t λ= → 1/ 2ln(2) / tλ =
→ ln(100) / ln(2) 4,605/ 0,693 6,645x = = =
→ after 6,645x = half life times the decay rate is
at 1% of its initial value, for 23592U : 9
1/ 2 4,72 10 axt = × ,
more than 1 billion years!!!
199
Chapter 7.4 Decay types
(a) β− decay
• occurs in nuclei that have too many neutrons for stability
• neutron inside nucleus decays to proton + free electron + free
anti-neutrino
(because of conservation of energy and momentum)
• equation of reaction: 1A AZ Z eX Y e
β
ν−
−+→ + +
FIG. 7-05. β- - decay with the emission of an electron and an anti - neutrino. • simplest example: decay of free neutron into proton,
electron and anti-neutrino:
- reaction: 1 10 1 en p e
β
ν−
−→ + +
- half life time: 1/ 2 10,8 mint ≈
- mass defect:
initial mass: 1,008665i nm m u= =
final mass: ... 1,007825f p em m m u= + = =
→ ... 0,00084i fm m m u∆ = − = =
200
- decay energy: 2 ... 0,0781 MeVdE mc= ∆ = =
• anti-neutrino:
- measurements of kinetic energy K of emitted electrons:
FIG. 7-06. Decay of neutrons. In general kinetic energy K of emitted electrons is smaller than decay energy Ed, only Kmax = Ed. Energy conservation leads to existence of so called anti-neutrino with nearly no mass, gets remaining energy Eν = Ed - K [7.03].
→ electrons have different kinetic energies K with
maximum energy max dK E=
→ for maxK K< , energy conservation: there must be
another particle that carries remaining energy
dE E K= −
→ anti-neutrino eν : nearly no mass, no charge, but
kinetic energy
• popular example: decay of 146C
- radioactive carbon dating, dating of former living
organisms (dinosaurs, plants, human beings, ...)
201
- reaction: 14 146 7 eC N e
β
ν−
−→ + + , 1/ 2 5730 at ≈
(b) β+ decay
• occurs in nuclei that have too few neutrons for stability
• proton inside nucleus changes into neutron + free
positron + free neutrino
(because of energy conservation)
• equation of reaction: 1A AZ Z eX Y e
β
ν+
+−→ + +
FIG. 7-07. β+ - decay with the emission of a positron and a neutrino.
• example: 52 5225 24 eMn Cr e
β
ν+
+→ + + , 1/ 2 5,7 dt ≈
(c) γ decay
• occurs when a nucleus in excited state decays to lower
energy state
• nuclear counterpart: spontaneous emission of radiation by
atoms or molecules - see chapter 5)
• radioactive nucleus remains the same after γ decay
• reaction equation: A AZ ZX X
γγ∗ → +
202
FIG. 7-08. γ - decay of an excited nucleus to a lower energy state with emission of a γ - photon. • energy of γ - photon: E hf=
• order of magnitude: spacing of energy levels in nucleus:
~ 1 MeV → / / 1,24 pmc f ch Eλ = = ≈
(for atoms & molecules: spacing of atomic energy levels
~ 1 eV , → 1,24 mλ ≈ )
• half life time: mostly very short, 111/ 2 ~ 10 st −
(d) α decay
• all heavy nuclei ( 83Z > ) are theoretically unstable to α decay
because mass of final products lower than initial mass
• reaction equation: 42
A AZ ZX Y
αα−
−→ +
FIG. 7-09. α - decay with the emission of an α - particle.
• example: Thorium decay
- reaction: 232 228
90 88Th Raα
α→ +
203
- mass defect:
initial mass: 232,038im u=
final mass: 228,031 4,0026 232,0336fm u u u= + =
→ ... 0,0044i fm m m u∆ = − = =
- decay energy: 2 ... 4,08 MeVdE mc= ∆ = =
almost all decay energy: kinetic energy of α - particle
(e) K - capture
• reaction not real radioactivity, but nuclear reaction
• occurs when an electron in K - shell (quantum
number 1n = , 0l = ) stays close to the nucleus
(remember: related probability density is low but not zero
in this orbit) and is captured by a proton
→ proton changes into neutron
→ electron shell is re- filled with electrons out of higher
shells or free electrons with accompanied photon
emission (characteristic spectrum)
• reaction equation: 1 1 2 ...A AZ Z
K capture
X Y hf hf−−
→ + + +
204
FIG. 7-10. K-capture of an electron of a proton in nucleus. Due to successive re-filling of the electron shell characteristic radiation is emitted (photons).
Chapter 7.5 Natural radioactivity
natural radioactivity: decay series
• remember α - decay: all heavy nuclei ( 83Z > ) are
theoretically unstable to α - decay
• daughter of radioactive nucleus is often itself radioactive
(β - or α - decay)
• intensive investigation: all naturally radioactive nuclei can be
assigned to 4 natural decay series: FIG. 7-11
- Uranium Radium series: initial radioactive nucleus: 24294 Pu ,
stable nucleus (product): 20682 Pb
- Uranium Actinium series: initial radioactive nucleus: 23992U ,
stable nucleus (product): 20782 Pb
205
- Thorium series: initial radioactive nucleus: 24092U ,
stable nucleus (product): 20882 Pb
- Neptunium series: initial radioactive nucleus: 24194 Pu ,
stable nucleus (product): 20983 Bi
- special case: nucleus has two different decay types
with different half life times, e.g. Thorium series:
20881
2122128384 e
Ti
BiPo e
α
β
α
ν−
−
→ + → + +
- Neptunium series can't be found on earth because of
low half life times: 61/ 2 2 10 at ≤ ×
(age of earth: 94,55 10 aT ≈ × )
• natural radioactivity: only α -, β- - decay!
206
FIG. 7-11. The four natural radioactive decay series. The neptunium series has already disappeared because of the low half life times of its components [7.04].
207
Chapter 7.6 Fusion & fission
(a) fusion, reaction inside sun
• temperature needed: 710 KT = , all atoms fully ionized!
→ no chemical processes to fabricate energy (no molecular
bonds, no chemical reactions!)
• reaction energy cannot be determined out of mass defect, depends
on reaction!
• example: energy fabrication in sun: p-p chain, 3 ways lead to a
change of 4 H-nuclei in one He nucleus:
FIG. 7-12. Energy fabrication of the sun. In 3 different ways the fission of 4
H nuclei to 1 He nucleus leads to a gain of energy, E(way I) ≥ 26,2 MeV (~
91 %), E(way II) ≥ 25,67 MeV
(~ 9 %), E(way III) ≥ 19,28 MeV (~ 0,1 %) [7.05].
208
- way 1: 91 %
(i) 1. reaction, 2 possibilities:
- 1 1 21 1 1 eH H H e ν++ → + +
fusion energy: ,1 1,19 MeVfE =
- 1 1 21 1 1 eH H e H ν−+ + → +
fusion energy: ,2 1,44 MeVfE =
(ii) 2. reaction: 2 1 31 1 2H H He γ+ → +
fusion energy: ,2 5,49 MeVfE =
(iii) 3. reaction: 3 3 4 12 2 2 12He He He H+ → +
fusion energy: ,3 12,85 MeVfE =
(iv) whole fusion energy:
,1 ,2 ,32 2 26,2 MeVf f f fE E E E= × + × + ≥
209
FIG. 7-13. Different steps in the fission of 236U [7.06].
210
(b) fission, nuclear reactor
• bombardment of heavy atoms with low energy neutrons
(see FIG. 7-13a)
• neutron is captured by nucleus, leaving it in an excited state (see
FIG. 7-13b)
• 2 possibilities:
(i) excited nucleus deexcites by emission of γ - photon
(~ 15 %)
(ii) heavy nuclei (Uranium, Plutonium) begin to oscillate and
finally split into other nuclei and several neutrons
(fission, ~ 85 %) (see FIG. 7-13c)
• generated neutrons can be used to cause further fissions,
producing chain reaction
• example: fission of 23592U
(first discovered by Strassmann, Hahn, 1939)
reaction: 235 236 141 9292 92 56 36 3n U U Ba Kr n∗+ → → + +
mass defect:
initial mass: 1,008665 235,044 236,052665im u u u= + =
final mass: 140.914411 91.926156 3 1,007825
235,866562fm u u u
u
= + + ×
=
mass defect: ... 0,186103i fm m m u∆ = − = =
fission energy: 2 ... 173,7 MeVfE mc= ∆ = =
• problem: control of chain reaction, 23692U is radioactive (see Thorium
series), what to do with the nuclear waste?
211
References: [0.1] W. Demtröder - Experimentalphysik 1, Springer, 2. ed. (2001), p. 27 [1.1] W. Demtröder - Experimentalphysik 1, Springer, 2. ed. (2001), p. 417 [1.2] W. Demtröder - Experimentalphysik 1, Springer, 2. ed. (2001), p. 418 [1.3] P.A. Tipler - Physics, 4. ed. (1999), p. 230 [1.4] P.A. Tipler - Physics, 4. ed. (1999), p. 232 [2.1] P.A. Tipler - Physics, 4. ed. (1999), p. 425 [3.1] P.A. Tipler - Physics, 4. ed. (1999), p. 442 [3.2] P.A. Tipler - Physics, 4. ed. (1999), p. 443 [3.3] W. Demtröder - Experimentalphysik 1, Springer, 2. ed. (2001), p. 362, 365 [3.4] P.A. Tipler - Physics, 4. ed. (1999), p. 488 [3.5] P.A. Tipler - Physics, 4. ed. (1999), p. 492 [3.6] P.A. Tipler - Physics, 4. ed. (1999), p. 453 [3.7] P.A. Tipler - Physics, 4. ed. (1999), p. 457 [3.8] P.A. Tipler - Physics, 4. ed. (1999), p. 464 [3.9] P.A. Tipler - Physics, 4. ed. (1999), p. 1009 [3.10] P.A. Tipler - Physics, 4. ed. (1999), p. 1004 [3.11] P.A. Tipler - Physics, 4. ed. (1999), p. 454 [3.12] P.A. Tipler - Physics, 4. ed. (1999), p. 1040 [3.13] P.A. Tipler - Physics, 4. ed. (1999), p. 1058 [3.14] P.A. Tipler - Physics, 4. ed. (1999), p. 1059 [3.15] P.A. Tipler - Physics, 4. ed. (1999), p. 1076 [3.16] P.A. Tipler - Physics, 4. ed. (1999), p. 1078 [3.17] P.A. Tipler - Physics, 4. ed. (1999), p. 1091 [3.18] P.A. Tipler - Physics, 4. ed. (1999), p. 1092 [3.19] P.A. Tipler - Physics, 4. ed. (1999), p. 1093 [3.20] P.A. Tipler - Physics, 4. ed. (1999), p. 1095 [3.21] P.A. Tipler - Physics, 4. ed. (1999), p. 1096 [3.22] P.A. Tipler - Physics, 4. ed. (1999), p. 1097 [3.23] P.A. Tipler - Physics, 4. ed. (1999), p. 1048 [3.24] P.A. Tipler - Physics, 4. ed. (1999), p. 1112 [3.25] P.A. Tipler - Physics, 4. ed. (1999), p. 1119 [5.01] P.A. Tipler - Physics, 4. ed. (1999), p. 1170 [5.02] P.A. Tipler - Physics, 4. ed. (1999), p. 1171 [5.03] P.A. Tipler - Physics, 4. ed. (1999), p. 1178 [5.03a] U.Leute [5.03] P.A. Tipler - Physics, 4. ed. (1999), p. 1179 [5.04] P.A. Tipler - Physics, 4. ed. (1999), p. 1193 [5.05] W. Demtröder - Experimentalphysik 3, Springer, 2. ed. (2001), p. 191 [6.01] P.A. Tipler - Physics, 4. ed. (1999), p. 1249 [7.01] P.A. Tipler - Physics, 4. ed. (1999), p. AP-6 [7.02] P.A. Tipler - Physics, 4. ed. (1999), p. 1286 [7.03] P.A. Tipler - Physics, 4. ed. (1999), p. 1291 [7.04] W. Demtröder - Experimentalphysik 4, Springer, 2. ed. (2001), p. 42 [7.05] W. Demtröder - Experimentalphysik 4, Springer, 2. ed. (2001), p. 301 [7.06] P.A. Tipler - Physics, 4. ed. (1999), p. 1299
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X EXEMPLARY PROBLEMS FOR THE WRITTEN TEST
X.1 units (2 points)
What is the pressure o a 10 cm high column of mercury (ρ = 13,6 g/cm3) in hPa?
X.2 motion (4 points) An aeroplane starts from airport A to airport B, which is 600km away. The course (N) over ground from A to B is 30°. The wind is blowing with vW = 40 km/h from NW. The relative velocity is 200 km/h. (a) Draw the vector diagram of velocities. (b) Which course will the pilot choose? (c) What is the velocity over ground? (d) How long will the flight take?
X.3 oscillations (3 points) A mathematical pendulum with the length l1 = 120 cm has a period of T1. What length l2 is needed to increase the period by 30 s?
X.4 waves (3 points) A plane wave is described with the formula y(x,t) = y0 sin (ωt – kx). What is the adequate expression using wavelength λ and period T?
X.5 accoustic(4 points) An acoustic intensity I0 is amplified by 6 dB to an intensity I1. What is the ratio of I1/I0?
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X.6 geometric optics (4 points) Given is a parallel glass plate with a thickness of 30 mm and a refractive index n = 1.65. A light beam hits the surface under an angle of β = 45°. (a) What is the parallel shift of the light beam? (b) What changes when the experiment is performed in water (n = 1.33)?
X.7 physical optics (4 points) Given is a white light source from 400 nm to 700 nm and an optical grating with d = 4 µm. The first deflected order is detected on a screen which is placed 1 m behind the grating. What is the width of the first order from blue (400 nm) to red (700 nm) on the screen?
X.8 quantum physics (4 points) The external photoeffect is measured in 2 experiments using light with λ1 = 330 nm and λ2 = 240 nm. For an opposite voltage of -3,9 V and -5,05 V the photocurrent could be compensated. Determine the value of h from these data!
X.9 atomic physics (4 points) Please tell the quantum numbers and describe their physical meaning.
X.10 nuclear physics (5 points) During the Tschernobyl accident in 1986 the isotope 137Cs was released. The half life time is 30 years. When will the activity of 137Cs reach 10 % of its initial activity?