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1
The course is supposed to be
BilingualInteractiveOpenScientific
Scientific
methodologymorality or ethicshistory and forthgoers’ innovation
thinking
pseudo-sciences, pathological science
伪科学、病态科学
道德
原创性
…
way and content of teaching
knowledge
All lecturenotes(pdf)Supplements for University PhysicsUphysics2nd.pdferrata
The following materials will be put on net: Textbook
(1) Resnick R, Halliday D, Krane K S. Physics5th ed. John Wiley & Sons, 2002
References
(2) Ford K W. Classical and Modern Physics. Xerox College Publishing, 1972
(3) Alonso M, Fin E J. Fundamental University Physics. Addison-Wesley Publishing Company, 1978
(4) Orear J. Physics MacMillian Publishing Co. Inc., 1979
(5) Feynman R P, Leighton R B, and Sands M.The Feynman Lectures on Physics. Addison-Wesley Publishing Company, 1975
1
University Physics
PERSON OF THE CENTURY
Philippe Halsman (1948)
HE WAS ANENTHUSIASTIC BUTNEVER BRILLIANTAMATURE MUSICIAN
MechanicsThermal PhysicsElectricity and MagnetismFundamental Modern Physics
Introduction
Chapter 1 Introduction Matter and Motion, and their Interplay
Order, symmetry, similarity, simplicity
Matter and Interaction
1.1 What is physics?Physics is the discipline of science most directly concerned with the fundamental law of nature.
The eternal motive is to seek
2
Matter:microscopic — fundamental particles
macroscopic —
cosmological — universe
mesoscopic — nm, cluster介观 团簇
宇观
monopole Cabrera et al (1982 Stanford)
dark matter90% matter invisible
fractional charge B. Fairbank et al (1965, 1977~1981)
to search for
antimatter, say, anti-hydrogen atomAMS
news: multi-quark particle;supersymmetry matter?
mechanicalthermalnucleus particles ...
Motion:
gravitational, weak, electromagnetic, strong
Interaction:
electroweak
Einstein: unified field theory“Is There a Fifth Fundamental Force?”
3
gravitational, weak, electromagnetic, strong
Interaction:
C.N.Yang
electroweak
supersymmetry
Grand Unification Theory
邓伟摄(1994)
spatial (taiji, Eight trigram..), temporal, matter-antimatter, gauge dynamic
Symmetry:
Day and Night by Escher M C (1898-1972)
1
What is physics
EM Interaction →Exchange of photonStrong Interaction → Exchange of gluon
Matter: Field propagates interaction.
This is even more distinct in advanced theory:It is a special form of matter.
Experiments provide enough variation and flexibility 4.2K, nK!
Experimental Theoretical
Computational
Physics
Five Great Theories in physics:
•Newtonian Mechanics (classical mechanics)•Thermodynamics•Electromagnetism•Relativity•Quantum mechanics
None will ever be completely overthrown. None will prove to be entirely correct. No theory is unique...
•Physics is the most fundamental science.
•Physics provides most and fundamental means of scientific research.
•Physics is the most developed science.
*Quantum information & computation
10 21 CC +=ψ
baχφ≠
qubits
entanglement 纠缠
( )babaab
11002
1+=ψ
quantum statesboolean states: 0 and 1;
Quantum computer !
2
•to obtain facts & data through observation, experiments, or computational simulation•to analyze facts & data in light of known and applicable principles•to form hypotheses that will explain the facts•to predict additional facts•to modify & update hypotheses by new evidence
Normal scientific method:Normal Abnormal?
Einstein? Michelson-Morley experiment
To ignore alleged facts:not really fact; irrelevant or inconsistent; mask other more important facts;complicating a situation.
Intuition(premonition)Dumb luck (serendipity)
At college level you must do more thanlearn facts, laws, equations and problem-solving techniques.• to seek to grasp the whole of physics, • to appreciate its generality, • to get an idea of the beauty, simplicity, harmony, and grandeur of some of the basic physical laws, • to see the interconnection of its parts and perceive its boundaries.
theory and application, physical idea and mathematical tools,general law and specific fact, dominant and irrelevant effect, traditional and modern reasoning.
• to distinguish between
1.2 Physical quantities
• a set of concepts, usually not directly perceptible• assumption about the mathematical
representation (physical quantities)• relationships among physical quantities
(equation)
A general theory of physics is made up of
Personal preference or historical accident
e.g., Temperature:Celsius(0o C), Fahrenheit, Kelvin
A certain mathematical representation of a physical quantity is never natural, assumptions are necessary.
3
Basic quantities are defined through measurement (operational definition)standard & unit, procedure
Basic or Derived —Arbitrary!
Derived quantities are defined in terms of other quantities 1N = 1kg·m/s2
Electric quantity Q or current I basic?
SI base unit
Le Système International d’Unites
cdcandelaLuminous intensitymolmoleAmount of substanceKkelvinTemperatureAampereElectric currentssecondTimekgkilogramMassmmeterLength
Length:
Moving body?
一尺之杵,日取其半,万世无穷
Typical lengths: table 1-2 6110
Space quantized—fundamental length
Kr86
sm458792299=c
standard meter1960 atomic standard1983
AU(Astronomical unit) 1.495 978 70 × 1011m
light year 9.460 530 × 1015m
pc, parsec, parallax second 3.261 633 l.y.
Å, Ångström m100.1 10−×
1
1AUpc1′′
=
Example 1.1
AU206265=
°π
′′°
18003601
Assignment: 1.1, 1.3/1.4 , 1.5
1 10 100 1000
Problem 1.5 ‘logarithmic scale’
1
Mass
Standard kilogram
C12121u1 m=unified atomic mass unit
( ) kg103173660538.1u1 27−×=
1965: 1kg + 0.271mg1989 ~ 1993: 1kg + 0.295mg
kg No.60 90% Pt + 10% Ir
39mm
39mm
12312
2121 rF
rmmG−=
Gravitational mass vs inertial massactive, passive
passivemactivem →→ 21 ;
Mass is motion-dependent:
Does rest mass conserve?
Particles moving with c m0 = 0
Fission and fusion
)(vmm =Rest mass )0(0 mm =
?0=νm
neutrino 中微子
Time1second = (1/86400) mean solar day
Cs133Atomic standard
1956 IBWM:1/31556925.9747 of
a standard tropical year—1900
Twin paradox
佯谬
A clock counts repeated events of motion.A moving clock runs slower.Unidirectivity of time, arrow,
Universal physical constants:
211
34
kgmN106.673(10)
sJ10)82(596571054.12
⋅×=
⋅×=π
=
−
−
G
hh
Gamow G, Stannard R. The New World of Mr. Tompkins. Cambridge University Press 1999.(中译本:伽莫夫,斯坦纳德.物理世界奇遇记(最新版).吴伯泽译.长沙:湖南教育出版社,2000)
2
Dimension and dimension analysis
Geometrically area~(length)2
volume ~(length)3
e.g. dim velocity~L1T–1
mechanical quantity~LαMβTγ
physical quantity~(basic quantity)d
Hubble constantH0 = 50~100km·s–1·Mpc–1
dim H0–1 = T
量纲、维度
Dimensionless quantity:
ηρvdeR =Reynold’s number
The slowdown rate ~ s/day, s/century
Dim ( ) = L0M0T0 = 1
dimension ? unit
Plasma parameter ae2β Γ =
TkB
1=β
Example 1.2Ldim =γβα cG h
11TL dim −=c
m106.1 352
1
3−×≈
=
cGlgh
TTMLdimdim 22 −=h22dimdim mFrG =
2222 MLTML −−= 214 TML −−=
Known gravitational system of m, R ——— energy
WU ~
The gravitational self-energy
model-dependent constant: uniform sphere 0.6, thin shell 0.5
RmGU
2−=
sphericalrF ⋅~ r
rmmG ⋅− 2
21~
1. spherical: single parameter R means spherical,
6. numerical factor is model-dependent, beyond the dimensional analysis.
5. physical systems are divided into layers, certain layer is refered.
4. meaning of the quantity: broken to sheet.
3. either negative sign or positive sign is ok.
2. G is introduced due to “gravitational” , different from Example 1.2.
Assignment:1.8, 1.9*
* optional
1
1.3 Approximation in physics
Correct and ingenious use of approximation is fundamental for scientific workers.
Exact solution model approximation
g = constant?
212
21
041
rqqF
επ=
numerical computation, math approximation
∞<<=→
CCxgxf
ax0,
)()(lim
))(()( xgOxf =→f,g are of same order of magnitude (10)
Math approximation
数量级
if
)( 2xO
)( 4xO
=+ 42 53 xx∞→x
0→x
0)()(lim =
→ xgxf
ax
f is infinitesimal relative to g
Usually we say a << b if a is two order smaller than b
1e 1 <<− means ...
))(()( xgoxf =
ifTaylor series:
L+′′+′+=+ 20000 )(
!21)()()( xxfxxfxfxxf
1)1(!2
11)1(
111
1
2
2
<+−++=+
<+++=−
xxqqqxx
xxxx
q L
L
For nonlinear system, problem is sensitive to initial values.
“Butterfly effect”:
Assignment: 1.12
The flap of a butterfly’s wings in Beijing might set off a tornado in New York .
1.4 Vectors
Scalars: magnitude (a number & a unit) e.g., m, l, t, ρ, E, T...
...,,, EΜa,v,rvrrvrv ω∆
Hand-writing
Vectors: have magnitude, direction,
...,,,,, EΜavr ω∆e.g.,combine according to specific rules
&
矢量、向量
标量
1
• geometric: directed line segment
Representations of vector
AA Ae =
• tensor of rank one
),,( 321 AAA=A• analytic e.g.
A=A),(),( rFrr WWt ==
Unit vector
解析的
一阶张量
Combination rules1.Addition :
parallelogram law or triangle law
Combination rules1.Addition :
+=+ BA• Commutative
B AA
A+BB
?xyyx θθθθ +=+
Combination rules1.Addition :
• CommutativeABBA +=+
AA =+ 0• null vector
CBACBA ++=++ )()(• associative
Zero, naught
2. Multiplication by a scalar
;CA =λ0>λ parallel to A
• associative AA )()( λµµλ =
)1( BABA ×−+=−
0<λ anti-parallel to A
BABAAAA
λλλµλµλ
+=++=+
)()(• distributive
AC λ=
2
3. Scalar (dot, inner) product
A • B = (a scalar function)θcosAB
ABBA ⋅=⋅
What does mean?0=⋅BA
CABACBA ⋅+⋅=+⋅ βαβα )(02 ≥=⋅ AAA
?=⋅ eAA
eθ
θcosA
projectioncomponent
Resolution
2211 eeA AA +=
coplanar;,, 21 eeA
*In three-dimensional space?
)( )( 2121111 eeeeAe ⋅+⋅=⋅ AA
collinear.not , 21 ee
)( 111 ee ⋅= A when e1·e2 = 0
1A= when e1·e1 = 1
4. Vector (cross, external) product
)0(sin π≤≤=× θθABBA
A, B, and C in right-handed screw
CBA =×
A B
BAC ×=
θ
4. Vector (cross, external) product
)0(sin π≤≤=× θθABBA
A, B, and C in right-handed screw
CBA =×
ABBA ×−=×( )
0=××+×=+×
AACABACBA βαβα
3
polar
polar & axial vectors
axial
5. Triple products
( ) ( ) ( )( ) CAB
ACBBACCBA⋅×−=
⋅×=⋅×=⋅× •
( ) ( ) ( )BACCABCBA ⋅−⋅=×××
( )( )222
111
coscos
ϕωϕω+=+=tAxtAxExample 1.3
iAiA ⋅+⋅=+ 2121 xx
1x2x
( ) iAA ⋅+= 21
( )δω +≡ tAcos
Example 1.4 ( ) ( ) ( ) L+++ ϕϕ 2coscos0cos AAA
22sin AR =
ϕ
2,21
2sin 21 =+=
⋅ nnR AAϕ
⋅
=+
2sin
22sin
21 ϕ
ϕ
AAA
∑−
=
+−
1
0
0 cosN
nNntkx
NA ϕω
−
+−
⋅
= ϕωϕ
ϕ
NNtkx
N
NN
NAA
21cos
2sin
2sin
0
*
1
Assignment: 1.14, 1.17, 1.19*, 1.22
( )DCBA
exp,,ln,1illegal: , A = B
See §1.5
1.5 Orthogonal Coordinate systems
Cartesian system(i, j, k)正交 basisbase vector
1.5 Orthogonal Coordinate systems
,1,0
=⋅=⋅=⋅=⋅=⋅=⋅
kkjjiiikkjji
Cartesian system(i, j, k)
jiki,kjk,ji
=×=×=×
orthogonal
normalized 归一化
right-handed screw
kjiA 321 AAA ++=kAjAiA ⋅=⋅=⋅= 321 , , AAA
• i • j • k
332211 BABABA ++=)()( 321321 kjikjiBA BBBAAA ++⋅++=⋅
321
321
BBBAAAkji
BA =× ( )321
321
321
CCCBBBAAA
=⋅× CBA
3331
232113
3331
232112
21
3332
232211
333231
232221
131211
)1(ccbb
accbb
accbb
acccbbbaaa
+−+= +
322333223332
2322 cbcbccbb
−=
Position vector r) , , ( zyx
coordinates of end point
components) , , ( zyx
2
Base vectors must be linearly independent and complete.
With zero vector in it, a set is linearly dependent.
holds except the trivial one with C1 = C2 = L = Cn = 0.
0=+++ nnCCC AAA L2211
*
A set of vectors are said to be linearly independent provided no equation
orthogonal or oblique? right-handed or left-handed?
Either OK
In a particular space there exist n linearly independent vectors but no set of n + 1 linearly independent ones, the space is said to be
n-dimensional.
E.g. for vectors A and B, C1A+C2B = 0, C1,C2 finite
they are linearly dependent. IndeedA = –(C2/C1) B
*
Completeness requires the number of base vectors equal to the dimensionality of space.
So two linearly dependent vectors are collinear.
Polar coordinate systems
0 ,1 =⋅=⋅⋅ ϕρϕϕρρ eeeeee =
ϕϕρρ eeA AA +=
planar ~
ρρ e =ρposition vector:
) , ( ϕρcoordinates of end point
)0 , (ρcomponents
jiejieϕϕ
ϕϕ
ϕ
ρ
cossin
sincos
+−=
+=
ρ
ϕ
eρeϕ
i
j
O
jiejieϕϕ
ϕϕ
ϕ
ρ
cossin
sincos
+−=
+=
i
j
ρ
eρeϕ
Cylindrical ~ ) , ,( kee ϕρ
keeA zAAA ++= ϕϕρρ
kr z+= ρPosition vector
), , ( zϕρcoordinates of end
),0 , ( zρcomponents
3
spherical ~ ) , ,( ϕθ eeerϕϕθθ eeeA AAA rr ++=
rPosition vector
), , ( ϕθrcoordinates of end
)0 ,0 , (rcomponents
)sin(coscossin)sincos(sincos jike
jikeϕϕθθ
ϕϕθθ
θ ++−=++=r
jie ϕϕϕ cossin +−=
( )
( )ϕϕϕ
ϕϕϕ
ϕ
ρ
&
&
jie
jie
sincosd
d
cossind
d
−−=
+−=
t
t ϕϕe&=
ρϕe&−=
jiejieϕϕ
ϕϕ
ϕ
ρ
cossin
sincos
+−=
+= )( )( tt ϕϕρρ == ,
Derivation against time
ϕρ ϕee d1d ⋅=
ϕρ ϕe
e&=
tdd
ρϕ
ϕρ
ϕ
ϕ
ee
ee
&
&
−=
=
t
t
ddd
dρρ e =ρ
ϕ
ϕ
ϕρϕ
ϕρ
e
e
&&&&&
&&
+−=
=
ρρ
ρ2
In circular motion
In uniform circular motion
ϕρ ϕρρ ee &&& +=ρ
ρe)(=ρ&& ϕe)(+
0== ρρ &&&
constant=ϕ&
2ϕρρ &&& − ϕρϕρ &&&& 2+
ϕρ
cos1 ep
+=Example 1.5 )10( << e
Find the radial acceleration at π= ,0ϕ2 ϕρρρ &&& −=a
1
( )0 ,
1
22min2
2 =−=+
−= ϕϕϕρ pr
epa
( )π=−=
−−= ϕϕϕρ ,
1
22max2
2 pr
epa
*intrinsic coordinate system tangential and normal directions
内秉
ttdd eev vts
==
ts
sv
tvv
t dd
dd
dd
dd)(
dd t
ttτ
τeeea +==
232
2
2
dd1
dd
1dd
+
==
xy
xy
s ρτ
v
curvaturen
2
tdd ee
ρv
tv
+=
radius of curvature
Assignment:1.21
1
Part 1 Mechanics
• position (distance and direction) • orientation• deformation
mechanical motion
spacetime
Kinematics & dynamics
description of motion law & reason of motion
Isaac Newton1687 Principia Methematica Philosophia
Naturalis
Isaac Newton1687 Principia Methematica Philosophia
Naturalis
Albert Einstein1905 special theory of relativity1916 general theory of relativity
Schrödinger, Heisenberg, Dirac et al1920s Quantum mechanics
Few-body and many-body
Thermodynamic system 2310
scale, size; two levels
1,2,3 ~ 18
Macroscopic and microscopic
2
deterministic vs uncertain
• quantum mechanical uncertainty of microscopic objects;
• statistical uncertainty of individual particle in many-particle system;
• unpredictability in nonlinear dynamic system.
Chapter 2 Kinematics
2.1Mechanical motion and moving object
translational motion — particle or mass point system of particlescontinuum or fluid
position change
rotational motion — rigid bodyorientation change
oscillation and deformation — elastic or plastic bodies
elastic or plastic body
earthquake, crust movement
?
rigid bodyrotation(mean radius)
6.4×106m
mass pointrevolution (mean orbital radius)
1.5×1011m
earthmotionscale
Motion of the earth
Reference frame, or coordinate system is needed.
Motion is relative.
Spacetime coordinates
2.2 Translation
t∆∆
=rv
tsv
∆∆
=
Average velocity
Average speed
速率
∆r displacement
位移∆s pathPosition change
3
instantaneous velocity and speed
ttt ddlim
0
rrv =∆∆
=→∆ t
stsv
t ddlim
0=
∆∆
=→∆
tddva =
acceleration
,
tsterrv ===
dd
dd v
ts
dd
∫+= td0 vrr∫+= td0 avv
0d vva −=∫ t
0d rrv −=∫ t
or
tt
B
A
gvvgv+=
=
0
Example 2.1
ttgtttt
dd00∫∫ −= kg
k2
21 gt−=
tt
B
A
gvvgv+=
=
0
Example 2.1
2
21 tghhA −=
20 2
1 tgtvhB −=
ttgtttt
dd00∫∫ −= kg k2
21 gt−=
kgkvg
v−=
=
BA hh =
−= 2
0C 2
1v
ghhh
Collision takes place when and
we have 0vht = .
Substituting it into the expression for hA or hB , we have
hA, hB are magnitude of position vectorsrather than paths.
Discussion:
hv ,0Assigned ghv212
0 ≥satisfy
In case ghv212
0 ≥ …...
4
= kv2
Example 2.2 2vka −=
zvv
tz
zv
tva
dd
dd
dd
dd
===
Find )(tvv =Analysis: one-dimension (1D) implied
finite v, finite a.Solution: Write
kvzv
−=ddWe have
Known
dim k?
∫∫ −=z
z
v
v
zkvv
00
dd
( )[ ]00 exp zzkvv −−=
)(ln 00
zzkvv
−−=
Discussion:
1Ldim −=k?)(tvv =
…...
Galileo Galilei (1564―1642)
independence of motioncomposition and decomposition of motion
Example 2.3
20b
20c
21
21
tt
t
gvr
grr
+=
+= clay pigeon
Projectile
20b
20c
21
21
tt
t
gvr
grr
+=
+=Hit-on condition:
bc rr =t00 vr =
collinear (aim)Discussion: kir 00 hd +=Write
+≥ 2
0
2
020 1
21
hdghv
20
20
2
0C 21
vhdghd +
−+= kkir
220
20
220 tvhdr =+=
021
20
20
2
0C ≥+
−=⋅v
hdghkr
…Assignment 2.2, 2.3, 2.5
1
Circular motion
ϕρ ϕρρ ee &&& +=ρ
ϕω &=ω,, ϕρ ee in right-handed screw relation
ϕϕρ e&Define an angular velocity vector so that ω
ρωρ ×=&
ϕϕρ e& ρρ e
ω
ρO
r′+′=→OOρ
rr ′×=′ ω&
ρωρ ×=&Q
For r’, we still have
r′×=× ωρωr&& ′=ρand
In UCM
)( ρωωρωρ ××=×= &&&
Centripetal accelerationρ2ω−
fT
π=π
== 22ϕω &
Acceleration:(ω = const. )
向心加速度
constant== ϕω &
ρωρ ×=&
1/s
rev/s2.3 Rotational motion
θd
Translation+rotation
tddθω =
Translation~revolution
tddϕω =
ϕd
boomerang
回旋标,飞去来器
platypus: duck-billed animalIt is an oviparous mammal.
2
BA ωω =
Example 2.4The angular velocity of the earth’s rotation
Tπ
=2ω 3652~ πϕ
s236==∆ ωϕT
?mean solar day
s10640.8 4×=T
s10616.8 4×=T
sidereal day Tπ2
ϕ
2.4 Oscillation
periodic motion )()( tATtA =+ fT ≡1
• orbital motion of planets• blood circulation• ecological cycle• economic depression• …...
frequencyperiod
==
tRytRx
ωω
sincos
Spatially back and forth
1
tωωθθ sin0−=&tωθθ cos0=
tt BAx
tAxtBtAx
ωω
ϕωωω
ii ee
)(coscossin
−+=
+=+=
A Amplitude 振幅
ϕωΦ +≡ t Phase 相位ϕ Initial phase 初相位
ω Circular frequency 角频率
( ) ( ) π=−+ 2ttT ΦΦ ωπ= 2T
iR ⋅=+= )cos( ϕωtAx
ϕω +t)cos()cos()cos( 2211 δωϕωϕω +=+++ tAtAtA
( )122122
21
2 cos2 ϕϕ −++= AAAAA
2211
2211
coscossinsintan
ϕϕϕϕδ
AAAA
++
=
Superposition of two oscillation
ωt
21 ϕϕ = , collinear. constructive 1A 2A
π=− 21 ϕϕ 21 AA −=A destructive
1A
2A
221 π=−ϕϕ 22
21 AAA += quadrature
1
==
)(cos)(cos
222
111
tAxtAx
ωω
Oscillations of different frequencies
021
21
====
ϕϕAAASet
( )
+
=
+=+
tA
ttAxx
2cos 2
coscos
21
2121
ωωωω
− t
2cos 21 ωω
?
amplitude modulation
Assignment: 2.7/2.8, 2.9
James Gleick, Chaos:Making a New Science
phase spaceFractals in
分形
Chaos混沌
2.5 Phase space
θ
t
tωθθ cos0=tωωθθ sin0−=&
t
θ&
2.5 Phase space
tωθθ cos0=tωωθθ sin0−=&
2
2.5 Phase space
tωθθ cos0=tωωθθ sin0−=&
θ
?
θ&
12
0
2
0=
+
ωθθ
θθ &
“trajectory”
tωθθ cos0=tωωθθ sin0−=&
θ
θ& 20ωθπ
?
Complete kinematics should include phase diagram. spiralsaddle
nodecenter
2.6 Galilean transformation
→′−=′ OOrr
urr −=′ &&
turr −=′
r r’u = constant
rr &&&& =′
uvv −=′
Solution:Take walker as S’.In walker’s view, the velocity of raindrop is:
Walking in rain
αvu
=αtan
kv v−=
u
ik uv −−=
3
Aberration of light
skm8.29=uSolution:
410~arctan −=cuα
光行差
roughly correct
u
v= – ck gedenkan experiment (thought experiment)
vcdtt
cdt
++=′<=′ outoutf
0fire =toutt
d vcdt+
+out
cdt =′fire
+−>
vccdt 11
out
Large enough dmay break the inequality.
So, Galilean transformation is not valid for light.
Assignment: *2.13, 2.14, 2.15
Fig.3.19&7.10
U
x
2.7 Coriolis acceleration
What is the trajectory of the virusrelative to the table surface?
Coordinate transformation between rotational systems
1
( ) ( ) ( ) tttt ddd AAAA ×=−+= ω
For A fixed in S', its end point is in CM relative to S.
td)(d~ ρωρρωρ ×=→×=&
ωwith respect to S′A′d
( ) tddd AAA ×+′= ω
AAA×+
=
′
ωSS d
ddd
tt
or
For additional
Symbolically
×+′
= ωtt dd
dd
r(t)
rrr×+
′= ωtt d
ddd
rrr&
&&×+
′= ωtt d
ddd
×+
′×+
×+
′′= rrrr ωωω
ttt dd
dd
dd
( )rrr××+
′×+
′= ωωω
tt dd2
dd
2
2
centripetal accelerationCoriolis acceleration
ω = const.
L=
×+
=
=
′ SSS
SSS2
2
dd
dd
dd
dd
dd
dd
ttt
ttt
rr
rr
ω
1
In S, F is centripetal force which causes centripetal acceleration .caIn S’, the ball is at rest, no dr/dt' , d2r/dt'2.
( )rvgg ××−′×−= ωωω mmmm 20
20 sm8.9≈g
srad10292.7 5−×=ω( ) 0
222 sm1039.3 gR <<×≈≤×× −ωrωω
vgg ′×−= ωmmm 2eff
Centrifugal ‘force’Coriolis “force”In S’
falling body v′×− ωm2
ϕθθθ
ekkek
sin)sin(cos
=+×=× ρr
)('2 rvm ek −×−≈ ω)('2 rvm ek ×= ω
ϕθω esin'2 vm=
ϕθωωω
eekekv
sin'2)('2
)('22
vmvm
vmm
r
r
=×=
−×−≈′×− ω
eastward
ϕθθθ
ekkek
sin)sin(cos
=+×=× ρr
0≠θif
2
falling bodyKinematic effect
On the surface
On the surface
( ) ( )kekeee ×+×=+×− ϕϕθθϕϕθθ ω vvvv 22ω
( )ρϕϕθ θω ee vv +−= cos2
ϕϕθθ eev vv +=
On the surface
( ) ( )kekeee ×+×=+×− ϕϕθθϕϕθθ ω vvvv 22ω
( ) rvvv eee θωθω ϕθϕϕθ sin2cos2 ++−=
( )ρϕϕθ θω ee vv +−= cos2
ϕϕθθ eev vv +=
θϕϕϕθ ee vvv →= ;0
ϕθθθϕ ee vvv −→= ;0 south → west
east → south
In southern hemisphere 0cos <θ
righthand?
Railway tracks
river-bank
Molecular rotation-oscillation spectrum
Whirling in satellite cloud diagram. It is counterclockwise in northern hemisphere, and clockwise in southern hemisphere.
Coriolis effect
4
Foucault pendulumAssignment 2.16
v′×− ωm2( )r××− ωωm
Up-thrown Quantitative aspects of qualitativediscussion about up-throw problem:
ω ×v ~ ω ×(ω × v )
ω ×(ω × r )
ω ×(ω × v ) ↔ ω ×(ω × r )
v < escape speed
1
Chapter 3Particle Dynamics
3.1 The law of inertia
Newton’s first law reads:
Every body persists in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by forces impressed on it.
A reference frame in which the first law is validis called inertial or Galilean system
A free particle moves with constant velocity.
force-freeForce is negligibleForces are cancelledlocally
v = constincluding zero
Not true for some frame!Inertial system? Stars?
All inertial systems are equivalent.
urr −=′ &&
constant constant ?
urr −=′ &&
constant constant?
For a free particle in S
( ) RTRRva 22
22π=== ω
Earth inertial system?
Copernicus (1473—1543) (Heliocentric theory) asserts a better inertial system.
3×10–10m/s2sun’s revolution5.9×10–3m/s2earth’s revolution3.4×10–2m/s2earth’s rotation
3.2 Newton’s second and third laws
• The operational definition of force• The definition of inertial mass
1122 aa mm = 2112 aamm =• fundamental equation of dynamics
Newton’s second law
aF m=
2
tddpF =aF m=
vp m= momentum
2
20
1cv
mm
−
=
pvtt
mdd
dd
≡ We may have )(vmm =
Newton’s third law
• contact forces• action-at-distance
Modern viewpoint:Interactions propagate through fields with finite speed.
超距作用
BAAB FF −=
• body• forces• reference frame (origin, axes);• free-body diagram• equation of motion;
geometric relation;• approximation and solution• discussion.
problem-solving Example 3.1Find a’ and nF
x
y
amFgmFF
mamgF
′′=−=′−−
=−
θθ
θ
sin0cos
cos
n
nN
n
θtanxy aa ′=
θtandd xy ′=
x
y
amFgmFF
mamgF
′′=−=′−−
=−
θθ
θ
sin0cos
cos
n
nN
n
θtanxy aa ′=
gmm
ma
ammgam
x
xx
θθ
θθ
tancot
tancot
+′−=′
′=−′′−
gmm
mmamF x θθθ
θ 22n sincoscos
sin +′′
=′′
−=
Example 3.2
0coscos T2T1 : =−− mgFF θθk
θcosT1T2mgFF −=
Solution:
N25m0.1,kg0.1
T1 ===
Flm
( )
sm7.4
N26sinT2T1 :
==
=+=−
mFrv
FFF θρe
Find vF ,T2
Known
3
There is no loss of the number in adding (subtracting).
When multiplying, dividing or calculating square root, the number of significant figure is determined by the factor with the smallestnumber.
significant figure 有效数字 Simple pendulum
• a cord (or rod)• inextensible• of negligible mass
• a particle of mass
( )θθ
θθ
θ
ρ&&
&
mLmgLmFmg
=−−=−
sin :cos: 2
T
ee
• small angular displacement
sin θθ ≈
with to be fixed0,, ϕωA
0=
+ θθLg&&
• initial conditions ( ) ( ) 00 ,0 0 ==== tt θθθ &
( ) 0sin 0 =− ϕωA) cos( 00 tωθθ =
Lg
=20ω
The equation in eθ direction becomes
( )0 cos ϕωθ += tAWe may write
π== 2tLgtω defines period
gLT π= 2
Lgf
π=
21
and frequency
• a cord (or rod)• inextensible• of negligible mass
• a particle of mass
• small angular displacement• initial conditions
( )02 coscos2 θθθ −=
Lg&
For general case (angular displacement )
θθ sinmgmL −=&&
θθθθ dsinddd mgt
mL −=&
⌡⌠−=⌡
⌠θ
θ
θ
θθθθ
0
dsinddd
0mg
tmL&
&
4
0=
+ θθLg&&
0sin=
+ θ
θθθ
Lg&&
20
2 sin~ ωθθω =≤
Lg
Lg“ ”
Comparing the equation
with small angle approximation equation
We may feel
Reliable conclusion comes from exact solution.
Assignment 3.1, 3.2 3.3 Forces
Gravitation
• planet motion, • redshift of light in gravitational field, • accretion, • tide and tidal disruption• ...
Chapter 4123
12
2121 rF
rmmG−=
吸积
Elastic or restoring force
Approximate, empirical
quark confinement? m10 15−
Hooke’s lawxF k−=
Intermolecular force 713
2~
−
rrF σσ
complicated
~ internal structure of molecules
1
Friction
reduce or increase?
complicated not fundamentalapproximate, empirical laws dissipative 耗散
industry and technology Tribology
fundamental studies dune model(de Gennes) self-organized criticality
• nature of materials• surface finish• surface (oxide) film• extent of contamination• temperature• …
Surface friction depends on many factors:
Numerically static friction ~ normal force
Nsf FF µ≤
Sliding friction ~ normal force
Nkf FF µ=
coefficient of static friction
coefficient of kinetic friction
cast iron~cast iron10.1s =µ 15.0k =µ
Teflon on Teflon04.0~~ ks µµ
0=++ gFF mfN
Example:Critical angle
0cos0sin
N
f
=−=−θθ
mgFmgF
In x, y direction
θµ tanN
fs =≥FF
Csarctan θµθ ≡≤i.e.
Nsf FF µ≤
2dd 2
1 vACF ρ=
Frictional drag and terminal speed
2d2
1 vACmgFz ρ−=
Example 3.3
2T
d
2 2 vACmgv ≡=ρ
0=zF
2Td2
1 vACmg ρ=
−= 2
T
2
z 1vvmgF
−= 2
T
2
1dd
vvmg
tvm
−= 2
T
2
1dd
dd
vvg
tz
zv zg
vvv d2
1d
2T
2
2=
−
( )ce12T
2 zzvv −−=
2T0
2T
2 )0(21lnvzg
vv
v−
=
−−
C2T
2T 2
2zz
gvz
vgz
≡=
2
tg
vvv d
1
d
2T
2=
−
tvg
vv
vv
vv d2d
1
1
1
1
TT
TT
=
++
−
tvg
vvvv
T
T
T 2
1
1ln =
+
−
12exp
12exp
T
T
T +
−
=t
vg
tvg
vv
−+
−−
=t
vgt
vg
tvgt
vg
TT
TT
expexp
expexp
tvgvvT
T tanh=
Viscosity force
rvF ηη π= 6
( )[ ] sPa1smm1N 1 ⋅=⋅ −
100~ −=ηρ dveR
Reynolds number
( ) m10~,exp1~ 14002
−− rrrr
F
Nuclear force (between nucleons)
Nucleons have structure.
3.4 Noninertial frame and inertial force
)()( tt uvv −=′
In S’ 1st law×
For a free particle in S,
S’ is called as noninertial frame.
OOt ′−=′ rr )( uaF &mm +′=
In S’ 2nd law ×
aFF ′=+ m)( in
uF &m−≡in
uar &&& +′=Secondly if F ≠ 0,
In S’ 2nd law √
Fin is called as inertial force
3
Example 3.4 Apparent weight 视重
)( ig ag mmP −−=
Apparent weight:ig mm =+ Pg
Equation of motion in Sa
0=−++ )( ig mm PgIn S’
a
acceleration of the elevator
0)( ig =−−= ag mmPWeightlessness
Gravitation can be canceled
We can not distinguish• gravitation and inertial force• gravitational field and accelerated frame
principle of equivalenceGeneral theory of relativity
ig mm = ag =when
In free-falling elevator
Example 3.5
Only in S’Complete solution needs equation of motion for wedge
ymmgFxmmF′=−
′=−&&
&&
θθ
cossin
n
n
θtandd xy ′=′−
x..In wedge frame ( )x.. Conic pendulum
ρϕρ egF 2T &mm −=+
0=++ ρϕρ egF 2T &mm
S
S’
νπ2
mg
rm 2)2( νπ
FN
Problem 3.4Coriolis force
)(2 rotinrot rraa ××−×−= ωωω mmmm &
inertial centrifugal forcekinematic effect
Newton’s second law validthird law not valid
Assignment: 3.4, 3.5, 3.7/*3.9
1
3.5 Momentum and Angular Momentum
vp m=
tddpF =
iii CpF ====
,0, CpF 0
Conservation of momentum
2
20
1cv
m
−
=vp
*
In 1956 C. L. Cowan Jr. and F. Reines et al. detected neutrinos
discoveriesepn +→
The conservation of momentum and energy requires a third particle.In 1927 Pauli proposed the existence of an “undetectable, light, and neutral” particle.
In 1930 E. Fermi named it as “neutrino”
ν+
∫ ∫==−f
i
f
i
ddif
t
t
tFpppp
p
impulse
I≡
冲量
( )if tt −×= F
∫−≡
f
i
d1
if
t
t
ttt
FF
prL ×=angular momentum with respect to O
( )( )ωω
ω
⋅−=
××=
rr
rrL
mmr
m2
In a circular motion
With respect to O
ω2mr=Lvp m=~
Bohr model of hydrogen atom
kg1011.9 31e
−×=m m10529.0 10−×=r116s1013.4 −×=ω
skgm1005.1 234−×=L h=Earth-moon system
s10616.83.27
60~m1008.3 kg1036.74
822
××=
×=×= ⊕
T
Rrm
kg/sm1084.2 230×=L
2
impact parameter
Example 3.6
bmvL =Direction?
Angular momentum exists for finite r, p.
ttt dd
dd
dd prprL
×+×=
torque with respect to O
Fr ×=
M=
力矩
ML=td
d
( ) ( )tmgbtmgrLmgbmM==
=×=
θsingr
∫ tMdθsinrp
With respect to O
3.6 Mechanical Work and Energy
rF dd ⋅=W
3.6 Mechanical Work and Energy
rF dd ⋅=W
∫ ⋅= rF dW ⌡⌠ ⋅⋅= t
ttm d
dd
dd rv
∫ ⋅= vv dm
work-energy relation
tt
tm
ddd
dd rv
⋅⋅⌡⌠=
⌡⌠ ⋅=
tm
ddd rv
3
∫ ⋅= rF dW
2i
2f 2
121 vmvm −=
TTT
∆=−≡ if
∫ ⋅= vv dm
vvv d2d2 2 =⋅=⋅ vvvv ,
kinetic energy
work-energy relation
increment of kinetic energy
∫=f
i
dv
v
vmvf
i
2
21 v
vmv=
2
21mvT = CmvT += 2
21
)()(21
21 2 uvuv +′⋅+′= mvm
Kinetic energy depends on coordinate system!
Does work depend on coordinate system?
uuv ⋅+′+′= )2(21
21 2 mvm
13
Consider a system S'
0fi ==→= papapa WWTT
0=⋅∫ dp
a
rF ∫ ⋅−p
a
drFconservative
∫ ⋅+a
p
drF
Along C1
1C
AlongC2
2C
perihelionaphelion
0d =⋅∫ rF
conservative
( ) rr
r
r
rrFi
i
d U−≡⋅∫CmghU +=
( )
CkxU
kAkxdxkxx
A
+=
−−=−∫
2
22
21
21
21
potential energy function
additive constant
( ) ( )rr UU −= i U∆−=
potential energy
( ) ( )rrrFr
r
UU −=⋅∫ i
i
d
xUFx d
d−=
U−∇=F
zU
yU
xUU
∂∂
+∂∂
+∂∂
≡∇ kjigradient
1D ( ) ( )xUxUxFx
x
−=⋅∫ i
i
d
3D
where
4
(if F is conservative.)U∆−=⋅∫ rF d
T∆=⋅∫ rF d
0=∆+∆ UTConservation of mechanical energy
02
21 mgzmgzvmE =+=
Simple pendulum
)cos1( θ−= Lz
( )02 coscos
21 θθ −= mgLvm
• a particle of mass
• a cord (or rod)
• inextensible
• of negligible mass
• small angular displacement
• initial conditions
Physical pendulum, variable mass
Saddle etc.
Elastic pendulum, nonlinear
Physical pendulum
Nonlinear
222
21
21
21 kxmvkA +=
SHO(Simple harmonic oscillator)
21
Model potential energy curve
At F 0dd <xU0>xF
0dd =xUAt A,B,C,D
0=xFequilibrium
0dd =xUAt A, C,
0dd 22 >xU 0dd <xFRestoring force
At B?At D?
Stable equilibrium
unstable equilibriumneutral equilibrium
1
Potential barrier and well
E?
势垒和势阱 Is potential energy height-dependent?Surely not in elastic potential energy case.
Ammonia molecule is an example that U….
Assignment 3.10, 3.13, 3.14
1
Saturn土 星
Chapter 4Gravitation
Historical review
Gravitation is closely related to the dynamic problem.
The description of planetary motion is a kinematic problem.
There were many “world models”(geocentric, heliocentric, Tycho’s model …). All of them need dynamic explanation.
The frontispiece to Galileo’s Dialogue Concerning the Two World Systems (1632).
According to the labels, Copernicus is to the right, with Aristotle and Ptolemy at the left; Copernicus was drawn with Galileo’s face, however
Claudius Ptolemy(127—152 working in Alexandria, Egypt )
Nicolaus Copernicus (1473—1543)
Ptolemy system(70 spheres)
均轮
本轮
2
均轮
本轮
偏心
Ptolemy system (70 circles)
Copernicus system (46 circles)
Galileo Galilei (1564—1642)
Tycho Brahe (1546—1601)
Johannes Kepler (1571—1630)
Issac Newton (1642 — 1727)
Galileo Galilei (1564—1642)
• establishment of scientific experimental procedures
• elimination of systematic errors (the flexure of his instruments, refraction)
• data with quoted the error: 2 arcmin( )
• systematically 21 years observation.
m1mm61.0<
Tycho’s merits to science
3
Kepler’s nested set: Saturn—Jupiter—Mars—Earth—Venus—Mercury
cube dodecahedron octahedrontetrahedron icosahedron
“Pythagorean” or “Platonic” solids
•Mysterium Cosmographorum (Cosmic Mystery)(1596) •Harmony of the World (1619).
(1) The orbit of each planet about the Sun is an ellipse with the Sun at one focus. (the law of orbit);
Kepler's laws
(2) The line joining any planet and the Sun sweeps out equal areas in equal times. (the law of areas);(3) The square of the period of revolution of a planet about the Sun is proportional to the cube of the planet's mean distance of the Sun. (the law of period or Harmonic law)
• planet motion — Kepler problem and scattering,
• Newton’s law of gravitation,• redshift of light in gravitational field, • accretion, • tide and tidal disruption,• ...
In this chapter we will discuss
12312
2121 rF
rmmG−=
4.1 The law of gravitation
2211 kgmN106.673(10) ⋅×= −G
22~ ωrrvF =
223 1~ ,~ rFrT
torsion balance
Kepler’s third law
22 ~~ TrrF ω
4
Halley's Comet appeared 1456, 1531, 1607, 1682 with a period about 76 years. Edmund Halley predicted 1758.
Successes of Newton's law
The last time Halley's comet visited Earth, in 1986. Comet Halley isn't officially scheduled to visit Earth again until 2061 when it swings through the inner solar system on its 76-year orbit . Photographed from Australia on March 13,1986(Akira Fujii)
The nucleus of Halley’s Comet and some of its dust jets.(© 1987 Max-Planck-Institut fur Aeronomie, H.U.Keller)
Neptune
John Adams (1845.10), ignored by Challis and Airy, the leading observational astronomers.
Le Verrier's work (1846.8) was taken seriously by Calle of the Berlin Observatory. On 9. 23
precession of Mercury's perihelion
海王星
Pierre Simon Laplace (1749—1827)18-body problem
Henri Poincaré (1854 — 1912)randomness in deterministic dynamic systemsoriginator of chaotic theory
( )∑ −−
−=i
ii
immG rrrr 3
( ) ( )ii i
ii mVG rrrr
−−
∆−= ∑ 3
ρ
( )⌡
⌠′−
′−
′− rr
rr 3)d( mVG ρ
∑=i
iFF
5
Shell~particletR
m24π′
=ρ
θθ d sin2d RtRV π=
∫π
⋅π
=0
2
2
cosdsin2 αθθρxtRmGF
dV
F = – Fk
∫π
′=0
2dcossin
21
xmmG θαθ
∫π
02
dcossinx
θαθ
θαθθ
θ
coscosdsin2d2
cos2222
RrxrRxx
rRrRx
−==
−+=
∫ xd?
?∫+
−
Rr
Rr
xd
( )kF −
−+−′= ∫
+
−
xrxRrr
rRxmmG
Rr
Rr
d2
121 222
2
θαθθ
θ
coscosdsin2d2
cos2222
RrxrRxx
rRrRx
−==
−+=
RrrmmG >′
− ,2 k
Rr < ,0=
∫ →
−⌡⌠ →
xxxx
x
d
1d
0
2
grF mrmmG =−= ⊕
3
Outside a sphere, say the earth,
rr
gg d2d
−=grmG =⊕
2
RrrmmG >′
− ,2 k
Rr < ,0F =
2
3
3
r
mRrm
GF⋅
′
−=
Rr
RmmG 2′
−=
Inside a sphere
1
rF 3RmmG
r′
−=
zRmmGFz 3′
−=
zRmmGzm 3′
−=&&
Assignment:4.4
Example 4.1 a chute in sphereExample 4.1
4.2 Gravitational potential energy
⌡⌠ ⋅
′−= rr d3r
mmGW
′−−
′−−=
irmmG
rmmG
m particle m’ particle, shell, sphere
U∆−=
rr
mmGr
r
d1
i
2∫′−=m outside of sphere
0UrmmGU +′
−=
mgzU =
?
+
−=⊕
⊕
⊕
⊕
zRR
RmGmU 1
( ) ( ) ( ) ⊕
⊕
⊕
⊕⊕ +
+−==
RmGm
zRmGmzURU ,0
( ) ( )rmmGrUU ⊕−==∞ ,0
⊕
⊕
⊕
⊕⊕
⊕
+=
+=
Rzz
RmGm
zRz
RmGm
12
⊕
+=
Rz
mgz1
1 [ ]⊕−≈ Rzmgz 1
0UrmmGU +′
−=
mgzU =
?
+
−=⊕
⊕
⊕
⊕
zRR
RmGmU 1
( ) ( ) ( ) ⊕
⊕
⊕
⊕⊕ +
+−==
RmGm
zRmGmzURU ,0
( ) ( )rmmGrUU ⊕−==∞ ,0
free-falling:
zRr += ⊕ ⊕R
?
−=
⊕Rzgzv 122
2
21 vm
RmmG
zRmmG =++
−⊕
⊕
⊕
⊕
1
)(d3d4 ),( 3
22
3
3shellr
Rmrrrcore
Rrm =πρ
⌡⌠−=
R
rRmr
rRmrGU
0
3
2
3
3
s d31
Self-energy
Core to all particles on the shell
RmG
2
53
−=
accretion
RmmGE′
=∆ a released
kgJ10~ 16kgJ1010
1021067.6~ 3
3011
××××∆ −
mE
In nuclear fusion
kgJ103.6 14f ×=∆mE
ppHe HeHe
HeH p
eH pp
433
32
2
++→+
γ+→+
ν++→+ +
p506
RmG
mE '=
∆⊙mm ≈′ km10~∗Rif
orbiting
rmGvrmmGUUEE
rmmGU
′=
<′
−==+=
′−=
022
1k
skm9.71 =≡= ⊕⊕
⊕ vgRRGm
UrmmGmvE
rmmG
rvm
21
21
21 2
k
2
2
−=′
==
′=
orbiting speed
"2648
2222213213
′=
π=
′
π≥
′
π=π
= ⊕⊕ gRmGR
mGr
vrT
Launched from the surface of the earth
rmmGvm
RmmGvm ⊕
⊕
⊕ −=− 22L 2
121
−=
⊕⊕ rR
Gmv21122
L
21vgR == ⊕
−≥
⊕
⊕
⊕
⊕
RR
RGm
212
rmmG
2⊕−=
∞
⊕∞
⊕
⊕ −=−rmmGvm
RmmGvm 22
21
21
Escape speed from the earth
skm2.1122 1E === ⊕⊕ vRGmv
1E <<cv
Weak gravitation
1E ≤cv Strong gravitation
逃逸速度
g22 rcGmR ≡≤cRGmv == 2EIf or
even light can not escape from it. Such a celestial body is called Laplacian black hole.
Laplacian black holegravitational radius
Gc
Rm
2
2≥
It is necessary for a black hole to have higher compactness
致密性
rather than higher density.
11
3
31817 mkg10~10~ρNeutron stars have very high density,
they are not the candidates for black holes.
223
6
3g
3
1~32
3
34
34 mmG
c
r
m
R
mπ
=π
≥π
=ρ
⊙m910
3kg/m20~ρ can still be blackhole
A galaxy of , its density is as low as
The condition for density is
12
4.3 Gravitational mass, redshift and *collapse
~ g Wm 1~ i am 1gi =mm ?
,i 2g am
Rmm
G =⊕
⊕1.
rgP && ig mm −=−2.
gRmGa ==⊕
⊕2
θθ sin gi gmLm −=&&Lg
mm
i
g2 =ω
Newton method: simple pendulum
R. Eötvös 1890+25
0cos cos :
0sincos sin :22
igT
2iT
=+−
=+−
λωα
λλωαθ
RmgmF
RmF
re
e
厄缶
( )g
i
g
2i ~
22sintan
mm
gmRm λωαα ≈≈
gPt
iPt
g
i
mm
mm
=R. Eötvös 1890+25
1964 R. H. Dicke et al
1971 Braginski
8105 −×
1110−
1210−
Gravitational red shiftphoton’s energy (section 22.2) νhE =
mass-energy relation (section 9.2) 2cmE =
zgch d2ν
=
⌡⌠−=⌡
⌠H
zcg
0
2 ddo
e
ν
ννν
2chm ν=Mass of photon
zmgh dd =− νConservation of energy
−≈
−= 2e2eo 1 exp
cgH
cgH ννν
1959, R. V. Pound and G. A. Rebka
1510)26.057.2( −×±
152 1046.2 −×==
∆cgH
νν
1
νdh−( )
zzRmmG d2+
′= ( )
zzRmG
ch d22 +
′=
ν
( ) HRH
RcmG
zRz
cmG
H
+′
−=⌡⌠
+
′−=⌡
⌠2
0
22dd
o
e
ν
ννν
′−≈
′−=
RcmG
RcmG
2e2eo 1 exp νννRH >>For
( )z
zRmmGU ddd 2+
′=⋅−= rF
′−≈
′−=
RcmG
RcmG
2e2eo 1 exp ννν
′−≈
′−=
RcmG
RcmG
2e
21
2eo 1 21 ννν (GTR)
12 g2
E2 <<=
=
′Rr
cv
RcmG
′−≈
′−=
RcmG
RcmG
2e2eo 1 exp ννν
′−≈
′−=
RcmG
RcmG
2e
21
2eo 1 21 ννν (GTR)
21g
e 1
−=Rr
ν 0g
→→rR blackhole
( )
( )observed106.6
calculated109.5
5
5
−
−
×−=∆
×−=∆
νννν
White dwarf:
20
*Gravitational collapse
a cold, diffuse cloud of dust or hydrogen atoms
drak EEU ∆+∆=∆−
hydrogen fuel exhausts
carbon, oxygen,... iron Fe56
K103 K107 nuclear fusioncontracts
outward pressure halts the contraction
collapses
helium “red giant”
Assignment: 4.5, 4.6
1
22+2(Sept 28)
11,rm
22,rm
21 rrr −=
O
4.4 Kepler problem and *scattering
( )rr 1 1 rfm −=&&
( )rr 2 2 rfm =&&
0=+ 2211 rr &&&& mm
( )rrr 11
212 1 rf
mm
+−=− &&&&
21 rrr −=
321~
rmmGf
0=+ 2211 rr &&&& mm
( )rr 11
21 rf
mm
+−=&&
1
0C =r&&rr )( rf−=&&µ
µ111
21
CC2 21 1
C21
≡+
≡+≡+
mm
mmmmmm
rrrDefine
reduced massWhich particle’s eq?
mass?Position vector?
force?
2 ( ) 0=×= rrM rf
CprL =×=
• r and p are co-planar
• areal “velocity”
( ) keer 2θµθµ θ&&& rrr r =+×=• L
Conservation of L
2 ( ) 0=×= rrM rf
CprL =×=
rr ⋅θd21
• r and p are co-planar
( ) keer 2θµθµ θ&&& rrr r =+×=• L
Conservation of L2 ( ) 0=×= rrM rf
CprL =×=
• areal “velocity”
constant2d
d 21
dd
===µ
θ Ltrr
tA
• can never change sign. θ&
• r and p are co-planar
( ) keer 2θµθµ θ&&& rrr r =+×=• L
Conservation of L
2
3( ) ( )rrfrr −=− 2 θµ &&&
ttr
trrr d
dd
ddd ⋅=⋅&
&&
=⋅− rr d2θ&
1st integrals
( ) Erkrr =−+ 2 22
21 θµ &&
v2
rr && ⋅= d
= 2
21d r&
rrLr d
2
2 ⋅
−
µ
=−= 22
2
32
2
2dd
rLr
rL
µµ
[ ] rFrrrf dd)( ⋅=⋅−
−−≡
−−=rk
rmmG dd 21Ud−=
( ) 0 2 =+ θθµ &&&& rr constant 2 =θµ &r L( )rU
rLrE ++= 2
22
2
21
µµ & ( )rUr
21
eff2 +≡ &µ
4
( )rk
rLrU −= 2
2
eff 2
µ
Effective potential
∞→
→
=−r
rrk
rL 0
?
?2 2
2
µ
2 2
21 θµ &r
r
hyperbola
parabola
ellipse
circle
Total energyKinetic energy?
9
?
5 )(tr )(θrr =
θµθ
θ dd
dd
dd
2r
rL
trr ==&
( )r
rUrLEr
L d
22
d 21
2
22
−−
=
µµµ
θ
r
rk
rLEr
L d
22
21
2
22
+−
=
µµµ
( )rUrLrE ++= 2
22
2
21
µµ &
( )
−− rU
rLE 2
2
22
µµ
r2.
r.
r
rk
rLEr
L d2
21
21
2
2
22
+−
=µµ
µµ
+−−+
−=r
Lk
rk
rL
LkE
L 1d2
2
21
2
22
2
2
2
22 µµµµ
=
−−
ax
xa
x arccosdd~22
−−+
−=r
Lk
rL
LkE
L 1d
2
212
2
22 µµµ
02
1
2
222
arccos θ
µµ
µ
θ +
+
−=
ELk
Lk
rL
( )erp
kLE
rp 1
21
1cos
22
20
−≡
+
−=−
µµ
θθ
)(cos1 0θθ −+=
epr
kLpµ
2=
( )pkeE
21 2−−=
3
)(cos1 0θθ −+=
epr
hyperbolaE > 0e > 1
parabolaE = 0e = 1
ellipse0 < e < 1
circlee = 0
orbitenergyeccentricity
pkE
2−=
022
<−<<−akE
pk
11 maxmin repr
epr ≡
−≤≤
+≡
2maxmin
12 eprra−
=+
=
For ellipse orbit
apb =
tLttAA
µ2d
dd
=⌡⌠= ( )ba
LT 2
π=µ pa
L232 π
=µ
mmGmmmma
mmGLa
LT
′+′′
π=′
π=
1)(
2 2 232
23
µµ
kLpµ
2=
For ellipse orbit
( )mmGaT
+′π=
1 2 23 )Kepler(1 2 23
mGa
′π≈
11 maxmin repr
epr ≡
−≤≤
+≡
2maxmin
12 eprra−
=+
=
apb =
tLttAA
µ2d
dd
=⌡⌠= ( )ba
LT 2
π=µ pa
L232 π
=µ
*Hyperbola orbit202
1 vE µ=
bvL 0µ=
e1cos −=∞θ
θcos1 epr
+=
π−≈π−= ∞θλϕ 22
Scattering angle
∞−= θϕ cot2
tanEbk
2=
1
111
1sincoscot
2
2−
=
−
=−
=−∞
∞∞
ee
eθθθ
µµ
µµ
220
2
22
222
1
bvE
k
kLE
==
Eq.(4.4.17)
2E
Assignment: 4.7, 4.12, *4.14
1
Taking retardation into account, …It is correct in Newtonian mechanics frame.
field propagator11,rm
22 , rm
O
r
4.5 Gravitational field
rF 3rmmG′
−=
rFg 3f rmG
m′
−== ∑−=i
ii
i
rGm rg 3f
fgm≡
fgFv == min Suvv −=′in S’
Gravitational field exists for certain mass (dis-tribution).Kinematic quantities , e.g. acceleration, may change with reference systems
Example 4.4Inside a sphere
( )
r
rg r
ρ
ρ
G
rrG
π−=
π−=
34
34
33
( ) ( )rgrgg ′−=
m'
lρGπ−=34
In the cavity
U−∇=F VmU
m−∇≡−∇==
Fgf
Gravitational potential
CrmGV +
′−=Around particle m'
Sphere or shell
potential energy?
Tidal force
( ) ( )arar
rg −−
−= 3fGm
( )zara
GmGmV222 −+
−=−
−=ar
ror find potential firstly
( )212
21−
+−−=
ar
az
aGmV r
( )
⋅⋅⋅+−
++−= 1cos3
21cos1 2
2
θθar
ar
aGm
+++≡ 210 VVV
2
( )rV ( )
⋅⋅⋅+−
++−= 1cos3
21cos1 2
2
θθar
ar
aGm
+++≡ 210 VVV
V−∇=g θθ eek ggaGm
rr ++= 2
( )
( )θθθ
θ
θ sincos31
1cos3
22
22
2
ar
aGmV
rg
ar
aGm
rVgr
−=∂
∂−=
−=∂∂
−=
k21
aGm
zV
=∂∂
−
83
24
223
1060.53.60
11098.51036.7~ −
⊕⊕
×=
××
=
ar
mm
ggm
3
11
6
24
303
SE 10496.11037.6
1098.51099.1~
××
××
=
⊕⊕ ar
mm
gg ⊙⊙
θθθ
θ sincos3~1cos3~ 2
−−
ggr
81057.2 −×=
* Tidal disruption
ρω 2m∆
maa
mG∆
′ ρ22
rrmmG ρ
2∆
π=θ1cos3 2 −θ
Disruption condition is
02 32
3 ≥
−+
′∆
rGm
amGm ωρ
3131
26.12
′
=
′
≤mmr
mmra0=ω
2323
)(a
mGa
mmG ′≈
+′== Ωω
313131
44.144.13
=
′
=
′
≤ ′
m
mRmmr
mmra
ρρ
Kepler’s 3rd law
Gravitational radiation
supernova SN 1987A
Assignment: 4.18
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