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