1 august 22 elementary approach 1. 1 definitions, elementary approach scalar quantities: quantities...

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1 August 22 Elementary approach 1. 1 Definitions, elementary approach Scalar quantities: Quantities having magnitude only. Length, mass, time, temperature, energy. Chapter 1 Vector analysis Vector quantities: Quantities having both magnitude and directions. Displacement, velocity, acceleration, momentum, angular momentum, electric field, magnetic field, dipoles. Tensor quantities: (Tensors of rank n) moment of inertia, electric permittivity, nonlinear susceptibility . Geometrical representation of a vector: An arrow. Addition and subtraction: Vector addition is commutative and associative.

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1

August 22 Elementary approach

1. 1 Definitions, elementary approach

Scalar quantities: Quantities having magnitude only.Length, mass, time, temperature, energy.

Chapter 1 Vector analysis

Vector quantities: Quantities having both magnitude and directions.Displacement, velocity, acceleration, momentum, angular momentum, electric field, magnetic field, dipoles.Tensor quantities: (Tensors of rank n) moment of inertia, electric permittivity, nonlinear susceptibility .

Geometrical representation of a vector:An arrow.

Addition and subtraction:

Vector addition is commutative and associative.

2

Algebraic representation of a vector:

332211 ˆˆˆ

ˆˆˆ

ˆˆˆ

)cos,cos,cos(

),,(

eeex

kjiA

zyxA

r

A

xxx

AAA

AAA

rrr

AAA

zyx

zyx

zyx

component, projection

direction cosine

unit vector, basis

more preferred form

most convenient for doing algebra

23

22

21

333222111

321

ˆ)(ˆ)(ˆ)(

),,(

xxx

yxyxyx

axaxaxa

x

eeeyx

x

magnitude (length, norm)

3

1. 2 Rotation of the coordinate axes

1e

'ˆ1e

2e'ˆ 2eV

ijji ee ˆˆ scalar product (more later)

Kronecker delta symbol

Einstein’s summation convention: Any repeated indices are summed over, unless otherwise specified. Greatly simplifies many expressions.

iii

ii AAAAA eeeeeA ˆˆˆˆˆ3

1332211

jiij

jijjjii

iiii

a

a

V

ee

eeeee

eeVeV

ˆ'ˆ

ˆˆ)ˆ'ˆ('ˆ

ˆ)ˆ(ˆ

Rotation matrix aij represents the transformation

between the two sets of coordinates. Its elements are the projections between the two sets of bases.

Orthogonality conditions:

ijjijkikjkikkjki

ijjijkikkjikjkik

aa

aaaa

eeeeeeeeee

eeeeee

ˆˆˆ'ˆ)ˆ'ˆ()ˆ'ˆ)(ˆ'ˆ(

'ˆ'ˆ'ˆˆˆ'ˆ Orthonormal transformation

ooo

ooo

oooaij

4

Components of a vector after transformation:

jijijijijjkk VaVVVVVi

' )ˆ'ˆ('ˆ'ˆ' 'ˆ

eeeeVe

Question: Is electric current (I) a vector?

Redefinition of a vector quantity:A quantity is a vector if its components transform as under an orthonormal transformation specified by aij.

jiji VaV '

For tensors: mnjnimij TaaT '

Generalization of vectors. Vectors can also be1)Complex quantities.2)Functions.3)Multi-dimensions or infinite dimensions.

5

Read: Chapter 1: 1-2Homework: 1.1.1, 1.2.1(a)Due: September 2

Geometric definition:Scalar product is commutative and distributive.

6

cosABBA

August 24,26 Scalar and vector products

1. 3 Scalar or dot product B

A

Algebraic definition:

iiijjijjii BABABA eeBA ˆˆ ( , orthonormal)ijji ee ˆˆ

Invariance of the scalar product under an orthogonal transformation:

mmnmmnnminimninmimii BABABAaaBaAaBA ''

We thus proved that A·B is indeed a scalar.

ii BABA

7

Algebraic definition:Levi-Civita symbol (antisymmetric tensor) ijk:

123. ofn permutatio oddan is if ,1

123. ofn permutatioeven an is if ,1

equal. are of any two if ,0

ijk

ijk

ijk

ijk1

23

+-

Look at your watch.

jiijkk

kijkjijijijjii

kijkji

BAC

BABABA

eeeeeBAC

eee

ˆ)ˆˆ(ˆˆ

ˆˆˆ

Geometric definition:

Meaning: area of the parallelogram formed by A and B.Cross product is anticommutative:

1. 4 Vector or cross product

rule. handright theand sin with , ABC BAC

ABBA B

A

C

kjikij

jikijk

BA

BA

eBA

BA

ˆ

)(

8

Matrix representation of cross product:

xyyx

zxxz

yzzy

z

y

x

xy

xz

yz

xy

xz

yz

BABA

BABA

BABA

B

B

B

AA

AA

AA

AA

AA

AA

0

0

0

0

0

0

BAA

Example:

z

y

x

yxyzxz

zyzxxy

zxyxzy

z

y

x

xy

xz

yz

xy

xz

yz

E

E

E

ssssss

ssssss

ssssss

E

E

E

ss

ss

ss

ss

ss

ss

22

22

22

0

0

0

0

0

0

Ess

Cross product in determinant form:

zyx

zyx

BBB

AAA

kji

BA

ˆˆˆ

Cross product is based on the nature of our 3-dimentional space.

9

Triple scalar product :

1. 5 Triple scalar product, triple vector product

zyx

zyx

zyx

jikkijikjjkikjiijk

kjijki

ii

CCC

BBB

AAA

BACACBCBA

CBA

A

B)(ACA)(CB

CB

C)(BA

)(

= the volume enclosed by the parallelepiped defined by A, B and C.

10

Triple cross product :

)()()(

)()(

)(

)()]([

BACCABCBA

BACA

CBCBA

ii

ijjjij

mljjlimjmil

mljklmkij

mljklmijk

mlklmjijk

kjijki

CB

CBACBA

CBA

CBA

CBA

CBA

A

jlimjmilklmkij

Can be proved by brute force (though not trivial).

C

B

B×C A

)( CBA

The “bac-cab” rule.

11

Read: Chapter 1: 3-5Homework: 1.3.3, 1.4.7,1.5.12Due: September 2

12

Reading: Proof that C=A×B is a vector:

kk

ij

mlmlmlmlml

mlmlmlml

mlmlmlml

mlmlml

mlmlmlml

mljmilij

mljmilkijmjmlilkijjikijk

Ca

almCaCaCa

ABBAaaaa

BAaaaaBAaaaa

mlBAaaaa

BAaaBAaa

BAaaC

BAaaBaAaBAC

3

131232333

1221

12211221

1221

1232121312

33

cofactor) its ;23 ,13 ,12 ( ))((

))((

)()(

0) then is term (the )(

'

'''

13

About matrices:

Inverse matrix A-1 :

Transpose matrix Ã:

For orthonormal matrices:

iklkiljkij aaaa 1111 ,1AAAA

jiij aa ~

About determinates:

Minor Mij: removing the ith row and the jth column

Cofactor Cij : (-1)i+j Mij

Expansion of a determinant:

Calculation of A-1:

For orthonormal matrices with |A|=1:

AA~~

~

111

1

jkjk

kjjk

kjjk

ikjkij

ikkjij

aa

aa

aaaa

aa

i

ijijj

ijij CaCaA

Aji

ij

Ca 1

ijij Ca 1221221133

33

2221

1211

..

.

.

For

aaaaa

a

aa

aa

14

Definition: In a 3D Cartesian coordinate system,

August 29, 31 Gradient, divergence and curl

1. 6 Gradient,

iixzyxekji ˆˆˆˆ

Example: VF

dr

dV

r

x

dr

dV

x

r

dr

dV

x

rVrV i

ii

ii

i

reee ˆˆˆˆ)(

)(

Central force:

Proof that is a vector:

jij

i

j

jii

iji

j

j

i

ijji

jii

jijij

ijj

ijiji

xa

x

x

xxx

ax

x

x

x

aa

ax

xxax

ax

xxax

'''

'

'

'

matrices lorthonorma

''

''

1

11

i

i xzyxekji ˆˆˆˆoperator

15

Geometrical interpretations:

1) cosrr dddzz

dyy

dxx

d

The function has the steepest change along the direction of its gradient.

0 ,)( surface On the ddC rr2)

is perpendicular to the surface of .)( Cr

16

Definition: In a 3D Cartesian coordinate system,

1. 7 Divergence ,·

i

izyx

x

V

z

V

y

V

x

V

V

Examples:

dr

dfrrf

r

x

dr

dfxrf

x

r

dr

dfxrf

x

rfxrfrfx

xrf

ii

ii

iii

i

)(3)(3

)(3)(

)(3)()(r

VVV

ffx

VfV

x

ffV

xf

i

ii

ii

i

Physical meaning:

dxdyjdxdzjdydzj

dxdydzz

j

y

j

x

jdxdydz

zyx

zyx

j dydzjjdydzjxxdxxxx

j is the net outflow flux of j per unit volume.

B is solenoidal if .0 B

17

Definition: In a 3D Cartesian coordinate system,

1. 8 Curl,

j

kijki

zyx

x

V

VVVzyx

eV

kji

V ˆ ,

ˆˆˆ

Example: VVeeV

ffx

VfV

x

ffV

xf

j

kk

jijkik

jijki ˆˆ

dxVdyV

ncirculatio

xy total

V is the total circulation of V per unit area. V is irrotational if .0 V

Physical meaning:Set the coordinate system so that z is along at an arbitrarily chosen point. Suppose the coordinates of that point is then (x0, y0, z0 ).

dxVVdyVV

dxVdyV

dxdyy

V

x

Vdxdy

yxdyyxxydxxy

xy

xyz

0000

V

V

18

BAABBAABBA

ABAB

BABABA

eeeBeBA

B

)(

rule cab"-bac" ,

ˆˆˆˆ

jj

jj

j

ij

i

jji

l

mjjlimjmili

l

mklmjijkikjijki

AB

BA

x

BA

x

BA

x

BA

x

BAA

1)

1. 9 Successive applications of

Laplacian , 222

2

i iii xxx

2) 0ˆ

kj

ijki xx

e

3) 0ˆ

j

kijki

i x

V

xeV

4) VVVVV 2

Example:

19

More about Laplacian:

0,-1.) when 0( )1(

have we,)(For

2

13

13

11

)()(

22

2

2

2

nrnnr

rrV

dr

Vd

dr

dV

r

dr

dV

rdr

dr

dr

dV

r

dr

dV

rdr

d

rdr

dV

r

dr

dV

rdr

dV

r

dr

dV

rrVrV

nn

n

rr

rr

r

20

Reading: Physical meaning of Laplacian:

)0(24

24)0(

2

1)0(

1

2

1)0(

20

2

0

222

0

2

2

33

0

2

0

a

xx

adxdydzx

xadxdydz

a

xxxx

xx

iii

i

jiji

ii

Laplacian measures the difference between the average value of the field around a point and the value of the field at that point.

If then cannot increase or decrease in all directions.

a

,02

21

Read: Chapter 1: 6-9Homework: 1.6.3,1.6.4,1.7.5,1.8.4,1.8.11,1.8.13,1.8.14,1.9.7Due: September 9

22

Line integral:

Circulation:

September 2,7 Gauss’ theorem and Stokes’ theorem

1. 10 Vector integration

B

AdW lF

If , the line integral is independent of the path between A and B:

F

lF d A

B

B

A

B

A ii

B

AABddx

xd )()( lF

The circulation of F around any loop is then 0: .0 lF d

(mostly used line integral)

Surface integral: σV d (mostly used surface integral, flux)

Volume integral: dVd ii eV ˆ

23

Integral definition of gradient, divergence and curl:

VσV

σVV

σ

d

d

d

lim

lim

lim

0

0

0

Proof: let dxdydzd

surfaces 6

ˆˆˆˆˆˆ 0

0

0

0

0

0σkjikji ddxdydxdzdydzdxdydz

zyxdxdydz

dzz

z

dyy

y

dxx

x

surfaces 6

0

0

0

0

0

0σVV ddxdyVdxdzVdydzVdxdydz

z

V

y

V

x

Vdxdydz

dzz

zz

dyy

yy

dxx

xxzyx

z

rhs.) expanding(by

surfaces 6

0

0

0

0

0

0

0

0

xz

dzz

zyy

dyy

yz

dzz

zy

dyy

yzyz

x

ddVdV

dxdyVdxdzVdxdydzz

V

y

Vdxdydz

V

dy

dz

y

x

dx

24

1. 11 Gauss’ theorem

Gauss’ theorem :

(Over a simply connected region.)

VV

dd VσV

Proof: For a differential cube,

Sum over all differential cubes, at all interior surfaces will cancel, only the contributions from the exterior surfaces remain.

σV d

.

surfaces 6

0

0

0

0

0

0

σVV

σV

V

dd

d

dxdyVdxdzVdydzV

dxdydzz

V

y

V

x

Vdxdydz

dzz

zz

dyy

yy

dxx

xx

zyx

25

Green’ theorem :

VVduvvuduvvu

22 σ

Proof:

),()(

)( 22

2

2

uvvuuvvuuvuvuv

vuvuvu

then use Gauss’ theorem.

Variant: VV

dvuvudvu 2

σ

Alternate forms of Gauss’ theorem :

VV

VV

VVddzyx

VddVzyxVdd

),,,(

),,,(

PPσPaV

σaVVσV

These can also be obtained from the integral definition of gradient, divergence and curl.

Example of Gauss’ theorem: Gauss’ law of electric field

. 0

0

Q

dddVVV

EσE

26

1. 12 Stokes’ theorem

Stokes’ theorem :

(Over a simply connected region. The surface does not need to be flat.)

Proof: Set the coordinate system so that x is along dat an arbitrarily chosen point on the surface. Suppose the coordinates of that point is then (x0, y0, z0 ).

Sum over all differential squares, at all interior lines will cancel, only the contributions from the exterior lines remain.

λV d

σVλV ddSS

sides 4

0

0

0

0λVVσV ddyVdzVdydz

z

V

y

Vdydzd

dzz

zy

dyy

yzyz

x

27

Alternate forms of Stokes’ theorem :

PσaaPσPaσσPa

PσPλPaV

σλaVσVλV

dddd

ddzyx

ddzyxdd

SS

SS

SS

)()()(

),,,(

),,,(

Example of Stokes’ theorem: Faraday’s induction law

. dt

dd

dt

ddd

SSS

σBσErE

Stokes’ theorem for a curve on the x-y plane: Green’s theorem

dxdyy

f

x

ggdyfdx

yxgyxfyx

SS

then)),,( ),,((),(Let

V

28

Read: Chapter 1: 10-12Homework: 1.10.5,1.11.2,1.11.6,1.12.3,1.12.10Due: September 9

29

September 9 Gauss’ law and Dirac delta function

1. 14 Gauss’s law

Electric field of a point charge:2

04

ˆ

r

q

r

E

Gauss’ law:

. includenot does if ,0

. includes if ,0

qS

qSq

dS

σE

q

S

S′

Proof:

1) If S does not include q, using Gauss’ theorem,

VS

drr

d0

ˆˆ22

rσr

dr

dfrrfrf )(3)(r

30

2) If S includes q, construct a small sphere S' with radius around the charge and a small hole connecting S and S’.

q

SS′

00

2

' 2' 2

' 22

44

4)ˆ(ˆ'ˆ

0'ˆˆ

qqd

r

d

ddd

d

r

d

S

S

SS

SS

σE

σr

rrσr

σrσr

If S includes multiple charges,

For a charge distribution,

Using Gauss’ theorem,

.0

i

i

iS iS

qdd

σEσE

.

0

0

E

EσE

σE

dd

dd

VS

VS

31

1. 15 Dirac delta function

0

4ˆˆ1122

2

VVVV r

dd

rd

rd

r

σrr

Using the Dirac delta function, )()()(4)(412 zyxr

r

Dirac delta function is defined such that

-).0()()(

0; if ,0)(

fdxxxf

xx

Especially,

-.1)( dxx

Example:

32

The sequence of integrals has a limit:

).0()()(lim fdxxfxn

n

We write it as implying that we are always doing the limit.

,)()( dxxfx

Dirac delta function (distribution) is the limit of a sequence of functions, such as

nx

nn

nx

nx

xn

2

1

2

1 ,

2

1or

2

1 ,0

)(

n

n

ixtn dte

x

nxx

2

1sin)(

33

i

x

xi i

iii

iii i

i

i

i xg

xfdxxfxgxxdxxfxg

xgxgxg

xxxg

|)('|

)()())(')(()())(( :Proof

.0)(' and 0)( where,|)('|

)())(( 4)

)(')(')()(')( 5) afdxxfaxdxaxxf

Properties of (x):

).()( )1 xδxδ

Proof: )0()()()()()()( fdyyfydyyfydxxfx

||

)0()(

||

1)()( :Proof

).(||

1)( 2)

a

fdy

a

yfy

adxxfax

xδa

axδ

)()()()( :Proof

).()()( 3)

afdyayfydxxfax

afdxxfax

34

Read: Chapter 1: 13-15Homework: 1.13.8,1.13.9,1.14.3,1.15.7Due: September 16

35

September 12 Helmholtz’s theorem

1. 16 Helmholtz’s theorem

The uniqueness theorem : A vector is uniquely specified within a simply connected region by given its 1) divergence , 2) curl, and 3) normal component on the boundary.

Proof:

region. wholein the 0 Therefore

.0 have we

,0 ,0 mindin Keep

.let , theoremsGreen'In

.0 have we,0With

). (e.g., . have we,0 Since

.0 ,0 ,0 then ,Let

. , , Suppose

21

22

2

2

2

21

212121

VVW

σWσ

σ

W

rWWW

WWVVW

VVVV

V V

n

VV

P

A

n

nn

dWd

dWdd

vudvuvudvu

d

W

VV

36

Corollary: The solution of Laplacian equation is unique at a given boundary condition.

0)(2 r

unique is

given

unique is

given

0

00

:Proof

2

S

S

37

Helmholtz’s theorem (the fundamental theorem of vector calculus): Any rapidly decaying vector field (faster than 1/r at infinity) can be resolved into the sum of an irrotational vector field and a solenoidal vector field.

Proof: Let us prove that any vector V can be decomposed as .AV

.

Let

)()()(

)()(

)(')'(4)'(4

1'

|'|

1)'(

4

1

then,|'|

')'(

4

1)(Let

2

22

AV

WA

W

WWrV

WWW

rVrrrVrr

rVW

rr

rVrW

dd

d

The explicit expressions of and A are given later.

38

'|'|

)'('

4

1

'|'|

)'('

4

1'

|'|

)'(

4

1

'|'|

)'('

|'|

)'('

4

1

')'(|'|

1)'(

4

1

')'(|'|

1

4

1

|'|

')'(

4

1

d

dd

d

d

d

d

rr

rV

rr

rVσ

rr

rV

rr

rV

rr

rV

rVrr

rVrr

rr

rVW

'|'|

)'('

4

1

'|'|

)'('

4

1

|'|

)'('

4

1

'|'|

)'('

|'|

)'('

4

1

')'(|'|

1)'(

4

1

')'(|'|

1

4

1

|'|

')'(

4

1

d

dd

d

d

d

d

rr

rV

rr

rV

rr

rVσ

rr

rV

rr

rV

rVrr

rVrr

rr

rVWA

VVdd

fff

fff

used We

PPσ

VVV

VVV

'|'|

)'(

4)(

'|'|

)'(

4

1

:Examples

0

0

d

d

rr

rJrA

rr

r

39

Read: Chapter 1: 16Homework: 1.16.1Due: September 23