vector analysis.ppt

Chapter 6 Vector analysis ( 벡터 해석 )

Mathematical methods in the physical sciences 3rd edition Mary L. Boas

Lecture 18 Basic vector analysis

1. Introduction

Vector function, Vector calculus, ex. Gauss’s law

2. Application of vector multiplication ( 벡터곱의 응용 )

zzyyxx BABABAABBA cos

sin , ABBA

BBB

AAA

kji

BA

zyx

zyx

BA

BA

a) Work

rdFdW

dFFdW

�

cos

b) Torque Fr

rv

v

sinrr

c) Angular velocity

1) Dot product

2) Cross product

- Example

1) Triple scalar product ( 삼중 스칼라곱 )

3. Triple products ( 삼중곱 )

zyx

zyx

zyx

CCC

BBB

AAA

CBA )(

cf. volume of unit cell for reciprocal vectors

321

323

321

132

321

321 2 ,2 ,2

aaa

aab

aaa

aab

aaa

aab

“Volume of the parallelepipe”

cosAheight

BCA

BAC

CBACBA

)(

)(

)()(

zyx

zyx

zyx

CCC

BBB

AAA

CBA )(

)()( CBACBA

So, it does not matter where the dot and cross are.

- An interchange of rows changes just the sign of a determinant.

)( CBA

CbBaCBA�

)(

CBABCA

)()(

some vector in the plane of B and C

2) Triple vector product ( 삼중 벡터곱 )

Prove this!

(Vector equation is true independently of the coordinate system.)

kji

ji

i

zyx

yx

x

AAAA

CCC

BB

3) Application of the triple scalar product

“Torque”

Fr

This question is in one special case, namely when r and F are in a plane perpendicular to the axis.

nFrn

4) Application of the triple vector product

v

sinrr

)( rrm

vrm

prL

Angular momentum

Centripetal acceleration )( ra

4. Differentiation of vectors ( 벡터의 미분 )

)( zyx AAAA

kjidt

dA

dt

dA

dt

dA

dt

Ad zyx

Example 1.

2

2

2

2

2

2

2

2

,

,

dt

zd

dt

yd

dt

xd

dt

rd

dt

vda

dt

dz

dt

dy

dt

dx

dt

rdv

zyxr

kji

kji

kji

1) Differentiation of a vector

)order! of (careful )(

,)(

,)(

Bdt

Ad

dt

BdABA

dt

d

Bdt

Ad

dt

BdABA

dt

d

dt

AdaA

dt

daAa

dt

d

2) Differentiation of product

Example 2. Motion of a particle in a circle at constant speed

.

.,2

2

constvvv

constrrr

Differentiating the above equations,

0or 02

,0or 02

avdt

vdv

vrdt

rdr

“two vectors are perpendicular”

r

va

avvrvar

var

vvar

vr

2

2

2

,0 & 0

0 this,atingDifferenti

,0

3) Other coordinates (e.g., polar)

coord.polar ),(

coord.r rectangula),(

θr ee

ji

:),( ji

:),( θr ee

constant in magnitude and direction

constant in magnitude, but directions changes

cossin

sincos

jie

jie

θ

r

.sincos

,cossin

dt

d

dt

d

dt

d

dt

ddt

d

dt

d

dt

d

dt

d

rθ

θr

ejie

ejie

Example 3. ? , dt

Ad AAA r

θr ee

.dt

dA

dt

dA

dt

dA

dt

dA

dt

Ad

dt

dA

dt

dA

dt

dA

dt

dA

dt

Ad

rr

rr

rθθr

θθ

rr

eeee

ee

ee

Chapter 6 Vector analysis


Lecture 19 Directional derivative; Gradient

5. Fields ( 장 )

Field: region + the value of physical quantity in the region ex) electric field, gravitational field, magnetic field

6. Directional derivative: gradient ( 방향 도함수 ; 기울기벡터 )

),,( zyxT

sT for The change of temperature with distance depends on the direction. directional derivative

functionscalar : ),,( zyx

u

kji

ds

d

zyx,grad

(directional derivative for u: directional unit vector)

ds

dT

1) definition of directional derivative

Example 1. Find the directional derivative

)1,2,2(direction

(1,2,-1)at 2

A

xzyx

)1,2,2(3

1

A

A

u

3

5

)1,1,3()1,2,1(

,)2( 2

u

kjikji

xy

xzxyzyx

2) Meaning of gradient : along it the change (slope) is fastest (steepest).

3) Relation between scalar function and gradient

“The vector grad. is perpendicular to the surface =const.”

u

0s

lim

0s

,0

0s

Example 3. surface x^3y^2z=12. find the tangent plane and normal line at (1,-2,3)

kjikji 4123623 23322

23

yxyzxzyxw

zyxw

1

3

3

2

9

1

,0)3()2(3)1(9

zyx

xyx

4) other coordinates (e.g., polar)

rryx

1θr eeji

cf. Cylindrical & Spherical coord.

φθr ˆsin

1ˆ1ˆ

T

r

T

rr

TT

zφr ˆˆ1

ˆz

TT

rs

TT

cylindrical

spherical

7. Some other expressions involving grad. ( 을 포함하는 다른 표현들 )

zyxzyx

zyx

kjikji

kji

)(

1) vector operator

2) divergence of V

z

V

y

V

x

V

VVVzyx

VV

zyx

zyx

),,(),,(div

3) curl of V

zyx

zyx

VVVzyx

VVVzyx

VV

kji

),,(),,(curl

4) Laplacian

2

2

2

2

2

2

2

),,(),,(

grad

zyx

zyxzyx

div

conductionheat ofequation or equation diffusion theis 1

equation. wave theis 1

equation.Laplace' is 0

22

2

2

22

2

ta

ta

5) and etc.

VV

VVV

2)(

)()()(

)()( VVV



Lecture 20 Line integral & Green’s theorem

8. Line integrals ( 선적분 )

integrating along a given curve. only one independent variable!

1) definition

rF d

Example 1. F=(xy)i-(y2)j, find the work from (0,0) to (2,1)

)( 2

2

dyyxydxW

dyyxydxrdF

dydxrd

rdFdW

ji

path 1 (straight line)

dxdyxy2

1 ,

2

1 1

2

1

2

1

2

12

0

22

0

21

dxxxdxxdyyxydxW

path 2 (parabola) xdxdyxy2

1 ,

4

1 2

3

2

2

1

4

1

4

12

0

222

2

0

22

xdxxdxxxdyyxydxW

path 3 (broken line)

0dx

0dy

1

0

2

3

1)00(

ydyyy

3

52

3

1 ,2)011( 3

2

0

Wdxxy

path 4 (parameter) x=2t^2, y=t^2

x: (0,2) t: (0,1)

6

72242

1

0

222224 tdtttdtttdyyxydxW

1)

2)

1)

2)

Example 2. Find the value of

22 yx

ydxxdyI

path 1 (polar coordinate ) r=1 (constant) so, only d may be considered.

d

dd

yx

ydxxdy

1

)sin(sin)(coscos22

1 ,cos ,sin

sin ,cos22

yxddyy

ddxx

0

1 dI

path 2

1) 2)

(0,1)

(-1,0) (1,0)

1xyxy 1

2)12arctan(

)1(

)1( 0

1

0

12222

x

xx

dxxxdx

yx

ydxxdy

2)12arctan(

)1(

)1( 1

0

1

02222

xxx

dxxxdx

yx

ydxxdy

2I

## Question: Would you compare between example 1 and 2?

2) Conservative fields (F or V) ( 보존장 )

- Example 1 : depends on the path. nonconservative field- Example 2 : does not depend on the path. conservative field

field veconservatifor condition sufficient andnecessary ,0curl F

0 this,From

,,similarly and ,

, Using

,,

2

22

F

x

F

z

F

y

F

z

F

x

F

xy

W

y

F

xy

W

yx

W

z

WF

y

WF

x

WF

z

W

y

W

x

WWF

zxzyyx

zyx

kji

.for which findcan we,0 if ,Conversely

0, If

WFWF

FWF

3) Potential () ( 퍼텐셜 )

potentialscalar :)(

field veconservati : ,

W

FF

B

A

sdF

for A: a proper reference point

cf. Electric field, gravitational field conservative

Example 3. Show that F is conservative, and find a scalar potential.

kji )13()2( 223 xzxzxyF

0

132 223

xzxzxyzyx

F

kji

1) F is conservative.

B

A

B

A

dzxzdyxdxzxyrdFW )13()2( 223

(0,0,0)

(x,y,z)

a. find the point where the field (or potential) is zero.b. do line integral to an arbitrary point along the path with which the

integration is easiest.

i) dxii) dy

iii) dz

(x,0,0)

(x,y,0)

i) only dx

0)(

0 ,0

0

x

x

W

dzdyzy

ii) only dy

y

yxdyx

dzdxzconstx

0

22

0,0.,

iii) only dz

z

zxzdzxz

dydxconstyx

0

32 )13(

0.,,zxzyxW 32

2) Scalar potential of F

Example 4. scalar potential for the electric field of a point charge q at the origin

r

q

r

q

r

rdrq

rdrrdrrrd

r

rdrqrdE

rr

q

r

r

r

q

r

qE

rr

3

r to3

r to

322

22)(For

re

AdAAdAcf

.

9. Green’s theorem in the plane ( 평면에서의 Green 정리 )

- The integral of the derivative of a function is the function.

)()()( afbfdxxfdx

db

a

1) Definition of Green theorem

sderivative partial first continuous withfunction a : ),( ),,( yxQyxP

dyyaQybQdxdyx

yxQdxdy

x

yxQ d

cy

d

c

b

axA

)],(),([),(),(

Area integral:

Line integral:

d

c

c

dC

d

c

dyyaQybQ

dyyaQdyybQdyyxQ

)],(),([

),(),(),(

CA

Qdydxdyx

Q

cf.

dxcxPdxPdxdyy

yxPdxdy

y

yxP d

cy

b

a

b

axA

)],(),([),(),(

Similarly,

d

a

a

bC

a

a

dxdxPcxP

dxdxPdxcxPdxyxP

)],(),([

),(),(),(

CA

Pdxdxdyy

P

theoremsGreen' ,)(

AC

dxdyy

P

x

QQdyPdx

This relation is valid even for an irregular shape!!

“Using Green’s theorem we can evaluate either a line integral around a closed path or a double integral over the area inclosed, whichever is easier to do.”

Example 1. F=xyi-y2j, find the work from (0,0) to (2,1) and back

)( 2

2

dyyxydxW

dyyxydxrdF

dydxrd

rdFdW

ji

For a closed path, 132 WW

path 2 (parabola) xdxdyxy2

1 ,

4

1 2

3

2

2

1

4

1

4

12

0

222

2

0

22

xdxxdxxxdyyxydxW

(previous section)

path 3 (broken line)

0dx

0dy

1

0

2

3

1)00(

ydyyy

3

52

3

1 ,2)011( 3

2

0

Wdxxy

1)

2)1)

2)

Example 1. F=xyi-y2j, find the work from (0,0) to (2,1) and back

)( 2

2

dyyxydxW

dyyxydxrdF

dydxrd

rdFdW

ji

For a closed path, 132 WW

Using Green’s theorem,

1

0

2

0

22

1

)()(

y

y

xA

AA

xdxdyxdxdy

dxdyxyy

yx

dyyxydxW

AC

dxdyy

P

x

QQdyPdxcf )(.

Example 2. dxdyy

F

x

FdyFdxFW x

A

y

A

yx )()(

0 If

y

F

x

Fxy ( z-component of curl F = 0),

then, W from one point to another point is independent of the path. (F : conservative field)

AC

dxdyy

P

x

QQdyPdxcf )(.

- Two useful way to apply Green’s theorem to the integration of vector functions

yxyx VVVPVQ jiV where, ,

0 with ,div

zyx V

y

V

x

V

y

P

x

QV

0

wherenormal), (outward

(tangent)

22

dsd

dydxdsdxdyds

dydxd

nr

jin

jir

dsdxdyVVdyVdxVQdyPdx yxxy nVjiji )()(

AA

dsdxdy nVVdiv

Divergence theorem

ddxdydz nVVdiv

a) Divergence theorem

AC

dxdyy

P

x

QQdyPdxcf )(.

b) Stoke’s theorem

yxxy VVVPVQ jiV where, ,

0 with ,)(curl

zxy V

y

V

x

V

y

P

x

QkV

rVjiji ddydxVVdyVdxVQdyPdx yxyx )()(

AA

ddxdy rVkV)(curl

Stoke’s theorem

rVnV dd)(curl

AC

dxdyy

P

x

QQdyPdxcf )(.



Lecture 21 Divergence and Divergence theorem

10. Divergence and divergence theorem ( 발산과 발산정리 )

z

V

y

V

x

V

VVVzyx

VV

zyx

zyx

),,(),,(div

flow of a gas, heat, electricity, or particles

vV

nV

coscos

cos

))()((

Vv

AvtAvt

Avt

: flow of water

amount of water crossing A’ for t

1) Physical meaning of divergence

),,( zyx VVVV

- Rate at which water flows across surface 1 dydzV x )1(

- Rate at which water flows across surface 2 dydzV x )2(

- Net outflow along x-axis dydzdxx

VdydzVV x

xx

)]1()2([

axis-z along ,

axis-y along ,

dxdydzz

V

dzdxdyy

V

z

y

In this way,

dxdydzdxdydzdxdydzz

V

y

V

x

V zyx VV

div

“Divergence is the net rate of outflow per unit volume at a point.”

(a) positive divergence for positive charge (or negative divergence for a negative charge)(b) zero divergence(c) positive divergence along the z-axis

cf. (from ‘Griffiths’)

Example 1.4 in Figure 1.18

s?divergenceir then the..............

ˆ

,ˆ

,ˆˆˆ

zv

zv

zyxrv

z

zyx

c

b

a

.1

0

3

c

b

a

v

v

v


= (source density) minus (sink density)

= net mass of fluid being created (or added via something like a minute

sprinkler system) per unit time per unit volume

= density of fluid = mass per unit volume

/t = time rate of increase of mass per unit volume

Rate of increase of mass in dxdydz = (rate of creation) minus (rate of outward flow)

V

V

t

dxdydzdxdydzdxdydzt

1) If there is no source or sinks, 0

continuity ofEquation ,0

t

V

V,0 If

t

cf. 0div ,div BD

2) Example of the divergence 1

2)

Consider any closed surface.

ddrd sin2

Mass of fluid flowing out through d is Vn d.

Total outflow: dnV

For volume element d=dxdydz, the outflow from d is dV

d

od

d

dd

dd

of surface

of surface

1lim nVV

nVV

another definition of divergence

3) Example of the divergence 2

inclosingsurface volume

dd nVV

4) Divergence theorem

5) Example of the divergence theorem

? , dzyx nVkjiV

haddd

z

z

y

y

x

x

dd

3

cylinder of volume

cylinder of surface

33

,3

VnV

V

VnV

If we directly evaluate dnV

rji

n

k

a

yx :surface curved

:top

6) Gauss’s law

electric field at r due to a point chage q at (0,0)

unit) mksin in vacuum 1091/4 constant, dielectric : (

law sCoulomb' 4

9

2

reE

r

q

reDED24

,r

qε

qddrr

qDdAdDd

dAr

ddd

4

1

4cos

1sin

22

2

nD

) inside ( 444

angle solidtotal

σ surface closed

qqπ

qd

π

qdσ

nD

i

ii

i

qdσdσ

σ surface closed

σ surface closed

nDnDFor multi-sources,

surface closed theinside charge total

σ surface closed

dσnD

by boundedvolume

σ surface closed

ddσ

nD Gauss’s Law

by boundedvolume

by boundedvolume

σ surface closed

ddσ

DnDUsing the divergence theorem,

ext D

7) Example of gauss’s law. E=?

a) For electrostatic problem, E=0 insideb) For symmetry, E should be vertical.

area) surface( , DnDDnD d

total charge inside is C(surface area) for C surface charge density.

C

C

D

D area) (surfacearea) surface(



Lecture 22 Curl and Stoke’s theorem

11. Curl and Stoke’s theorem ( 회전이론과 Stoke’s 정리 )

zyx

zyx

VVVzyx

VVVzyx

VV

kji

),,(),,(curl

v

sinrr

0 since

)()(

3)(

)()()(

x

z

x

y

zyxzyx

z

z

y

y

x

x

zyxzyx ωkjikjikjirω

ωωrω

rωωrrωv

rωv

ωrωv 2)( “Curl v gives the angular velocity.”

Ex.

rV dn circulatio

curl V 0

curl V 0

curl V 0

curl V = 0

curl V = 0

1) meaning of curl

vs.


Example 1.5

curls? their then,................

1.19b) (Fig. ˆ

1.19a) (Fig. ˆˆ

yv

yxv

x

xy

b

a

.ˆ

,ˆ2

zv

zv

a

a


dd

ddd

dd

nV

ddxdyd

around0

around

1lim)(

,)(curl)(curl

rV

nVkVrV

surface

boundingcurve

)( dd nVrVStoke’s theorem

goodbad

Example 1.

.0, hemisphere over the

?)( ,24

2222

zazyx

dzxy nVkjiV

kV 3

(a) integrate the expression at it stands

(b) use Stoke’s theorem and evaluate the line integral around the circle

(c) use Stoke’s theorem to say that the integral is the same over any

surface bounded by the circle, for example, the planar area inside the

circle.

33

kknV

kn

Ampere’s law IdC

rH

H : magnetic fieldC : closed curveI : current

r

I

IrrddC

2

22

0

H

HHrH

JH

nJnH

nHrH

nJJnJrH

dd

dd

dIdId

C

C

)(

)(

cf.densitiycurrent :,

: one of the Maxwell equations

For a specific case,

Conservative fields

‘simply connected’ if a simple closed curve in the region can be shrunk to a point without encountering any points not in the region.

If the components of F have continuous first partial derivatives in a simple connected region, any one implies all the others.

a) curl F = 0b) closed line integral = 0c) F conservatived) F = grad W, W single valued

Vector potential

0 ,definitionBy

0potential vector : for ,0div

0 ,definitionBy

0potentialscalar :for ,0curl

A

BAAVV

EVV

W

WW

Example 2. ),2(2)( 22 zxzyzyzx kjiV ?, AAV

02222

)2()2()( 22

xzzx

zxxz

yzy

yzzx

divV

zyx AAAzyx

kji

AV

There are many A’s to satisfy this equation. For convenience, set one component A_x =0.

kji

kji

AV

x

A

x

A

z

A

y

A

AAzyx

yzyz

zy0

i)

ii)

x

Azxz

x

Ayz yz

2 ,2 2

),(2

),(

2

122

zyfxyzA

zyfzxxzA

z

y

,2

z

A

y

Ayzx yz

z

fx

y

f

z

fxzx

y

fxz

z

A

y

Ayzx yz

1221222 22

There are many ways to select f1 and f2.

)2

12()( then,

,2

1,0 takingIf

222

221

zyxyzzxxz

zyff

kjA

Generalization for A potential vector : for ,0div AAVV

For A_x=0,x

AV

x

AV y

zz

y

, ,AV

),,( ),,( zygdxVAzyfdxVA yzzy

),( ,

,0divFor

),(

zyhdxx

VV

x

V

z

V

y

V

zyhdxz

V

y

V

z

A

y

AV

xx

xzy

zyyzx

V

cf. When we know one A, all others are of the form,

AVA 0u , u

2

2

2

2

22 11

ˆˆˆ1

11

ˆˆ1

ˆ

z

TT

ss

Ts

ssT

Laplacian

vrvvzs

s

r

Curl

z

vv

ssv

ss

Divergence

z

TT

ss

TT

Gradient

zs

zs

zφs

v

v

zφs

cf. Cylindrical coordinate

.sin

1sin

sin

11

sin

ˆsinˆˆ

sin

1

sin

1sin

sin

11

ˆsin

1ˆ1ˆ

2

2

2222

22

2

22

T

r

T

rr

Tr

rrT

Laplacian

vrrvvr

rr

r

Curl

v

rv

rvr

rr

Divergence

T

r

T

rr

TT

Gradient

r

r

φθr

v

v

φθr

cf. Spherical coordinate

Homework

Chapter 6

3-17, 4-5, 6-10, 10-14, 11-16

vector analysis.ppt

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