Chapter I Vector Analysis

.

1.1 Vector Algebra

kAjAiAA zyxˆˆˆ

It is well-known that any vector can be written as

Vectors obey the following algebraic rules:

scalaraisˆˆˆ)( kAjAiAAiv zyx

zzyyxx BABABAABBAv cos)(

eABBAii Commutativ)(

ssociativeCBACBAiii A)(

2)( AAAvi

kBAjBAiBABAi zzyyxxˆ)(ˆ)(ˆ)()(

(vii) The law of cosines

BACLet

B

A C

BBBAAABABACC

2

cos2222 ABBAC

ecommutativnotABBA(ix)

Triple Product

ACBBACCBAi

)(

B

A

C

Acos

volume of parallelepiped

rulecabbac)( BACCABCBAii

zyx

zyx

BBB

AAABA(viii)kji

Position, Displacement, & Separation Vectors

x

y

z

P(x,y,z)

The position vector of a point in 3-D is expressed in

Cartesian coordinates as

kzjyixr ˆˆˆ

The infinitesimal displacement vector is

kdzjdyidxld ˆˆˆ

A source point r is the point where an e. charge is located.

A field point r is the point at which you are calculating electric or magnetic filed.

The separation vector from the source point

to the field point is defined as

rrd

rr

r

r

Source point

field point

Differential Calculus

Ordinary Derivatives It is known that dxdx

dfdf

that is if x is changed by dx, the function changes by df with is the

proportionality factor. dx

df

In other words we say that the derivative is small if the function varies slowly

with x and large if the function varies rapidly with x.

Geometrically we say that is the slope of the graph f(x) vs. x. dx

df

Gradient (Directional Derivative)

Let T be a scalar function of 3-variables, i.e., ),,( zyxTT

)1(dzz

Tdy

y

Tdx

x

TdT

This tells us how T varies as we go a small distance (dx, dy, dz) away from the point (x, y, z). Let us rewrite Eq. (1) as

)2(ˆˆˆˆˆˆ kdzjdyidxkz

Tj

y

Ti

x

TdT

)3(ldTdTor

)4(ˆˆˆ kz

Tj

y

Ti

x

TTwith

is called the gradient of the function T.

The symbol is called the del operator. kz

jy

ix

ˆˆˆ

Like any vector the gradient has a magnitude and a direction. Now Eq. (3) can

be written as

)5(cosldTdT

For fixed , dT is maximum when , that is dT is maximum when we move in the same direction as .

ld

0T

The gradient points in the direction of maximum increase of the function T.

The magnitude gives the slope along this maximal direction. T

is directed normal to the level surface of T through the point being considered, i.e., is perpendicular to the surface T=constant.

T

T

Example Find the gradient of the position vector.

Soution In Cartesian coordinates the magnitude of the position vector is

kz

rj

y

ri

x

rr ˆˆˆ

r

r

rk

r

zj

r

yi

r

xˆˆˆˆ

This means that the distance from the origin increases most rapidly in the

radial direction.

222 zyxr

It is also defined as the net rate of flow per unit volume, i.e.,

densitysourcetheA

Example Find the divergence of the vectors

Soution: Since

kzkr ˆand,ˆ,

kzjyixr ˆˆˆ

Divergence

The divergence of a vector is defined as A

z

A

y

A

x

AkAjAiAk

zj

yi

xA z

yxzyx

ˆˆˆˆˆˆ

The divergence is a measure of how much the vector spreads out (diverge)

from the point in question.

3

z

z

y

y

x

xr

0100ˆ

zyxk

100ˆ

z

z

yxkz

The Curl

The curl of a vector is defined as A

zyx AAA

zyx

kji

A

ˆˆˆ

The curl of a vector is a measure of how much the vector curl around the point

in question.

Example Find the curl of the vectors

Soution:

jxBjxiyA ˆandˆˆ

kky

y

x

xj

xz

yi

z

x

y

xy

zyx

kji

A ˆ2ˆˆ0ˆ0

0

ˆˆˆ

kkyx

xj

xzi

z

x

y

x

zyx

kji

B ˆˆ0ˆ00ˆ0

00

ˆˆˆ

The Del Operator Operations

2

2

2

2

2

22)(

z

T

y

T

x

TTTi

0)( Tii

0)( Aiii

AAAiv 2)(

Integral Calculus

Line Integral The line integral is expressed as

b

aldA

where is a vector function and is an infinitesimal displacement vector

along a path from point a to point b. A

ld

If the path forms a closed loop, a circle is put on the integral sign, i.e.,

ldA

If the line integral is independent on the path followed, the vector is called

conservative A

Example Let find

Soution:

jyxiyA ˆ12ˆ2

b

aldA

from point a=(1,1,0) to point b=(2,2,0) along the solid path (path 1) and along

the dashed path (path 2).

a

x

y b

c

kdzjdyidxld ˆˆˆ

we have for the first path

b

a zyxb

adzAdyAdxAldA

b

adyyxdxy 122

dyyxdxyb

c

c

a 122

Now along the path ac y=1, and along the path cb x=2. This leads to

11101142

1

2

1 dyydxldA

b

a

For the second path we have x=y and this gives

10212 21

22

122

1

2

12 dyyydxxdyyxdxyldA

b

a

102 21

22

12 dyyydxx

Note that the two results are different, i.e., the vector is not conservative.

Surface Integral The surface integral is expressed as

S SdA

where is a vector function and is an infinitesimal element of area. A

Sd

Again if the surface is closed we put a circle on the integral sign, that is

SdA

The direction of is perpendicular to the surface an directed outward for

closed surfaces and arbitrary for open surfaces. Sd

Example Let

Soution:

Find over the 5-sides of a cube of side 2,

as shown in the figure (excluding the bottom).

kzyjxixzA

3ˆ2ˆ2 2

S SdA

dxdz dxdz

dxdy

y

z

x

For the top side kdydxSd topˆ

dxdyzySdAtop 32

But on the top side z=2 42

0

2

0 ydydxSdAtop

For the right side jdxdzSd right ˆ

dxdzxSdAright 2

122 20

2

0 dzdxx For the left side

jdxdzSd left ˆ

dxdzxSdAleft 2

122 20

2

0 dzdxx

For the front side idzdySd ˆ

xzdydzSdAfront 2

But on the top side x=2 1642

0

2

0 zdzdySdAfront

For the back side idzdySd ˆ

xzdydzSdAback 2

But on the top side x=0 0back SdA

Volume Integral The volume integral is expressed as

V dT

where T is a vector function and d is an infinitesimal element of volume.

The Fundamental Theorem of Calculus

)()( afbfdxdx

dfba

The Fundamental Theorem of Gradient

where T =T(x,y,z) be a scalar function of three variables, then

)()( aTbTldTba

Since the right side of the last equation depends only on the end points and not

on the path followed we conclude that

Corollary 1: is independent on the path followed from a to b. ldTba

Corollary 2 0ldT

Example:

Soution:

Check the fundamental theorem of gradient by taking two paths from point a (0,0,0) to point b (2,1,0).

2xyTLet

The first path is 2-steps: step (i) along the x-axis and then up step (ii).

(i) x

y

b

(ii)

a

(iii)

kdzjdyidxldNow ˆˆˆ

jxyiyTand ˆ2ˆ2

)ii()i( ˆˆ jdyTidxTldTb

a

1

0

2

02 2xydydxy

Bur for step (i) y=0, and for step (ii) x=2 220 ldTb

a

For the 1st path

For the 2nd path dxdyxy21

21

22 20

2432

212

02

41

)iii(2 dxxdxxdxxxydydxyldT

b

a

The Fundamental Theorem of Divergence

It states that the integral of a divergence over a volume is equal to the value of

the function at the boundary.

In another world, the divergence theorem states that the outward flux of a

vector field through a surface is equal to the triple integral of the divergence on

the region inside the surface.

Sv SdAdA

Example: Check the divergence theorem using the vector

kyzjzxyiyA ˆ2ˆ2ˆ 22 Over the unit cube situated at the origin.

Solution:

yx

z

A

y

A

x

AA z

yx 22

1

0

1

0

1

0

222 dxdydzyxdAv

http://en.wikipedia.org/wiki/Fluxhttp://en.wikipedia.org/wiki/Divergence

dxdz dxdz

dxdy

y

z

x dxdy

S SdA

To find we have to calculate the integral over all

the faces:

yzdxdySdAtop 2

But for the top side z=1, so we have

121

0

1

0 ydydxSdAtop

yzdxdySdAbottom 2

But for the top side z=0, so we have 0bottom SdA

dxdzzxySdAright22

But for the right side y=1, so we have 341

021

02 dzzdxxSdAright

dxdzzxySdAleft22

But for the left side y=0, so we have 311

02 dzzSdAleft

dydzySdAfront2

311

0

1

02 dzdyy

dydzySdAback2

311

0

1

02 dzdyy

20131

31

31

34

S

SdA

Stokes' Theorem

ldASdAS

Since the boundary line for any closed surface shrink down to a point, then

0 SdAS

Example: Check the Stokes' theorem using the vector

kyzjyxzA ˆ4ˆ32 22

Over the square surface shown.

Solution: kzixzA ˆ2ˆ24 2

idydzSd ˆ

1

0

21

0

24 dydzxzSdA

S

But on the surface x=0, so we have

34

1

0

21

0

4 dydzzSdAS

lefttoprightbottom

kdzAjdyAkdzAjdyAldA ˆˆˆˆ

132ˆ 10

2 dyyxzjdyAbottom

Along the bottom side x=z=0, so we have

Along the top side x=0, z=1 so we have

132ˆ 10

2 dyyxzjdyAtop

Along the right side x=0, y=1 so we have 341

024ˆ dzyzkdzA

right

Along the left side x=0, y=0 so we have 04ˆ 102 dzyzkdzA

left

34

34 011 ldA

Integration by Parts

It is known that dx

dfg

dx

dgffg

dx

d

Integrating both sides we get dxdx

dfgdx

dx

dgfdxfg

dx

d b

a

b

a

b

a

Using the fundamental theorem of calculus we get

dxdx

dfgfgdx

dx

dgf

b

a

ba

b

a

Example: Evaluate the integral

0

dxxe x

Solution: It is known that

00

dxedx

dxdxxe xx

100

xx edxxe

Curvilinear Coordinates

Spherical coordinates (r, ,)

r: is the distance from the origin (from 0 to ∞)

: the polar angle, is the angle between r and the z-axis (from 0 to )

: the azimuthal angle is the angle between the projection of r to the x-y plane and the x-axis (from 0 to 2)

The relation between the Cartesian coordinates and

the spherical coordinates can be written as

cossinrx

sinsinry cossinrz

The unit vectors associated with the spherical coordinates are related to the

corresponding unit vectors in the Cartesian coordinates as

kjir ˆcosˆsinsinˆcossinˆ

kji ˆsinˆsincosˆcoscosˆ

ji ˆcosˆsinˆ

The infinitesimal displacement vector in spherical coordinates is expressed as

ˆsinˆˆ drrdrdrld

The volume element is expressed as

drddrdldldld r sin2

For the surface elements we have

constantisˆˆ1 drdrdldlSd r

constantisˆsinˆ2 drdrdldlSd r

constantˆsinˆ2

3 isrrddrrdldlSd

To find the volume of a sphere of radius R we have

3342

02

00sin RdddrrdV

R

To find the gradient in spherical coordinates let T=T (r, , ) so

)1(sin drTrdTdrTldTdT r

)2(

dT

dT

drr

TdT

Equating the above two equations we get

rr

TT r ˆ

ˆ1

r

T

rT

,

ˆsin

1

r

T

rT

ˆ

sin

1ˆ1ˆorr

T

rr

T

rr

r

TT

ˆsin

1ˆ1ˆrrrr

rr

Similarly, one can find the divergence and the curl in spherical coordinates

A

rA

rAr

rrA r

sin

1sin

sin

11 22

ArrAA

r

rrr

rA

r sin

ˆsinˆˆ

sin

12

2

2

2222

22

sin

1sin

sin

11

T

r

T

rr

Tr

rrT

The Laplacian is defined as

Cylindrical coordinates (, , )

: is the distance from the z-axis (from 0 to ∞)

: the azimuthal angle is the angle between and the x-axis (from 0 to 2)

z: the distance from the x-y plane (from -∞ to ∞)

The relation between the Cartesian coordinates and

the cylindrical coordinates can be written as

cosx siny zz

The unit vectors associated with the cylindrical coordinates are related to the

corresponding unit vectors in the Cartesian coordinates as

ji ˆsinˆcosˆ

ji ˆcosˆsinˆ

kz ˆˆ

The infinitesimal displacement vector in cylindrical coordinates is expressed as

zdzddld ˆˆˆ

The volume element is expressed as

dzdddldldld z

For the surface elements we have

constantiszˆˆ1 zddzdldlSd

constantisˆˆ2 dzddldlSd zr

constantisˆˆ3 dzddldlSd z

zzˆˆ

1ˆ

The Del

The Divergence z

AAAA z

11

The Curl

zAAAz

A

ˆsinˆˆ

1

2

2

2

2

2

2 11

z

TTTT

The Laplacian

The Dirac Delta Function

Consider the function rr

A ˆ12

011

Now2

22

rr

rrA

2

0

22

0

ˆsinˆ1

and rddrrr

SdAS

But from the divergence theorem we know that

4sin2

00

dd

Sv SdAdA

Her we have a contradiction. The problem is the point r=0, where the vector blows up.

Her we have a contradiction. The problem is the point r=0, where the vector blows up. So we write

rr

r 32

4ˆ

Where (x) is called the Dirac delta function with the following properties:

axax

ax

0

Using theses properties we have

1

dxax

aFdxaxxF

zyxr 3

44 3 vv drdA

As expected