Vector Analysis

VECTOR ANALYSIS

Vector Analysis 2Vector Analysis 2Vector Analysis 2

Contents

1. Vector algebra• Addition, Subtraction, & Multiplication of Vectors.

2. Orthogonal coordinate system• Cartesian, cylindrical, & spherical coordinates.

3. Vector calculus• Differentiation and integration of vectors; line,

surface and volume integrals; “del” operator,

gradient, divergence, and curl operations.


Vector Addition and Subtraction

• A vector A can be written as:

– A=aAA

– Where A is the magnitude of A and has the

unit and dimension.

– A=|A|

– aA is a dimensionless unit vector with a unity

magnitude having the direction of A.

– aA=A / |A| = A / AA=|A|

A=aAA


• Two vectors A and B can be added in two

ways. C=A+B

– Parallelogram rule

– Head to tail rule


A

B

C

A

BC



• Vector addition obeys the commutative

and associative laws

– Commutative law

• A + B = B + A

– Associative law

• A + (B + C) = (A + B) + C



• Vector subtraction is defined in following

way:

A – B = A + (-B)

Where –B has the same magnitude as B but

the direction is opposite to that of B.

-B = (-aB)B

A

B

-B A-B


• Multiplication of a vector A by a scalar k

changes only the magnitude of A by a

factor k.

kA = aA (kA)

• Scalar or Dot Product:

A . B = AB cos θAB

Product of Vectors

A

ABcosθAB

B

θAB


Product of Vectors

A . A = A2

A = √ A . A

– Commutative law

A . B = B . A

– Distributive law

A . (B + C) = A . B + A . C


Product of Vectors

• Vector or Cross Product

A x B = an|AB sin θAB|

– Cross Product is not commutative

B x A = - A x B

– Cross Product obeys the distributive law

A x (B + C) = A x B + A x C

– Cross Product is not associative

A x (B x C) = (A x B) x C

A

B

BsinθABθABan

A x B


Product of Vectors

• Product of Three Vectors

– Scalar Triple Product

A . (B x C) = B . (C x A) = C . (A x B)

A . (B x C) = -A . (C x B)

= -B . (A x C)

= -C . (B x A)


Product of Vectors

Magnitude is equal

to volume of the

parallelepiped formed

by vectors A, B, and C.

Base area is |B x C| = |BC sin θ1|

Height is |A cos θ2|

Hence the volume is |ABC sin θ1 cos θ2|

B

C

A

B x C

θ1

θ2


Product of Vectors

• Vector Triple Product

Back-cab rule

A x (B x C) = B(A . C) – C(A . B)

A = A|| + A┴

A┴ x (B x C) = 0 as both are parallel.

We are left with D = A|| x (B x C)

A||

A┴

CB

B(A|| . C)

-C(A|| . B)

DaD

θ1

θ2


Product of Vectors

– Figure shows the plane containing B, C, A||.

– D also lies in the same plane and is normal to A||.

– Magnitude of (B x C) is BC sin (θ1 – θ2).

– Magnitude of A|| x (B x C) is A||BC sin (θ1 – θ2).

D = D . aD = A||BC sin (θ1 – θ2)

= (B sin θ1)(A||C cos θ2) - (C sin θ2) (A||B cos θ1)

= [B(A|| . C) – C(A|| . B)] . AD

– It is not guaranteed that quantity inside the brackets is

equal to D; as it may contain a vector that is normal to

D; ie parallel to A||. Hence


Product of Vectors

B(A|| . C) – C(A|| . B) = D + kA||

Multiplying both sides by A||; we get

(A|| . B)(A|| . C) – (A|| . C)(A|| . B) = A||. D + kA||2

0 = A||. D + kA||2

Since A||. D = 0 ( as D is normal to A||), so k = 0

Hence

D = B(A|| . C) – C(A|| . B)

This proves the Back-Cab rule.

A|| . C = A . C and A|| . B = A . B


Division of Vectors

• Division of Vectors is not defined

• Expressions such as k/A and B/A are

meaningless.


Orthogonal Coordinate Systems

• We need position of the source and the location

of this point in coordinate system to determine

the Electric Field at a certain point in space.

• In three dimensional space a point can be

located as the intersection of three surfaces u1,

u2, u3.

• If these three surfaces are perpendicular to one

another; we have the Orthogonal Coordinate

System.


• Let au1, au2, and au3 be the unit vectors called theBase Vectors in the three coordinate system;then in a general right handed, orthogonal,curvilinear coordinate system:

• au1 x au2 = au3,

• au2 x au3 = au1,

• au3 x au1 = au2.

• Above three equations are not all independent,as the specification of one automatically impliesthe other two



• au1 . au2 = au2 . au3 = au3 . au1 = 0

• au1 . au1 = au2 . au2 = au3 . au3 = 1


• A = au1Au1 + au2Au2 + au3Au3

• Magnitude of vector A is

• A = |A| = (Au12+Au2

2+Au32)1/2.



• EXAMPLE:

– Given three vectors A, B, and C, obtain the

expressions of:

(a) A . B (b) A x B (c) C . (A x B) in the

orthogonal curvilinear coordinate system

(u1,u2,u3).



• SOLUTION:

A = au1Au1 + au2Au2 + au3Au3

B = au1Bu1 + au2Bu2 + au3Bu3

C = au1Cu1 + au2Cu2 + au3Cu3

a) A . B = (au1Au1 + au2Au2 + au3Au3) . (au1Bu1 +

au2Bu2 + au3Bu3)

= Au1Bu1 + Au2Bu2 + Au3Bu3



b) A x B = (au1Au1 + au2Au2 + au3Au3) x (au1Bu1 +

au2Bu2 + au3Bu3)

= au1(Au2Bu3 – Au3Bu2) + au2(Au3Bu1 – Au1Bu3)

+ au3(Au1Bu2 – Au2Bu1)

au1 au2 au3

= Au1 Au2 Au3

Bu1 Bu2 Bu3



c) C . (A x B)

= Cu1(Au2Bu3 – Au3Bu2) + Cu2(Au3Bu1 – Au1Bu3)

+ Cu3(Au1Bu2 – Au2Bu1)

Cu1 Cu2 Cu3

= Au1 Au2 Au3

Bu1 Bu2 Bu3



• Differential change in length corresponds to the change in one of the coordinates and a factor is needed for such a change.

dli = hi dui, (i = 1, 2, or 3)

Where hi is called metric coefficient and may itself be a function of ui

• e.g: In a two coordinate system (u1, u2) = (r, Ø) a differential change dØ (=du2) in Ø (=u2) corresponds to a differential length change dl2 = rdØ (h2 = r = u1) in the aØ (=au2) direction.



• A directed differential length change in an

arbitrary direction can be written as vector

sum of component length changes;

dl = au1 dl1 + au2 dl2 + au3 dl3

dl = au1 (h1 du1) + au2 (h2 du2) + au3 (h3 du3)

Magnitude of dl is

dl = [(dl1)2 + (dl2)

2 + (dl3)2]1/2

= [(h1 du1)2 + (h2 du2)

2 + (h3 du3)2]1/2



• The differential volume formed by differential

coordinate changes du1, du2, and du3 in

directions au1, au2, and au3 respectively is (dl1dl2 dl3), or

dv = h1h2h3 du1du2du3

• In order to express the current or flux flowing

through a differential area, cross-sectional area

perpendicular to the current or flux is to be used

ds = ands



• Let current density J is not perpendicular

to a differential area ds, the current dI,

flowing through ds must be the component

of J normal to the area, multiplied by the

area.

dI = J . ds

=J . ands



• In general orthogonal curvilinearcoordinate system the differential area ds1

normal to the unit vector au1 is:

• ds1 = dl2 dl3• ds1 = h2h3du2du3

• Similarly differential areas normal tovectors au2 and au3 are respectively

• ds2 = h1h3du1du3

• ds3 = h1h2du1du2



• Main orthogonal coordinate systems are:

• Cartesian (or Rectangular) Coordinates

• Cylindrical Coordinates

• Spherical Coordinates



Cartesian Coordinates

• (u1, u2, u3) = (x, y, z)

• Point P(x1, y1, z1) is

Intersection of three

Planes x = x1, y = y1,

z = z1

Base vectors are ax,

ay, az in the

respective Directions.

z=z1 plane

y=y1 plane

X=x1 plane

http://upload.wikimedia.org/wikipedia/commons/f/fd/Rectangular_coordinates.svg


• Base vectors satisfy following relations:

ax x ay = az,

ay x az = ax,

az x ax = ay.

• Position vector to point P P(x1, y1, z1) is:

OP = axx1 + ayy1 + azz1.


A = axAx + ayAy + azAz.



• The dot product of two vectors A and B is:

A . B = AxBx + AyBy + AzBz

• The cross product of A and B is:

A x B =

ax(AyBz-AzBy) + ay(AzBx-AxBz) + az(AxBy-AyBx)

ax ay az

= Ax Ay Az

Bx By Bz



• Since x, y, and z are lengths so all three matric

coefficients are unity ie, h1 = h2 = h3 = 1. The

expressions for differential length, differential

area, and differential volume are:

dl = axdx + aydy + azdz.

dsx = dydz,

dsy = dxdz,

dsz = dxdy.

dv = dxdydz



x

y

z

dx

dy

dz

dsx =dydz dsz = dxdy

Dsy = dxdz

o


A differential volume in Cartesian Coordinates



http://en.wikipedia.org/wiki/Image:Cartesian_coordinates_3D.svg


• EXAMPLE: Given A = ax5 – ay2 + az, find

the expression of a unit vector B such

that:

a) B||A

b) B┴A, if B lies in the xy-plane.

• SOLUTION:

– Let B = axBx + ayBy + azBz. We know that

– B = (Bx2 + By

2 + Bz2)1/2 = 1



a) B||A requires B x A = 0, hence we have

-2Bz – By = 0,

Bx – 5Bz = 0,

5By + 2Bx = 0.

Solving above equations along with magnitude equation; we get:

Bx = 5/√30, By = -2/√30, Bz = 1/√30

Therefore

B = (ax5 – ay2 + az)/√30



b) B┴A requires B . A = 0, hence we

have 5Bx – 2By = 0.

Bz = 0, since B lies in the xy-plane

Solution of above equation along with

magnitude equation yields:

Bx = 2/√29, By = 5/√29

Hence

B = (ax2 + ay5)/√29



• EXAMPLE:

– (a) Write the expression of the vector going

from point P1(1,3,2) to point P2(3,-2,4) in

Cartesian coordinates.

– (b) What is the length of this line?

• SOLUTION:



From Figure, we see

P1P2 = OP2 – OP1

= (ax3-ay2+az4) –

(ax+ay3+az2)

= ax2 – ay5 + az2

The length of the line is

P1P2 =|P1P2| = √22 + (-5)2 + 22

= √33


P1(1,3,2)

P2(3,-2,4)

x y

z


• (u1, u2, u3) = (r, Ø, z)

• Point P(r1, Ø1, z1) is

the intersection of a

cylindrical surface

r=r1, a half plane

containing the z axis

and making an angle

Ø=Ø1 with the xz-

plane, and a plane

parallel to xy plane at

z=z1.

Cylindrical Coordinates

z

y

x

r1az

aØ

ar

z1

y1

x1

o

Ø1

Ø=Ø1 plane

z=z1 plane

r=r1 cylinder



• Angle Ø is measured from +ve x-axis, and

base vector aØ is tangential to the

cylindrical surface.

• Following right handed relations apply.

ar x aØ = az

aØ x az = ar

az x ar = aØ

Vector Analysis 42Vector Analysis 42


Vector Analysis 42

,0

,0

,0

,1

,1

,1

rz

z

r

zz

rr

aa

aa

aa

aa

aa

aa

,0

,0

,0

zz

rr

aa

aa

aa



• A vector in cylindrical coordinates is

written as:

A = arAr + aØAØ + azAz

• Dot and cross product of two vectors in

cylindrical coordinates follow the equations

as discussed on slides 20,21.

• Two of the coordinats, r and z (u1 and u3)

are lengths; hence h1 = h3 = 1.


• However Ø is an angle requiring a metric

co-efficient h2 = r to convert dØ to dl2.

• General expression for a differential length

in cylindrical coordinates is then:

dl = ardr + aØrdØ + azdz

• Expressions for differential areas and

differential volume are:

dsr = r dØ dz,



• dsØ = dr dz,

• dsz = r dr dØ,

• dv = r dr dØ dz.



• A vector given in cylindrical coordinates

i.e A = arAr + aØAØ + azAz

can be transformed into Cartesian coordinates ie

A = axAx + ayAy + azAz.

• Z component remains un-altered.

• To find Ax, we equate dot product of above both

expressions of A with ax. Thus:

Ax = A . ax = arAr . ax + aØAØ . ax

• az . ax = 0, hence Az disappears.



From figure:

ar . ax = cos Ø

aØ . ax = cos(π/2 + Ø)

= - sin Ø

Ax = Ar cos Ø – AØ sin Ø

Similarly Ay = A . ay

= arAr . ay + aØAØ . ay

ar . ay = cos(π/2 - Ø)

= sin Ø

aØ . ay = cos Ø

Ay = Ar sin Ø + AØ cos Ø


ar

aØ



Vector Analysis 48

z

y

x

z

r

a

a

a

a

a

a

100

0cossin

0sincos


• Conversion Matrix is:

Ax cosØ -sinØ 0 Ar

Ay = sinØ cosØ 0 AØ

Az 0 0 1 Az


z

y

x

z

r

A

A

A

A

A

A

100

0cossin

0sincos


• Conversions formulas are:

Cartesian Cylindrical

x = r cos Ø r = √ x2 + y2

y = r sin Ø Ø= tan-1 y/x

z = z z = z



• EXAMPLE: The cylindrical coordinates ofan arbitrary point P in the z = 0 plane are(r, Ø, 0). Find the unit vector that goesfrom a point z = h on z-axis toward P.

• SOLUTION:

QP = OP – OQ

= (arr) – (azh)

aQP = QP/|QP|

= (1/√r2 + h2) (arr – azh)



• EXAMPLE: Express the vector

A = ar(3cosØ) – aØ2r + az5 in cartesian

coordinates:

Ax cosØ -sinØ 0 3cosØ

Ay = sinØ cosØ 0 -2r

Az 0 0 1 5

• A = ax (3cos2 Ø + 2r sin Ø) + ay (3sin Øcos Ø

– 2r cos Ø) + az 5



• Cos Ø = x / √x2 + y2

• Sin Ø = y / √x2 + y2

• Therefore:

A = ax (3x2/(x2 + y2) + 2y)

+ ay (3xy/(x2 + y2) - 2x)

+ az 5



Spherical Coordinates

• (u1, u2, u3) = (R, θ, Ø)

• Point P(R1, θ1, Ø1) is the intersection of aspherical surface centered at the originwith a radius R=R1, a right circular conewith it’s apex at the origin, it’s axiscoincides with the + z-axis and having ahalf angle θ=θ1 and a half plane containingthe z axis and making an angle Ø=Ø1 withthe xz-plane.





• The base vector aR at P is radial from the

origin and is quite different from ar in

cylindrical coordinates, as the latter is

perpendicular to the z-axis. The base

vector aθ lies in the Ø=Ø1 plane and is

tangential to the spherical surface,

whereas the base vector aØ is the same as

in the cylindrical coordinates.


0

0

0

1

1

1

aa

aa

aa

aa

aa

aa

R

R

RR




• For a right-handed system we have

aR x aθ = aØ,

aθ x aØ= aR,

aØ x aR = aθ

• Spherical coordinates are important forproblems involving point sources andregions with spherical boundaries.

• Spherical coordinates are used in solvingantenna problems in the far field.



• A vector in spherical coordinates is written as:

A = aRAR + aθAθ + aØAØ.

• Expressions for dot and cross products of two vectors in spherical coordinates are similar to those shown on slide 31.

• In spherical coordinates only R (u1) is a length. The other two coordinates θ and Ø (u2 and u3) are angles.




• Metric coefficients h2=R and h3=R sinθ are

required to convert dθ and dØ into dl2 and

dl3 respectively.

• From equation on page 24 the general

expression for differential length is:

• dl = aR dR + aθR dθ + aØR sinθ dØ




• Differential areas and differential volume

resulting from differential changes dR, dθ,

dØ are:

• dsR = R2 sinθ dθ dØ,

• dsθ = R sinθ dR dØ,

• dsØ = R dR dθ,

• dv = R2 sinθ dR dθ dØ.


• A vector in spherical coordinates can be

transformed into Cartesian coordinates as:

x = R sinθ cosØ,

y = R sinθ sinØ,

z = R cosθ,




• Cartesian coordinates can be converted to

spherical coordinates as:

R = √x2 + y2 + z2,

θ = tan-1 √(x2 + y2)/z,

Ø = tan-1 y/x



,cos

,sinsin

,cossin

zR

yR

xR

aa

aa

aa

,sin

,sincos

,coscos

z

y

x

aa

aa

aa

,0

,cos

,sin

z

y

x

aa

aa

aa

z

y

xR

a

a

a

a

a

a

0cossin

sinsincoscoscos

cossinsincossin



• Transformation of Vector

.sincoscoscossin

,

,

AAA

aaAaaAaaAaAA

aAaAaAA

R

xxxRRxx

RR



A

A

A

A

A

A R

z

y

x

0sincos

cossincossinsin

sincoscoscossin

z

y

xR

A

A

A

A

A

A

0cossin

sinsincoscoscos

cossinsincossin


• Example: The position of a point P in

spherical coordinates is (8, 120, 330).

Specify it’s location (a) in Cartesian

coordinates (b) in cylindrical coordinates.

• Solution: Coordinates of the point P are

R=8, θ=120°, Ø=330°.



• a) Let us use the equations on page 60.

• x = 8 sin120° cos330° = 6,

• y = 8 sin120° sin330° = -2√3,

• z = 8 cos120° = -4.

• Hence the location of point is P(6, -2√3, -4).

• And the position vector is:

• OP = ax6 – ay2√3 – az4.



• b) The cylindrical coordinates of point P can be obtained by applying equations on page: 48; but these can also be calculated directly from the spherical coordinates by using following equations:

– r = R sinθ,

– Ø = Ø,

– z = R cos θ.

• Hence we get the point P(4√3, 330, -4).



• Position vector in cylindrical coordinates is:OP = ar4√3 – az4

• We note that position vector does not contain Ø=330°; however exact direction of ar depends on Ø.

• In spherical coordinates position vector contains only one term:

OP = aR8.

• Here the direction of ar changes with the θ and Ø coordinates of point P.



• Example: Convert the vector A = aRAR + aθAθ + aØAØ into Cartesian coordinates.

• Solution: In this problem we want to write A in the form of A = axAx + ayAy + azAz.

1) We assume that the expression of the given vector A holds for all points of interest and that all three given components AR, Aθ, and AØ may be functions of coordinate variables.

2) At a given point AR, Aθ, and AØ will have definite numerical values, but these values



that determine the direction of A will, in general, be

entirely different from the coordinate values of the

point.

• Taking dot product of A with ax, we get:

Ax = A . ax

= ARaR . ax + Aθaθ . ax + AØaØ . ax

• aR . ax, aθ . ax, and aØ . ax yield respectively, the

component of unit vectors aR, aθ, and aØ in the

direction of ax, we find from fig on page 57 and

equations on page 60:



• aR . ax = sinθ cosØ = x/√(x2 + y2 + z2)

• aθ . ax = cosθ cosØ = xz/ √((x2 + y2)

(x2 + y2 + z2))

• aØ . ax = - sinØ = -y/√(x2 + y2)

• Thus Ax = AR sinθ cosØ + Aθ cosθ cosØ - AØ sinØ

• = ARx/√(x2 + y2 + z2) + Aθxz/√((x2 + y2)

(x2 + y2 + z2)) - AØy/√(x2 + y2)



• Similarly Ay = AR sinθ sinØ + Aθ cosθ sinØ

+ AØ cosØ

• = ARy/√(x2 + y2 + z2) + Aθyz/√((x2 + y2)

(x2 + y2 + z2)) + AØx/√(x2 + y2)

• AZ = AR cosθ + Aθ sinθ

= ARz/√(x2 + y2 + z2) - Aθ√(x2 + y2) /

√(x2 + y2 + z2))



• Example: Assuming that a cloud of

electrons confined in a region between two

spheres of radii 2 and 5 cm has a charge

density of -3x10-8 cos2Ø / R4 C/m3.

• Solution: ρ = -3x10-8 cos2Ø / R4,

Q = ∫ ρdv.

Q = 0∫2π

0∫π

0.02∫0.05 ρR2 sinθ dR dθ dØ



• Q = -3x10-80∫

2π 0∫

π0.02∫

0.05 (1/R2) cos2Ø

sinθ dR dθ

dØ

• = -3x10-80∫

2π 0∫

π(-1/0.05 + 1/0.02) cos2Ø

sinθdθ dØ

• = -0.9x10-60∫

2π (-cosθ)0|π cos2ØdØ

• = -1.8x10-6 (Ø/2 + (sin2Ø)/4)0|2π

• = -1.8π (µC).



Integrals Containing Vector

Functions

• Above integral can be evaluated as the sum of

three scalar integrals by first resolving the vector

F into it’s three components in the appropriate

coordinate system.

• dv represents the differential volume.

• This is the triple integral over three dimensions

shown in a shorthand way.

VFdv



Functions

CVdl

1

2

P

PVdl

•This integral is a scalar function of space.

•dl shows the differential increment of length.

•C is the path of integration.

•If the integral is from P1 to P2, we write

•If it is for a closed path C, we write CVdl



Functions

• In Cartesian coordinates Integral can be written

as:

Cz

CCyx

C

Czyx

C

dzzyxVadyzyxVadxzyxVaVdl

dzadyadxazyxVVdl

),,(),,(),,(

)[,,(

•Three integrals on right hand side are ordinary

scalar integrals. These can be evaluated for a

given V(x,y,z) around a path C.



Functions

• Example: Evaluate the integral , where

___________, from the origin to the point P(1,1):

• a) Along the direct path OP.

• b) Along the path OP1P, and

• c) Along the path OP2P.

P

Odrr 2

222 yxr



Functions

• Solution:

• a) Along the direct path OP

3

2

3

2

)45sin45cos(3

22

3

222

0

22

yx

yx

P

Orr

aa

aa

adrradrr



Functions• Solution:

• b) Along the path OP1P

.3

1

3

4

)3

1(

3

1

)1()(

1

0

31

0

3

2222

1

1

yx

xy

P

Px

P

Oy

P

O

aa

xxaya

dxxadyyadryx


• Along the path OP2P

.3

4

3

1

)3

1(

3

1

)1()(

1

0

31

0

3

2222

2

2

yx

yx

P

Py

P

Ox

P

O

aa

yyaxa

dyyadxxadryx


Functions



Functions

C

dlF

•Above is a line integral, in which integrand

represents the component of F along the path of

integration.

•If F is a force, the integral is work done by the

force in moving an object from a point P1 to P2

along a specified path C.

•If F is replaced by E, then the integral is work

done by electric field in moving a unit charge from

P1 to P2.



Functions

• Example: Given , evaluate the scalar line integral , along the quarter circle shown in figure.

xaxyaF yx 2

B

AdlF



Functions

• Solution:

• a) In Cartesian coordinates:

)2

1(9

3sin99)9(

3

1

929

)3,0(9

2

3

0

120

3

23

2

3

0

20

3

2

22

yyyx

dyydxxxdlF

yxyx

xdyxydxdlF

B

A



Functions

• Solution:

• b) In cylindrical coordinates:

)cos2sin()sin2cos(

0

2

100

0cossin

0sincos

100

0cossin

0sincos

xxyaxxyaF

x

xy

F

F

F

A

A

A

A

A

A

r

z

r

z

y

x

z

r



Functions• Path of integration is along a quarter-circle of a

radius 3. There is no change in r or z along thepath (dr=0 and dz=0); hence equation dl = ardr +aØrdØ + azdz simplifies to:

).2

1(9

)cossin(sin9

)cos6cossin9(3

)cos2sin(3

3

2

0

3

2

0

22

B

AddlF

dxxydlF

dadl



Functions

• This is a surface integral. It is actually a double

integral over two dimensions.

• The integral measures the flux of the vector field

A flowing through the area S.

• Vector differential surface element ds=ands has

a magnitude ds and the direction shown by an.

• The conventions for the +ve direction of ds are

as follows:

s

dsA



Functions• If the surface of integration S is a closed surface

enclosing a volume, then the +ve direction of an

is always is the outward direction.

• Positive direction of an depends on the location of ds.

• Further closed surface integral requires a small circle added over the integration sign.

s

ns

dsaAdsA .


• If S is an open surface, the +ve direction of andepends on the direction in which the perimeterof the open surface is traversed.

• Acc to right hand rule if the fingers follows thedirection of travel around the perimeter then thethumb points in the direction of +ve an.

• Again the +ve direction of an depends on thelocation of ds.


Functions



Functions

• Example: Given , evaluate the

scalar surface integral over the surface

of a closed cylinder about the z-axis specified

by z=±3 and r=2.

zkarkaF zr 21

s

dsF

•Solution: The specified

surface of integration is

that of closed cylinder as

shown. It has three

surfaces: The top face,

the bottom face, and the

side wall.



Functions

sidewalln

topface bottomfacenn

sn

dsaFdsaFdsaF

dsaFdsF

...

•Where an is a unit vector normal outwards from

the respective surfaces.

•Three integrals on the right side can be

evaluated separately.



Functions

• a) Top face z = 3, an = az,

TopFacen

n

krdrdkdsaF

rdrdds

kzkaF

2

0

2

022

22

123

;

,3



Functions

• b) Bottom Face: z = -3, an = -az,

BottomFacen

n

krdrdkdsaF

rdrdds

kzkaF

2

0

2

022

22

123

;

,3



Functions

• C) Side Wall r = 2, an = ar,

SideWalln

n

kdzdkdsaF

dzddzrdds

k

r

kaF

.12

;2

,2

3

3

2

011

11



Functions

• Therefore

)2(12

121212

21

122

kk

kkkdsFs

•This surface integral gives the net outward flux

of the vector F through the closed cylindrical

surface.


Gradient of a Scalar Field

• We encounter scalar and vector fields that are

functions of four variables: (t, u1, u2, u3).

• Method is required for describing the space rate

of change of a scalar field at a given time.

• Consider a scalar function of space coordinates

V(u1, u2, u3) which represents say, the

temperature distribution in a building, the altitude

of a mountainous terrain, or the electric potential

in a region



• Magnitude of V depends on the position of the

point in space, but it may be constant along

certain lines or surfaces as shown in figure two

surfaces having constant magnitudes V1 and

V1+dV.

• Point P1 is on the surface V1; P2 is the

corresponding point on surface V1+dV along the

normal vector dn; and P3 is a point close to P2

along another vector dl ≠ dn.

Vector Analysis 100



Vector Analysis 101

•For the same change dV in V the space rate of

change, dV/dl, is greatest along dn as dn is the

shortest distance b/w the two surfaces.

•Since the magnitude

of dV/dl depends on

the direction of dl,

dV/dl is a directional

derivative


• “We define the vector that represents both

the magnitude and the direction of the

maximum space rate of increase of a

scalar as the gradient of that scalar.”

Vector Analysis 102


dn

dVaV

dn

dVagradV

n

n

gradV of placein V writeand

symbol by the drepresente del,operator employ tocustomary isIt

.


• We have assumed that dV is +ve if increase in

V; if dV is –ve (a decrease in V from P1 to P2) ,

V will be –ve in an direction.

• Directional derivative along dl is

Vector Analysis 103


lln aVaadn

dV

dn

dV

dl

dn

dn

dV

dl

dV

)(

cos

•This equation states that the space rate of increase

of V in the al direction is equal to the projection of

the gradient of V in that direction


• We can also write

Vector Analysis 104


,

:scoordinatein changes aldifferenti theof terms

in expressed becan it hence );P toP (fromposition in change a

ofresult a as V of aldifferenti total theis dV Now , Where

,)(

3

3

2

2

1

1

31

dll

Vdl

l

Vdl

l

VdV

dladl

dlVdV

l

•Where dl1, dl2, and dl3 are the components of the

vector differential displacement dl in a chosen

coordinate system.


• In terms of general orthogonal coordinates (u1,

u2, u3), dl is:

Vector Analysis 105


dll

Va

l

Va

l

Va

dladladlal

Va

l

Va

l

VadV

duhaduhaduha

dladladladl

uuu

uuuuuu

uuu

uuu

)(

)()(

:follows as vectors

twoofproduct dot as written becan dV

)()()(

3

3

2

2

1

1

332211

3

3

2

2

1

1

333222111

332211



• Comparing above equation with the equation on top of slide 104.

33

3

22

2

11

1

3

3

2

2

1

1

uh

Va

uh

Va

uh

VaV

l

Va

l

Va

l

VaV

uuu

uuu

• Above equation is useful for computing gradient of

a scalar, when the scalar is given as a function of

space coordinates.



• In Cartesian coordinates, (u1, u2, u3) = (x, y, z)

and h1 = h2 = h3 = 1, hence we have:

za

ya

xa

Vz

ay

ax

aV

z

Va

y

Va

x

VaV

zyx

zyx

zyx

operator. aldifferenti vector a as

scoordinateCartesian in consider toconvenient isIt

)(



• We see that we can define in general

orthogonal coordinates as:

)(33

3

22

2

11

1uh

auh

auh

a uuu



• Example: The Electrostatic field intensity E isderivable as the –ve gradient of a scalar electricpotential V; that is, E= - V. Determine E at thepoint (1, 1, 0) if

cos )

,4

sin )

REVb

yeVVa x



• We use Cartesian Coordinates for part (a) and spherical coordinates for part (b) to solve E= - V.

• a)

).4

()16(1

1

,)16

1(2

1

,2

)4

()0,1,1(

.)4

cos44

sin(

4sin][

2

2

yxE

Eyx

x

yx

x

zyx

aaa

EEwhere

EaE

aaEThus

eEy

ay

a

yeE

za

ya

xaE



b)

.)sincos(

cos]sin

[

Eaa

RER

aR

aR

aE

R

R

In view of following equation:

AZ = AR cosθ + Aθ sinθ

= ARz/√(x2 + y2 + z2) - Aθ√(x2 + y2) /

√(x2 + y2 + z2))

the result of above converts to E = -azEo in

Cartesian coordinates.



• This is not surprising, as careful examination

of the given V reveals that EoRcosθ is infact

equal to Eoz. Hence in Cartesian coordinates:

EazEz

aVE zz

)(


Divergence of a Vector Field

• Flux lines or streamlines are directed lines or

curves indicating at each point the direction of

the vector field.

• Magnitude of the field at a point is depicted

either by the density or by the length of the

directed lines in the vicinity of the point.

• This figure shows that the field in the

region A is stronger than that in region

B, as there is higher density of equal

length directed lines in region A.

Vector Analysis 114


Vector Analysis 114

• This figure indicate a radial field

that is strongest in the region

closest to the point q and

decreasing arrow lengths show

the weaker field away from the

charge q.

• This figure depicts a uniform field.

• Vector Field Strength is measured by the

number of flux lines passing through a unit

surface normal to the vector.

Vector Analysis 115


• The flux of vector field is analogous to the flow of

an incompressible fluid such as water.

• Net +ve divergence indicates the presence of a

source of fluid inside the volume.

• Net –ve divergence indicates the presence of

sink inside the volume.

• In the uniform field, there is an equal amount of

inward and outward flux going through any

closed volume containing no source or sink,

resulting in a zero divergence.Vector Analysis 115

Vector Analysis 116


• We define the divergence of a vector field at a

point, (abbreviated div A) as the net outward flux

of A per unit volume as the volume about the

point tends to zero:

Vector Analysis 116

v

dsAvdivA s

0lim

• The numerator represents the net outward flux,

is an integral over the entire surface S that

bounds the volume

Vector Analysis 117


• Div A is a scalar quantity whose magnitude may

vary from point to point.

• Consider a differential volume of sides Δx, Δy,

and Δz centered about a point P(xo, yo, zo) in the

field of a vector A; and we wish to find div A at

the point (xo, yo, zo).

• Since the differential volume has six faces, the

surface integral can be decomposed into six

parts.

Vector Analysis 117

Vector Analysis 118


Vector Analysis 118

zyzyx

xA

zyaASAdsA

dsAdsA

x

xfrontfacefrontfacefrontface

frontface

s bottomfacetopfaceleftfacerightfacebackfacefrontface

),,2

(

)(

facefront On the

Vector Analysis 119


• The quantity can be expanded as a

Taylor series about its value at (xo, yo, zo), as

follows:

Vector Analysis 119

),,2

( zyx

xAx

,__2

),,(),,2

(),,(

termsorderhigherx

AxzyxAzy

xxA

zyx

xxx

Where the higher order terms (H.O.T) contain

the factors (Δx/2)2,(Δx/2)3, etc.

Vector Analysis 120


Vector Analysis 120

.)..(][

:get wefaceback andfront for equations Combining

..2

),,(),,2

(

:is ),,2

( ofexpansion series-Taylor The

),,2

(

)(

),,(

),,(

zyxTOHx

AdsA

TOHx

AxzyxAzy

xxA

zyx

xA

zyzyx

xA

zyaASAdsA

zyx

x

BackfaceFrontface

zyx

xxx

x

x

xbackfacebackfacebackface

backface

Vector Analysis 121


• Following the same procedure for the right and

the left faces, where the coordinate changes are

+Δy/2 and –Δy/2, respectively and Δs=ΔxΔz, we

find:

Vector Analysis 121.z)(z, factors thecontains H.O.T theHere

.)..(][

:have wefaces bottom

and topFor the ,y)(y, factors thecontains H.O.T theHere

.)..(][

2

),,(

2

),,(

zyxTOHz

AdsA

zyxTOHy

AdsA

zyx

z

BottomfaceTopface

zyx

y

Leftfacerightface

Vector Analysis 122


• Now combining the results of all the sides:

Vector Analysis 122

.,,)(),,(

zyxintermsorderhigherzyxz

A

y

A

x

AdsA

zyx

zy

s

x

• Since Δv=ΔxΔyΔz substituting above equation

in div A equation in Cartesian coordinates we

get:

z

A

y

A

x

AdivA zyx

•The higher order terms vanish as the differential

volume ΔxΔyΔz approaches zero.

Vector Analysis 123


• Value of div A depends on the position of the

point at which it is evaluated.

• We have dropped notation (xo, yo, zo) in above

equation because it applies to every point at

which A and its partial derivates are defined.

Vector Analysis 123

.)()()(1

:have we),,,( scoordinate orthogonal generalIn

321

3

231

2

132

1321

321

Ahhu

Ahhu

Ahhuhhh

A

uuu

divAA

Vector Analysis 124


• Example: Find the divergence of the position

vector to an arbitrary point.

• Solution: We will find the solution in Cartesian as

well as in spherical coordinates.

a) Cartesian coordinates:

Expression for a position vector to an arbitrary point

(x, y, z) is:

Vector Analysis 124

.3)(

.

z

z

y

y

x

xOP

zayaxaOP zyx

Vector Analysis 125


b) Spherical coordinates: Here the position

vector is simply:

Its divergence in spherical coordinates (R, θ, Ø)

can be obtained from equation on page: 123:

Vector Analysis 125

.RaOP R

3

:get eequation w abovein OP of value thengSubstituti

sin

1)sin(

sin

1)(

1 2

2

OP

A

RA

RAR

RRA R

Vector Analysis 126


• Example: The magnetic flux density B outside a

very long current-carrying wire is circumferential

and is inversely proportional to the distance to

the axis of the wire. Find div B.

• Solution: Let the long wire be coincident with the

z-axis in a cylindrical coordinate system. The

problem states that:

Vector Analysis 126

.r

kaB

Vector Analysis 127

Divergence of a Vector Field• The divergence of a vector field in cylindrical

coordinates (r, Ø, z) can be found from equation

on page: 123.

Vector Analysis 127

.0

:givesequation above Hence ,0 and ,

.1

)(1

B

BBrkB

z

BB

rrB

rrB

zr

zr

• Above vector is not a constant but its divergence is

zero. Hence magnetic flux lines close upon

themselves and there are no sources or sinks. A

divergence less field is called a solenoidal field.

Vector Analysis 128

Divergence Theorem

• The volume integral of the divergence of a

vector field equals the total outward flux of the

vector through the surface that bounds the

volume; that is,

• This identity is called the divergence theorem,

also known as Gauss’s theorem.

• The direction of ds is always outward

perpendicular to the surface ds and directed

away from the volume.Vector Analysis 128

SV

dsAAdv

Vector Analysis 129

Divergence Theorem

• For a very small differential volume element Δvj

bounded by a surface sj, the definition of in

previous equation gives directly:

• Let an arbitrary volume V, subdivided into many

say N, small differential volumes of which Δvj is

typical as shown in figure.

Vector Analysis 129

A

Sj

jj dsAvA .)(

Vector Analysis 130

Divergence Theorem

• Combine the contribution of all these differential

volumes to both sides of previous equation:

• Left side of above equation is by definition the

volume integral of :

Vector Analysis 130

N

jSj

j

N

j

jjj dsAvvAv11

0lim)(0lim

A

dvAvAvV

N

j

jjj )()(0lim1

Vector Analysis 131

Divergence Theorem• The surface integrals on the right side of

equation on the top of page 130 are summed

over all the faces of all the differential volume

elements.

• The contributions from the internal surfaces of

adjacent elements will cancel each other,

because at a common internal surface the

outwards normals of the adjecent elements point

in opposite directions.

• Hence the net contribution is due to only that of

external surface S bounding the volume V.

Vector Analysis 131

Vector Analysis 132

Divergence Theorem

• The last three equations yield the divergence

theorem.

• Validity of the limiting processes leading to the

proof of the divergence theorem requires that

the vector field A, as well as its first derivatives,

exist and be continuous both in V and on S.

• The Divergence theorem converts a volume

integral of the divergence of a vector to a closed

surface integral of the vector, and vice versa.Vector Analysis 132

S

N

jSj

j dsAdsAv1

0lim

Vector Analysis 133

Divergence Theorem

• Example: Given A=axx2+ayxy+azyz, verify the

divergence theorem over a cube one unit on

each side. The cube is situated in the first octant

of the Cartesian coordinate system with one

corner at the origin.

Vector Analysis 133

• Refer to figure. We first

evaluate the surface integral

over the six faces.

Vector Analysis 134

Divergence Theorem

1. Front face: x=1, ds=axdydz;

2. Back face: x=0, ds=-axdydz;

3. Left face: y=0, ds=-aydxdz;

Vector Analysis 134

Frontface

dydzdsA1

0

1

01

Frontface

dsA 0

Leftface

dsA 0

Vector Analysis 135

Divergence Theorem

4. Right face: y=1, ds=aydxdz;

5. Top face: z=1, ds=azdxdy;

6. Bottom face: z=0, ds=-azdxdy;

Vector Analysis 135

Rightface

xdxdzdsA1

0

1

0 2

1

Topface

ydxdydsA1

0

1

0 2

1

Bottomface

dsA 0

Vector Analysis 136

Divergence Theorem• Adding above six values:

• Now the divergence of A is:

• Hence:

• Results are same; so divergence theorem is

therefore verified.Vector Analysis 136

202

1

2

1001 S dsA

yxyzz

xyy

xx

A

3)()()( 2

V

dxdydzyxAdv1

0

1

0

1

02)3(

Vector Analysis 137

Divergence Theorem

• Example: Given F=aRkR, determine whether the

divergence theorem holds for the shell region

enclosed by spherical surfaces at R=R1 and

R=R2(R2>R1) centered at the origin, as shown in

figure:

Vector Analysis 137

• Solution: Here the

region has two

surfaces at R=R1 and

R=R2.

Vector Analysis 138

Divergence Theorem

• At the outer surface: R=R2, ds=aRR22sinθdθdØ;

• At the inner surface: R=R1, ds=-aRR12sinθdθdØ;

.4sin)( 3

2

2

2

2

0

2

02 kRddRKRdsF

ceOuterSurfa

)(4

:have weresults, two theAdding

.4sin)(

3

1

3

2

3

1

2

1

2

0 01

RRkdsF

kRddRKRdsF

S

ceInnerSurfa

138Vector Analysis

Vector Analysis 139

Divergence Theorem

• To find the volume integral, we first determine

• for an F that has only FR component:

• Since is a constant, its volume integral

equals the product of the constant and the

volume. The volume of the shell region between

the two spherical surfaces with radii R1 and R2

is

F

kkRRR

FRRR

F R 3)(1

)(1 3

2

2

2

F

.3

)(4 3

1

3

2 RR

139Vector Analysis

Vector Analysis 140

Divergence Theorem

• Therefore:

• This is the same result as in surface integral.

• This example shows that the divergence

theorem holds even when the volume has holes

inside.

),(4)( 3

1

3

2 RRkVFdvFV

140Vector Analysis

Vector Analysis 141

Curl of a Vector Field

• There is a kind of source called Vortex Source,

which causes a circulation of a vector field

around it.

• The net circulation of a vector field around a

closed path is defined as the scalar line integral

of the vector over the path. We have:

• The physical meaning of circulation depends on

what kind of field the vector A represents.

Vector Analysis 141

C

dlACcontour aroundA ofn Circulatio

Vector Analysis 142


• If A is a force acting on an object, its circulation

will be the work done by the force in moving the

object once around the contour.

• If A represents an Electric Field Intensity, then

the circulation will be an Electromotive Force

around the closed path.

• The familiar phenomenon of water whirling down

a sink drain is an example of a vortex sink

causing a circulation of fluid velocity.

• A circulation of A may exist even when div A=0.Vector Analysis 142

Vector Analysis 143


• As circulation is a line integral of a dot product,

its value obviously depends on the orientation of

the contour C relative to the vector A.

• To define a point function, which is the measure

of the strength of a vortex source, we must make

C very small and orient it in such a way that the

circulation is a maximum. We define:

Vector Analysis 143

.1

0lim curlmax

Cn dlAa

ssAA

Vector Analysis 144

Curl of a Vector Field• The curl of a vector field A, denoted by curl A

or , is a vector whose magnitude is the

maximum net circulation of A per unit area as

the area tends to zero and whose direction is

the normal direction of the area when the area

is oriented to make the net circulation

maximum.

• Normal to an area can point in two

opposite directions, we stick to the

right hand rule that when fingers

follow the direction of dl, the thumb points to

the an direction Vector Analysis 144

A

Vector Analysis 145


• Curl A is a vector point function its component in

any other direction au is , which can be

determined from the circulation per unit area

normal to au as the area approaches zero.

• Here the direction of the line integration is

around the contour Cu bounding the area Δsu

and the direction au follow the right hand rule.

Vector Analysis 145

)( Aau

)(1

0lim)()(

uC

u

uuu dlAs

sAaA

Vector Analysis 146


• Let us find the three components of in

Cartesian coordinates. Differential rectangular

area parallel to the yz-plane and having sides Δy

and Δz is drawn about a typical point P(xo, yo,

zo). We have au=ax and

Δsu = ΔyΔz, and the

contour Cu consist of the

four sides 1,2,3, and 4.

Thus:

Vector Analysis 146

A

Vector Analysis 147


Vector Analysis 147

).(1

0lim)(4,3,2,1

sidesdlA

zyzyA

•In Cartesian coordinates A=axAx+ayAy+azAz. The

contribution of the four sides to the line integral are

as follows:

TOHy

AyzyxAz

yyxA

zy

yxA

zzy

yxAdlAzadl

Side

zyx

zzz

z

zz

..2

),,(),2

,(

:seriesTaylor a as expanded becan ),2

,( where

,),2

,(,

1_

),,(

Vector Analysis 148


• Where H.O.T (higher order terms) contain the

factors (Δy)2, (Δy)3, etc. Thus:

Vector Analysis 148

).(..2

),,(

..2

),,(),2

,(

:,),2

,(,

3_

...2

),,(

3_),,(

),,(

1_),,(

zTOHy

AyzyxAdlA

TOHy

AyzyxAz

yyxA

wherezzy

yxAdlAzadl

Side

zTOHy

AyzyxAdlA

sidezyx

zz

zyx

zzz

zz

sidezyx

zz

Vector Analysis 149


• Combining equations of side 1 and side 3 we

have:

Vector Analysis 149

.)..(

:shown that be

canit Similarly y. of powerscontain stillequation abovein H.O.T

.)..(

),,(4&2

),,(3&1

zyTOHz

AdlA

zyTOHy

AdlA

zyx

y

sides

zyx

z

sides

Vector Analysis 150


Vector Analysis 150

).()()(

A. of components-z and -y write tous enable and z and y, x,

inorder cyclic a reveal illequation w above ofn examinatio closeA

)(

:A ofcomponent

- xobtain the we0,y as zero to tendH.O.T that thenoting and

147 page of topon theequation in the equations above ngSubstituti

y

A

x

Aa

x

A

z

Aa

z

A

y

AaA

z

A

y

AA

xy

zzx

y

yzx

yzx

Vector Analysis 151


• can be remembered easily by arranging it

in the determinantal form in the manner of the

cross product.

A

zyx

zyx

AAA

zyx

aaa

A

Vector Analysis 152

Curl of a Vector Field• The expression for in general orthogonal

curvilinear coordinates (u1, u2, u3) is as below:A

332211

321

332211

321

1

AhAhAh

uuu

hahaha

hhhA

uuu

• The expression of cylindrical and sphericalcoordinates can be easily obtained from aboveequation by using the appropriate u1, u2, and u3

and their metric coefficients h1, h2, and h3.

A

Vector Analysis 153


• Example: Show that = 0 if

a) A = aØ(k/r) in cylindrical coordinates.

b) A= aRf(R) in spherical coordinates, where

f(R) is any function of the radial distance R.

• Solution:

a) In cylindrical coordinates the following apply:

(u1, u2, u3) = (r, Ø, z); h1 = 1, h2 = r, h3 = 1.

We have:

A

Vector Analysis 154


.0

00

1

A,given for the yieldswhich

,1

kzr

araa

rA

ArAAzr

araa

rA

zr

zr

zr

Vector Analysis 155


a) In spherical coordinates the following apply:

(u1, u2, u3) = (R, θ, Ø); h1 = 1, h2 = R, h3 = R

sinθ. Hence:

,

sin

sin

sin

12

ARRAA

R

RaRaa

RA

R

R

Vector Analysis 156


• And, for the given A,

• A curl-free vector field is called an Irrotational or

a Conservative field.

0

00)(

sin

sin

12

RfR

RaRaa

RA

R

Vector Analysis 157

Stokes’s Theorem

• For a very small differential area Δsj bounded bya contour Cj, the definition of in aboveequation leads to:

• For an arbitrary surface S, we can subdivide itinto many, say N, small differential areas. Figureon next page shows such a scheme with Δsj as atypical differential element

.1

0lim curlmax

Cn dlAa

ssAA

A

jC

jj dlAsA )()(

Stokes’s Theorem

• Left side of above equation is the flux of the

vector through the area Δsj. Adding the

contributions of all differential areas to the flux,

we have:

Vector Analysis 158

A

S

N

j

jjj

dsA

sAs

)(

)()(0lim1

Stokes’s Theorem• Now we sum up the line integrals around the

contours of all the differential elements

represented by the right side of equation on

page 157.

• Since the common parts of the contours of two

adjacent elements is traversed in opposite

directions by two contours, the net contribution

of all the common parts in the interior to the total

line integral is zero, and only the contribution

from the external contour C bounding the entire

area S remains after the summation:

Vector Analysis 159

Stokes’s Theorem

• The Stokes’s theorem states that the surface

integral of the curl of a vector field over an open

surface is equal to the closed line integral of the

vector along the contour bounding the surface.

Vector Analysis 160

CS

N

jC C

j

dlAdsA

dlAdlAsj

)(

theorem.sStokes'obtain weequations, twoprevious Combining

.)(0lim1

Stokes’s Theorem• As with the divergence theorem, the validity of the

limiting processes leading to Stokes’s theorem

requires that the vector field A, as well as its first

derivatives, exist and be continuous both on S and

along C.

• Stokes’s theorem converts a surface integral of the

curl of a vector to a line integral of the vector and

vice versa.

• Like the divergence theorem, Stokes’s theorem is

an important identity in vector analysis, and we use

it frequently in estabilishing other theorems and

relations in electromagnetics.Vector Analysis 161

Stokes’s Theorem

• If the surface integral of is carried over a

closed surface, there will be no surface

bounding external contour, and previous

equation tells us that:

• The geometry in figure on page 158 is chosen

deliberately to emphasize the fact that a

nontrivial application of Stokes’s theorem always

implies an open surface with a rim.Vector Analysis 162

A

S

dsA S. surface closedany for ,0)(

Stokes’s Theorem

• Example: Given F=axxy-ay2x, verify Stokes’s

theorem over a quarter circular disk with a

radius 3 in the first quadrant as shown in figure.

• Solution: Let us first find the

surface integral of

Vector Analysis 163

F

),2(

02

xa

xxy

zyx

aaa

F z

zyx

Stokes’s Theorem

• Therefore

Vector Analysis 164

.2

19

62

9

3sin99

)9(2

192

)2(

)()()(

3

0

312

3

0

22

3

0

9

0

3

0

9

0

2

2

yy

yyy

dyyy

dydxx

dxdyaFdsF

y

S

y

z

Stokes’s Theorem

• For the line integral around ABOA, we have

already evaluated the part around the arc from A

to B in example on page 86-89.

• Hence Stokes’s Theorem is verified.

Vector Analysis 165

,2

19

89-86 pageon exampleper as

.0)( and 0,y :A toO From

.02)( and 0, x:O toB From

ABOA

B

A

x

y

dlFdlF

Hence

xydxdxaFdlF

xdydyaFdlF

Two Null Identities• Identity 1:

“The curl of the gradient of any scalar field is

identically zero.”

• As per Stokes’s theorem:

• The combination of above two equations states

that the surface integral of over any

surface is zero.Vector Analysis 166

0 V

0

104 page of topon theequation per asHowever

CC

CS

dVdlV

dlVdsV

V

Two Null Identities

• A converse statement of Identity 1 can be made

as follows:

“If a vector field is curl-free, then it can be

expressed as the gradient of a scalar field.”

• Let a vector field be E. Then, if , we can

define a scalar field V such that:

• The –ve sign is unimportant as far as Identity 1

is concerned, as it is used in a future concept.

Vector Analysis 167

0 E

VE

Two Null Identities• Identity II

“ The divergence of the curl of any vector field is

identically zero.”

• Taking volume integral of above equation on the

left side and applying divergence theorem:

• Let us choose the arbitrary volume V enclosed

by a surface S in figure on next page. The

closed surface S can be split into two open

surfaces S1 and S2 connected by a common

boundary that has been drawn twice as C1 and

C2.Vector Analysis 168

0 A

SV

dsAdvA

Two Null Identities

Vector Analysis 169

• We than apply Stokes’s

theorem to surface S1

bounded by C1 and

surface S2 bounded by

C2, and we write the

right side of above

equation as:

21

2121

CC

nS

nSS

dlAdlA

dsaAdsaAdsA

• The normals an1 and an2 to surfaces S1 and S2

are outward normals and their relation with the

Two Null Identities

path directions of C1 and C2 follow the right hand

rule.

• As contours C1 and C2 are one and the same

common boundary between S1 and S2, the two

line integrals on the right side of above equation

traverse the same path in opposite direction.

Their sum is therefore zero, and the volume

integral of on the left side of equation

on slide 168 vanishes.

• As this is true for any arbitrary volume , the

integrand itself must be zero, as indicated by the

Identity II.Vector Analysis 170

A

Two Null Identities• A converse statement of Identity II is:

• “If a vector field is divergence-less, then it can

be expressed as the curl of another vector field.”

• Let the vector field be we can define a

vector field A such that:

• A divergence-less field is also called a

solenoidal field. Solenoidal fields are not

associated with flow sources or sinks.

• The net outward flux of a solenoidal field through

any closed surface is zero, and the flux lines

close upon themselves.Vector Analysis 171

,0 B

.AB

Helmholtz’s Theorem

• We may classify vector fields in accordance with

their being solenoidal and / or irrotational.

1. Solenoidal and irrotational if:

e.g: A static electric field in a charge free region.

2. Solenoidal but not irrotational if:

e.g: A steady magnetic field in a current carrying

conductor.

Vector Analysis 172

.0 and 0 FF

.0 and 0 FF

Helmholtz’s Theorem3. Irrotational but not solenoidal if:

e.g: A static electric field in a charged region.

4. Neither solenoidal nor irrotational if:

e.g: An electric field in a charged medium with a

time varying magnetic field.

• The most general vector field then has both a

nonzero divergence and a nonzero curl, and

can be considered as the sum of a solenoidal

field and an irrotational field.

Vector Analysis 173

.0 and 0 FF

.0 and 0 FF

Helmholtz’s Theorem• Helmholtz’s theorem states that:

“A vector field (vector point function) is determined

to within an additive constant if both its divergence

and its curl are specified everywhere.”

• In an unbounded region we assume that both the

divergence and the curl of the vector field vanish at

infinity.

• If a vector field is confined within a region bounded

by a surface, then it is determined if its divergence

and curl throughout the region, as well as the

normal component of the vector over the bounding

surface are given.

Vector Analysis 174


• Here we assume that the vector function is

single-valued and that its derivatives are finite

and continuous.

• We remind that the divergence of a vector is a

measure of the strength of the flow source and

that the curl of a vector is a measure of the

strength of the vortex source.

• When the strength of both the flow source and

vortex source are specified, we expect that the

vector field will be determined.

Vector Analysis 175


• We can decompose a general vector field F into

an irrotational part Fi and a solenoidal part Fs:

Vector Analysis 176

.

:have Weknown. be toassumed areG and g Where

0

0

,

GFF

gFF

GF

F

gF

F

FFF

s

i

s

s

i

i

si


• Helmholtz’s theorem asserts that g and G are

specified, the vector function F is determined.

• The fact that Fi is irrotational enables us to

define a scalar (potential) function V, in view of

Identity-I discussed earlier.

• Similarly Identity-II and equations on previous

page allow the definition of a vector (potential)

function A such that:

Vector Analysis 177

VFi

.AFs


• Hence according to Helmholtz’s theorem that a

general vector function F can be written as the

sum of the gradient of a scalar function and the

curl of a vector function.

Vector Analysis 178

.AVF


• Example: Given a vector function:

a) Determine the constants c1, c2, and c3 if F is

irrotational.

b) Determine the scalar potential function V whose –ve

gradient equals F.

Vector Analysis 179

.23 321 zycazxcazcyaF zyx


• Solution:

a) For F to be irrotational that is:

Vector Analysis 180

;0 F

.2c ,3c ,0c

:Hence h.must vanis F ofcomponent Each

.0)3()2(

23

321

213

321

cacaca

zyczxczcyzyx

aaa

F

zyx

zyx

Helmholtz’s Theoremb) Since F is irrotational, it can be expressed as the

negative gradient of a scalar function V; that is:

Vector Analysis 181

zyz

V

zxy

V

yx

V

zyazxaya

z

Va

y

Va

x

VaVF

zyx

zyx

2

,23

,3

:obtained are equations Three

.2233

Helmholtz’s Theorem• Integrating above three equations with respect

to x, y, and z respectively; we get:

• Examination of above three equations enable us

to write the scalar potential function as:

• Addition of any constant would still make V an

answer.

Vector Analysis 182

).,(2

2

,,23

,,3

3

2

2

1

yxfz

yzV

zxfyzxyV

zyfxyV

223

2zyzxyV

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