Cheng; 3/4/2007; 2-1

Chapter 2. Vector Analysis 2.1 Overview At a given position and time a scalar function → a magnitude, a vector function → a magnitude and a direction Function conversion between different coordinates Physical laws should be independent of the coordinates. Coordinate system is chosen by convenience Three main topics (1) Vector algebra : addition, subtraction, multiplication (2) Orthogonal coordinate system : Cartesian, Cylindrical, Spherical (3) Vector calculus : differentiation, integration(gradient, divergence, curl) 2.2 Vector Addition and Subtraction A vector has a magnitude and a direction A AaA= ↑ ↑ magnitude, A

unit vector, AA

Graphical representation

Two vectors are equal if they have the same magnitude and direction, even though they may be displaced in space. • Vector addition, C A B= + Two vectors, A B and , form a plane Parallelogram rule : C is the diagonal of the parallelogram Head-To-Tail rule : The head of A touches the tail of B . C is drawn from the tail of A to the head of B .

Cheng; 3/4/2007; 2-2

Note C A B B A= + = + Commutative law ( ) ( )A B F A B F+ + = + + Associative law • Vector subtraction A B A B− = + −e j , where ˆ( )BB a B− = −

2.3 Vector Multiplication Multiplication of a vector by a positive scalar kA kA aA= b g A. Scalar or Dot Product A B AB AB• ≡ cosθ

Note that (1) A B AB• ≤ (2) A B A B• ≤ • ≥0 0 or (3) A B A• = × Projection of B onto A = B × Projection of A onto B (4) A B• = 0 when A B and are perpendicular to each other Note that

A A A• = 2 → A A A= • A B B A• = • : Commutative law ( )A B F A B A F• + = • + • : Distributive law

Cheng; 3/4/2007; 2-3

Example Law of Cosine Use vectors to prove C A B AB2 2 2 2= + − cosα

C B A= −

( ) ( )2

2 2 2 cos

C C C B A B A

B B A A A B B AA B AB α

= • ⇒ − • −

⇒ • + • − • − •

⇒ + −

B. Vector or Cross Product A B a ABn AB× ≡ sin θ ↑ unit vector normal to A B and (the right hand rule)

A B× = Area of the parallelogram formed by A B and

Note that A B B A× = − × NOT Commutative ( ) ( )A B F A B F× × ≠ × × NOT Associative

( )A B F A B A F× + = × + × Distributive law C. Product of Three Vectors Scalar triple product A B C A a BCn• × = •e j e j sin α

↑ ↑ Area of parallelogram Height of the parallelepiped → Volume of the parallelepiped

Vector triple product ( ) ( ) ( )A B C B A C C A B× × = • − • back-cab rule

Cheng; 3/4/2007; 2-4

2.4 Orthogonal Coordinate Systems A point in space is represented by three surfaces, u1 = const., u2 = const. and u3 = const. If they are mutually perpendicular, they form an orthogonal coordinate system. Unit vectors along 1 2 3, ,u u u → Base vectors, 1 2 3ˆ ˆ ˆ, , u u ua a a

A vector in ( )1 2 3, ,u u u coordinate system → = + +1 1 2 2 3 3ˆ ˆ ˆu u u u u uA A a A a A a Differential change idu → Differential length change =i i idl h du

↑ metric coefficient ih Vector differential length ( ) ( ) ( )= + +1 1 1 2 2 2 3 3 3ˆ ˆ ˆu u udl a h du a h du a h du Differential volume ( ) ( ) ( )= 1 1 2 2 3 3dv h du X h du X h du Vector differential area = ˆnds ds a , ˆna : surface normal Note =1 1 2 3 2 3 1ˆ ˆu uds a h h du du a =2 2 1 3 1 3 2ˆ ˆu uds a h h du du a =3 3 1 2 1 2 3ˆ ˆu uds a h h du du a A. Cartesian Coordinate System u u u x y z1 2 3, , , ,b g b g=

A point x y z1 1 1, , b g : The intersection of three planes x x= 1 , y y= 1, z z= 1 Base vectors : , ,a a ax y z

Properties of the base vectors a a ax y z× = ,

a a ay z x× = ,

a a az x y× =

a a a a a ax y y z x z• = • = • = 0

a a a a a ax x y y z z• = • = • = 1

The position vector to the point P x y z1 1 1, ,b g → OP x a y a z ax y z

→= + +1 1 1

A vector A in Cartesian coord. A A a A a A ax x y y z z= + +

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Vector differential length

dl dx a dy a dz ax y z

→= + +

Differential area

=

=

=

x

y

z

ds dydzds dxdzds dxdy

Differential volume =dv dxdydz The dot product

A B A a A a A a B a B a B a A B A B A Bx x y y z z x x y y z z x x y y z z• = + + • + + ⇒ + +d i d i

The cross product A B A a A a A a B a B a B ax x y y z z x x y y z z× = + + × + +d i d i

A Ba a aA A AB B B

A B A B a A B A B a A B A B ax y z

x y z

x y z

y z z y x z x x z y x y y x z× = ⇒ − + − + −d i b g d i

Example

A straight line 1L is given by 2 4x y+ = . (a) Find the unit normal to 1L starting from the origin (b) Fine the normal line to 1L passing through P(0,2) Solution: (a) A vector along 1L : ˆ ˆ2 4x y− + (1) A vector from the origin to a point Q on 1L : ˆ ˆ(4 2 )xx x y+ − (2) From (1) and (2) ( ) ( )ˆ ˆ ˆ ˆ2 4 (4 2 ) 0x y xx x y− + + − =i → Q(1.6, 0.8)

The unit normal is ( )1 ˆ ˆ25

x y+

(b) The straight line parallel to the unit normal is 12

y x c= +

This line should pass through P(0,2) → 1 22

y x= +

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B. Cylindrical coordinates u u u r z1 2 3, , , ,b g b g= φ A point P r z1 1 1, ,φb g : a cylindrical surface with radius of r1, a half-plane rotated by φ1 from x-axis, a plane cutting z-axis at z1 . Three base vectors : , ,a a ar z φ

↑ ↑ directions change with P a a ar z× =φ ,

a a az rφ × = ,

a a az r× = φ

Differential length

dl dr a rd a dz ar z= + + φ φ

↑ Differential areas

r

z

ds rd dzds drdzds rdrd

φ

φ

φ

==

=

Differential volume dv rdrd dzφ=

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A vector in cylindrical coords. A A a A a A ar r z z= + +φ φ

Transformation of A into Cartesian coord. A A a A a a A a a A a a A Ax x r r x x z z x r= • = • + • + • ⇒ −cos sinφ φ φφ φ

↑ ↑ ↑ =0 = cosφ = −sin φ

Similarly using sina ar y• = φ , cosa ayφ φ• =

A A Ay r= +sin cosφ φφ

In matrix form

AAA

AAA

x

y

z

r

z

L

NMMM

O

QPPP=

−L

NMMM

O

QPPP

L

NMMM

O

QPPP

cos sinsin cos

φ φφ φ φ

00

0 0 1

A point in cylindrical coord is transformed into Cartesian coord.

x ry rz z

===

cossin

φφ

Conversion from Cartesian to cylindrical coords.

2 2

1tan

r x yyx

z z

φ −

= +

=

=

Exercise Convert a vector in Cartesian coords., OQ a a ax y z

→= + +3 4 5 , to cylindrical coords.

OQ a a a a a a a a a a a a a a ax y z r r x y z x y z z z

→= + + • + + + • + + + •3 4 5 3 4 5 3 4 5d i d i d iφ φ

↑ ↑ ↑ 3 4cos sinφ φ+ 3 4− +sin cosφ φb g 5 From Cartesian Coords. we know that

cosφ =35

and sin φ =45

Therefore in cylindrical Coords.

OQ a ar z

→= +5 5

Can you convert this vector back to Cartesian Coords.?

Cheng; 3/4/2007; 2-8

C. Spherical Coordinates u u u R1 2 3, , , ,b g b g= θ φ A point P R1 1 1, ,θ φb g : a sphere with radius 1R a cone with a half-angle of 1θ a half-plane rotated by 1φ from x-axis

aR

aθ

aφ

φ1

θ1

x

y

zplanecone

sphereR R= 1

φ φ= 1θ θ= 1

P

Three base vectors a a aR × =θ φ ,

a a aRθ φ× = ,

a a aRφ θ× =

A vector in spherical coord A A a A a A aR R= + +θ θ φ φ

Vector differential length

dl dR a Rd a R d aR= + + sinθ θ φθ φ

↑ ↑ Differential areas

2 sinsin

Rds R d dds R dRdds RdRd

θ

φ

θ θ φθ φθ

===

Differential volume 2 sindv R dRd dθ θ φ=

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• Conversion of a point

x Ry Rz R

===

sin cossin sincos

θ φθ φθ

Conversion from Cartesian to spherical coords.

2 2 2

1 2 2

1

tan /

tan

R x y z

x y z

yx

θ

φ

−

−

= + +

⎡ ⎤= +⎣ ⎦

=

Integrals Containing Vector Functions

V

C

C

S

Fdv

Vdl

F dl

A ds

∫∫∫∫

i

i

→

( )( )

ˆ ˆ ˆ

ˆ ˆ ˆ

x y zV

C

C

S

F x F y F z dv

V xdx ydy zdz

F dl

A ds

+ +

+ +

∫∫∫∫

i

i

Sign convention for the surface integral

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Cheng; 3/4/2007; 2-11

2.5 Gradient of a Scalar Field

A scalar field V t u u u, , , 1 2 3b g

Move from 1V V= surface to 1V V dV= + surface P P1 2→ → dn

P P1 3→ → dl

↑ ↑ same dV different distances

The directional derivative

dVdl

: The space rate of change of V along l

The gradient of a scalar function V is defined as

grad V V dVdn

an ≡ ∇ ≡ : The maximum directional derivative of V

(an is perpendicular to V=const. plane) • The directional derivative

dVdl

dVdn

dndl

dVdn

dVdn

a a V an l l= ⇒ ⇒ • ⇒ ∇ •cosα b g : projection of gradV onto al direction

→ dV V dl= ∇ •b g , (1) In Cartesian coord.

dV Vx

dx Vy

dy Vz

dz Vx

a Vy

a Vz

a dx a dy a dz ax y z x y z= + + ⇒ + +FHG

IKJ • + +

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

d i ↑ (2) = dl From (1) and (2)

∇ = + + ⇒ + +FHG

IKJV V

xa V

ya V

za

xa

ya

za Vx y z x y z

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

↑ ≡ ∇ , “del” operator In u u u1 2 3, , b g coordinates

∇ = + +1 1 1

1 1 2 2 3 31 2 3h u

ah u

ah u

au u u∂

∂∂

∂∂

∂

Cheng; 3/4/2007; 2-12

2.6 Divergence of a Vector Field Flux is defined as a quantity per unit area per unit time A vector field is represented by flux lines Its direction → Direction of the flux line Its magnitude → Density of the flux lines

• Divergence of A The net outward flux of A per unit volume

divAA dS

VV≡

•

→

zlimΔ Δ0

S : dS dS an=

Find divA at P x y zo o o, ,b g For a small volume Δ Δ Δx y z

A dS A dS A dS A dS A dS A dS A dSS front back right left top botom

• = • + • + • + • + • + •z z z z z z z

(1) On the front surface

A dS A S A x x y z y zfront

front front x o o o• = • ⇒ +z Δ Δ Δ Δ( , , )12

↑ ↑

Δ Δy z axb g , A x y z x Axx o o ox

x y zo o o

( , , ), ,

+Δ2

∂∂ b g

, Taylor series

(2) On the back surface

A dS A S A y za A x x y z y zback

back back back x x o o o• = • ⇒ • − ⇒ − −z Δ Δ Δ Δ Δ Δ( , , )b g 12

↑

A x y z x Axx o o ox

x y zo o o

( , , ), ,

−Δ2

∂∂ b g

Combining

A dS A dS Ax

x y zfront back

x

x y zo o o

• + • =z z ∂∂ , ,b g

Δ Δ Δ (1)

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Similarly, combining contributions from the right and left surfaces

A dS A dSAy

x y zright left

y

x y zo o o

• + • =z z ∂

∂, ,b g

Δ Δ Δ (2)

Similarly, combining contributions from the top and bottom surfaces

A dS A dS Az

x y ztop bottom

z

x y zo o o

• + • =z z ∂∂ , ,b g

Δ Δ Δ (3)

From (1), (2) and (3)

A dS Ax

Ay

Az

x y zS

x y z

x y zo o o

• = + +FHG

IKJz ∂

∂

∂

∂∂∂

, ,b gΔ Δ Δ

→ divA Ax

Ay

Az

x y z= + +∂∂

∂

∂∂∂

We define ∇ • ≡A divA

In general

∇ • = + +LNM

OQP

Ah h h u

h h Au

h h Au

h h A1

1 2 3 12 3 1

21 3 2

31 2 3

∂∂

∂∂

∂∂

b g b g b g

2.7 Divergence Theorem The volume integral of the divergence of a vector field equals the total outward flux of the vector through the surface

∇ • = •z zAdV A dSV S

Proof: Consider a differential volume ΔV j bounded by S j

From the definition of divA ∇ • = •zA V A dS

j j s je j Δ

Add all ΔV j ’s

lim limΔ Δ

ΔV j j

j

N

V sj

N

Sj j jA V A dS A dS

→=

→=

∇ •L

NMM

O

QPP = •

L

NMM

O

QPP⇒ •∑ z∑ z0 1 0 1

e j

↑ ↑ ↑ Integral at the external surface ∇ •z A dv

V e j An internal surface is shared by two adjacent volumes.

(Opposite surface normals) • ∇ • ≠A 0 means the existence of flow sources in the given volume

Cheng; 3/4/2007; 2-14

2.8 Curl of a Vector Field Divergence measures flow source Curl measures vortex source The circulation of a vector field around a contour C A dl

C•z

The definition of curl : The maximum circulation of A per unit area with the direction normal to the loop area (Right hand rule)

curlA As

A dl as C n= ∇ × ≡ •LNM

OQP→ zlimmaxΔ Δ0

1

The component of curl A in the direction of au

∇ × = • ∇ × = •→ zA a A

sA dl

u u s u cu ue j e j lim

Δ Δ0

1

The x-component of ∇ × A

∇ × = •LNMOQP→ zA

y zA dl

x y z sidese j lim

, , , ,Δ Δ Δ Δ0 1 2 3 4

1

At side 1:

A dl A x y z y Ay

zside

z o o oz

x y zo o o

• = + +L

NMM

O

QPPz 1 2

, , ..., ,

b gb g

ΔΔ

∂∂

At side 3 :

A dl A x y z y Ay

zside

z o o oz

x y zo o o

• = − +L

NMM

O

QPP −z 3 2

, , ..., ,

b g b gb g

ΔΔ

∂∂

Combine the two

A dl Ay

y zside

z

x y zo o o

• = +L

NMM

O

QPPz 1 3, , ,

...∂∂ b g

Δ Δ , (1)

Similarly, combining contributions from sides 2 and 4

A dlAz

y zsides

y

x y zo o o

• = − +L

NMM

O

QPPz 2 4, , ,

...∂

∂ b gΔ Δ (2)

From (1) and (2)

∇ × = •LNMM

OQPP⇒ −

→ zAy z

A dl Ay

Azx y z sides

z ye j lim, , , ,Δ Δ Δ Δ0 1 2 3 4

1 ∂∂

∂

∂ : Note the cyclic order in x, y, and z

Cheng; 3/4/2007; 2-15

In Cartesian coordinates

∇ × = −FHG

IKJ + −FHG

IKJ + −FHG

IKJA A

yAz

a Az

Ax

aAx

Ay

az yx

x zy

y xz

∂∂

∂

∂∂∂

∂∂

∂

∂∂∂

∇ × =A

a a a

x y zA A A

x y z

x y z

∂∂

∂∂

∂∂

In general

∇ × =Ah h h

a h a h a h

u u uh A h A h A

u u u

x y z

1

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

∂∂

∂∂

∂∂

If ∇ × =A 0 , A is called an irrotational field, or a conservative field

2.9 Stokes’s Theorem Consider a differential area ΔS j bounded by a contour C j

From the definition of ∇ × A

∇ × • = •zA S a A dlj j j

C je j d iΔ

Adding all the contributions The left side:

limΔ

ΔS j j j

j

N

Sj

A S a A dS→

=

∇ × • ⇒ ∇ × •∑ z01e j d i e j

The right side:

limΔS C

j

N

Cj j

A dl A dl→

=

•FHG

IKJ ⇒ •z∑ z0

1

↑ External contour C bounding S

Stokes’s theorem is defined

∇ × • = •z zA dS A dlS Ce j : Right-hand rule for dS dl and

→ The surface integral of the curl of a vector over a surface is equal to the closed line integral of the vector along the bounding contour.

Note ∇ × • =z A dS

Se j 0

Cheng; 3/4/2007; 2-16

2.10 Two Null Identities Identity I The curl of the gradient of a scalar field is always zero ∇ × ∇ =Vb g 0 Proof: (1) Direct operations of the gradient and curl (2) Using Stokes’s theorem ∇ × ∇ • = ∇ • ⇒ ⇒z z zV dS V dl dV

S C Cb g 0

• The converse statement If a vector field is curl-free, then it can be expressed as the gradient of a scalar field ∇ × =E 0 → E V= −∇

Identity 2 The divergence of the curl of any vector field is identically zero

∇ • ∇ × =Ae j 0

Using divergence theorem ∇ • ∇ × = ∇ × •z zA dV A dS

V Se j e j

Split the surface, S S S= +1 2 and apply Stokes’s theorem The right side ∇ × • ⇒ ∇ × • + ∇ × • ⇒z z zA dS A a dS A a dS

Sn

Sn

Se j e j e j1 2

1 2

0

↑ ↑ A dl

C•z

1

A dlC

•z2

• The converse statement If a vector field is divergenceless, then it can be expressed as the curl of another vector field ∇ • =B 0 → B A= ∇ × 2.11 Helmholtz’s Theorem 1. A static electric field in a charge-free region ∇ • = ∇ × =E E0 0, and : solenoidal, irrotational 2. A steady magnetic field in a current-carrying conductor ∇ • = ∇ × ≠H H0 0, and : solenoidal, not irrotational 3. A static electric field in a charged region ∇ • ≠ ∇ × =E E0 0, and : not solenoidal, irrotational 4. An electric field in a charge medium with a time-varying magnetic field ∇ • ≠ ∇ × ≠E E0 0, and : not solenoidal, not irrotational Helmholtz’s Theorem A vector field is determined if both its divergence and its curl are specified everywhere