chapter 2. vector analysis - skkuicc.skku.ac.kr/~yeonlee/electromagnetics/cheng_2.pdf · a point...
TRANSCRIPT
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= + +
Cheng; 3/4/2007; 2-5
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= +
Cheng; 3/4/2007; 2-6
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=
Cheng; 3/4/2007; 2-7
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=
Cheng; 3/4/2007; 2-9
• 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
Cheng; 3/4/2007; 2-10
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)
Cheng; 3/4/2007; 2-13
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