Lecture 2
Vector Analysis
Jang, Min Seok
EE204 Fall 2021
Maxwell’s Equations
Differential form
Integral form
𝛻 ⋅ 𝐃 = 𝜌𝑓
ර 𝐃 ⋅ 𝑑𝐒 = න 𝜌𝑓 𝑑𝑣
𝑆
𝛻⋅𝐁=0
𝑣
ර 𝐁 ⋅ 𝑑𝐒 = 0
𝑆
𝜕𝐁
𝛻×𝐄=−
𝜕𝑡
𝜕
ර 𝐄 ⋅ 𝑑𝐥 = − න 𝐁 ⋅ 𝑑𝐒
𝜕𝑡 𝑆
𝐿
𝜕𝐃
𝛻 × 𝐇 = 𝐉𝑓 +
𝜕𝑡
𝜕𝐃
ර 𝐇 ⋅ 𝑑𝐥 = න 𝐉𝑓 +
⋅ 𝑑𝐒
𝜕𝑡
𝐿
𝑆
𝐃 = electric flux density
𝐄 = electric field
𝜌𝑓 = free charge density
𝐁 = magnetic flux density
𝐇 = magnetic field
𝐉𝑓 = free current density
Vector analysis is a mathematical language for EM theory
Jang, Min Seok | EE204 Electromagnetics I
2
Mathematical Concepts
• Scalar and vector
• Vector algebra: addition, subtraction, and multiplications
• Vector differential operations
(gradient, divergence, curl, and Laplacian)
• Line, surface and volume integrals
• Divergence theorem and Stokes’ theorem
• Coordinate systems (Cartesian, cylindrical, spherical) and
coordinate transformations
Jang, Min Seok | EE204 Electromagnetics I
3
Scalars and Vectors
• A scalar is a quantity that has only magnitude
(ex) length, area, volume, speed, mass, density, pressure,
temperature, energy, entropy, work, power
volume
• A vector is a quantity that has both magnitude and
direction
(ex) displacement, velocity, acceleration, momentum, force
velocity
Jang, Min Seok | EE204 Electromagnetics I
4
Fields
• A field is a physical quantity that has a value for each point in
space (and time)
Examples of scalar field:
Examples of vector field:
- temperature distribution
- charge density profile
- electric potential
- gravitational force
- wind flow distribution
- electric/magnetic fields
Jang, Min Seok | EE204 Electromagnetics I
5
Standard Basis
• Standard basis {𝐚𝑥 , 𝐚𝑦 , 𝐚𝑧 }: a set of unit vectors (i.e. 𝐚𝑖 = 1)
pointing in the 𝑥, 𝑦, 𝑧 axes of Cartesian coordinate system
• Every vector 𝐀 in 3D space can be represented uniquely as
𝐀 = 𝐴𝑥 𝐚𝑥 + 𝐴𝑦 𝐚𝑦 + 𝐴𝑧 𝐚𝑧
𝐴𝑥 , 𝐴𝑦 , 𝐴𝑧 : components of 𝐀 in the 𝑥, 𝑦, 𝑧 directions
Jang, Min Seok | EE204 Electromagnetics I
6
Vector Addition and Subtraction
• Addition:
𝐂=𝐀+𝐁
= 𝐴𝑥 + 𝐵𝑥 𝐚𝑥 + 𝐴𝑦 + 𝐵𝑦 𝐚𝑦 + 𝐴𝑧 + 𝐵𝑧 𝐚𝑧
• Subtraction: 𝐃 = 𝐀 − 𝐁 = 𝐀 + (−𝐁)
= 𝐴𝑥 − 𝐵𝑥 𝐚𝑥 + 𝐴𝑦 − 𝐵𝑦 𝐚𝑦 + 𝐴𝑧 − 𝐵𝑧 𝐚𝑧
Jang, Min Seok | EE204 Electromagnetics I
7
Vector Addition and Subtraction
• Basic algebra:
Addition
Multiplication
𝐀+𝐁=𝐁+𝐀
𝑘𝐀 = 𝐀𝑘
associative
𝐀+ 𝐁+𝐂 = 𝐀+𝐁 +𝐂
𝑘 𝑙𝐀 = 𝑘𝑙 𝐀
distributive
𝑘 𝐀 + 𝐁 = 𝑘𝐀 + 𝑘𝐁
commutative
𝐀, 𝐁, 𝐂: vectors, 𝑘, 𝑙: scalars
Jang, Min Seok | EE204 Electromagnetics I
8
Vector Addition and Subtraction
• Exercise: Given points 𝑄(2,4,6) and 𝑅 0,3,8 , find
(a) the position vectors of 𝑄 (𝐫𝑄 ), and 𝑅(𝐫𝑅 ).
(b) 𝐫𝑄 + 𝐫𝑅
(c) the displacement vector 𝐫𝑄𝑅
(d) the distance between 𝑄 and 𝑅
Answer:
(a) 𝐫𝑄 = 𝑂𝑄 = 2𝐚𝑥 + 4𝐚𝑦 + 6𝐚𝑧 , 𝐫𝑅 = 𝑂𝑅 = 3𝐚𝑦 + 8𝐚𝑧
(b) 𝐫𝑄 + 𝐫𝑅 = 2 + 0 𝐚𝑥 + 4 + 3 𝐚𝑦 + 6 + 8 𝐚𝑧
= 2𝐚𝑥 + 7𝐚𝑦 + 14𝐚𝑧
(c) 𝐫𝑄𝑅 = 𝑄𝑅 = 𝐫𝑅 − 𝐫𝑄 = 0 − 2 𝐚𝑥 + 3 − 4 𝐚𝑦 + 8 − 6 𝐚𝑧
= −2𝐚𝑥 − 𝐚𝑦 + 2𝐚𝑧
(d) 𝐫𝑄𝑅 =
−2
2
+ −1
2
+ 2
2
=3
Jang, Min Seok | EE204 Electromagnetics I
9
Vector Multiplication: Scalar Product
• Scalar product (dot product):
𝐀 ⋅ 𝐁 = 𝐴𝐵 cos 𝜃𝐴𝐵 = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧
𝐴 = 𝐀 , 𝐵 = 𝐁 , 𝜃𝐴𝐵 = smaller angle between 𝐀 and 𝐁.
• Scalar projection (or component) of 𝐀 in the direction of 𝐁:
𝐁
𝐀 ⋅ 𝐚𝐵 = 𝐀 ⋅
= 𝐴 cos 𝜃𝐴𝐵
𝐵
• Basic algebra:
commutative
distributive
𝐀⋅𝐁=𝐁⋅𝐀
𝐀⋅ 𝐁+𝐂 =𝐀⋅𝐁+𝐁⋅𝐂
Jang, Min Seok | EE204 Electromagnetics I
10
Vector Multiplication: Scalar Product
• Usage in Physics: work(𝑊), force(𝐅), and displacement (𝐒)
𝑊 =𝐅⋅𝐒
𝑊=0
• Exercise: application to the law of cosines:
𝑐=
𝑎2 + 𝑏 2 − 2𝑎𝑏 cos 𝜃
Answer: 𝑐 2 = 𝐜 ⋅ 𝐜 = 𝐚 − 𝐛 ⋅ 𝐚 − 𝐛
=𝐚⋅𝐚−𝐚⋅𝐛−𝐛⋅𝐚+𝐛⋅𝐛
= 𝑎2 − 𝐚 ⋅ 𝐛 − 𝐚 ⋅ 𝐛 + 𝑏 2
= 𝑎2 + 𝑏 2 − 2𝐚 ⋅ 𝐛
= 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos 𝜃
Jang, Min Seok | EE204 Electromagnetics I
11
Vector Multiplication: Vector Product
• Vector product (cross product):
𝐚𝑥
𝐀 × 𝐁 = 𝐴𝐵 sin 𝜃𝐴𝐵 𝐚𝑛 = 𝐴𝑥
𝐵𝑥
𝐚𝑦
𝐴𝑦
𝐵𝑦
𝐚𝑧
𝐴𝑧
𝐵𝑧
Direction: 𝐚𝑛 is a unit vector normal to both 𝐀 and 𝐁.
Magnitude: area of the parallelogram having 𝐀 and 𝐁 as sides.
• Basic algebra:
anticommutative
not associative
distributive
𝐀 × 𝐁 = −𝐁 × 𝐀
𝐀× 𝐁×𝐂 ≠ 𝐀×𝐁 ×𝐂
𝐀× 𝐁+𝐂 =𝐀×𝐁+𝐁×𝐂
Jang, Min Seok | EE204 Electromagnetics I
12
Vector Multiplication: Vector Product
• Usage in Physics: Lorentz force (𝐅) on a moving charge (𝑞) with
velocity (𝐯) under magnetic flux density (𝐁)
𝐅 = 𝑞𝐯 × 𝐁
Jang, Min Seok | EE204 Electromagnetics I
13
Vector Multiplication: Triple Products
• Scalar triple product:
𝐀 ⋅ 𝐁 × 𝐂 = 𝐁 ⋅ 𝐂 × 𝐀 = 𝐂 ⋅ (𝐀 × 𝐁)
• Geometrical interpretation of scalar triple product:
Volume of parallelepiped having 𝐀, 𝐁, and 𝐂 as edges
• Vector triple product (BAC-CAB rule):
𝐀 × 𝐁 × 𝐂 = 𝐁 𝐀 ⋅ 𝐂 − 𝐂(𝐀 ⋅ 𝐁)
Jang, Min Seok | EE204 Electromagnetics I
14
Vector Multiplication
• Exercise: Let 𝐏 = 2𝐚𝑥 − 𝐚𝑧 and 𝐐 = 2𝐚𝑥 − 𝐚𝑦 + 2𝐚𝑧 . Determine
(a) scalar projection of 𝐏 along 𝐐
(b) area of the triangle having 𝐏 and 𝐐 as sides
(c) unit vector perpendicular to both 𝐏 and 𝐐
Answer:
𝐐
2 ⋅ 2 + 0 ⋅ −1 + −1 ⋅ 2
2
(a) 𝐏 ⋅
=
=
2
2
2
𝐐
3
2 + −1 + 2
𝐚𝑥
(b) 𝐏 × 𝐐 = 2
2
Area = 𝐏 × 𝐐
𝐚𝑦 𝐚 𝑧
0 −1 = −𝐚𝑥 − 6𝐚𝑦 − 2𝐚𝑧 ,
−1 2
/2 = (−1)2 + (−6)2 + −2 2 /2 = 41/2
𝐏×𝐐
1
=±
(𝐚𝑥 + 6𝐚𝑦 + 2𝐚𝑧 )
(c) ±
𝐏×𝐐
41
Jang, Min Seok | EE204 Electromagnetics I
15
Ordinary Derivatives
• A function of one variable: 𝑓(𝑥)
𝑑𝑓
• Ordinary derivative:
𝑑𝑥
• Geometrical interpretation: slope of the graph of 𝑓 vs 𝑥
𝑓(𝑥)
𝑥
Jang, Min Seok | EE204 Electromagnetics I
16
Del Operator: ∇
• Del operator is the vector differential operator
𝜕
𝜕
𝜕
𝛻=
𝐚𝑥 +
𝐚𝑦 + 𝐚𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
• Three ways the operator 𝛻 can act:
1. On a scalar or vector field 𝑉: 𝛻𝑉 (gradient)
2. On a vector field 𝐀 via the dot product: 𝛻 ⋅ 𝐀 (divergence)
3. On a vector field 𝐀 via the cross product: 𝛻 × 𝐀 (curl)
Jang, Min Seok | EE204 Electromagnetics I
17
Gradient
• Motivation:
How to define slope of a multi-variable function?
𝑓(𝑥, 𝑦)
Jang, Min Seok | EE204 Electromagnetics I
18
Gradient
• A scalar field defined in three dimensional space: 𝑉(𝑥, 𝑦, 𝑧)
• Gradient:
𝜕𝑉
𝜕𝑉
𝜕𝑉
𝛻𝑉 =
𝐚𝑥 +
𝐚𝑦 +
𝐚𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
• Geometrical interpretation:
Direction: direction of the maximum increase of 𝑉
Magnitude: maximum slope (rate of increase) of 𝑉
Jang, Min Seok | EE204 Electromagnetics I
19
Gradient
• Usage in Physics: relation between a force field (𝐅) and a scalar
potential field (𝑈)
(ex) gravitational and electrostatic fields
𝐅 = −𝛻𝑈
Why minus sign? Because water flows downhill rather than uphill
Jang, Min Seok | EE204 Electromagnetics I
20
Gradient
• Exercise: Let 𝑉 = 𝑥 2 − 2𝑥 + 𝑦 2 . Find
(a) gradient of 𝑉
(b) maximum directional derivative at point 𝑃(2,1,2)
(c) directional derivative at 𝑃 toward the origin
Answer:
𝜕𝑉
𝜕𝑉
𝜕𝑉
(a) 𝛻𝑉 =
𝐚𝑥 +
𝐚𝑦 +
𝐚𝑧 = 2𝑥 − 2 𝐚𝑥 + 2𝑦 𝐚𝑦
𝜕𝑥
𝜕𝑦
𝜕𝑧
(b) 𝛻𝑉 =
2⋅2−2
2
+ 2⋅1
2
=2 2
(c) 𝑃𝑂 = −2𝐚𝑥 − 1𝐚𝑦 − 2𝐚𝑍
𝑃𝑂
−2, −1, −2
𝛻𝑉 ⋅
= 2,2,0 ⋅
= −2
3
𝑃𝑂
Jang, Min Seok | EE204 Electromagnetics I
21
Divergence and Curl
• Motivation:
How to describe generation and rotation of a vector field?
Jang, Min Seok | EE204 Electromagnetics I
22
Divergence
• A vector field defined in three dimensional space: 𝐀(𝑥, 𝑦, 𝑧)
• Cartesian coordinate:
𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧
𝛻⋅𝐀=
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
• Coordinate independent definition:
‫𝐒𝑑 ⋅ 𝐀 𝑆ׯ‬
𝛻 ⋅ 𝐀 = lim
Δ𝑣→0
Δ𝑣
Δ𝑣 is the small volume enclosed by the closed surface 𝑆 in which
the point of interest 𝑃 is located
Jang, Min Seok | EE204 Electromagnetics I
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Divergence (Optional)
Let 𝑃 = (𝑥0 , 𝑦0 , 𝑧0 )
ර 𝐀 ⋅ 𝑑𝐒 = න
+න
𝑆
front
න
𝐀 ⋅ 𝑑𝐒 ≈ 𝐴𝑥 ቚ
front
න
𝑥0 +
back
front
+න
+න +න
left
top
back
𝐀 ⋅ 𝑑𝐒 ≈ −𝐴𝑥 ቚ
න
+න +න
𝑑𝑦𝑑𝑧
𝑑𝑥
,𝑦 ,𝑧
2 0 0
𝑥0 −
𝑑𝑥
,𝑦 ,𝑧
2 0 0
𝐀 ⋅ 𝑑𝐒 ≈ 𝐴𝑥 ቚ
back
right
𝐀 ⋅ 𝑑𝐒
bottom
(𝑑𝐒 = 𝑑𝑦𝑑𝑧 𝐚𝑥 )
𝑑𝑦𝑑𝑧
𝑥0 +
𝑑𝑥
,𝑦 ,𝑧
2 0 0
− 𝐴𝑥 ቚ
𝑥0 −
𝑑𝑥
,𝑦 ,𝑧
2 0 0
𝜕𝐴𝑦
𝑑𝑥𝑑𝑦𝑑𝑧,
𝜕𝑦
𝑑𝑦𝑑𝑧 =
𝜕𝐴𝑥
𝑑𝑥𝑑𝑦𝑑𝑧
𝜕𝑥
𝜕𝐴𝑧
𝑑𝑥𝑑𝑦𝑑𝑧
𝜕𝑧
Similarly,
න +න
Therefore,
‫𝐒𝑑 ⋅ 𝐀 𝑆ׯ‬
1
𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧
𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧
=
+
+
𝑑𝑥𝑑𝑦𝑑𝑧 =
+
+
Δ𝑣→0
Δ𝑣
𝑑𝑥𝑑𝑦𝑑𝑧 𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
left
right
𝐀 ⋅ 𝑑𝐒 ≈
න +න
top
bottom
𝐀 ⋅ 𝑑𝐒 ≈
lim
Jang, Min Seok | EE204 Electromagnetics I
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Divergence
‫𝐒𝑑 ⋅ 𝐀 𝑆ׯ‬
𝛻 ⋅ 𝐀 = lim
Δ𝑣→0
Δ𝑣
• Geometrical interpretation:
𝛻 ⋅ 𝐀 at a point 𝑃 is a measure of how much the vector 𝐀
spreads out (diverges) from 𝑃
𝛻⋅𝐀>0
𝛻⋅𝐀<0
Jang, Min Seok | EE204 Electromagnetics I
𝛻 ⋅ 𝐀 = 0 (solenoidal)
25
Divergence
• Usage in Physics: relation between gravitational field (𝐠) and its
source (mass density, 𝜌𝑚 )
𝛻 ⋅ 𝐠 = −4𝜋𝐺𝜌𝑚
Similarly, electric field (𝐄) and its source (charge density, 𝜌)
𝜌
𝛻⋅𝐄=
𝜖0
𝜌>0
𝜌<0
Jang, Min Seok | EE204 Electromagnetics I
26
Curl
• A vector field defined in three dimensional space: 𝐀(𝑥, 𝑦, 𝑧)
• Cartesian coordinate:
𝐚𝑥
𝛻 × 𝐀 = 𝜕/𝜕𝑥
𝐴𝑥
𝐚𝑦
𝜕/𝜕𝑦
𝐴𝑦
𝐚𝑧
𝜕/𝜕𝑧
𝐴𝑧
• Coordinate independent definition:
‫𝐥𝑑 ⋅ 𝐀 𝐿ׯ‬
𝛻 × 𝐀 ⋅ 𝐚𝑛 = lim
Δ𝑆→0
Δ𝑆
Δ𝑆 is the small area bounded by the closed curve 𝐿 and 𝐚𝑛 is the
unit vector normal to Δ𝑆 determined by using the right-hand rule.
Jang, Min Seok | EE204 Electromagnetics I
27
Curl (Optional)
Let 𝑃 = (𝑥0 , 𝑦0 , 𝑧0 ) and 𝐚𝑛 = 𝐚𝑥 (i.e. 𝛥𝑆 on 𝑦𝑧 plane)
ර 𝐀 ⋅ 𝑑𝐥 = න + න + න + න
𝐿
𝑎𝑏
𝑏𝑐
න 𝐀 ⋅ 𝑑𝐥 ≈ 𝐴𝑦 ቚ
𝑎𝑏
𝑥0 ,𝑦0 ,𝑧0 −
න 𝐀 ⋅ 𝑑𝐥 ≈ −𝐴𝑦 ቚ
𝑐𝑑
න +න
𝑎𝑏
𝑐𝑑
𝑑𝑧
2
𝑥0 ,𝑦0 ,𝑧0 +
𝐀 ⋅ 𝑑𝐥 ≈ 𝐴𝑦 ቚ
𝑐𝑑
Similarly,
Therefore,
න +න
𝑏𝑐
𝑑𝑎
𝐀 ⋅ 𝑑𝐥
𝑑𝑎
𝑑𝑦
𝑑𝑧
2
(𝑑𝐥 = 𝑑𝑦 𝐚𝑦 )
𝑑𝑦
𝑥0 ,𝑦0 ,𝑧0 −
𝑑𝑧
2
𝐀 ⋅ 𝑑𝐥 ≈
− 𝐴𝑦 ቚ
𝑥0 ,𝑦0 ,𝑧0 +
𝑑𝑧
2
𝑑𝑦 = −
𝜕𝐴𝑦
𝑑𝑦𝑑𝑧
𝜕𝑧
𝜕𝐴𝑧
𝑑𝑦𝑑𝑧,
𝜕𝑦
‫𝐥𝑑 ⋅ 𝐀 𝐿ׯ‬
1
𝜕𝐴𝑧 𝜕𝐴𝑦
lim
=
−
Δ𝑆→0
Δ𝑆
𝑑𝑦𝑑𝑧 𝜕𝑦
𝜕𝑧
𝐚𝑥
𝜕
𝜕𝐴𝑧 𝜕𝐴𝑦
𝑑𝑦𝑑𝑧 =
−
=
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝐴𝑥
Jang, Min Seok | EE204 Electromagnetics I
𝐚𝑦
𝜕
𝜕𝑦
𝐴𝑦
𝐚𝑧
𝜕
⋅ 𝐚𝑥
𝜕𝑧
𝐴𝑧
28
Curl
‫𝐥𝑑 ⋅ 𝐀 𝐿ׯ‬
𝛻 × 𝐀 ⋅ 𝐚𝑛 = lim
Δ𝑆→0
Δ𝑆
• Geometrical interpretation:
𝛻 × 𝐀 at a point 𝑃 is a measure of how much the vector 𝐀 curls
around 𝑃 (Direction: perpendicular to the plane of circulation)
𝛻⋅𝐀=0
𝛻×𝐀=0
𝛻⋅𝐀≠0
𝛻×𝐀=0
𝛻⋅𝐀=0
𝛻×𝐀≠0
Jang, Min Seok | EE204 Electromagnetics I
𝛻⋅𝐀≠0
𝛻×𝐀≠0
29
Curl
• Usage in Physics: relation between magnetic flux density (𝐁) and
vector potential field (𝐀)
𝐁=𝛻×𝐀
Relation between a magnetic flux density (𝐁) and its source
(current density 𝐉)
𝛻 × 𝐁 = 𝜇0 𝐉
Jang, Min Seok | EE204 Electromagnetics I
30
Divergence and Curl
• Exercise: Find the divergence and the curl of the following vector
fields:
(a) 𝐀1 = 3𝐚𝑥 (b) 𝐀 2 = 𝑥𝐚𝑥 + 𝑦𝐚𝑦 (c) 𝐀 3 = −𝑦𝐚𝑥 + 𝑥𝐚𝑦
Answer: (a) 𝛻 ⋅ 𝐴1 = 0, 𝛻 × 𝐴1 = 0
(b) 𝛻 ⋅ 𝐴2 = 2, 𝛻 × 𝐴2 = 0
(c) 𝛻 ⋅ 𝐴3 = 0, 𝛻 × 𝐴3 = 2𝐚𝑧
Jang, Min Seok | EE204 Electromagnetics I
31
Laplacian
• Laplacian of a scalar field: 𝑉 𝑥, 𝑦, 𝑧
2𝑉
2𝑉
2𝑉
𝜕
𝜕
𝜕
𝛻 2 𝑉 = 𝛻 ⋅ 𝛻𝑉 = 2 + 2 + 2
𝜕𝑥
𝜕𝑦
𝜕𝑧
• Laplacian of a vector field: 𝐀 𝑥, 𝑦, 𝑧
𝛻 2 𝐀 = 𝛻 2 𝐴𝑥 𝐚𝑥 + 𝛻 2 𝐴𝑦 𝐚𝑦 + 𝛻 2 𝐴𝑧 𝐚𝑧
• Usage in Physics: relation between electric potential field (𝑉)
and its source (charge density 𝜌)
𝜌
𝛻 𝑉 = 𝛻 ⋅ 𝛻𝑉 = −𝛻 ⋅ 𝐄 = −
𝜖0
2
Jang, Min Seok | EE204 Electromagnetics I
32
Divergence Theorem
• Divergence theorem:
ර 𝐀 ⋅ 𝑑𝐒 = න 𝛻 ⋅ 𝐀 𝑑𝑣
𝑆
𝑣
Total outward flux of 𝐀 through a closed surface 𝑆 is equal to the
volume integral of 𝛻 ⋅ 𝐀 over the volume 𝑣 enclosed by 𝑆
• Proof: Subdivide volume 𝑣 into arbitrarily large number of
tiny cubes. For each cube
෍
𝐀 ⋅ 𝑑𝐒 = 𝛻 ⋅ 𝐀𝑑𝑣
six surfaces
𝐀 ⋅ 𝑑𝐒 terms cancel (pairwise) for all interior faces and
only the contribution of the exterior surfaces survive.
෍ 𝐀 ⋅ 𝑑𝐒 =
exterior
surfaces
෍ 𝛻 ⋅ 𝐀𝑑𝑣 ↔ ර 𝐀 ⋅ 𝑑𝐒 = න 𝛻 ⋅ 𝐀 𝑑𝑣
volumes
𝑆
𝑣
Jang, Min Seok | EE204 Electromagnetics I
33
Stokes’ Theorem
• Stokes’ theorem:
ර 𝐀 ⋅ 𝑑𝐥 = න 𝛻 × 𝐀 ⋅ 𝑑𝐒
𝐿
𝑆
Circulation of 𝐀 around a closed path 𝐿 is equal to the surface
integral of 𝛻 × 𝐀 over the surface 𝑆 bounded by 𝐿.
• Proof: Subdivide area 𝑆 into arbitrarily large number of
tiny rectangles. For each rectangle
෍
𝐀 ⋅ 𝑑𝐥 = 𝛻 × 𝐀 ⋅ 𝑑𝐒
four sides
𝐀 ⋅ 𝑑𝐥 terms cancel for all interior line segments and only
the contribution of the exterior line segments survive.
෍
exterior line
segments
𝐀 ⋅ 𝑑𝐥 =
෍
rectangles
𝛻 × 𝐀 ⋅ 𝑑𝐒 ↔
ර 𝐀 ⋅ 𝑑𝐥 = න (𝛻 × 𝐀) ⋅ 𝑑𝐒
𝐿
Jang, Min Seok | EE204 Electromagnetics I
𝑆
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Divergence and Stokes’ Theorem
• Usage in Physics: relation between differential and integral forms
of Maxwell’s equations
Divergence theorem:
𝛻 ⋅ 𝐃 = 𝜌𝑓
ර 𝐃 ⋅ 𝑑𝐒 = න 𝜌𝑓 𝑑𝑣
𝛻⋅𝐁=0
ර 𝐁 ⋅ 𝑑𝐒 = 0
𝑆
𝑣
𝑆
Stokes’ theorem:
𝜕𝐁
𝛻×𝐄=−
𝜕𝑡
𝜕
ර 𝐄 ⋅ 𝑑𝐥 = − න 𝐁 ⋅ 𝑑𝐒
𝜕𝑡 𝑆
𝐿
𝜕𝐃
𝛻 × 𝐇 = 𝐉𝑓 +
𝜕𝑡
𝜕𝐃
ර 𝐇 ⋅ 𝑑𝐥 = න 𝐉𝑓 +
⋅ 𝑑𝐒
𝜕𝑡
𝐿
𝑆
Jang, Min Seok | EE204 Electromagnetics I
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Classification of Vector Fields
M. C. Escher
Non-conservative?
• Irrotational: 𝛻 × 𝐀 = 0, ‫ = 𝐥𝑑 ⋅ 𝐀 𝐿ׯ‬0, 𝐀 = −𝛻𝑉 (𝑉: scalar potential)
• Solenoidal: 𝛻 ⋅ 𝐀 = 0, ‫ = 𝐒𝑑 ⋅ 𝐀 𝑆ׯ‬0, 𝐀 = 𝛻 × 𝐅 (𝐅: vector potential)
• Helmholtz theorem: Any sufficiently smooth and rapidly decaying vector
field 𝐀 can be decomposed into the sum of an irrotational vector field and
a solenoidal vector field.
𝐀 = −𝛻𝑉 + 𝛻 × 𝐅
Jang, Min Seok | EE204 Electromagnetics I
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Useful Vector Identities
• 𝐀 and 𝐁 are vector fields and 𝑈 and 𝑉 are scalar fields
𝛻 𝑈 + 𝑉 = 𝛻𝑈 + 𝛻𝑉
𝛻⋅ 𝐀+𝐁 =𝛻⋅𝐀+𝛻⋅𝐁
𝛻× 𝐀+𝐁 =𝛻×𝐀+𝛻×𝐁
𝛻 𝑈𝑉 = 𝑈𝛻𝑉 + 𝑉𝛻𝑈
𝛻 ⋅ 𝑉𝐀 = 𝑉𝛻 ⋅ 𝐀 + 𝐀 ⋅ 𝛻𝑉
𝛻 × 𝑉𝐀 = 𝑉 𝛻 × 𝐀 + 𝛻𝑉 × 𝐀
𝛻 𝐀 ⋅ 𝐁 = 𝐀 ⋅ 𝛻 𝐁 + 𝐁 ⋅ 𝛻 𝐀 + 𝐀 × 𝛻 × 𝐁 + 𝐁 × (𝛻 × 𝐀)
𝛻⋅ 𝐀×𝐁 =𝐁⋅ 𝛻×𝐀 −𝐀⋅ 𝛻×𝐁
𝛻× 𝐀×𝐁 =𝐀 𝛻⋅𝐁 −𝐁 𝛻⋅𝐀 + 𝐁⋅𝛻 𝐀− 𝐀⋅𝛻 𝐁
𝛻 ⋅ 𝛻𝑉 = 𝛻 2 𝑉
𝛻 × 𝛻𝑉 = 0
𝛻⋅ 𝛻×𝐀 =0
𝛻 × 𝛻 × 𝐀 = 𝛻 𝛻 ⋅ 𝐀 − 𝛻2𝐀
Jang, Min Seok | EE204 Electromagnetics I
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