2.Vector analysis

Lecture 2 Vector Analysis Jang, Min Seok EE204 Fall 2023 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+𝐀⋅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 𝐀 × 𝐁 = −𝐁 × 𝐀 not associative 𝐀× 𝐁×𝐂 ≠ 𝐀×𝐁 ×𝐂 distributive 𝐀× 𝐁+𝐂 =𝐀×𝐁+𝐀×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 𝐚𝑧 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 23 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 24 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) 𝛻 2𝑉 𝜌 = 𝛻 ⋅ 𝛻𝑉 = −𝛻 ⋅ 𝐄 = − 𝜖0 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 ↔ volumes ර 𝐀 ⋅ 𝑑𝐒 = න 𝛻 ⋅ 𝐀 𝑑𝑣 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 34 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 35 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 36 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 37

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