ECE 2317 Applied Electricity and Magnetism Spring 2014 Prof. David R. Jackson ECE Dept. Notes 13 1 Divergence – The Physical Concept Start by considering a sphere of uniform volume charge density The electric field is calculated using Gauss's law: z r < a: v = v0 D nˆ dS Q encl S y r x a 4 r D r v 0 2 Dr 4 3 r 3 v0 r 3 C /m 2 Sphere of uniform charge density 2 Divergence -- Physical Concept (cont.) r > a: 4 2 3 4 r D r v 0 a 3 z Dr v0 a 3r 2 3 C /m 2 v = v0 y a r x 3 Divergence -- Physical Concept (cont.) Flux through a spherical surface: D nˆ dS 4 r D r 2 S Dr Recall that Dr v0 r 3 v0 a 3r 2 C /m (r < a) C /m 2 (r > a) 2 3 Hence 4 3 4 3 r v0 (r < a) a v0 (r > a) 3 3 (increasing with distance inside sphere) (constant outside sphere) 4 Divergence -- Physical Concept (cont.) Observation: More flux lines are added as the radius increases (as long as we stay inside the charge). D nˆ dS 0 S The net flux out of a small volume V inside the charge is not zero. V S D nˆ dS 0 S Divergence is a mathematical way of describing this. 5 Gauss’s Law -- Differential Form Definition of divergence: div D lim V 0 1 V D nˆ dS S V Note: The limit exists independent of the shape of the volume (proven later). A region with a positive divergence acts as a “source.” A region with a negative divergence acts as a “sink.” 6 Gauss’s Law -- Differential Form Apply divergence definition to a small volume inside a region of charge: div D lim V 0 V S v (r) 1 V D nˆ dS S Gauss's law: r D nˆ dS Q encl v r V S Hence div D r lim V 0 1 V r V v v r 7 Gauss’s Law -- Differential Form (cont.) The electric Gauss law in point (differential) form: d iv D r v r This is one of Maxwell’s equations. 8 Calculation of Divergence z div D lim x / 2, 0, 0 V 0 (0,0,0) z y S x x y 1 x y z D nˆ dS S x D nˆ d S D x , 0, 0 y z 2 x Dx , 0, 0 y z 2 y D y 0, , 0 x z 2 Assume that the point of interest is at the origin for simplicity (the center of the cube). The integrals over the 6 faces are approximated by “sampling” at the centers of the faces. y D y 0, , 0 x z 2 z D z 0, 0, 2 x y z D z 0, 0, x y 2 9 Calculation of Divergence (cont.) div D lim V 0 S x D nˆ d S D x , 0, 0 y z 2 x Dx , 0, 0 y z 2 y D y 0, , 0 x z 2 y D y 0, , 0 x z 2 z D z 0, 0, 2 x y z D z 0, 0, x y 2 1 x y z D nˆ dS S 1 x y z D nˆ dS S x Dx , 0, 0 D x 2 x x , 0, 0 2 y D y 0, ,0 Dy 2 y z D z 0, 0, Dz 2 z y 0, ,0 2 z 0, 0, 2 10 Calculation of Divergence (cont.) div D lim V 0 x Dx , 0, 0 D x 2 x x , 0, 0 2 y D y 0, ,0 Dy 2 y y 0, ,0 2 z D z 0, 0, Dz 2 z z 0, 0, 2 Hence div D D x x D y y D z z 11 Calculation of Divergence (cont.) Final result in rectangular coordinates: div D D x x D y y D z z 12 The “del” operator xˆ x yˆ y zˆ Examples of derivative operators: scalar d d : dx dx sin x cos x z This is a vector operator. Note: The del operator is only defined in rectangular coordinates. ˆ rˆ vector xˆ d dx Note: Any vector can be multiplied by a scalar, or multiplied by another vector in two different ways (dot and cross). : xˆ d dx sin x d xˆ dx ˆ zˆ r z ˆ ˆ r xˆ cos x d ˆ ˆ ˆ x sin x x x sin x cos x dx d d ˆ ˆ ˆ ˆ x y sin x x y sin x zˆ cos x dx dx 13 Example V x , y , z xˆ sin x yˆ 3 y zˆ xy F in d V x, y, z ˆ ˆ V x, y, z x y zˆ xˆ sin x yˆ 3 y zˆ xy y z x V x, y , z sin x 3 y xy y z x V x , y , z cos x 3 0 3 cos x 14 Divergence Expressed with del Operator Now consider: D xˆ yˆ zˆ xˆ D x yˆ D y zˆ D z y z x xˆ xˆ Hence D x x D yˆ yˆ D x We therefore notice that x D y y D y y zˆ zˆ D z z D z z D d iv D 15 Divergence with del Operator (cont.) D d iv D N o te th a t th e d o t a fte r th e d e l o p e ra to r is im p o rta n t; a n y sym b o l fo llo w in g it te lls u s h o w it is to b e u se d a n d h o w it is re a d : "g ra d ie n t" "d iv e rg e n ce " "cu rl" 16 Summary of Divergence Formulas Rectangular: D D x x D y y See Appendix A.2 in the Hayt & Buck book for a general derivation that holds in any coordinate system. D z z Cylindrical: D 1 D 1 D D z z Spherical: D 1 r r 2 r 2 Dr 1 r sin D sin 1 D r sin 17 Maxwell’s Equations (point or differential form) E B t H J D D v B 0 t Faraday’s law Ampere’s law Electric Gauss law Magnetic Gauss law Divergence appears in two of Maxwell’s equations. 18 Example Evaluate the divergence of the electric flux vector inside and outside a sphere of uniform volume charge density, and verify that the answer is what is expected from the electric Gauss law. r<a: v0r D rˆ 3 z v = v0 y D D v0 2 2 v0 1 3 r a r r r Dr 1 2 v0r 2 r r r 3 r x 1 1 r 2 2 v0r r r 3 2 Note: This agrees with the electric Gauss law. 19 Example (cont.) r>a: z v0a 3 D rˆ 2 3r D v = v0 r 2 r r 2 Dr 3 1 2 v0a 2 r 2 r r 3r y x 1 3 1 v0a 2 0 r r 3 a D 0 Note: This agrees with the electric Gauss law. 20 Divergence Theorem S nˆ V A dV V A nˆ dS S In words: The volume integral of "flux per volume" equals the total flux! 21 Divergence Theorem (cont.) Proof The volume is divided up into many small cubes. V rn N A dV V lim V 0 A n 1 rn V Note: The point rn is the center of cube n. 22 Divergence Theorem (cont.) From the definition of divergence: Ar lim V 0 n rn 1 V S n surface of cube n 1 V A nˆ dS Sn A nˆ dS Sn Hence: N A dV V lim V 0 V A n 1 1 lim V V 0 V n 1 N rn Sn N A nˆ dS lim A nˆ dS V 0 n 1 Sn 23 Divergence Theorem (cont.) V N A dV rn V lim V 0 A nˆ dS n 1 S n Consider two adjacent cubes: A nˆ nˆ 2 1 nˆ 1 is opposite on the two faces 2 Hence, the surface integral cancels on all INTERIOR faces. 24 Divergence Theorem (cont.) N A dV V lim V 0 V lim rn V 0 A nˆ dS A nˆ dS n 1 S n outside S n face s Hence A dV lim V 0 V As V 0 : outside S n f aces A dV V A nˆ dS A nˆ dS S S A nˆ dS (proof complete) 25 Example Given: A xˆ 3 x z Verify the divergence theorem using this box. A nˆ dS xˆ xˆ 3 3 2 xˆ xˆ 3 0 2 S 1 18 y 3 2 A Ax x x x A dV 3 dV V Note: A is constant on the front and back faces. Ay y Az z 3x 3 3 V 3 1 2 3 18 V 26 Note on Divergence Definition div D lim V 0 1 V Is this limit independent of the shape of the volume? D nˆ dS S nˆ V Use the divergence theorem for RHS: r S D nˆ dS S D dV V Small arbitrary-shaped volume div D lim V 0 lim V 0 1 V D dV V D V 1 r V D r Hence, the limit is the same regardless of the shape of the limiting volume. 27 Gauss’s Law (Differential to integral form) We can convert the differential form into the integral form by using the divergence theorem. D v Gauss’s law (differential form): Integrate both sides over a volume: D dV V Apply the divergence theorem to the LHS: D nˆ v dV V D nˆ dS S Hence v dV V dS Q encl S 28 Gauss’s Law (Summary of two forms) D nˆ dS Q encl Integral (volume) form of Gauss’s law S Definition of divergence + Gauss's law Divergence theorem D v Differential (point) form of Gauss’s law Note: All of Maxwell’s equations have both a point and an integral form. 29