Notes 6 Coordinate Systems

Transcript

1 ECE 3318 pplied Electricit and Magnetism Spring 018 Prof. David R. Jackson Dept. of ECE Notes 6 Coordinate Sstems Notes prepared b the EM Group Universit of Houston 1

2 Review of Coordinate Sstems P(,, ) n understanding of coordinate sstems is important for doing EM calculations.

3 Kinds of Integrals That Often Occur Line integrals : Volume integrals : Q V E Q E C C ρ d E dr C ρ Rˆ 4πε R V V 0 v ρ dv ρ ˆ v R 4πε R 0 (scalar integral, scalar result) d (vector integral, scalar result) (vector integral, vector result) (scalar integral, dv scalar result) (vector integral, vector result) Q I E Surface integrals : S s J nˆ ds S S ρ ds ρ ˆ s R 4πε R 0 (scalar integral, scalar result) (vector integral, scalar result) ds (vector integral, vector result) We wish to be able to perform all of these in various coordinates. 3

4 Rectangular Coordinates Position vector: r ˆ + ˆ + ˆ ẑ r P(,, ) Short hand notation: r (,, ) ˆ ŷ Note: We have the tip to tail rule when adding vectors. Note: unit vector direction is defined b increasing one coordinate variable while keeping the other two fied. Note: Different notations are used for vectors in the books. 4

5 Rectangular Coordinates Differentials d d ds dd d ds dd ds dd dv d d d We increase,, or starting from an initial point (blue dot). Note: ds ma be in three different forms. 5

6 Rectangular (cont.) Path Integral (we need dr) C r dr r + dr dl dr r ˆ + ˆ + ˆ dr d ( r ) dr dr ˆd + ˆd + ˆ d Note on notation: The smbol dl is often used instead of dr. 6

7 Clindrical Coordinates ρ ρ P ( ρφ,, ) φ φ. ρcosφ ρsinφ ρ + φ ( ) 1 tan / 7

8 Clindrical (cont.) ẑ Unit Vectors ρ. ˆρ φ Note: unit vector direction is defined b increasing one coordinate variable while keeping the other two fied. φ ˆρ Note: and ˆ φ depend on (, ) ˆρ φ This is wh we often prefer to epress them in terms of ˆ and ˆ 8

9 Clindrical (cont.) Epressions for unit vectors (illustrated for ˆρ ) φ ˆρ ssume ˆ ρ ˆ+ ˆ 1 φ Solve for 1 : Similarl, so ˆ ρ ˆ ˆ ˆ+ ˆ ˆ 1 1 ˆ ρ 1 ˆ ˆ ρ ˆ cosφ cosφ ˆ ˆ ρ π cos φ sinφ Hence, we have ˆ ρ ˆcos φ+ ˆsin φ 9

10 Clindrical (cont.) Summar of Results ˆ ρ ˆcos φ+ ˆsin φ ˆ φ ˆ sinφ + ˆcos φ ˆ ˆ ( ) ˆ ˆ ρcosφ+ ˆ φ sinφ ( ) ˆ ˆ ρsinφ+ ˆ φcosφ ˆ ˆ 10

11 Clindrical (cont.) ẑ ρ. φ ˆρ r Eample: Epress the r vector in clindrical coordinates. r ˆ + ˆ + ˆ Substituting from the previous tables of unit vector transformations and coordinate transformations, we have r ( ρcosφ ˆ φ( sinφ) )( ρcosφ) ( ρsinφ ˆ φcosφ)( ρsinφ) ˆ + + ˆ + + ˆ ( ˆ ρρ )( cos φ sin φ ) + + ˆ ˆ ρρ + ˆ 11

12 Clindrical (cont.) ẑ ρ. φˆ r ˆρ ẑ ˆρρ r ˆ ρρ + ˆ Note: r ˆ ρρ + ˆ φφ + ˆ 1

13 Clindrical (cont.) Differentials ds ρd ρdφ dφ Note: ds ma be in three different forms. ρ d ds dρd dρ ρdφ ds ρdφd dv ρdρdφd We increase ρ, φ, or starting from an initial point (blue dot). Note: The angle φ must be in radians here. 13

14 Clindrical (cont.) Path Integrals First, consider differential changes along an of the three coordinate directions. dρ dφ ρ ρdφ dφ d dr ˆ ρdρ dr ˆ φ( ρdφ) dr ˆ d Note: The angle φ must be in radians here. 14

15 Clindrical (cont.) In general: Note: change in is not shown, but is possible. ˆ( ) dr ˆ ρdρ + φ ρdφ + ˆ d C ˆ ρd ρ dr If we ever need to find the length along a contour: ( ρ) ( ρ φ) ( ) d dr d + d + d ( ρdφ) φ 15

16 Spherical Coordinates φ θ r. P( r, θφ, ) φ ρ θ r. P( r, θφ, ) Note: 0 θ π Note: ρ r sin θ 16

17 Spherical (cont.) φ ρ θ r. P( r, θφ, ) rsinθ cosφ rsinθ sinφ rcosθ r + + θ φ ( r) ( ) 1 cos / 1 tan / Note: ρ r sin θ 17

18 Spherical (cont.) Unit Vectors θ ˆr φˆ Note: unit vector direction is defined b increasing one coordinate variable while keeping the other two fied. Note: ( ) rˆ, ˆ θ, ˆ φ depend on,, 18

19 Spherical (cont.) Transformation of Unit Vectors ˆr rˆ ˆsinθcosφ+ ˆsinθsinφ+ ˆ cosθ ˆcos cos + ˆcos sin + ˆ ( sin ) θ θ φ θ φ θ θ φˆ ( sin ) ˆ φ ˆ φ + ˆcos φ ˆ( ) ˆ rˆsin θcosφ+ θcosθcosφ+ φ sinφ ˆ rˆsinθsinφ+ θcosθsinφ+ ˆ φcosφ ( ) ˆ rˆ cosθ + θ sinθ 19

20 Spherical (cont.) θ ˆr φˆ Eample: Epress the r vector in spherical coordinates. r ˆ + ˆ + ˆ Substituting from the previous tables of unit vector transformations and coordinate transformations, we have: ( ˆ sinθ cosφ θ cosθ cosφ ˆ φ( sinφ) )( sinθ cosφ) r r + + r + ( rˆ sinθ sinφ+ θ cosθ sinφ+ ˆ φ cosφ)( rsinθ sinφ) ( r ( )) + ˆ cosθ + θ sinθ rcosθ 0

21 Spherical (cont.) fter simplifing: ˆrr ˆr φˆ θ r rr ˆ Note : r rr ˆ + ˆ θθ + ˆ φφ 1

22 Spherical (cont.) Differentials ( sin ) ρdφ r θ dφ We increase r, θ, or φ starting from an initial point (blue dot). dφ r ρ ( sinθ) ds r dθdφ dθ dr rdθ dv r sinθ dr dθ dφ Note: ds ma be in three different forms (onl one is shown). The other two are: ds r drdθ ds r sinθdrdφ Note: The angles θ and φ must be in radians here.

23 Spherical (cont.) Path Integrals dr r dθ dr dφ ρ dr r ρdφ rsinθ dφ dr rˆ dr dr ˆ θ( rdθ) dr ˆ φ ( rsinθdφ) dr rˆ dr + ˆ θ rdθ + ˆ φr sinθdφ Note: The angles θ and φ must be in radians here. 3

24 Note on dr Vector Note that the formula for the dr vector never changes, no matter which direction we go along a path (we never add a minus sign!). Eample: Integrating along a horiontal radial path in clindrical coordinates. ˆ( ) dr ˆ ρdρ + φ ρdφ + ˆ d V E dr dr ˆ ρdρ The limits take care of the sign of dr. E ˆ ρe + ˆ φe + E V ρ ρ ρ Edρ This form does not change, regardless of which limit is larger. φ ρ ˆ C dr ρ < ρ d ρ > 0 dr +ρˆ dρ C dr ρ > ρ dρ < 0 dr ρˆ dρ 4

25 Eample Given: J ( ) ˆ [/m ] Find the current I crossing a hemisphere ( > 0) of radius a, in the outward direction. Hemisphere nˆ rˆ I J nˆ ds S J 5

26 Eample (cont.) I J nˆ ds S J rˆ ds S S S S S a ( ˆJ ) rˆds J ( sinθcosφ) ( )( sinθcosφ) ( a sinθcosφ)( sinθcosφ) ( sin θcos φ) S ds ds ds ds rˆ ˆsinθcosφ+ ˆsinθsinφ+ ˆ cosθ ˆcos cos + ˆcos sin + ˆ ( sin ) θ θ φ θ φ θ ( sin ) ˆ φ ˆ φ + ˆcos φ ˆ( ) ˆ rˆsin θcosφ+ θcosθcosφ+ φ sinφ ˆ rˆsinθsinφ+ θcosθsinφ+ ˆ φcosφ ( ) ˆ rˆ cosθ + θ sinθ rsinθ cosφ rsinθ sinφ rcosθ 6

27 Eample (cont.) S ( sin θcos φ) I a ds ππ/ a sin cos a sin d d a a a 0 0 ( ) ππ/ ( ) π / 3 ( ) ( sin ) ( ) ( sin ) ( π ) 0 π / a θ φ θ θ φ sin θcos φ sinθ dθdφ π θ sinθ dθ π θ dθ 3 I π a 3 3 [] ds r sinθ dθdφ Note : π 0 π / 0 1 φ φ ( π) π cos d Note : sin 3 θdθ 3 7

28 ppendi Here we work out some more eamples. 8

29 Eample ( ) ( ) o P1 4,60,1 Given: Clindrical coordinates (ρ, φ, ) o P 3,180, 1 with distances in meters Find d distance between points ( ) ( ) ( ) d This formula onl works in rectangular coordinates! ρcosφ ρsinφ ( ) ( ) 4cos 60 4sin ( ) ( ) 3cos sin d [m] 9

30 Eample Given: ( φ ) [ ] ρv r < r < cos / C/m, 5 m Find Q Solution: Note: The integrand is separable and the limits are fied. a a b [ ] b [ ] m, 5m Q V ρ dv v π π 0 0 b a v ( ) ρ r φ r θdrdθdφ, sin π π b cos sin r 0 0 a φ θ drdθdφ sphere with a hole in it π b π 8 1 r 0 a cos φdφ dr sinθdθ 30

31 Eample (cont.) π π cos 0 0 b a φdφ 1+ cos φ dφ π 1 sin φ φ + π dr r r 1 1 3/10 a b b a 0 Note: The average value of cos φ is 1/. π 0 1 φ φ π π cos d π 0 [ ] sinθ dθ cosθ π 0 Q [C] 31

32 Eample Derive ˆ rˆ cosθ + θ( sinθ) Let ẑ r ˆ + θ + ˆ φ 1 3 Dot multipl both sides with rˆ, θ, ˆ φ Then ˆ 1 ˆ r θ ˆ 3 ˆ ˆ φ 3

33 Eample (cont.) ẑ r ˆ + θ + ˆ φ 1 3 ẑ θ ˆr ẑ φˆ ẑ θ θ θ θ θ 1 1 ˆ rˆ ˆ rˆ cosθ cosθ ˆ θ ˆ π θ cos + θ sinθ 3 ˆ φ 0 3 ˆ ˆ φ 0 Result: ( ) ˆ rˆ cosθ + θ sinθ 33

34 Eample Derive rˆ ˆsinθcosφ+ ˆsinθsinφ+ ˆ cosθ rˆ ˆ + ˆ + ˆ Let 1 3 Dot multipl both sides with ˆ, ˆ, ˆ θ ˆr 1 3 ( component of ˆ) ( component of ˆ) ( component of ˆ) rˆ ˆ r rˆ ˆ r rˆ ˆ r φ L ˆρ n illustration of finding the component of ˆr (We use a two-step process.) 34

35 Eample (cont.) θ θ ˆr π / θ π L cos θ sin θ Hence φ L ˆρ rˆ ˆ Lcosφ sinθcosφ Similarl, rˆ ˆ Lsinφ sinθsinφ lso, rˆ ˆ cosθ Result: rˆ ˆsinθcosφ+ ˆsinθsinφ+ ˆ cosθ 35

36 Eample (Part 1) ( 3 ) ( ) ( 1) E ˆ + ˆ + ˆ + (This is not an electrostatic field.) Find V using path C shown below. ( 1,0,0) C E(,, ). ( 0,1, 0) Top view ( ) (( 3) d ( ) d) 1 d d V E dr E d + E d + E d ( 3 )( 1 ) ( 1 ) V d 36

37 Eample (cont.) Completing the calculus: ( 3 )( 1 ) ( 1 ) V d d d 3 ( ) + d V [ ] 5 /1 V 37

38 Eample (cont.) lternative calculation (we parameterie differentl): 0 1 ( ) V E dr E d + E d + E d ( 3 ) ( ) d + d ( 3 )( 1 ) ( 1 )( ) V d + d ( 3 3 ) ( ) d + d V [ ] 5 /1 V 38

39 Eample (Part ) ( 3 ) ( ) ( 1) E ˆ + ˆ + ˆ + (same field as in Part 1) Find V using path C shown below. ( 1,0,0) C E(,, ) ( 0,1, 0) ( ) E dr E d + E d + E d (( 3) d ( ) d) + 0 ( ( 3 ) ( ) ) ( 3 ) ( ) 0 1 0d + 0d ( ) d + d + d + d V 0 [ V] 39

40 Eample ( 3 ) ( ) ( ) E ˆ + ˆ + ˆ (This is a valid electrostatic field.) ( 1,0,0) Find V using an arbitrar path C in the plane. C E(,, ) ( 0,1, 0) Note: The path does not have to be parameteried. Hence, onl the endpoints are important. The integral is path independent! ( ) E dr E d + E d + E d ( 3 ) ( ) d + d 0 1 ( 3 ) ( ) d + d ( 3 ) ( ) V d + d V /6 [ ] 7/6 V 40

41 Eample Note: If we have an electric field of the form: ( ( )) ( ) ( ) ( ( )) E ˆ f + ˆ g + ˆ h E 0 (discussed later) V is path independent. 41

42 Eample ( ) ( ) E ˆ + ˆ Find V using path C shown below. V E dr ( dφ) ( ) dr ˆ ρdρ + ˆ φ ρdφ + d ˆ ˆ φ 3 ρ cosφ 3cosφ ρ sinφ 3sinφ 3[ m] C ( ) ˆ ˆ ρ cosφ + ˆ φ sinφ ˆ ˆ ρ sinφ+ ˆ φcosφ 4

43 Eample (cont.) ( ) ( ) E ˆ + ˆ ˆcos ˆsin ( 3cos ) ˆsin ˆcos ( ( 3sin )) dr ˆ φ ( 3dφ ) E ρ φ φ φ φ + ρ φ+ φ φ φ V π / π π / π 9sinφcosφdφ 9 sin ( φ) ( φ ) 9 cos dφ π / π ( ) E dr sinφcosφdφ 9sinφcosφdφ Note: The angle φ must change continuousl along the path. If we take the angle φ to be π / at point, then the angle φ must be -π at point. V 9/ [ V] 43

44 Eample (cont.) Let s eamine this same electric field once again: ( ) ( ) E ˆ + ˆ V E dr Question: Is this integral path independent? 3[ m] C Note: The answer is es because the curl of the electric field is ero, but we will talk about curl later. 44

45 Eample (cont.) ( ) ( ) E ˆ + ˆ Question: Is this integral path independent? V E dr Let s find out from the calculus: ( ) E dr E d + E d + E d ( ) d ( ) + d 3[ m] C 0 3 ( ) d ( ) / d Yes, it is path independent! V 9/ [ V] 45