Cheng; 3/2/2009; 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 AAa= A ↑ ↑ magnitude, A A unit vector, A Graphical representation Two vectors are equal if they have the same magnitude and direction, even though they may be displaced in space. • Vector addition, CA= + B Two vectors, AB and , form a plane Parallelogram rule : is the diagonal of the parallelogram C 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; 3/2/2009; 2-2 Note C =+=+A B B A Commutative law ABF++=++()() ABF Associative law • Vector subtraction ˆ AB−=+− Aej B, where −=−BaB()B 2.3 Vector Multiplication Multiplication of a vector by a positive scalar kA= bg kA a A A. Scalar or Dot Product AB•≡ ABcosθAB Note that •≤ (1) A B AB (2) •≤ •≥ A B 00 or A B •=× (3) A B A Projection of B onto A =B × Projection of A onto B (4) A •=B 0 when A and B are perpendicular to each other Note that A •=A A2 → AAA=• A •=•B B A : Commutative law ABF•+=•+•() ABAF : Distributive law Cheng; 3/2/2009; 2-3 Example Law of Cosine Use vectors to prove CAB222=+−2 ABcos α CBA=− CCCBABA2 =•⇒( −) •( −) ⇒•+•−•−•BB AAAB BA ⇒+−AB222cos AB α B. Vector or Cross Product AB×≡ aABnABsin θ ↑ unit vector normal to A and B (the right hand rule) AB× = Area of the parallelogram formed by A and B Note that AB×=−× BA NOT Commutative ABF××≠××()() ABF NOT Associative ABF×+=×+×() ABAF Distributive law C. Product of Three Vectors Scalar triple product ABC•×=•ejej AaBCn sin α ↑ ↑ Area of parallelogram Height of the parallelepiped → Volume of the parallelepiped Vector triple product ABCBACCAB××=( ) ( •−) ( •) back-cab rule Cheng; 3/2/2009; 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. y Unit vectors along uuu,, → Base vectors, aaaˆˆˆ, , 123 uu123 u A vector in (uuu123,,) coordinate system → A =+Aauu11ˆˆˆ Aa u 2 u 2 + Aa u 3 u 3 y Differential change dui → Differential length change dliii= h du ↑ metric coefficient hi Vector differential length dl=+ aˆˆuu111( h du) a 2( h 2 du 2) + a ˆ u 3( h 3 du 3) Differential volume dv= ( h11 du) X( h 2 du 2) X( h 3 du 3) Vector differential area ds= ds aˆn , aˆn : surface normal Note ds11 aˆˆuu= h 232 h du du 31 a ds22 aˆˆuu= h 131 h du du 32 a ds33 aˆˆuu= h 121 h du du 23 a A. Cartesian Coordinate System bgbguuu123,,= xyz ,, A point bgxyz111,, : The intersection of three planes xx= 1 , yy= 1 , zz= 1 Base vectors : aaa xyz, , Properties of the base vectors aaxy×= a z, aayz×= a x, aazx×= a y aaxy•=•=•= aa yz aa xz0 aaxx•=•=•= aa yy aa zz1 The position vector to the point Pxbg111,, y z → → OP=++ x111 axyz y a z a A vector A in Cartesian coord. AAaAaAa=++xx yy zz Cheng; 3/2/2009; 2-5 y Vector differential length → dl=++ dx axyz dy a dz a Differential area dsx = dydz dsy = dxdz dsz = dxdy Differential volume dv= dxdydz y The dot product A•= Bdidi Aaxx + Aa yy + Aa zz • Ba xx + Ba yy + Ba zz ⇒ AB x x + AB y y + AB z z The cross product A×= Bdidi Aaxx + Aa yy + Aa zz × Ba xx + Ba yy + Ba zz aaa xyz AB×=AAAxyz ⇒−diAByz AB zy a x +−bg AB zx AB xz a y +−di AB xy AB yx a z BBBxyz Example A straight line L1 is given by 2xy+= 4 . (a) Find the unit normal to L1 starting from the origin (b) Fine the normal line to L1 passing through P(0,2) Solution: (a) A vector along L1 : −+24xyˆˆ (1) A vector from the origin to a point Q on L1 : xxˆˆ+−(42 x) y (2) From (1) and (2) (−+24xyxxˆˆˆ)i( +−(42 xy) ˆ) =0 → Q(1.6, 0.8) 1 The unit normal is ()2xyˆˆ+ 5 1 (b) The straight line parallel to the unit normal is yxc= + 2 1 This line should pass through P(0,2) → yx= + 2 2 Cheng; 3/2/2009; 2-6 B. Cylindrical coordinates bgbguuu123,,=φ r ,, z A point Prbg111,,φ z : 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 : aaarz, φ , ↑ ↑ directions change with P aarz×=φ a , aaφ ×=zr a , aaazr×=φ y Differential length dl=+ dr arz rdφ aφ + dz a ↑ Differential areas dsr = rdφ dz dsφ = drdz dsz = rdrdφ Differential volume dv= rdrdφ dz Cheng; 3/2/2009; 2-7 y A vector in cylindrical coords. AAaAaAa=++rrφφ zz Transformation of A into Cartesian coord. AAaAaaAaaAaaxxrrxxzzxr=•=•+•+•⇒φφ Acosφφ − A φ sin ↑ ↑ ↑ =0 = cos φ = − sin φ Similarly using aary• =φsin , aaφ • y = cos φ AAyr=+sinφφ Aφ cos In matrix form MLAx PO MLcosφφ− sin 0POMLAr PO MAy P = Msinφφ cos 0PMAφ P NMAz QP NM 001QPNMAz QP y A point in cylindrical coord is transformed into Cartesian coord. x = r cos φ yr= sin φ zz= Conversion from Cartesian to cylindrical coords. rxy=+22 y φ = tan−1 x zz= → Exercise Convert a vector in Cartesian coords., OQ=++345 axyz a a , to cylindrical coords. → OQa=dididi345xyzrrxyz ++• a aaaa + 345 ++• a aaaaφφ + 345 xyzzz ++• a aaa ↑ ↑ ↑ 34cosφφ+ sin 34bg−+sinφφ cos 5 From Cartesian Coords. we know that 3 4 cos φ= and sin φ= 5 5 Therefore in cylindrical Coords. → OQ=+55 arz a Can you convert this vector back to Cartesian Coords.? Cheng; 3/2/2009; 2-8 C. Spherical Coordinates bgbguuu123,,=θφ R ,, A point PRbg111,,θφ : a sphere with radius R1 a cone with a half-angle of θ1 a half-plane rotated by φ1 from x-axis z φφ= 1 plane θθ= 1 cone θ1 aR P a φ a θ φ1 y x RR= 1 sphere Three base vectors aaaR ×=θφ, aaθφ×= a R , aaφθ×=R a A vector in spherical coord AAaAaAa=++RRθθ φφ y Vector differential length dl=+ dR aR Rdθθφ aθφ + Rsin d a ↑ ↑ Differential areas 2 dsR = Rsinθ dθφ d dsθ = Rsinθ dRdφ dsφ = RdRdθ Differential volume dv= R2 sinθ dRdθφ d Cheng; 3/2/2009; 2-9 • Conversion of a point xR= sinθφ cos yR= sinθφ sin zR= cos θ Conversion from Cartesian to spherical coords. Rxyz=++222 θ =+tan−122⎡⎤xyz / ⎣⎦ y φ = tan−1 x Integrals Containing Vector Functions Fdv Fxˆˆˆ++ Fy Fz dv ∫V ∫V ( xyz) Vdl V() xdxˆˆˆ++ ydy zdz ∫C ∫C → Fdli Fdli ∫C ∫C Adsi Adsi ∫S ∫S Sign convention for the surface integral Cheng; 3/2/2009; 2-10 Cheng; 3/2/2009; 2-11 2.5 Gradient of a Scalar Field A scalar field Vtbg,,, u123 u u Move from VV= 1 surface to VV=+1 dV surface PP12→ → dn PP13→ → dl ↑ ↑ same dV different distances The directional derivative dV : The space rate of change of V along l dl The gradient of a scalar function V is defined as dV grad V≡∇ V ≡ a : The maximum directional derivative of V dn n (an is perpendicular to V=const. plane) • The directional derivative dV dV dn dV dV =⇒cos α ⇒•⇒∇•aanlbg Va l : projection of gradV onto al direction dl dn dl dn dn → dV=∇bg V • dl , (1) In Cartesian coord. ∂ ∂ ∂ ∂ ∂ ∂ V V V F V V V I dV =++⇒dx dy dz axyz ++a adxadyadza•++di xyz ∂x ∂y ∂z HG ∂x ∂y ∂z KJ ↑ (2) = dl From (1) and (2) ∂V ∂V ∂V F ∂ ∂ ∂ I ∇=V a +a +a ⇒a +a + aV ∂x xyz∂y ∂z HG ∂x xyz∂y ∂z KJ ↑ ≡ ∇ , “del” operator In bguuu123,, coordinates 11∂ ∂ 1∂ ∇=a +a + a uu123 u hu11∂ hu 22∂ hu 3∂ 3 Cheng; 3/2/2009; 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 AdS• zS divA ≡ lim : dS= dS an ΔV →0 ΔV Find divA at Pxbgooo,, y z For a small volume ΔxΔΔy z A•= dS A •+ dS A •+ dS A •+ dS A •+ dS A •+ dS A • dS zSz frontz backz rightz leftz topz botom (1) On the front surface 1 AdSA•=front •ΔΔΔΔ S front ⇒ Ax x(,,) o +2 xy o z o yz zfront ↑ ↑ ΔxA∂ x bgΔΔyzax , Axxo(,,) y o z o+ , Taylor series 2 ∂x bgxyzooo,, (2) On the back surface 1 A•= dSAback •ΔΔΔΔΔΔ S back ⇒ A back •−bg yza x ⇒−− A x(,,) x o2 x y o z o yz zback ↑ ΔxA∂ Ax(,,) y z − x xo o o 2 ∂x bgxyzooo,, Combining ∂A AdS•+ AdS •=x ΔΔΔxyz (1) zfrontz back ∂x bgxyzooo,, Cheng; 3/2/2009; 2-13 Similarly, combining contributions from the right and left surfaces ∂A AdS•+ AdS •=y ΔΔΔxyz (2) zrightz left ∂y bgxyzooo,, Similarly, combining contributions from the top and bottom surfaces ∂A AdS•+ AdS •=z ΔΔΔxyz (3) ztopz bottom ∂z bgxyzooo,, From (1), (2) and (3) F ∂A ∂Ay ∂A I AdS•=x + +z ΔΔΔxyz zS HG ∂x ∂y ∂z KJ bgxyzooo,, ∂A ∂Ay ∂A → divA =++x z ∂x ∂y ∂z We define ∇•AdivA ≡ In general 1 L ∂ ∂ ∂ O ∇•A =M bgbgbghhA231 +hh13 A 2 + hh12 A 3P hhh123N∂ u 1 ∂u2 ∂u3 Q 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 ∇•AdV = A • dS zzVS Proof: Consider a differential volume ΔV j bounded by S j From the definition of divA ej∇•AVΔ j = AdS • j zs j Add all ΔV j ’s L N O L N O limM ∇•AVΔ j P =• lim M AdSAdSP ⇒• ∑ejj ∑ ΔΔV jj→0M P V →0M zs j P zS N j =1 Q N j =1 Q ↑ ↑ ↑ Integral at the external surface ∇•Adv An internal surface is shared by two adjacent volumes.