Chapter 2. Vector nalysis Cheng; 3/4/2007; 2-2. verview t 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 () Vector algebra : addition, subtraction, multiplication (2) rthogonal coordinate system : Cartesian, Cylindrical, pherical (3) Vector calculus : differentiation, integration(gradient, divergence, curl) 2.2 Vector ddition and ubtraction vector has a magnitude and a direction = a magnitude, unit vector, Graphical representation Two vectors are equal if they have the same magnitude and direction, even though they may be displaced in space. Vector addition, C = + B Two vectors, and B, form a plane Parallelogram rule : C is the diagonal of the parallelogram Head-To-Tail rule : The head of touches the tail of B. C is drawn from the tail of to the head of B.
Cheng; 3/4/2007; 2-2 Note C = + B = B + Commutative law + ( B + F) = ( + B) + F ssociative law Vector subtraction B = + e Bj, where B = ( aˆ B ) B 2.3 Vector Multiplication Multiplication of a vector by a positive scalar k = bkg a. calar or Dot Product B B cos θ B Note that () B B (2) B 0 or B 0 (3) B = Projection of B onto =B Projection of onto B (4) B = 0 when and B are perpendicular to each other Note that = 2 = B = B : Commutative law ( B + F) = B + F : Distributive law
Example Law of Cosine 2 2 2 Use vectors to prove C = + B 2B cos α C = B 2 C = C C ( B ) ( B ) B B + B B 2 2 B 2Bcosα + Cheng; 3/4/2007; 2-3 B. Vector or Cross Product B a Bsin θ n B unit vector normal to and B (the right hand rule) B = rea of the parallelogram formed by and B Note that B = B ( B F) ( B) F ( B + F) = B + F NT Commutative NT ssociative Distributive law C. Product of Three Vectors calar triple product B C = a BCsin α e j e nj rea of parallelogram Height of the parallelepiped Volume of the parallelepiped Vector triple product B C = B C C B ( ) ( ) ( ) back-cab rule
Cheng; 3/4/2007; 2-4 2.4 rthogonal Coordinate ystems point in space is represented by three surfaces, u = const., u 2 = const. and u 3 = const. If they are mutually perpendicular, they form an orthogonal coordinate system. Unit vectors along u, u2, u 3 Base vectors, aˆu, aˆu2, a ˆu3 u, u, u coordinate system = aˆ + aˆ + a ˆ vector in ( ) 2 3 Differential change du i Differential length change dli = hidu i metric coefficient h i Vector differential length dl = aˆ h du + aˆ h du + aˆ h du ( ) ( ) ( ) u u2 2 2 u3 3 3 Differential volume dv h du X h du X h du ( ) ( ) ( ) = 2 2 3 3 Vector differential area ds = ds a ˆn, a ˆn : surface normal u u u2 u2 u3 u3 Note ds aˆ = h h du du a ˆ u 2 3 2 3 u ds aˆ = h h du du a ˆ 2 u2 3 3 u2 ds aˆ = h h du du a ˆ 3 u3 2 2 u3. Cartesian Coordinate ystem u, u, u = x, y, b 2 3g b g point bx, y, g : The intersection of three planes x x Base vectors Properties of the base vectors a a = a, x y a a = a, y x a a = a x y : a, a, a x y a a = a a = a a = 0 x y y x a a = a a = a a = x x y y The position vector to the point Px, y, b g P = x a + y a + a x y =, y = y, = vector in Cartesian coord. = a + a + a x x y y
Vector differential length dl = dx a + dy a + d a x y Cheng; 3/4/2007; 2-5 Differential area ds = dyd ds ds x y = dxd = dxdy Differential volume dv = dxdyd The dot product B = a + a + a B a + B a + B a B + B + B d i d i x x y y x x y y x x y y The cross product B= a + a + a Ba + Ba + Ba d x x y y i d x x y y i a a a x y B = B B a + B B a + B B a x y B B B x y d i b g d i y y x x x y x y y x Example straight line L is given by 2x + y = 4. (a) Find the unit normal to L starting from the origin (b) Fine the normal line to L passing through P(0,2) olution: (a) vector along L : 2xˆ + 4yˆ () vector from the origin to a point Q on L : xxˆ + (4 2 x) yˆ (2) From () and (2) ( 2xˆ + 4 yˆ) i ( xxˆ + (4 2 x) yˆ) = 0 Q(.6, 0.8) The unit normal is ( 2xˆ + yˆ) 5 (b) The straight line parallel to the unit normal is y = x + c 2 This line should pass through P(0,2) y = x + 2 2
B. Cylindrical coordinates u, u, u = r,, b 2 3g b g Cheng; 3/4/2007; 2-6 point Pr,, b g : a cylindrical surface with radius of r, Three base vectors : a, a, a r directions change with P a half-plane rotated by from x-axis, a plane cutting -axis at. a a = a r, a a a = r, a a = a r Differential length dl = dr a + rd a + d a r Differential areas ds = rdd ds ds r = drd = rdrd Differential volume dv = rdrdd
vector in cylindrical coords. = a + a + a r r Cheng; 3/4/2007; 2-7 Transformation of into Cartesian coord. x = ax = a r r ax + a ax + a a x r cos sin =0 = cos = sin imilarly using a r ay = sin, a a y = cos = sin + cos y r In matrix form Lx r M Lcos sin 0 y P = M L sin cos 0 P 0 0 M NM Q NM QN QP point in cylindrical coord is transformed into Cartesian coord. x = r cos y = r sin = Conversion from Cartesian to cylindrical coords. 2 2 r = x + y y = tan x = Exercise Convert a vector in Cartesian coords., Q = 3 a + 4 a + 5 a, to cylindrical coords. x y d i d i d i Q = 3 a + 4 a + 5 a a a + 3 a + 4 a + 5 a a a + 3 a + 4 a + 5 a a a x y r r x y x y 3cos + 4sin 3 sin + 4cos From Cartesian Coords. we know that b g 5 cos = 3 5 and sin = 4 5 Therefore in cylindrical Coords. Q = 5 a + 5 a r Can you convert this vector back to Cartesian Coords.?
C. pherical Coordinates u, u, u = R, θ, b 2 3g b g Cheng; 3/4/2007; 2-8 point PR, θ, b g : a sphere with radius R a cone with a half-angle of θ a half-plane rotated by from x-axis θ = θ cone = plane θ P a R a a θ y x R = R sphere Three base vectors a a a R θ =, a θ a = a R, a a = a R vector in spherical coord = a + a + a R R θ θ θ Vector differential length dl = dr a + Rdθ a + R sin θd a R Differential areas ds = R 2 sinθdθd R ds = R sinθdrd ds θ = RdRdθ Differential volume dv = R 2 sinθdrdθd θ
Cheng; 3/4/2007; 2-9 Conversion of a point x = Rsin θcos y = Rsin θsin = Rcos θ Conversion from Cartesian to spherical coords. 2 2 2 R = x + y + 2 2 θ = tan x + y / y = tan x Integrals Containing Vector Functions Fdv V Vdl C Fdl i C ds i ( Fxˆ ˆ ˆ x + Fy y + F ) dv V V ( xdx ˆ + ydy ˆ + d ˆ ) C Fdl i C ds i ign convention for the surface integral
Cheng; 3/4/2007; 2-0
2.5 Gradient of a calar Field scalar field Vt, u, u, u b 2 3 g Cheng; 3/4/2007; 2- Move from V = V surface to V = V + dv surface P P2 dn P P dl 3 same dv different distances The directional derivative dv dl : The space rate of change of V along l The gradient of a scalar function V is defined as grad V V dv dn a n : The maximum directional derivative of V ( a n is perpendicular to V=const. plane) The directional derivative dv dv dn dv dv = cos dl dn dl dn dn a n a l V a l dv = V dl, () b g α b g : projection of gradv onto a l direction In Cartesian coord. dv F HG V x dx V y dy V d V x a V y a V = + + + + a dx a + dy a + d a KJ From () and (2) V F I V = + + + + x a V y a V a x a x y x y a y a V In bu, u2, u3g coordinates = + + h u a h u a h u a u u2 u3 2 2 3 3 HG I d x y x y, del operator KJ (2) = dl i
2.6 Divergence of a Vector Field Flux is defined as a quantity per unit area per unit time Cheng; 3/4/2007; 2-2 vector field is represented by flux lines Its direction Direction of the flux line Its magnitude Density of the flux lines Divergence of The net outward flux of per unit volume d div lim : d = d a ΔV 0 ΔV n Find div at Px, y, b o o o g For a small volume ΔxΔyΔ d = d + d + d + d + d + d front back right left top botom () n the front surface d = front Δfront x( xo + Δx, yo, o ) ΔyΔ 2 front x x bδyδg a x, x( xo, yo, o) + Δ 2 x bxo, yo, og, Taylor series (2) n the back surface d = back Δback back ΔyΔa b xg x ( xo Δx, yo, o ) ΔyΔ 2 back x x x( xo, yo, o) Δ 2 x x front back x,, Combining d + d = bxo, yo, og x y bxo yo og Δ Δ Δ ()
imilarly, combining contributions from the right and left surfaces y d + d = x y right left y bxo, yo, og Δ Δ Δ (2) Cheng; 3/4/2007; 2-3 imilarly, combining contributions from the top and bottom surfaces d + d = x y top bottom b g Δ Δ Δ (3) xo, yo, o From (), (2) and (3) F d = + + x y HG x y div = + + x y x y I KJ b xo, yo, o g ΔxΔyΔ We define div In general L = M hh + + hhh u u hh 2 3 3 2 u hh N 2 3 b g b g b 2 3g Q P 2 3 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 = V dv d Proof: Consider a differential volume ΔV j bounded by j From the definition of div V = d e j Δ dd all ΔV j s L M j j e j s j P L M P N N lim ΔVj lim d d ΔV j j ΔV j s j NM j = QP = NM j = QP 0 0 Integral at the external surface dv V e j n internal surface is shared by two adjacent volumes. (pposite surface normals) 0 means the existence of flow sources in the given volume
2.8 Curl of a Vector Field Divergence measures flow source Curl measures vortex source The circulation of a vector field around a contour C dl C Cheng; 3/4/2007; 2-4 The definition of curl : The maximum circulation of per unit area with the direction normal to the loop area (Right hand rule) curl = L lim dl a n Δs 0 Δs C max NM QP The component of curl in the direction of a u e j = a u e j = lim dl u Δsu 0 Δs c u The x-component of L e j = lim dl x ΔyΔ ΔyΔ sides side t side : L NM b NM 0,, 234,, g Δy dl = xo, yo, o + +... Δ 2 y side3 t side 3 : L M N b u QP xo, yo, o b g Pb g bxo, yo, og Q Δy dl = xo, yo, o +... M 2 y P Δ g Combine the two L dl = +... y side M Δ Δ, () 3, y P x, y, N b g o o o Q QP imilarly, combining contributions from sides 2 and 4 L y dl = +... y sides M Δ Δ (2) 24, P x, y, N b o o og Q From () and (2) L = M P e j lim dl x ΔyΔ 0 ΔyΔ sides,, 234,, y N Q y : Note the cyclic order in x, y, and
In Cartesian coordinates F I F I = + HG KJ HG K J F I + y a x a y x y x x y HG x y KJ a Cheng; 3/4/2007; 2-5 = a a a x y x y x y In general = hhh 2 3 a h a h a u h u2 u3 u u2 u3 h h h 2 3 x 2 y 3 If = 0, is called an irrotational field, or a conservative field 2.9 tokes s Theorem Consider a differential area Δ j bounded by a contour C j From the definition of e j dδjia j = dl j C j dding all the contributions The left side: N lim e j dδ jiaj e j d Δ j = j 0 j 0 HG KJ The right side: N F I lim dl dl Δ j C j j = tokes s theorem is defined d = dl C C External contour C bounding e j : Right-hand rule for d and dl The surface integral of the curl of a vector over a surface is equal to the closed line integral of the vector along the bounding contour. Note d = e j 0
2.0 Two Null Identities Identity I The curl of the gradient of a scalar field is always ero V = b g 0 Cheng; 3/4/2007; 2-6 Proof: () Direct operations of the gradient and curl (2) Using tokes s theorem V d = V dl dv 0 b g C C The converse statement If a vector field is curl-free, then it can be expressed as the gradient of a scalar field E = 0 E = V Identity 2 The divergence of the curl of any vector field is identically ero = e j 0 Using divergence theorem dv = d V e j e j plit the surface, = + 2 and apply tokes s theorem The right side d a d + a d n e j e j e j n2 2 dl C dl C2 The converse statement If a vector field is divergenceless, then it can be expressed as the curl of another vector field B = 0 B = 0 2. Helmholt s Theorem. static electric field in a charge-free region E = 0, and E = 0 : solenoidal, irrotational 2. steady magnetic field in a current-carrying conductor H = 0, and H 0 : solenoidal, not irrotational 3. static electric field in a charged region E 0, and E = 0 : not solenoidal, irrotational 4. n electric field in a charge medium with a time-varying magnetic field E 0, and E 0 : not solenoidal, not irrotational Helmholt s Theorem vector field is determined if both its divergence and its curl are specified everywhere