Vector alculus Vector Fields Reading Trim 14.1 Vector Fields Assignment web page assignment #9 hapter 14 will eamine a vector field. For eample, if we eamine the temperature conditions in a room, for ever point P in the room, we can assign an air temperature, T, where P T = f(,, ) This is a scalar function or scalar field. However, suppose air is moving around in the room and at ever point P, we can assign an air velocit vector, V (,, ), where V (,, ) = î u(,, ) + ĵ v(,, ) + ˆk w(,, ) where the right side of the above equation consists of,, components at a point, with each component being a function of (,, ). To describe V in the room, we need to keep track of the 3 scalar function, u, v, w at each f(,, ). V (,, ) is a vector function or a vector field. 1
Gradient, Divergence and url Operations The basic operator is the del operator given as 2D 3D = î + ĵ = î + ĵ + ˆk The del operator operates on a scalar function, such as T = f(,, ) or on a vector function, such as F = îp + ĵq + ˆkR (force) or V = îu + ĵv + ˆkw (velocit). We will eamine three operations is more detail 1. gradient: operating on a scalar function Eamples include the temperature in a 2D plate, T = f(, ) T = î f + ĵ f or for a 3D volume, such as a room T = î f + ĵ f + ˆk f The phsical meaning was given in hapter 13. T is a vector perpendicular to the T contours that points uphill on the contour plot or level surface plot. 2. divergence: dotted with a vector function V We can define the velocit in a 3D room as V (,, ) = îu(,, ) + ĵv(,, ) + ˆku(,, ) DIVERGENE of V = div V = V V = ( î + ĵ + ˆk ) (îu + ĵv + ˆkw ) = u + v + w The DIVERGENE of a vector is a scalar. 2
3. curl: the vector product of the del operator and a vector, V, produces a vector curl of V = curl V = V V = ( î + ĵ + ˆk ) ( îu + ĵv + ˆkw ) î ĵ ˆk u v w w = î v + ĵ u w + ˆk v }{{}}{{} u }{{} V 1 V 2 V 3 = î V 1 + ĵ V 2 + ˆk V 3 The 3D case gives w curl V = î v + ĵ u w + ˆk v }{{}}{{} u }{{} V 1 V 2 V 3 The 2D case gives curl V = ˆk ( v u ) 4. Laplacian: ( ) operation. A primar eample of the Laplacian operator is in determining the conduction of heat in a solid. Given a 3D temperature field T (,, ), the Laplacian is 2 T + 2 T 2 + 2 T 2 = 0 2 3
2D eample onsider a 2D heat flow field, q(, ). Look at a small differential element,. In stead state, heat flows in is equivalent to heat flow out. Therefore q = 0 (1) But q is related to temperature b Fourier s law q = k T (2) ombining (1) and (2) ( k T ) = 0 2 T = 0 This is Laplace s equation in 2D, which gives the stead state temperature field. Eample 4.1 how for the 3D case f(,, ) that curl grad f = 0 f = î f + ĵ f + ˆk f holds for an function f(,, ), for instance f(,, ) = 2 + 2 + sin + 2 4
onservative Force Fields and the curl grad f = 0 identit uppose we have a conservative field F (,, ). We know F is irrotational, i.e. F = 0 (1) (ero work in a closed path in the field) But the identit sas ( φ) = 0 (2) alwas holds when φ(,, ) is a scalar function. omparing (1) and (2) for a onservative Force Field, we can alwas find a scalar function φ(,, ) such that F = φ where φ is called the scalar potential function. ometimes, for convenience, we introduce a negative sign F = u The following statements are equivalent F is a conservative net work when a particle moves through force field F around a closed path in space is ero F = 0 irrotational a scalar function φ(,, ) can be found such that F = φ or a function u(,, ) can be found such that F = u 5
Line Integrals of calar Functions Reading Trim 14.2 Line Integrals Assignment web page assignment #10 One place that line integrals often come up is in the computation of averages of a function. 2D ase f f = 1 L L =0 f()d 2 L 2 3D ase The temperature in a room is given b T = f(,, ). A curve in the room is given b (t), (t) and (t). If we measure temperature along curve, what is the average temperature T? T = 1 L value of f along { }}{ arc length of {}}{ f(,, ) d } {{ } line integral along t=0 (=0) L t= (=L) alculation of the line integral along α t=0 ) ( d 2 ( ) d 2 ( ) d 2 f[(t), (t), (t)] + + }{{} dt dt dt f values along }{{} d along dt 6
Eample 4.2 Given a 3D temperature field T = f(,, ) = 8 + 6 + 30 find the average temperature, T along a line from (0, 0, 0) to (1, 1, 1). 1. if we have an eplicit equation for a planar curve : = g() we can reduce fd to fnc of d or fnc of d we do not have to alwas use the parametric equations. 2. value of (i) function f f d depends on (ii) curve in space (iii) direction of travel B A A f d = f d B 3. notation - sometimes is a closed loop in space W W evaluate once around the loop W or W f d c f d c evaluation method is the same as the eample 7
Eample: 4.3a uppose the temperature near the floor of a room (sa at = 1) is described b T = f(, ) = 20 2 + 2 3 where 5 5 4 4 What is the average temperature along the straight line path from A(0, 0) to B(4, 3). Eample: 4.3b What is the average room temperature along the walls of the room? T = f(, )d d where the closed curve is defined in 4 sections 1 = 4 = t 5 t 5 2 = 5 = t 4 t 4 3 = 4 = 5 t 0 t 10 4 = 5 = 8 t 0 t 8 Eample: 4.3c What is the average temperature around a closed circular path 2 + 2 = 9? where is a closed circular path T = f(, )d d where (t) = 3 cos t (t) = 3 sin t for 0 t 2π. 8
Line Integrals of Vector Functions Reading Trim 14.3 Line Integrals Involving Vector Functions 14.4 Independence of Path Assignment web page assignment #10 Let s eamine work or energ in a force field. W = F d Now consider a particle moving along a curve is a 3D force field. W = F (,, ) }{{} force value evaluated along d r }{{} displacement along (1) This is a line integral of the vector F along the curve in 3D space. In component form: F = îp + ĵq + ˆkR d r = îd + ĵd + ˆkd W = P d + Qd + Rd (2) Equations (1) and (2) are equivalent. Use the equation of curve (either in an eplicit form, i.e. = f() etc., or in a parametric form) to reduce (2) to Eample 4.4 α t=0 Given a force field in 3D: g(t)dt or β =n h()d etc. F = î(3 2 6) + ĵ(2 + 3) + ˆk(1 4 2 ) What is the work done b F on a particle (i.e. energ added to the particle) if it moves in a straight line from (0, 0, 0) to (1, 1, 1) through the force field. 9
Notes 1. If we have an eplicit equation for the curve, we can sometimes reduce to a form (fnc of ) d etc. There is no need for parametric equations. 2. The work term W can be either + ve or - ve. In a + ve form, F and d r, are in the same direction, where the work done b the force energ is added to the object b F. In the - ve form, F opposes the displacement. The energ is removed from the object. + ve W = F d r 3. closed path notation W = W = W W F d r F d r once W around loop once W around loop 4. A special case is the conservative force field F d r = 0 We know that F = 0. There is no work in a closed loop. We also know that φ = F, which is integrated gives F d r = φ 1 φ 2 Therefore for a conservative force field, the work is a function of the end points not the path. This is the same for an connecting the same 2 end points. W is alwas the same number if F is conservative. 10
Eample: 4.5a The gravitational force on a mass, m, due to mass, M, at the origin is F = G Mm r r 3 = K r r 3 where K = GMm The vector field is given b: F (,, ) = îp + ĵq + ˆkR K where P = ( 2 + 2 + 2 ) 3/2 K Q = ( 2 + 2 + 2 ) 3/2 K R = ( 2 + 2 + 2 ) 3/2 ompute the work, W, if the mass, m moves from A to B along a semi-circular path in the (, ) plane: 2 + 2 = 16 or = 16 2 and = 0 From A(0, 0.1, 1.261) to B(0, 16, 0) m curve in the =0 plane ( plane) - circle, radius 8 center (0,8,0) A M B Eample: 4.5b Find the work to move through the same field, but following a straight line path from A(0, 0.1, 1.261) to B(0, 16, 0). m A curve M B 11
onservative Force Fields Reading Trim 14.5 Energ and onservative Force Fields Assignment web page assignment #10 Given a flow field in 3D space F = î (2 3 + 6) +ĵ (6 2) }{{}}{{} P Q +ˆk (3 2 2 2 ) }{{} R Part a: Is the force field, F, conservative? heck to see if F = 0 (i.e. irrotational F?) Part b: ompute the work done on an object if it goes around a W circular path of radius 1 for center point P (2, 0, 3) with form Part c: Find the scalar potential function φ(,, ) F = φ Part d: Use φ to verif part b). W = F d r = φ d r Part e: Find the work done if the object moves along from A(0, 0, 0) to B(3, 4, 2) within F. 12
urface Integrals of calar Functions Reading Trim 14.7 urface Integrals Assignment web page assignment #11 surface =g(,) surface area d projection da = d d surface area of = g(, ) = R ( ) g 2 ( ) g 2 1 + + dd Now suppose the surface = g(, ) is within a 3D temperature field T = f(,, ) projection in (,) plane What is the average temperature measured over the surface = g(, )? Add up T for each d area element and divide b the total area of to get the average. T = f(,, )d Area of The numerator is called the surface integral of f(,, ) over the surface (i.e. = g(, )). To evaluate fd = f[,, g(, )] }{{} R T values on the surface ( ) g 2 ( ) g 2 1 + + dd }{{} d area element This becomes F (, )dd R 13
Notes 1. sometimes it is easier if we switch to polar coordinates R F (, )dd R H(r, θ)rdrdθ where = r cos θ and = r sin θ. 2. notation = we have a closed surface in space, i.e. a sphere surface g(, ) = f(,, )d a 2 2 2 Eample: 4.6 uppose the temperature variation (same for all (, )) in the atmosphere near the ground is T () = 40 2 5 where T is in and is in m. Look at a clindrical building roof as follows: 10 m high 30 m long 10 m What is the air temperature in contact with the roof? 14
urface Integrals of Vector Functions Reading Trim 14.8 urface Integrals Involving Vector Fields Assignment web page assignment #11 Given a full 3D velocit field, V (,, ) in space (i.e. air flow in a room). Given some surface = g(, ) within the flow calculate the flow rate Q (m 3 /s) crossing the surface, we can write G = g(, ), where G is a constant since the surface is a level surface or a contour. surface =g(,) or G(,,) = - g(,) = 0 d at (,,) V The basic idea is to consider the element d of the surface at some arbitrar (,, ). Then compute the unit normal vector to d ˆn = (±) G G This will var over. The (±) will be controlled b the direction of the flow. The flow across d is ( V ˆn) }{{} component normal to surface d. Add up over all d elements to get the total flow across the surface Q = V ˆn d 15
This is called the surface integral of the vector field V over the surface defined b = g(, ). The actual evaluation is similar to the last eample. V ˆn will end up giving some integrand function f project d onto (, ) plane ( ) g 2 ( ) g 2 d = 1 + + dd ( ) g 2 ( ) g 2 Q = f(, ) 1 + + dd R, then proceed as before. Eample: 4.7 Given a velocit field in 3D space V = î(2 + ) + ĵ( 2 ) + ˆk() find a) the flow rate Q (m 3 /s) across the surface = 1 for 0 1 and 0 1 in the + ve direction b) the average velocit across the surface 1 d surface 1 1 16
Integral Theorems Involving Vector Functions We will eamine vector functions in 3D: Force Field: F (,, ) = îp + ĵq + ˆkR Vector Field: V (,, ) = îu + ĵv + ˆkw We have defined 2 tpes of integrals for such functions. Line Integrals F d r = P d + Qd + Rd F This can be interpreted as work done b F field on an object that moves along within the field. urface Integrals V V ˆn d This can be interpreted as the flow across surface in the ˆn direction due to the V field. There are three theorems which state identities involving these tpes of integrals. 17
1. Divergence Theorem V field in 3D space closed surface enclosing volume (on RH) Also called Green s theorem in space - this is the 2nd vector form of Green s theorem. V n d = V ( V )dv where V = velocit field V = volume 2. tokes Theorem Also called 1st vector from of Green s theorem. F d r = ( F ) ˆn d where the surface is an surface in 3D with as a boundar. closed curve in 3D space F defined in 3D space 3. Greens Theorem This is essentiall a 2D statement of toke s theorem, where in 2D F = îp + ĵq F = ˆk ( Q P ) P d + Qd = R F defined in 2D space ( Q P ) dd 18
Divergence Theorem Reading Trim 14.9 The Divergence Theorem Assignment web page assignment #11 Trim in section 14.9 has a detailed proof of the Divergence Theorem. The tr to interpret the meaning of V ˆnd = V ( V )dv This equation applies for an vector function V, but is used most for velocit fields in fluids. When we consider V, the theorem concerns net outflow to inflow (m 3 /s) for a region in space (like a sphere). The left side of the equation is a surface integral of V over a closed surface, in 3-D space with ˆn being the outward normal to each d. We recall that V ˆnd gives the flow rate (m 3 /s). When we add this up over the entire surface (as in the LH of the equation) we obtain the net flow rate crossing the closed surface. i.e. net outward flow(m 3 /s) net inward flow(m 3 /s) The right hand side of the equation is a calculation of V for each differential volume, dv inside the surface. We then add them all up. For the differential volume, V inflow (m 3 /s) = u() area + v()( )( ) outflow (m 3 /s) = u( + ) ( )( ) + v( + )( )( ) The net flow is then outflow inflow = ( )( )( ) 19 [ u( + ) u() + ] v( + ) v()
= (dv) = ( V )dv ( u + v ) Therefore the RH ( V )dv gives the net outflow minus inflow for a volume element dv. The integral V, adds up the differential flow for all volume elements, dv inside of surface. There is a cancellation of terms because the outflow from one V becomes the inflow to the net volume. When we sum over all V, we are left with the difference between the inflow and the outflow at the boundaries of the volume. V nd = V ( V )dv outflow inflow(m 3 /s) across boundar surface of volume V in 3D space triple sum of all outflow inflow for V volumes inside the volume V in 3D space Eample: 4.8 Given V = î(1 + ) + ĵ(1 + 2 ) + ˆk(1 + 3 ) verif the divergence theorem for a cube, where 0,, 1 V ˆnd = ( V )dv V i.e. show that where = cube surface (closed) V = interior volume of the cube 20
toke s Theorem Reading Trim 14.10 toke s Theorem Assignment web page assignment #11 The formal proof is offered in Trim 14.10. F d r = ( F ) ˆn d This has a similar meaning to Green s theorem but now in 3D space instead of a plane. F F (,, ) is a 3D force field. The LH of the equations is the work done when the object moves once in a W direction along the path in 3D. The RH is the work computed over a surface integral. Like Greens theorem, it works because of the interior cancellations of work, when we move around a surface element d inside. Note: the unit normal, ˆn for the LH is based on a right hand rule as follows. ^n an surface in 3D with as the boundar ^n d d d ^n if W on if W on The work on all internal surfaces cancel, leaving onl the surface work in the W direction. 21
Eample: 4.9 Given: F = î + ĵ2 + ˆk (a force field in 3D). The closed path is given b the intersection of: 2 + 2 = 4 = 4 The object moves once in a W direction around starting at (2, 0, 2). Verif toke theorem: F d r = ( F ) ˆn d 22
Green s Theorem Reading Trim 14.6 Green s Theorem Assignment web page assignment #10 Q F The theorem involves work done on an object b a 2D force field. The 2D force field is given b P F (, ) = îp (, ) + ĵq(, ) We will eamine an object that moves once W around a loop in F. The region inside the loop is defined as R. The normal vector for the region R is ˆk (outwards) to the right for a W motion of. The theorem states P d + Qd = R ( Q P ) dd The left hand side of the equation, F d r in 2D is the work done b the field F on the object as it moves on. This consists of force times distance for each d r added up over. dr F The right hand side of the equation, R ( F ) ˆkdd in 2D gives the direction of travel on for the work given b the LH. Maps movement of an element as it moves around from A to A. + A + ^ k out Q F P 23
P at + The force component times the distance for each side gives the work done. Q at Q at + P at A W ork = Q at + P at + Q at + P at = ( ) [ Q at + Q at P at + P ] at In the limit, the work done b F to move the object W around dd area is ( Q P ) dd Now we can sum up for all dd elements inside. There are some cancellations (+) (-) for all interior, paths. Therefore when we do we are left with the work terms on the boundaries of R. P d + Qd = R ( Q P ) dd R, work when we move once W around boundar curve sum of all work if move once W around all (dd) area elements of R inside 24
Eample: 4.10 Given a 2D force field, F (, ) = î( 3 ) + ĵ( 2 ) and a path in the field: Verif Green s theorem with P d + Qd = P = 3 Q = 2 R ( Q P ) dd 25