Vector Algebra and Calculus

Transcription

1 Vector Algebra and Calculus 1. Revision of vector algebra, scalar product, vector product 2. Triple products, multiple products, applications to geometry 3. Differentiation of vector functions, applications to mechanics 4. Scalar and vector fields. Line, surface and volume integrals, curvilinear co-ordinates 5.Vectoroperators grad,divandcurl 6. Vector Identities, curvilinear co-ordinate systems 7. Gauss and Stokes Theorems and extensions 8. Engineering Applications

2 4. Line, Surface and Volume Integrals I Westartedoff beingconcernedwithindividualvectorsa,b,c,andsoon. Wewenton toconsiderhowsinglevectorsvaryovertimeoroversomeotherparametersuchasarclength Inrestofthecourse,wewillbeconcernedwith scalars and vectors which are defined over regions in space Inthislectureweintroduce line, surface and volume integrals definition in curvilinear coordinates

3 Reminder about Scalar and vector fields 4.2 Ifascalarfunctionu(r)isdefinedateachrinsomeregion uisascalarfieldinthatregion. Examples: temperature,pressure,altitude,co 2 concentration Similarly, if a vector function v(r) is defined at each point, then visavectorfieldinthatregion. Examples: wind velocity, magnetic field, traffic flows, optical flow, electric fields Infieldtheoryouraimistoderivestatementsaboutbulkpropertiesofscalarandvectorfields(ratherthantodeal with individual scalars or vectors)

4 Line integrals through fields 4.3 Line integrals are concerned with measuring theintegratedinteractionwithafieldasyoumovethroughitonsomedefinedpath. Muche Filthe Greate Stench Vile Odours do pour out Eg,givenamapshowingthepollution density field in Oxford how much pollution would you breath in when cycling from college to the Department on different routes? Miasma in ye Turl Streete Vapours

5 Vector line integrals 4.4 PathLchoppedintovectorsegmentsδr i. Eachsegmentismultipliedbythefieldvalueatthatpoint in space, Products are summed. Three types 1:IntegrandU(r)isascalarfield. Integral is a vector. I= U(r)dr L r F(r) δ r 2:Integranda(r)isavectorfielddottedwithdr.Integral isascalar: I= a(r) dr L 3:Integranda(r)isavectorfieldcrossedwithdr.Integral is vector. I= a(r) dr L

6 Examples 4.5 TotalworkdonebyforceFasitmovespointfromAtoBalongpathC. InfinitessimalworkdoneisdW=F.dr,hencetotalworkis W C = F.dr Ampère slawrelatingmagneticfieldbtolinkedcurrentcanbewrittenas B.dr=µ 0 I C ForceonanelementofwirecarryingcurrentI,placedinamagneticfieldofstrengthB,isdF=Idr B. SototalforceonloopofwireC: F=I C C dr B Note: expressions above are beautifully compact in vector notation, and are all independent of coordinate system

7 Examples 4.6 Question:AforceF=x 2 yî+xy 2 ĵactsonabodyatitmovesbetween(0,0)and(1,1). Findworkdonewhenpathis 1.alongtheliney=x. 2.alongthecurvey=x n. 3.alongthexaxistothepoint(1,0)andthenalongthe linex=1 1 1,1 2 3 Answer: InplanarCartesians dr=îdx+ĵdy Thentheworkdoneis F dr= (x 2 yî+xy 2 ĵ) (îdx+ĵdy)= L L L 0,0 0,1 (x 2 ydx+xy 2 dy).

8 Example Path , ,0 0,1 PATH1:Forthepathy=xwefindthatdy=dx.Soitiseasiesttoconvertallyreferencestox. (1,1) (0,0) (x 2 ydx+xy 2 dy) = = x=1 x=0 x=1 NB! Although x, y involved these are NOT double integrals. Why not? x=0 (x 2 xdx+xx 2 dx) 2x 3 dx = [ x 4 /2 x=1 x=0 = 1/2.

9 Example Path ,1 1 PATH2:Forpathy=x n finddy=nx n 1 dx Again convert y references to x ,0 0,1 (1,1) (0,0) (x 2 ydx+xy 2 dy) = = = x=1 x=0 x=1 x=0 (x n+2 dx+nx n 1.x.x 2n dx) (x n+2 dx+nx 3n dx) 1 n+3 + n 3n+1

10 Example Path , PATH3:notsmooth,sobreakintotwo. Alongthefirstsection,y=0anddy=0, alongsecondsectionx=1anddx=0: 0,0 0,1 B A (x 2 ydx+xy 2 dy) = x=1 x=0 (x 2 0dx)+ = 0+ [ y 3 /3 y=1 y=0 = 1/3. y=1 y=0 Line integral depends on path taken 1.y 2 dy

11 Example Question2:Repeatpath(2),butnowusingthe ForceF=xy 2 î+x 2 yĵ. Answer 2: F (îdx+ĵdy)=xy 2 dx + x 2 ydy. Forthepathy=x n wefindthatdy=nx n 1 dx,so Thisisindependentofn,so (1,1) (0,0) (xy 2 dx+x 2 ydy) = = = x=1 x=0 x=1 x=0 (x 2n+1 dx+nx n 1 x 2 x n dx) (x 2n+1 dx+nx 2n+1 dx) 1 2n+2 + n 2n+2 = 1 2 This line is independent of path! Can we understand why?

12 Line integrals in Conservative fields 4.11 Write Then the perfect differential is g(x,y)=x 2 y 2 /2 dg = g x dx+ g y dy = y 2 xdx+x 2 ydy Soourlineintegral F dr= B A (y 2 xdx+yx 2 dy)= B A dg=g B g A Itdependssolelyonthevalueofgatthestartandendpoints,andnotatallonthepath Avectorfieldwhichgivesrisetolineintegralswhichareindependentofpathsiscalled a conservative field

13 Some questions about conservative fields 4.12 Onesortoflineintegralperformstheintegrationaroundacompleteloop.Itisdenoted 1.IfEisaconservativefield,whatisthevalueof E dr? 2.IfE 1 ande 2 isconservative,ise 1 +E 2 conservative? 3.Laterwewillshowthattheelectricfieldaroundapointchargeq is conservative. Are all electric fields conservative? E=Kqˆr r 2 K=1/4πǫ r ǫ 0 4.IfEistheelectricfield,thepotentialfunctionis φ= E dr. So are all electric fields conservative?

14 Line integrals involving scalar arc-length 4.13 I= L F(x,y,z)ds, wherepathlalongacurveisdefinedasx=x(p),y=y(p),z=z(p) These integrals don t appear to involve vectors, but they could be reformulated to! First,convertthefunctiontoF(p),writing where(lecture 3) ds dp = I= pend [ (dx ) 2 + dp Then do the(now straightforward) integral w.r.t. p. p start F(p) ds dp dp ( ) 2 ( ) ] 2 1/2 dy dz +. dp dp

15 Line integrals involving scalar arc-length: Special cases 4.14 I= L F(x,y,z)ds 1:Iftheparameterisarc-lengthsandthepathLisx=x(s),y=y(s),z=z(s). Convert the function to F(s), writing I= send s start F(s)ds 2:Ifpisx soy=y(x)andz=z(x)(orsimilarforp=yorp=z) [ xend ( ) 2 ( ) ] 2 1/2 dy dz I= F(x) 1+ + dx. x start dx dx

16 Surface integrals 4.15 Surface S is divided into infinitesimal vector elements of area ds: thedirnofthevectordsisthesurfacenormal ds its magnitude represents the area of the element. Again there are three possibilities: 1: S UdS scalarfieldu; vector integral. 2: S a ds vectorfielda; scalar integral. 3: S a ds vectorfielda; vector integral.

17 Physical example of surface integral 4.16 Physicalexamplesofsurfaceintegralsofteninvolvetheideaoffluxofa vector field through a surface a ds MassoffluidcrossingasurfaceelementdSatrintimedtis S dm=ρv dsdt Totalrateofgainofmasscanbeexpressedasasurfaceintegral: dm dt = ρ(r)v(r) ds Note again that expression is coordinate free. S

18 Example 4.17 Question:Evaluate F dsoverthex=1sideofthecubeshowninthefigurewhenf=yî+zĵ+xˆk. Answer: ds is perp to the surface. Often, the surface will enclose a volume, and the surface direction is everywhere outofthevolume Forthex=1faceofthecube S F ds = = ds=dydzî (yî+zĵ+xˆk) dydzî y=1 z=1 y=0 z=0 = 1 2 y2 1 0 z 1 0 = 1 2. ydydz 1 z 1 y 1 d S = dy dz i x d S

19 Volume integrals 4.18 Thedefinitionofthevolumeintegralisagaintakenasthelimitofasumofproductsasthesizeofthevolume element tends to zero. Oneobviousdifferencethoughisthattheelementofvolumeisascalar. The possibilities are: 1: V 2: V U(r)dV scalarfield;scalarintegral(1p1stuff!) a(r)dv vectorfield;vectorintegral.inthiscaseonecantreateachcomponentseparately. adv = a 1 (x,y,z)îdv+ a 2 (x,y,z)ĵdv+ a 3 (x,y,z)ˆkdv V V V V = î a 1 (x,y,z)dv+ĵ a 2 (x,y,z)dv+ˆk a 3 (x,y,z)dv So,3 1P1stuff. V V V

20 Changing variables: curvilinear coordinates 4.19 Before dealing with further examples of line, surface and volume integrals......itisimportanttounderstandhowtoconvertanintegralfromonesetofcoordinatesintoanothermoresuited to the geometry or symmetry of the problem. Yousawhowtodothisforscalarvolumeintegralsin1P1(andwe veseenthatvolumeintegralscanalwaysbe handledasscalars)......butweneedtounderstandwherejacobianscamefrom,andhowwecanapplythemechanismmoregenerally. Youmayfindthegeneralproblemslightlyheavygoing......thegoodnewsisthatwewilldealwithsomespecialcasesfirstthatshouldbeatleastvaguelyfamiliar

21 Changing variables: curvilinear coordinates 4.20 A line integral in Cartesian coordinates used r=xî+yĵ+zˆk and dr=dxî+dyĵ+dzˆk Youcanbesurethatlengthscalesareproperlyhandledbecause dr =ds= dx 2 +dy 2 +dz 2. But often symmetry screams at you to use another coordinate system: likely to be plane, cylindrical, or spherical polars, butcanbesomethingmoreexoticlike u,v,w A u,v,w, r,φ,θ systemisacurvilinearcoordinatesystem Buthere sthebadnews:lengthscalesarescrewedup r uû+vˆv+wŵ dr duû+dvˆv+dwŵ dr =ds du 2 +dv 2 +dw 2.

22 Cylindrical polar coordinates 4.21 z ê z ê φ z Lines of constant r P r ê r Lines of constant z ˆk î ĵ r y r y x φ x Lines of constant φ r = xî+yĵ+zˆk=rcosφî+rsinφĵ+zˆk r r represents direction in which(instantaneously) r changing while other two coords stay const. It is tangent to surfacesofconstantφandz

23 z ê z ê φ z Lines of constant r P r ê r Lines of constant z ˆk î ĵ r y r y x φ x Lines of constant φ r = rcosφî+rsinφĵ+zˆk e r = r r =cosφî+sinφĵ e φ = r φ = rsinφî+rcosφĵ r ˆ form a basis set for infinitessimal vector displacements in the positionofp,dr.

24 Cylindrical polars: scale factors 4.22 More usual to normalise the basis vectors ê r,ê φ,ê z dr = r r dr+ r φ dφ+ r z dz = dre r +dφe φ +dze z = drê r +rdφê φ +dzê z NOTE: In cylindrical polars, small change dφ keeping r and z constant results in displacement of ds= dr = r 2 (dφ) 2 =rdφ THUS: size of(infinitessimal) displacement depends on value of r risscalefactorormetriccoefficient Incylindricalpolarsscalefactorsare1,rand1.

25 Example: line integral in cylindrical polars 4.23 Question:Evaluate C a dl,wherea=x3 ĵ y 3 î+x 2 yˆkandcisthecircleofradiusρinthez=0plane,centred on the origin. z Answer: a=ρ 3 ( sin 3 φî+cos 3 φĵ+cos 2 φsinφˆk) and(sincedz=dr=0onthepath) x φ ρ y dl=ρdφê φ so that C a dl= dl = ρdφê φ = ρdφ( sinφî+cosφĵ) 2π 0 ρ 4 (sin 4 φ+cos 4 φ)dφ= 3π 2 ρ4

26 Volume integrals in cylindrical polars 4.24 z y dxî dv =dxdydz dv =rdrdφdz x dyĵ dzˆk dzê z drê r rdφê φ y In Cartesians, volume element given by x z φ dv =dxî.(dyĵ dzˆk)=dxdydz dφ rdφ In cylindrical polars, volume element given by dv =drê r.(rdφê φ dzê z )=rdφdrdz Note: Volume is scalar triple product, hence can be written as a determinant: ê r dr x y z dv = ê φ rdφ ê z dz = r r r x y z φ φ φ drdφdz x z y z z z

27 Surface integrals in cylindrical polars 4.25 z ds z rdφêφ drê r In Cartesians, for surface element with normal î we have ds = dyĵ dzˆk=dydzî=îds dzê z dzê z ds r Cylindrical polars: surface area elements given by: ds z = drê r rdφê φ =rdrdφê z drê r rdφê φ ds r = rdφê φ dzê z =rdφdzê r ds φ y x

28 Spherical polars 4.26 î z ˆk r θ P ĵ ê r ê θ ê φ y Lines of constant (longitude) φ z Lines of constant r y x φ x Lines of constant (latitude) θ r = xî+yĵ+zˆk=rsinθcosφî+rsinθsinφĵ+rcosθˆk e r = r r = sinθcosφî+sinθsinφĵ+cosθˆk =ê r e θ = r θ = rcosθcosφî+rcosθsinφĵ rsinθˆk =rê θ e φ = r φ = rsinθsinφî+rsinθcosφĵ =rsinθê φ

29 Spherical polars: scale factors 4.27 e r = r r = sinθcosφî+sinθsinφĵ+cosθˆk =ê r e θ = r θ =rcosθcosφî+rcosθsinφĵ rsinθˆk=rê θ e φ = r φ = rsinθsinφî+rsinθcosφĵ =rsinθê φ Small displacement dr given by: dr = r r dr+ r θ dθ+ r φ dφ = dre r +dθe θ +dφe φ = drê r +rdθê θ +rsinθdφê φ Thus,metriccoefficientsare1,r,rsinθ.

30 Volume integrals in spherical polars 4.28 z drê r rsinθ rsinθdφê φ dφ dv =r 2 sinθdrdθdφ rdθê θ θ r dθ y x φ dφ rsinθdφ Volume element given by dv =drê r.(rdθê θ rsinθdφê φ )=r 2 sinθdrdθdφ Noteagainthatthisvolumecouldbewrittenasadeterminant

31 Surface integrals in spherical polars 4.29 Three possibilities, but most useful are surfaces of constant r ThesurfaceelementdS r isgivenby ds r = rdθê θ rsinθdφê φ = r 2 sinθdθdφê r z rsinθdφê φ ds r =r 2 sinθdθdφê r rdθê θ y x

32 Example: Surface integral in spherical polars 4.30 QEvaluate S a ds,wherea=z3ˆkandsisthesphereofradiusacentredontheorigin. A In general: z=rcosθ ds=r 2 sinθdθdφê r Onsurfaceofthesphere,r=A,sothat a=a 3 cos 3 θˆk ds=a 2 sinθdθdφê r Hence S a ds = 2π φ=0 π θ=0 2π = A 5 dφ 0 = 2πA 51 5 A 3 cos 3 θa 2 sinθ[ê r ˆk]dθdφ π 0 cos 3 θsinθ[cosθ]dθ [ cos 5 θ ] π 0 =4πA5 5

33 General curvilinear coordinates 4.31 Suppose So and [similarlyforpartialswithrespecttovandw] x=x(u,v,w), y=y(u,v,w), z=z(u,v,w) r=x(u,v,w)î+y(u,v,w)ĵ+z(u,v,w)ˆk r u = x uî+ y uĵ+ z uˆk dr= r u du+ r r dv+ v w dw

34 Curvilinear coords/ctd 4.32 Postionvector: Infinitessimal displacement: Local basis: r=x(u,v,w)î+y(u,v,w)ĵ+z(u,v,w)ˆk dr= r u du+ r r dv+ v w dw e u = r u = e v = r v = e w = r w = r u êu=h u ê u r v êv=h v ê v r w êw=h w ê w

35 Curvilinear coords/ctd 4.33 Metric coefficients h u = r u,h v= r v and h w= r w Volume element dv =h u duê u.(h v dvê v h w dwê w ) andsimplifiesifthecoordinatesystemisorthonormal(sinceê u.(ê v ê w )=1)to dv =h u h v h w dudvdw Surfaceelement(normaltoconstantw,say) ds=h u duê u h v dvê v and simplifies if the coordinate system is orthogonal to ds=h u h v dudvê w

36 Summary 4.34 We introduced line, surface and volume integrals involving vector fields. We defined curvilinear coordinates, and realized that metric coefficient were necessary to relate change in an arbitrary coordinate to a length scale. We showed in detail how line, surface and volume elements are derived, and how the results specialized for orthogonal curvilinear system, in particular plane, cylindrical and spherical polar coordinates. TheoriginoftheJacobianwasclarified.