3 VECTOR CALCULUS I. 3.1 Divergence and curl of vector fields. 3.2 Important identities

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1 3 VECTOR CALCULU I 3.1 Divegence and cul of vecto fields Let = ( x, y, z ). eveal popeties of φ fo scala φ wee intoduced in section 1.2. The gadient opeato may also be applied to vecto fields. Let F = (F 1,F 2,F 3 ) = F 1 i+f 2 j+f 3 k, whee F 1 = F 1 () = F 1 (x,y,z) etc. y B e.g. wind velocity F A The divegence is a scala measuing net flux of the field fom each point. x div(f) = F = x F 1 + y F 2 + z F 3 (3.1) Fo point A of the figue, F > 0. Fo conveging vectos, F < 0. Nea B, F 0. The cul is a vecto giving the magnitude and axis of otation about each point. i j k cul(f) = F = x y z F 1 F 2 F 3 = ( yf 3 z F 2, z F 1 x F 3, x F 2 y F 1 ) (3.2) The otation nea B defines a vecto F pointing out of the page. The vecto would point into the page fo otation in the opposite diection. Nea A, F 0. NOTE: Remembe that is a deivative opeato. Just like x f f x, also F F. x f and F ae scalas, wheeas f x and F ae opeatos. The latte is simila to the diectional deivative of section 1.2. Woked Example 3.1 (A) Conside F =. how that = 3 and = 0. (B) Let F = xj. how that F = 0 and F = k. 3.2 Impotant identities Fo any scala field φ, and fo any vecto field F, φ = 0. (3.3) ( F) = 0. (3.4) 10

2 Futhe, it can be shown that fo a volume V, and F = 0 in V ψ in V s.t. F = ψ, (3.5) B = 0 in V A in V s.t. B = A. (3.6) If F = 0, then F is said to be iotational and may be defined in tems of a scala potential ψ. If B = 0, then B is said to be solenoidal o divegence-fee, and may be defined via a vecto potential A. Example (Poof of (3.3)) By definition, φ = ( x φ, y φ, z φ). Applying the cul (3.2), φ = ( y z φ z y φ,...,...). But y z yz 2 y z 2 z y φ = The Laplacian opeato The Laplacian opeato is defined The Laplace equation, 2 φ ( φ) = ( xx + yy + zz )φ. (3.7) 2 φ = 0. (3.8) fequently aises in applications. The Laplacian may be applied to a vecto. In Catesian coodinates we have that 2 F = ( 2 F 1, 2 F 2, 2 F 3 ) = ( 2 F 1 )i+( 2 F 2 )j +( 2 F 3 )k. (3.9) Example Veify that φ = ln(x 2 +y 2 ) satisfies the Laplace equation. xx φ = Also z φ = 0. Togethe, x φ = 1 x 2 +y 2 2x, 1 x 2 +y x (x 2 +y 2 ) 2 2x = 2(x2 +y 2 ) 4x 2 (x 2 +y 2 ) 2 = 2(y2 x 2 ) (x 2 +y 2 ) 2. By symmety yy φ = 2(x2 y 2 ) (x 2 +y 2 ) 2 yy φ = xx φ. 2 φ = xx φ+ yy φ+ zz φ = 0. Woked Example 3.3 how that 2 f = 1 d 2 d (2 d d f) if f = f(). Hence find the most geneal f = f() s.t. 2 f = 0. 11

3 3.4 Othe impotant opeatos The constuction ( F) is sometimes called the double cul, and in Catesian coodinates it can eadily be shown that ( F) = ( F) 2 F. (3.10) The advective deivative, often just called v dot gad, is given by v = v 1 x +v 2 y +v 3 z. (3.11) It is simila to the diectional deivative, ˆv (see section 1.2), and, like the Laplacian, can be applied to both scala and vecto fields: (v )φ = (v 1 x +v 2 y +v 3 z )φ, (v )u = [(v )u 1 ]i+[(v )u 2 ]j +[(v )u 3 ]k. Woked Example 3.4 Find (u )u fo the case u = (U/L)(x, y, 0), whee U and L ae constants. 3.5 uffix notation and the summation convention Definitions The Konecke delta, δ ij is defined fo i = 1,2,3 and j = 1,2,3: { 1 if i = j δ ij = 0 if i j The altenating symbol, ǫ ijk, is defined +1 if (i,j,k) = (1,2,3), (2,3,1), (3,1,2) ǫ ijk = 1 if (i,j,k) = (3,2,1), (2,1,3), (1,3,2) 0 fo any othe (i,j,k) (when 2 o moe indices equal) Fo example, δ 32 = 0, δ 22 = 1, ǫ 132 = 1, ǫ 221 = 0, ǫ iik = 0. Obseve that switching the ode of indices on ǫ eveses the sign, e.g. ǫ jki = ǫ ikj The summation convention Let x 1 = x, x 2 = y, x 3 = z, so that i = x i. imilaly, let e 1 = i, e 2 = j, e 3 = k. When an index, e.g. i, occus only once in a tem of an equation it is called a fee suffix, and it epesents all possibilities fom 1 to 3. Examples: x i = 5 x 1 = x 2 = x 3 = 5. a ij = δ ij p a ij = 0 fo i j and a ij = p fo i = j. When an index occus twice in a tem of an equation, the tem is summed fom 1 to 3. Example: a k b k = a 1 b 1 +a 2 b 2 +a 3 b 3 = a b. 12

4 Example (Impotant summation popeties) (a) δ jj = 3 (b) ǫ ijk ǫ ijk = 6 (c) a b = a i b i (d) = x i e i (e) (a b) i = ǫ ijk a j b k (f) a b c = a i ǫ ijk b j c k Thee ae a couple of ules to check that the notation makes sense! : No index may occu moe than twice in a single tem of an equation. E.g. a i b i c i is meaningless. If fee suffices appea, then they must be the same in evey tem of the equation. E.g. a j = c j and b ij a i +c j = 0 ae ok, but c j = b ij a j is meaningless. Using this convention, we may wite new foms fo the opeatos of the pevious sections. (1) ( φ) i = φ x i Let i = x i = i. (2) F = F i x i (4) 2 φ = 2 φ x j x j φ (6) (v )φ = v j x j (3) ( F) i = ǫ ijk F k x j (5) ( 2 F) i = 2 F i x j x j (7) [(v )u] i = v j u i x j Woked Example 3.5 (A) Use the suffix notation to show that (φv) = φ v + φ v The substitution popety of δ ij Conside the tem δ ij a j, whee summation ove j is implied. Now, δ ij is non-zeo only fo one case, j = i. Theefoe we may simplify: δ ij a j = a i. (3.12) In othe wods, if a delta has a summed index, then the delta may be omitted, and in the est of the tem the summed suffix is substituted by the othe index of the delta. Example: δ ip ǫ ijk = ǫ pjk. (i summed, eplace by p) Woked Example 3.5 how that (B) δ ip δ jq c p d q = c i d j (C) δ ij b ij = b ii = b jj (D) δ ij ǫ ijk = ǫ iik = 0 13

5 3.5.4 ome useful esults ǫ ijk ǫ klm = δ il δ jm δ im δ jl (3.13) Note the epeated suffix k in the popety. The emaining suffices i,j,l,m ae fee in geneal, none of them occus moe than once in the same tem. The double-epsilon usually aises in the pesence of two vecto-poducts. Example {a (b c)} i = ǫ ijk a j (b c) k = ǫ ijk a j ǫ klm b l c m = ǫ ijk ǫ klm a j b l c m = (δ il δ jm δ im δ jl )a j b l c m = δ il }{{} l i δ jm a j b l c m δ }{{}}{{} im m j m i i = 1,2 o 3 a (b c) = (a c)b (a b)c. δ jl a j b l c m = a j b i c j a j b j c i = (a c)b i (a b)c i }{{} l j Othe useful esults include: x i / x j = δ ij, x i x i = 2, ǫ ijk δ jk = 0, ǫ ijk x j x k = 0, ǫ ijk (a j b k +a k b j ) = 0. (3.14) Fo the fist two, ecallx 1 =x, x 2 =y,x 3 =z. The thid esult follows fom the substitution popety. The penultimate esult states ( ) i = 0 and the last that (a b) i +(b a) i = 0. Woked Example 3.5 (E) how that (Ω ) = 2Ω fo a constant vecto Ω. Example Find 2 φ when φ = (a )f(), whee a is a constant vecto. φ = a j x j f(), φ x j df = a j f +a j x j x i x }{{} i d x }{{} i δ ij j i x i / j i = a i f +a j x j x i f 2 φ = 2 φ df f = a i +a j δ ij x i x i x i d x i }{{} +a f jx j δ }{{} ii +a d jx j x i d =3 On the last tem, x i x i = x 2 i = 2. Then 2 φ = 5a i x i f +a jx j d d ( ) f = (a ) ) (f +4 f ( ) f x i }{{} x i / 14

6 4 VECTOR CALCULU II 4.1 Cylindical pola and pheical pola coodinates Fo many poblems the Catesian coodinate system is not the most natual. Fo example a full sphee is easily descibed by limits on, as 0 < < R, but the sphee is cumbesome to descibe in tems of limits on x, y and z. The following section descibes the two most common altenatives to the Catesian system. It is impotant to be familia with the following diagams: z P z P O z O θ (CPs) x θ y (Ps) x φ y Fo Cylindical Polas (CPs), (, θ, z): x = cosθ, y = sinθ, z = z, = (x 2 +y 2 ) 2, 1 θ = tan 1 y, x 0, 0 θ 2π (4.1) Fo pheical Polas (Ps), (, θ, φ): x = sinθcosφ, y = sinθsinφ, z = cosθ, { } = (x 2 +y 2 +z 2 ) 2, 1 θ = cos 1 z, φ = tan 1 y x (x 2 +y 2 +z 2 ) 1 2 0, 0 θ π, 0 φ 2π (4.2) In Ps, is the distance fom the oigin O, as usual. In CPs, is the distance fom the z-axis, the cylindical-adius. = ˆ (Ps). = ˆ +zẑ (CPs). Note that θ (CPs) φ (Ps). When confusion might be possible, CPs ae sometimes witten (s,φ,z) athe than (,θ,z). Let the coodinates (α 1, α 2, α 3 ) denote e.g. (x,y,z) o (,θ,z), o one of the many othogonal coodinate systems. ˆα 1, ˆα 2, ˆα 3 denote unit vectos, e.g. CPs ˆα 1 = ˆ, ˆα 2 = ˆθ, ˆα 3 = ẑ. At evey point the unit vectos fo the coodinates ae othogonal, i.e. ˆα i ˆα j = 0 if i j, and they ae odeed such that ˆα 1 ˆα 2 = ˆα 3, etc., e.g. in Catesian coodinates i j = k. Woked Example 4.1 (A) ketch coodinate sufaces, coesponding to α i = const, fo CPs and Ps. (B) ketch the unit vectos fo CPs. 15

7 4.2 The line element evisited Let P and P be neighbouing points, with OP = and OP = +δl. The notations δ and δs ae also commonplace. o fa we ve used the fome, but if thee is potential fo ambiguity, when is a coodinate diection, then δl is safe. In Catesians we wite the small change in position δl = δxi + δyj + δzk. In geneal we wite, δl = δl 1 ˆα 1 +δl 2 ˆα 2 +δl 3 ˆα 3 = h 1 δα 1 ˆα 1 +h 2 δα 2 ˆα 2 +h 3 δα 3 ˆα 3. (4.3) As δl is a change in position, each of δl 1 = h 1 δα 1 etc. must be a length. This defines the scale factos h i, and whee δα i is an angle h i must be a length. Fo Catesians we have h 1 = h 2 = h 3 = 1. It can be shown that fo Woked Example 4.2 Check (4.4) via sketches. CPs (,θ,z): h 1 = 1, h 2 =, h 3 = 1, Ps (,θ,φ): h 1 = 1, h 2 =, h 3 = sinθ. (4.4) Example (Line integal on a cuved path in CPs) y δl B F = 2i a A x Path: θ = α 2 δl = h 2 δα 2 αˆ 2 = δθ ˆθ, with = a. Then B A π π F δl = 2a i ˆθdθ = 2a sinθdθ = 4a Genealised gadient opeatos The physical intepetations emain unchanged fo gad, div and cul. The divegence gives the same scala, gad and cul give the same vecto, but the epesentation will look diffeent in anothe coodinate system. Fo cuvilinea othogonal coodinates (α 1,α 2,α 3 ), not all of the α i have the dimension of length and the unit vectos ˆα i ae not constant. Apat fom the genealised gadient, poofs of the following ae lengthy. Gadient: The gadient gives the change in the scala φ fo a change in position δl. Theefoe ( φ) 1 δφ = δφ δl 1 h 1 δα 1 φ = ( 1 h 1 φ α 1, 1 h 2 φ α 2, 1 h 3 φ α 3 ) (4.5) Divegence: Cul: F = { 1 (h 2 h 3 F 1 )+ (h 1 h 3 F 2 )+ } (h 1 h 2 F 3 ) h 1 h 2 h 3 α 1 α 2 α 3 1 h 1 ˆα 1 h 2 ˆα 2 h 3 ˆα 3 F = h 1 h 2 h 3 α 1 α 2 α 3 h 1 F 1 h 2 F 2 h 3 F 3 (4.6) (4.7) 16

8 cala Laplacian: By definition, 2 φ = ( φ). This can be calculated fom the above to be { ( ) 2 1 h2 h 3 φ φ = + ( ) h1 h 3 φ + ( )} h1 h 2 φ h 1 h 2 h 3 α 1 h 1 α 1 α 2 h 2 α 2 α 3 h 3 α 3 (4.8) Vecto Laplacian: The following expansion fomulae, which can be shown to hold fo Catesians, we insist holds fo all coodinate systems: 2 F ( F) ( F), (4.9) The RH can be calculated fom fomulae above. The esult 2 F = ( 2 F i ) ˆα i holds fo Catesians but not in geneal. 4.4 Volume and suface elements h 2 δα 2 δv δ h 1 δα 1 h 3 δα 3 Fom the pevious sections, the line element is given byδl = (h 1 δα 1, h 2 δα 2, h 3 δα 3 ). It follows diectly that a volume element is given by δv = h 1 h 2 h 3 δα 1 δα 2 δα 3. The volume integal fo f = f(α 1,α 2,α 3 ) is then f dv = f h 1 h 2 h 3 dα 1 dα 2 dα 3. (4.10) V Note that δ denotes a small but finite element, and that the infinitesimal is denoted d in the integal. Conside a suface defined by α 1 = const. A suface element is geneated by vaiations in the othe two vaiables in this case. Hence, δ = h 2 h 3 δα 2 δα 3. The suface integal fo scala g = g(α 1,α 2,α 3 ) in this case is gd = g h 2 h 3 dα 2 dα 3. (4.11) (In pevious couses the pefactos h 2 h 3 etc. may have been deived via the Jacobian deteminant. This is easie.) 17

9 Often we want to know the flux of a vecto field passing though a suface. δ F The vecto suface element δ = ˆnδ, whee ˆn is a nomal to the suface. We may now conside the component of a vecto field F in the diection of ˆn and wite F d = F ˆnd = F cosθd. (4.12) The ightmost fom should be used with cae thee θ is the angle between F and the unit vecto ˆn, NOT to be confused with θ of a coodinate system. Convention is to take δ to be an outwads nomal fo a closed suface, i.e. when encloses a volume V. If is not closed then the diection should be specified. Example (Volume integal) V is a cylinde bounded by = a, z = 0 and z = h. Find I = V 2z(z2 + 2 )dv. V cylinde CPs: h 1 = h 3 = 1, h 2 =. δv = δδθδz. I = = 4π 2z(z )dv = 2 V h z=0 ( h 4 = 4π [z ] a 4 +z 4.a2 2 + h2 2.a4 4 4 ) =0 h z=0 dz δθ 2π a θ=0 =0 = π 2 h2 a 2 (h 2 +a 2 ). δ δθ δz (z 3 +z 2 )ddθdz Example (uface integal) Evaluate d whee is the spheical suface define by = a. = a Ps: α 1 = const, h 2 =, h 3 = sinθ. a δ δ = ˆ, δ = 2 sinθδθδφ ˆ, so is paallel to δ in this case. Theefoe, δ = a 3 sinθδθδφ on. d = 2π π 0 0 a 3 sinθdθdφ π = 2πa 3 sinθdθ = 4πa

10 Woked Example 4.4 (A) Find the suface aea of a sphee of adius a. (B) A spheical shell has inne adius a, oute adius b and density pofile ρ() = ρ 0 a/, whee ρ 0 is a constant. Find its mass m. Example (The hell Theoem) z O m d θ a δ Conside a thin spheical shell of adiusaand unifom mass pe unit aea σ. total mass of shell M = 4πa 2 σ. mass of suface element δm = σ δ. Let mass m be a distance z fom the cente of the shell. Then with z = zk, a = aˆ and ˆ k = cosθ, we have M z +d = a d = a z d 2 = (a z) (a z) = a a+a a 2a z = a 2 +z 2 2azcosθ. The contibution to the gavitational potential fo each small suface element is δv = GmδM d, and δm = σδ, δ = 2 sinθδθδφ, with = a. The total potential is then V = δv = Gmσ (δ/d) V = Gmσ 2π π Let u = cosθ, du = sinθdθ 0 0 a 2 sinθdθdφ d π = 2πa 2 sinθdθ σgm 0 a 2 +z 2 2azcosθ. 1 [ V = 1 2 MGm du 1 1 a 2 +z 2 2azu = 1 2 GMm 1 a az 2 +z 2azu] 2 1 a 2 +z 2 2az = (a z) 2 = a z, a,z > 0 Finally we have the esult! : V = GMm 2az { a z (a+z)} If z a, then a z = (a z) {(a z) (a+z)} = 2z. If z a, then a z = (z a) {(z a) (a+z)} = 2a. If z a, then V = GMm a If z a, then V = GMm z = const inside the sphee, foce F = V = 0. outsize the sphee, F same as fo point mass M at oigin. 19

11 4.5 Gauss Theoem (o the Divegence Theoem) Let F be a vecto field, and be a closed suface, enclosing a volume V. Then Gauss Theoem states that F d = F dv. (4.13) As is closed, d is the outwads nomal. The theoem descibes the balance between the flux though the suface, on the LH, and the total divegence within a volume, on the RH. Woked Example 4.5 Veify the theoem holds whee (A) V is a sphee of adius a, cented on the oigin, and F =. (B) V is the cylinde bounded by = a, z = 0, z = h, and F = xz 3 i+yz 3 j +z 2 (x 2 +y 2 )k. V Coollay of Gauss Theoem Let F = φa, whee φ is a scala field and a is an abitay constant vecto. Expanding F = a φ+φ a, but a = 0 as a is constant. ubstitution into Gauss Theoem gives a φd = a φdv. As a is abitay we have φd = φ dv (4.14) Note that this is a set of thee equations, one fo each component of d and φ. V 4.6 tokes Theoem d C dl Let be an open suface, bounded by the closed cicuit C. Then tokes Theoem states that F dl = F d. (4.15) Woked Example 4.6 Veify tokes Theoem fo Example C Example how that wok done B A F dl is independent of the path fomatob fo a consevative foce using tokes Theoem. As F = V, a consevative foce is iotational: F = V = 0. Let C 1 and C 2 be two diffeent paths fom A to B, let C be the closed path C 1 C 2 and be a suface enclosed by C. Then F dl = F d = 0 F dl = F dl F dl = 0 F dl = F dl C 1 C 2 C 1 C 2 C C Hence the integals along diffeent pathsc 1 andc 2 fom A tob ae the same. Given that the paths ae abitay, the wok done must be independent of the path. 20

As is natural, our Aerospace Structures will be described in a Euclidean three-dimensional space R 3.

As is natural, our Aerospace Structures will be described in a Euclidean three-dimensional space R 3. Appendix A Vecto Algeba As is natual, ou Aeospace Stuctues will be descibed in a Euclidean thee-dimensional space R 3. A.1 Vectos A vecto is used to epesent quantities that have both magnitude and diection.