3 VECTOR CALCULUS I 3.1 Divergence and curl of vector fields Let ∇ =(∂x, ∂y, ∂z). Several properties of ∇φ for scalar φ were introduced in section 1.2. The gradient operator may also be applied to vector fields. Let F =(F1, F2, F3)= F1i + F2j + F3k, where F1 = F1(r)= F1(x, y, z) etc. y B ∇ F e.g. wind velocity × A x The divergence is a scalar measuring net flux of the field from each point. div(F )= ∇ F = ∂ F + ∂ F + ∂ F (3.1) · x 1 y 2 z 3 For point A of the figure, ∇ F > 0. For converging vectors, ∇ F < 0. Near B, ∇ F 0. · · · ≈ The curl is a vector giving the magnitude and axis of rotation about each point. i j k curl(F )= ∇ F = ∂ ∂ ∂ =(∂ F ∂ F , ∂ F ∂ F , ∂ F ∂ F ) (3.2) × x y z y 3 − z 2 z 1 − x 3 x 2 − y 1 F F F 1 2 3 The rotation near B defines a vector ∇ F pointing out of the page. The vector would point into the page for rotation in the opposite× direction. Near A, ∇ F 0. × ≈ NOTE: Remember that ∇ is a derivative operator. Just like ∂ f = f∂ , also ∇ F = F ∇. x 6 x · 6 · ∂xf and ∇ F are scalars, whereas f∂x and F ∇ are operators. The latter is similar to the directional derivative· of section 1.2. · Worked Example 3.1 (A) Consider F = r. Show that ∇ r = 3 and ∇ r = 0. · × (B) Let F = x j. Show that ∇ F = 0 and ∇ F = k. · × 3.2 Important identities For any scalar field φ, ∇ ∇φ = 0 . (3.3) × and for any vector field F , ∇ (∇ F )=0 . (3.4) · × 10 Further, it can be shown that for a volume V , ∇ F = 0 in V ψ in V s.t. F = ∇ψ , (3.5) × ⇐⇒ ∃ and ∇ B =0 in V A in V s.t. B = ∇ A . (3.6) · ⇐⇒ ∃ × If ∇ F = 0, then F is said to be ‘irrotational’ and may be defined in terms of a scalar potential× ψ. If ∇ B =0, then B is said to be ‘solenoidal’ or ‘divergence-free’, and may be defined via a vector· potential A. Example 3.2.1 (Proof of (3.3)) By definition, ∇φ = (∂xφ, ∂yφ, ∂zφ) . Applying the curl (3.2), ∇ ∇φ = (∂ ∂ φ ∂ ∂ φ, ..., ... ) . × y z − z y But ∂2 ∂2 ∂ ∂ ∂ ∇ ∇φ = 0 . y z ≡ yz ≡ ∂y∂z ≡ ∂z∂y ⇒ × 3.3 The Laplacian operator The Laplacian operator is defined 2φ ∇ (∇φ)=(∂ + ∂ + ∂ ) φ . (3.7) ∇ ≡ · xx yy zz The Laplace equation, 2φ =0 . (3.8) ∇ frequently arises in applications. The Laplacian may be applied to a vector. In Cartesian coordinates we have that ∇2F = ( 2F , 2F , 2F )=( 2F ) i +( 2F ) j +( 2F ) k . (3.9) ∇ 1 ∇ 2 ∇ 3 ∇ 1 ∇ 2 ∇ 3 Example 3.3.1 Verify that φ = ln(x2 + y2) satisfies the Laplace equation. 1 ∂ φ = 2x , x x2 + y2 · 1 2x 2(x2 + y2) 4x2 2(y2 x2) ∂ φ = 2+ − 2x = − = − . xx x2 + y2 · (x2 + y2)2 · (x2 + y2)2 (x2 + y2)2 2(x2 y2) By symmetry ∂ φ = − ∂ φ = ∂ φ . yy (x2 + y2)2 ⇒ yy − xx Also ∂zφ = 0. Together, 2φ = ∂ φ + ∂ φ + ∂ φ = 0 . ∇ xx yy zz Worked Example 3.3 Show that 2f = 1 d (r2 d f) if f = f(r). Hence find the most general f = f(r) s.t. 2f = 0. ∇ r2 dr dr ∇ 11 3.4 Other important operators The construction ∇ (∇ F ) is sometimes called the ‘double curl’, and in Cartesian coor- dinates it can readily× be shown× that ∇ (∇ F )= ∇(∇ F ) ∇2F . (3.10) × × · − The advective derivative, often just called ‘v dot grad’, is given by v ∇ = v ∂ + v ∂ + v ∂ . (3.11) · 1 x 2 y 3 z It is similar to the directional derivative, vˆ ∇ (see section 1.2), and, like the Laplacian, can be applied to both scalar and vector fields:· (v ∇) φ = (v ∂ + v ∂ + v ∂ ) φ , · 1 x 2 y 3 z (v ∇) u = [(v ∇) u ] i + [(v ∇) u ] j + [(v ∇) u ] k . · · 1 · 2 · 3 Worked Example 3.4 Find (u ∇)u for the case u = (U/L)(x, y, 0), where U and L are constants. · − 3.5 Suffix notation and the summation convention 3.5.1 Definitions The Kronecker delta, δij is defined for i =1, 2, 3 and j =1, 2, 3: 1 if i = j δ = ij 0 if i = j 6 The alternating symbol, ǫijk, is defined +1 if (i, j, k)=(1, 2, 3), (2, 3, 1), (3, 1, 2) ǫ = 1 if (i, j, k)=(3, 2, 1), (2, 1, 3), (1, 3, 2) ijk − 0 for any other (i, j, k) (when 2 or more indices equal) For example, δ =0, δ =1, ǫ = 1, ǫ =0, ǫ =0. Observe that switching the order 32 22 132 − 221 iik of indices on ǫ reverses the sign, e.g. ǫ = ǫ . jki − ikj 3.5.2 The summation convention Let x = x, x = y, x = z, so that r = x . Similarly, let e = i, e = j, e = k . • 1 2 3 i i 1 2 3 When an index, e.g. i, occurs only once in a term of an equation it is called ‘a free • suffix’, and it represents all possibilities from 1 to 3. Examples: x =5 x = x = x =5. i ⇒ 1 2 3 a = δ p a =0 for i = j and a = p for i = j. ij ij ⇒ ij 6 ij When an index occurs twice in a term of an equation, the term is summed from 1 to 3. • Example: a b = a b + a b + a b = a b. k k 1 1 2 2 3 3 · 12 Example 3.5.1 (Important summation properties) (a) δjj = 3 (b) ǫijkǫijk = 6 (c) a b = a b · i i (d) r = xiei (e) (a b) = ǫ a b × i ijk j k (f) a b c = a ǫ b c · × i ijk j k There are a couple of rules to check that the notation makes sense! : No index may occur more than twice in a single term of an equation. • E.g. aibici is meaningless. If free suffices appear, then they must be the same in every term of the equation. • E.g. aj = cj and bijai + cj =0 are ok, but cj = bijaj is meaningless. Using this convention, we may write new forms for the operators of the previous sections. ∂ Let ∇i = = ∂i . ∂xi ∂φ ∂Fi ∂Fk (1) (∇φ)i = (2) ∇ F = (3) (∇ F )i = ǫijk ∂xi · ∂xi × ∂xj 2 2 2 ∂ φ 2 ∂ Fi (4) φ = (5) (∇ F )i = ∇ ∂xj ∂xj ∂xj ∂xj ∂φ ∂ui (6) (v ∇) φ = vj (7) [(v ∇)u]i = vj · ∂xj · ∂xj Worked Example 3.5 (A) Use the suffix notation to show that ∇ (φ v)= φ∇ v + ∇φ v. × × × 3.5.3 The substitution property of δij Consider the term δ a , where summation over j is implied. Now, δ is non-zero only • ij j ij for one case, j = i. Therefore we may simplify: δijaj = ai . (3.12) In other words, if a delta has a summed index, then the delta may be omitted, and in the rest of the term the summed suffix is substituted by the other index of the delta. Example: δip ǫijk = ǫpjk . (i summed, replace by p) Worked Example 3.5 Show that (B) δip δjq cp dq= ci dj (C) δij bij= bii = bjj (D) δijǫijk= ǫiik = 0 13 3.5.4 Some useful results ǫ ǫ = δ δ δ δ (3.13) ijk klm il jm − im jl Note the repeated suffix k in the property. The remaining suffices i, j, l, m are free in general, none of them occurs more than once in the same term. The double-epsilon usually arises in the presence of two vector-products. Example 3.5.2 a (b c) = ǫ a (b c) = ǫ a ǫ b c = ǫ ǫ a b c = (δ δ δ δ ) a b c { × × }i ijk j × k ijk j klm l m ijk klm j l m il jm − im jl j l m = δ δ a b c δ δ a b c = a b c a b c = (a c) b (a b) c il jm j l m − im jl j l m j i j − j j i · i − · i l i m j m i l j → → → → i = 1, 2 or 3 |{z}a (b c) = (a |{z}c) b (a b) c. ∀ ⇒ ×|{z} × · |{z}− · Other useful results include: 2 ∂xi/∂xj = δij, xixi = r , ǫijk δjk =0, ǫijk xj xk =0, ǫijk(aj bk + ak bj)=0. (3.14) For the first two, recall x1=x, x2=y, x3=z. The third result follows from the substitution property. The penultimate result states (r r) =0 and the last that (a b) +(b a) =0. × i × i × i Worked Example 3.5 (E) Show that ∇ (Ω r) = 2 Ω for a constant vector Ω. × × Example 3.5.3 Find 2φ when φ = (a r) f(r), where a is a constant vector. ∇ · φ = ajxj f(r) , ∂φ ∂xj df ∂r f ′ = aj f + ajxj = aif + ajxjxi ∂xi ∂xi dr ∂xi r δij j i x /r → i 2 2 ∂ φ df ∂r|{z} f ′ |{z} f ′ d f ′ ∂r φ = = ai + aj δij xi + ajxj δii + ajxjxi ∇ ∂xi ∂xi dr ∂xi r r dr r ∂xi j i =3 → xi/r 2 2 |{z} |{z} On the last term, xixi = xi = r . Then |{z} P 2 f ′ d f ′ f ′ φ = 5a x + a x r = (a r) f ′′ + 4 ∇ i i r j j dr r · r 14 4 VECTOR CALCULUS II 4.1 Cylindrical polar and Spherical polar coordinates For many problems the Cartesian coordinate system is not the most natural.