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 ×

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 coordinates 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. For example a full sphere is easily described by limits on r, as 0

z z P P r z θ O O θ φ x r y x y (CPs) (SPs)

For Cylindrical Polars (CPs), (r, θ, z): x = r cos θ, y = r sin θ, z = z, 2 2 1 1 y 2 (4.1) r =(x + y ) , θ = tan− x , r 0, 0 θ 2π ≥ ≤ ≤ For Spherical Polars (SPs), (r,θ,φ): x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ,

2 2 2 1 1 z 1 y r =(x + y + z ) 2 , θ = cos− 1 , φ = tan− (4.2) (x2+y2+z2) 2 x r 0, 0 θ π, 0 φ 2π ≥ ≤ ≤ ≤ ≤ In SPs, r is the distance from the origin O, as usual. r = r rˆ (SPs). In CPs, r is the distance from the z-axis, the cylindrical-radius. r = r rˆ + zzˆ (CPs). Note that θ (CPs) φ (SPs). When confusion might be possible, CPs are sometimes written ≡ (s,φ,z) rather than (r, θ, z).

Let the coordinates (α1, α2, α3) denote e.g. (x, y, z) or (r, θ, z), or one of the many orthogonal ˆ coordinate systems. αˆ1, αˆ2, αˆ3 denote unit vectors, e.g. CPs αˆ1 = rˆ, αˆ2 = θ, αˆ3 = zˆ.

At every point the unit vectors for the coordinates are orthogonal, i.e. αˆi αˆj =0 if i = j, and they are ordered such that αˆ αˆ = αˆ , etc., e.g. in Cartesian coordinates· i j = 6k. 1 × 2 3 ×

Worked Example 4.1 (A) Sketch coordinate surfaces, corresponding to αi = const, for CPs and SPs. (B) Sketch the unit vectors for CPs.

15 4.2 The line element revisited

Let P and P’ be neighbouring points, with −→OP = r and OP−−→′ = r + δl. The notations δr and δs are also commonplace. So far we’ve used the former, but if there is potential for ambiguity, when r is a coordinate direction, then δl is safer. In Cartesians we write the small change in position δl = δx i + δy j + δz k. In general we write, δl = δl1αˆ1 + δl2αˆ2 + δl3αˆ3 = h1 δα1 αˆ1 + h2 δα2 αˆ2 + h3 δα3 αˆ3 . (4.3)

As δl is a change in position, each of δl1 = h1 δα1 etc. must be a length. This deﬁnes the scale factors hi, and where δαi is an angle hi must be a length.

For Cartesians we have h1 = h2 = h3 =1. It can be shown that for CPs (r, θ, z): h =1, h = r, h =1, 1 2 3 (4.4) SPs (r,θ,φ): h1 =1, h2 = r, h3 = r sin θ .

Worked Example 4.2 Check (4.4) via sketches.

Example 4.2.1 (Line integral on a curved path in CPs) y δl

Path: θ = α δl = h δα αˆ2 = rδθ θˆ, with r = a . F = 2i 2 ⇒ 2 2 − B π π Then F δl = 2a i θˆ dθ = 2a sin θ dθ = 4a . a · − · B A x ZA Z0 Z0

4.3 Generalised gradient operators The physical interpretations remain unchanged for grad, div and curl. The divergence gives the same scalar, grad and curl give the same vector, but the representation will look different in another coordinate system. For curvilinear orthogonal coordinates (α1,α2,α3), not all of the αi have the dimension of length and the unit vectors αˆi are not constant. Apart from the generalised gradient, proofs of the following are lengthy. Gradient: • The gradient gives the change in the scalar φ for a change in position δl. Therefore δφ δφ (∇φ)1 = ≈ δl1 h1 δα1 ⇒ 1 ∂φ 1 ∂φ 1 ∂φ ∇φ = , , (4.5) h ∂α h ∂α h ∂α 1 1 2 2 3 3 Divergence: • 1 ∂ ∂ ∂ ∇ F = (h h F )+ (h h F )+ (h h F ) (4.6) · h h h ∂α 2 3 1 ∂α 1 3 2 ∂α 1 2 3 1 2 3 1 2 3 Curl: h αˆ h αˆ h αˆ • 1 1 1 2 2 3 3 ∇ F ∂ ∂ ∂ (4.7) = ∂α ∂α ∂α × h1 h2 h3 1 2 3 h1 F1 h2 F2 h3 F3

16 Scalar Laplacian: • By deﬁnition, 2φ = ∇ (∇φ). This can be calculated from the above to be ∇ · 1 ∂ h h ∂φ ∂ h h ∂φ ∂ h h ∂φ 2φ = 2 3 + 1 3 + 1 2 (4.8) ∇ h h h ∂α h ∂α ∂α h ∂α ∂α h ∂α 1 2 3 1 1 1 2 2 2 3 3 3 Vector Laplacian: • The following expansion formulae, which can be shown to hold for Cartesians, we insist holds for all coordinate systems:

∇2F ∇(∇ F) ∇ (∇ F), (4.9) ≡ · − × × 2 2 The RHS can be calculated from formulae above. The result ∇ F =( Fi) αˆi holds for Cartesians but not in general. ∇

4.4 Volume and surface elements

δV δS h2 δα2

h1 δα1

h3 δα3

From the previous sections, the line element is given by δl =(h1 δα1, h2 δα2, h3 δα3). It follows directly that a volume element is given by

δV = h1 h2 h3 δα1 δα2 δα3 .

The volume integral for f = f(α1,α2,α3) is then

f dV = f h1 h2 h3 dα1 dα2 dα3. (4.10) ZV ZZZ Note that δ denotes a small but ﬁnite element, and that the inﬁnitesimal is denoted d in the integral.

Consider a surface S deﬁned by α1 = const. A surface element is generated by variations in the other two variables in this case. Hence,

δS = h2 h3 δα2 δα3 .

The surface integral for scalar g = g(α1,α2,α3) in this case is

g dS = g h2 h3 dα2 dα3. (4.11) ZS ZZ

(In previous courses the prefactors h2 h3 etc. may have been derived via the Jacobian deter- minant. This is easier.)

17 Often we want to know the ﬂux of a vector ﬁeld passing through a surface. δS F

The vector surface element δS = nˆ δS, where nˆ is a normal to the surface S. We may now consider the component of a vector ﬁeld F in the direction of nˆ and write

F dS = F nˆ dS = F cos θ dS. (4.12) · · | | ZS ZS ZS The rightmost form should be used with care – there θ is the angle between F and the unit vector nˆ, NOT to be confused with θ of a coordinate system. Convention is to take δS to be an outwards normal for a closed surface S, i.e. when S encloses a volume V . If S is not closed then the direction should be speciﬁed.

Example 4.4.1 (Volume integral) 2 2 V is a cylinder bounded by r = a, z = 0 and z = h. Find I = V 2z(z + r )dV . R rδθ

V cylinder CPs: h1 = h3 = 1, h2 = r. δθ ≡ → δz δV = rδrδθδz. ⇒ δr

h 2π a I = 2z(z2 + r2)dV = 2 (z3 + zr2) r dr dθ dz ZV Zz=0 Zθ=0 Zr=0 h r2 r4 a = 4π z3 + z dz 2 4 Zz=0 r=0 h4 a2 h2 a4 π = 4π . + . = h2a2(h2 + a2) . 4 2 2 4 2

Example 4.4.2 (Surface integral) Evaluate r dS where S is the spherical surface deﬁne by r = a. S · R r = a SPs: α = const, h = r, h = r sin θ. → 1 2 3 2 r δS r = r rˆ, δS = r sin θδθδφ rˆ, || S so r is parallel to δS in this case. Therefore, r δS = a3 sin θδθδφ on S. a δS · 2π π r dS = a3 sin θ dθ dφ S · 0 0 Z Z Z π = 2π a3 sin θ dθ = 4 π a3 . Z0

18 Worked Example 4.4 (A) Find the surface area of a sphere of radius a.

(B) A spherical shell has inner radius a, outer radius b and density proﬁle ρ(r)= ρ0 a/r, where ρ0 is a constant. Find its mass m.

Example 4.4.3 (The Shell Theorem)

z m Consider a thin spherical shell of radius a and uniform mass d per unit area σ. total mass of shell M = 4πa2 σ. ⇒ mass of surface element δM = σδS. ⇒ θ δS Let mass m be a distance z from the centre of the shell. a O Then with z = z k, a = a rˆ and rˆ k = cos θ, we have · M z + d = a d = a z ⇒ − d2 = (a z) (a z)= a a+a a 2a z = a2+z2 2az cos θ . ⇒ − · − · · − · −

The contribution to the gravitational potential for each small surface element is

GmδM δV = , and δM = σδS, δS = r2 sin θδθδφ, with r = a . − d The total potential is then V = δV = Gmσ (δS/d) − 2π Pπ a2 sin θ dθ dφ P π sin θ dθ V = Gmσ = 2πa2σ Gm . 2 2 ⇒ − 0 0 d − 0 √a + z 2az cos θ Z Z Z − Let u = cos θ, du = sin θ dθ − 1 du 1 1 V = 1 MGm = 1 GMm a2 + z2 2azu 2 2 2 2 ⇒ − 1 √a + z 2azu az − 1 Z− − p − a2 + z2 2az = (a z)2 = a z , a, z > 0 ∓ ∓ | ∓ | p GMm V = p a z (a + z) ⇒ 2az {| − |− } 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. ≥ | − | − ⇒ { − − } − Finally we have the result! :

GMm If z a, then V = = const inside the sphere, force F = ∇V = 0 . ≤ − a ⇒ − GMm If z a, then V = outsize the sphere, F same as for point mass M at origin. ≥ − z ⇒

19 4.5 Gauss’ Theorem (or the Divergence Theorem)

Let F be a vector ﬁeld, and S be a closed surface, enclosing a volume V . Then Gauss’ Theorem states that F dS = ∇ F dV. (4.13) · · ZS ZV As S is closed, dS is the outwards normal. The theorem describes the balance between the ﬂux through the surface, on the LHS, and the total divergence within a volume, on the RHS.

Worked Example 4.5 Verify the theroem holds where (A) V is a sphere of radius a, centred on the origin, and F = r. (B) V is the cylinder bounded by r = a, z = 0, z = h, and F = xz3i + yz3j + z2(x2 + y2)k.

Corollary of Gauss’ Theorem Let F = φa, where φ is a scalar ﬁeld and a is an arbitrary constant vector. Expanding ∇ F = a ∇φ + φ ∇ a, but ∇ a = 0 as a is constant. Substitution into Gauss’ Theorem gives· a φ· dS = a ·∇φ dV . As· a is abitrary we have · · R R φ dS = ∇φ dV (4.14) ZS ZV Note that this is a set of three equations, one for each component of dS and ∇φ.

4.6 Stokes’ Theorem dS

C dl S

Let S be an open surface, bounded by the closed circuit C. Then Stokes’ Theorem states that F dl = ∇ F dS. (4.15) · × · IC ZS Worked Example 4.6 Verify Stokes’ Theorem for Example 3.2.2 .

Example 4.6.1 Show that work done B F dl is independent of the path from A to B for a conservative force using A · Stokes’ Theorem. R As F = ∇V , a conservative force is irrotational: ∇ F = ∇ ∇V = 0. − × − × Let C and C be two different paths from A to B, let C be the closed path C C and S be a 1 2 1 − 2 surface enclosed by C. Then

F dl = ∇ F dS = 0 F dl = F dl F dl = 0 F dl = F dl · × · ⇒ · · − · ⇒ · · IC ZS IC ZC1 ZC2 ZC1 ZC2 Hence the integrals along different paths C1 and C2 from A to B are the same. Given that the paths are arbitrary, the work done must be independent of the path.