Notes on Vector Calculus

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1 The University of New South Wales Math2111 Notes on Vector Calculus Last updated on May 17,

2 1. Rotations Einstein Summation Convention Basis vectors: i = e 1 = 1 0, j = e 2 = 0 0 1, k = e 3 = Abbreviation: if an index occurs exactly twice in a term then sum that index from 1 to 3. E.g., if v = then v = v i e i = 3 i=1 v 1 v 2 v 3 v i e i = v 1 e 1 + v 2 e 2 + v 3 e 3. 2

3 Position vector: write r = x i + y j + z k or x = x i e i = x 1 e 1 + x 2 e 2 + x 3 e 3. Define the Kronecker delta δ ij = and Levi-Civita symbol ɛ ijk = so that 1 if i = j, 0 if i j, +1 if (i, j, k) is (1, 2, 3) or (2, 3, 1) or (3, 1, 2), 1 if (i, j, k) is (2, 1, 3) 0 otherwise, or (1, 3, 2) or (3, 2, 1), e i e j = δ ij and e i e j = ɛ ijk e k. 3

4 Notice δ ij is symmetric in i and j, δ ij = δ ji, whereas ɛ ijk is skew-symmetric with respect to odd permutations, ɛ ijk = ɛ jik = ɛ ikj = ɛ kji. Use δ ij and ɛ ijk in calculations involving dot products, v w = (v i e i ) (w j e j ) = v i w j e i e j = v i w j δ ij = v i w i, and cross products, v w = (v i e i ) (w j e j ) = v i w j e i e j = v i w j ɛ ijk e k. Remember that v w = w v whereas v w = w v, and that v v = v 2 whereas v v = 0. 4

5 Recall that if the angle between u and v is θ (0 θ π), then and u v = u v cos θ (1) u v = u v sin θ e (2) where e is a unit vector orthogonal to u and v with direction given by the right-hand rule. In particular, for non-zero vectors u and v we have u v = 0 iff θ = π/2, i.e., iff u and v are orthogonal. 5

6 We call u (v w) the scalar triple product of the vectors u, v, w. 1.1 Exercise. Show that u (v w) = ɛ ijk u i v j w k = det Thus, u 1 v 1 w 1 u 2 v 2 w 2 =: det[u, v, w]. u 3 v 3 w 3 u (v w) = w (u v) = v (w u) and u (v w) = (u v) w. 1.2 Exercise. Show that u (v w) is the volume of the parallelepiped generated by u, v, w. Given a matrix A = [a ij ] R 3 3, w = Av w i = a ij v j. 6

7 1.3 Exercise. Show that (Av) w = v (A T w). 1.4 Exercise. Show that Au (Av Aw) = (det A) u (v w). The above exercise implies that if det A = 1 then the volumes of the parallelepipeds generated by {u, v, w} and {Au, Av, Aw} respectively are the same. In other words, the linear map x Ax is volume-preserving if det A = 1. 7

8 We call u (v w) the vector triple product of u, v, w. 1.5 Lemma. For i, j, m, n {1, 2, 3}, 3 k=1 ɛ ijk ɛ mnk = +1 if i = m and j = n and i j, 1 if i = n and j = m and i j, 0 otherwise, or, more succinctly, ɛ ijk ɛ mnk = δ im δ jn δ in δ jm. 1.6 Theorem. u (v w) = (u w)v (u v)w. 8

9 Orthogonal Matrices A matrix Q R n n is orthogonal if Q T = Q 1, or in other words if Q T Q = I = QQ T. The set of all orthogonal matrices, O(n) = { Q R n n : Q T = Q 1 }, is called the orthogonal group of order n. 1.7 Theorem. For Q R n n, the following are equivalent: 1. Q O(n); 2. (Qv) (Qw) = v w for all v, w R n ; 3. Qv = v for all v R n. Thus, orthogonal transformations preserve angles and distances. 9

10 1.8 Example. It is readily verified that [ ] cos θ sin θ Q = O(2). sin θ cos θ In fact, the action of Q on a vector v represents the rotation of v about the origin by the angle θ since where w = Qv. w = v, w v = w v cos θ, 1.9 Exercise. Show that if Q O(3) then for all v, w R 3, Qv Qw = (det Q) Q(v w) Exercise. Show that if Q O(n) then det Q = ±1. 10

11 The set SO(n) = { Q O(n) : det Q = +1 } is called the special orthogonal group of order n. When Q SO(n) we say that v Qv is a proper orthogonal transformation. The preceding results show that for Q SO(3), Qv Qw = v w and Qv Qw = Q(v w). (3) It turns out that a transformation from R 3 to R 3 is proper orthogonal, i.e., belongs to SO(3), iff it represents a rotation about an axis through the origin. (We will prove the if part in the next theorem. The only if part is one of the tutorial problems.) Given this fact, we see that (3) also follows from (1) and (2). 11

12 The identities (3) help to explain why many geometric and physical quantities are expressed in terms of dot and cross products: such quantities should not depend on the choice of a right-handed, Euclidean coordinate system for physical space Theorem. Let a, v R 3 with a = 1 and v 0. The vector w obtained by rotating v through an angle θ about a is w = (1 cos θ)(a v)a + (cos θ)v + (sin θ)a v. Morover, w = Rv where R = [r ij ] SO(3) has entries r ij = (1 cos θ)a i a j + (cos θ)δ ij (sin θ)ɛ ijk a k Corollary. Given any two vectors v, w R 3 with v = w, there exists a rotation matrix Q SO(3) such that w = Qv. 12

13 2. Line Integrals Arc Length Let C be an oriented curve (or path) with parametric representation x = x(u), a u b. We will always assume that x(u) is continuous and piecewise continuously differentiable with respect to the parameter u. Choose M + 1 points in the parameter interval [a, b], a = u 0 < u 1 < u 2 < < u M = b and define corresponding points on C, x k = x(u k ), 0 k M. Also choose ū k [u k, u k+1 ] and put x k = x(ū k ), 0 k M 1. 13

14 The length of the line segment from x k to x k+1 is s k = x k where x k = x k+1 x k. 2.1 Motivation. The line integral of f : C R with respect to arc length should be C f ds = lim M 1 taking the limit as M and k=0 f( x k ) s k max s k 0. 0 k M 1 [One may think of C as a thin wire of mass density f(x) (= mass per unit length). Then, total mass = C f ds.] 14

15 2.2 Definition. C f ds = b a f dx b du du = f ( x(u) ) x (u) du. a Notice that by taking f 1 we have length of C = lim M 1 k=0 s k = C 1 ds = b a x (u) du. 2.3 Exercise. Find the length of the parabolic curve y = 1 x 2, 1 < x < Theorem. The value of the line integral C f ds does not depend on the choice of parametric representation for the oriented curve C. 15

16 Integration of Vector Fields 2.5 Motivation. The line integral of the vector field should be F : R n R n C F (x) dx = lim M 1 taking the limit as M and k=0 max 0 k M 1 x k 0. F ( x k ) x k, [One may think of C as the path of a particle which moves under the influence of a force F (x). Then, work done = C F (x) dx.] 2.6 Definition. C F (x) dx = b a F ( x(u) ) dx du du. 16

17 2.7 Exercise. For F = (log x) i y k, find if C is described by C F (x) dx, x = (x, y, z) = (1 + u) i u 2 j + 2u k, 0 u Theorem. The value of the line integral F (x) dx C does not depend on the choice of parametric representation for the oriented curve C. 17

18 Alternative notation for a line integral: ( ) P dx + Q dy + R dz = where and C C F = P i + Q j + R k F (r) dr r(u) = x(u) i + y(u) j + z(u) k. The motive for this notation is that the dot product of F with dr = dx i + dy j + dz k gives the differential form P dx + Q dy + R dz. Notice that, e.g., so C P dx = lim M 1 C k=0 P ( x k, ȳ k, z k ) x k = ( P dx + Q dy + R dz ) = b a ( b a P ( x(u), y(u), z(u) ) dx du du P dx du + Q dy du + R dz du ) du. 18

19 Properties of Line Integrals If C 1 ends at the point where C 2 begins, then we can combine the two curves into a single curve denoted by C 1 + C 2. Let C denote the curve obtained by parameterizing C in the opposite direction. Thus, if C is given by x = φ(u) for a u b then C is given by x = φ(a + b u) for a u b. 2.9 Theorem F (x) dx = C 1 +C 2 C F (x) dx = C C 1 F (x) dx + F (x) dx. C 2 F (x) dx. 19

20 Similarly, but f ds = f ds + f ds C 1 +C 2 C 1 C 2 C f ds = f ds. C 2.10 Theorem. For any vector fields F, G and any scalar constants λ, µ, we have ( ) λf (x) + µg(x) dx = λ C F (x) dx + µ G(x) dx. C C Similarly, C (λf + µg) ds = λ C f ds + µ C g ds. 20

21 For line integrals, the fundamental theorem of calculus takes the following form Theorem. If the curve C starts at x a = x(a) and finishes at x b = x(b), then C grad f(x) dx = f(x b) f(x a ). We say that C is a closed curve if it finishes at the same point where it starts, i.e., if x(b) = x(a). It is customary to put a circle on the integral sign to indicate that the curve is closed: F (x) dx. C The theorem above shows that grad f(x) dx = 0 for any scalar field f. C 21

22 Work and Energy The work W done by a constant force F on an object that moves in a straight line from r to r + r is defined by W = F r. Thus, the work done by a variable force F (r) on an object that moves along a curve C is W = lim M k=1 F ( r k ) r k = C F dr. If a force field F may be derived from a scalar field U, that is F = grad U, then we call U a potential for F. Notice that we may add an arbitrary constant to U and it remains a potential for F. 22

23 If F is derived from a potential and if C is any path from r a to r b, then C F dr = C grad U dr = [U(r b) U(r a )] or in words: work done by force = (change in potential) so that, in particular, C F (r) dr = 0. Consider a particle of mass m with position r(t) velocity v = dr dt speed v = v. The kinetic energy T of m is defined by T = T (v) = 1 2 mv2. 23

24 2.12 Theorem. If F = F (r) is the net force on m, so that m dv dt = F, and if C is the trajectory of m for a t b, then C F dr = T (v b) T (v a ). In words, work done by force = change in kinetic energy Corollary. In the case of a net force F = grad U there is a constant E, called the total energy, such that T +U = E throughout the motion, i.e., T ( v(t) ) + U ( r(t) ) = E for a t b. 24

25 2.14 Exercise. Consider the central force field F = f( r )r. Show that F may be derived from a potential by explicitly constructing an associated potential energy function. Determine the total energy and establish how the speed v depends on the radial distance r. 25

26 3. Surface Integrals Surface Area Recall that the area of the parallelogram { ua + vb : 0 < u < 1, 0 < v < 1 }. equals a b = a b sin θ b θ a O 26

27 Consider a surface S in R 3 with parametric representation S : x = x(u, v), (u, v) U, for some closed, bounded region U in R 2. We will always assume that x(u, v) is continuously differentiable on U. It may be necessary (or convenient) to dissect the surface S into several pieces, each with its own parametric representation. Construct a rectangular mesh in R 2 that covers U and write with x ij = x(u i, v j ), u i = u i+1 u i, u x ij = x i+1,j x ij, v j = v j+1 v j, v x ij = x i,j+1 x ij. We choose ū i [u i 1, u i ] and v j [v j 1, v j ] and put x ij = x(ū i, v j ). 27

28 The vector S ij = ( u x ij ) ( v x ij ) is approximately normal to S at x ij and its length S ij is approximately the area of the piece of S corresponding to u i < u < u i+1 and v j < v < v j Motivation. The integral of f : S R with respect to surface area should be S f ds = lim (ū i, v j ) U f( x ij ) S ij. taking the limit as the number of grid points and max( u i, v j ) 0. [One may think of S as a thin material sheet of mass density f(x) (= mass per unit area). Then, total mass = S f ds.] 28

29 3.2 Definition. S f ds = U f ( x(u, v) ) x u x v du dv 3.3 Theorem. Any two parametric representations x(u, v) and x(u, v ) of a surface S give rise to the same surface integral S f ds. 3.4 Theorem. If S is given by z = ϕ(x, y) for (x, y) U, for some function ϕ, then S f ds = U f ( x, y, ϕ(x, y) ) 1 + grad ϕ(x, y) 2 dx dy. 29

30 Notice that taking f 1 gives area of S = lim (ū i, v j ) U S ij = S f ds = U 1 + grad ϕ(x, y) 2 dx dy. 3.5 Exercise. Find the surface integral S z ds, where S is the upper hemisphere defined by x 2 + y 2 + z 2 = 1, z 0. 30

31 Orientation A vector field N is said to be normal to the surface S if, for every point x S, N(x) tangent plane of S at x. If N(x) 0 then we can define a unit normal vector n(x) = 1 N(x) N(x). The partial derivatives x/ u and x/ v are vectors tangent to the surface S given by x(u, v). Hence, the cross product is normal to S. N(x) = x u x v 31

32 3.6 Example. If S is given by then is a normal vector field for S. z = ϕ(x, y) for (x, y) U, N = ϕ x i ϕ y j + k A surface S is orientable if it has a continuous unit normal vector field n, in which case we speak of the orientation defined by n. An oriented surface is a surface S equipped with a chosen orientation, i.e., a chosen unit normal n. In this case, S denotes the oriented surface with unit normal n. Later, we will generalise the above discussion to cover cases when S has sharp corners or edges and so n is discontinuous. 32

33 3.7 Example. Let 0 < b < a. The Möbius band S described by for x = (a + w cos 1 2θ) cos θ, y = (a + w cos 1 2θ) sin θ, z = w sin 1 2 θ, b w b, π θ π, is not orientable. Although the normal vector field N = r w r θ is a continuous function of (w, θ) on [ b, b] [ π, π], it is not a continuous function of (x, y, z) on S because lim θ π N = a w/2 but lim N = θ π 0 a w/

34 Moebius Band 0.5 z y x 0 34

35 Flux 3.8 Motivation. The flux (integral) of the vector field F through the oriented surface S should be S F ds = lim (ū i, v j ) U taking the limit as max( u i, v j ) 0. F ( x ij ) S ij, [One may think of S as an imaginary membrane in a fluid flow of velocity F (x). Then, flux through membrane = S F ds.] 3.9 Definition. S F ds = U F ( x(u, v) ) ( x u x ) du dv v 35

36 3.10 Exercise. Calculate the flux of the vector field F (x, y, z) = (3x y)i + (x 2 + z)j 3xyk through the oriented surface S : x(u, v) = (u v)i + (u + v)j 2uk, for 1 u 1 and 2 v Theorem. The flux of F through S equals the integral of F n with respect to surface area: S F ds = S F n ds. 36

37 Note that S f ds makes sense even when S is not orientable. Moreover, if S is orientable then S f ds = S f ds but S F ds = S F ds Exercise. Consider the vector field F = f( x )x. Determine the flux through the sphere of radius R centred at the origin. What does one conclude in the case f x 3? 37

38 Consider a fluid with velocity field u = u(x, t) passing through a fixed, oriented surface S given by x(ξ, η) for (ξ, η) U. Assume for simplicity that u n > 0 everywhere on S. Let V (t) be the volume of fluid that passes through S between time 0 and time t. Let P ij be a small parallelogram spanned by ξ x ij and η x ij. At time t + t, the fluid particles that passed through P ij since time t make up a (slightly distorted) parallelepiped with sides ξ x ij, η x ij and t u. The volume of these particles is therefore given (to first order in t, ξ and η) by the scalar triple product ( t u) ( ξ x ij η x ij ) = t u S ij. 38

39 Summing the contributions from all parallelograms, V = V (t + t) V (t) = t i,j u S ij ( 1 + O( t + ξ + η)), and in the limit as ξ and η tend to zero, V = t u ds + S O( t2 ), so V t = u ds + O( t). S Thus, in the limit t 0, we obtain dv u ds = S dt. Conclusion: the flux of the fluid velocity equals the instantaneous rate at which fluid volume passes through the surface. 39

40 4. The Operator Grad, curl and div The vector differentiation operator, called nabla or del, is defined by = e i x i = e 1 x 1 + e 2 + e 3. x 2 x 3 Acting on a scalar function, gives the gradient: Abbreviations: grad f = f = f x 1 e 1 + f x 2 e 2 + f x 3 e 3 = i = f/ x 1 f/ x 2 f/ x 3 x i, = e i i, f = e i i f = i f e i.. 40

41 4.1 Definition. The curl of the vector field is the vector field F (r) = F 1 e 1 + F 2 e 2 + F 3 e 3 curl F = F = ɛ ijk i F j e k = e i i F = det Written out explicitly, ( F3 curl F = F 2 x 2 x 3 e 1 e 2 e = det F 1 F 2 F 3 ) e 1 + ( F1 F ) 3 e 2 + x 3 x 1 e 1 e 2 e 3 x 1 x 2 x 3 F 1 F 2 F 3 ( F2 F ) 1 e 3. x 1 x 2. 41

42 4.2 Exercise. Find curl F if F = e xy i + (x z) j + 2xy k. 4.3 Definition. The divergence of the vector field is the scalar field F (r) = F 1 e 1 + F 2 e 2 + F 3 e 3 div F = F = i F i = F i x i = F 1 x 1 + F 2 x 2 + F 3 x Exercise. Find div F if F = (yz + sin πx) i + e x cos πz j + (x + 2y 3z) k. 42

43 The following lemma is pivotal in potential theory. 4.5 Lemma. curl(grad f) = 0, div(curl F ) = 0. If F is really a 2D vector field, say F (x, y, z) = P (x, y) i + Q(x, y) j, then and curl F = ( Q x P ) k y div F = P x + Q y. 43

44 For any vector field A we define the scalar differential operator A in the obvious way. For a scalar function f, and for a vector field F, (A )f = A i i f = A i f x i = A ( f), (A )F = A i i F = A i F x i = A i F j x i e j = (A F j )e j. 4.6 Exercise. Show that (A B) (C D) = (A C)(B D) (A D)(B C) = B [(A C)D] A [(B C)D]. 4.7 Exercise. Show that (A B) curl F = B [(A )F ] A [(B )F ]. 44

45 Interpretation If n denotes a unit vector then n f measures the change of f in the direction n at any point P. Let v denote the velocity field of a fluid. Then, the fluid tends to accumulate in or move away from a small neighbourhood of a point P if div v < 0 or div v > 0 at that point respectively. curl v measures the rotational tendency of a fluid, that is, if the fluid in a small neighbourhood of a point P were suddenly solidified and the surrounding fluid simultaneously annihilated then this solid neighbourhood would rotate with an angular velocity 1 2 curl v at P. These statements may be made precise by means of the integral theorems discussed in the following sections. 45

46 4.8 Exercise. Determine the divergence and the curl of the vector fields Interpret the results. v = αr, v = ωy i + ωx j. 4.9 Exercise. Show that curl(curl A) = 2 A + (div A). 46

47 5. Integral Theorems Boundary of a Set Let U R n. We say that x R n is a boundary point of U if every neighbourhood of x intersects both U and its complement U c = R n U. The set of all boundary points is called the topological boundary of U, denoted Bdry(U). It is easy to see that R n is the disjoint union of the interior of U, the topological boundary of U and the interior of U c. 5.1 Example. In R 1, if U = (a, b) or [a, b) or (a, b] or [a, b] then Bdry(U) = {a, b}. 47

48 5.2 Example. In R 2, the boundary of a disc U = { (x, y) : x 2 + y 2 < a 2 } is a circle Bdry(U) = { (x, y) : x 2 + y 2 = a 2 }. 5.3 Example. In R 3, the boundary of a ball U = { (x, y) : x 2 + y 2 + z 2 < a 2 } is a sphere Bdry(U) = { (x, y, z) : x 2 + y 2 + z 2 = a 2 }. 48

49 Let 1 k n. We say that S is a simple k-dimensional surface in R n if S has a parametric representation x = ψ(u), u U R k, (4) such that the mapping ψ is one-to-one and continuously differentiable on the closed set U. In this case, we define the boundary of S to be the k 1-dimensional set Notice that S = { ψ(u) : u Bdry(U) }. S = Bdry(S) only if k = n. 5.4 Example. If C is a simple curve x(u) (a u b) in R n from x a = x(a) to x b = x(b), then x a x b and C = {x a, x b } but Bdry(C) = C unless n = Exercise. Find S and Bdry(S) if S = { (x, y, z) : x 2 + y 2 a 2, z = 0 }. 49

50 Green s Formula We say that a function is C k if all its (partial) derivatives or order k exist and are continuous. Let U be a closed region in R 2 whose topological boundary Bdry(U) consists of finitely many continuous, piecewise C 1 curves. The standard orientation for U = Bdry(U) is determined by the requirement that as you move along any piece of U the region U is always on the left. Thus, in the simplest case when U consists of a single, closed curve, the standard orientation goes counterclockwise. The following result is a two-dimensional version of the fundamental theorem of calculus. 50

51 5.6 Theorem (Green s Formula). If P and Q are C 1 on U, then ( Q U x P ) ( dx dy = P dx + Q dy). y U If F is a 2D vector field, say then curl F = ( F (x, y, z) = P (x, y) i + Q(x, y) j, i x + j y ) ( P (x, y) i + Q(x, y) j ) = and so we can write Green s formula as U curl F ds = U F dr ( Q x P ) k y if we give U the usual orientation for a region in R 2, i.e., with unit normal n = k. 51

52 5.7 Exercise. Verify Green s formula for P = xy + y 2, Q = 6xy + 5x 2 with C being the square with vertices (0, 0), (a, 0), (a, a), (0, a). 52

53 Stokes s Formula Let S be a simple oriented surface (4) with unit normal n, and assume that the parameter set U satisfies the assumptions for Green s formula. The standard orientation for its boundary curve S is defined as follows: if we imagine walking around S with the up direction given by n, then S is always on the left. 5.8 Theorem (Stokes Formula). If F is C 1 on a neighbourhood of S in R 3, then S curl F ds = F (x) dx. S 5.9 Exercise. Verify Stokes s theorem for F = z 2 i 2x j + y 3 k, where S is the hemisphere defined by x 2 + y 2 + z 2 = 1, z 0. 53

54 Let us say that two oriented curves C 1 and C 2 cancel if C 2 = C 1 and so, for any vector field F, F dx + F dx = 0. C 1 C 2 Consider a surface S consisting of several pieces S 1,..., S r, where each piece is a simple, oriented surface. We say that S is orientable if wherever a part of S i coincides with a part of S j, for i j, the two curves cancel. In this case, we define the boundary S to consists of all parts of S 1,..., S r that do not cancel. The above definition ensures that Stokes s theorem is valid for a non-simple, oriented surface. We say that S is a closed surface if S =. If S is a closed surface then S curl F ds = 0. 54

55 We will use the next lemma to characterise curl F in a way that is clearly independent of the choice of Euclidean coordinates Lemma. Let S ɛ R 2 be the disc with centre 0 and radius ɛ. If f : R 2 R is continuous at (0, 0), then 1 lim ɛ 0 + area(s ɛ ) S ɛ f ds = f(0). In words: the average value of f over S ɛ converges to f(0, 0) as ɛ tends to zero Theorem. If S ɛ R 3 is the disc with centre x 0, radius ɛ and unit normal n 0, then n 0 curl F (x 0 ) = 1 lim ɛ 0 + area(s ɛ ) S ɛ F (x) dx. 55

56 Circulation and Vorticity Consider a fluid with velocity u = u(x, t). The vorticity ω is defined by ω = curl u = u. For any closed, oriented curve C within the flow region, we call C u dx the circulation around C. Stokes s formula gives or in words, ω ds = u dx, S S flux of vorticity through S = circulation around S. 56

57 If S ɛ is the oriented disc with centre x 0, radius ɛ and unit normal n 0, then n 0 ω(x) = 1 lim ɛ 0 + area(s ɛ ) S ɛ u(x) dx circulation around S ɛ = lim. ɛ 0 + area(s ɛ ) 5.12 Exercise. Consider the velocity field u = ωy i + ωx j. Use the (vorticity flux) = (velocity circulation) identity to show that the area enclosed by a planar simple curve C with position vector r is given by A = 1 2 C r dr. 57

58 Divergence Theorem Let U be a closed region in R 3 such that its boundary U = Bdry(U) is an orientable, two-dimensional surface. The standard orientation of U is the one obtained by choosing n to be the outward unit normal for U. We will use the notation U f dv = for the volume integral of a function. U f(x, y, z) dx dy dz 5.13 Divergence Theorem. If F is C 1 on U, then U div F dv = U F ds = U F n ds. 58

59 5.14 Exercise. Use the divergence theorem to find the total flux out of the region U defined by 0 x 2 + y 2 1, 0 z 1 for F = x i + 2y 2 j + 3z 2 k. As with curl F, we can characterise div F in a way that is clearly independent of the choice of Euclidean coordinates Lemma. Let U ɛ R 3 be the ball with centre 0 and radius 0. If f : R 3 R is continuous at 0, then 1 lim ɛ 0 + volume(u ɛ ) U ɛ f dv = f(0) Corollary. If U ɛ R 3 is the ball with centre x 0 and radius ɛ then 1 div F (x 0 ) = lim F ds. ɛ 0 + volume(u ɛ ) U ɛ 59

60 Consider a fluid with velocity field u = u(x, t) and density ϱ = ϱ(x, t). For a fixed region U, the fluid mass inside U at time t is M(t) = U ϱ dv = ϱ(x, t) dv x. U On the one hand, dm dt U = ϱ t dv, and on the other hand, conservation of mass implies that so dm dt = rate at which fluid mass leaves U through U = U ϱu ds = (ϱu) dv, U U ( ) ϱ t + (ϱu) dv = 0. 60

61 Taking U = U ɛ, dividing by volume(u ɛ ) and letting ɛ 0 +, we obtain the continuity equation ϱ t + (ϱu) = 0. (5) 5.17 Exercise. Let φ : R 3 R. Show that (φf ) = ( φ) F + φ F. A flow is said to be incompressible if the density is constant, i.e., independent of x and t. Then, the continuity equation (5) reduces to u = 0. 61

62 6. Curvilinear Coordinates Examples Cartesian coordinates (x 1, x 2 ) = (x, y) for R 2. Cartesian coordinates (x 1, x 2, x 3 ) = (x, y, z) for R 3. Polar coordinates (r, θ) for R 2 : (x 1, x 2 ) = (r cos θ, r sin θ) Cylindrical coordinates (r, θ, z) for R 3 : x 1 = r cos θ, x 2 = r sin θ, x 3 = z. Spherical coordinates (ρ, φ, θ) for R 3 : x 1 = ρ sin φ cos θ, x 2 = ρ sin φ sin θ, x 3 = ρ cos φ. 62

63 Basis Vectors Let (x 1, x 2, x 3 ) = Φ(ξ 1, ξ 2, ξ 3 ). We say that (ξ 1, ξ 2, ξ 3 ) are curvilinear coordinates for x = (x 1, x 2, x 3 ) if the transformation Φ is such that (x 1, x 2, x 3 ) (ξ 1, ξ 2, ξ 3 ) 0. (6) By the inverse function theorem, Φ is locally invertible so there is a one-to-one correspondence between a point and its curvilinear coordinates, provided the ξ i are suitably restricted. 63

64 Assumption (6) holds iff the vectors b ξi = x ξ i = x 1 / ξ i x 2 / ξ i, i = 1, 2, 3, x 3 / ξ i form a basis for R 3. orthogonal if We say that the curvilinear coordinates are b ξi b ξj = 0 whenever i j. In this case, we define the scale factors and the unit basis vectors h i = h ξi = b ξi = e ξi = 1 h i b ξi = 1 h i x ξ i x ξ i > 0 (no sum over i). 64

65 Thus, if (ξ 1, ξ 2, ξ 3 ) are orthogonal (curvilinear) coordinates then e ξ1, e ξ2, e ξ3 form an orthonormal basis for R 3 : e ξi e ξj = δ ij. We will always order the coordinates ξ 1, ξ 2, ξ 3 so that the orthonormal frame is right-handed; thus, e ξi e ξj = ɛ ijk e ξk and (x 1, x 2, x 3 ) (ξ 1, ξ 2, ξ 3 ) = b ξ 1, b ξ2, b ξ3 > Lemma. (x 1, x 2, x 3 ) (ξ 1, ξ 2, ξ 3 ) = h 1h 2 h 3. 65

66 6.2 Example. Cartesian coordinates (x 1, x 2, x 3 ) = (x, y, z) generate the standard basis e 1 = e x = i, e 2 = e y = j, e 3 = e z = k for R 3. In this case the scale factors are just h 1 = h 2 = h 3 = 1 or, with the alternative notation, h x = h y = h z = 1. 66

67 6.3 Example. For cylindrical coordinates (r, θ, z) we have the orthogonal basis vectors b r = the scale factors cos θ sin θ 0, b θ = r sin θ cos θ, b z = 0 h r = 1, h θ = r, h z = 1, 0 0 1, and the orthonormal basis vectors e r = cos θ sin θ, e θ = 0 sin θ cos θ, e z =

68 6.4 Example. For spherical coordinates (ρ, φ, θ) we have othogonal basis vectors b ρ = sin φ cos θ sin φ sin θ cos φ the scale factors, b φ = ρ cos φ cos θ cos φ sin θ sin φ, b θ = ρ sin φ h ρ = 1, h φ = ρ, h θ = ρ sin φ, and the orthonormal basis vectors e ρ = sin φ cos θ sin φ sin θ, e φ = cos φ cos φ cos θ cos φ sin θ, e θ = sin φ sin θ cos θ 0 sin θ cos θ 0., 68

69 Line Integrals Consider a curve C given in terms of orthogonal coordinates, ξ 1 = ξ 1 (u) ξ 2 = ξ 2 (u) ξ 3 = ξ 3 (u) for a u b. If f is a scalar function of ξ 1, ξ 2, ξ 3, we can compute by observing that C f ds = b a f dx du du so that dx du = x ξ i dξ i du = dx du ( = 3 i=1 3 i=1 h i dξ i du dξ i du h ie ξi ) 2. 69

70 The (purely formal) expression ds 2 = 3 i=1 h 2 i dξ2 i is called the metric for the curvilinear coordinate system, and provides a convenient way to list the scale factors. Cartesian coordinates: ds 2 = dx 2 + dy 2 + dz 2. Cylindrical coordinates: ds 2 = dr 2 + r 2 dθ 2 + dz 2. Spherical coordinates: ds 2 = dρ 2 + ρ 2 dφ 2 + ρ 2 sin 2 φ dθ 2. 70

71 6.5 Exercise. Find the length of the helix given by r = 1, z = θ, 0 θ 4π. Since e ξ1, e ξ2, e ξ3 are orthonormal, if we define then F ξi (ξ) = F e ξi F = F ξi e ξi = F ξ1 e ξ1 + F ξ2 e ξ2 + F ξ3 e ξ3. To compute use b C C F dr = F dx = F dx a du du F dx du = 3 i=1 F ξi dξ i du h i. 71

72 Alternatively, write C F dx = C 3 i=1 F ξi h i dξ i. 6.6 Exercise. Find C F dx if C is the curve ρ = 1, θ = φ, 0 φ π 2 and if F = ρ e ρ 2ρ cos θ e φ + ρ sin φ e θ. 72

73 Surface Integrals Consider a surface S given in terms of orthogonal coordinates, ξ 1 = ξ 1 (u, v) ξ 2 = u 2 (u, v) ξ 3 = ξ 3 (u, v) for (u, v) U. We want to compute or S f ds = S F ds = U f U F x u x v du dv ( x u x ) du dv. v 73

74 Since it follows that x u = x ξ i ξ i u = x v = x ξ i ξ i v = x u x v = det 3 i=1 3 i=1 ξ i u h ie ξi, ξ i v h ie ξi, e ξ1 e ξ2 e ξ3 ξ h 1 1 u h 2 h 1 ξ 1 v ξ 2 u h 3 h 2 ξ 2 v ξ 3 u h 3 ξ 3 v. 74

75 In particular, if the surface has the form ξ 3 = g(ξ 1, ξ 2 ), (ξ 1, ξ 2 ) U, then x ξ 1 x ξ 2 = det e ξ1 e ξ2 e ξ3 g h 1 0 h 3 ξ 1 g 0 h 2 h 3 ξ 2 g g = h 2 h 3 e ξ1 h 1 h 3 e ξ2 + h 1 h 2 e ξ3. ξ 1 ξ Exercise. Find the mass of a hemispherical dome with radius a and surface density µ(1 + sin φ) per unit area. 75

76 Volume Integrals If U = Φ(R) = { x : x = Φ(ξ), ξ R } then U f dv = f dx 1 dx 2 dx 3 = U f R (x 1, x 2, x 3 ) (ξ 1, ξ 2, ξ 3 ) dξ 1 dξ 2 dξ 3 Thus, for orthogonal coordinates, U f dv = R fh 1h 2 h 3 dξ 1 dξ 2 dξ Exercise. Let 0 < α < π. Find the volume of the region T given by 0 ρ 1, 0 φ α, 0 θ 2π. 76

77 Grad, curl and div 6.9 Theorem. grad f = 1 h 1 f ξ 1 e ξ1 + 1 h 2 f ξ 2 e ξ2 + 1 h 3 f ξ 3 e ξ3, curl F = 1 h 1 h 2 h 3 det div F = 1 h 1 h 2 h 3 ( ξ 1 h 1 e ξ1 h 2 e ξ2 h 3 e ξ3 ξ 1 ξ 2 ξ 3 h 1 F ξ1 h 2 F ξ2 h 3 F ξ3 ( h2 h 3 F ξ1 ) + ξ 2, ( h3 h 1 F ξ2 ) + ξ 3 ( h1 h 2 F ξ3 ) ) Example. In spherical coordinates, grad f = f ρ e ρ + 1 ρ f φ e φ + 1 ρ sin φ f θ e θ. 77

78 6.11 Exercise. Verify in cylindrical coordinates that div F = 0, curl F = 2ω e z for the vector field F = ωy i + ωx j. 78

79 7. Potential Theory The Laplace Operator The Laplace operator (or Laplacian) 2 is defined by In Cartesian coordinates, so or 2 f = ( f) = div(grad f). f = f x 1 e 1 + f x 2 e 2 + f x 3 e 3 2 f = 2 f x f x f x f = 2 f x f y f z 2. 79

80 7.1 Exercise. Show that if (ξ 1, ξ 2, ξ 3 ) are orthogonal curvilinear coordinates with metric ds 2 = h 2 1 dξ2 1 + h2 2 dξ2 2 + h2 3 dξ2 3 then 2 f = 1 h 1 h 2 h 3 [ ξ 1 ( h2 h 3 h 1 f ξ 1 ) + ( h3 h 1 ξ 2 h 2 f ξ 2 ) + ( h1 h 2 ξ 3 h 3 f ξ 3 )]. 7.2 Exercise. Verify that for spherical coordinates (ρ, φ, θ) we have 2 f = 1 ρ 2 ρ ( ρ 2 f ρ ) + 1 ρ 2 sin φ φ ( sin φ f φ ) + 1 ρ 2 sin 2 φ 2 f θ 2 80

81 Green s Identities Let U be a region in R 3 satisfying our assumptions for the Divergence Theorem. (Note: U is a closed set so U U.) We say that a function ϕ is C k on U if all partial derivatives of ϕ of order k exist and are continuous on a neighbourhood of U. Thus, for ϕ to be C 2 we require continuity of ϕ, ϕ x, 2 ϕ x y = 2 ϕ y x, ϕ y, ϕ z, 2 ϕ x 2, 2 ϕ y z = 2 ϕ z y, 2 ϕ y 2, 2 ϕ z 2, 2 ϕ z x = 2 ϕ x z. If ϕ is C 1 on U then we define its normal derivative by ϕ = n(x) ϕ(x) for x U, n where, as usual, n is the outward unit normal for U. 81

82 The following identitites are consequences of the Divergence Theorem. 7.3 Theorem (1st Green identity). If ϕ is C 2 on U and if ψ is C 1 on U then U ψ 2 ϕ dv = ψ ϕ U n ds ψ ϕ dv. U 7.4 Theorem (2nd Green identity). If ϕ and ψ are C 2 on U, then U ( ψ 2 ϕ ϕ 2 ψ ) dv = U ( ψ ϕ n ϕ ψ n ) ds. 82

83 Whether the Divergence Theorem may be applied to singular functions depends crucially on the nature of the singularities. 7.5 Theorem. Let ϕ be a continuous function and B(x 0, ɛ) be the ball of radius ɛ about a point x 0. Then lim ϕ(x) ɛ 0 + B ( 1 x x 0 ) ds = 4πϕ(x 0 ). Note that if the Divergence Theorem were applicable then the corresponding volume integral would be zero! This is a consequence of the following observation: 7.6 Exercise. Show that 2 ( 1 x ) = 0, x 0. 83

84 7.7 Corollary (3rd Green identity). If ϕ is C 2 on U, then U ( 1 x y ϕ (y) ϕ(y) n 1 n y x y = ) ds y U 1 x y 2 ϕ(y) dv y 4πϕ(x), if x interior of U, 0, if x / U. In particular, if ϕ and its derivatives decay sufficiently fast at then R 3 1 x y 2 ϕ(y) dv y = 4πϕ(x), 84

85 Application: The Poisson Equation The Dirichlet problem is to find a function ϕ satisfying the Poisson equation subject to the boundary condition 2 ϕ = f in U, (7) ϕ = g on U, for given functions f and g. The first Green identity immediately gives rise to the following uniqueness result. 7.8 Theorem. The Dirichlet problem has at most one C 2 solution. 85

86 7.9 Exercise. In terms of spherical coordinates, the C 2 solution of the boundary value problem 2 ϕ = 1 in U, ϕ = 1 on U, where U is the closed unit ball, is given by ϕ = 5 + ρ

87 The function G(x) = 1 4π x, x 0, is called the fundamental solution of the Laplace equation 2 G(x) = 0, x 0. If ϕ is a solution of Poisson s equation (7) which, along with its derivatives, decays sufficiently fast at then the 3rd Green identity implies that ϕ(x) = 1 4π Conversely, the following key theorem holds: R 3 f(y) x y dv y. (8) 7.10 Theorem. If f : R 3 R is C 2 with sufficiently rapid decay at, then a solution of the Poisson equation is given by the C 2 function (8). 2 ϕ = f on R 3 (9) 87

88 Scalar and Vector Potentials 7.11 Theorem. In a convex region, the following are equivalent: 1. F = ϕ for some scalar potential ϕ; 2. F is irrotational or curl-free, i.e., F = 0; 3. F is conservative, i.e., C F dr = 0 for any closed curve C; 4. F dr = F dr whenever the curves C 1 and C 2 share the C 1 C 2 same starting point and the same finishing point. 88

89 7.12 Exercise. Verify that (φf ) = ( φ) F + φ F Exercise. Show that the central force F = f( x )x is conservative in a convex region. 89

90 7.14 Corollary. If v = 0 and v = 0, then v = ϕ for some scalar field ϕ satisfying Laplace s equation: 2 ϕ = Theorem. In a convex region, the following are equivalent: 1. F = A for some vector potential A; 2. F is solenoidal or divergence-free, i.e., F = Exercise. Show that any two vector potentials A 1 and A 2 for the same solenoidal vector field F are necessarily related by A 1 A 2 = ϕ, where ϕ is a scalar potential. Demonstrate that the frequently adopted gauge div A = 0 is admissible. 90

91 The following integral representations of irrotational or solenoidal vector fields are of importance, for instance, in electromagnetic theory Theorem (Helmholtz; particular case). Let F : R 3 R 3 be a vector field which is C 2 and, together with its derivatives, decays sufficiently rapidly at. 1. If F = 0 then F = ϕ, ϕ(x) = 1 4π R 3 F (y) x y dv y. 2. If F = 0 then and div A = 0. F = A, A(x) = 1 4π R 3 F (y) x y dv y 91

92 8. Electromagnetic Theory Static Electric Fields In a vacuum, a stationary particle with electric charge q 1 located at x 1 creates an electric field E(x) = 1 4πɛ 0 q 1 x x 1 3 (x x 1), x x 1, where ɛ 0 is a constant called the permittivity of free space. A second stationary particle with charge q located at x experiences an electrostatic force qe(x). This force law was discovered experimentally by Coulomb around

93 The electric field due to n charges q 1, q 2,..., q n located at x 1, x 2,..., x n, respectively, is a simple vector sum of the fields due to the individual charges: Since we have E(x) = 1 4πɛ 0 n i=1 q i x x i 3 (x x i). ( ) x y 1 x y 3 = x x y E = grad Φ where Φ, the electrostatic (scalar) potential generated by the n charges, is given by Φ(x) = 1 4πɛ 0 n i=1 q i x x i. In particular, the electrostatic field E is conservative. 93

94 Recall that curl(grad Φ) = ( Φ) = 0 for any scalar function Φ, so the electrostatic field is curl-free: curl E = Theorem (Gauss s Law for discrete charges). For any region U satisfying the assumptions of the divergence theorem, ɛ 0 provided no charges lie on U. U E ds = x i U q i, In words, ɛ 0 (flux of electric field through U) = total charge enclosed by U. 94

95 Now consider a continuous charge distribution with volume density ϱ. The electric field due to the charge in a small region of volume V y around a point y is approximately 1 ϱ(y) V y (x y), 4πɛ 0 x y 3 so the electric field due to all of the electric charge is E(x) = 1 ϱ(y) 4πɛ 0 x y 3 (x y) dv y and once again E = grad Φ, but with the electrostatic potential given by [since x = Φ(x) = 1 4πɛ 0 x is admissible]. ϱ(y) x y dv y 95

96 8.2 Lemma. Throughout R 3, ɛ 0 div E = ϱ. 8.3 Theorem (Gauss s Law). For any region U satisfying the assumptions of the divergence theorem, ɛ 0 E ds = ϱ dv. U U 8.4 Exercise. Show that if the charge distribution is radially symmetric, i.e., if ϱ = ϱ( x ), and ϱ = 0 for x > R then the electrostatic potential Φ is given by Φ = 1 4πɛ 0 Q x where Q is the total charge. for x > R, 96

97 Electric Currents Suppose there are electric charges with density ϱ = ϱ(x, t) moving with velocity v = v(x, t). The current density of these charges is J = ϱv. In time t the charge flowing through a small oriented parallelogram with area vector S = n S is Thus, the flux integral ϱ(v t) n S = (J S) t. S J ds gives the rate at which charge flows through the oriented surface S, i.e., the current flowing through S. 8.5 Theorem (Continuity Equation). Conservation of electric charge implies that ϱ t + div J = 0. 97

98 Static Magnetic Fields If a current flows along a wire with a (very small) cross-section S then we call J S simply the current flowing in the wire. Consider a steady electric current I 1 flowing in a stationary wire loop in the shape of a closed curve C 1 and surrounded by a vacuum. Orient C 1 in the direction of the current. The Biot Savart Law (1820) states that the current creates a magnetic flux density B(x) = µ 0 4π C 1 I 1 dy (x y) x y 3, where µ 0 is a constant called the permeability of free space, and that a second steady current I 2 flowing in a second stationary wire loop C 2 experiences a force F = C 2 I 2 dx B. 98

99 If the current is not confined to a wire then we replace I 1 dy by J dv y to obtain the magnetic flux density Since we have B(x) = µ 0 4π B(x) = µ 0 4π x J(y) x y J(y) (x y) x y 3 dv y. = J(y) (x y) x y 3 x J(y) x y dv y = x µ 0 4π J(y) x y dv y. 99

100 Thus, B = curl A where the magnetic vector potential is Recalling that A(x) = µ 0 4π J(y) x y dv y. div(curl A) = ( A) = 0 for any vector field A, we see that the magnetostatic flux density is divergence-free: div B = Theorem. The magnetostatic flux density B due to the steady motion of charges with current density J satisfies curl B = µ 0 J. 100

101 8.7 Corollary (Ampère s Law). For any oriented surface S, B dx = µ 0 J ds, S S If the current is confined to a wire linking S B dx = µ 0I. S, then 8.8 Example. An axially-symmetric current density J = f(r)e z generates a magnetic flux density where I(r) = B = µ 0 2π I(r) r e θ, r<r J ds = 2π r 0 rf( r) d r. 101

102 Electromagnetic Induction Around 1830, Faraday discovered that a time-varying (i.e., nonsteady) magnetic flux density B induces an electric field E satisfying E dx = d S dt B ds S for any (stationary) oriented surface S. Since S E dx = curl E ds and d S dt S B ds = we see that ( ) B S t + curl E ds = 0. The surface S is arbitrary and hence S B t ds B + curl E = 0. t In particular, a non-steady electric field is generally not curl-free. 102

103 Maxwell s Equations In a vacuum, the electric displacement D and the magnetic field H are given by D = ɛ 0 E and H = 1 µ 0 B. (10) For static fields, we have seen that div D = ϱ (Gauss s law), (11) curl E = 0 (Existence of electrostatic potential), (12) div B = 0 (Existence of magnetic vector potential), (13) curl H = J (Ampère s law). (14) Thus, if E = φ and B = A with the gauge div A = 0 then Gauss s and Ampère s laws reduce to 2 φ = ϱ, 2 A = J, where ɛ 0 = µ 0 = 1 without loss of generality. 103

104 For time-varying fields, (12) is no longer true because Faraday s law of induction implies that B + curl E = 0. (15) t Also, (14) cannot hold in general because it implies that div J = 0 which violates the continuity equation ϱ + div J = 0. (16) t In 1865, Maxwell realised that if (11) were valid for time-varying fields then ( ) ϱ D + div J = div t t + J must vanish by (16). Thus, we can make (14) consistent with (16) by adding a term, D/ t, called the displacement current, to the right hand side, so that curl H = J + D t. 104

105 In this way, we arrive at Maxwell s equations: div D = ϱ, curl E + B t = 0, div B = 0, curl H D t = J. 8.9 Theorem. Maxwell s equations may be brought into the compact form 2 φ t 2 2 φ = ϱ, subject to the Lorentz condition φ t 2 A t 2 2 A = J (17) + div A = 0, (18) where E = φ A t and B = A with ɛ 0 = µ 0 =

106 8.10 Exercise. Show that the wave equation 2 ϕ t 2 = 2 ϕ is invariant under the Lorentz transformation x (cosh λ) x + (sinh λ) t y y z z where λ is an arbitrary constant. t (cosh λ) t + (sinh λ) x, The above exercise implies that Maxwell s equations are compatible with Einstein s Theory of Special Relativity (1905)! 106