Vector Calculus Theorems with Proof

Home , Divergence theorem, Quotient rule, Vector calculus

Part IA — Vector Calculus Theorems with proof

Based on lectures by B. Allanach Notes taken by Dexter Chua Lent 2015

These notes are not endorsed by the lecturers, and I have modiﬁed them (often signiﬁcantly) after lectures. They are nowhere near accurate representations of what was actually lectured, and in particular, all errors are almost surely mine.

3 Curves in R 3 Parameterised curves and arc length, tangents and normals to curves in R , the radius of curvature. [1]

2 3 Integration in R and R Line integrals. Surface and volume integrals: deﬁnitions, examples using Cartesian, cylindrical and spherical coordinates; change of variables. [4]

Vector operators Directional derivatives. The gradient of a real-valued function: definition; interpretation as normal to level surfaces; examples including the use of cylindrical, spherical *and general orthogonal curvilinear* coordinates. Divergence, curl and ∇2 in Cartesian coordinates, examples; formulae for these operators (statement only) in cylindrical, spherical *and general orthogonal curvilinear* coordinates. Solenoidal fields, irrotational fields and conservative fields; scalar potentials. Vector derivative identities. [5]

Integration theorems Divergence theorem, Green’s theorem, Stokes’s theorem, Green’s second theorem: statements; informal proofs; examples; application to ﬂuid dynamics, and to electromagnetism including statement of Maxwell’s equations. [5]

Laplace’s equation 2 3 Laplace’s equation in R and R : uniqueness theorem and maximum principle. Solution of Poisson’s equation by Gauss’s method (for spherical and cylindrical symmetry) and as an integral. [4]

3 Cartesian tensors in R Tensor transformation laws, addition, multiplication, contraction, with emphasis on tensors of second rank. Isotropic second and third rank tensors. Symmetric and antisymmetric tensors. Revision of principal axes and diagonalization. Quotient theorem. Examples including inertia and conductivity. [5]

1 Contents IA Vector Calculus (Theorems with proof)

Contents

0 Introduction 4

1 Derivatives and coordinates 5 1.1 Derivative of functions ...... 5 1.2 Inverse functions ...... 5 1.3 Coordinate systems ...... 5

2 Curves and Line 6 2.1 Parametrised curves, lengths and arc length ...... 6 2.2 Line integrals of vector ﬁelds ...... 6 2.3 Gradients and Diﬀerentials ...... 6 2.4 Work and potential energy ...... 6

3 Integration in R2 and R3 7 3.1 Integrals over subsets of R2 ...... 7 3.2 Change of variables for an integral in R2 ...... 7 3.3 Generalization to R3 ...... 8 3.4 Further generalizations ...... 8

4 Surfaces and surface integrals 9 4.1 Surfaces and Normal ...... 9 4.2 Parametrized surfaces and area ...... 9 4.3 Surface integral of vector ﬁelds ...... 9 4.4 Change of variables in R2 and R3 revisited ...... 9 5 Geometry of curves and surfaces 10

6 Div, Grad, Curl and ∇ 11 6.1 Div, Grad, Curl and ∇ ...... 11 6.2 Second-order derivatives ...... 11

7 Integral theorems 12 7.1 Statement and examples ...... 12 7.1.1 Green’s theorem (in the plane) ...... 12 7.1.2 Stokes’ theorem ...... 12 7.1.3 Divergence/Gauss theorem ...... 12 7.2 Relating and proving integral theorems ...... 12

8 Some applications of integral theorems 17 8.1 Integral expressions for div and curl ...... 17 8.2 Conservative ﬁelds and scalar products ...... 17 8.3 Conservation laws ...... 18

9 Orthogonal curvilinear coordinates 19 9.1 Line, area and volume elements ...... 19 9.2 Grad, Div and Curl ...... 19

2 Contents IA Vector Calculus (Theorems with proof)

10 Gauss’ Law and Poisson’s equation 21 10.1 Laws of gravitation ...... 21 10.2 Laws of electrostatics ...... 21 10.3 Poisson’s Equation and Laplace’s equation ...... 21

11 Laplace’s and Poisson’s equations 22 11.1 Uniqueness theorems ...... 22 11.2 Laplace’s equation and harmonic functions ...... 23 11.2.1 The mean value property ...... 23 11.2.2 The maximum (or minimum) principle ...... 23 11.3 Integral solutions of Poisson’s equations ...... 24 11.3.1 Statement and informal derivation ...... 24 11.3.2 Point sources and δ-functions* ...... 24

12 Maxwell’s equations 25 12.1 Laws of electromagnetism ...... 25 12.2 Static charges and steady currents ...... 25 12.3 Electromagnetic waves ...... 25

13 Tensors and tensor ﬁelds 26 13.1 Deﬁnition ...... 26 13.2 Tensor algebra ...... 26 13.3 Symmetric and antisymmetric tensors ...... 26 13.4 Tensors, multi-linear maps and the quotient rule ...... 26 13.5 Tensor calculus ...... 26

14 Tensors of rank 2 28 14.1 Decomposition of a second-rank tensor ...... 28 14.2 The inertia tensor ...... 28 14.3 Diagonalization of a symmetric second rank tensor ...... 28

15 Invariant and isotropic tensors 29 15.1 Deﬁnitions and classiﬁcation results ...... 29 15.2 Application to invariant integrals ...... 30

3 0 Introduction IA Vector Calculus (Theorems with proof)

0 Introduction

4 1 Derivatives and coordinates IA Vector Calculus (Theorems with proof)

1 Derivatives and coordinates 1.1 Derivative of functions Proposition. 0 0 F (x) = Fi (x)ei. Proposition. d df dg (fg) = g + f dt dt dt d dg dh (g · h) = · h + g · dt dt dt d dg dh (g × h) = × h + g × dt dt dt Note that the order of multiplication must be retained in the case of the cross product. Theorem. The gradient is ∂f ∇f = ei ∂xi Theorem (Chain rule). Given a function f(r(u)),

df dr ∂f dx = ∇f · = i . du du ∂xi du Theorem. The derivative of F is given by

∂yj Mji = . ∂xi

Theorem (Chain rule). Suppose g : Rp → Rn and f : Rn → Rm. Suppose that p n m the coordinates of the vectors in R , R and R are ua, xi and yr respectively. By the chain rule, ∂y ∂y ∂x r = r i , ∂ua ∂xi ∂ua with summation implied. Writing in matrix form,

M(f ◦ g)ra = M(f)riM(g)ia.

Alternatively, in operator form, ∂ ∂x ∂ = i . ∂ua ∂ua ∂xi

1.2 Inverse functions 1.3 Coordinate systems

5 2 Curves and Line IA Vector Calculus (Theorems with proof)

2 Curves and Line 2.1 Parametrised curves, lengths and arc length Proposition. Let s denote the arclength of a curve r(u). Then

ds dr 0 = ± = ±|r (u)| du du with the sign depending on whether it is in the direction of increasing or decreasing arclength. Proposition. ds = ±|r0(u)|du

2.2 Line integrals of vector ﬁelds 2.3 Gradients and Diﬀerentials Theorem. If F = ∇f(r), then Z F · dr = f(b) − f(a), C where b and a are the end points of the curve. In particular, the line integral does not depend on the curve, but the end points only. This is the vector counterpart of the fundamental theorem of H calculus. A special case is when C is a closed curve, then C F · dr = 0. Proof. Let r(u) be any parametrization of the curve, and suppose a = r(α), b = r(β). Then

Z Z Z dr F · dr = ∇f · dr = ∇f · du. C C du So by the chain rule, this is equal to

Z β d β (f(r(u))) du = [f(r(u))]α = f(b) − f(a). α du Proposition. If F = ∇f for some f, then ∂F ∂F i = j . ∂xj ∂xi

2 This is because both are equal to ∂ f/∂xi∂xj. Proposition.

d(λf + µg) = λdf + µdg d(fg) = (df)g + f(dg)

2.4 Work and potential energy

6 3 Integration in R2 and R3 IA Vector Calculus (Theorems with proof)

3 Integration in R2 and R3 3.1 Integrals over subsets of R2 Proposition. Z Z Z ! f(x, y) dA = f(x, y) dx dy. D Y xy with xy ranging over {x :(x, y) ∈ D}. Theorem (Fubini’s theorem). If f is a continuous function and D is a compact (i.e. closed and bounded) subset of R2, then ZZ ZZ f dx dy = f dy dx.

While we have rather strict conditions for this theorem, it actually holds in many more cases, but those situations have to be checked manually. Proposition. dA = dx dy in Cartesian coordinates. Proposition. Take separable f(x, y) = h(y)g(x) and D be a rectangle {(x, y): a ≤ x ≤ b, c ≤ y ≤ d}. Then

Z Z b ! Z d ! f(x, y) dx dy = g(x) dx h(y) dy D a c

3.2 Change of variables for an integral in R2 Proposition. Suppose we have a change of variables (x, y) ↔ (u, v) that is smooth and invertible, with regions D,D0 in one-to-one correspondence. Then Z Z f(x, y) dx dy = f(x(u, v), y(u, v))|J| du dv, D D0 where ∂x ∂x

∂(x, y) ∂u ∂v J = = ∂(u, v) ∂y ∂y ∂u ∂v is the Jacobian. In other words, dx dy = |J| du dv. Proof. Since we are writing (x(u, v), y(u, v)), we are actually transforming from (u, v) to (x, y) and not the other way round. Suppose we start with an area δA0 = δuδv in the (u, v) plane. Then by Taylors’ theorem, we have ∂x ∂x δx = x(u + δu, v + δv) − x(u, v) ≈ δu + δv. ∂u ∂v We have a similar expression for δy and we obtain δx ∂x ∂x δu ≈ ∂u ∂v δy ∂y ∂y δv ∂u ∂v

7 3 Integration in R2 and R3 IA Vector Calculus (Theorems with proof)

Recall from Vectors and Matrices that the determinant of the matrix is how much it scales up an area. So the area formed by δx and δy is |J| times the area formed by δu and δv. Hence

dx dy = |J| du dv.

3.3 Generalization to R3 Proposition. dV = dx dy dz. Proposition. Z Z f dx dy dz = f|J| du dv dw, V V with

∂x ∂x ∂x

∂u ∂v ∂w

∂(x, y, z) ∂y ∂y ∂y J = = ∂(u, v, w) ∂u ∂v ∂w

∂z ∂z ∂z

∂u ∂v ∂w Proposition. In cylindrical coordinates,

dV = ρ dρ dϕ dz.

In spherical coordinates

dV = r2 sin θ dr dθ dϕ.

Proof. Loads of algebra.

3.4 Further generalizations Proposition. Z Z f(x1, x2, ··· xn) dx1 dx2 ··· dxn = f({xi(u)})|J| du1 du2 ··· dun. D D0

8 4 Surfaces and surface integrals IA Vector Calculus (Theorems with proof)

4 Surfaces and surface integrals 4.1 Surfaces and Normal Proposition. ∇f is the normal to the surface f(r) = c.

4.2 Parametrized surfaces and area Proposition. The vector area element is ∂r ∂r dS = × du dv. ∂u ∂v The scalar area element is

∂r ∂r dS = × du dv. ∂u ∂v

4.3 Surface integral of vector ﬁelds 4.4 Change of variables in R2 and R3 revisited

9 5 Geometry of curves and surfacesIA Vector Calculus (Theorems with proof)

5 Geometry of curves and surfaces

Theorem (Theorema Egregium). K is intrinsic to the surface S. It can be expressed in terms of lengths, angles etc. which are measured entirely on the surface. So K can be deﬁned on an arbitrary surface without embedding it on a higher dimension surface. Theorem (Gauss-Bonnet theorem). Z θ1 + θ2 + θ3 = π + K dA, D integrating over the area of the triangle.

10 6 Div, Grad, Curl and ∇ IA Vector Calculus (Theorems with proof)

6 Div, Grad, Curl and ∇ 6.1 Div, Grad, Curl and ∇ Proposition. Let f, g be scalar functions, F, G be vector functions, and µ, λ be constants. Then

∇(λf + µg) = λ∇f + µ∇g ∇ · (λF + µG) = λ∇ · F + µ∇ · G ∇ × (λF + µG) = λ∇ × F + µ∇ × G.

Proposition. We have the following Leibnitz properties:

∇(fg) = (∇f)g + f(∇g) ∇ · (fF) = (∇f) · F + f(∇ · F) ∇ × (fF) = (∇f) × F + f(∇ × F) ∇(F · G) = F × (∇ × G) + G × (∇ × F) + (F · ∇)G + (G · ∇)F ∇ × (F × G) = F(∇ · G) − G(∇ · F) + (G · ∇)F − (F · ∇)G ∇ · (F × G) = (∇ × F) · G − F · (∇ × G) which can be proven by brute-forcing with suﬃx notation and summation convention.

6.2 Second-order derivatives Proposition.

∇ × (∇f) = 0 ∇ · (∇ × F) = 0

Proof. Expand out using suﬃx notation, noting that

∂2f εijk = 0. ∂xi∂xj since if, say, k = 3, then

∂2f ∂2f ∂2f εijk = − = 0. ∂xi∂xj ∂x1∂x2 ∂x2∂x1

Proposition. If F is deﬁned in all of R3, then ∇ × F = 0 ⇒ F = ∇f

for some f.

Proposition. If H is deﬁned over all of R3 and ∇ · H = 0, then H = ∇ × A for some A.

11 7 Integral theorems IA Vector Calculus (Theorems with proof)

7 Integral theorems 7.1 Statement and examples 7.1.1 Green’s theorem (in the plane) Theorem (Green’s theorem). For smooth functions P (x, y), Q(x, y) and A a bounded region in the (x, y) plane with boundary ∂A = C,

Z ∂Q ∂P Z − dA = (P dx + Q dy). A ∂x ∂y C Here C is assumed to be piecewise smooth, non-intersecting closed curve, tra- versed anti-clockwise.

7.1.2 Stokes’ theorem Theorem (Stokes’ theorem). For a smooth vector ﬁeld F(r), Z Z ∇ × F · dS = F · dr, S ∂S where S is a smooth, bounded surface and ∂S is a piecewise smooth boundary of S. The direction of the line integral is as follows: If we walk along C with n facing up, then the surface is on your left.

7.1.3 Divergence/Gauss theorem Theorem (Divergence/Gauss theorem). For a smooth vector ﬁeld F(r), Z Z ∇ · F dV = F · dS, V ∂V where V is a bounded volume with boundary ∂V , a piecewise smooth, closed surface, with outward normal n.

7.2 Relating and proving integral theorems Proposition. Stokes’ theorem ⇒ Green’s theorem Proof. Stokes’ theorem talks about 3D surfaces and Green’s theorem is about 2D regions. So given a region A on the (x, y) plane, we pretend that there is a third dimension and apply Stokes’ theorem to derive Green’s theorem. Let A be a region in the (x, y) plane with boundary C = ∂A, parametrised by arc length, (x(s), y(s), 0). Then the tangent to C is

dx dy t = , , 0 . ds ds

Given any P (x, y) and Q(x, y), we can consider the vector ﬁeld

F = (P, Q, 0),

12 7 Integral theorems IA Vector Calculus (Theorems with proof)

So ∂Q ∂P ∇ × F = 0, 0, − . ∂x ∂y Then the left hand side of Stokes is Z Z Z F · dr = F · t ds = P dx + Q dy, C C C and the right hand side is Z Z ∂Q ∂P (∇ × F) · kˆ dA = − dA. A A ∂x ∂y Proposition. Green’s theorem ⇒ Stokes’ theorem. Proof. Green’s theorem describes a 2D region, while Stokes’ theorem describes a 3D surface r(u, v). Hence to use Green’s to derive Stokes’ we need ﬁnd some 2D thing to act on. The natural choice is the parameter space, u, v. Consider a parametrised surface S = r(u, v) corresponding to the region A in the u, v plane. Write the boundary as ∂A = (u(t), v(t)). Then ∂S = r(u(t), v(t)). We want to prove Z Z F · dr = (∇ × F) · dS ∂S S given Z Z ∂Fv ∂Fu Fu du + Fv dv = − dA. ∂A A ∂u ∂v Doing some pattern-matching, we want

F · dr = Fu du + Fv dv

for some Fu and Fv. By the chain rule, we know that ∂r ∂r dr = du + dv. ∂u ∂v So we choose ∂r ∂r F = F · ,F = F · . u ∂u v ∂v This choice matches the left hand sides of the two equations. To match the right, recall that ∂r ∂r (∇ × F) · dS = (∇ × F) · × du dv. ∂u ∂v Therefore, for the right hand sides to match, we want ∂F ∂F ∂r ∂r v − u = (∇ × F) · × . (∗) ∂u ∂v ∂u ∂v Fortunately, this is true. Unfortunately, the proof involves complicated suﬃx notation and summation convention: ∂Fv ∂ ∂r ∂ ∂xi ∂Fi ∂xj ∂xi ∂xi = F · = Fi = + Fi . ∂u ∂u ∂v ∂u ∂v ∂xj ∂u ∂v ∂u∂v

13 7 Integral theorems IA Vector Calculus (Theorems with proof)

Similarly, ∂Fu ∂ ∂r ∂ ∂xj ∂Fj ∂xi ∂xj ∂xi = F · = Fj = + Fi . ∂u ∂u ∂u ∂u ∂u ∂xi ∂v ∂u ∂u∂v So ∂F ∂F ∂x ∂x ∂F ∂F v − u = j i i − j . ∂u ∂v ∂u ∂v ∂xj ∂xi This is the left hand side of (∗). The right hand side of (∗) is ∂r ∂r ∂Fj ∂xp ∂xq (∇ × F) · × = εijk εkpq ∂u ∂v ∂xi ∂u ∂v ∂Fj ∂xp ∂xq = (δipδjq − δiqδjp) ∂xi ∂u ∂v ∂F ∂F ∂x ∂x = j − i i j . ∂xi ∂xj ∂u ∂v

So they match. Therefore, given our choice of Fu and Fv, Green’s theorem translates to Stokes’ theorem. Proposition. Greens theorem ⇔ 2D divergence theorem. Proof. The 2D divergence theorem states that Z Z (∇ · G) dA = G · n ds. A ∂A with an outward normal n. Write G as (Q, −P ). Then ∂Q ∂P ∇ · G = − . ∂x ∂y Around the curve r(s) = (x(s), y(s)), t(s) = (x0(s), y0(s)). Then the normal, being tangent to t, is n(s) = (y0(s), −x0(s)) (check that it points outwards!). So dx dy G · n = P + Q . ds ds Then we can expand out the integrals to obtain Z Z G · n ds = P dx + Q dy, C C and Z Z ∂Q ∂P (∇ · G) dA = − dA. A A ∂x ∂y Now 2D version of Gauss’ theorem says the two LHS are the equal, and Green’s theorem says the two RHS are equal. So the result follows. Proposition. 2D divergence theorem. Z Z (∇ · G) dA = G · n ds. A C=∂A

14 7 Integral theorems IA Vector Calculus (Theorems with proof)

Proof. For the sake of simplicity, we assume that G only has a vertical component, noting that the same proof works for purely horizontal G, and an arbitrary G is just a linear combination of the two. Furthermore, we assume that A is a simple, convex shape. A more complicated shape can be cut into smaller simple regions, and we can apply the simple case to each of the small regions. Suppose G = G(x, y)ˆj. Then

∂G ∇ · G = . ∂y Then Z Z Z ∂G ∇ · G dA = dy dx. A X Yx ∂y

Now we divide A into an upper and lower part, with boundaries C+ = y+(x) and C− = y−(x) respectively. Since I cannot draw, A will be pictured as a circle, but the proof is valid for any simple convex shape. y

C+

Yx C−

x dy

We see that the boundary of Yx at any speciﬁc x is given by y−(x) and y+(x). Hence by the Fundamental theorem of Calculus,

Z ∂G Z y+(x) ∂G dy = dy = G(x, y+(x)) − G(x, y−(x)). Yx ∂y y−(x) ∂y To compute the full area integral, we want to integrate over all x. However, the divergence theorem talks in terms of ds, not dx. So we need to ﬁnd some way to relate ds and dx. If we move a distance δs, the change in x is δs cos θ, where θ is the angle between the tangent and the horizontal. But θ is also the angle between the normal and the vertical. So cos θ = n · ˆj. Therefore dx = ˆj · n ds. In particular, G dx = Gˆj · n ds = G · n ds, since G = Gˆj. ˆ However, at C−, n points downwards, so n · j happens to be negative. So, actually, at C−, dx = −G · n ds.

15 7 Integral theorems IA Vector Calculus (Theorems with proof)

Therefore, our full integral is

Z Z Z ∂G ∇ · G dA = dY dx A X yx ∂y Z = G(x, y+(x)) − G(x, y−(x)) dx X Z Z = G · n ds + G · n ds C+ C− Z = G · n ds. C

16 8 Some applications of integral theoremsIA Vector Calculus (Theorems with proof)

8 Some applications of integral theorems 8.1 Integral expressions for div and curl Proposition. 1 Z (∇ · F)(r0) = lim F · dS, diam(V )→1 vol(V ) ∂V where the limit is taken over volumes containing the point r0. Proposition. 1 Z n · (∇ × F)(r0) = lim F · dr, diam(A)→0 area(A) ∂A where the limit is taken over all surfaces A containing r0 with normal n.

8.2 Conservative ﬁelds and scalar products R Proposition. If (iii) ∇ × F = 0, then (ii) C F · dr is independent of C. Proof. Given F(r) satisfying ∇ × F = 0, let C and C˜ be any two curves from a to b. C b

C˜

If S is any surface with boundary ∂S = C − C˜, By Stokes’ theorem, Z Z Z Z ∇ × F · dS = F · dr = F · dr − F · dr. S ∂S C C˜ But ∇ × F = 0. So Z Z F · dr − F · dr = 0, C C˜ or Z Z F · dr = F · dr. C C˜ R Proposition. If (ii) C F · dr is independent of C for fixed end points and orientation, then (i) F = ∇f for some scalar field f. R 0 0 Proof. We fix a and define f(r) = C F(r ) · dr for any curve from a to r. Assuming (ii), f is well-defined. For small changes r to r + δr, there is a small extension of C by δC. Then Z f(r + δr) = F(r0) · dr0 C+δC Z Z = F · dr0 + F · dr0 C δC = f(r) + F(r) · δr + o(δr).

17 8 Some applications of integral theoremsIA Vector Calculus (Theorems with proof)

So δf = f(r + δr) − f(r) = F(r) · δr + o(δr). But the deﬁnition of grad is exactly

δf = ∇f · δr + o(δr).

So we have F = ∇f.

8.3 Conservation laws

18 9 Orthogonal curvilinear coordinatesIA Vector Calculus (Theorems with proof)

9 Orthogonal curvilinear coordinates 9.1 Line, area and volume elements 9.2 Grad, Div and Curl Proposition. 1 ∂f 1 ∂f 1 ∂f ∇f = eu + ev + ew. hu ∂u hv ∂v hw ∂w Proposition. 1 ∂ 1 ∂ 1 ∂ ∇ = eu + ev + ew . hu ∂u hv ∂v hw ∂w Proposition.

1 ∂ ∂ ∇ × F = (hwFw) − (hvFv) eu + two similar terms hvhw ∂v ∂w

hueu hvev hwew 1 ∂ ∂ ∂ = ∂u ∂v ∂w huhvhw huFu hvFv hwFw

and 1 ∂ ∇ · F = (hvhwFu) + two similar terms . huhvhw ∂u Proof. (non-examinable)

(i) Apply ∇· or ∇× and diﬀerentiate the basis vectors explicitly. (ii) First, apply ∇· or ∇×, but calculate the result by writing F in terms of ∇u, ∇v and ∇w in a suitable way. Then use ∇×∇f = 0 and ∇·(∇×f) = 0. (iii) Use the integral expressions for div and curl. Recall that 1 Z n · ∇ × F = lim F · dr. A→0 A ∂A

So to calculate the curl, we first find the ew component. Consider an area with W fixed and change u by δu and v by δv. Then this has an area of huhvδuδv with normal ew. Let C be its boundary.

δv C

u δu

We then integrate around the curve C. We split the curve C up into 4 parts (corresponding to the four sides), and take linear approximations by assuming F and h are constant when moving through each horizontal/vertical

19 9 Orthogonal curvilinear coordinatesIA Vector Calculus (Theorems with proof)

segment. Z F · dr ≈ Fu(u, v)hu(u, v) δu + Fv(u + δu, v)hv(u + δu, v) δu C − Fu(u, v + δv)hu(u, v + δv) δu − Fv(u, v)hv(u, v) δv ∂ ∂ ≈ h F − (h F ) δuδv. ∂u v v ∂v u u

Divide by the area and take the limit as area → 0, we obtain

1 Z 1 ∂ ∂ lim F · dr = hvFv − (huFu) . A→0 A C huhv ∂u ∂v So, by the integral deﬁnition of divergence,

1 ∂ ∂ ew · ∇ × F = (hvFv) − (huFu) , huhv ∂u ∂v and similarly for other components. We can ﬁnd the divergence similarly.

20 10 Gauss’ Law and Poisson’s equationIA Vector Calculus (Theorems with proof)

10 Gauss’ Law and Poisson’s equation 10.1 Laws of gravitation Law (Gauss’ law for gravitation). Given any volume V bounded by closed surface S, Z g · dS = −4πGM, S where G is Newton’s gravitational constant, and M is the total mass contained in V . Law (Gauss’ Law for gravitation in diﬀerential form).

∇ · g = −4πGρ.

10.2 Laws of electrostatics Law (Gauss’ law for electrostatic forces).

Z Q E · dS = , S ε0 where ε0 is the permittivity of free space, or electric constant. Law (Gauss’ law for electrostatic forces in diﬀerential form). ρ ∇ · E = . ε0

10.3 Poisson’s Equation and Laplace’s equation

21 11 Laplace’s and Poisson’s equationsIA Vector Calculus (Theorems with proof)

11 Laplace’s and Poisson’s equations 11.1 Uniqueness theorems Theorem. Consider ∇2ϕ = −ρ for some ρ(r) on a bounded volume V with S = ∂V being a closed surface, with an outward normal n. Suppose ϕ satisﬁes either

(i) Dirichlet condition, ϕ(r) = f(r) on S

∂ϕ(r) (ii) Neumann condition ∂n = n · ∇ϕ = g(r) on S. where f, g are given. Then (i) ϕ(r) is unique

(ii) ϕ(r) is unique up to a constant.

Proof. Let ϕ1(r) and ϕ2(r) satisfy Poisson’s equation, each obeying the boundary 2 conditions (N) or (D). Then Ψ(r) = ϕ2(r) − ϕ1(r) satisﬁes ∇ Ψ = 0 on V by linearity, and

(i) Ψ = 0 on S; or

∂Ψ (ii) ∂n = 0 on S. ∂Ψ Combining these two together, we know that Ψ ∂n = 0 on the surface. So using the divergence theorem, Z Z ∇ · (Ψ∇Ψ) dV = (Ψ∇Ψ) · dS = 0. V S But ∇ · (Ψ∇Ψ) = (∇Ψ) · (∇Ψ) + Ψ ∇2Ψ = |(∇Ψ)|2. |{z} =0 So Z |∇Ψ|2 dV = 0. V Since |∇Ψ|2 ≥ 0, the integral can only vanish if |∇Ψ| = 0. So ∇Ψ = 0. So Ψ = c, a constant on V . So

(i) Ψ = 0 on S ⇒ c = 0. So ϕ1 = ϕ2 on V .

(ii) ϕ2(r) = ϕ1(r) + C, as claimed. Proposition (Green’s ﬁrst identity). Z Z Z (u∇v) · dS = (∇u) · (∇v) dV + u∇2v dV, S V V Proposition (Green’s second identity). Z Z (u∇v − v∇u) · dS = (u∇2v − v∇2u) dV. S V

22 11 Laplace’s and Poisson’s equationsIA Vector Calculus (Theorems with proof)

11.2 Laplace’s equation and harmonic functions 11.2.1 The mean value property Proposition (Mean value property). Suppose ϕ(r) is harmonic on region V containing a solid sphere deﬁned by |r−a| ≤ R, with boundary SR = |r−a| = R, for some R. Deﬁne 1 Z ϕ¯(R) = 2 ϕ(r) dS. 4πR SR Then ϕ(a) =ϕ ¯(R). Proof. Note that ϕ¯(R) → ϕ(a) as R → 0. We take spherical coordinates (u, θ, χ) centered on r = a. The scalar element (when u = R) on SR is

dS = R2 sin θ dθ dχ.

dS So R2 is independent of R. Write 1 Z dS ϕ¯(R) = ϕ . 4π R2

Diﬀerentiate this with respect to R, noting that dS/R2 is independent of R. Then we obtain Z d 1 ∂ϕ ϕ¯(R) = 2 dS dR 4πR ∂u u=R But ∂ϕ ∂ϕ = e · ∇ϕ = n · ∇ϕ = ∂u u ∂n

on SR. So Z Z d 1 1 2 ϕ¯(R) = 2 ∇ϕ · dS = 2 ∇ ϕ dV = 0 dR 4πR SR 4πR VR by divergence theorem. So ϕ¯(R) does not depend on R, and the result follows.

11.2.2 The maximum (or minimum) principle Proposition (Maximum principle). If a function ϕ is harmonic on a region V , then ϕ cannot have a maximum at an interior point of a of V . Proof. Suppose that ϕ had a local maximum at a in the interior. Then there is an ε such that for any r such that 0 < |r − a| < ε, we have ϕ(r) < ϕ(a). Note that if there is an ε that works, then any smaller ε will work. Pick an ε suﬃciently small such that the region |r − a| < ε lies within V (possible since a lies in the interior of V ). Then for any r such that |r − a| = ε, we have ϕ(r) < ϕ(a).

1 Z ϕ¯(ε) = 2 ϕ(r) dS < ϕ(a), 4πR SR which contradicts the mean value property.

23 11 Laplace’s and Poisson’s equationsIA Vector Calculus (Theorems with proof)

11.3 Integral solutions of Poisson’s equations 11.3.1 Statement and informal derivation Proposition. The solution to Poisson’s equation ∇2ϕ = −ρ, with boundary conditions |ϕ(r)| = O(1/|r|) and |∇ϕ(r)| = O(1/|r|2), is

Z 0 1 ρ(r ) 0 ϕ(r) = 0 dV 4π V 0 |r − r | For ρ(r0) non-zero everywhere, but suitably well-behaved as |r0| → ∞, we can also take V 0 = R3.

11.3.2 Point sources and δ-functions*

24 12 Maxwell’s equations IA Vector Calculus (Theorems with proof)

12 Maxwell’s equations 12.1 Laws of electromagnetism Law (Lorentz force law). A point charge q experiences a force of

F = q(E + r˙ × B).

Law (Maxwell’s equations). ρ ∇ · E = ε0 ∇ · B = 0 ∂B ∇ × E + = 0 ∂t ∂E ∇ × B − µ ε = µ j, 0 0 ∂t 0 where ε0 is the electric constant (permittivity of free space) and µ0 is the magnetic constant (permeability of free space), which are constants determined experimentally.

12.2 Static charges and steady currents 12.3 Electromagnetic waves

25 13 Tensors and tensor ﬁelds IA Vector Calculus (Theorems with proof)

13 Tensors and tensor ﬁelds 13.1 Deﬁnition 13.2 Tensor algebra 13.3 Symmetric and antisymmetric tensors 13.4 Tensors, multi-linear maps and the quotient rule

Proposition (Quotient rule). Suppose that Ti···jp···q is an array deﬁned in each coordinate system, and that vi···j = Ti···jp···qup···q is also a tensor for any tensor up···q. Then Ti···jp···q is also a tensor. Proof. We can check the tensor transformation rule directly. However, we can reuse the result above to save some writing. Consider the special form up···q = cp ··· dq for any vectors c, ··· d. By assumption, vi···j = Ti···jp···qcp ··· dq is a tensor. Then

vi···jai ··· bj = Ti···jp···qai ··· bjcp ··· dq

is a scalar for any vectors a, ··· , b, c, ··· , d. Since Ti···jp···qai ··· bjcp ··· dq is a scalar and hence gives the same result in every coordinate system, Ti···jp···q is a multi-linear map. So Ti···jp···q is a tensor.

13.5 Tensor calculus Proposition. ∂ ∂ ··· T , (∗) ∂x ∂x ij ··· k p q | {z } | {z } n m is a tensor of rank n + m. Proof. To show this, it suﬃces to show that ∂ satisﬁes the tensor transfor- ∂xp mation rules for rank 1 tensors (i.e. it is something like a rank 1 tensor). Then by the exact same argument we used to show that tensor products preserve tensorness, we can show that the (∗) is a tensor. (we cannot use the result of tensor products directly, since this is not exactly a product. But the exact same proof works!) 0 Since xi = Riqxq, we have

0 ∂xi = Rip. ∂xp

∂xp (noting that = δpq). Similarly, ∂xq

∂xq 0 = Riq. ∂xi

Note that Rip,Riq are constant matrices.

26 13 Tensors and tensor ﬁelds IA Vector Calculus (Theorems with proof)

Hence by the chain rule, ∂ ∂xq ∂ ∂ 0 = 0 = Riq . ∂xi ∂xi ∂xq ∂xq

So ∂ obeys the vector transformation rule. So done. ∂xp Theorem (Divergence theorem for tensors).

Z Z ∂ Tij···k`n` dS = (Tij···k`) dV, S V ∂x` with n being an outward pointing normal. Proof. Apply the usual divergence theorem to the vector field v defined by v` = aibj ··· ckTij···k`, where a, b, ··· , c are fixed constant vectors. Then ∂v` ∂ ∇ · v = = aibj ··· ck Tij···k`, ∂x` ∂x` and n · v = n`v` = aibj ··· ckTij···k`n`. Since a, b, ··· , c are arbitrary, therefore they can be eliminated, and the tensor divergence theorem follows.

27 14 Tensors of rank 2 IA Vector Calculus (Theorems with proof)

14 Tensors of rank 2 14.1 Decomposition of a second-rank tensor 14.2 The inertia tensor 14.3 Diagonalization of a symmetric second rank tensor

28 15 Invariant and isotropic tensorsIA Vector Calculus (Theorems with proof)

15 Invariant and isotropic tensors 15.1 Deﬁnitions and classiﬁcation results Theorem. (i) There are no isotropic tensors of rank 1, except the zero tensor.

(ii) The most general rank 2 isotropic tensor is Tij = αδij for some scalar α.

(iii) The most general rank 3 isotropic tensor is Tijk = βεijk for some scalar β.

(iv) All isotropic tensors of higher rank are obtained by combining δij and εijk using tensor products, contractions, and linear combinations.

Proof. We analyze conditions for invariance under speciﬁc rotations through π or π/2 about coordinate axes.

(i) Suppose Ti is rank-1 isotropic. Consider a rotation about x3 through π:

−1 0 0 (Rij) =  0 −1 0 . 0 0 1

We want T1 = RipTp = R11T1 = −T1. So T1 = 0. Similarly, T2 = 0. By consider a rotation about, say x1, we have T3 = 0.

(ii) Suppose Tij is rank-2 isotropic. Consider

 0 1 0 (Rij) = −1 0 0 , 0 0 1

which is a rotation through π/2 about the x3 axis. Then

T13 = R1pR3qTpq = R12R33T23 = T23

and T23 = R2pR3qTpq = R21R33T13 = −T13

So T13 = T23 = 0. Similarly, we have T31 = T32 = 0. We also have T11 = R1pR1qTpq = R12R12T22 = T22.

So T11 = T22.

By picking a rotation about a diﬀerent axis, we have T21 = T12 and T22 = T33.

Hence Tij = αδij.

(iii) Suppose that Tijk is rank-3 isotropic. Using the rotation by π about the x3 axis, we have

T133 = R1pR3qR3rTpqr = −T133.

29 15 Invariant and isotropic tensorsIA Vector Calculus (Theorems with proof)

So T133 = 0. We also have

T111 = R1pR1qR1rTpqr = −T111.

So T111 = 0. We have similar results for π rotations about other axes and other choices of indices.

Then we can show that Tijk = 0 unless all i, j, k are distinct. Now consider  0 1 0 (Rij) = −1 0 0 , 0 0 1

a rotation about x3 through π/2. Then

T123 = R1pR2qR3rTpqr = R12R21R33T213 = −T213.

So T123 = −T213. Along with similar results for other indices and axes of rotation, we ﬁnd that Tijk is totally antisymmetric, and Tijk = βεijk for some β.

15.2 Application to invariant integrals Theorem. Let Z Tij···k = f(x)xixj ··· xk dV. V where f(x) is a scalar function and V is some volume. Given a rotation Rij, consider an active transformation: x = xiei is mapped 0 0 0 to x = xiei with xi = Rijxi, i.e. we map the components but not the basis, and V is mapped to V 0. Suppose that under this active transformation, (i) f(x) = f(x0), (ii) V 0 = V (e.g. if V is all of space or a sphere).

Then Tij···k is invariant under the rotation. Proof. First note that the Jacobian of the transformation R is 1, since it is 0 0 ∂xi simply the determinant of R (x = Ripxp ⇒ = Rip), which is by deﬁnition i ∂xp 1. So dV = dV 0. Then we have Z 0 0 0 RipRjq ··· RkrTpq···r = f(x)xixj ··· xk dV V Z 0 0 0 0 = f(x )xixj ··· xk dV using (i) V Z 0 0 0 0 0 = f(x )xixj ··· xk dV using (ii) V 0 Z 0 = f(x)xixj ··· xk dV since xi and xi are dummy V = Tij···k.