part ia | vector calculus...part ia | vector calculus theorems with proof based on lectures by b....
TRANSCRIPT
Part IA — Vector Calculus
Theorems with proof
Based on lectures by B. AllanachNotes taken by Dexter Chua
Lent 2015
These notes are not endorsed by the lecturers, and I have modified them (oftensignificantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Curves in R3
Parameterised curves and arc length, tangents and normals to curves in R3, the radiusof curvature. [1]
Integration in R2 and R3
Line integrals. Surface and volume integrals: definitions, examples using Cartesian,cylindrical and spherical coordinates; change of variables. [4]
Vector operatorsDirectional derivatives. The gradient of a real-valued function: definition; interpretationas normal to level surfaces; examples including the use of cylindrical, spherical *andgeneral orthogonal curvilinear* coordinates.
Divergence, curl and ∇2 in Cartesian coordinates, examples; formulae for these oper-ators (statement only) in cylindrical, spherical *and general orthogonal curvilinear*coordinates. Solenoidal fields, irrotational fields and conservative fields; scalar potentials.Vector derivative identities. [5]
Integration theoremsDivergence theorem, Green’s theorem, Stokes’s theorem, Green’s second theorem:statements; informal proofs; examples; application to fluid dynamics, and to electro-magnetism including statement of Maxwell’s equations. [5]
Laplace’s equationLaplace’s equation in R2 and R3: uniqueness theorem and maximum principle. Solutionof Poisson’s equation by Gauss’s method (for spherical and cylindrical symmetry) andas an integral. [4]
Cartesian tensors in R3
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 51.1 Derivative of functions . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Coordinate systems . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Curves and Line 62.1 Parametrised curves, lengths and arc length . . . . . . . . . . . . 62.2 Line integrals of vector fields . . . . . . . . . . . . . . . . . . . . 62.3 Gradients and Differentials . . . . . . . . . . . . . . . . . . . . . 62.4 Work and potential energy . . . . . . . . . . . . . . . . . . . . . . 6
3 Integration in R2 and R3 73.1 Integrals over subsets of R2 . . . . . . . . . . . . . . . . . . . . . 73.2 Change of variables for an integral in R2 . . . . . . . . . . . . . . 73.3 Generalization to R3 . . . . . . . . . . . . . . . . . . . . . . . . . 83.4 Further generalizations . . . . . . . . . . . . . . . . . . . . . . . . 8
4 Surfaces and surface integrals 94.1 Surfaces and Normal . . . . . . . . . . . . . . . . . . . . . . . . . 94.2 Parametrized surfaces and area . . . . . . . . . . . . . . . . . . . 94.3 Surface integral of vector fields . . . . . . . . . . . . . . . . . . . 94.4 Change of variables in R2 and R3 revisited . . . . . . . . . . . . . 9
5 Geometry of curves and surfaces 10
6 Div, Grad, Curl and ∇ 116.1 Div, Grad, Curl and ∇ . . . . . . . . . . . . . . . . . . . . . . . . 116.2 Second-order derivatives . . . . . . . . . . . . . . . . . . . . . . . 11
7 Integral theorems 127.1 Statement and examples . . . . . . . . . . . . . . . . . . . . . . . 12
7.1.1 Green’s theorem (in the plane) . . . . . . . . . . . . . . . 127.1.2 Stokes’ theorem . . . . . . . . . . . . . . . . . . . . . . . . 127.1.3 Divergence/Gauss theorem . . . . . . . . . . . . . . . . . 12
7.2 Relating and proving integral theorems . . . . . . . . . . . . . . . 12
8 Some applications of integral theorems 178.1 Integral expressions for div and curl . . . . . . . . . . . . . . . . 178.2 Conservative fields and scalar products . . . . . . . . . . . . . . . 178.3 Conservation laws . . . . . . . . . . . . . . . . . . . . . . . . . . 18
9 Orthogonal curvilinear coordinates 199.1 Line, area and volume elements . . . . . . . . . . . . . . . . . . . 199.2 Grad, Div and Curl . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2
Contents IA Vector Calculus (Theorems with proof)
10 Gauss’ Law and Poisson’s equation 2110.1 Laws of gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . 2110.2 Laws of electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . 2110.3 Poisson’s Equation and Laplace’s equation . . . . . . . . . . . . . 21
11 Laplace’s and Poisson’s equations 2211.1 Uniqueness theorems . . . . . . . . . . . . . . . . . . . . . . . . . 2211.2 Laplace’s equation and harmonic functions . . . . . . . . . . . . . 23
11.2.1 The mean value property . . . . . . . . . . . . . . . . . . 2311.2.2 The maximum (or minimum) principle . . . . . . . . . . . 23
11.3 Integral solutions of Poisson’s equations . . . . . . . . . . . . . . 2411.3.1 Statement and informal derivation . . . . . . . . . . . . . 2411.3.2 Point sources and δ-functions* . . . . . . . . . . . . . . . 24
12 Maxwell’s equations 2512.1 Laws of electromagnetism . . . . . . . . . . . . . . . . . . . . . . 2512.2 Static charges and steady currents . . . . . . . . . . . . . . . . . 2512.3 Electromagnetic waves . . . . . . . . . . . . . . . . . . . . . . . . 25
13 Tensors and tensor fields 2613.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2613.2 Tensor algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2613.3 Symmetric and antisymmetric tensors . . . . . . . . . . . . . . . 2613.4 Tensors, multi-linear maps and the quotient rule . . . . . . . . . 2613.5 Tensor calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
14 Tensors of rank 2 2814.1 Decomposition of a second-rank tensor . . . . . . . . . . . . . . . 2814.2 The inertia tensor . . . . . . . . . . . . . . . . . . . . . . . . . . 2814.3 Diagonalization of a symmetric second rank tensor . . . . . . . . 28
15 Invariant and isotropic tensors 2915.1 Definitions and classification results . . . . . . . . . . . . . . . . 2915.2 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.F′(x) = F ′i (x)ei.
Proposition.
d
dt(fg) =
df
dtg + f
dg
dtd
dt(g · h) =
dg
dt· h + g · dh
dtd
dt(g × h) =
dg
dt× h + g × dh
dt
Note that the order of multiplication must be retained in the case of the crossproduct.
Theorem. The gradient is
∇f =∂f
∂xiei
Theorem (Chain rule). Given a function f(r(u)),
df
du= ∇f · dr
du=
∂f
∂xi
dxidu
.
Theorem. The derivative of F is given by
Mji =∂yj∂xi
.
Theorem (Chain rule). Suppose g : Rp → Rn and f : Rn → Rm. Suppose thatthe coordinates of the vectors in Rp,Rn and Rm are ua, xi and yr respectively.By the chain rule,
∂yr∂ua
=∂yr∂xi
∂xi∂ua
,
with summation implied. Writing in matrix form,
M(f ◦ g)ra = M(f)riM(g)ia.
Alternatively, in operator form,
∂
∂ua=∂xi∂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
du= ±
∣∣∣∣ dr
du
∣∣∣∣ = ±|r′(u)|
with the sign depending on whether it is in the direction of increasing or decreasingarclength.
Proposition. ds = ±|r′(u)|du
2.2 Line integrals of vector fields
2.3 Gradients and Differentials
Theorem. If F = ∇f(r), then∫C
F · dr = f(b)− f(a),
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 ofcalculus. 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 ∫
C
F · dr =
∫C
∇f · dr =
∫∇f · dr
dudu.
So by the chain rule, this is equal to∫ β
α
d
du(f(r(u))) du = [f(r(u))]βα = f(b)− f(a).
Proposition. If F = ∇f for some f , then
∂Fi∂xj
=∂Fj∂xi
.
This is because both are equal to ∂2f/∂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. ∫D
f(x, y) dA =
∫Y
(∫xy
f(x, y) dx
)dy.
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∫∫
f dx dy =
∫∫f dy dx.
While we have rather strict conditions for this theorem, it actually holds in manymore 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∫
D
f(x, y) dx dy =
(∫ b
a
g(x) dx
)(∫ d
c
h(y) dy
)
3.2 Change of variables for an integral in R2
Proposition. Suppose we have a change of variables (x, y) ↔ (u, v) that issmooth and invertible, with regions D,D′ in one-to-one correspondence. Then∫
D
f(x, y) dx dy =
∫D′f(x(u, v), y(u, v))|J | du dv,
where
J =∂(x, y)
∂(u, v)=
∣∣∣∣∣∣∣∂x
∂u
∂x
∂v∂y
∂u
∂y
∂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 δA′ = δuδv in the (u, v) plane. Then byTaylors’ theorem, we have
δx = x(u+ δu, v + δv)− x(u, v) ≈ ∂x
∂uδu+
∂x
∂vδv.
We have a similar expression for δy and we obtain(δxδy
)≈(∂x∂u
∂x∂v
∂y∂u
∂y∂v
)(δuδv
)
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 howmuch it scales up an area. So the area formed by δx and δy is |J | times the areaformed by δu and δv. Hence
dx dy = |J | du dv.
3.3 Generalization to R3
Proposition. dV = dx dy dz.
Proposition. ∫V
f dx dy dz =
∫V
f |J | du dv dw,
with
J =∂(x, y, z)
∂(u, v, w)=
∣∣∣∣∣∣∣∣∣∣∣∣
∂x
∂u
∂x
∂v
∂x
∂w∂y
∂u
∂y
∂v
∂y
∂w∂z
∂u
∂z
∂v
∂z
∂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.∫D
f(x1, x2, · · ·xn) dx1 dx2 · · · dxn =
∫D′f({xi(u)})|J | du1 du2 · · · dun.
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
dS =∂r
∂u× ∂r
∂vdu dv.
The scalar area element is
dS =
∣∣∣∣ ∂r
∂u× ∂r
∂v
∣∣∣∣ du dv.
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 beexpressed in terms of lengths, angles etc. which are measured entirely on thesurface. So K can be defined on an arbitrary surface without embedding it on ahigher dimension surface.
Theorem (Gauss-Bonnet theorem).
θ1 + θ2 + θ3 = π +
∫D
K dA,
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, λbe 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 summationconvention.
6.2 Second-order derivatives
Proposition.
∇× (∇f) = 0
∇ · (∇× F) = 0
Proof. Expand out using suffix notation, noting that
εijk∂2f
∂xi∂xj= 0.
since if, say, k = 3, then
εijk∂2f
∂xi∂xj=
∂2f
∂x1∂x2− ∂2f
∂x2∂x1= 0.
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 = ∇×Afor 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 abounded region in the (x, y) plane with boundary ∂A = C,∫
A
(∂Q
∂x− ∂P
∂y
)dA =
∫C
(P dx+Q dy).
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),∫S
∇× F · dS =
∫∂S
F · dr,
where S is a smooth, bounded surface and ∂S is a piecewise smooth boundaryof S.
The direction of the line integral is as follows: If we walk along C with nfacing 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),∫V
∇ · F dV =
∫∂V
F · dS,
where V is a bounded volume with boundary ∂V , a piecewise smooth, closedsurface, 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 about2D regions. So given a region A on the (x, y) plane, we pretend that there is athird dimension and apply Stokes’ theorem to derive Green’s theorem.
Let A be a region in the (x, y) plane with boundary C = ∂A, parametrisedby arc length, (x(s), y(s), 0). Then the tangent to C is
t =
(dx
ds,
dy
ds, 0
).
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
∇× F =
(0, 0,
∂Q
∂x− ∂P
∂y
).
Then the left hand side of Stokes is∫C
F · dr =
∫C
F · t ds =
∫C
P dx+Q dy,
and the right hand side is∫A
(∇× F) · k dA =
∫A
(∂Q
∂x− ∂P
∂y
)dA.
Proposition. Green’s theorem ⇒ Stokes’ theorem.
Proof. Green’s theorem describes a 2D region, while Stokes’ theorem describesa 3D surface r(u, v). Hence to use Green’s to derive Stokes’ we need find some2D 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 inthe u, v plane. Write the boundary as ∂A = (u(t), v(t)). Then ∂S = r(u(t), v(t)).
We want to prove ∫∂S
F · dr =
∫S
(∇× F) · dS
given ∫∂A
Fu du+ Fv dv =
∫A
(∂Fv∂u− ∂Fu
∂v
)dA.
Doing some pattern-matching, we want
F · dr = Fu du+ Fv dv
for some Fu and Fv.By the chain rule, we know that
dr =∂r
∂udu+
∂r
∂vdv.
So we choose
Fu = F · ∂r
∂u, Fv = F · ∂r
∂v.
This choice matches the left hand sides of the two equations.To match the right, recall that
(∇× F) · dS = (∇× F) ·(∂r
∂u× ∂r
∂v
)du dv.
Therefore, for the right hand sides to match, we want
∂Fv∂u− ∂Fu
∂v= (∇× F) ·
(∂r
∂u× ∂r
∂v
). (∗)
Fortunately, this is true. Unfortunately, the proof involves complicated suffixnotation and summation convention:
∂Fv∂u
=∂
∂u
(F · ∂r
∂v
)=
∂
∂u
(Fi∂xi∂v
)=
(∂Fi∂xj
∂xj∂u
)∂xi∂v
+ Fi∂xi∂u∂v
.
13
7 Integral theorems IA Vector Calculus (Theorems with proof)
Similarly,
∂Fu∂u
=∂
∂u
(F · ∂r
∂u
)=
∂
∂u
(Fj∂xj∂u
)=
(∂Fj∂xi
∂xi∂v
)∂xj∂u
+ Fi∂xi∂u∂v
.
So∂Fv∂u− ∂Fu
∂v=∂xj∂u
∂xi∂v
(∂Fi∂xj− ∂Fj∂xi
).
This is the left hand side of (∗).The right hand side of (∗) is
(∇× F) ·(∂r
∂u× ∂r
∂v
)= εijk
∂Fj∂xi
εkpq∂xp∂u
∂xq∂v
= (δipδjq − δiqδjp)∂Fj∂xi
∂xp∂u
∂xq∂v
=
(∂Fj∂xi− ∂Fi∂xj
)∂xi∂u
∂xj∂v
.
So they match. Therefore, given our choice of Fu and Fv, Green’s theoremtranslates to Stokes’ theorem.
Proposition. Greens theorem ⇔ 2D divergence theorem.
Proof. The 2D divergence theorem states that∫A
(∇ ·G) dA =
∫∂A
G · n ds.
with an outward normal n.Write G as (Q,−P ). Then
∇ ·G =∂Q
∂x− ∂P
∂y.
Around the curve r(s) = (x(s), y(s)), t(s) = (x′(s), y′(s)). Then the normal,being tangent to t, is n(s) = (y′(s),−x′(s)) (check that it points outwards!). So
G · n = Pdx
ds+Q
dy
ds.
Then we can expand out the integrals to obtain∫C
G · n ds =
∫C
P dx+Q dy,
and ∫A
(∇ ·G) dA =
∫A
(∂Q
∂x− ∂P
∂y
)dA.
Now 2D version of Gauss’ theorem says the two LHS are the equal, and Green’stheorem says the two RHS are equal. So the result follows.
Proposition. 2D divergence theorem.∫A
(∇ ·G) dA =
∫C=∂A
G · n ds.
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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 isjust a linear combination of the two.
Furthermore, we assume that A is a simple, convex shape. A more complicatedshape can be cut into smaller simple regions, and we can apply the simple caseto each of the small regions.
Suppose G = G(x, y)j. Then
∇ ·G =∂G
∂y.
Then ∫A
∇ ·G dA =
∫X
(∫Yx
∂G
∂ydy
)dx.
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.
C+
C−
dy
Yx
x
y
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,∫
Yx
∂G
∂ydy =
∫ y+(x)
y−(x)
∂G
∂ydy = G(x, y+(x))−G(x, y−(x)).
To compute the full area integral, we want to integrate over all x. However, thedivergence theorem talks in terms of ds, not dx. So we need to find some wayto relate ds and dx. If we move a distance δs, the change in x is δs cos θ, whereθ is the angle between the tangent and the horizontal. But θ is also the anglebetween the normal and the vertical. So cos θ = n · j. Therefore dx = j · n ds.
In particular, G dx = G j · n ds = G · n ds, since G = G j.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∫A
∇ ·G dA =
∫X
(∫yx
∂G
∂ydY
)dx
=
∫X
G(x, y+(x))−G(x, y−(x)) dx
=
∫C+
G · n ds+
∫C−
G · n ds
=
∫C
G · n ds.
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.
(∇ · F)(r0) = limdiam(V )→1
1
vol(V )
∫∂V
F · dS,
where the limit is taken over volumes containing the point r0.
Proposition.
n · (∇× F)(r0) = limdiam(A)→0
1
area(A)
∫∂A
F · dr,
where the limit is taken over all surfaces A containing r0 with normal n.
8.2 Conservative fields and scalar products
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 ato b.
a
b
C
C
If S is any surface with boundary ∂S = C − C, By Stokes’ theorem,∫S
∇× F · dS =
∫∂S
F · dr =
∫C
F · dr−∫C
F · dr.
But ∇× F = 0. So ∫C
F · dr−∫C
F · dr = 0,
or ∫C
F · dr =
∫C
F · dr.
Proposition. If (ii)∫C
F · dr is independent of C for fixed end points andorientation, then (i) F = ∇f for some scalar field f .
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 smallextension of C by δC. Then
f(r + δr) =
∫C+δC
F(r′) · dr′
=
∫C
F · dr′ +
∫δC
F · dr′
= f(r) + F(r) · δr + o(δr).
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8 Some applications of integral theoremsIA Vector Calculus (Theorems with proof)
Soδf = 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.
∇f =1
hu
∂f
∂ueu +
1
hv
∂f
∂vev +
1
hw
∂f
∂wew.
Proposition.
∇ =1
hueu
∂
∂u+
1
hvev
∂
∂v+
1
hwew
∂
∂w.
Proposition.
∇× F =1
hvhw
[∂
∂v(hwFw)− ∂
∂w(hvFv)
]eu + two similar terms
=1
huhvhw
∣∣∣∣∣∣hueu hvev hwew∂∂u
∂∂v
∂∂w
huFu hvFv hwFw
∣∣∣∣∣∣and
∇ · F =1
huhvhw
[∂
∂u(hvhwFu) + two similar terms
].
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
n · ∇ × F = limA→0
1
A
∫∂A
F · dr.
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. Thenthis has an area of huhvδuδv with normal ew. Let C be its boundary.
uδu
v
δv C
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 assum-ing F and h are constant when moving through each horizontal/vertical
19
9 Orthogonal curvilinear coordinatesIA Vector Calculus (Theorems with proof)
segment.∫C
F · dr ≈ Fu(u, v)hu(u, v) δu+ Fv(u+ δu, v)hv(u+ δu, v) δu
− Fu(u, v + δv)hu(u, v + δv) δu− Fv(u, v)hv(u, v) δv
≈[∂
∂uhvFv −
∂
∂v(huFu)
]δuδv.
Divide by the area and take the limit as area → 0, we obtain
limA→0
1
A
∫C
F · dr =1
huhv
[∂
∂uhvFv −
∂
∂v(huFu)
].
So, by the integral definition of divergence,
ew · ∇ × F =1
huhv
[∂
∂u(hvFv)−
∂
∂v(huFu)
],
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 closedsurface S, ∫
S
g · dS = −4πGM,
where G is Newton’s gravitational constant, and M is the total mass containedin V .
Law (Gauss’ Law for gravitation in differential form).
∇ · g = −4πGρ.
10.2 Laws of electrostatics
Law (Gauss’ law for electrostatic forces).∫S
E · dS =Q
ε0,
where ε0 is the permittivity of free space, or electric constant.
Law (Gauss’ law for electrostatic forces in differential form).
∇ ·E =ρ
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 withS = ∂V being a closed surface, with an outward normal n.
Suppose ϕ satisfies either
(i) Dirichlet condition, ϕ(r) = f(r) on S
(ii) Neumann condition ∂ϕ(r)∂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 boundaryconditions (N) or (D). Then Ψ(r) = ϕ2(r) − ϕ1(r) satisfies ∇2Ψ = 0 on V bylinearity, and
(i) Ψ = 0 on S; or
(ii) ∂Ψ∂n = 0 on S.
Combining these two together, we know that Ψ∂Ψ∂n = 0 on the surface. So using
the divergence theorem,∫V
∇ · (Ψ∇Ψ) dV =
∫S
(Ψ∇Ψ) · dS = 0.
But∇ · (Ψ∇Ψ) = (∇Ψ) · (∇Ψ) + Ψ∇2Ψ︸︷︷︸
=0
= |(∇Ψ)|2.
So ∫V
|∇Ψ|2 dV = 0.
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).∫S
(u∇v) · dS =
∫V
(∇u) · (∇v) dV +
∫V
u∇2v dV,
Proposition (Green’s second identity).∫S
(u∇v − v∇u) · dS =
∫V
(u∇2v − v∇2u) dV.
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 Vcontaining a solid sphere defined by |r−a| ≤ R, with boundary SR = |r−a| = R,for some R. Define
ϕ(R) =1
4πR2
∫SR
ϕ(r) dS.
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.
So dSR2 is independent of R. Write
ϕ(R) =1
4π
∫ϕ
dS
R2.
Differentiate this with respect to R, noting that dS/R2 is independent of R.Then we obtain
d
dRϕ(R) =
1
4πR2
∫∂ϕ
∂u
∣∣∣∣u=R
dS
But∂ϕ
∂u= eu · ∇ϕ = n · ∇ϕ =
∂ϕ
∂n
on SR. So
d
dRϕ(R) =
1
4πR2
∫SR
∇ϕ · dS =1
4πR2
∫VR
∇2ϕ dV = 0
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 isan 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 εsufficiently small such that the region |r− a| < ε lies within V (possible since alies in the interior of V ).
Then for any r such that |r− a| = ε, we have ϕ(r) < ϕ(a).
ϕ(ε) =1
4πR2
∫SR
ϕ(r) dS < ϕ(a),
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 boundaryconditions |ϕ(r)| = O(1/|r|) and |∇ϕ(r)| = O(1/|r|2), is
ϕ(r) =1
4π
∫V ′
ρ(r′)
|r− r′|dV ′
For ρ(r′) non-zero everywhere, but suitably well-behaved as |r′| → ∞, we canalso take V ′ = 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).
∇ ·E =ρ
ε0
∇ ·B = 0
∇×E +∂B
∂t= 0
∇×B− µ0ε0∂E
∂t= µ0j,
where ε0 is the electric constant (permittivity of free space) and µ0 is themagnetic constant (permeability of free space), which are constants determinedexperimentally.
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 eachcoordinate system, and that vi···j = Ti···jp···qup···q is also a tensor for any tensorup···q. Then Ti···jp···q is also a tensor.
Proof. We can check the tensor transformation rule directly. However, we canreuse the result above to save some writing.
Consider the special form up···q = cp · · · dq for any vectors c, · · ·d. Byassumption,
vi···j = Ti···jp···qcp · · · dqis 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 ascalar and hence gives the same result in every coordinate system, Ti···jp···q is amulti-linear map. So Ti···jp···q is a tensor.
13.5 Tensor calculus
Proposition.∂
∂xp· · · ∂
∂xq︸ ︷︷ ︸m
Tij · · · k︸ ︷︷ ︸n
, (∗)
is a tensor of rank n+m.
Proof. To show this, it suffices to show that ∂∂xp
satisfies the tensor transfor-
mation rules for rank 1 tensors (i.e. it is something like a rank 1 tensor). Thenby the exact same argument we used to show that tensor products preservetensorness, we can show that the (∗) is a tensor. (we cannot use the result oftensor products directly, since this is not exactly a product. But the exact sameproof works!)
Since x′i = Riqxq, we have
∂x′i∂xp
= Rip.
(noting that∂xp
∂xq= δpq). Similarly,
∂xq∂x′i
= Riq.
Note that Rip, Riq are constant matrices.
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13 Tensors and tensor fields IA Vector Calculus (Theorems with proof)
Hence by the chain rule,
∂
∂x′i=
(∂xq∂x′i
)∂
∂xq= Riq
∂
∂xq.
So ∂∂xp
obeys the vector transformation rule. So done.
Theorem (Divergence theorem for tensors).∫S
Tij···k`n` dS =
∫V
∂
∂x`(Tij···k`) dV,
with n being an outward pointing normal.
Proof. Apply the usual divergence theorem to the vector field v defined byv` = aibj · · · ckTij···k`, where a,b, · · · , c are fixed constant vectors.
Then
∇ · v =∂v`∂x`
= aibj · · · ck∂
∂x`Tij···k`,
andn · v = n`v` = aibj · · · ckTij···k`n`.
Since a,b, · · · , c are arbitrary, therefore they can be eliminated, and the tensordivergence 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 εijkusing tensor products, contractions, and linear combinations.
Proof. We analyze conditions for invariance under specific rotations through πor π/2 about coordinate axes.
(i) Suppose Ti is rank-1 isotropic. Consider a rotation about x3 through π:
(Rij) =
−1 0 00 −1 00 0 1
.
We want T1 = RipTp = R11T1 = −T1. So T1 = 0. Similarly, T2 = 0. Byconsider a rotation about, say x1, we have T3 = 0.
(ii) Suppose Tij is rank-2 isotropic. Consider
(Rij) =
0 1 0−1 0 00 0 1
,
which is a rotation through π/2 about the x3 axis. Then
T13 = R1pR3qTpq = R12R33T23 = T23
andT23 = R2pR3qTpq = R21R33T13 = −T13
So T13 = T23 = 0. Similarly, we have T31 = T32 = 0.
We also haveT11 = R1pR1qTpq = R12R12T22 = T22.
So T11 = T22.
By picking a rotation about a different axis, we have T21 = T12 andT22 = T33.
Hence Tij = αδij .
(iii) Suppose that Tijk is rank-3 isotropic. Using the rotation by π about thex3 axis, we have
T133 = R1pR3qR3rTpqr = −T133.
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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 andother choices of indices.
Then we can show that Tijk = 0 unless all i, j, k are distinct.
Now consider
(Rij) =
0 1 0−1 0 00 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 ofrotation, we find that Tijk is totally antisymmetric, and Tijk = βεijk forsome β.
15.2 Application to invariant integrals
Theorem. Let
Tij···k =
∫V
f(x)xixj · · ·xk dV.
where f(x) is a scalar function and V is some volume.Given a rotation Rij , consider an active transformation: x = xiei is mapped
to x′ = x′iei with x′i = Rijxi, i.e. we map the components but not the basis, andV is mapped to V ′.
Suppose that under this active transformation,
(i) f(x) = f(x′),
(ii) V ′ = 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
simply the determinant of R (x′i = Ripxp ⇒ ∂x′i∂xp
= Rip), which is by definition
1. So dV = dV ′.Then we have
RipRjq · · ·RkrTpq···r =
∫V
f(x)x′ix′j · · ·x′k dV
=
∫V
f(x′)x′ix′j · · ·x′k dV using (i)
=
∫V ′f(x′)x′ix
′j · · ·x′k dV ′ using (ii)
=
∫V
f(x)xixj · · ·xk dV since xi and x′i are dummy
= Tij···k.
30