Embed Size (px)
Citation preview
Justin Shaw 1
Invariant Vector Calculus
Justin Shaw
June 27, 2014
Contents
Preface 3
1 A review of submanifold theory 4
2 AMATH 231 62.1 1 dimensional submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.1.1 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.1.2 Line Integral of a Scalar Field . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1.3 Line Integral of a Vector Field . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1.4 First and Second Fundamental Theorems for Line Integrals . . . . . . . . . . 10
2.2 2 dimensional submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2.1 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2.2 Surface Integral of a Scalar Field . . . . . . . . . . . . . . . . . . . . . . . . 152.2.3 Surface Integral of a Vector Field . . . . . . . . . . . . . . . . . . . . . . . . 16
2.3 Grad, Div, Curl, and the Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3.1 The Gradient of a Scalar Field . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3.2 The Divergence of a Vector Field . . . . . . . . . . . . . . . . . . . . . . . . 172.3.3 The Curl of a Vector Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3.4 Properties of Grad, Div, and Curl . . . . . . . . . . . . . . . . . . . . . . . . 202.3.5 The Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.4 Stokes’ Theorem in Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.4.1 Gauss’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.4.2 The Generalized Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . 232.4.3 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.4.4 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.5 Summary of AMATH 231 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3 AMATH 361 273.1 What is Continuum Mechanics? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.1.1 The Continuum Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.2.1 Particles and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.2.2 Pathlines and Streamlines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.2.3 The Material Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2.4 Material Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1
Justin Shaw 2
3.2.5 The Transport Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.3 Derivation of the Governing Equations: Mass, Momentum, and Energy . . . . . . . 37
3.3.1 Conservation of Mass and the Continuity Equation . . . . . . . . . . . . . . 383.3.2 Conservation of Linear Momentum and the Stress Tensor . . . . . . . . . . . 403.3.3 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.3.4 Summary of Equations so far . . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.4 Newtonian Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.4.1 The Velocity Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.4.2 Pressure and the Deviatoric Stress Tensor . . . . . . . . . . . . . . . . . . . 503.4.3 The Momentum Equation for a Newtonian Fluid . . . . . . . . . . . . . . . 523.4.4 The Energy Equation for a Newtonian Fluid . . . . . . . . . . . . . . . . . . 533.4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4 The Kamchatnov Paper 564.1 Hopf Fibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.1.1 Spheres and Stereographic projection . . . . . . . . . . . . . . . . . . . . . . 574.1.2 Which spheres are groups? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.1.3 Projective Spaces and Hopf Fibrations . . . . . . . . . . . . . . . . . . . . . 604.1.4 The Quaternions and Rotations . . . . . . . . . . . . . . . . . . . . . . . . . 624.1.5 The Octonians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.1.6 The Classical Hopf Fibration . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4.2 Kamchatnov’s Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
5 Translating to 7 dimensions 735.1 Div, Curl, Grad, and Laplacian Revisited . . . . . . . . . . . . . . . . . . . . . . . . 73
5.1.1 Forms on Quaternions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735.1.2 The Exterior Derivative on Quaternions . . . . . . . . . . . . . . . . . . . . 755.1.3 Newtonian Navier-Stokes Revisited . . . . . . . . . . . . . . . . . . . . . . . 77
5.2 Quaternionic Navier-Stokes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 785.2.1 Quaternionic Formulation of Maxwell’s Equations . . . . . . . . . . . . . . . 78
5.3 Maxwell’s Equations in Differential Forms . . . . . . . . . . . . . . . . . . . . . . . 795.3.1 Some Preliminary Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805.3.2 The Electromagnetic Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 805.3.3 The Current One-Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 825.3.4 The Charge Conservation Equation . . . . . . . . . . . . . . . . . . . . . . . 86
Bibliography 87
2
Justin Shaw 3
The purpose of these notes
Often, those who take differential geometry at Waterloo are in Pure Mathematics, and so have nottaken the fluid dynamics sequence in Applied Mathematics. This is unfortunate because fluiddynamics in three dimensions has, historically, motivated the study of differential geometry to agreat extent. Because fluid dynamics has so many real world applications and ramifications thereare many many people who spend their whole lives studying it. As a result there is a welldeveloped set of notations for this field of mathematics, but unfortunately it is not the same oneas we’ve developed for differential geometry. These notes are an attempt to bridge the notationalgap between the disciplines of fluid dynamics and differential geometry, and so to give PureMathematics students a short introduction in how to read the Applied Mathematics notations,and to show the equivalence of both languages.
Amath 231 is Calculus 4 here at Waterloo, and unsurprisingly has only calculus 3 as aprerequisite. Therefore 231, and the follow up course AMATH 361 Continuum Mechanics do notmention many of the connections of their material to differential geometry. One of the goals ofthese notes is to illuminate these connections. These notes should then also give those who areonly familiar with the theoretical side of differential geometry a better understanding of how tocalculate standard integrals, and an appreciation of the physical interpretations behind many ofthe definitions in differential geometry.
3
Justin Shaw 4
1 A review of submanifold theory
Definition 1.0.0.1. Let Mn and Lk be smooth manifolds. A smooth map ι : Lk →Mn is calledan injective immersion if
1. ι is injective
2. ι is an immersion, meaning (ι∗)p : TpL→ Tι(p)M is injective ∀p ∈ L
Notice here that if ι is an injective immersion, we can use (ι∗)p to identify each TpL with asubspace of Tι(p)M . If ι : L→M is an injective immersion then L is called an immersedsubmanifold of M. The name of this definition is suggestive of its physical meaning, because wecan think of L as sitting inside M. The notation ι is also suggestive because perhaps the mostnaturally arising case of an injective immersion is the inclusion map from L to M when L ⊂Mwhere L is also a smooth manifold.
If L is oriented, then condition 2 ensures that orientation is preserved or reversed, but there still isone, since ι is injective at every point of L and so has either positive determinant of itspushforward at every point, or negative determinant of its pushforward at every point. Which oneit is will be irrelevant for us, because we will be using ι = inclusion, which is clearly an injectiveimmersion and orientation preserving:
The coordinate representation of ι is
(x1, . . . , xk) 7→ (x1, . . . , xk, 0, . . . , 0)
where there are n− k zeros on the right. Taking the Jacobian of this map gives us thepushforward in local coordinates, an n× k matrix which looks like the k × k identity matrixsitting on top of a block of zeroes, which is injective.
Using the inclusion gives L the metric ι∗(gM), which is the induced metric on L from M . Bydefinition ι∗(gM) = gM
∣∣L, and so we also have that the restriction of any smooth map on M is also
smooth on L, because its pullback by the inclusion is then the composition of smooth maps and sosmooth.
The astute reader may object that in general, if X is a smooth vector field on M , then X∣∣L
maynot be a smooth vector field on L, because if p ∈ L, Xp ∈ TpM \ TpL is possible. This is a concernif we use a map other than the inclusion, but the definitions from vector calculus implicitly usethe inclusion, so for these notes we don’t have to worry about these concerns. For a proof that theinclusion map allows us to restrict see [smooth lee lemma 5.39]. Because the pullback by a smoothmap of a form is a smooth form, we can always restrict forms to L.
Disclaimer:Throughout this document we assume everything is as smooth as it needs to be so that we canapply our various definitions. Many of the vector calculus results and identities only require C1 orC2, but we want to view things as immersed manifolds, and will assume smoothness where it isconvenient.
Since parameterizations are inverses of charts, by definition most of the immersed submanifoldsstudied below have a global parameterization, but technically not a global chart. The reason for
4
Justin Shaw 5
this is that some of the submanifolds have a boundary. However we are in large part discussingintegrals over submanifolds, and the boundary always has codimension 1, which is a zero set, andso does not effect the integral. We will therefore omit any further discussion of the actual requirednumber of charts to cover the manifolds below, realizing that for our purposes the question isirrelevant.
5
Justin Shaw 6
2 AMATH 231
In this section we survey the course notes for AMATH 231 to show how all the definitions andtheorems from the vector calculus portion of that course can be rewritten invariantly (that is ”indifferential geometry terms”). Unless otherwise stated, every vector calculus result from thissection is taken directly from the AMATH 231 coursenotes [1], and every differential geometryresult is taken from the differential geometry courses I’ve taken at Waterloo.
6
Justin Shaw 7
2.1 1 dimensional submanifolds
We will study 1 dimensional immersed submanifolds of Rn in the form of smooth curves γ = γ(t)where a ≤ t ≤ b. Then γ is a smooth manifold with boundary of its own endpoints by the globalsmooth chart induced by its own parameterization, meaning γ is diffeomorphic to [a, b] throughthe global chart φ = γ−1. The inclusion ι, which is an injective immersion as mentioned, makes γan immersed submanifold of Rn.
It should be noted here that technically γ is a manifold with boundary, and so if γ is not closed itcannot be covered with a single chart. However in what follows we will be concerned withintegrals, and the boundary of a manifold always has one less dimension than the manifold (wesay it has codimension 1), and so is a set of measure zero. Therefore the integral’s value is notchanged by the boundary, and so we will assume one chart is sufficient.
Recall that if ω is a top form on a manifold M with a global chart φ, then∫M
ω =
∫φ(M)
(φ−1)∗(ω)
by definition. Taking M = γ, φ = γ−1 as in the preceding paragraph we get∫γ
ω =
∫φ(γ)
(φ−1)∗(ω) =
∫γ−1(γ([a,b]))
γ∗ω =
∫ b
a
(γ∗ω)t
where (γ∗ω)t is the pullback of ω by γ, paramaterized by t.
Using the inclusion map, we can identify γ(t) with γ(t) = (γ1(t), ... , γn(t)) ∈ Rn. Recall also thatif t0 ∈ [a, b] then we define γ′(t0) = (γ∗)t0(
∂∂t
∣∣t0
). So the tangent spaces Tγ(t)γ are completely
described as scalar multiples of γ′(t). Suppose f is a scalar field, then γ′(t0)f is
(γ∗)t0(∂
∂t
∣∣t0
)f =∂
∂t
∣∣t0
(f(γ(t))) =∂
∂t
∣∣t0
(f(γ1(t), ... , γn(t))) =∂f
∂xi(γ(t0))
∂γi∂t
(t0) = (γ′i(t0)∂
∂xi∣∣γ(t0)
)f
so γ′(t) = (γ′1(t), ... , γ′n(t)) which agrees with the vector calculus definition.
2.1.1 Arc Length
Definition 2.1.1.1. The arclength of the curve γ = γ(t) in Rn, a ≤ t ≤ b is∫γ
ds =
∫ b
a
‖γ′(t)‖ dt
As with most definitions in vector calculus, this definition makes physical sense. If γ(t) is the pathfollowed by a particle, then γ′(t) is the velocity of that particle, and so ‖γ′(t)‖ is the speed of thatparticle at a given time t. Then arclength is the distance travelled, and we can interpret distancetravelled as the integral of the speed with respect to time, which is exactly our definition.
Now in Rn the standard metric is g = (dx1)2 + · · ·+ (dxn)2 = dx1 ⊗ dx1 + · · ·+ dxn ⊗ dxn wherereally this last term on the right is what we should always write, but the middle sum is often
7
Justin Shaw 8
written as a light abuse of notation. From section 2 we know that we can think of γ as animmersed submanifold of Rn with metric ι∗(g), where ι : γ → Rn is the inclusion, an injectiveimmersion preserving the orientation of the positive t direction being the ”outside.” Now we have
gγ := ι∗(g) = g∣∣γ
= (dx1)∣∣2γ
+ · · ·+ (dxn)∣∣2γ
Now if µ is the volume form of γ, to find∫γµγ, we need to find (γ∗µ)t by definition of the integral.
The manifold γ is diffeomorphic to [a, b] through the global parameterization γ(t). Writeγ(t) = (γ1(t), ... , γn(t)) in components, so that xi(γ(t)) = γi(t) on γ, ∀i = 1, ... , n. Therefore
d(xi(γ(t))
)= d
(γi(t)
)= γ′i(t) dt
for each i. Therefore we have
(γ∗gγ)t = γ∗[(dx1)∣∣2γ
+ · · ·+ (dxn)∣∣2γ]t
= (dx1(γ(t)))2 + · · ·+ (dxn(γ(t)))2
= (γ′1(t))2 (dt)2 + · · ·+ (γ′n(t))2 (dt)2
= [(γ′1(t))2 + · · ·+ (γ′n(t))2] (dt)2
= ‖γ′(t)‖2 (dt)2
Now we can express gγ in global coordinates as
gγ = (γ−1)∗γ∗gγ = (γ−1)∗(‖γ′(t)‖2 (dt)2
)and γ−1 = φ by definition, so
gγ = ‖γ′(t φ)‖2 (d(t φ))2
But recall that the Riemannian Volume form in local coordinates y1, · · · , yk for an orientedmanifold (Mk, g) is given by
√det(g)dy1 ∧ · · · ∧ dyk. In this case our manifold is γ with the metric
gγ. We have√
det(gγ) =√gγ since k = 1, so the volume form of gγ is
µ(gγ) =√‖γ′(t φ)‖2 d(t φ) = ‖γ′(t φ)‖ d(t φ)
But we want γ∗(µ(gγ)), and this is easy to calculate:
γ∗(µ(gγ)) = γ∗(‖γ′(t φ)‖ d(t φ)) = ‖γ′(t φ φ−1)‖ d(t φ φ−1)) = ‖γ′(t)‖dt
since γ = φ−1. Therefore we have shown that (γ∗µ)t = ‖γ′(t)‖dt. But now we see that thearclength definition from vector calculus is just the integral over the manifold γ of the volumeform µ of γ, so ∫ b
a
‖γ′(t)‖ dt =
∫ b
a
(γ∗µ)t =
∫γ
µ = vol(γ)
by definition of volume of a manifold. Intuitively, this is exactly what you would expect. So wehave shown
8
Justin Shaw 9
∫γ
ds =
∫γ
µ (1)
2.1.2 Line Integral of a Scalar Field
A scalar field is just a function defined on some subset of Rn. This term is analogous to thedefinition of vector field, in that a vector field gives a vector at each point, and a scalar field givesa scalar at each point.
Definition 2.1.2.1. Consider a curve γ = γ(t), a ≤ t ≤ b in Rn which is at least C1, and a scalarfield f which is at least continuous on γ. Then the line integral of f along γ is defined to be∫
γ
f ds =
∫ b
a
f(γ(t))‖γ′(t)‖ dt
Physically, the line integral’s meaning depends on f ’s physical meaning. If f is a mass densityfunction, the line integral is the total mass of γ. If f is the height of fence whose base is on γ,then the line integral is the area of one side of the fence. There are many examples.
From the previous section we immediately recognize the volume form of γ in global coordinates(γ∗µ)t = ‖γ′(t)‖ dt. We can also write f(γ(t)) = (γ∗f)t Therefore∫
γ
f ds =
∫ b
a
f(γ(t))(γ∗µ)t =
∫ b
a
(γ∗f)t(γ∗µ)t =
∫ b
a
(γ∗fγ∗µ)t =
∫ b
a
(γ∗[fµ])t =
∫γ
fµ
because fµ is a top form, so this last equality is just the definition of integration on the manifoldγ, as we saw above. Note that if f ≡ 1 on γ then arclength is the result of taking the line integralof f . Therefore we have shown ∫
γ
f ds =
∫γ
f µ (2)
2.1.3 Line Integral of a Vector Field
Definition 2.1.3.1. Consider a curve γ = γ(t) = x, a ≤ t ≤ b in Rn which is at least C1, and avector field F which is at least continuous on γ. Then the line integral of F along γ is defined to be∫
γ
F · dx =
∫ b
a
F(γ(t)) · γ′(t) dt
Physically, if we interpret F as a force field, then the line integral is the work done on a particletraversing γ in the direction of positive parameterization. It can also be thought of as a measure ofthe tendency for γ to align with F because by looking at the integrand, we can see that the valueF(γ(t)) · γ′(t) = ‖F(γ(t))‖‖γ′(t)‖cos(θ) , where θ is the angle between the vectors, is greatest
9
Justin Shaw 10
when θ = 0. Therefore a positive result indicates a tendency to follow the field, and a negativeresult indicates a tendency to go against the field, but these are weighted sums remember.
Write F = (f 1, · · · , fn) where the f i are continuous functions on Rn. We also haveγ′(t) = (γ′1(t), · · · , γ′n(t)). Therefore
(F(γ(t)) · γ ′(t)) dt = f i(γ(t))γi′(t) dt = [f i(γ(t))γi
′(t)] dt
Note to each vector v in Rn we can associate the map w 7→ v ·w which is by definition a 1-form.Since v ·w = g(v,w) = (v)[w we see that this identification is just identifying v with v[. Thiswas done all for a fixed vector v, so for our vector field F we have F[ = gijf
idxj =∑
i fidxi since
gij = δji . This is why the identification between F and F[ is so often made in Rn: you can use thesame coefficient functions for both the vector field and its associated 1 form. However this is onlyvalid if the metric is δji .
Now since (γ∗F[)t =∑i
f i(γ(t))d (xi(γ(t))) = [f i(γ(t))γi′(t)] dt because d (xi(γ(t))) = γ′i(t) dt we
know that
(F(γ(t)) · γ ′(t)) dt = (γ∗F[)t
and so ∫γ
F · dx =
∫ b
a
F(γ(t)) · γ′(t) dt =
∫ b
a
(γ∗F[)t =
∫γ
F[∣∣γ
This shows us that our definitions agree, giving us∫γ
F · dx =
∫γ
F[∣∣γ
(3)
2.1.4 First and Second Fundamental Theorems for Line Integrals
Fix x0,x1 ∈ Rn. Recall that if γ is a curve joining x0,x1 then the line integral∫γF · dx is said to
be path independent if the value of this integral is independent of the curve γ joining x0,x1. Animportant point of interest in vector calculus is to determine when F has the property that its lineintegral between two points is path independent, largely because it makes calculation easier:
First Fundamental Theorem for Line Integrals 2.1.4.1. Let U ⊂ Rn be open and connected.Let F : U → Rn be a continuous vector field whose line integral is path independent in U . Fixx0 ∈ U . Define φ(x) =
∫ x
x0F · dx, then ∇φ = F on U
Proof. (Sketch) We wish to show that ∂φ∂xi
= f i where f i is the ith component of F. Since theintegral φ is path independent, we are free to choose a path that traverses from x0 to x in such away that only one of x is changing at a time. Then by the definition of the line integral and theFundamental Theorem of Calculus we can write φ as a sum of integrals each of which aredependent on a single xi, and the partial derivative of that integral with respect to that xi isexactly f i, and zero otherwise.
10
Justin Shaw 11
Lets consider this theorem invariantly. It says that if φ(x) =∫ x
x0F · dx, then ∇φ = F. Let γ(x) be
some curve joining x0 to x, then γ(x) is an immersed submanifold of Rn with the inclusion, andwe can write φ(x) =
∫γ(x)
F[. By definition ∇φ = (dφ)], so ∇φ = F⇔ (dφ)] = F⇔ df = F[.
Now the theorem says that if φ(x) =∫γ(x)
F[ then dφ = F[.
So really this theorem gives a way to construct a primitive of a differential form on a subset of Rn.This theorem is about cohomology! Lets try the converse theorem.
Second Fundamental Theorem for Line Integrals 2.1.4.1. Let U ⊂ Rn be open andconnected. Let F : U → Rn be a continuous vector field, and let x1,x2 be two fixed points in U . IfF = ∇φ, where φ : U → Rn is a C1 scalar field, and γ is any curve in U joining x1 to x2, then∫
γ
F · dx = φ(x2)− φ(x1)
Proof. Chain rule and fundamental theorem of calculus
Invariantly, if F = ∇φ, then F[ = dφ, so using the fact that∫γF · dx =
∫γF[ we proved above this
theorem says: ∫γ
dφ = φ(γ(b))− φ(γ(a))
Which is a result that’s usually proven right after integration of forms along curves is defined in adifferential geometry class.
Putting these two results together we have that if F is a continuous vector field, then
∫γ
F · dx is path independent in F’s domain⇔ F = ∇φ for some scalar field φ on F’s domain.
We’ve also shown that this is equivalent to:∫γ
F[ is path independent in F[’s domain⇔ F[ is exact on its domain
If γ is a closed curve in F[’s domain, then we can pick fix two points t1, t2 on gamma and break γinto two smooth curves, γ1 from t1 to t2, and γ2 from t2 to t1.
∫γ
=∫γ1
+∫γ2
=∫γ1−∫−γ2 because
reversing the orientation of γ2 gives minus the integral. If we have path independence then sinceγ1 and −γ2 are both curves from t1 to t2,
∫γ1
=∫−γ2 . Therefore
∫γ
= 0
Conversely if∫γF[ = 0 for every closed curve γ in F[’s domain, then given any two points x1,x2 in
F[’s domain, take any two smooth curves γ1 and γ2 in F[’s domain which join x1 to x2 and whichdo not intersect. Let γ = γ1 ∪ (−γ2) is a simple closed curve, and so0 =
∫γ
=∫γ1
+∫−γ2 =
∫γ1−∫γ2⇒∫γ1
=∫γ2
. Note if the two chosen curves do intersect, theintersection points again form closed curves, and the argument can be repeated. We havetherefore shown that: ∫
γ
F[ is path independent in F[’s domain
11
Justin Shaw 12
m
F[ is exact on its domain
m∫γ
F[ = 0 for every closed curve γ in F[’s domain
12
Justin Shaw 13
2.2 2 dimensional submanifolds
We now begin our study of surfaces in Rn. We will denote our surfaces by Σ and assume we canwrite Σ = p(u, v) = (p1(u, v), p2(u, v), p3(u, v)) where p : Duv
∼−−→ Σ is a diffeomorphism andDuv ⊂ R2. So letting φ = p −1, φ is a global chart for Σ. Therefore Σ is a 2 manifold withboundary, and the inclusion is an orientation preserving injective immersion making Σ animmersed submanifold of Rn. As with the one dimensional case Σ is a manifold with boundarymeans one chart may not be sufficient, but for the purposes of integration one chart is sufficient.
Note the similarity of this construction with that of the one dimensional case. In particular wehave that, in general, parameterizations are inverses of charts on the manifold, because charts arethought of as mapping from the manifold, and parameterizations are thought of as mapping to themanifold. By assuming a smooth parameterization we are also assuming that our surface can beoriented. We will assume, as is standard, that closed surfaces have an outward pointing normal,and that the boundary of surfaces with boundary have the orientation such that if one werewalking along the boundary the side with the normal pointing up would be on your left. Thismeans if we’re looking at the side where the normal vector points up, the boundary is orientedcounter clockwise.
Taking our chart φ and the manifold Σ, by definition we have, for a 2 form ω∫Σ
ω =
∫φ(Σ)
(φ−1)∗(ω) =
∫p−1(p(Duv))
p∗ω =
∫∫Duv
(p∗ω)(u,v)
We remember this fact for use below, and by analogy with our study of curves, we being with thearea of a surface.
2.2.1 Surface Area
The definition of Surface Area does not require p to be smooth, only C1.
Definition 2.2.1.1. If Σ = p(u, v), (u, v) ∈ Duv where p is C1, then the surface area of Σ isdefined as
S(Σ) =
∫∫Duv
‖∂p
∂u× ∂p
∂v‖ du dv
Note that when we write du dv in calculus we actually mean du ∧ dv.Lets find the volume form for Σ, because we’ll need it for the rest of the 2 dimensional case. Firstwe find the induced metric on Σ, call it gΣ, and pull it back to Duv. We know that
gΣ = g∣∣Σ
= (dx1)∣∣2Σ
+ (dx2)∣∣2Σ
+ (dx3)∣∣2Σ
To find (p∗gΣ)(u,v), we use the fact that
(x, y, z) = (p1(u, v), p2(u, v), p3(u, v))
on Σ, so d(xi(p(u, v))) = ∂pi∂udu+ ∂pi
∂vdv for each i. Therefore we can pullback gΣ:
13
Justin Shaw 14
(p∗gΣ)(u,v) =∑i
(∂pi∂u
du+∂pi∂v
dv)2
but now
(∂pi∂u
du+∂pi∂v
dv)2 = (∂pi∂u
du+∂pi∂v
dv)⊗ (∂pi∂u
du+∂pi∂v
dv)
=∂pi∂u
2
(du)2 + 2∂pi∂u
∂pi∂v
du⊗ dv +∂pi∂v
2
(dv)2
so summing over i we have
(p∗gΣ)(u,v) = (∑i
∂pi∂u
2
)(du)2 + 2(∑i
∂pi∂u
∂pi∂v
) du⊗ dv + (∑i
∂pi∂v
2
)(dv)2
Now we can write
gΣ = (p−1)∗p∗gΣ
= φ∗p∗gΣ
= φ∗
((∑i
∂pi∂u
2
)(du)2 + 2(∑i
∂pi∂u
∂pi∂v
) du⊗ dv + (∑i
∂pi∂v
2
)(dv)2
)
= (∑i
[∂pi∂u φ]2)(d(u φ))2 + 2(
∑i
∂pi∂u
∂pi∂v φ) d(u φ)⊗ d(v φ) + (
∑i
[∂pi∂v φ]2)(d(v φ))2
and now to find the volume form we need to take the determinant. Suppressing the ”φ” that’s inall the terms, taking the determinant gives:
det(gΣ) = (∑i
∂pi∂u
2
)(∑i
∂pi∂v
2
)− (∑i
∂pi∂u
∂pi∂v
)2
= (∂p1
∂u
∂p2
∂v− ∂p2
∂u
∂p1
∂v)2 + (
∂p1
∂u
∂p3
∂v− ∂p3
∂u
∂p1
∂v)2 + (
∂p2
∂u
∂p3
∂v− ∂p3
∂u
∂p2
∂v)2
where this last equality follows by expanding everything out, collecting terms, and recognizing thesquared terms. We also recognize that this last line is in fact (∂p
∂u× ∂p
∂v) · (∂p
∂u× ∂p
∂v) = ‖∂p
∂u× ∂p
∂v‖2,
so that writing in the ”φ” that’s been in every term all along we have
det(gΣ) =
(‖∂p
∂u× ∂p
∂v‖ φ
)2
Now we can define µ(gΣ), the volume form for Σ with the metric gΣ by
µ(gΣ) =√
det(gΣ) d(u φ) ∧ d(v φ) =
(‖∂p
∂u× ∂p
∂v‖ φ
)d(u φ) ∧ d(v φ)
But now we have
14
Justin Shaw 15
p∗µ(gΣ) = (φ−1)∗((‖∂p
∂u× ∂p
∂v‖ φ
)d(u φ) ∧ d(v φ)
)=
(‖∂p
∂u× ∂p
∂v‖ φ φ−1
)d(u φ φ−1) ∧ d(v φ φ−1)
= ‖∂p
∂u× ∂p
∂v‖ du ∧ dv
Now if we suppress the dependence on the metric, and express the dependence on u, v, we haveshown that
(p∗µ)(u,v) = ‖∂p
∂u× ∂p
∂v‖ du ∧ dv
And now we can express the surface area of Σ as
S(Σ) =
∫∫Duv
‖∂p
∂u× ∂p
∂v‖ du dv =
∫∫Duv
(p∗µ)(u,v) =
∫Σ
µ = vol(Σ)
So again, this is exactly what we would expect.
2.2.2 Surface Integral of a Scalar Field
Definition 2.2.2.1. If Σ = p(u, v), (u, v) ∈ Duv where p is C1, and if f is a continuous scalarfield on Σ, then the surface integral of f over Σ is defined as∫∫
Σ
f dS =
∫∫Duv
f(p(u, v))‖∂p
∂u× ∂p
∂v‖ du dv
Physically, f will often represent the surface density of a physical quantity, and so the surfaceintegral of f over Σ gives the total mass of that physical quantity on the surface Σ.
We see that little has changed from the case of surface area here. We know from the previoussection that (p∗µ)(u,v) = ‖∂p
∂u× ∂p
∂v‖ du ∧ dv, and we have
f(p(u, v)) = (p∗f)(u,v)
so we have
∫∫Σ
f dS =
∫∫Duv
f(p(u, v))‖∂p
∂u× ∂p
∂v‖ du dv =
∫∫Duv
(p∗f)(u,v)(p∗µ)(u,v) =
∫∫Duv
(p∗[fµ])(u,v) =
∫Σ
fµ
Therefore we have shown ∫∫Σ
f dS =
∫Σ
fµ (4)
15
Justin Shaw 16
2.2.3 Surface Integral of a Vector Field
Definition 2.2.3.1. Let Σ = p(u, v), (u, v) ∈ Duv be an oriented surface with unit normal n. If pis C1, and if F is a continuous vector field on Σ, then the surface integral of f over Σ is defined as∫∫
Σ
F · n dS =
∫∫Duv
F(p(u, v)) · (∂p
∂u× ∂p
∂v) du dv
Physically, if F is a force field of some kind, then the surface integral gives the net flux through Σcaused by F, where flux is ”stuff per unit time”
F(p(u, v)) · (∂p∂u× ∂p
∂v) du ∧ dv is
F(p1, p2, p3) · (∂p2
∂u
∂p3
∂v− ∂p3
∂u
∂p2
∂v,∂p1
∂u
∂p3
∂v− ∂p3
∂u
∂p1
∂v,∂p1
∂u
∂p2
∂v− ∂p2
∂u
∂p1
∂v) du ∧ dv
Lets compute ∗(F[):
∗(F[) = ∗(fi dxi) = fi ∗ (dxi) = f1dy ∧ dz + f2dz ∧ dx+ f3dx ∧ dy
Lets pull back this form by p(u, v). Recall from the previous section that
d(xi(p(u, v))) =∂pi∂u
du+∂pi∂v
dv
for each i. Lets pull back f1dy ∧ dz using this formula
(p∗(f1dy ∧ dz))(u,v) = f1(p(u, v))d(y(p(u, v))) ∧ d(z(p(u, v)))
= f1(p(u, v))(∂p2
∂udu+
∂p2
∂vdv) ∧ (
∂p3
∂udu+
∂p3
∂vdv)
= f1(p(u, v))(∂p2
∂u
∂p3
∂v− ∂p3
∂u
∂p2
∂v)du ∧ dv
which is the first term in F(p(u, v)) · (∂p∂u× ∂p
∂v) du ∧ dv. Pulling back the other two terms in ∗(F[)
gives the second and third terms respectively. Putting it all together by linearity we have that
F(p(u, v)) · (∂p
∂u× ∂p
∂v) du ∧ dv = (p∗[∗(F[)])(u,v)
Therefore∫∫Σ
F · n dS = F(p(u, v)) · (∂p
∂u× ∂p
∂v) du ∧ dv =
∫∫Duv
= (p∗[∗(F[)])(u,v) =
∫Σ
∗(F[)
So now we have shown that ∫∫Σ
F · n dS =
∫Σ
∗(F[) (5)
16
Justin Shaw 17
2.3 Grad, Div, Curl, and the Laplacian
Before we discuss the major theorems from AMATH 231, we must show the equivalence of someconcepts from vector calculus with definitions from differential geometry.
2.3.1 The Gradient of a Scalar Field
We define the vector differential operator ∇ as
∇ = i∂
∂x+ j
∂
∂y+ k
∂
∂z= (
∂
∂x,∂
∂y,∂
∂z)
and define the action of ∇ on a scalar field f as ∇(f) = ∇f the gradient of f .
Recall that in differential geometry the gradient of f is defined to be the vector field (df)]. Inlocal coordinates, then,
∇f = (df)] = (∂f
∂xidxi)] = (gij
∂f
∂xi)∂
∂xj
However we’re in the case (R3, g), so we have global coordinates, so we can write
∇f = δji∂f
∂xi∂
∂xj= (
∂f
∂x,∂f
∂y,∂f
∂z)
so the two definitions agree.
Physically, the gradient of a scalar field is a vector field that points in the direction of greatestincrease of that function.
2.3.2 The Divergence of a Vector Field
We define the divergence of a vector field by using the symbol ∇· to be the operator that acts on avector field F as
∇ · F = (∂
∂x,∂
∂y,∂
∂z) · (f1, f2, f3) =
∂f1
∂x+∂f2
∂y+∂f3
∂z
where this last equality is by analogy with the operation of the inner product on R3.
Recall that in differential geometry if (M,µ) is a manifold with volume form, we define thedivergence div : Γ(TM)→ C∞(M) by div(F)µ = d(Fyµ) and that the formula for divergence of avector field F in local coordinates is
div(F) =∂fk
∂xk+ fk
∂
∂xk(log
√det(g))
But in (R3, g), we have global coordinates and g = δji , so log√
det(g) = log√
1 = 0, and therefore
∇ · F =∂f 1
∂x+∂f 2
∂y+∂f 3
∂z= div(F)
17
Justin Shaw 18
so again the two formulas agree, but we can do better:
Lemma 2.3.2.1. If F = (f1, f2, f3) is a vector field in R3, then
div(F) = ∇ · F = ∗d(∗F[)
Proof.∗F[ = ∗(fidxi) = fi ∗ (dxi) = f1dy ∧ dz + f2dz ∧ dx+ f3dx ∧ dy
by linearity of ∗. So now
d(∗F[) = (∂f1
∂x+∂f2
∂y+∂f3
∂z) dx ∧ dy ∧ dz
but now ∗dx ∧ dy ∧ dz = 1, so we have our result.
Physically, the divergence measures the tendency of a vector field to diverge from a given point.
2.3.3 The Curl of a Vector Field
We also define ∇× F to be the vector field
∇× F = (∂
∂x,∂
∂y,∂
∂z)× (f1, f2, f3) = (
∂f3
∂y− ∂f2
∂z,∂f1
∂z− ∂f3
∂x,∂f2
∂x− ∂f1
∂y)
again by analogy with the cross product.
We want to write this invariantly, but the cross product is not usually defined in a differentialgeometry course, and this is just an analogy anyway. Again we can do better:
Lemma 2.3.3.1. If F = (f1, f2, f3) is a vector field in R3, then
curl(F) = ∇× F = [∗(d(F[))]]
Proof.
d(F[) = d(fidxi) = (
∂fi∂x
dx+∂fi∂y
dy +∂fi∂z
dz) ∧ dxi
=∂f1
∂ydy ∧ dx+
∂f1
∂zdz ∧ dx+
∂f2
∂xdx ∧ dy +
∂f2
∂zdz ∧ dy +
∂f3
∂xdx ∧ dz +
∂f3
∂ydy ∧ dz
= (∂f2
∂x− ∂f1
∂y) dx ∧ dy + (
∂f3
∂y− ∂f2
∂z) dy ∧ dz + (
∂f3
∂x− ∂f1
∂z) dx ∧ dz
now we can write
∗d(F[) = (∂f3
∂y− ∂f2
∂z) dx+ (
∂f1
∂z− ∂f3
∂x) dy + (
∂f2
∂x− ∂f1
∂y) dz
and now taking the sharp gives the result.
18
Justin Shaw 19
Notice here we could define another operator
curl : Ω1(R3)→ Ω1(R3)
α = αidxi 7→ ∗dα = ∗
(∂αi∂xj
dxj ∧ dxi)
Where we’re simply using the calculation we did above in the proof, rewriting the second line inthe proof in summation notation.
In this definition we’ve seen the first case of the use of the cross product so far. Everything elsewe’ve encountered has meant something in differential geometry terms. The cross product is onlydefined in R3 and R7 as we shall see. We need a lemma for the 3 dimensional case:
Lemma 2.3.3.1. If X and Y are two vector fields in R3, then X × Y = (∗(X[ ∧ Y [))]
Proof. Let X =∑i
Xi∂∂xi
and Y =∑j
Yj∂∂xj
. As we’ve seen before, in Rn this means that
X[ = Xi dxi, Y [ = Yj dx
j. We’re in R3, so we have, by definition,
X × Y = (X2Y3 −X3Y2 , X3Y1 − X1Y3, X1Y2 −X2Y1)
We also have
X[ ∧ Y [ = Xi dxi ∧ Yj dxj = Xi Yj dx
i ∧ dxj
Now take ∗, ∗(Xi Yj dxi ∧ dxj) = Xi Yj ∗ (dxi ∧ dxj) by linearity of ∗. Recall that
∗2 = (−1)k(3−k)Id on a k form, so on 1 and 2 forms in R3, ∗2 = Id If i = j, dxi ∧ dxj = 0 and ifi 6= j we have, using ∗2 = Id:
• ∗(dx ∧ dy) = dz
• ∗(dz ∧ dx) = dy
• ∗(dy ∧ dz) = dx
Therefore
∗(Xi Yj dxi ∧ dxj) = X1 Y2 dz −X1 Y3 dy +X2 Y3 dx−X2 Y1 dz +X3 Y1 dy −X3 Y2 dx
= (X2 Y3 −X3 Y2 )dx+ (X3 Y1 −X1 Y3 )dy + (X1 Y2 −X2 Y1 )dz
Now take the sharp of this to get the result
Note: We will write ∗(X[ ∧ Y [)] for (∗(X[ ∧ Y [))]. A common abuse of notation is to write thingslike X ∧ Y for X[ ∧ Y [ because the identification of vector fields and one forms is so common, andstraightforward, in R3.
19
Justin Shaw 20
2.3.4 Properties of Grad, Div, and Curl
All three of these operations are linear, and all three have an associated product rule with a scalarfunction. Suppose F,G are vector fields, and suppose f, g are scalar fields, then
∇(fg) = g∇f + f∇gdiv(fF) = fdiv(F) +∇f · F
curl(fF) = fcurl(F) +∇f × F
along with ”zero identities” such as
curl(∇f) = 0
div(curl(F)) = 0
We leave the product rules for an exercise. Note that since we proved equivalence with ourinvariant formulas above and these are formulas in R3, we are free to use either notation to provethe results, but lets prove the zero identities with our invariant formulas.
curl(∇f) = curl((df)])) = [∗(d((df)])[)]] = [∗(d2f)]] = 0
since both ] and ∗ are linear isomorphisms. Similarly
div(curl(F)) = div([∗(dF[)]]) = ∗d ∗ [([∗(dF[)]])[] = ∗d ∗2 (dF[)
but dF[ is a 2 form, and ∗2 = Id, so we have d2 again, so this is zero.
There are many, many vector calculus identities, seemingly all of which can be proved by simplycranking it out. For this reason we will mention them when we need them, but not prove all ofthem.
2.3.5 The Laplacian
The Laplacian is defined as
∇2 = ∇ · ∇ =∂2
∂x2+
∂2
∂y2+
∂2
∂z2
and can act on both scalar and vector fields. Notice that as it acts on functions ∇2 = −∆ where∆ is the Laplacian defined in differential geometry. We have the following identities:
∇(∇2f) = ∇2(∇f)
∇ · (∇2F) = ∇2(∇ · F)
∇× (∇2F) = ∇2(∇× F)
and we can rewrite this invariantly:
Lemma 2.3.5.1. If f is a scalar field in R3, then
∇2f = ∗d ∗ df
20
Justin Shaw 21
Proof. This is pure application of the definitions:
∗d ∗ df = ∗d ∗ (∂f
∂xidxi)
= ∗d(∂f
∂xdy ∧ dz +
∂f
∂ydz ∧ dx+
∂f
∂zdx ∧ dy)
= ∗((∂2f
∂x2+∂2f
∂y2+∂2f
∂z2)dx ∧ dy ∧ dz)
=∂2f
∂x2+∂2f
∂y2+∂2f
∂z2
21
Justin Shaw 22
2.4 Stokes’ Theorem in Vector Calculus
We will see that the meat of the theoretical component of AMATH 231 comes from special casesof Stokes’ Theorem. Now that we’ve laid the groundwork, we finish our review of vector calculus 1with the major theorems presented in that course. The proofs of these theorems given in thatcourse are proofs of special cases, so we do not repeat them here, because we will see that the onlyspecial case we need is that of a smooth boundary.
2.4.1 Gauss’ Theorem
Gauss’ Theorem 2.4.1.1. Let Ω be a bounded subset of R3 whose boundary ∂Ω is a singlepiecewise smooth oriented closed surface. If F is of class C1 on Ω ∪ ∂Ω, then∫∫∫
Ω
∇ · F dV =
∫∫∂Ω
F · n dS
where n is the outward normal to ∂Ω.
Proof. Assume the boundary is smooth, and note that we can also assume Ω ⊃ ∂Ω since theboundary is a zero set so the integral is unaffected. This means Ω is a closed and bounded subsetof R3 and so compact by Heine-Borel, which means any form on Ω is compactly supported.
On the left side of the equality in the integrand we have
∇ · F dV = div(F)dx ∧ dy ∧ dz
because on R3 the volume form is dx∧ dy ∧ dz, so Ω has the same volume form restricted to it. Byan earlier identity we proved
div(F)dx ∧ dy ∧ dz = ∗d(∗F[)(dx ∧ dy ∧ dz).
but ∗(dx ∧ dy ∧ dz) = Id, and ∗2 = Id so we can write
div(F)dx ∧ dy ∧ dz = ∗2d(∗F[) ∗ (dx ∧ dy ∧ dz) = d(∗F[)
On the right side, by work above we know∫∫∂Ω
F · n dS =
∫∂Ω
∗(F[)
Therefore this theorem says ∫Ω
d(∗F[) =
∫∂Ω
∗F[
viewing both Ω and ∂Ω as manifolds. But since Ω is orientable and compact, this is exactlyStokes’ Theorem.
22
Justin Shaw 23
This theorem gives a little more insight into divergence. Suppose A is some physical quantity in Ωsuch as charge, gas, etc., whose flux density is given by F. This thoerem then says that the rate atwhich the amount of A in Ω is decreasing is equal the rate at which A is leaving Ω through ∂Ω.
Gauss’ Theorem is also known as the Divergence Theorem, in both differential geometry andvector calculus.
2.4.2 The Generalized Divergence Theorem
Generalized Divergence Theorem 2.4.2.1. Let Ω be a bounded subset of R3 whose boundary∂Ω is a single piecewise smooth oriented closed surface. If F is of class C1 on Ω ∪ ∂Ω except at apoint a ∈ Ω, then take a sufficiently small ball H centered at a. We can still apply Gauss’Theorem in a modified form:∫∫
∂Ω
F · n dS =
∫∫∫Ω−H
∇ · F dV +
∫∫∂H
F · n dS
where n is the outward normal pointing outward on ∂Ω and ∂H.
Proof. Consider Ω−H satisfies Gauss’ Theorem with boundary ∂Ω ∪ ∂H. If we take the unitnormal to point outward on all of the boundary, this reverses the direction of the normal on ∂H,which negates the integral. Therefore by Gauss’ Theorem∫∫∫
Ω−H
∇ · F dV =
∫∫∂Ω∪∂H
F · n dS =
∫∫∂Ω
F · n dS −∫∫∂H
F · n dS
It should be noted there is also a generalized divergence theorem for tensors, and that this is not it.
2.4.3 Green’s Theorem
Green’s Theorem 2.4.3.1. Let D be a bounded subset of R2 (with an interior). Suppose ∂D ispiecewise C1, a simple closed curve oriented counter-clockwise. Let F = (f 1, f 2) be a C1 vectorfield on D, then ∫
∂DF · dx =
∫∫D
(∂f 2
∂x− ∂f 1
∂y) dx dy
Proof. Notice here that D is a 2 dimensional immersed submanifold of R2. Let µ be its volumeform. The metric on D is g = g
∣∣D = (dx)2 + (dy)2, so then µ =
√det(g) dx ∧ dy identifying x and
y as the same coordinates on D under inclusion. However g = I2x2 as a matrix, so√
det(g) = 1Therefore µ = dx ∧ dy. As a standard abuse of notation, dx ∧ dy is written as dx dy in doubleintegrals, with a similar abuse for triple integrals.
23
Justin Shaw 24
Using what we’ve shown so far, we can write Green’s Theorem in terms of integrals over manifolds:∫∂D
F[ =
∫D
(∂f 2
∂x− ∂f 1
∂y)µ
But now F[ = f 1dx+ f 2dy, so
dF[ = df 1 ∧ dx+ df 2 ∧ dy =∂f 1
∂ydy ∧ dx+
∂f 2
∂xdx ∧ dy = (
∂f 2
∂x− ∂f 1
∂y)dx ∧ dy = (
∂f 2
∂x− ∂f 1
∂y)µ
Therefore Green’s Theorem says∫∂D
F[ =
∫D
(∂f 2
∂x− ∂f 1
∂y)µ =
∫D
dF[
But now since a condition on D was that it was closed and bounded in R2 by Heine-Borel D iscompact, and so (F
∣∣D)[ is compactly supported. Therefore Green’s Theorem is just Stokes’
Theorem in 2 dimensions.
The quantity ∂f2
∂x− ∂f1
∂yis often defined as the vorticity of F = (f 1, f 2) when F is the velocity field
of a fluid flow in 2 dimensions, meaning a fluid flow in 3 dimensions where one of the dimensionsdoes not change. The vorticity in some sense describes the tendency of the vector field to rotatesomething dropped in the fluid.
Therefore Green’s Theorem says that the work done by a vector field F on a particle traversingthe boundary of a region in R2 is equal to the sum over the whole region of all the infinitesimalmeasures of the tendency of F to rotate (infinitesimally small) objects. This gives us a glimpseinto what the exterior derivative does to 1 forms.
2.4.4 Stokes’ Theorem
Recall that
curl(F) = (∂f3
∂y− ∂f2
∂z,∂f1
∂z− ∂f3
∂x,∂f2
∂x− ∂f1
∂y)
and that the vorticity of F is ∂f2∂x− ∂f1
∂y, so we see that the vorticity is the third component of the
curl. Letting k = (0, 0, 1) we can write this as
curl(F) · k =∂f2
∂x− ∂f1
∂y
Green’s Theorem says ∫∂D
G · dx =
∫∫D
(∂g2
∂x− ∂g1
∂y) dx dy
for the planar region D. We can think of D as a surface in R3 with unit normal n = k. Now thinkof G as a vector field in R3 by writing G = (g1, g2, 0). Now we can rewrite Green’s Theorem as
24
Justin Shaw 25
∫∂D
G · dx =
∫∫D
curl(G) · n dS
because we know the volume form of D in terms of it’s parameterization is
‖∂p
∂u× ∂p
∂v‖ du ∧ dv
However D is in the x− y plane, so here p(x, y) = (x, y, 0), so that x = u, y = v and
‖∂p
∂u× ∂p
∂v‖ = ‖(1, 0, 0)× (0, 0, 1)‖ = ‖(0, 0, 1)‖ = 1
and so dS = dx ∧ dy. This is the heuristic justification of the statement of Stokes’ Theorem fromVector Calculus.
Stokes’ Theorem (Vector Calculus) 2.4.4.1. Let Σ be a piecewise C1 orientable surface, with∂Σ a simple piecewise C1 closed curve. If F is of class C1 on some open set in R3 containingΣ ∪ ∂Σ, then ∫∫
Σ
(∇× F) · ndS =
∫∂Σ
F · dx
where the boundary is oriented counter-clockwise when viewed from the side on which the unitnormal points.
Proof. Note that the requirement that the boundary be simple means the boundary is a manifold,and since the boundary is closed, we may assume that it is included in Σ because we’re dealingwith integrals, so that in fact Σ is compact. We will assume that in fact the boundary is smooth.
We know that for a vector field G we have∫∫Σ
G · n dS =
∫Σ
∗(G[)
so taking G = ∇× F = [∗(dF[)]] we have∫∫Σ
(∇× F) · n dS =
∫Σ
∗(G[) =
∫Σ
∗(([∗(dF[)]])[) =
∫Σ
∗2dF[) =
∫Σ
dF[
and on the right side of the equation we have∫∂Σ
F · dx =
∫∂Σ
F[
so the vector calculus version of Stokes’ Theorem says∫Σ
dF[ =
∫∂Σ
F[
which is exactly Stokes’ Theorem from differential geometry.
25
Justin Shaw 26
2.5 Summary of AMATH 231
We have seen that with the proper natural identifications and conventions, that all the notationand theory from AMATH 231 agrees with that of standard differential geometry notation andtheory.
• Integration of immersed submanifolds with a global parameterization is carried out throughthe equations
∫γω =
∫ baω(γ(t)) and
∫Σ
ω =∫∫Duv
ω(p(u, v)).
• The volume forms of 1 and 2 dimensional immersed submanifolds are given in globalcoordinates by µ(γ(t)) = ‖γ′(t)‖ dt and µ(p(u, v)) = ‖∂p
∂u× ∂p
∂v‖ du ∧ dv respectively.
• The line integral of a scalar field is given by∫γf ds =
∫γfµ where µ is the volume form of γ.
With f = 1 we get vol(γ) as a manifold.
• The line integral of a vector field is given by∫γF · dx =
∫γF[∣∣γ.
26
Justin Shaw 27
3 AMATH 361
In this section we survey the course notes for AMATH 361 in the same way we did AMATH 231,rewriting definitions and theorems invariantly. Unless otherwise stated, every vector calculusresult from this section is taken directly from the AMATH 361 coursenotes for winter 2014 [2],and every differential geometry result is taken from the differential geometry courses I’ve taken atWaterloo. We will not cover the entirety of [2] because we specifically need the results for fluiddynamics. We begin by justifying our models.
3.1 What is Continuum Mechanics?
Continuum mechanics is concerns with the deformation of matter at scales large compared tointermolecular distances. Typical problems studied include fluid flow and deformation of solids.We will be more concerned with fluids as we make our way towards the Navier-Stokes equations.A fluid is matter which tends to take the shape of the container it’s in. Fluids cannot withstandshearing forces and so deform as long as those forces act on them. In contrast solids will deform toa certain degree but then the internal forces that hold the solid together will balance the shearand the deformation will stop. Moreover if the shearing force stops, the solid has a tendency torevert to its previous shape. Liquids do not have this property. For this reason sometimes we saythat a fluid has no memory.
My mother is not a mathematician, and when I said that I would be studying fluid mechanics thissummer she asked ”how is it possible to use math to study fluids?” This is a totally validquestion. How would you answer it?
The naive idea is that if we knew the position and velocity of every particle in a fluid we could intheory determine the fluid’s flow over time. By the Heisenberg uncertainty principle this isimpossible, and besides, such a model would be hopelessly complicated. This is an example wheresimplifications we assume actually lead us to a better qualitative understanding of a physicalmodel. There’s a balance to be struck here: we want a model complex enough to give us a goodapproximation of the qualities of fluids we wish to study, but simple enough that we can in somesense solve for what we want.
3.1.1 The Continuum Hypothesis
What are some of the fluid flows we might wish to study? Examples include: atmosphericcirculation, ocean circulation, surface waves, flow around an object or through a pipe, heating andair conditioning systems, flows of water in rivers, flows of liquid hot magma, blood flow, and evenflows of rock in the Earth’s mantle over very long time scales. All the models used to study thesefluid flows are related to the Navier-Stokes equations, whose derivation is the goal of the firstportion of these notes.
All of the problems we just mentioned consider phenomena that occur at scales much larger thanintermolecular distances. That is we are not concerned with how molecules move, but how thecontinuous matter made up of massive aggregates of molecules move. For example a thimble ofwater contains on the order of 1022 molecules of water, and we’re studying flows that include morethan a thimble full. In continuum mechanics we ignore the molecular structure of matter and
27
Justin Shaw 28
treat the matter as continuous, meaning made up of infinitely small pieces. This simplificationallows us to pose problems we can actually solve.
We call this assumption the Continuum Hypothesis.
Note that physically speaking, we know for a fact that this assumption is false, but it acts as agood approximation because molecules are so small. We will see that this strikes the balance wewere looking for: it makes our equations simple enough that we can study them, but does notsimplify so much that we lose the ability to describe the essential physics of fluid flows. This ismainly because the continuum hypothesis allows us to define functions on the fluid, and so toapply calculus methods.
3.2 Kinematics
Kinematics is the study of motion without regard for the causes of that motion. The forcescausing motion are not considered in this section, and so apply to all continuous matter.
3.2.1 Particles and Functions
We will take the continuum hypothesis as the justification for why we can use calculus on all ofthese problems, but we will no longer talk about it. In fact, now that we have these functions,which are defined continuously, and indeed smoothly whenever we find it convenient, we wish toknow what they mean physically. This may seem circular but it is not. We have assumed thecontinuum hypothesis, which is an approximation of reality, and now we consider how to thinkabout these approximations so that they still make physical sense.
To talk about the deformation and motion of matter, we want to be able to discuss physicalquantities like density ρ(x, t), velocity u(x, t), and pressure p(x, t). What do these functions meanphysically?
We can talk about the average density of matter easily: if we have a chunk of matter of volume Vand it has mass m, then the average density is just
ρ =m
V
But this is the average density over the volume V , and the functions we’ve just defined take valuesat a point, which by definition have no volume. So what do we mean by density at a point x?Consider the following thought experiment:
Take a cube of some material with sides of length l centered at x and let Ml(x) be the mass of thematerial in this cube. By definition, the average density of the matter in the cube is then
ρl(x) =Ml(x)
l3
Now imagine what happens as l→ 0. By continuity ρl varies as l decreases, and the morehomogeneous the material the slower it varies. When l is small ρl is almost constant because itvaries so little. However when l is so small that it is comparable to intermolecular distances, ρl
28
Justin Shaw 29
varies violently because the number of molecules inside the volume is small, so a change in l thatexcludes a molecule can result in a massive difference to the average density.
Definition 3.2.1.1. A particle is a volume of matter small enough that it can be regardedmathematically as a point, but large enough to contain a large number of molecules.
This definition was given by Euler. What it says is that a particle is a physical point, not just amathematical point. In particular a particle large enough that we can discuss its mass andconsider forces acting on it, and it also makes sense to discuss its motion through space. Usingthis concept we can now make sense of our functions.
Definition 3.2.1.2. The density field ρ(x, t) is the average density of a particle centered at x attime t.
Similarly
Definition 3.2.1.3. The velocity field u(x, t) is the average velocity of a particle centered at x attime t.
We will describe pressure later.
3.2.2 Pathlines and Streamlines
If the position of a particle is given by x(t) at time t, then its velocity is given by
dx
dt= u(x, t) (6)
If we are given u(x, t), then 6 gives us a set of n ODEs to solve, which is usually impossible to doanalytically.
Definition 3.2.2.1. A pathline is a solution to dxdt
= u(x, t)
Physically, a pathline is the trajectory traced out by a particle as it moves. In differentialgeometry we call these integral curves of the velocity field.
A pathline x(t) is a curve in space, and we can use the fact that its components are solutions toODEs to represent x in terms of an initial point a and t. That is, we can write x = x(a, t), andthe pathline x(a, t) describes the point a’s position at time t. Then different values of a will ingeneral give different pathlines, because we are selecting a different particle and watching ittraverse the field over time. In contrast we have
Definition 3.2.2.2. A streamline is a solution to dxdt
(s) = u(x(s), t) for a fixed time t.
Here s can be any parameter, not necessarily length along the curve. Physically, a streamline is acurve which is everywhere tangent to the velocity field u at a fixed time t. The difference betweena pathline and streamline is that time varies for the pathline and for the streamline it does not.Therefore a streamline is like a path drawn on a photograph of the velocity field such that it iseverywhere tangent to the field, and a pathline would require an animation or video as it describesthe motion of a particle over time such that at every instant the tangent to it’s trajectory agreeswith the velocity field. A pathline is tangent to the velocity field at x at time t, but notnecessarily any other time. If the velocity field is time-independent then steamlines and pathlinesare the same concept.
29
Justin Shaw 30
Definition 3.2.2.3. A flow in Rn is steady if all flow quantities are independent of time withrespect to global coordinates.
So not just the velocity field, but all scalar fields, and every other flow quantity do not vary overtime. Of course the same definition applies to a manifold.
3.2.3 The Material Derivative
Consider a fixed fluid particle a which follows the integral curve x(t) of the velocity field. We canthink of x as a function of a and time because x is the solution to an ODE, and so a gives theinitial conditions for this ODE, meaning without loss of generality we can assume that x(0) = a.Conceptually, we can think of a as a moving particle. Suppose f(x(t), t) is a physical property ofthe flow such as density or pressure. Then the value of f at the location of the moving particle a,which we’ll denote fL(a, t), is
fL(a, t) = f(x(a, t), t) = f(x1(a, t), x2(a, t), x3(a, t), t)
Now by the chain rule
∂fL∂t
=∂f
∂t+∂f
∂x1
dx1
dt+∂f
∂x2
dx2
dt+∂f
∂x3
dx3
dt
=∂f
∂t+ (
dx1
dt,dx2
dt,dx3
dt) · ( ∂f
∂x1
,∂f
∂x2
,∂f
∂x3
)
=∂f
∂t+dx
dt· ∇f
However recall that by definition of a pathline, dxdt
= u(x(a, t), t), so explicitly showing thedependence of variables, we have
∂fL(a, t)
∂t=∂f(x(a, t), t)
∂t+ u(x(a, t), t) · ∇f(x(a, t), t)
That is, this is a derivative along an integral curve of the velocity field. Suppressing the variables,we get
∂fL∂t
=∂f
∂t+ u · ∇f
Definition 3.2.3.1. The material derivative is the differential operator
D
Dt=
∂
∂t+ u · ∇
We define the material derivative to act on vector fields by acting on each component by theabove formula. That is, for a vector field v = (v1, v2, v3),
Dv
Dt=
(Dv1
Dt,Dv2
Dt,Dv3
Dt
)What does this mean physically? Recall from AMATH 231 that ∇f is a vector pointing in thedirection of greatest increase of f. We also have that if k is a unit vector, then ∇f · k is thedirectional derivative of f in direction k. Since u
‖u‖ is a unit vector, we have
30
Justin Shaw 31
u · ∇f =
((
u
‖u‖) · ∇f
)‖u‖
so u · ∇f is the rate of change of f with respect to time in the direction of motion multiplied bythe speed of the particle a. But then if f is independent of time, this is the rate of change of fwith respect to time as measured by the observer who is moving with velocity v.
Then the material derivative is the total rate of change of f with respect to time as measured byan observer moving with the fluid, but expressed in global coordinates. This means that Df
Dtcan
be nonzero even if f does not change in time, because the particle is following a pathline and somoving over time.
Conceptually the two contributions to the material derivative are ∂fdt
, which gives us the rate ofchange of f at the observer a’s position, and the rate of change of f with respect to time due tomotion through space, u · ∇f .
u · ∇v for vector fields u,v looks ambiguous, but it is not. First note that
(u · ∇)v = ui∂
∂xiv
using the standard abuse of treating nabla as a vector of differential operators. Really u · ∇ iswhat we call u in differential geometry. We also have ∇v = ∂vi
∂xjei ⊗ ej is the gradient of the vector
field v by definition. Then
u · (∇v) = u · ( ∂vi
∂xjei ⊗ ej)
where here · is contraction through the metric, meaning
(u · (∇v))j = gik∂vj
∂xiuk
but here gik = δki , so
(u · (∇v))j =∂vj
∂xiui
so we can write
u · (∇v) = (ui∂v1
∂xi, . . . , ui
∂vn
∂xi)
so in fact u · (∇v) = (u · ∇)v = u · ∇v, so we can unambiguously write
u · ∇v = ui∂
∂xiv = (ui
∂v1
∂xi, . . . , ui
∂vn
∂xi)
Similarly, for a function f ,
(u · ∇)f = ui∂f
∂xi= u · (∇f)
Therefore u · ∇f and u · ∇u are unambiguous expressions. Also note that we have proved thatour definition of Dv
Dtas
31
Justin Shaw 32
Dv
Dt=
(Dv1
Dt,Dv2
Dt,Dv3
Dt
)is now justified, because
Dv
Dt=∂v
∂t+ u · ∇v
=
(∂v1
∂t, . . . ,
∂vn
∂t
)+
(ui∂v1
∂xi, . . . , ui
∂vn
∂xi
)=
(∂v1
∂t, . . . ,
∂vn
∂t
)+(u · ∇v1, . . . ,u · ∇vn
)=
(Dv1
Dt,Dv2
Dt,Dv3
Dt
)
u · ∇ can be rewritten invariantly, but we must be careful. Recall that the Euclidean connectionon Rn is also denoted ∇ and is a covariant derivative, meaning it takes as input two vector fields,and gives as output another vector field. The Euclidean connection for two vector fieldsX = X i ∂
∂xiand Y = Y j ∂
∂xjis by definition
∇XY = (XY j)∂
∂xj= (X i∂Y
1
∂xi, . . . , X i∂Y
n
∂xi)
Therefore u · ∇v = ∇uv where the nabla on the left is the gradient operator, and the nabla on theright is the euclidean connection on Rn. We also know that by definition for f a function,
∇uf = uf = ui∂
∂xif = ui
∂f
∂xi= u · ∇f
Therefore we have shown that
D
Dt=
∂
∂t+∇u (7)
So we’ve written the material derivative invariantly, and we see that in the case of timeindependent vectors the material derivative is simply the covariant derivative with respect to u.Above we defined the material derivative to operate on vectors simply by operating on thecomponent functions and this works because the Christoffel symbols for the Euclidean connectionin Rn vanish, but in general the Christoffel symbols for a connection do not vanish. So for ageneral manifold we must insist that the material derivative is applied to vector fields using thedefinitions from differential geometry. That is, in general
Dv
Dt=
(Dv1
Dt,Dv2
Dt,Dv3
Dt
)does not hold. Explicitly,
∇uv = ∇u(vi∂
∂xi) = (uvi)
∂
∂xi+ vi∇u
(∂
∂xi
)
32
Justin Shaw 33
and ∇uvi = uvi. So unless ∇u
(∂∂xi
)vanishes, the componentwise formula is incorrect.
Another important point to note is that the material derivative is nonlinear.
3.2.4 Material Volumes
Definition 3.2.4.1. A material volume is a fixed piece of matter in R3 which moves with a fluidflow. It is comprised of the same particles for all time. So in coordinates relative to a framemoving with the fluid, a material volume is a fixed region.
Physically, we can imagine a patch of water in a river dyed red to be our material volume. Ofcourse diffusion will cause this volume to dissipate, but for short time scales where the diffusion isnot significant, the red patch moves and deforms with the flow. Therefore in Cartesiancoordinates the volume and dimensions of the red patch do change, but as measured from eachindividual particle, the material volume is a set of particles following their trajectory through theflow, and so remains a constant region. Particles as described above are sometimes called materialparticles because they consist of the same matter for all time.
Mathematically we can think of a material volume as a fixed set of particles which are movingalong their pathlines together. To avoid confusion we repackage some notation:
Definition 3.2.4.2. Let Φ(·, t) : R3 → R3 be the mapping that takes an initial point a of a particleto its position at time t, meaning Φ(a, t) = x(a, t). We will also write Φt(a) or Φ(a)(t) for Φ(a, t)as alternate forms.
This is simply the repackaging of all pathlines into a single global time evolution map. Indifferential geometry terms, this is the flow of the vector field u, with integral curves Φ(a)(t),where Φ(a)(0) = a. We assume that Φ is a diffeomorphism in terms of both x and t. Physicallythis means two different particles of matter can’t be at the same place at a later time, or that oneparticle can’t split into two.
Notice that since the material derivative is taken along integral curves, the assumption that Φ is adiffeomorphism also means that we can treat any point as a point on a trajectory, and so we cantake derivatives at any point in space. There is always a solution to an ODE for small times, butthere is no guarantee that solutions exist for all time. We aren’t going to worry about this toomuch, but it’s worth mentioning. Unless otherwise stated, we’ll assume our flows exist for all time.
Now Φ(a, t) = Φa(t) = x(a, t), and when we derived the material derivative we pointed out thatthere was a dependence on x(a, t) in each term. Therefore
Df(Φa(t), t)
Dt=∂f(Φa(t), t)
∂t+ u(Φa(t), t) · ∇f(Φa(t), t)
Lets rewrite the right hand side in a different way. The spacial components are all in terms ofΦa(t), so we have function composition in the spatial components. Holding a fixed, then, we canmake the right hand side a function of t only. Define Ψ(t) = (Φa(t), t), then we can write
∂f(Ψ(t))
∂t+ u(Ψ(t)) · ∇f(Ψ(t)) = (Ψ)∗
(∂f
∂t+ u · ∇f
)
33
Justin Shaw 34
which is a function of t. Since a is fixed, this is a function of t at a which gives the total rate ofchange of f at a’s position, at time t. Therefore we have recovered the Lagrangian interpretationof the derivative.
Finally, since we’re assuming Φ is a diffeomorphism, we are justified in taking the materialderivative at any point, because every point is in the image of Φ, and so is on an integral curve ofthe fluid flow.
3.2.5 The Transport Theorem
Now take a material volume which is time dependent, say W (t). By definition of a materialvolume, since material volumes are fixed sets of particles, every particle in W (0) gets mapped to aparticle in W (t) and every particle in W (t) has a preimage in W (0). We will write W for W (0)sometimes, so we can write things like W (t) = Φt(W (0)) = Φt(W ). If W (0) is a curve, then W (t)is called a material curve, and if W (0) is a surface then W (t) is called a material surface.
Intuitively, you can think of a material curve as a string moving through the fluid, and a materialsurface as a chunk of material of codimension 1 moving through the fluid.
For now, we will be interested in the time evolution of material volumes, and integrals of the form
I(t) =
∫∫∫W (t)
f(x, t) dV (8)
Here the volume of integration is a time dependent material volume. We wish to find thederivative of I(t). We will use the diffeomorphism Φ and the change of variables theorem. Letx = Φ(a, t) = (Φ1(a, t),Φ2(a, t),Φ3(a, t)). Then by the change of variables theorem∫∫∫
W (t)
f(x, t) dV =
∫∫∫W (0)
f(Φ, t) det(J(a, t)) dV
Where here J(a, t) is the Jacobian matrix of Φ with respect to the variables in a, so J(a, t) = ∂Φi
∂aj.
Notice at time t = 0, Φ(·, 0) is the identity map, and so has Jacobian determinant 1. Since we’reassuming Φ is invertible, the Jacobian determinant of Φ is never zero, for any t, and thereforeJ(a, t) has positive determinant for all t by continuity of J . This is why we did not have theabsolute value of the determinant above, only the determinant, because we knew it would bepositive. This also shows us that Φ is an orientation preserving map.
Lemma 3.2.5.1. Suppose Φ is a diffeomorphism and x = Φ(a, t) = (Φ1(a, t),Φ2(a, t),Φ3(a, t)) ,then
∂ det(J)
∂t= div(u) det(J)
where u = dxdt
.
Notice here that div(u) is actually div(u)(Φ(a, t)) = (Φt)∗ div(u) which will be important for us
later, but not now.
34
Justin Shaw 35
Proof. We have u = (u1, u2, u3), so
div(u) =∂u1
∂x+∂u2
∂y+∂u3
∂z
We have J(a, t) = ∂Φi
∂aj, write Φi,j = ∂Φi
∂ajso in 3 dimensions we have
det(J(a, t)) = det(Φi,j)
=∑σ∈S3
sgn(σ)3∏i=1
Φi,j
Not take the partial derivative with respect to t:
∂ det(J)
∂t=∑σ∈S3
sgn(σ)∂
∂t
(3∏i=1
Φi,σj
)
=∑σ∈S3
sgn(σ)
((∂
∂tΦ1,σ1
)Φ2,σ2Φ3,σ3 + Φ1,σ1
(∂
∂tΦ2,σ2
)Φ3,σ3 + Φ1,σ1Φ2,σ2
(∂
∂tΦ3,σ3
))But now
∂
∂tΦi,j =
∂2
∂t∂aj=
∂
∂aj∂
∂tΦi =
∂
∂ajui(Φ(a, t)) =
∂ui∂xk
∂Φk
∂aj=∂ui∂xk
Φk,j
so in particular ∂∂t
Φi,σi = ∂ui∂xk
Φk,σi . Putting it all together we have:
∂ det(J)
∂t=∑σ∈S3
sgn(σ)
(∂u1
∂xiΦi,σ1Φ2,σ2Φ3,σ3 +
∂u2
∂xjΦ1,σ1Φj,σ2Φ3,σ3 +
∂u3
∂xkΦ1,σ1Φ2,σ2Φk,σ3
)
Look at the first term∑σ∈S3
sgn(σ)(∂u1∂xi
Φi,σ1Φ2,σ2Φ3,σ3
). expanding the sum we get
∂u1∂x
( ∑σ∈S3
sgn(σ)Φ1,σ1Φ2,σ2Φ3,σ3
)= ∂u1
∂xdet(J) plus these two terms:
∂u1
∂y
(∑σ∈S3
sgn(σ)Φ2,σ1Φ2,σ2Φ3,σ3
),∂u1
∂z
(∑σ∈S3
sgn(σ)Φ3,σ1Φ2,σ2Φ3,σ3
)These bracketed terms correspond to the det(J) with the first column replaced by the second, anddet(J) with the first column replaced by the second, respectively. However the determinant of amatrix with repeated columns (or rows) is zero, so these two terms vanish, so we get ∂u1
∂xdet(J) in
the first term of the equation for ∂ det(J)∂t
above. Expanding the other terms gives similar results,and we find that
∂ det(J)
∂t=∂u1
∂xdet(J) +
∂u2
∂ydet(J) +
∂u3
∂zdet(J) = div(u) det(J)
35
Justin Shaw 36
The Transport Theorem 3.2.5.1. If Φ is invertible, u(x, t) is C1, and f(x, t) is C1, then
dI
dt=
d
dt
∫∫∫W (t)
f(x, t) dV =
∫∫∫W (t)
(Df
Dt+ f∇ · u
)dV
Proof. We have
dI
dt=
d
dt
∫∫∫W (t)
f(x, t) dV =d
dt
∫∫∫W (0)
f(Φ(a, t), t) det(J(a, t)) dV
=
∫∫∫W (0)
([∂f
∂t+∂f
∂xi∂Φi
∂t
]det(J) + f
∂ det(J)
∂t
)dV
But now by definition of Φ, Φi(a, t) = xi(a, t), so
∂Φi
∂t(a, t) = ui(Φ(a, t), t)
Now we use this fact and the lemma we just proved gives us ∂ det(J)∂t
= div(u) det(J), so we have
dI
dt=
∫∫∫W (0)
([∂f
∂t+∂f
∂xiui
]det(J) + f div(u) det(J)
)dV
=
∫∫∫W (0)
(Df
Dtdet(J) + f div(u) det(J)
)dV
=
∫∫∫W (0)
(Df
Dt+ f div(u)
)det(J) dV
=
∫∫∫W (t)
(Df
Dt+ f div(u)
)dV
Where here the suppressed variables in the integrand are just (x, t)
Recall thatdiv(fu) = u · ∇f + f div(u)
so thatDf
Dt+ f div(u) =
∂f
∂t+ u · ∇f + f div(u) =
∂f
∂t+ div(fu)
36
Justin Shaw 37
Now we can use the Divergence Theorem (Stokes’ Theorem) to get
dI
dt=
∫∫∫W (t)
(Df
Dt+ f∇ · u
)dV
=
∫∫∫W (t)
(∂f
∂t+ div(fu)
)dV
=
∫∫∫W (t)
∂f
∂tdV +
∫∫∂W (t)
fu · n dS
Physically, this says that the rate of change of I is equal to the integral of ∂f∂t
over the constant intime region W (t) plus the flux of f out of W (t).
Lets repackage this section in differential geometry terms.
The Transport Theorem 3.2.5.2. Suppose (M,µ) is a manifold with volume form, andX ∈ Γ(TM). Let θt be the flow of X. Suppose f ∈ C∞(M × R) and write ft(p) for f(p, t). Thenif U ⊂M is open
d
dt
∫θt(U)
ft µ =
∫θt(U)
(∂f
∂t+ divµ(ftX)
)µ
This is the version given in [6]. Here the subscript on the divergence is stressing its the divergencewith respect to the volume form µ.
We can see immediately that this is just a change of notation from the vector calculus version ofthe theorem, by our work done in the AMATH 231 section. We have θt(U) instead ofΦt(W ) = W (t). Because an open set must have ”full dimension”, the volume form on θt(U) is justµ∣∣θt(U)
which we can just write as µ. The only difference between this version and the version
given above, is that this version chooses not to talk about the material derivative, and rewrites theintegrand as we did above.
3.3 Derivation of the Governing Equations: Mass, Momentum, andEnergy
We will now start developing the equations we need to derive and understand the Navier-Stokesequations. We will make some physical assumptions, and will justify those assumptions. [6]
1. Mass is neither created nor destroyed
2. Newton’s 2nd Law: F = ma = ddt
p = p where here p = mu
3. Energy is neither created nor destroyed
37
Justin Shaw 38
3.3.1 Conservation of Mass and the Continuity Equation
Our first assumption is that mass is neither created nor destroyed. In our integral (8)
I(t) =
∫∫∫W (t)
f(x, t) dV
if we take f = ρ, the density field we defined above, then this integral is the mass ofW (t) = Φt(W ). Define
M(t) =
∫∫∫W (t)
ρ(x, t) dV
the mass of the material volume W (t). Conservation of mass implies M(t) is constant. Therefore,by the transport theorem
0 =dM
dt=
∫∫∫W (t)
(∂ρ
∂t+ div(ρu)
)dV
and this is for any material volume W (t), so by the Dubois-Reymond lemma we have
∂ρ
∂t+ div(ρu) = 0 (9)
Which is called the continuity equation, or the differential form of the law of conservation of mass.Note that there is a smoothness assumption required to get to this equation. If we don’t have theintegrand smooth enough, we must leave it in the integral form.
As an aside, notice that
V (t) =
∫∫∫W (t)
dV
is the volume of W (t), and the time derivative is
dV
dt=
∫∫∫W (t)
(∂1
∂t+ div(1u)
)dV =
∫∫∫W (t)
div(u) dV
This means div(u) is the rate of change of volume per unit volume.
Getting back to the continuity equation, we can rewrite it as
0 =∂ρ
∂t+ div(ρu) =
Dρ
Dt+ ρ div(u)
which implies, since ρ > 0, that we can write
1
ρ
Dρ
Dt= − div(u)
so when div(u) > 0, the material volume is expanding, and density is decreasing because 1ρ
ispositive here, as always.
38
Justin Shaw 39
We have a product rule of sorts too, provided mass is conserved:
d
dt
∫∫∫W (t)
ρf dV =
∫∫∫W (t)
(∂(ρf)
∂t+ div((ρf)u)
)dV
=
∫∫∫W (t)
(ρ∂f
∂t+ f
∂ρ
∂t+ u · ∇(ρf) + ρf div(u)
)dV
=
∫∫∫W (t)
(ρ∂f
∂t+ f
∂ρ
∂t+ u · f∇ρ+ u · ρ∇f + ρf div(u)
)dV
=
∫∫∫W (t)
(ρ
[∂f
∂t+ u · ∇f
]+ f
[∂ρ
∂t+ u · ∇ρ+ ρ div(u)
])dV
but since we’re assuming mass is conserved, by the continuity equation the second term in theintegrand is zero, and the first is ρDf
Dtby definition. Therefore we have shown that
d
dt
∫∫∫W (t)
ρf dV =
∫∫∫W (t)
ρDf
DtdV
but only when mass is conserved. Confusingly, this is also sometimes called the TransportTheorem. Note that this formula applies when f is vector valued as well, since D
Dtacts
componentwise.
Notice that so far the metric has not really been needed because we showed we can write thematerial derivative in terms of the connection, so all the results above really only required avolume form so that we could define divergence.
Lets derive the continuity equation in differential geometry terms:
Lie Derivative Formula for Functions 3.3.1.1. Suppose f ∈ Cr(M), X ∈ Γr−1(TM), and letX have a flow θt, then
d
dt(θt)
∗f = (θt)∗£Xf
This is a theorem from [6]. Note that if we identify X as a partial derivative as we normally do,then £Xf = Xf by definition.
Proof. Fix a point p ∈M . Now we apply the chain rule:
d
dt((θt)
∗f) (p) =d
dt(f θt)(p) =
∂f
∂xi(θt(p))
dθitdt
(p) = ∇f(θt(p)) ·d
dtθt(p)
but by definition ddtθt(p) = Xθt(p). Therefore
d
dt((θt)
∗f) (p) = ∇f(θt(p)) ·Xθt(p) = df(θt(p))Xθt(p) = (df(X))(θt(p)) = (Xf)(θt(p)) = £Xf(θt(p))
applying the musical isomorphism definition to ∇f = (df)]. Finally, by definition of the pullback£Xf(θt(p)) = ((θt)
∗£Xf)(p)
39
Justin Shaw 40
Now by the differential geometry version of change of variables, we have that by our observationabove, Φ is an orientation preserving diffeomorphism, so for any top form ω on W (t) = Φt(W ), wehave ∫
Φt(W )
ω =
∫W
(Φt)∗ω
in particular, conservation of mass says∫Φt(W )
ρtµ =
∫W
ρ0µ
for some initial value ρ0 of ρ(x, t). So now by the change of variable theorem we have∫W
(Φt)∗ρtµ =
∫Φt(W )
ρtµ =
∫W
ρ0µ
so (Φt)∗ρtµ = ρ0µ which is constant. By Du Bois - Reymond and the definition of the pullback
this is equivalent to the statement
(Φ∗tρt) det(J) = ρ0
where det(J) is the determinant of the jacobian matrix for Φ as before. Now
0 =d
dt((Φ∗tρt) det(J)) = det(J)
d
dt(Φ∗tρt) + (Φ∗tρt)
d
dtdet(J)
= det(J)((Φt)∗∂ρt∂t
+ Φ∗t (uρt)) + Φ∗t (ρt div(u)) det(J)
using our previous lemmas. We also have that det(J) is never zero, so we have
0 = Φ∗t
(∂ρt∂t
+ u · ∇ρt + ρt div(u)
)= Φ∗t
(∂ρt∂t
+ div(ρtu)
)which means we’ve recovered the continuity equation.
3.3.2 Conservation of Linear Momentum and the Stress Tensor
We could also call this section ”Balance of Momentum” which is the same concept. Whichevername is chosen, the concept is analagous to Newton’s second law. Consider that for a solid, rigidbody of mass m, considered as a point with velocity v, Newton’s second law states
d
dt(mv) = F
where here F is the net force acting on the object. This formula does not apply to a continuum,so we need an analogue we can use. Fortunately, in 1776 Euler came up with just such ananalogue [2].
1. The total force acting upon a body equals the rate of change of the total linear momentum
40
Justin Shaw 41
2. The total torque acting upon a body equals the rate of change of the total moment ofmomentum, where both the torque and the moment are taken with respect to the same fixedpoint.
The first statement is also known as balance of momentum. Euler said these laws apply to allbodies or systems of bodies and to every part of every body whether or not they are treated as acontinuum or whether or not they are deforming. It turns out that the second law is not valid forall continuums, but is valid for all the ones we will discuss. The first law is more relevant for usanyway.
∫∫∫W (t)
ρu dV is the total linear momentum of the material volume W (t) by definition and
this is a vector valued integral as it should be, since momentum is a vector. The first law then says
d
dt
∫∫∫W (t)
ρu dV = net force acting on W(t)
Note that vector valued integrals are evaluated componentwise.
Now we know that the rate of change of momentum equals the net force acting on a materialvolume, but what forces can act on a material volume W (t)? There are three types
1. Body forces act throughout the volume. One example is the gravitational force
Fg =
∫∫∫W (t)
ρg dV
but we could also have a magnetic field or something. These are external forces. In generalwe will assume that the body forces can be codified by a force density vector field b withunits of force per unit mass, so that the body forces can be written∫∫∫
W (t)
ρb dV
2. Surface forces are forces exerted on ∂W (t) by matter outside of W (t). These forces are dueto short range forces between molecules and to movement of molecules across ∂W (t). Theseforces are called forces of stress as well.
3. Line forces, also known as surface tension, acts on the interface between a liquid and a gasor between two immiscible liquids. We won’t worry about these forces because they do notappear in the equations of motion. They do arise in boundary conditions though.
Definition 3.3.2.1. The stress vector t(x, t, n) is the force per unit area acting on a surfaceelement at (x, t) with unit outward normal n. The force is exerted by the material into which npoints and acts on the material from which n points.
So if dA was a surface element at (x, t), then t(x, t, n)dA is the force exerted by the matteroutside W (t) on dA. This definition allows us to write down the total surface forces as∫∫
∂W (t)
t(x, t, n) dS
41
Justin Shaw 42
and since we have the total body forces above, this gives us the balance of momentum equation
d
dt
∫∫∫W (t)
ρu dV =
∫∫∫W (t)
ρb dV +
∫∫∂W (t)
t(x, t, n) dS (10)
Cauchy’s Fundamental Theorem for Stress 3.3.2.1. Balance of momentum implies thatt(x, t, n) is linear in the components of n, meaning
tj =∑i
τijni
for some scalars τij(x, t).
We omit the proof of this theorem because although it is straightforward, it is somewhat tedious.It’s in the course notes. What this theorem says is that if t exists and satisfies the balance ofmomentum equation, then it must be of the form τ · n for a two-tensor τ . Here the notation using· is slightly confusing. Suppose τ has components τ ij, and n has components nk. τ · n is acontraction through the metric g, as follows [6]:
(τ · n)i = gjkτijnk
but in Rn, gjk = δkj , so we have
(τ · n)i =∑j
τ ijnj
Similarly
(n · τ)i = gjkτjink =
∑j
τ jinj = ti
by Cauchy’s Theorem above. Therefore we have shown that for a two tensor τ ,
t(x, t, n) = n · τ
Definition 3.3.2.2. Let τjk(x, t) = tk(x, t, ij), then τjk form the components of the stress tensorwe’ll call τ . This tensor is called the Cauchy stress tensor.
Physically, τjk is the kth component of the force per unit area acting on a surface with outwardnormal ij. Therefore the material into which ij points is exerting the force on the material fromwhich ij points.
This brings us to the derivation of the momentum equations. The balance of linear momentum is
d
dt
∫∫∫W (t)
ρu dV =
∫∫∫W (t)
ρb dV +
∫∫∂W (t)
t(x, t, n) dS
We can use t(x, t, n) = n · τ in the second integral on the right, and the fact thatddt
∫∫∫W (t)
ρf dV =∫∫∫W (t)
ρDfDtdV we derived above on the left, to write
42
Justin Shaw 43
∫∫∫W (t)
ρDu
DtdV =
∫∫∫W (t)
ρb dV +
∫∫∂W (t)
n · τ dS
but now∫∫
∂W (t)
n · τ dS is a vector valued integral whose kth component is∫∫∂W (t)
τikni dS
since (n · τ)k =∑j
τjknj as we just showed above. Notice also that if we write τk for the kth
column of τ , that∑j
τjknj = τk · n where this · is the dot product. Therefore∫∫∂W (t)
τikni dS =
∫∫∂W (t)
τk · n dS =
∫∫∫W (t)
div(τk) dV
by the divergence theorem. We can define the divergence of a two tensor to be
div(τ) = (div(τ1), . . . , div(τn))
Note that really what we’ve done here is contract τ , which is a 1− 1 tensor, with the covariantderivative [11]
(div τ)j ≡ ∇iτij
This extends our definition invariantly. With this convention we can write∫∫∫W (t)
ρDu
DtdV =
∫∫∫W (t)
ρb dV +
∫∫∫W (t)
div(τ) dV
so by the Du Bois - Reymond lemma, since W (t) is arbitrary, we have
ρDu
Dt= ρb + div(τ) (11)
and this is called the momentum equation. This is the second basic equation of continuummechanics after the continuity equation. The ith component of this equation is then
ρDuiDt
= ρbi + div(τi)
Note that the momentum equation applies to all continuums for which mass is conserved. Theform of τ is dependent on the type of matter. It turns out that if the fluid is such that Euler’ssecond law holds, then the stress tensor is symmetric [2]. Therefore there are 6 rather than 9independent components of the stress tensor. It also turns out that angular momentum isconserved in this case.
Since τ is symmetric, so is it’s matrix representation. Recall from linear algebra that a realsymmetric matrix is diagonalizable, and as a result the eigenvectors of τ form a basis for R3. Wecall the eigenvectors of a symmetric second-order tensor the principal axes. Specifically for thestress tensor τ , the eigenvalues are called the principal stresses.
43
Justin Shaw 44
Now what does this mean physically? Take a reference frame in which τ is diagonal at a fixedpoint, so
τ =
τ11 0 00 τ22 00 0 τ33
Recall that our stress vector t is the force per unit area acting on a surface with outward unitnormal n, and that by definition of τ , tk = τikni. Now since τ is diagonal, this means tk = τkknkwith no sum over k. This means in this coordinate system where τ is diagonal, surfaces parallel toa coordinate plane have only normal surface forces, so an infinitesimal volume is subjected tonormal forces of either compression or tension along the direction of the coordinate axes in whichτ is diagonal.
Another fact from linear algebra, is that for a symmetric matrix the eigenvectors corresponding todifferent eigenvalues are orthogonal, and so by taking an orthonormal basis in each eigenspace weget an orthonormal basis of eigenvectors. This is also intuitively clear because the change of basismatrices that diagonalize a symmetric matrix are orthogonal, meaning they correspond torotations.
Now in R3 with τ , consider the surface forces acting on an infinitesimal volume. The surface forcesare made up of three compressive or tensile forces acting in three orthogonal directions. Since theeigenvectors of τ , the principal axes, form an orthonormal basis, we see where the name comesfrom. The principal stresses, the eigenvalues, give us the force per unit area in the direction of thecorresponding principal axis. A positive stress indicates a tensile or stretching surface force in thedirection of that principal axis, and a negative stress indicates a compressive or ”squishing”surface force in the direction of that principal axis.
This whole construction was done pointwise, so the stress tensor can be thought of as a matrix ofscalar fields symmetric at every point, and so diagonalizable at every point. The matricesdiagonalizing the stress tensor at a given point correspond to rotations, and in this rotated framewe get a simple description in three independent directions of the forces on a particle at that point.
3.3.3 Conservation of Energy
We now cover the third of our assumptions, the conservation of energy. Taking the dot product ofthe momentum equation with the velocity gives∑
i
ρuiDuiDt
=∑i
[ρuibi + ui div(τi)]
We know that DDt
= ∂∂t
+∇v so DDt
clearly has a product rule. Now we have
∑i
ρuiDuiDt
=∑i
[ρD
Dt(1
2uiui)] =
∑i
[D
Dt(ρ
2uiui)−
1
2uiui
Dρ
Dt] =
∑i
[D
Dt(ρ
2uiui) + (
ρ
2uiui) div(u)]
by the continuity equation. By linearity of DDt
we have
44
Justin Shaw 45
∑i
ρuiDuiDt
=D
Dt(ρ
2u · u) + (
ρ
2u · u) div(u)
Let K = ρ2u · u, called the kinetic energy density, which is the kinetic energy per unit volume.
Notice that this value is positive or zero since the dot product is positive definite, and the densityis always positive. Now we’ve shown that
ρu · b + u · div(τ) =∑i
ρuiDuiDt
=DK
Dt+K div(u)
There are lots of physical interpretations to make here:
1. DKDt
is the rate of change of the kinetic energy density of a particle.
2. Recall that div(u) being positive indicates an increase in the material volume. Therefore−K div(u) is a negative quantity in this case, and the kinetic energy density , DK
Dt,
decreases. If div(u) is negative then DKDt
increases.
3. ρb is the body force per unit volume, so ρu · b is the rate at which work is done, per unitvolume, by the body force on the material volume.
4. u · div(τ) is the rate work is done by the surface forces.
Lets define our energy terms, assuming b = −∇b for b a scalar field which is constant in time:
Definition 3.3.3.1. Define K = ρ2u · u as the kinetic energy density, P = ρb as the potential
energy density, and the total mechanical energy density E as E = K + P .
Now that we have this definition DEDt
= DKDt
+ DPDt
, and from above we know that
DK
Dt= ρu · b + u · div(τ)−K div(u)
and we can calculate
DP
Dt=D(ρb)
Dt= ρ
Db
Dt+ b
Dρ
Dt= ρ
∂b
∂t+ ρu · ∇b− bρ div(u)
using the continuity equation. Now P = bρ, ∇b = −b, and ∂b∂t
= 0 so we have
DE
Dt=DK
Dt+DP
Dt= ρu · b + u · div(τ)−K div(u)− ρu · b− P div(u)
= −E div(u) + u · div(τ)
so we have proven
DE
Dt= −E div(u) + u · div(τ) (12)
However the total mechanical energy is not conserved because friction causes some of the energyto be transformed into heat. We give heat the name internal energy. It is the energy associatedwith molecular motion, of which there are 3 types: translational, rotational, and vibrational.
45
Justin Shaw 46
Definition 3.3.3.2. Let e(x, t) be the internal energy per unit mass. Then we define ρe to be theinternal energy per unit volume.
This makes sense because in the integral ρe dV corresponds to energy. Now we can discuss balanceof energy, because the internal energy plus the total mechanical energy is conserved, meaning thatwork done by the body force remains constant, because the body force converts potential energyto kinetic energy, leaving E unaffected. Therefore
d
dt
∫∫∫W (t)
(E + ρe) dV = rate at which work is done by surface and thermodynamic forces
First the surface forces: We have t giving force per unit area, and u is our velocity field. The dotproduct’s units are the product of the units of each vector, so u · t has units of force timesdistance over time times area, meaning the rate of change of work per unit area. Therefore wewant to integrate u · t over ∂W (t) to get the rate at which work is done by the surface forces:
∫∫∂W (t)
u · t dV =
∫∫∂W (t)
u ·
(∑j
τjinj
)dV
=
∫∫∂W (t)
(∑i,j
uiτjinj
)dV
However∑i,j
uiτjinj = (uiτ1i, uiτ2i, uiτ3i) · (n1, n2, n3) where we’re summing over i in each term.
Therefore by the divergence theorem
∫∫∂W (t)
u · t dV =
∫∫∂W (t)
(∑i,j
uiτjinj
)dV
=
∫∫∫W (t)
div((uiτ1i, uiτ2i, uiτ3i)) dV
but (u · τ)k =∑i
τikni =∑i
τkini by symmetry of τ , so
u · τ = (uiτ1i, uiτ2i, uiτ3i)
so we have
∫∫∂W (t)
u · t dV =
∫∫∫W (t)
div((uiτ1i, uiτ2i, uiτ3i)) dV
=
∫∫∫W (t)
div(u · τ) dV
For the thermodynamic forces, consider that internal energy, or heat, is transferred from hot tocold. Let q be the local internal flux density. Therefore q · ndA is the heat flux through a surfaceelement of area dA in the direction of the unit normal n. The rate at which work is done by
46
Justin Shaw 47
thermodynamic forces is the rate at which heat is transferred through the surface by definition,represented by the integral ∫∫
∂W (t)
−q · n dS = −∫∫∫
W (t)
div(q) dV
where the minus sign is because we want heat flux into W (t), which is in the opposite direction tothe normal. Now we can combine our integrals to get
d
dt
∫∫∫W (t)
(E + ρe) dV =
∫∫∫W (t)
div(−q + u · τ) dV
We can rewrite ddt
∫∫∫W (t)
(E + ρe) dV using the transport theorem:
d
dt
∫∫∫W (t)
(E + ρe) dV =
∫∫∫W (t)
(D
Dt(E + ρe) + (E + ρe) div(u)
)dV
so by Du Bois - Reymond lemma
D
Dt(E + ρe) + (E + ρe) div(u) = div(−q + u · τ) (13)
but we can also rewrite this as
∂
∂t(E + ρe) + div((E + ρe)u + q− u · τ) = 0
which is the equation written in what’s called conservative form. In general if we have a PDE ofthe form
∂R
∂t+ div(F(R)) = 0
integrating this equation over the region indicates that the total amount of the function R isconserved. This equation says that R can change only if the amount of R leaving a region isdifferent from that entering the region, meaning the divergence must be zero.
47
Justin Shaw 48
3.3.4 Summary of Equations so far
By assuming conservation of mass, linear momentum, angular momentum, and energy, we havederived the following equations:
1. The Continuity Equation (9) (conservation of mass):
Dρ
Dt+ ρ div(u) = 0
2. The Momentum Equation (11) (balance of momentum):
ρDu
Dt= ρb + div(τ)
3. The Energy Equation (13) (conservation of energy)
D
Dt(E + ρe) + (E + ρe) div(u) = div(−q + u · τ)
This is five equations in fourteen unknowns:
• the density ρ
• the three velocity components ui of u
• the six components of τ , since τ is symmetric
• the internal energy density e
• the three components of q, the local internal flux density
3.4 Newtonian Fluids
Up to now, we have been working in the greatest generality possible given our assumptions. Tomake further progress towards the Navier-Stokes equations we now consider the special case ofNewtonian fluids. Experimentally many common liquids such as water and air can be consideredNewtonian, which makes this a very important special case.
3.4.1 The Velocity Gradient
Definition 3.4.1.1. The velocity gradient tensor, denoted ∇u for the velocity field u(x, t), is the(2, 0) tensor with components
(∇u)ij =∂ui∂xj
48
Justin Shaw 49
Notice this is simply the Jacobian matrix of u thought of as a map from R3 to R3. As such we canthink of it as the linear approximation of u. By definition we know the (∇u)ij gives us the rate ofchange of the component function ui in the direction ∂
∂xj. Notice also that the trace of ∇u is
div(u).
Lets break up ∇u into symmetric and antisymmetric parts:
(∇u)ij =∂ui∂xj
=1
2
(∂ui∂xj
+∂uj∂xi
)+
1
2
(∂ui∂xj− ∂uj∂xi
)Definition 3.4.1.2. We call the symmetric part of ∇u, namely 1
2
(∂ui∂xj
+∂uj∂xi
)the strain rate
tensor, and denote it e. This is also called the Cauchy strain tensor.
Note that the trace of e is div(u). We can think of this in a different way. We call the term ∂ui∂xi
the linear strain rate of u in the direction ∂∂xi
. Physically, this represents the rate of change of
length per unit length in the ∂∂xj
direction. Therefore the divergence is the sum of the rates of
change of length per unit length in three orthogonal directions.
The off diagonal entries of e have a different physical meaning. The shear strain rate is the rate atwhich the angle between two initially perpendicular lines decreases. This is a measure of thedeformation caused by the velocity field, and so describes the rate at which a material volume ischanging shape, rather than changing volume, or being translated. If we assume our two lines areparallel to the x1 and x2 axes, the shear strain rate is ∂u1
∂x2+ ∂u2
∂x1. This makes intuitive sense
because these are the off diagonal terms in a 2 dimensional rotation matrix in the xi, xj plane. We
can form a shear tensor by taking sij = ∂ui∂xj
+∂uj∂xi
for i 6= j and sij = 0 for i = j
Notice for i 6= j, sij = 2eij. Now we have a decomposition of the strain rate tensor e in terms of∇(u) and the shear s.
e =1
2s+
(∇u)11 0 00 (∇u)22 00 0 (∇u)33
So far we have decomposed the symmetric part of ∇u into physically meaningful components: swhich measures deformation not including volume changes, and the matrix, which measures therate of change of volume. Therefore e measures the total deformation of a material volume underu.
The antisymmetric part of ∇u is 12
(∂ui∂xj− ∂uj
∂xi
)and this is directly related to another important
vector field.
Definition 3.4.1.3. The vorticity ω is ω = curl(u)
If we define rij = ∂ui∂xj− ∂uj
∂xi, then by definition of the curl we have the rotation tensor
r =
0 −ω3 ω2
ω3 0 −ω1
−ω2 ω1 0
49
Justin Shaw 50
This is called a rotation tensor because if we imagine that u is a pure rotation, then there is noshear and no change in volume, so e is zero. Therefore
(∇u)ij =∂ui∂xj
=1
2
(∂ui∂xj− ∂uj∂xi
)so ∇u = 1
2r. Note this also means that in the case of a pure rotation, ∂ui
∂xj= −∂uj
∂xi. Now, in
general, we can write
(∇u)ij =1
2
(∂ui∂xj
+∂uj∂xi
)+
1
2
(∂ui∂xj− ∂uj∂xi
)so we have
∇u = e +1
2r =
1
2(s+ r) +
(∇u)11 0 00 (∇u)22 00 0 (∇u)33
Therefore we can think of ∇u as the sum of a shear rate, a rotation rate, and a volume dilationrate. If we choose to keep e intact, then ∇u is composed of a component that keeps track of rateof orthogonal rotation, and another that keeps track of the total deformation rate. These rates areper unit distance, not per unit time.
3.4.2 Pressure and the Deviatoric Stress Tensor
At this point you may be wondering, in all our talk of fluids so far, there has been no discussion ofpressure. We define pressure, as in [6], as a scalar field p = p(x, t) such that if S is any surface in afluid with unit normal n, then the force of stress per unit area across S at x at time t is simply−pn. This is a special form of our stress vector, where the stress is always exerted in the directionof n. So we have t(x, t,n) = −pn, then in particular when n = ij, t = −pij. Then in thedefinition of τ we have
τjk = tk(x, t, ij) = −(pij)k = −pδkjPhysically this makes sense, because τjk is the kth component of the force per unit area acting ona a surface with outward normal ij. By definition of p we know that this is 0 if i 6= j, and −p ifi = j. A fluid with a scalar field p such that τjk = pgkj is called an ideal fluid [6].
This was a special case, but now we can define τ in terms of the pressure in general. Recall thatour definition of a fluid was not rigorous. Now that we have defined the stress tensor we can makeit more exact. Recall the off diagonal terms in the stress tensor correspond to stresses acting on asurface element which are perpendicular to the normal of that surface, so tangent to the surface.These are our shear forces as we saw above. A fluid, then, is a continuum of material that deformscontinuously over time whenever shear forces are applied, for as long as they are applied [12]. Thisis in contrast to a solid, which will deform under shear to a certain extent, but then resist thedeformation as it tries to ”spring back” to its original shape. Note that both solids and fluidsrespond to compression forces similarly, but they respond very differently to shear forces.
If a fluid is at rest, then it must not be the subject of any shear forces, because if it were, it woulddeform continuously by definition, and so would not be at rest. Therefore the Cauchy stress tensor
50
Justin Shaw 51
τ must be diagonal. Moreover τ must be diagonal in any basis we choose to represent it in,because the fluid doesn’t care what basis we’re looking at it in, and must be at rest no matter howit is described. Recall from linear algebra that the only matrices which are diagonal in every basisare scalar multiples of the identity, so τ = −pId for some scalar p. Therefore we can redefinepressure to be the scalar p at each point [13].
So for fluid at rest, τ + pId= 0, but in motion this will not be zero. Therefore we can define asymmetric (2, 0) tensor σ by
σij = τij + pδij
We call σ the deviatoric stress tensor. Now for any τ we can write
τij = −pδij + σij
Physically, by definition the deviatoric stress tensor is that portion of the stress tensor that is notcaused by pressure. Pressure is constant in every direction by construction, but the deviatoricstress tensor encodes all the forces which are not related to compression, which for fluids are forceswhich cause deformation. Note that the deviatoric stress tensor may still have nonzero entries onthe diagonal, because the Cauchy stress tensor may not have −p for its values on the diagonal.
Definition 3.4.2.1. A Newtonian fluid is a fluid for which the deviatoric stress tensor σ islinearly related to the strain rate tensor e. That is, there exists a (2, 2) tensor K = K kl
ij (x, t) suchthat
σij = K klij ekl
Note that this is an assumption, justified by experiments using simple fluids, meaning fluids withsimple molecular structure. This assumption leads to very accurate models for important fluidssuch as air, water, and many oils. We are now losing generality, but the importance of air andwater to human existence can hardly be overestimated, so it makes sense to begin our study ofspecial cases with one that models these well.
That said, there are many non-Newtonian fluids including airplane fuel, ketchup, and saladdressing. There are many ways a fluid can be non-Newtonian. If the deviatoric and strain ratetensor are nonlinearly related then by definition the fluid is non-Newtonian. Another way is if afluid has ”memory”, meaning its past state influences its time evolution.
We make an additional assumption now that the fluid is isotropic, meaning that its properties arethe same in each direction. This implies the stress-strain relationship is independent of theorientation of the orthogonal coordinate system, so that K’s components have the same value inevery Cartesian coordinate system. It turns out that that any fourth order isotropic tensor K kl
ij
has the form [10]
K klij = λδji δ
lk + µδki δ
lj + γδliδ
kj
but now since σ is symmetric, K klij is symmetric in i, j as well. Permuting i and j in the above
equation makes δki δlj = +γδliδ
kj , so that γ = µ. Therefore
51
Justin Shaw 52
K klij = λδji δ
lk + µ(δki δ
lj + δliδ
kj )
For scalar fields λ(x, t) and µ(x, t). These two scalar fields are regarded as known because theyare the result of experiment, as measurable properties of the fluid in question. Now
σij = K klij ekl
=[λδji δ
lk + µ(δki δ
lj + δliδ
kj )]ekl
= λδji δlkekl + µ(δki δ
ljekl + δliδ
kj ekl)
= λδji ekk + µ(δki ekj + δliejl)
= λδji∑k
∂uk∂xk
+ µ(δki ekj + δliejl)
= λ(div(u))δji + µ(eij + eji)
but now e is symmetric, so we can write
σij = λ(div(u))δji + 2µeij
and therefore
τij = −pδji + λ(div(u))δji + 2µeij (14)
Now consider the trace of τ divided by the dimension 3, so the average of the diagonal terms,which gives
1
3
∑i
τii = −p+ λ div(u) +2
3µ∑i
eii
but recall again that∑i
eii = div(u), so we have
1
3
∑i
τii = −p+
(λ+
2
3µ
)div(u)
µ is often called the shear viscosity, and is easy to measure experimentally. λ+ 23µ is called the
bulk viscosity and is difficult to measure because div(u) is often very small, and the bulk viscosityis multiplied by it.
3.4.3 The Momentum Equation for a Newtonian Fluid
We will now use (14) to rewrite the momentum equation under our assumption that the fluid inquestion is Newtonian. Recall from (11) that the ith component of the momentum equation is
ρDuiDt
= ρbi + div(τi)
where τi is the ith column of τ . Therefore div(τi) = ∂τki∂xk
where there is a sum over k here. Now we
can use (14) to write, for our Newtonian fluid,
52
Justin Shaw 53
∂τki∂xk
=∂
∂xk
(−pδik + λ(div(u))δik + 2µeki
)= − ∂p
∂xi+
∂
∂xi(λ div(u)) + 2
∂
∂xk(µeki)
Therefore the general momentum equation for a Newtonian fluid is
ρDuiDt
= ρbi −∂p
∂xi+
∂
∂xi(λ div(u)) + 2
∂
∂xk(µeki) (15)
This equation has now eliminated τ altogether, expressing the equation in terms of pressure andthe two viscosities λ and µ. As we mentioned, these quantities are regarded as known becausethey are observable properties of the fluid, determined by experiment. They are functions oftemperature and pressure, but not of the fluid velocity u(x, t). So really, we have introduced onlyone new variable, the pressure. In practice it often happens that the temperature and pressurevariations are sufficiently small that λ and µ can be treated as constant. In this case we can pullout the scalars , and we have
2∂
∂xk(µeki) = 2µ
∂
∂xkeki = µ
∂
∂xk(∂ui∂xk
+∂uk∂xi
) = µ(∂2ui∂x2
k
+∂uk
∂xk∂xi) = µ
∂
∂xidiv(u) + µ∇2ui
So now if we have constant viscosities, we can rewrite 15 as
ρDuiDt
= ρbi −∂p
∂xi(λ+ µ)
∂
∂xidiv(u) + µ∇2ui
and so in vector form we get
ρDu
Dt= ρb−∇p+ (λ+ µ)∇(div(u)) + µ∇2u
where this last expression is because we’re in Cartesian coordinates. Otherwise the laplacian of avector field must be written in terms of the cross product.
3.4.4 The Energy Equation for a Newtonian Fluid
Recall that we have two energy equations:
D
Dt(E + ρe) + (E + ρe) div(u) = div(−q + u · τ)
and
DE
Dt= −E div(u) + u · div(τ)
Subtracting the second from the first gives
D
Dt(ρe) + (ρe) div(u) = div(−q + u · τ)− u · div(τ))
We can rewrite the left hand side:
53
Justin Shaw 54
D
Dt(ρe) + (ρe) div(u) = e
Dρ
Dt+ ρ
De
Dt+ (ρe) div(u)
= e
(Dρ
Dt+ ρ div(u)
)+ ρ
De
Dt
which by the continuity equation means the left hand side is just ρDeDt
. Now since τ is symmetric,
div(u · τ)− u · div(τ) = div(τ · u)− u · div(τ)
= div(τ 1i u
i, τ 2i u
i, τ 3i u
i)− ui div(τi)
=∂
∂xj(τ ji u
i)− ui∂τ
ji
∂xj
=∂ui
∂xjτ ij
=
(eij +
1
2rij
)τ ji
But now τ is symmetric and r is antisymmetric, so their product is zero. For a Newtonian fluid weknow
τij = −pδji + λ(div(u))δji + 2µeij
and so
τijeij = −pδji eij + λ(div(u))δji eij + 2µeijeij
= −p div(u) + λ(div(u))2 + 2µeijeij
because δji eij is the trace of e as a matrix, which we saw above was div(u). So we have
ρDe
Dt= − div(q)− p div(u) + λ(div(u))2 + 2µeijeij
Assumption: Experimental evidence suggests that we can write q = −k∇T where T istemperature, and k(x, t) is the thermal conductivity. With this assumption we get the internalenergy equation:
ρDe
Dt= div(k∇T )− p div(u) + λ(div(u))2 + 2µeijeij (16)
3.4.5 Summary
The continuity equation remains unchanged because it has no τ terms, but we get two newequations under the Newtonian assumption. Using conservation of mass, linear momentum, andangular momentum. The assumption that the fluid is Newtonian and isotropic gives us:
54
Justin Shaw 55
1. The Continuity Equation (9) :Dρ
Dt+ ρ div(u) = 0
2. The Momentum Equation (15) for Newtonian fluids :
ρDuiDt
= ρbi −∂p
∂xi+
∂
∂xi(λ div(u)) + 2
∂
∂xk(µeki)
and if λ and µ can be treated as constant,
ρDu
Dt= ρb−∇p+ (λ+ µ)∇(div(u)) + µ∇2u
3. The Internal Energy Equation (16) for Newtonian fluids:
ρDe
Dt= div(k∇T )− p div(u) + λ(div(u))2 + 2µeijeij
We are down to five equations in seven unknowns: ρ, p, e, T , and u. The scalar fields k, λ, and µare assumed known because they are determined experimentally. We now need somethermodynamics in order to have a well posed system.
If we assume that the State of the fluid is determined by two state variables, meaning there existsa function f such that
f(ρ, p, T ) = 0
where here T is temperature in Kelvin, a scalar field. In this case for some known scalars k, λ, µ,Cp and β which are functions of the thermodynamic variables, the energy equation can berewritten as:
ρCpDT
Dt− βT Dρ
Dt= div(k∇T ) + λ(div(u))2 + 2µeijeij
Now including f , this is six equations in the six unknowns ρ, p, T , and u.
55
Justin Shaw 56
4 The Kamchatnov Paper
We now turn our attention to the paper ”Topological solitons in magnetohydrodynamics” [7]. Thefirst part of this paper, after the introduction, gives a method for associating a vector field to theHopf fibration from S3 to S2. This vector field turns out to satisfy a special case of theNavier-Stokes equations. However, there is a great deal to discuss before we delve into this paperif we wish to understand any of the meaning behind the relatively straight-forward calculations itinvolves.
56
Justin Shaw 57
4.1 Hopf Fibrations
4.1.1 Spheres and Stereographic projection
We begin by reviewing some differential geometry. Recall that as a set
Sn =x ∈ Rn+1 : ‖x‖ = 1
⊂ Rn+1
by definition, and Sn can be given the structure of an n dimensional manifold by stereographicprojection, which will feature heavily throughout our study of [7]. The point (0, . . . , 0,−1) ∈ Sn iscommonly referred to as the ”south pole” of Sn because its on the ”bottom” of the sphereSn ⊂ Rn+1. Similarly (0, . . . , 0, 1) ∈ Sn is called the north pole.
By removing each pole we get an atlas for Sn consisting of just two maps: stereographicprojection from the north pole, and stereographic projection from the south pole. Lets look atthese maps. Stereographic projection from the south pole, calling the south polse s, is defined as
ψs : Sn \ s → Rn
(x1, . . . , xn, xn+1) 7→ 1
1 + xn+1(x1, . . . , xn)
and recall this is a diffeomorphism with inverse
ψ−1s : Rn → Sn \ s
(y1, . . . , yn) 7→(
2y1
1 + ‖y‖2,
2y2
1 + ‖y‖2, . . . ,
2yn
1 + ‖y‖2,1− ‖y‖2
1 + ‖y‖2
)Therefore we can view Sn as Sn = Rn ∪ s where s represents a single point at infinity.Therefore, topologically speaking Sn is the one point compactification of Rn.
Now lets call the north pole n. We can define
ψn : Sn \ n → Rn
(x1, . . . , xn, xn+1) 7→ 1
1− xn+1(x1, . . . , xn)
and recall this is a diffeomorphism with inverse
ψ−1n : Rn → Sn \ n
(y1, . . . , yn) 7→(
2y1
1 + ‖y‖2,
2y2
1 + ‖y‖2, . . . ,
2yn
1 + ‖y‖2,‖y‖2 − 1
1 + ‖y‖2
)Now to show this forms an atlas we need only show that ψs ψ−1
n : Rn \ 0 → Rn \ 0 is adiffeomorphism. We have Rn \ 0 here because
ψs(Sn \ n ∪ Sn \ s) = ψs(S
n \ n, s) = Rn \ 0because ψs(n) = 0 ∈ Rn. Similarly the domain is also Rn \ 0. A straightforward calculationshows that
57
Justin Shaw 58
ψs ψ−1n (y) =
1
‖y‖2(y1, . . . , yn)
which is clearly smooth and is its own inverse, because
y 7→ y
‖y‖27→ 1
‖ y‖y‖2‖2
y
‖y‖2=‖y‖4
‖y‖2
y
‖y‖2= y
Therefore Sn is an n dimensional manifold through stereographic projection.
4.1.2 Which spheres are groups?
It should be noted that the next few sections are modelled on a talk given by Spiro Karigiannis,my Master’s supervisor, at CUMC in 2010 [19]Recall [16]:
An n - dimensional normed real division algebra A satisfies:
1. A is an n dimensional real vector space, with a norm | · |
2. A has a ring structure with identity, for which any non-zero element is invertible
3. the norm is multiplicative, so for all a, b ∈ A
|ab| = |a||b|
Sn =x ∈ Rn+1 : |x|2 = 1
⊂ Rn+1
Therefore:
S0 =x ∈ R : |x|2 = 1
= 1,−1 ∼= Z/2Z
S1 =x ∈ R2 : |x|2 = 1
⊂ R2 ∼= C
Question: These are both groups. Are all spheres groups?Answer: No. The reason these two are groups is because the norms are multiplicative, and theseare the sets of unit norm elements, which is a group since the identity is included.
Given a normed division algebra A, if A is associative, we can form
SA =x ∈ A : |x|2 = 1
m
SR =x ∈ R : |x|2 = 1
= S0
mSC =
x ∈ C : |x|2 = 1
= S1
mSH =
x ∈ H : |x|2 = 1
= S3
all of which are groups. What about if Sn−1 is a group?
58
Justin Shaw 59
Lemma: 4.1.2.1. Suppose Sn−1 is a group, then we can induce an associative normed realdivision algebra structure on Rn.
Proof.Sn−1 =
x ∈ Rn : |x|2 = 1
⊂ Rn
take a, b ∈ Sn−1, then by the group structure ab ∈ Sn−1. Therefore
|a||b| = 1 · 1 = 1 = |ab|
However every point x ∈ Rn is λa for a ∈ Sn−1 and λ ∈ R. Therefore, given x, y ∈ Rn we candefine, assuming y = γb for b ∈ Sn−1
xy = (λγ) ab
Where the ab is the group product of a and b in Sn−1. Then for the norm conditions:
‖x‖‖y‖ = |λ|‖a‖|γ|‖b‖ = |λγ|‖ab‖ = ‖xy‖
This shows both that we have a norm, and that it is multiplicative. We take the identity for Rn asthe identity e that must exist because Sn−1 is assumed to be a group. Now take x ∈ Rn \ 0, wemust show it has an inverse. Assume x = λa, then λ 6= 0 so it has an inverse since λ ∈ R \ 0 anda has an inverse, so let y = λ−1a−1, then
xy = yx = λλ−1aa−1 = e
Notice here that the scalars were not passing through any vectors.
A theorem of Hurwitz [16], says that up to isomorphism the only real normed division algebras areR,C,H, and O. This lemma says that if S2 were a group, then R3 would have a real divisionalgebra structure. By Hurwitz then, since R3 does not have a real division algebra structure, S2 isnot a group, and indeed any Sk for k 6= 0, 1, 3 is not a group.
Now we have
A is an associative real normed division algebra of dimension n+1
m
SA = Sn is a group
Spiro showed last week that there are only three associative real normed division algebras, namelyR,C and H. We also have that the only reason O fail to be a group is because of the failure to beassociative.
Therefore: The only spheres that are groups are S0, S1, S3 and almost S7, corresponding tonormed real division algebras R,C,H, and O on R,R2,R4, and R8
59
Justin Shaw 60
4.1.3 Projective Spaces and Hopf Fibrations
Def: Let A be an associative real normed division algebra. We define A projective space, AP n tobe the set of all A-lines through the origin in An+1, a compact, smooth n dimensional manifold.
An A-line through the origin is of the form kp for k ∈ A, and a fixed p ∈ An+1 as k varies.
Formally we define ∼ to be an equivalence relation on An+1 \ 0 where
x ∼ y ⇔ x = λy, for some λ ∈ A∗
Then AP n = (An+1 \ 0) / ∼
However there’s a nicer way: Each A-line through the origin intersects a sphere Sk where kdepends both on which A and which n we’re in. For example a C line through the origin inCn+1 ∼= R2n+2 intersects S2n+1. Therefore we can restrict ∼ to Sk so that as an equivalencerelation on Sk
x ∼ y ⇔ x = λy, λ ∈ A, |λ| = 1
Now if π : Sk → AP n is the map that takes each point to it’s equivalence class under ∼, call itπ(x) = [x], then what is the fiber at each point [x] in AP n? Exactly [x] as a set in Sk:
π−1([x]) =y ∈ Sk : π(y) = [x]
=y ∈ Sk : [y] = [x]
=y ∈ Sk : y ∼ x
= [x]
=y ∈ Sk : ∃λ ∈ A, |λ| = 1 : y = λx
= λx ∈ H : |λ| = 1∼= λ ∈ A : |λ| = 1= SA
Therefore the fibres of each of these maps are themselves groups , copies of SAassociated to each x. Also since quotient maps are surjective, we’ve shown that each Sk can bethought of as a union of fibres, meaning we’ve shown that Sk can be thought of as AP n with acopy of SA attached at every point. Or equivalently a family of SAs parameterized by AP n.
We call Sk a fibration over AP n with fiber SA These ARE fiber bundles , trivializations are
π−1(U)→ U × SA
z 7→ ([z], z)
It is natural to try to define OP n in an analogous way, but the non-associativity is an obstruction.It is possible to define OP 1 and OP 2, but OP n is not defined for n ≥ 3. For n = 1 there’s afibration π : S15 → OP 1 with fibre S7 = SO
60
Justin Shaw 61
Its unsatisfying to have the k in Sk so we can write explicitly the maps we call the HopfFibrations:
S0 Sn
RP n
π
S1 S2n+1
CP n
π
S3 S4n+3
HP n
π
S1 S15
OP 1
π
We make use of some exceptional isomorphisms: It can be shown using stereographic projectionthat
RP 1 ∼= S1 CP 1 ∼= S2 HP 1 ∼= S4 OP 1 ∼= S8
So for n = 1 we have
S0 S1
S1
π
S1 S3
S2
π
S3 S7
S4
π
S1 S15
OP 1
π
61
Justin Shaw 62
4.1.4 The Quaternions and Rotations
Sidenote: You may be wondering if the cross product gives the structure of a normed divisionalgebra to R3. The answer is no, because for example |a× b| 6= |a||b| in general.
This shows the cross product doesn’t work, but we showed above that NO productwould work.
Construction: Denote e1, e2, e3, e4 = 1, i, j, k, and define 1q = q1 = q, ∀q ∈ H with thefollowing diagram for rules of multiplication.
Conjugation is defined by a+ bi+ cj + dk = a− bi− cj − dk and it turns out that pq = qp whenyou do the calculation, which we leave as an exercise for the reader.
Properties:
1. H is a normed division algebra
2. H is not commutative, but is associative
3. therefore SH = S3 is a group, known as Sp(1) or SU(2).
We can write each q ∈ H as q = (t,v) for t ∈ R and v ∈ R3 Then the sum of two quaternions isstill carried out componentwise, but the product formula is given by:
(t1,v1)(t2,v2) = (t1t2 − v1 · v2, t1v2 + t2v1 + v1 × v2)
This formula is not a definition, but a direct consequence of the construction above. Here, sincereal numbers commute with vectors, we have t2v1 in the second component, but actually if we donot allow ourselves to commute, then the calculation yields v1t2. This will be important laterwhen we are defining our differential operators on the quaternions.
Note this formula also means the multiplication of two purely imaginary quaternions is given by
(0,v1)(0,v2) = (−v1 · v2,v1 × v2)
Notice also that we can identify scalar multiplication by reals as
k(t,v) = (k, 0)(t,v) = (kt, kv) = (t,v)(k, 0) = (t,v)k
Now consider the following product. Take r = (t,v) ∈ H \ 0, and q = (0,w) ∈ Im(H), thensince |r|2 = rr
r−1 =r
|r|2
Therefore
62
Justin Shaw 63
rqr−1 =1
|r|2(t,v)(0,w)(t,−v)
=1
|r|2(−v ·w, tw + v ×w)(t,−v)
=1
|r|2(−tv ·w + v · (tw + v ×w), (v ·w)v + t(tw + v ×w)− (tw + v ×w)× v)
=1
|r|2(v · (v ×w), (v ·w)v + t(tw + v ×w) + v × (tw + v ×w))
=1
|r|2(v · (v ×w), (v ·w)v + t2w + 2tv ×w + v × (v ×w))
Now since v · (v ×w) = w · (v × v) = 0, we see that rqr−1 ∈ Im(H), and recall that the vectortriple product is
a× (b× c) = (a · c)b− (a · b)c
so
v × (v ×w) = (v ·w)v − (v · v)w
and so finally
rqr−1 =1
|r|2(0, 2(v ·w)v + (t2 − v · v)w + 2tv ×w)
properties of Ωr
1. We have rqr−1 ∈ Im(H).
2. Since we are identifying R3 ∼= Im(H), we can use fact 1 to define the following map as inLyons-elem: Let r ∈ H, and let
Ωr : R3 → R3
(x, y, z) 7→ xi+ yj + zk 7→ r(xi+ yj + zk)r−1 = x′i+ y′j + z′k 7→ (x′, y′, z′)
by the above construction, we have shown explicitly that if r = (t,v) is a unit lengthquaternion, then
Ωr(w) = 2(v ·w)v + (t2 − v · v)w + 2tv ×w
3. If we replace r with kr for k ∈ R \ 0 we see immediately that we get the same answer,meaning we get the same map for the whole line through the origin in R4. For this reasonwe may assume that |r| = 1 because every r′ ∈ H is a real multiple of a unit lengthquaternion r. This also means that on the line kr in H ∼= R4 we have two representatives inS3 that give the same map, antipodal points where the line intersects S3.
63
Justin Shaw 64
4.Ωr Ωs = Ωrs
Ωr Ωs(p) = π(rspsr) = π((rs)p(rs)) = Ωrs(p)
5. If r ∈ H \ 0, then by definition of a real division algebra, r has an inverse. We claim thatΩr is also invertible, and that (Ωr)
−1 = Ωr−1 . Well, by fact 3, we may assume |r| = 1, inwhich case r−1 = r. Now
Ωr (Ωr−1(x)) = π(r(r−1(0,x)r
)r−1)
= x
where π projects onto the second factor. The other direction is almost the same.
6. By the ring structure of H we see that Ωr is linear over R, where here r = (t,v):
Ωr(ax + by) = 2(v · (ax + by))v + (t2 − v · v)(ax + by) + 2tv × (ax + by)
= a2(v · x)v + 2b(v · y)v + a(t2 − v · v)x + b(t2 − v · v)y
+ a2tv × x + b2v × y
= aΩr(x) + bΩr(y)
Moreover we can calculate Ωr’s matrix representation by calculating it’s value on the basisi, j, k:
7.
|Ωr(p)| = |rpr−1| = |r||p|| r|r|2| = |p|
since |r| = |r|.
8. Each Ωr is an isometry. This can be seen by the fact that Ωr is linear, so|Ωr(p)− Ωr(q)| = |Ωr(p− q)| = |p− q|, and this implies that 〈Ωr(x),Ωr(y)〉 = 〈x, y〉,assuming |z|2 = 〈z, z〉. Therefore, thinking of Ωr as a matrix under our identification ofe1, e2, e3 = i, j, k we have
〈x, y〉 = 〈Ωr(x),Ωr(y)〉 = 〈x,ΩtrΩr(y)〉
and this is for every x, y, so ΩtrΩr = I3×3, and Ωr is orthogonal, and so has determinant ±1.
What is this map Ωr for? Note that every rotation in R3 can be encoded by a scalarrepresenting the angle of rotation, and a vector indicating the axis of rotation. (In fact you couldjust use three numbers because you could use the length of the vector to indicate the angle, butwe’re not going to.)
Question: What would you expect the eigenstructure of a rotation to be?It turns out [18] for r = (t,v), Ωr has eigenvalue 1, with eigenvector v.
64
Justin Shaw 65
But we showed Ωr is an isometry with invariant subspace spanned by the imaginary part of r.This means that its a rotation, but what is the angle of rotation θ? Take any vector wperpendicular to Im(r), then
cos(θ) =w · Ωr(w)
|w|2= 2Re(r)2 − 1⇒ |Re(r)| = |cos(θ
2)|
Therefore the real part keeps track of the angle, and the imaginary part keeps trackof the axis of rotation. It seems clear that every rotation can be written this way.However this does not constitute a proof. For an actual proof that every rotation is of this form,see [18]
Recall that SO(3) = A ∈ SL(3,R) : AtA = I. Therefore with what’s given above, we can seethat the set
Ωr : r ∈ SH, det(Ωr) = 1
is exactly SO(3) if we give it the group operation of composition.
Alternatively: By everything above, we have seen that to every r ∈ H we can identify a map Ωr,and this map has an inverse. Moreover quaternion multiplication corresponds to mapcomposition, so
S3 → O(3)
r 7→ Ωr
is a continuous (see the Reeder notes on quaternions for explicit construction of the matrix, thisshows that the map is continuous.) group homomorphism, and our remark above that everyrotation is Ωr for some r corresponds to this map being surjective. Since S3 is connected, itsimage under this map is connected. However O(3) is not connected, and this maps ±1 7→ I, sincecos(0) = 1, so the image of this map is in SO(3) ⊂ O(3). This shows that all of our maps Ωr havedeterminant 1. Moreover we claim that the kernel of this map is just ±1.
Suppose Ωr = I, r = (t,v) Let w ∈ R3. Since Ωr(w) = w for every w, if we suppose v 6= 0,without loss of generality we can take w nonzero and not colinear with v. Now by definition of Ωr,
(0,w) = (t,v)(0,w)(t,−v)
(0,w)(t,v) = (t,v)(0,w)
(−v ·w, tw + w × v) = (−v ·w, tw + v ×w)
Now v 6= 0, and w × v = v ×w, which means w × v = 0 which is a contradiction since we chosew not perpendicular or colinear with v, and both vectors are nonzero.
Therefore v = 0, so r = ±1. Therefore, since we remarked above that this map is surjective, bythe first isomorphism theorem
S3/Z2∼= SO(3)
65
Justin Shaw 66
But we saw above that this is exactly RP 3 in the above Hopf fibrations
Now
Ωr : r ∈ SH ∼= SO(3)
because we have shown that every Ωr has positive determinant.
Theorem: 4.1.4.1. SO(3) ∼= RP 3
Proof. omitted, but clear.
4.1.5 The Octonians
Arthur Cayley constructed the Octonians. The method we used last week is called the Cayley -Dickson construction, in which, given an element p ∈ O ∼= R8, we associate a pair of quaternionsby noting that p = (s, t) for s, t ∈ R4 ∼= H, so taking s, t ∈ H in the standard way we did above inthe previous section, we can think of p as a pair of quaternions. Then by the product we derivedlast week, we have, for p = (s, t), q = (u, v) ∈ O,
pq = (s, t)(u, v) = (st− vu, sv + ut)
where everything inside the brackets on the right is done with the quaternion product andconjugation. Basically what we’ve done here is map
e1, . . . , e8 7→ (1, 0), (i, 0), (j, 0), (k, 0), (0, 1), (0, i), (0, j), (0, k)
There is another crazy diagram, but its crazy.
4.1.6 The Classical Hopf Fibration
Now consider S3 ∈ R4. As above we can write
S3 =x ∈ R4 : ‖x‖ = 1
⊂ R4
or we can identify R4 ∼= C2 by (a, b, c, d) 7→ (a+ bi, c+di) in the usual way. The norm on C2 here is
‖(a+ bi, c+ di)‖ =
√(a+ bi)(a+ bi) + (c+ di)(c+ di) =
√a2 + b2 + c2 + d2
So that we have
‖(a, b, c, d)‖ =√a2 + b2 + c2 + d2 = ‖(a+ bi, c+ di)‖
so that under our identification we may rewrite S3 as
S3 =
(z1, z2) ∈ C2 : z1z1 + z2z2 = 1
Consider, as in [9], the following equivalence relation on S3 ⊂ C2:
(z1, z2) ∼ (w1, w2)⇔ (z1, z2) = λ(w1, w2) for some λ ∈ α ∈ C : |α| = 1
66
Justin Shaw 67
Now define CP 1 as the set of equivalence classes of C2 \ 0 under ∼, which is a smooth manifoldof complex dimension 2. Let [z1 : z2] be the equivalence class of (z1, z2) under ∼, and letπ : S3 → CP 1 be defined by π((z1, z2)) = [z1 : z2]. π is a quotient map, and so surjective.
Recall that CP 1 ∼= S2. To see this, recall that one the standard atlas on CP 1 is [15]:
[z1 : z2] 7→ z1
z2
, [z1 : z2] 7→ z2
z1
where the first map is on the open set where z2 6= 0 and the second map is on the open set wherez1 6= 0. Consider the image of the first map, we have:
z1
z2
=z11 + z12i
z21 + z22i=z11 + z12i
z21 + z22i
z21 − z22i
z21 − z22i=
1
|z2|2[(z11z21 + z12z22) + i(z12z21 − z11z22)]
Which we can identify as an element of R2, so
z1
z2
7→ 1
|z2|2[z11z21 + z12z22, z12z21 − z11z22]
Now that we’re back in R2, we can use stereographic projection to get to S2, so understeregographic projection from the north pole
1
|z2|2(z11z21 + z12z22, z12z21 − z11z22) 7→
2(z11z21 + z12z22)
|z2|2(1 +(|z1||z2|
)2
),2(z12z21 − z11z22)
|z2|2(1 +(|z1||z2|
)2
),
|z1|2|z2|2 − 1
1 + |z1|2|z2|2
7→(2(z11z21 + z12z22), 2(z12z21 − z11z22), |z1|2 − |z2|2
)because we started on S3, so |z1|2 + |z2|2 = 1. So now starting from S3 and tracing our progress toS2, we have:
(a, b, c, d) 7→ (a+ bi, c+ di) 7→ [(a+ bi) : (c+ di)] 7→ a+ bi
c+ di
7→ 1
c2 + d2(ac+ bd, bc− ad) 7→ (2(ac+ bd), 2(bc− ad), a2 + b2 − c2 − d2)
The keen observer may object that we had a choice here, namely that we used the first of the twocharts for CP 1, which requires z2 6= 0. Suppose z2 = 0, then z1 6= 0 by definition of CP 1. Then
(z1, z2) 7→ z2
z1
=1
|z1|2[(z11z21 + z12z22) + i(z11z22 − z21z12)]
7→(2(z11z21 + z12z22), 2(z11z22 − z12z21), |z1|2 − |z2|2
)which is the same map with the second coordinate multiplied by negative one, where here we usedstereographic projection from the south pole.
The point, is that if we define the Hopf fibration
f : S3 → S2
(a, b, c, d) 7→ (2(ac+ bd), 2(bc− ad), a2 + b2 − c2 − d2)
67
Justin Shaw 68
This is perfectly well defined for every point on S3. Moreover for z2 = c+ di = 0,f(z1, z2) = (0, 0, 1). This means that f is surjective because π : S3 → CP 1 is surjective and thecharts for CP 1 are diffeomorphisms from CP 1 \ p onto R2 after identification, where p is either[z1 : 0] or [0 : z2]. If p = [z1 : 0] then p corresponds to a point at infinity, using stereographicprojection from the north pole means we map back to S2 \ n from R2, which means n is thepoint at infinity on S2. Therefore composing the chart for CP 1 with (the inverse of) stereographicprojection from the north pole on S2 we have a bijection CP 1 \ [z1 : 0] ↔ S2 \ n which weextend to a bijection CP 1 ↔ S2 by mapping [z1 : 0] 7→ n. f is clearly smooth since all of itscomponent functions are smooth.
68
Justin Shaw 69
4.2 Kamchatnov’s Algorithm
Kamchatnov’s paper does not deal directly with the Navier-Stokes equations. Instead, thesolution he induces to a special case of the Navier-Stokes equations arises in the form of a vectorpotential for a magnetic field, by which we mean a vector field B ∈ Γ(TR3) such that
H = curl(B)
is the magnetic field with the properties he desires. We will not discuss magnetic fields, and ratherthan H, we will soon show that it is this vector potential B that interests us. Lets followKamchatnov’s Algorithm for constructing this vector potential.
Lemma 4.2.0.1. Let A ∈ Ω1(R3), ψ : S3 \ s → R3 be stereographic projection from the south
pole, and let A = ψ∗A. Assume that A] is tangent to S3. Then the components of A =4∑i=1
Aidui
and A =3∑j=1
Ajdxj satisfy:
Ai =1
2(1 + |x|2)Ai − xi
3∑j=1
xjAj for i = 1, 2, 3
A4 = −3∑j=1
xjAj
Ai = (1 + u4)Ai − uiA4 for i = 1, 2, 3
where here u1, u2, u3, u4 are the the standard coordinates on R4 and x1, x2, x3 are the standardcoordinates on R3.
Proof. From the previous section we know ψ is a diffeomorphism we can pull back A by ψ:
ψ∗A = ψ∗(3∑j=1
Ajdxj) =3∑j=1
(Aj ψ) d(xj ψ)
Let L = (u1, u2, u3,−1) ∈ R4 : ui ∈ R which is clearly closed by taking sequences. As a mapfrom R4 → R3, we actually have ψ : R4 \ L → R3 given by
ψ(u1, u2, u3, u4) =1
1 + u4(u1, u2, u3)
We have that R4 \ L =: M is an open subset of R4 since L is closed, so u1, u2, u3, u4 are thenatural coordinates to take on M. Notice L ∩ S3 = S3 \ s. So we know ψ∗A ∈ Ω1(M), so we canwrite
ψ∗A =3∑j=1
Ajduj + A4du
4
for Bi ∈ C∞(M). Now we wish to restrict this form to S3, but restriction is by definition pullbackby the inclusion, so we need the inclusion explicitly. Well because
69
Justin Shaw 70
S3 =u ∈ R4 : ‖u‖ = 1
⊂ R4
we can take u1, u2, u3 as independent coordinates on S3, and writeu4 = ±
√1− (u1)2 − (u2)2 − (u3)2. Then the inclusion map has the form
ι : S3 → R4
(u1, u2, u3) 7→ (u1, u2, u3,±√
1− (u1)2 − (u2)2 − (u3)2)
depending which square root we pick. Now
ι∗(ψ∗A) =3∑j=1
Ajduj + A4d(u4 ι)
because ι does nothing to the first three coordinates by construction. For the fourth we have
d(u4 ι) = ±d(√
1− (u1)2 − (u2)2 − (u3)2)
= ±3∑i=1
∂
∂ui(√
1− (u1)2 − (u2)2 − (u3)2)dui
= ±3∑i=1
1
2
(1− (u1)2 − (u2)2 − (u3)2
)− 12 (−2ui)dui
= ∓3∑i=1
ui
u4dui
Therefore
ι∗(ψ∗A) =3∑i=1
Aidui − A4
3∑i=1
ui
u4dui
=3∑i=1
(Ai − A4
ui
u4
)dui
on the other hand, we can calculate ι∗(ψ∗A) directly, taking the first three ui as the independentcoordinates:
ι∗(ψ∗A) = ι∗(ψ∗3∑i=1
Aidxi)
=3∑i=1
(Ai ψ ι) d(xi ψ ι
)We suppress the ψ ι for the component functions Ai in what follows for clarity of notation. Now
70
Justin Shaw 71
d(xi ψ ι
)=
3∑j=1
∂
∂uj
(ui
1 + u4
)duj
=3∑j=1
(δji
1 + u4+
uiuj
u4(1 + u4)2
)duj
where here u4 is not a coordinate, but a function of the first three ui. Putting it all together, wehave
ι∗(ψ∗A) = ι∗(ψ∗3∑i=1
Aidxi)
=3∑i=1
Ai
3∑j=1
(δji
1 + u4+
uiuj
u4(1 + u4)2
)duj
=3∑j=1
(Aj
1 + u4+
3∑i=1
Aiuiuj
u4(1 + u4)2
)duj
And now we equate our two versions of ι∗(ψ∗A) which of course must be equal:
3∑j=1
(Aj − A4
uj
u4
)duj = ι∗(ψ∗A) =
3∑j=1
(Aj
1 + u4+
3∑i=1
Aiuiuj
u4(1 + u4)2
)duj
Equating coefficients gives us:
Aj − A4uj
u4=
Aj1 + u4
+uj
u4
3∑i=1
Aiui
(1 + u4)2, for j = 1, 2, 3
We still haven’t used tangency of A] to S3. Notice the normal of S3 at a point (u1, u2, u3, u4) ∈ S3
is exactly the vector (u1, u2, u3, u4) ∈ R4. Therefore we can check that A is tangent to S3 simplyby checking that
4∑i=1
(Aiu
i)
= 0
at every point on S3. The keen observer will now notice that we have 4 equations, and so we canrewrite the Ai in terms of the Aj and vice versa. Multiply both sides of our equation by uj and
sum over j = 1, 2, 3, then we use the fact that4∑j=1
(uj)2 = 1 and the tangency condition:
71
Justin Shaw 72
Ajuj − A4(uj)2
u4=
Ajuj
1 + u4+
(uj)2
u4
3∑i=1
Aiui
(1 + u4)2
3∑j=1
Ajuj − A41− (u4)2
u4=
3∑j=1
Ajuj
1 + u4+
1− (u4)2
u4
3∑i=1
Aiui
(1 + u4)2
4∑j=1
Ajuj − A4u4 − A41− (u4)2
u4=
3∑j=1
Ajuj
1 + u4+
1− (u4)2
u4
3∑i=1
Aiui
(1 + u4)2
Now the first term on the left is 0 by tangency. Multiply both sides by u4 to get
−A4(u4)2 − A4(1− (u4)2) =3∑j=1
Ajuj
(u4
1 + u4+
1− (u4)2
(1 + u4)2
)
A4 = −3∑j=1
Ajuj
(u4
1 + u4+
1− (u4)2
(1 + u4)2
)
A4 = −3∑j=1
Ajuj
(u4 + (u4)2
(1 + u4)2+
1− (u4)2
(1 + u4)2
)
A4 = −
3∑j=1
Ajuj
(1 + u4)
but now xi ψ ι = ui
1+u4, so suppressing the ψ ι for both Aj and xj we have
A4 = −3∑j=1
Ajxj
We still need two equations. Consider again our coordinate equations
Aj − A4uj
u4=
Aj1 + u4
+uj
u4
3∑i=1
Aiui
(1 + u4)2
We can now use our equation for A4, and the fact that a short calculation shows that1
1+u4= 1
2(1 + |x|2) and some simplification to get
72
Justin Shaw 73
Aj +3∑i=1
Aixiu
j
u4=
1
2
(1 + |x|2
)Aj +
3∑i=1
Aiui
(uj
u4
1
(1 + u4)2
)
Aj +3∑i=1
Aixiu
j
u4=
1
2
(1 + |x|2
)Aj +
3∑i=1
Aixi
(uj
u4
1
(1 + u4)
)
Aj =1
2
(1 + |x|2
)Aj +
3∑i=1
Aixiu
j
u4
(1
(1 + u4)− 1
)
Aj =1
2
(1 + |x|2
)Aj +
3∑i=1
Aixiu
j
u4
(−u4
1 + u4
)
Aj =1
2
(1 + |x|2
)Aj − xj
3∑i=1
Aixi
Which is another equation we wanted. For the third equation, we again use that A4 = −3∑j=1
Ajxj,
this time to rewrite the right hand side of our coordinate equation:
Aj − A4uj
u4=
Aj1 + u4
+uj
u4
3∑i=1
Aiui
(1 + u4)2
Aj − A4uj
u4=
Aj1 + u4
− uj
u4
3∑i=1
Aixi
(1 + u4)
Aj(1 + u4)− A4uj
u4(1 + u4) = Aj − A4
uj
u4
Aj = Aj(1 + u4) + A4
(uj
u4− uj
u4(1 + u4)
)Aj = Aj(1 + u4)− ujA4
Which is the last equation we wanted.
5 Translating to 7 dimensions
5.1 Div, Curl, Grad, and Laplacian Revisited
5.1.1 Forms on Quaternions
In this section we use what was developed in section 2.3 to define analogous operators on forms,and to rewrite the exterior derivative in terms of those operators for each of the sets of 0, 1, 2 and3 forms.
For time dependent forms on R3, what we really mean is forms on the product manifold R× R3
where the first copy of R is a global time coordinate. Note this distinction is important, because if
73
Justin Shaw 74
we passed to a general manifold M , by time dependent vector field on M , we mean a vector fieldon R×M , so that the first coordinate is global, while the coordinates on M are local, and we takethe subset of forms or vector fields which take a zero coefficient on terms containing dt or ∂
∂t.
Denote time dependent vector fields on M by Γt(TM) and time dependent forms on M by Ωkt (M)
In our case, sticking with R× R3 = R4 ∼= H, we can think of every time dependent form as a formon the quaternions H. Therefore
Ω0(H) 3 f = f(t, x, y, z)
Ω1(H) 3 α = f(t, x, y, z)dt+ αi(t, x, y, z)dxi
Ω2(H) 3 α = αi(t, x, y, z)dt ∧ dxi + βij(t, x, y, z)dxi ∧ dxj
Ω3(H) 3 α = f(t, x, y, z)dx ∧ dy ∧ dz + βij(t, x, y, z)dt ∧ dxi ∧ dxj
Ω4(H) 3 α = f(t, x, y, z)dt ∧ dx ∧ dy ∧ dz
It is clear that all of these addition signs correspond to L2 orthogonal decompositions of the spacebecause for example in the sum in Ω1(H), ∗dxi contains a dt, so wedging with dt in the innerproduct will give zero, and similarly for the other two.
We can also immediately pick out some isomorphisms.
1. Clearly through ∗ we have
Ω4(H) ∼= Ω0(H) = C∞(H)
2. In Ω1(H), since there’s no sum in f(t, x, y, z)dt, all 1-forms of this type are isomorphic toΩ0(H) as well. Clearly the set of all forms like αi(t, x, y, z)dxi is just Ω1
t (R3), giving us
Ω1(H) ∼= C∞(H)⊕⊥
Ω1t (R3)
3. In Ω2(H), we claim that
Ω1t (R)→ Ω2(H)
α 7→ dt ∧ α
is a linear injective map. Linear is clear because ∧ is bilinear. For injectivity, supposedt ∧ α = dt ∧ β, then dt ∧ (α− β) = 0, but neither α or β has a factor which includes dt, soα = β.
Now the set of all forms of the type αi(t, x, y, z)dt ∧ dxi is clearly the image of this map, andsince injective linear maps are isomorphic to their images, the set of all forms of this type isisomorphic to Ω1
t (R3).
Through ∗ for R3 we see that the set of forms of the type βij(t, x, y, z)dxi ∧ dxj is isomorphicto Ω1
t (R3). Putting it all together we have
74
Justin Shaw 75
Ω2(H) ∼= Ω1t (R3)⊕
⊥Ω1t (R3)
4. In Ω3(H) clearly the set of forms of the type f(t, x, y, z)dx ∧ dy ∧ dz is isomorphic to Ω0(H).
For forms of the type βij(t, x, y, z)dt ∧ dxi ∧ dxj, using ∗ for R4 gives us Ω1t (R3). Therfore
Ω3(H) ∼= C∞(H)⊕⊥
Ω1t (R3)
Now to summarize:
Ω0(H) ∼= Ω4(H) = C∞(H)
Ω1(H) ∼= C∞(H)⊕⊥
Ω1t (R3)
Ω2(H) ∼= Ω1t (R3)⊕
⊥Ω1t (R3)
Ω3(H) ∼= C∞(H)⊕⊥
Ω1t (R3)
5.1.2 The Exterior Derivative on Quaternions
Now we can write the exterior derivative
d : Ωk(H)→ Ωk+1(H)
in terms of div, grad, curl on time dependent forms of R3, which we’ve repackaged as H. Note thatall the coefficient functions below are functions of t, x, y, z, but we’ve suppressed this so we can fitit all in one pdf.
1. Let f ∈ Ω0(H), then
df =∂f
∂tdt+
∂f
∂xidxi
2. Let α = fdt+ αidxi ∈ Ω1(H), then
75
Justin Shaw 76
dα = d(fdt+ αidxi)
= df ∧ dt+ dαi ∧ dxi
=
(∂f
∂tdt+
∂f
∂xidxi)∧ dt+
(∂αi∂t
dt+∂αi∂xj
dxj)∧ dxi
=∂f
∂xidxi ∧ dt+
∑i
∂αi∂t
dt ∧ dxi +∑i
∂αi∂xj
dxj ∧ dxi
= grad f ∧ dt+ dt ∧(∂
∂t=(α)
)+∑i
∂αi∂xj
dxj ∧ dxi
=
(grad f − ∂
∂t=(α)
)∧ dt+
∑i
∂αi∂xj
dxj ∧ dxi
=
(grad f − ∂
∂t=(α)
)∧ dt+ ∗ curl=(α)
Because for the second term, we mentioned above that this sum is ∗ curl=(α)
3. Let α = αidt ∧ dxi + βijdxi ∧ dxj ∈ Ω2(H), then we can rewrite this as α′ = αidx
i ∈ Ω1t (R3),
β = βijdxi ∧ dxj ∈ Ω2
t (R3), so α = dt ∧ α′ + β
dα = d(αidt ∧ dxi + βijdxi ∧ dxj)
= (dαi) ∧ dt ∧ dxi + (dβij) ∧ dxi ∧ dxj
=
(∂αi∂t
dt+∂αi∂xk
dxk)∧ dt ∧ dxi +
(∂βij∂t
dt+∂βij∂xk
dxk)∧ dxi ∧ dxj
=∂αi∂xk
dxk ∧ dt ∧ dxi +∂βij∂t
dt ∧ dxi ∧ dxj +∂βij∂xk
dxi ∧ dxj ∧ dxk
= dt ∧(∂αi∂xk
dxi ∧ dxk)
+ dt ∧(∂βij∂t
dxi ∧ dxj)
+∑σ∈S3
sgn(σ)∂βσ(1)σ(2)
∂xσ(3)dx1 ∧ dx2 ∧ dx3
= dt ∧(∗ curl(α′) +
∂
∂tβ
)+
(∑σ∈S3
sgn(σ)∂βσ(1)σ(2)
∂xσ(3)
)dx1 ∧ dx2 ∧ dx3
But now
∑σ∈S3
sgn(σ)∂βσ(1)σ(2)
∂xσ(3)= div(β12 − β21, β23 − β32, β31 − β13)dx1 ∧ dx2 ∧ dx3
Let β′ = (β12 − β21, β23 − β32, β31 − β13), then we can write
dα = dt ∧(∗ curl(α′) +
∂
∂tβ
)+ div(β′)dx1 ∧ dx2 ∧ dx3
76
Justin Shaw 77
4. Let α = fdx ∧ dy ∧ dz + βijdt ∧ dxi ∧ dxj ∈ Ω3(H) then
dα = d(fdx ∧ dy ∧ dz + βijdt ∧ dxi ∧ dxj
)= d (fdx ∧ dy ∧ dz) + d
(βijdt ∧ dxi ∧ dxj
)= df ∧ dx ∧ dy ∧ dz + (dβij) ∧ dt ∧ dxi ∧ dxj
=
(∂f
∂tdt+
∂f
∂xidxi)∧ dx ∧ dy ∧ dz +
(∂βij∂t
dt+∂βij∂xk
dxk)∧ dt ∧ dxi ∧ dxj
=∂f
∂tdt ∧ dx ∧ dy ∧ dz +
∂βij∂xk
dt ∧ dxi ∧ dxk ∧ dxj
=∂f
∂tdt ∧ dx ∧ dy ∧ dz − dt ∧
(∂βij∂xk
dxi ∧ dxj ∧ dxk)
=∂f
∂tdt ∧ dx ∧ dy ∧ dz − dt ∧ (div(β′)dx ∧ dy ∧ dz)
= dt ∧((
∂f
∂t− div(β′)
)dx ∧ dy ∧ dz
)5.1.3 Newtonian Navier-Stokes Revisited
Recall we had the following equations for Navier-Stokes, assuming we have a Newtonian fluid withconstant viscocities :
1. The Continuity Equation (9) :Dρ
Dt+ ρ div(u) = 0
2. The Momentum Equation (15) for Newtonian fluids :
ρDu
Dt= ρb−∇p+ (λ+ µ)∇(div(u)) + µ∇2u
3. The Internal Energy Equation (16) for Newtonian fluids:
ρDe
Dt= div(k∇T )− p div(u) + λ(div(u))2 + 2µeijeij
Recall also that DρDt
= ∂ρ∂t
+ u · ∇ρ for our fixed velocity field u, so the continuity equation is just∂ρ∂t
+ div(ρu) = 0.
Let w = (ρ,u) ∈ H, then we can write the continuity equation as ∂∂t<w + div(<w=w) = 0
For the momentum equation we use the fact that DuDt
= ∂u∂t
+ u · ∇u. Recall that for a vector fieldA,
1
2∇(A · A) = A× curl(A) + A · ∇A⇒ A · ∇A =
1
2∇(A · A) + curl(A)× A
77
Justin Shaw 78
5.2 Quaternionic Navier-Stokes
5.2.1 Quaternionic Formulation of Maxwell’s Equations
In this section we review the content of [5], reformulating the Maxwell equations ofelectromagnetism in terms of the complex quaternions. Unless otherwise noted, all information inthis section is from that book. We wish to study this reformulation in order to show theinspiration for adapting this method to reformulate the Navier-Stokes equations in terms of thecomplex quaternions.
Let
A = H⊗ C = H + iH
where the tensor product is over R. We must be careful what we mean here. This i is not the ifrom the basis for H as a vector space, but the i from the copy of C we tensored with. For thisreason we will adopt the convention that if we need to mention the basis 1, i, j, k for H, we willwrite these as 1, v1, v2, v3, and we’ll call spanv1, v2, v3 = V ∼= R3
We must define the interaction of i with H, but we do so by declaring that
i(qr) = (iq)(r) = (q)(ir) = (qr)i
We also define
B = R⊕ iV
Compare this with H = R⊕ V , and it is clear B is a linear subspace of A over R. Note also that1, iv1, iv2, iv3 is a basis for B.
We define the complex conjugate of a = q + ir ∈ A as
a = q − ir
and we note that ad = ad, whose proof is left as an exercise.
Take coordinates (t, x1, x2, x3) = (t, x, y, z) on R4. Define the operator
D =
(∂
∂t, i∇
)where
∇ = (∂
∂x,∂
∂y,∂
∂z) = v1
∂
∂x+ v2
∂
∂y+ v3
∂
∂z
is the differential operator familiar from vector calculus. We’ll call D the Hamilton-Maxwelloperator. Clearly D can act on mappings of the form θ : R4 → A.
In particular, let E,H : R4 → V ⊂ B ⊂ A be electric and magnetic fields respectively. Setθ = E + iH, then θ : R4 → V + iV ⊂ A. This means we can write θ = (0, E + iH) in complexquaternionic form, because E,H take values in V .
78
Justin Shaw 79
Recall that the product on H is defined as
(t1,v1)(t2,v2) = (t1t2 − v1 · v2, t1v2 + v1t2 + v1 × v2)
Now suppose D(θ) = 0, then
0 = D(θ)
=
(∂
∂t, i∇
)(0, E + iH)
=
(−i divE − i2 divH,
∂E
∂t+ i
∂H
∂t+ i curlE + i2 curlH
)=
(−i divE + divH,
∂E
∂t+ i
∂H
∂t+ i curlE − curlH
)But now we can equate real and imaginary parts in both the scalar and vector components to get
∂E
∂t= curl(H)
∂H
∂t= − curl(E)
divE = 0
divH = 0
which are the free Maxwell equations. Lets add a source term, that is, consider D(θ) = (−iρ,−J)for ρ the charge density and J the electric current. Then comparing real and imaginary parts in(
−i divE + divH,∂E
∂t+ i
∂H
∂t+ i curlE − curlH
)= (−iρ,−J)
gives us
∂E
∂t= curl(H)− J
∂H
∂t= − curl(E)
divE = ρ
divH = 0
which are the maxwell equations with an electric, but no magnetic, source.
5.3 Maxwell’s Equations in Differential Forms
In this section we look at Maxwell’s equations as described by differential forms, as discussed in[4]. Unless otherwise stated ever result we use in this section is from that excellent book.
79
Justin Shaw 80
5.3.1 Some Preliminary Facts
Take a manifold (M, η) where η is a metric of index (3, 1), called a Lorentz metric. In particularwe take η of the form
η =
−1 0 0 00 1 0 00 0 1 00 0 0 1
which is called the Minkowski metric. Note that it is not positive definite, so in fact its not ametric at all, but it is still called a metric everywhere, and we do not break with that regrettabletradition here. Now (M, η) is called a Lorentz manifold. We now list some facts from [4] that wewill need.
Lemma: 1. If M is a Lorentz manifold of dimension n, then
1. Let d : Ωk−1(M)→ Ωk(M) be the exterior derivative, then the adjoint of d is
d∗ : Ωk(M)→ Ωk−1(M)
d∗ = (−1)n(k+1) ∗ d∗
2. The hodge star, ∗, has the property that
∗2 = (−1)1+k(n−k)Id
We now specialize to the case where M = R4, with coordinates (t, x, y, z). In this case our lemmagives us
1. d∗ = ∗d∗
2. ∗2 = (−1)k+1Id
because n = 4, and one can check easily that (−1)k+1 = (−1)1+k(4−k) for k = 0, 1, 2, 3, 4.
5.3.2 The Electromagnetic Tensor
Now let A = φdt+ Aidxi ∈ Ω1(R4), where i = 1, 2, 3. Define F = dA, the electromagnetic tensor,
so
F = dA
= d(φdt+ Aidx
i)
= dφ ∧ dt+ dAi ∧ dxi
=
(∂φ
∂tdt+
∂φ
∂xidxi)∧ dt+
(∂Ai∂t
dt+∂Ai∂xj
dxj)∧ dxi
= −dt ∧(∂φ
∂xidxi)
+ dt ∧(∂Ai∂t
dxi)
+∂Ai∂xj
dxj ∧ dxi
= dt ∧([
∂Ai∂t− ∂φ
∂xi
]dxi)
+∂Ai∂xj
dxj ∧ dxi
80
Justin Shaw 81
Now use the curl map on one forms defined above, and define
B = curl(A) = ∗3d3A = ∗3
(∂Ai∂xj
dxj ∧ dxi)
since ∗23 = Id, we have
∗3B =∂Ai∂xj
dxj ∧ dxi
Let E = −[∂Ai
∂t− ∂φ
∂xi
]dxi, so E] = ∇φ− ∂A
∂t. Then we can rewrite F as
F = dA = −dt ∧ E + ∗3B
but now since d2 = 0, dF = d2A = 0. On the other hand
dF = d (−dt ∧ E + ∗3B)
= dt ∧ dE + d
(∂Ai∂xj
dxj ∧ dxi)
= dt ∧ dE +∂
∂t
∂Ai∂xj
dt ∧ dxj ∧ dxi +∂
∂xk∂Ai∂xj
dxk ∧ dxj ∧ dxi
but notice in the first term that d = d4 has a dt term which will drop out since we’re wedging withdt, so dt ∧ dE = dt ∧ d3E. We also have that the last term is d3 ∗3 B. Now
0 = dF = dt ∧ (d3E) + dt ∧(∂
∂t
∂Ai∂xj
dxj ∧ dxi)
+ d3 ∗3 B
= dt ∧(∂ ∗3 B
∂t+ d3E
)+ d3 ∗3 B
Which means
∂ ∗3 B
∂t+ d3E = 0
d3 ∗3 B = 0
Notice that ∂∂t∗3 = ∗3
∂∂t
because ∗3 is linear. Therefore we can take ∗3 of the first equation to get
∂B
∂t= − ∗3 d3E = − curl(E)
So we can identify these 1 forms with their musical isomorphism dual vector fields.
In the second equation, take ∗3 of both sides, and use the fact that for a vector field X,div(X) = ∗3d3 ∗3 X
[, we see that
div(B]) = 0
However since we’re in R3, we can identify B with B] since their coefficient functions are identical,and so div(B) = 0.
81
Justin Shaw 82
Therefore, since d2 = 0, dF = 0 implies
div(B) = 0
∂B
∂t= − curl(E)
which are two of the four Maxwell equations.
5.3.3 The Current One-Form
We want the other two of course. Define the current one-form J = −ρdt+ jidxi ∈ Ω1(R4) with ρ
the electric charge density and (j1, j2, j3) the electric current density. Suppose that d∗F = J ,which is really a condition on A, then since J is a 1 form, we know that ∗2
4J = J by the lemma.Therefore
d∗F = J ⇔ ∗4d ∗4 F = J ⇔ d ∗4 F = ∗4J
where we’ve used the fact that d ∗ F is a 3 form, and so again ∗24 = Id. Now we calculate:
d ∗4 F = d ∗4 (−dt ∧ E + ∗3B)
= d ∗4
(−dt ∧ E +
∂Ai∂xj
dxj ∧ dxi)
= d
(− ∗4 dt ∧ E + ∗4
∂Ai∂xj
dxj ∧ dxi)
Lets do each term separately:
∗4dt ∧([
∂Ai∂t− ∂φ
∂xi
]dxi)
=
[∂Ai∂t− ∂φ
∂xi
]∗4 dt ∧ dxi
Now a quick calculation shows that
∗4dt ∧ dx = dz ∧ dy = − ∗3 dx
∗4dt ∧ dy = dx ∧ dz = − ∗3 dy
∗4dt ∧ dz = dy ∧ dx = − ∗3 dz
so that ∗4dt ∧ dxi = − ∗3 dxi Therefore if α = αidx
i,
∗4(dt ∧ α) = ∗4(αidt ∧ dxi)= αi ∗4 dt ∧ dxi
= αi(− ∗3 dxi)
= − ∗3 α
Now we can use this fact:
82
Justin Shaw 83
∗4dt ∧ E = − ∗3 E
So now we have the first term. For the second, another calculation shows that
∗4dx ∧ dy = dt ∧ dz∗4dy ∧ dz = dt ∧ dx∗4dx ∧ dz = dy ∧ dt
Therefore
∗4∂Ai∂xj
dxj ∧ dxi
=
(∂A2
∂x1− ∂A1
∂x2
)∗4 dx ∧ dy +
(∂A3
∂x1− ∂A1
∂x3
)∗4 dx ∧ dz +
(∂A3
∂x2− ∂A2
∂x3
)∗4 dy ∧ dz
=
(∂A2
∂x1− ∂A1
∂x2
)dt ∧ dz +
(∂A3
∂x1− ∂A1
∂x3
)dy ∧ dt+
(∂A3
∂x2− ∂A2
∂x3
)dt ∧ dx
= dt ∧((
∂A3
∂x2− ∂A2
∂x3
)dx+
(∂A1
∂x3− ∂A3
∂x1
)dy +
(∂A2
∂x1− ∂A1
∂x2
)dz
)However we also have that
∗3∂Ai∂xj
dxj ∧ dxi
=
(∂A2
∂x1− ∂A1
∂x2
)∗3 dx ∧ dy +
(∂A3
∂x1− ∂A1
∂x3
)∗3 dx ∧ dz +
(∂A3
∂x2− ∂A2
∂x3
)∗3 dy ∧ dz
=
(∂A2
∂x1− ∂A1
∂x2
)dz +
(∂A3
∂x1− ∂A1
∂x3
)(−dy) +
(∂A3
∂x2− ∂A2
∂x3
)dx
Therefore
∗4 ∗3 B = ∗4∂Ai∂xj
dxj ∧ dxi = dt ∧(∗3∂Ai∂xj
dxj ∧ dxi)
= dt ∧B
since ∗23 = Id. Putting it all together, we have shown that
∗4F = ∗3E + dt ∧B
Now we need to take d = d4 of this.
d ∗4 F = d ∗3 E + d(dt ∧B)
= d ∗3 E − dt ∧ d3B
where the second term is using d3 since a term with dt is zero when wedged with dt. For the firstterm, we can explicitly write out d ∗3 E:
83
Justin Shaw 84
d ∗3 E = d ∗3
(Eidx
i)
= d(Ei ∗3 dx
i)
= d (E1dy ∧ dz + E2dz ∧ dx+ E3dx ∧ dy)
Now consider
d(E1dy ∧ dz) =
(∂E1
∂tdt+
∂E1
∂xidxi)∧ dy ∧ dz
=∂E1
∂tdt ∧ dy ∧ dz +
∂E1
∂xdx ∧ dy ∧ dz
= dt ∧ ∗3∂E1
∂tdx+
∂E1
∂xdx ∧ dy ∧ dz
d(E2dz ∧ dx) =
(∂E2
∂tdt+
∂E2
∂xidxi)∧ dz ∧ dx
=∂E2
∂tdt ∧ dz ∧ dx+
∂E2
∂ydx ∧ dy ∧ dz
= dt ∧ ∗3∂E2
∂tdy +
∂E2
∂ydx ∧ dy ∧ dz
d(E3dx ∧ dy) =
(∂E2
∂tdt+
∂E3
∂xidxi)∧ dx ∧ dy
=∂E3
∂tdt ∧ dx ∧ dy +
∂E3
∂zdx ∧ dy ∧ dz
= dt ∧ ∗3∂E3
∂tdz +
∂E3
∂zdx ∧ dy ∧ dz
Therefore
d ∗3 E = dt ∧ ∗3∂E
∂t+
(∂E1
∂x+∂E2
∂y+∂E3
∂z
)dx ∧ dy ∧ dz
However
∗3
(∂E1
∂x+∂E2
∂y+∂E3
∂z
)dx ∧ dy ∧ dz = div(E) = ∗3d3 ∗3 E
so the second term is d3 ∗3 E, and so
d ∗3 E = dt ∧ ∗3∂E
∂t+ d3 ∗3 E
Now we have shown that
d ∗4 F = dt ∧ ∗3∂E
∂t+ d3 ∗3 E − dt ∧ d3B
We have assumed that d ∗4 F = ∗4J , so what s ∗4J?
84
Justin Shaw 85
∗4J = −ρ ∗4 dt+ ji ∗4 dxi
= ρdx ∧ dy ∧ dz + j1dt ∧ dz ∧ dy + j2dt ∧ dx ∧ dz + j3dt ∧ dy ∧ dx= ρdx ∧ dy ∧ dz − dt ∧ j1 ∗3 dx− dt ∧ j2 ∗3 dy − dt ∧ j3 ∗3 dz
= ρdx ∧ dy ∧ dz − dt ∧ (∗3jidxi)
Now we have
dt ∧ ∗3∂E
∂t+ d3 ∗3 E − dt ∧ d3B = ρdx ∧ dy ∧ dz − dt ∧ (∗3jidx
i)
which gives us
∗3∂E
∂t− d3B = − ∗3 jidx
i
−∂E∂t
+ ∗3d3B = jidxi
but recall that∗3d3B = curl(B), so we have the Maxwell equation
−∂E∂t
+ curl(B) = (j1, j2, j3)
and we also have
d3 ∗3 E = ρdx ∧ dy ∧ dz
and taking ∗3 of both sides gives us the other Maxwell equation we’re missing, namely
div(E) = ρ
So now lets summarize: We let A = φdt+Aidxi ∈ Ω1(R4). We let F = dA = −dt∧E + ∗3B. Since
F is exact its closed, so the equation dF = 0 gave us the first two of Maxwell’s equations below.
We then demanded that A was chosen such that d∗F = J = −ρdt+ jidxi ∈ Ω1(R4). This equation
gave us the second two of the Maxwell’s equation below.
div(B) = 0
∂B
∂t= − curl(E)
−∂E∂t
+ curl(B) = (j1, j2, j3)
div(E) = ρ
85
Justin Shaw 86
5.3.4 The Charge Conservation Equation
Notice that d∗F = J means that d∗J = (d∗)2F = 0. We know from the lemma above thatd∗ = ∗d∗, so we have
∗d ∗ J = 0⇔ d ∗ J = 0
Well we found ∗J = ρdx ∧ dy ∧ dz − dt ∧ (∗3jidxi) in the previous section. Now we take d = d4
0 = d ∗ J= d
(ρdx ∧ dy ∧ dz − dt ∧ (∗3jidx
i))
=∂ρ
∂tdt ∧ dx ∧ dy ∧ dz + dt ∧ (d ∗3 jidx
i)
However in the second term we can take d3 instead of d4 because we’re wedging with dt.
d3 ∗3 jidxi = d3(j1dy ∧ dz + j2dz ∧ dx+ j3dx ∧ dy)
=∂j1
∂xdx ∧ dy ∧ dz +
∂j2
∂ydx ∧ dy ∧ dz +
∂j3
∂zdx ∧ dy ∧ dz
= div((j1, j2, j3))dx ∧ dy ∧ dz
Therefore
0 = d ∗ J
=∂ρ
∂tdt ∧ dx ∧ dy ∧ dz + div((j1, j2, j3))dt ∧ dx ∧ dy ∧ dz
Now we take ∗4 of both sides to get
∂ρ
∂t+ div((j1, j2, j3)) = 0
which is the continuity equation from fluid dynamics if we let (j1, j2, j3) = ρu for u the fluid flowand ρ the density. In electromagnetism this is called the charge conservation equation.
86
Justin Shaw 87
Bibliography
References
[1] Wainwright, J. and Tenti, G. (2005), Calculus 4: Course Notes for AMATH 231 (spring 2012ed.), University of Waterloo.
[2] Lamb, K. (2014), AMATH 361: Continuum Mechanics Course Notes (winter 2014 ed.),University of Waterloo.
[3] Chorin, A.J. and Marsden, J.E. (1993), A Mathematical Introduction to Fluid Mechanics (3rded.), New York: Springer-Verlag.
[4] Nakahara, M (2003), Geometry, Topology, and Physics, Philadelphia: Institute of PhysicsPublishing.
[5] Hermann, R. (1974), Interdisciplinary Mathematics Vol VII: Spinors, Clifford and CayleyAlgebras Brookline, MA: Math Sci Press.
[6] Marsden, J.E. and Ratiu, T. (2007), Manifold, Tensor Analysis, and Applications (3rd ed.),New York: Springer-Verlag.
[7] Kamchatnov, A.M., Topological Solitons in Magnetohydrodynamics, Zh. Eksp. Teor. Fiz. 82(1982), 117-124.
[8] Matthews, J (2006) The Quaternionic Structure of the Equations of Geophysical FluidDynamics, University of Reading submitted thesis.
[9] Urbantke, H.K., The Hopf fibration - seven times in Physics, Journal of Geometry and Physics46 (2003), 125-150.
[10] Aris, Rutherford (1962), Vectors, Tensors, and the Basic equation of Fluid Mechanics , NewYork: Dover.
[11] Clarke, David A. (2011), A Primer on Tensor Calculus, St Mary’s University, Halifax NS,Canada.
[12] Kundu, P.K. and Cohen, I.M. (2008), Fluid Mechanics (4th ed.), Oxford: Elsevier.
[13] Woods Hole Oceanographic Institute. (2006), Fluid Dynamics of the Atmosphere and Ocean:Coursenotes Woods Hole Oceanographic Institute.
[14] Spivak, M (2005), A Comprehensive Introduction to Differential Geometry, Houston: Publishor Perish.
[15] Bott, R. and Tu, L. W. (1982), Differential Forms in Algebraic Topology, New York:Springer-Verlag.
[16] Reese, H. (1990), Spinors and Calibrations, San Diego: Academic Press.
[17] Lyons, D.W. An Elementary Introduction to the Hopf Fibration, Annville, PA: LebanonValley College.
87
Justin Shaw 88
[18] Reeder, M. (2010) Math 845: Introduction to Lie Groups Course Notes, Boston College,Chestnut Hill, MA.
[19] Karigiannis, S. (2010) Algebra and Geometry: Spheres and Projective Spaces, University ofWaterloo.
88
