vector analysis

1

VECTOR ANALYSIS

SOLO HERMELIN

http://www.solohermelin.com

http://www.solohermelin.com/

2

Vector AnalysisSOLOTABLE OF CONTENT

Algebras HistoryVector Analysis HistoryVector Algebra

Reciprocal Sets of Vectors Vector Decomposition The Summation Convention The Metric Tensor or Fundamental Tensor Specified by .

3321 ,, Eeee

Change of Vector Base, Coordinate Transformation

Vector Space

Differential Geometry

Osculating Circle of C at P

Theory of CurvesUnit Tangent Vector of path C at a point PCurvature of curve C at P

Osculating Plane of C at P

Binormal TorsionSeret-Frenet EquationsInvoluteEvolute

Vector Identities Summary Cartesian Coordinates

3

Vector AnalysisSOLO

TABLE OF CONTENT (continue – 1)


Conjugate Directions

Surfaces in the Three Dimensional SpacesFirst Fundamental Form:Arc Length on a Path on the SurfaceSurface AreaChange of CoordinatesSecond Fundamental FormPrincipal Curvatures and Directions

Asymptotic Lines

Scalar and Vector FieldsVector Differentiation

Ordinary Derivative of Scalars and VectorsPartial Derivatives of Scalar and VectorsDifferentials of VectorsThe Vector Differential Operator Del (, Nabla)Scalar DifferentialVector DifferentialDifferential Identities

4

Vector AnalysisSOLO


Scalar and Vector FieldsCurvilinear Coordinates in a Three Dimensional Space

Covariant and Contravariant Components of a Vector in Base .

321,, uuu rrr

Coordinate Transformation in Curvilinear Coordinates Covariant Derivative Covariant Derivative of a Vector .

A

Vector Integration

Ordinary Integration of Vectors Line Integrals Surface Integrals Volume Integrals Simply and Multiply Connected Regions Green’s Theorem in the PlaneStoke’s TheoremDivergence TheoremGauss’ Theorem Variations Stokes’ Theorem Variations

Green’s IdentitiesDerivation of Nabla ( ) from Gauss’ TheoremThe Operator .

5

Vector AnalysisSOLO


Scalar and Vector Fields

Fundamental Theorem of Vector Analysis for a Bounded Region V (Helmholtz’s Theorem) Reynolds’ Transport TheoremPoisson’s Non-homogeneous Differential EquationKirchhoff’s Solution of the Scalar Helmholtz Non-homogeneous Differential Equation

Derivation of Nabla ( ) from Gauss’ TheoremThe Operator . Orthogonal Curvilinear Coordinates in a Three Dimensional SpaceVector Operations in Various Coordinate Systems

Applications

Laplace FieldsHarmonic Functions

Rotations

6

Synthetic GeometryEuclid 300BC

Algebras History SOLO

Extensive AlgebraGrassmann 1844

Binary AlgebraBoole 1854

Complex AlgebraWessel, Gauss 1798

Spin AlgebraPauli, Dirac 1928

Syncopated AlgebraDiophantes 250AD

QuaternionsHamilton 1843

Tensor CalculusRicci 1890

Vector CalculusGibbs 1881

Clifford AlgebraClifford 1878

Differential FormsE. Cartan 1908

First Printing1482

http://modelingnts.la.asu.edu/html/evolution.html

Geometric Algebraand Calculus

Hestenes 1966

Matrix AlgebraCayley 1854

DeterminantsSylvester 1878

Analytic Geometry Descartes 1637

Table of Content

7

Vector Analysis History SOLO

John Wallis1616-1703

1673

Caspar Wessel1745-1818

“On the Analytic Representationof Direction; an Attempt”, 1799

bia

Jean Robert Argand1768-1822

18061i

Quaternions1843

William Rowan Hamilton 1805-1865

3210 qkqjqiq Extensive Algebra

1844

Herman Günter Grassmann1809-1877

“Elements of VectorAnalysis” 1881

Josiah Willard Gibbs 1839-1903

Oliver Heaviside1850-1925

“ElectromagneticTheory” 1893

3 .R.S. Elliott, “Electromagnetics”,pp.564-568

http://www-groups.dcs.st-and.ac.uk/~history/index.html

Table of Content

Edwin Bidwell Wilson1879-1964

“Vector Analysis”1901

8

Vector AnalysisSOLO

ba

Vector Algebra

b

a

a

Addition of Vectors ParallelogramLaw of addition

Subtraction of Vectors baba

1

ParallelogramLaw of subtraction

b

a

b

b a

ba

Multiplication of Vector by a Scalar am

aa

am

b

a

ba

ba

b

Geometric Definition of a Vector

A Vector is defined by it’s Magnitude and Directiona a

a

9

Vector AnalysisSOLO

bababa

,sin

b

a

ba

,

b

a

ba

,

ba

abba

Scalar product of two vectors ba

,

Vector product of two vectors ba

,

2/1

2/1 ,cos

aaaaaaaMagnitude of Vector a

Unit Vector (Vector of Unit Magnitude)aa 1ˆ aa

aa

1

:1:ˆ

Zero Vector (Vector of Zero Magnitude)0

aa

0

0:0

00 a

ababbababa ba

,cos, ,

Vector Algebra (continue – 1)

10

Vector AnalysisSOLO

n||ˆˆˆˆ aanannana n

n

a

n

an

a

ˆ

ˆn||

na

n

an

ˆ

ˆ

Vector decomposition in two orthogonal directions nn ,||

Vector decomposition in two given directions (geometric solution)

1n

a

2n

A B

C

Given two directions and , and the vector a

1n 2n

anBCnCA

21ˆˆ

1n

a

2n

A B

Draw lines parallel to those directions passingthrough both ends A and B of the vector .The vectors obtained are in the desired directionsand by rule of vector addition satisfy

a

Vector Algebra (continue – 2)

Table of Content

11

SOLO

Triple Scalar Product

Vectors & Tensors in a 3D Space

3321 ,, Eeee

are three non-coplanar vectors, i.e.

1e

2e

3e

0:,, 321321 eeeeee

0,,

,,,,

123123213

132132132321

eeeeeeeee

eeeeeeeeeeee

Reciprocal Sets of Vectors The sets of vectors and are called Reciprocal Sets or Systems of Vectors if:

321 ,, eee 321 ,, eee

DeltaKroneckertheisji

jiee j

i

j

i

j

i

1

0

Because is orthogonal to and then

2e

3e

1e

321

321321

1

132

1

,,

1,,1

eeekeeekeeekeeeeke

and in the same way and are given by:2e 3e

1e

321

213

321

132

321

321

,,,,,, eee

eee

eee

eee

eee

eee

12

SOLO Vectors & Tensors in a 3D Space

Reciprocal Sets of Vectors (continue)

By using the previous equations we get:

321

3

2

321

13323132

2

321

133221

,,,,,, eee

e

eee

eeeeeeee

eee

eeeeee

321

213

321

132

321

321

,,,,,, eee

eee

eee

eee

eee

eee

0,,

1

,,,,

321321

3

3321321

eeeeee

eeeeeeee

Multiplying (scalar product) this equation by we get:

3e

In the same way we can show that:

Therefore are also non-coplanar, and:

321 ,, eee 1,,,, 321

321 eeeeee

321

21

3321

13

2321

32

1 ,,,,,, eee

eee

eee

eee

eee

eee

1e

2e

3e

1e

2e

3e

Table of Content

1e

2e

3e

13


Vector Decomposition Given we want to find the coefficients and such that:

3EA

321 ,, AAA 321 ,, AAA

3

1

3

3

2

2

1

1

3

1

3

3

2

2

1

1

j

j

j

i

i

i

eAeAeAeA

eAeAeAeAA

3,2,1, iee i

i

are two reciprocalvector bases

Let multiply the first row of the decomposition by :

je

Let multiply the second row of the decomposition by :

ie

j

i

j

i

i

i

j

i

ij AAeeAeA

3

1

3

1

i

j

ij

j

j

i

j

ji AAeeAeA

3

1

3

1

Therefore:

ii

jj eAAeAA

&

Then:

3

1

3

3

2

2

1

1

3

1

3

3

2

2

1

1

j

j

j

i

i

i

eeAeeAeeAeeA

eeAeeAeeAeeAA

Table of Content

1e

2e

3e

14


The Summation Convention

j

j

j

j

j eAeAeAeAeA

3

1

3

3

2

2

1

1

The last notation is called the summation convention, j is called the dummy index or the umbral index.

i

i

i

i

j

j

j

j

j

j

j

j

j

i

i

i

i

i

eAeeAeeAeeA

eAeeAeeAeeAA

3

1

3

1

Instead of summation notation we shall use the shorter notation first adopted by Einstein

3

1j

j

j eA j

j eA

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15


Let define:

The Metric Tensor or Fundamental Tensor Specified by .

3321 ,, Eeee

jiijjiij geeeeg

3321 ,, Eeee

the metric covariant tensors of

By choosing we get:

j

ijiii

j

jiiiii

egegegeg

eeeeeeeeeeeee

3

3

2

2

1

1

3

3

2

2

1

1

ieA

or:

j

iji ege

For i = 1, 2, 3 we have:

3

2

1

332313

322212

312111

3

2

1

333231

232221

131211

3

2

1

e

e

e

eeeeee

eeeeee

eeeeee

e

e

e

ggg

ggg

ggg

e

e

e

1e

2e

3e

16


We want to prove that the following determinant (g) is nonzero:


3321 ,, Eeee

332313

322212

312111

333231

232221

131211

detdet:

eeeeee

eeeeee

eeeeee

ggg

ggg

ggg

g

g is the Gram determinant of the vectors 321 ,, eee

Jorgen Gram1850 - 1916

Proof:

Because the vectors are non-coplanars the following equations:

321 ,, eee

03

3

2

2

1

1 eee

is true if and only if 0321 Let multiply (scalar product) this equation, consecutively, by :321 ,, eee

0

0

0

0

0

0

3

2

1

332313

322212

312111

33

3

23

2

13

1

32

3

22

2

12

1

31

3

21

2

11

1

eeeeee

eeeeee

eeeeee

eeeeee

eeeeee

eeeeee

Therefore α1= α2= α3=0 if and only if g:=det {gij}≠0 q.e.d.

17


Because g ≠ 0 we can take the inverse of gij and obtain:


3321 ,, Eeee

where Gij = minor gij having the following property: ji

kj

ik ggG

3

2

1

333231

232221

131211

3

2

1

333231

232221

131211

3

2

1

1

e

e

e

ggg

ggg

ggg

e

e

e

GGG

GGG

GGG

ge

e

e

and:g

G

g

gminorg

ijijij

Therefore:g

gg

g

Ggg

ji

kj

ik

kj

ik j

i

kj

ik gg

Let multiply the equation by gij and perform the summation on ij

iji ege

jj

ij

ij

i

ij eeggeg

Therefore: i

ijj ege

Let multiply the equation byk

kjj ege

ie iji

k

kjijji geegeeee

jiijjiij geeeeg

i

jkj

ik ggG

The Operator .

18



3321 ,, Eeee

Let find the relation between g and 321321 :,, eeeeee

We shall write the decomposition of in the vector base32 ee

321 ,, eee

3

3

2

2

1

1

32 eeeee

Let find λ1, λ2, λ3. Multiply the previous equation (scalar product) by . 1e

i

i ggggeeeeee 113

3

12

2

11

1

321321 ,,

Multiply this equation by g1i: ii

i

ii ggeeeg

1

1

1321

1 ,,

Therefore: 321

1 ,, eeeg ii

Let compute now: 321

1

0

323

3

0

322

2

321

1

3232 eeeeeeeeeeeeeeee

321

11

321

11

321

2

233322

321

3322332

321

3232

321

32321

,,,,,,,,

,,,,

eee

gg

eee

G

eee

ggg

eee

eeeeeee

eee

eeee

eee

eeee

From those equations we obtain: 321

111

,, eee

gg

Finally: geeeeee 321

2

321

1 ,,,,

We can see that if are collinear than and g are zero. 321 ,, eee 321 ,, eee

Table of Content

19


Change of Vector Base, Coordinate Transformation

Let choose another base and its reciprocal 321 ,, fff 321 ,, fff

3

2

1

3

2

1

3

3

2

3

1

3

3

2

2

2

1

2

3

1

2

1

1

1

3

2

1

e

e

e

L

e

e

e

f

f

f

ef j

j

ii

where j

i

j

i ef

By tacking the inverse of those equations we obtain:

3

2

1

1

3

2

1

3

3

2

3

1

3

3

2

2

2

1

2

3

1

2

1

1

1

3

2

1

f

f

f

L

f

f

f

e

e

e

fe i

j

ij

wherej

ij

i ef

Because are the coefficients of the inverse matrix with coefficients :j

i j

i

i

j

i

k

k

j

20


Change of Vector Base, Coordinate Transformation (continue – 1)

Let write any vector in those two bases: A

3

2

1

3

2

1

3

3

2

3

1

3

3

2

2

2

1

2

3

1

2

1

1

1

3

2

1

e

e

e

L

e

e

e

f

f

f

ef

ef

ji

j

i

j

j

ii

then:

3

2

1

1

3

2

1

3

3

3

2

3

1

2

3

2

2

2

1

1

3

1

2

1

1

3

2

1

E

E

E

L

E

E

E

F

F

F

ef

EFT

j

ii

j

i

j

ji

i

i

j

j fFeEA

iijj fAFeAE

&i

j

ji

i

i

j

j

j

j

i

i EFfEeEfF

or:

But we remember that:

We can see that the relation between the components F1, F2, F3 to E1, E2, E3 is not similar, contravariant, to the relation between the two bases of vectorsto . Therefore we define F1, F2, F3 and E1, E2, E3 as the contravariant

components of the bases and .

321 ,, fff

321 ,, eee

321 ,, fff 321 ,, eee

where

21



3

2

1

3

2

1

3

3

2

3

1

3

3

2

2

2

1

2

3

1

2

1

1

1

3

2

1

E

E

E

L

E

E

E

F

F

F

EF j

iji

i

i

j

j fFeEA

iijj fAFeAE

&

then:

Let write now the vector in the two bases andA 321 ,, fff

321 ,, eee

where

and j

ijef

ijjjjii EfeEeEAfAFj

iji

We can see that the relation between the components F1, F2, F3 to E1, E2, E3 is similar, covariant, to the relation between the two bases of vectorsto . Therefore wew define F1, F2, F3 and E1, E2, E3 as the covariant

components of the bases and .

321 ,, fff

321 ,, eee

321 ,, fff 321 ,, eee

22


We have:


j

j

j

j

i

i

i

i

eeAeA

eeAeAA

Ai contravariant component

Aj covariant component

Let find the relation between covariant and the contravariant components:

j

j

j

ij

i

egei

i eAegAeAAj

iji

i

i

i

ij

jege

j

j eAegAeAAi

ijj

Therefore: ij

j

i

ij

i

j gAAgAA &

Let find the relation between gij and gij defined in the bases and to and defined in the bases and .

ie

ie

if

if

ijg ijg

m

m

kkj

j

ii efef

&

jm

m

k

j

imj

m

k

j

ikiik geeffg

Hence: jm

m

k

j

iik gg

This is a covariant relation of rank two, (similar, two times, to relation between to .if

je

23



3

2

1

3

2

1

3

3

2

3

1

3

3

2

2

2

1

2

3

1

2

1

1

1

3

2

1

e

e

e

L

e

e

e

f

f

f

ef

ef

ji

j

i

j

j

ii

Since we have:k

iki fgf

m

jm

j

iege

j

j

ief

k

iki egefgfm

jmjjj

ii

and: m

jm

j

i

km

kjm

j

igg

k

ik egfgfgmjm

mk

jiik

Therefore, by equalizing the terms that multiply we obtain:

3

2

1

3

2

1

3

3

3

2

3

1

2

3

2

2

2

1

1

3

1

2

1

1

3

2

1

f

f

f

L

f

f

f

e

e

e

fe Tkm

k

m

jm

j

i g

We found the relation:

24



Therefore:

3

2

1

1

3

2

1

3

3

3

2

3

1

2

3

2

2

2

1

1

3

1

2

1

1

3

2

1

e

e

e

L

e

e

e

f

f

f

efTmj

m

j

Let take the inverse of the relation by multiplying by and summarize on m:j

mkm

k

m fe

jkj

k

km

k

j

m

mj

m fffe

From the relation: mk

m

kjj

i

i efef

&

we have: jmk

m

j

i

mjk

m

j

i

kiik geeffg

or: jmk

m

j

i

ik gg This is a contravariant relation of rank two.

From the relation:m

m

kk

jj

i

i efef

&

we have: m

j

m

k

i

jm

jm

k

i

j

i

kk

i eeff

or: This is a relation once covariant and once contravariant of rank two.

m

j

m

k

i

j

i

k Table of Content

25


Cartesian Coordinates

Three dimensional cartesian coordinates are define as coordinates in a orthonormalbasis such that: zyxorkjieee 1,1,1ˆ,ˆ,ˆ,, 321

ix ˆ1

jy ˆ1

O

kz ˆ1

x1

y1z1

0111111

1111111

zyzxyx

zzyyxx

yzxxyzzxy

yxzxzyzyx

111111111

111111111

11111,1,12 zyxzyxg

The reciprocal set is identical to the original set

ji

jiggg ij

ijj

iij 0

1

Given

z

y

x

zyx

A

A

A

zyxzAyAxAA 111111

z

y

x

A

A

A

AMatrix notationof a vector

26


Cartesian Coordinates (continue – 1)

ix ˆ1

jy ˆ1

O

kz ˆ1

Given

z

y

x

zyx

A

A

A

zyxzAyAxAA 111111

z

y

x

A

A

A


z

y

x

zyx

B

B

B

zyxzByBxBB 111111

ABABBABABABA

B

B

B

AAAzByBxBzAyAxABA

T

zzyyxx

T

z

y

x

zyxzyxzyx

111111

z

y

x

B

B

B

B

27



zyx

zyxxyyxzxxzyzzy

zyxzyx

BBB

AAA

zyx

zBABAyBABAxBABA

zByBxBzAyAxABA

111

det111

111111

ABAB

A

A

A

BB

BB

BB

BA

B

B

B

AA

AA

AA

BA

z

y

x

xy

xz

yz

z

y

x

xy

xz

yz

0

0

0

0

0

0

ix ˆ1

jy ˆ1

O

kz ˆ1

Given

z

y

x

zyx

A

A

A

zyxzAyAxAA 111111

z

y

x

A

A

A


z

y

x

zyx

B

B

B

zyxzByBxBB 111111

z

y

x

B

B

B

B

28



Table of Content

ACBACBCBACBCBACBCB

BACBACACBACACBACAC

CBABACBABACBABA

CCC

BBB

AAA

BBB

AAA

CCC

zCyCxC

BBB

AAA

zyx

CBA

zxyyxyzxxzxyzzy

zxyyxyzxxzxyzzy

zxyyxyzxxzxyzzy

zyx

zyx

zyx

zyx

zyx

zyx

zyx

zyx

zyx

detdet111

111

det

ABCABCBAC

B

B

B

AA

AA

AA

CCCBAC TT

z

y

x

xy

xz

yz

zyx

0

0

0

Given

z

y

x

zyx

A

A

A

zyxzAyAxAA 111111

z

y

x

A

A

A


z

y

x

zyx

B

B

B

zyxzByBxBB 111111

z

y

x

B

B

B

B

z

y

x

zyx

C

C

C

zyxzCyCxCC 111111

z

y

x

C

C

C

C

29

Vector AnalysisSOLO

bacbacacbacbcbacba

,,,,:,,

cbabcacba

cbdadbca

dcbcdba

dcbadcba

adcbbdca

dcbacdbadcba

,,,,

,,,,

2,, cbaaccbba

feabdcfebadc

dcefbadcfeba

fedcbafecdbafedcba

,,,,,,,,

,,,,,,,,

,,,,,,,,

Vector Identities Summary

0 bacacbcba

Table of Content

30

SOLO

VECTOR SPACE

Given the complex numbers .C ,,

A Vector Space V (Linear Affine Space) with elements over C if its elements satisfy the following conditions:

Vzyx

,,

I. Exists a operation of Addition with the following properties:

xyyx

Commutative (Abelian) Law for Addition 1

zyxzyx

Associative Law for Addition 2

xx

0 Exists a unique vector 0

3

II. Exists a operation of Multiplication by a Scalar with the following properties:

0..

yxtsVyVx4 Inverse

xx

15

xx Associative Law for Multiplication 6

xxx Distributive Law for Multiplication 7

yxyx Commutative Law for Multiplication 8

00101010 3

575 xxxxxxxxWe can write:

Vector Analysis

31

SOLO

Scalar Product in a Vector Space

The Scalar Product of two vectors is the operation with the symbol with the following properties:

Vyx

, Cyx

,

xyyx

,, yxyx

,,

zyzxzyx

,,,

00,&0,

xxxxx

Distance Between Two Vectors The Distance between two Vectors is defined by the following properties:

Vyx

, yxyxd

,

00,&0,

xxxdxxd

xydyxd

,,

yzdzxdyxd

,,,

Vector Analysis

Table of Content

32

SOLO

Differential Geometry is the study of geometric figures using the methods of Calculus.

Here we present the curves and surfaces embedded in a three dimensional space.

Properties of curves and surfaces which depend only upon points close to a particular point of the figure are called local properties.. The study of local properties is called differential geometry in the small.

Those properties which involve the entire geometric figure are called global properties. The study of global properties is called differential geometry in the large.

Hyperboloidof RotationToroyd

MobiusMovement


Differential Geometry in the 3D Euclidean Space

Table of Content

33

SOLODifferential Geometry in the 3D Euclidean Space

A curve C in a three dimensional space is defined by one parameter t, tr

ur

rd

P

O

a

b

C

Theory of Curves

Regular Parametric Representation of a Vector Function:

parameter t, defined in the interval I and:

Ittrr ,

tr

(i) is of class C1 (continuous and 1st order differentiable) in I

Arc length differential: tdtd

rdtd

td

rd

td

rdtrdtrdsd

2/1

2/1:

We also can define sdtrdtrdsd 2/1* :

(ii)

Ittd

trd0

Iinconstantnottr

Arc length as a parameter: t

t

tdtd

rds

0

Regular Curves:

A real valued function t = t (θ), on an interval Iθ, is an allowable change of parameter if:

(i) t (θ) is of class C1 in Iθ (ii) d t/ d θ ≠ 0 for all θ in Iθ

A representation on Is is a representation in terms of arc length or a natural representation

srr

Table of Content

34



ur

rd

P

O

a

b

C

- arc length differential tdtd

rd

td

rdtrdtrdsd

2/1

2/1:

td

rd

td

rdr

sd

rdt /:: - unit tangent vector of path C at P

(tangent to C at P)

1x

2x

3x

td

rdr ' - tangent vector of path C at P

(tangent to C at P)

0,0,sincos 321 baetbetaetar

Example: Circular Helix

0,0,cossin' 321 baebetaetatd

rdr

2/122

2/1

batd

rd

td

rd

td

rd

321

2/122 cossin/: ebetaetabatd

rd

td

rdt

Theory of Curves (continue – 1)

We also can define sdtrdtrdsd 2/1* :

tsd

rd

sd

rd

*

Unit Tangent Vector of path C at a point P

Table of Content

35


The earliest investigations by means of analysis were made by René Descartes in 1637.

tr

ur

rdP

O

a

b

C

René Descartes1596 - 1650

Pierre Fermat1601 - 1665

Christian Huyghens1629 - 1695

Gottfried Leibniz1646 - 1716

The general concept of tangent was introduced in seventeenth century, in connexion with the basic concepts of calculus. Fermat, Descartes and Huyghens made important contributions to the tangent problem, and a complete solution was given by Leibniz in 1677.

The first analytical representation of a tangent was given by Monge in 1785.

Gaspard Monge1746 - 1818


36



- arc length differential tdtd

rd

td

rdtrdtrdsd

2/1

2/1:

'/'/:: rrtd

rd

td

rdr

sd

rdt

- unit tangent vector of path C at P (tangent to C at P)

Normal Plane to at P: t 00 trr

We also can define - arc length differential sdtrdtrdsd 2/1* :

tsd

rd

sd

rd

*

O

a

C

t

P

r

b

0r

Normal Plane 00 trr


Return to Table of Contents

37


O

a

C

t

P

r

b

Normal Plane 00 trr

0r

Curvature of curve C at P: rtsd

tdk

:

Since 01 tktttsd

tdtt

Define nnkkkkkkn

1

/1:&/:

ρ – radius of curvature of C at P k – curvature of C at P

A point on C where k = 0 is called a point of inflection and the radius of curvatureρ is infinite .

'' sttd

sd

sd

rd

td

rdr

"'"'"''

''''' 22 stskststststd

sd

sd

td

td

sdts

td

tdst

td

dr

td

dr

32 '"''''' skntstskstrr

'' sr

3

1

'''' skntrr

3'

'''

r

rrk

Let compute k as a function of and :'r

''r


38


1x

2x

3x

t

k


Example 2: Circular Helix

0,0,cossin' 321 baebetaetatd

rdr

2/1222/122

2/1

bardsdbatd

rd

td

rd

td

rd

321

2/122 cossin/: ebetaetabatd

rd

td

rdt

2122sincos/ etet

ba

a

td

sd

td

rdt

sd

tdk

1x

2x

3x

t

k

0,sincos 21 aetaetar

Example 1: Circular Curve

0,cossin' 21 aetaetatd

rdr

2/1222/122

2/1

bardsdbatd

rd

td

rd

td

rd

21 cossin/: etaetaatd

rd

td

rdt

21 sincos1

/ etetatd

sd

td

rdt

sd

tdk


Table of Content

39


O

a

C

t

Pb

ntk

1

Normal Plane

OsculatingPlane

00 trr

0r

00 ktrrOsculating Plane of C at P is the plane that contains

and P: 00 ktrrkt

,

The name “osculating plane” was introduced by D’Amondans Charles de Tinseau (1748-1822) in 1780.

O

a

C

t

Pb

ntk

1

Normal Plane

OsculatingPlane

00 trr

0r

00 ktrr

The osculating plane can be also defined as the limiting position of a plane passingthrough three neighboring points on the curve as the points approach the given point.

If the curvature k is zero along a curve C then:

tarrconstartt 00

The curve C is a straight line. Conversely if C is a straight line:

0//0 tkaatd

rd

td

rdttarr

C a regular curve of class ≥2 (Cclass) is a straight line if and only if k = 0 on C


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40


Osculating Circleof C at P is the plane that contains and P kt

,


The osculating circle of a curve C at a given point P is the circle that has the same tangent as C at point P as well as the same curvature. Just as the tangent line is the line best approximating a curve at a point P, the osculating circle is the best circle that approximates the curve at P.

http://mathworld.wolfram.com/OsculatingCircle.html

O

a

C

t

Pb

ntk

1

Normal Plane

OsculatingPlane

00 trr

0r

00 ktrr

OsculatingCircle

Osculating Circles on the Deltoid

The word "osculate" means "to kiss."

41


Osculating Circleof C at P is the plane that contains and P

kt

,

Theory of Curves (continue – 6a)

O

a

C

t

Pb

ntk

1

Normal Plane

OsculatingPlane

00 trr

0r

00 ktrr

OsculatingCircle

3xy

xy /1

xy cos xy sin http://curvebank.calstatela.edu/osculating/osculating.htm

xy tan

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42


O

a

C

t

Pb

ntk

1

Normal Plane

OsculatingPlane

00 trr

0r

00 ktrrb

RectifyingPlane

00 krr

Binormal ntb

:

Tangent Line:

Principal Normal Line:

Binormal Line:

Normal Plane:

Rectifying Plane:

Osculating Plane:

tmrr

0

nmrr

0

bmrr

0

00 trr

00 nrr

00 brr

The name binormal was introduced bySaint-Venant

Jean Claude Saint-Venant1797 - 1886

Fundamental Planes:Fundamental Lines:


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43


Torsion

Suppose that is a regular curve of class ≥ 3 (Cclass) along which is ofclass C1. then let differentiate to obtain:

srr

sn

snstsb

snstsnstsnsnksnstsnstsb

Since 001 snsnsnsnsnsnsnsn

Therefore is normal to , meaning that is in the rectifying plane, or that is a linear combination of and .

nn

t

b

sbsstssn

snssbsstsstsnstsb

O

a

C

t

Pb

n

0r

b

The continuous function τ (s) is called the second curvature or torsion of C at P.

snsbs


44


Torsion (continue – 1)

Suppose that the torsion vanishes identically (τ ≡0) along a curve , then srr

00 bsbsnssb

O

a

C

t

Pb

n

0r

0b

Since and are orthogonal st

sb

constbsrbtbsrsd

dbsr

sd

d 0000 0

Therefore is a planar curve confined to the plane srr

constbsr 0

C a regular curve of class ≥3 (Cclass) is a planar curve if and only if τ = 0 on C

1x

2x

3x

t

k


Example 2: Circular Helix

321

2/122 cossin ebetaetabat 21 sincos etetn

321

2/122

21321

2/122

cossin

sincoscossin

eaetbetbba

etetebetaetabantb

21

1222/122 sincos etbetbbabatd

bd

sd

td

td

bd

sd

bdb

122 babnb


45


Torsion (continue – 2)

Let compute τ as a function of and :'',' rr

'''r

ttrsd

td

td

rd

sd

rdr

' tbknkttrtrtrsd

dtrtr

sd

dr

2"''''

tkbkbknkbktnkbktbktbkbk

trttrtrtrttrttrtrtrtrsd

dr

2

332 '''"3''''"2"'"'

2

0

3

1

2

0

26

6

0

3

0

4

0

22

5232

32

,,,,,,'''",'

'''",'',",'''',','",','3

'''"'"''''"'3'

'''"3'"'',,

ktntkbntknntkktkbknknktrrrt

rrrtrrrttrrrttrrrtt

rrtrrttrrttrrtttr

trttrtrtrtrtrrrr

'

1

/

1

rtdsdsd

tdt

3'

'''

r

rrk

We also found:

6

2

2

6'

'''

'

'''",',,

r

rrk

r

rrrrrr

2'''

'''",'

rr

rrr


Table of Content

46


Seret-Frenet Equations


We found and snssb snskst

Let differentiate stsbsn

stsksbssnsbskstsnsstsbstsbsn

We obtain

sbsnskstst 00

sbsbsnstsksn 0

sbsnsstsb

00

or

sb

sn

st

s

ssk

sk

sb

sn

st

00

0

00

Jean Frédéric Frenet1816 - 1900

Those are the Serret – Frenet Equations of a curve.

Joseph Alfred Serret1819 - 1885

47

SOLO

Let compute:


Seret-Frenet Equations (continue – 1)


Let show that if two curves C and C* have the same curvature k (s) = k* (s) andtorsion τ (s) = τ*(s) for all s then C and C* are the same except for they position in space. Assume that at some s0 the triads and coincide.

999 ,, sbsnst 999 *,*,* sbsnst

*********

nttnknkttnkttttttsd

d kk

*************

*bnnbnttnkbtknnbtknnnnnn

sd

d kk

*********

nbbnnbbnbbbbbbsd

d

Adding the equations, we obtain: 0*** bbnnttsd

d

Integrating we obtain: 30

****** sbbnnttconstbbnntt

Since: and1,,1 *** bbnntt 3*** bbnntt

we obtain: 1*** bbnntt

Finally since: constsrsrsd

rdstst

sd

rd *

**

48

SOLO

Existence Theorem for Curves




Let k (s) and τ (s) be continuous functions of a real variable s for s0 ≤ s ≤ sf.Then there exists a curve , s0 ≤ s ≤ sf, of class C2 for which k is the curvature,τ is the torsion and s is a natural parameter.

srr

332211332211332211 ,, ebebebbenenennetetett

tnkttttttsd

d 2 nbntknnnnnn

sd

d 22

bnbbbbbbsd

d

2

with:

Proof: Consider the system of nine scalar differential equations:

3,2,1,,, isnssbsbsstsksnsnskst iiiiiii

and initial conditions: 302010 ,, esbesnest

btttktnkntntntsd

d nnbbbtkbnbnbn

sd

d

ntbnkbtbtbtsd

d

and initial conditions:

1,0,1,0,0,1000000

ssssss bbbnnnbtnttt

49

SOLO

Existence Theorem for Curves (continue – 1)




bnbbsd

dntbnkbt

sd

d

nnbbbtkbnsd

dbtttktnknt

sd

d

nbntknnsd

dtnktt

sd

d

2

222

Proof (continue – 1):

and initial conditions: 1,0,1,0,0,1000000

ssssss bbbnnnbtnttt

We obtain:

The solution of this type of differential equations with given initial conditions has a unique solution and sinceis a solution, it is unique.

1,0,1,0,0,1 bbbnnnbtnttt

The solution is an orthonormal triad.bnt

,,

We now define the curve: s

s

dtsrr0

:

We have: and , therefore k (s) is the curvature. 1 tr 1& snsnskst

Finally since: nbtttknnkntntbntb

Therefore τ (s) is the torsion of srr

q.e.d.

50

SOLO

From the previous development we can state the following theorems:




A curve is defined uniquely by the curvature and torsionas functions of a natural parameter. The equations k = k (s), τ = τ (s), which give the curvatureand torsion of a curve as functions of s are called the naturalor intrinsec equations of a curve, for they completely definethe curve. O

0s

C

t

P

n

0r

b

k

1 fs

Fundamental Existence and Uniqueness Theorem of Space Curves

Let k (s) and τ (s) be arbitrary continuous functions on s0≤s≤sf. Then there exists, for position in space, one and only one space curve C for which k (s) is the curvature, τ (s) is the torsion and s is a natural parameter along C. O

0s

C

t

P

n

b

fs

*C

0r

*

0r

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51

SOLO

Let consider a space curve C. We construct the tangent lines to every point on C and define an involute Ci as anycurve which is normal to every tangent of C.


Involute


From the Figure we can see that the equation of theInvolute is given by:

turr

1

Differentiating this equation we obtain:

11

1

1

1

sd

sdt

sd

udnkut

sd

sdt

sd

ud

sd

tdu

sd

rdt

sd

rd

Scalar multiply this equation by and use the fact that and from the definition of involute :

t

0nt

01 tt

1101

10sd

sdtt

sd

udntkutttt

01 sd

udscu

stscsrsr

1

C

iC

Or 1r

t

1t

Involute

Curve

52


Involute (continue – 1)


C

iC

Or 1r

t

1t

Involute

Curve

stscsrsr

1

n

sd

sdksc

sd

sdt

sd

tdsc

sd

rd

sd

rdt

t

111

11

and are collinear unit vectors, therefore:1t

n

kscsd

sd

sd

sdksc

11

11

The curvature of the involute, k1, is obtained from:

ksc

btk

kscsd

nd

sd

sd

sd

tdnk

sd

td nt

kscsd

sd

11

1

11

111

1

1

Hence: 22

222

1ksc

kk

For a planar curve (τ=0) we have: tsc

nk

1011

53


Involute (continue – 3)


C

iC

Or 1r

t

1t

Involute

Curve

http://mathworld.wolfram.com/Involute.html

Table of Content

54

SOLO

The curve Ce whose tangents are perpendicular to agiven curve C is called the evolute of the curve.


Evolute


11 twrbvnurr

Differentiating this equation we obtain:

11

1

1

1

sd

sdb

sd

vdn

sd

udnvbtkut

sd

sdb

sd

vdn

sd

ud

sd

bdv

sd

ndu

sd

rdt

sd

rd

Scalar multiply this equation by and use the fact that and from the definition of evolute :

t

0 btnt

01 tt

111

10sd

sdttkutttt

01 ku

ku

1

C

eCO

r

1r

t1t

Evolute

Curve

The tangent to Ce, , must lie in the plane of and since it is perpendicular to . Therefore:

n

b

t1t

1

1 sd

sdn

sd

udvb

sd

vdut

55


Evolute (continue – 1)


ccuv tantan

k

u1

C

eCO

r

1r

t1t

Evolute

Curve

We obtained:1

1 sd

sdn

sd

udvb

sd

vdut

111 // wbvnuwrrt

But:

Therefore:v

vsd

ud

u

usd

vd

or:

u

v

sd

d

vu

sd

udv

sd

vdu

1

22tan

cu

vds

s

s

1tan

0

and: bcnrr

tan1

We have one parameter family that describes the evolutes to the curve C.

56


Evolute (continue – 2)


http://math.la.asu.edu/~rich/MAT272/evolute/ellipselute.html

Evolute of Ellipse

Evolute of Logarithmic Spiralalso a Logarithmic Spiral

Evolute of Parabola

Table of Content

C

eCO

r

1r

t1t

Evolute

Curve

57


The vector defines a surface in E3 vur ,

vu

vu

rr

rrN

vur ,

vdvudur ,

rd 2rdr

udru

v

d

Nd

P

O

vudur ,

22

2

22

22

2

2

,22

1

,2

1,,,

vdudOvdrvdudrudrvdrudr

vdudOrdrdvurvdvudurvur

vvvuuuvu

The vectors and define the tangent plane to the surface at point P.

P

u u

rr

P

v v

rr

Define: Unit Normal Vector to the surface at Pvu

vu

rr

rrN

:

First Fundamental Form:

2222 22: vdGvdudFudEvdrrvdudrrudrrrdrdI vvvuuu

0

2 0,0,00:

GFFEforConditionSylvester

FEGGEvd

ud

GF

FEvdudrdrdI

Surfaces in the Three Dimensional Spaces

Table of Contents

58

SOLO

Arc Length on a Path on the Surface:

b

a

b

a

vuvu

b

a

tdvdrudrvdrudrtdtd

rd

td

rdtd

td

rdL 2/1

2/1

b

a

b

a

td

td

vd

td

ud

GF

FE

td

vd

td

udtd

td

vdG

td

vd

td

udF

td

udEL

2/1

2/122

2

Surface Area:

vur ,

rd

d

P

O

vdudFGEvdudGE

FGE

vdudrr

rrrrvdudrrrr

vdudrrrrvdudrrvdrudrd

vu

vuvuvuvu

vuvuvuvu

2

2/12

2/12

2/12

1

1,cos1

,sin

vdudFGEd 2

vur ,

rdP

O

a

b


Table of Contents

59

SOLO

Change of Coordinates vur ,

rd

d

P

O

vdrudrvdrudrd vuvu

vdudFGEvdudvu

vuJFGEvdudFGEd 222

,

,

vurvurr ,,

Change of coordinates from u,v to θ,φ

vuvv

vuuu

,

,

The coordinates are related by

v

u

vv

uu

vd

ud

vu

vu

Ivd

ud

GF

FEvdud

vd

ud

vv

uu

GF

FE

vu

vuvdud

vd

ud

GF

FEvdudI

vu

vu

vv

uu

td

td

vd

td

ud

GF

FE

td

vd

td

udtd

td

vd

td

ud

GF

FE

td

vd

td

udtd

td

rd

td

rdLd

2/12/1

2/1

vu

vuJFGE

vv

uuFGE

vv

uu

GF

FE

vu

vu

GF

FEFGE

vu

vu

vu

vu

vv

uu

,

,detdetdetdetdet 22

**

**

2

Arc Length on a Path on the Surface and Surface Area are Invariant of the Coordinates:

First Fundamental Form is Invariant to Coordinate Transformation


Table of Contents

60

SOLO

vu

vu

rr

rrN

vur ,

vdvudur ,

rd 2rdr

udru

vdr v

d

Nd

P

O

vudur ,

Second Fundamental Form: NdrdII :

22

2

2

2

2

:

vdNvdudMudL

vdNrvdudNrNrudNr

vdNudNvdrudrNdrdII

N

vv

M

uvvu

L

uu

vuvu

vdNudNNdNNdNN vu

01

NrNrNrNrNrvd

d

NrNrNrNrNrud

d

Nr

vuvuvuvuu

uuuuuuuuu

u

0

0

0

NrNrNrNrNrvd

d

NrNrNrNrNrud

d

Nr

vvvvvvvvv

vuuvuvvuv

v

0

0

0


61

SOLO

vu

vu

rr

rrN

vur ,

vdvudur ,

rd 2rdr

udru

vdr v

d

Nd

P

O

vudur ,


2

2

2: vdNrvdudNrNrudNrNdrdIIN

vv

M

uvvu

L

uu

NrNr uuuu

NrNr vuuv

NrNr

NrL

uuuu

uu

uvvu

vuuv

vuvu

NrNrM

NrNr

NrNr

NrNr

NrN

vvvv

vv

NrNr vuvu

NrNr vvvv

22 2: vdNvdudMudLNdrdII

NrL uu

NrM vu

NrN vv


62

SOLO

vu

vu

rr

rrN

vur ,

O

vdvudur ,udru

vdrv

rd


33

3

3223

22

33

3

32

,336

1

22

1

,6

1

2

1,,,

vdudOvdrvdudrvdudrudr

vdrvdudrudrvdrudr

vdudOrdrdrdvurvdvudurvur

vvvvuvvuuuuu

vvvuuuvu

IINvdudOvdNvdudMudL

NvdudOvdNrvdudNrudNr

NvdudONrdNrdNrdNr

vvvuuu

2

1,2

2

1

,22

1

,6

1

2

1

22

2

22

22

2

22

33

3

32

0


63

SOLO

N


N

N

(i) Elliptic Case (ii) Hyperbolic Case (iii) Parabolic Case

02 MNL 02 MNL

0

&0222

2

MNL

MNL


64

SOLO

Differential Geometry in the 3D Euclidean Space (continue – 6a)

vur ,

vP

ON

1nr

2nr

u

2M

1M

02 MNL

Dupin’s Indicatrix

N

1nr

2nr

P

2M

1M

02 MNL

N

1nr2nr

P

1M

2M

0

0222

2

MNL

MNL

http://www.mathcurve.com/surfaces/inicatrixdedupin/indicatrixdedupin.html

Pierre Charles François Dupin

1784 - 1873

We want to investigate the curvature propertiesat a point P.

IINvdudOvdNvdudMudLNr2

1,2

2

1 22

2

22

The expression 12 2

221

2

1 xNxxMxL

was introduced by Charles Dupin in 1813 in “Développmentsde géométrie”, to describe the local properties of a surface.


http://www.groups.dcs.st-and.ac.uk/~history/Biographies/Dupin.html


65

SOLO

N


N

(iv) Planar Case

0 MNL

3223

33

3

3223

6

1

,336

1

vdDvdudCvdudBudA

vdudOvdrNvdudrNvdudrNudrNNr vvvvuvvuuuuu

DxCxBxA 23

has 3 real rootsMonkey Saddle

DxCxBxA 23

has one real root


66

SOLO


vurvurr ,,


vuvv

vuuu

,

,


v

u

vv

uu

vd

ud

vu

vu

2222 22 uuuuuvvuuvuuuuuu vNvuMuLNvrvururNrL

vuvuvuvuvuvvvuvuvuuvvuuuvu vvNvuuvMuuLNvvrvuruvruurNrM

2222 2 vvvvvvvvvuvvvvuvuuvv vNvuMuLNvruvrvururNrN

Unit Normal Vector to the surface at Pvu

vu

vu

vu

rr

rr

rr

rrN

:

uvuuvuu vruru

vr

u

urr

vvvuvuv vrurv

vr

v

urr

IIvd

ud

NM

MLvdud

vd

ud

vv

uu

NM

ML

vu

vuvdud

vd

ud

NM

MLvdudII

vu

vu

vv

uu

Second Fundamental Form is Invariant (unless the sign) to Coordinate Transformation

Differential Geometry in the 3D Euclidean SpaceTable of Contents

67

SOLO

N

OsculatingPlane of C

at P

Principal NormalLine of C at P

Surface

t

P

k

n1

vur ,

Normal Curvature

- Length differential 2/1

rdrdrdsd

tvturr ,

Given a path on a surface of classCk ( k ≥ 2) we define:

td

rd

td

rd

sd

rdt /: - unit vector of path C at P

(tangent to C at P)

td

rd

td

td

sd

tdk /:

- curvature vector of path C at P

curvatureofradius

nnnnk

sd

tdk

111

11

1

NNkkn

: - normal curvature vector to C at P

/coscos1

:

kNnk

Nkkn

- normal curvature to C at P


68

SOLO

N

OsculatingPlane of C

at P

Principal NormalLine of C at P

Surface

t

P

k

n1

vur ,

Normal Curvature (continue – 1)

N Because C is on the surface, is on the tangent

plan normal to .t

td

NdtN

td

td

td

NdtN

td

tdNt

td

dNt

00

and

vdrudrvdrudrvdNudNvdrudr

td

rd

td

rd

td

Nd

td

rd

td

rd

td

Nd

td

rd

td

rd

td

Ndt

td

rdN

td

tdN

sd

tdNkk

vuvuvuvu

n

/

/

///

2

Gvd

udF

vd

udE

Nvd

udM

vd

udL

I

II

vdGvdudFudE

vdNvdudMudL

td

vdG

td

vd

td

udF

td

udE

td

vdN

td

vd

td

udM

td

udL

kn

2

2

2

2

2

2

2

2

22

22

22

22


69

SOLO

Normal Curvature (continue – 2)

Gvd

udF

vd

udE

Nvd

udM

vd

udL

I

II

vdGvdudFudE

vdNvdudMudL

td

vdG

td

vd

td

udF

td

udE

td

vdN

td

vd

td

udM

td

udL

kn

2

2

2

2

2

2

2

2

22

22

22

22

- kn is independent on dt therefore on C.

- kn is a function of the surface parameters L, M, N, E, F, G and of the direction .vd

ud

- Because I = E du2 + 2 F du dv + G dv2 > 0 → sign kn=sign II

- kn is independent on coordinates since I and II are independent. vur ,

rdP

O

N1Ck

2Ck

1C

2C


Table of Contents

70

SOLO

Principal Curvatures and Directions

Gvd

udF

vd

udE

Nvd

udM

vd

udL

I

II

vdGvdudFudE

vdNvdudMudLkn

2

2

2

22

2

22

22

- kn is a function of the surface parameters L, M, N, E, F, G and of the direction .vd

ud

Let find the maximum and minimum of kn as functions of the directions d u/ d v.

vur ,

rdP

O

N1Ck

2Ck

1C

2C

If this occurs for d u0/ d v0 we must have:

0&000

00

0000

00

00 ,

2

,,

2

,

vdud

vdvd

vdud

n

vdud

udud

vdud

n

I

IIIIII

v

k

I

IIIIII

u

k

0&000

00

00

00

00

0000

00

00

00

,,,

0

,,,

0

vdudvdnvd

vdud

vdvd

vdud

n

vdududnud

vdud

udud

vdud

n IkIII

IIIII

v

kIkIII

I

IIII

u

k

Multiply by I and use 00 ,

0

vdud

n I

IIk


71

SOLO

Principal Curvatures and Directions (continue – 1)

vur ,

rdP

O

N1Ck

2Ck

1C

2C

0&000

00

00

0000

00

,,

0

,,

0

vdudvdnvd

vdud

n

vdududnud

vdud

n IkIIv

kIkII

u

k

22 2: vdNvdudMudLNdrdII

22 2: vdGvdudFudErdrdI

00 220

vdFudEI ud 00 220

vdGudFI vd

00 220

vdMudLII ud 00 220

vdNudMII vd

000

00

00

,,

0

vdududnud

vdud

n IkIIu

k

000

00

00

,,

0

vdudvdnvd

vdud

n IkIIv

k

00000 0 vdFudEkvdMudL n

00000 0 vdGudFkvdNudM n


72

SOLO

We found:


vur ,

rdP

O

N1Ck

2Ck

1C

2C

0

0

0000

0000

0

0

vdGudFkvdNudM

vdFudEkvdMudL

n

n

or:

0

0

0

0

00

00

vd

ud

GkNFkM

FkMEkL

nn

nn

This equation has non-trivial solution if:

0det00

00

GkNFkM

FkMEkL

nn

nn

or expending: 02 222

00 MNLkMFLGNEkFGE nn


73

SOLO

Study of the quadratic equation:


vur ,

rdP

O

N1Ck

2Ck

1C

2C

The discriminant of this equation is:

02 222


222 42 MNLFGEMFLGNE

222

222

2

22222

2

2

22222424

E

LFLG

E

LFMFLGNEENLLFMELF

E

FGELFMELFME

E

FGE

NLFNLGEE

MLFLMGF

E

LF

E

LGFLFME

E

FLGNELFME

E

FGE 23

2

24222

2

2

2

44884424

E

LGFLG

E

LGFLMGFLG

NLGEE

LF

E

MLF

E

LGFNLF

E

LF

2222

2222

2

243222

2

24

84884

488444

024

42

2

2

2

0

2

222

LFMEE

FLGNELFME

E

FGE

MNLFGEMFLGNE


74

SOLO

Study of the quadratic equation (continue – 1):


vur ,

rdP

O

N1Ck

2Ck

1C

2C

The discriminant of this equation is:

02 222


024

42

2

2

2

0

2

222

LFMEE

FLGNELFME

E

FGE

MNLFGEMFLGNE

The discriminant is greater or equal to zero, therefore we always obtain two real solutionsthat give extremum for kn: 21

, nn kk Those two solutions are called Principal Curvatures and the corresponding two directionsare called Principal Directions 2211 ,,, vdudvdud

0&0 LGNELFME G

N

F

M

E

L

The discriminant can be zero if: 02&0 LFMEE

FLGNELFME

In this case:G

N

F

M

E

L

vdGvdudFudE

vdNvdudMudLkn

22

22

2

2 This point in which kn is constant in all directions is called anUmbilical Point.


75

SOLO

Gaussian and Mean Curvatures


vur ,

rdP

O

N1Ck

2Ck

1C

2C

Rewrite the equation:

02 222


as:

0

22

2

2

2

00

FGE

MNLk

FGE

MFLGNEk nn

We define:

2

2:

21 FGE

MFLGNEkkH nn

2

2

21:

FGE

MNLkkK nn

Mean Curvature

Gaussian Curvature

Karl Friederich Gauss1777-1855


76

SOLO

Gaussian and Mean Curvatures (continue – 1) Principal Curvatures and Directions (continue – 6)

vur ,

rd

P

O

N1Ck

2Ck

1C

2C

2

2

21:

FGE

MNLkkK nn

Gaussian Curvature

vurvurr ,,


vuvv

vuuu

,

,


v

u

vv

uu

vd

ud

vu

vu

IIvd

ud

NM

MLvdud

vd

ud

vv

uu

NM

ML

vu

vuvdudII

vu

vu

vv

uu

We found: Ivd

ud

GF

FEvdud

vd

ud

vv

uu

GF

FE

vu

vuvdudI

vu

vu

vv

uu

vu

vu

vv

uu

vv

uu

GF

FE

vu

vu

GF

FE

vu

vu

vv

uu

vv

uu

NM

ML

vu

vu

NM

ML

2

2

2

2 detdetdetdet

vu

vu

vu

vu

vv

uuFGE

vv

uu

GF

FE

GF

FEFGE

2

2

2

2 detdetdetdet

vu

vu

vu

vu

vv

uuMNL

vv

uu

NM

ML

NM

MLMNL

Therefore: invariant to coordinate changes

2

2

2

2

21:

FGE

MNL

FGE

MNLkkK nn


77

SOLO


vur ,

rdP

O

N1Ck

2Ck

1C

2CStart with:

0

0

0000

0000

0

0

vdGudFkvdNudM

vdFudEkvdMudL

n

n

rewritten as :

0

01

00000

0000

nkvdGudFvdNudM

vdFudEvdMudL

that has a nontrivial solution (1,-kn0) only if:

0det0000

0000

vdGudFvdNudM

vdFudEvdMudL

or: 02

000

2

0 vdNFMGvdudNEGLudMEFL

or: 0

0

0

2

0

0

NFMG

vd

udNEGL

vd

udMEFL


78

SOLO


vur ,

rdP

O

N1Ck

2Ck

1C

2C

We obtained:

This equation will define the two Principal Directions 2211 21& vdrudrrvdrudrr vunvun

021

21

2

2

1

1

2

2

1

1

2112212121

vdvdGMEFL

NEGLF

MEFL

NFMGE

vdvdGvd

ud

vd

udF

vd

ud

vd

udE

vdvdrrvdudvdudrrududrrrr vVvuuunn

00

0

2

0

0

NFMG

vd

udNEGL

vd

udMEFL

From the equation above we have:

MEFL

NFMG

vd

ud

vd

ud

MEFL

NEGL

vd

ud

vd

ud

2

2

1

1

2

2

1

1

Let compute the scalar product of the Principal Direction Vectors:

The Principal Direction Vectorsare perpendicular.


79

SOLO


vur ,

rdP

O

N1Ck

2Ck

1C

2C

Since the two Principal Directions are orthogonal

21 21& vdrrudrr vnun

they must satisfy the equation:

Let perform a coordinate transformation to the PrincipalDirection: vu ,

02

000

2

0 vdNFMGvdudNEGLudMEFL

21 ,0&0, vdud

or:

02

1 udMEFL

02

2 vdNFMG 0 NFMG

01

ud

0 MEFL

02

vd

0E

0G0

0

NrM

rrF

vu

vu

at P

Definition: A Line of Curvature is a curve whose tangent at any point has a directioncoinciding with a principal direction at that point. The lines of curvatureare obtained by solving the previous differential equation


80

SOLO


vur ,

rdP

O

N1Ck

2Ck

1C

2C

Suppose (du0,dv0) is a Principal Direction, then they must satisfy the equations:

Rodriguez Formula

NrNrL uuuu

NrNrNrM vuuvvu

NrNrN vvvv

0

0

0000

0000

0

0

vdGudFkvdNudM

vdFudEkvdMudL

n

n

0

0

0000

0000

0

0

vdrrudrrkvdNrudNr

vdrrudrrkvdNrudNr

vvvunvvuv

vuuunvuuu

uu rrE

vu rrF

vv rrG

0

0

0000

0000

0

0

vvunvu

uvunvu

rvdrudrkvdNudN

rvdrudrkvdNudN

0

0

0

0

vn

un

rrdkNd

rrdkNd

But are in the tangent plane at P since and are, and the vectors and are independent, therefore:

rdkNd n

0

Nd

rd

vr ur

00

rdkNd n

The direction (du0,dv0) is a Principal Direction on a point on a surface if and only iffrom some scalar k, and satisfy:00 vdNudNNd vu

00 vdrudrrd vu

rdkNd

Rodriguez Formula

We found:


Table of Contents

81

SOLO

Conjugate Directions

vur ,

rdP

O

N

Q

NdN

l

Let P (u,v) and Q (u+du,v+dv) neighboring points on a surface. The tangent planes to the surface at p and Q intersect along a straight line L. Now let Q approach P along a given direction (du/ dv=const= PQ), then the line l will approach a limit LC. The directions PQ and LC are called Conjugate Directions.

Let be the normal at P and the normal at Q.N

NdN

Let the direction of LC be given by: vrurr vu

Since LC is in both tangential planes at P and at Q we have:

0&0 NdNrNr

0 vdNudNvrurNdr vuvu

0 vdvNrvduNrudvNruduNr vvvuuvuu

We foundvvuvvuuu NrNNrNrMNrL

&&

The previous relation becomes: 0 vdvNvduudvMuduL Given (du,dv) there is only one conjugate direction (δu,δv) given by the previous equation.


Table of Contents

82

SOLO

Asymptotic Lines

The directions which are self-conjugate are called asymptotic directions.

becomes:

0 vdvNvduudvMuduL

We see that the asymptotic directions are those for which the second fundamental form vanishes. Moreover, the normal curvature kn vanishes for this direction.

Those curves whose tangents are asymptotic directions are called asymptotic lines.

v

u

vd

ud

If a direction (du,dv) is self-conjugate than and the equation of conjugate lines

02 22 vdNvdudMudL

The conjugat and asymptotic lines were introduced by Charles Dupin in 1813 in “Dévelopments de Géométrie”.

Pierre Charles François Dupin

1784 - 1873

http://www.groups.dcs.st-and.ac.uk/~history/Biographies/Dupin.html


Table of Contents

83


Scalar and Vector Fields

Let express the cartesian coordinates (x, y, z) of any point, in a three dimensional space as a function of three curvilinear coordinates (u1, u2, u3), where:

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

zyxuuzyxuuzyxuu

uuuzuuuyuuuxx

,,,,,,,,

,,,,,,,,

332211

321321321

Those functions are single valued with continuous derivatives and the correspondence between (x,y,z) and (u1,u2,u3) is unique (isomorphism).

kzjyixr

or3

3

2

2

1

1

udu

rud

u

rud

u

rrd

321 ,,,, uuuzyx 321 ,,,, uuuAzyxAA

Assume now that the scalars Φ and vectors are functions of local coordinates, cartesian (x,y,z) or general, curvilinear (u1,u2,u3)

A

In general ( we can not assume that Φ and are functions of position).

A

rAAr

,

Table of Contents

84


Vector Differentiation

tAA

Let a vector function of a single parameter tA

Ordinary Derivative of Scalars and Vectors

The Ordinary Derivative of the Vector is defined as

t

tAttA

td

tAdt

0lim

tAttAA

If the limit exists we say that is continuous and differentiable in t. tA

Differentiation FormulasIf are differentiable vector functions of a scalar t and φ is a differentiable scalar of t, then

CBA

,,

td

Bd

td

AdBA

td

d

td

BdAB

td

AdBA

td

d

td

AdA

td

dA

td

d

td

CdBAC

td

BdACB

td

AdCBA

td

d

td

BdAB

td

AdBA

td

d

td

CdBAC

td

BdACB

td

AdCBA

td

d

Table of Contents

85



Partial Derivatives of Scalar and Vectors

321 ,,,, uuuzyx 321 ,,,, uuuAzyxAA

Assume now that the scalars Φ and vectors are functions of local coordinates, cartesian (x,y,z) or general, curvilinear (u1,u2,u3)

A

The partial derivatives are defined as follows

1

3213211

01

321 ,,,,lim

,,1 u

uuuuuuu

u

uuuu

2

3213221

02

321 ,,,,lim

,,2 u

uuuuuuu

u

uuuu

3

3213321

03

321 ,,,,lim

,,3 u

uuuuuuu

u

uuuu

1

3213211

01

321 ,,,,lim

,,1 u

uuuAuuuuA

u

uuuAu

2

3213221

02

321 ,,,,lim

,,2 u

uuuAuuuuA

u

uuuAu

3

3213321

03

321 ,,,,lim

,,3 u

uuuAuuuuA

u

uuuAu

Higher derivatives are also defined

2

3

2

1

2

31

3

1212

2

2121

2

33

2

3

2

22

2

2

2

11

2

1

2

&&

&&

u

A

uuu

A

u

A

uuu

A

u

A

uuu

A

u

A

uu

A

u

A

uu

A

u

A

uu

A

Table of Contents

86



Differentials of Vectors

3

3

2

2

1

1

udu

Aud

u

Aud

u

Azd

z

Ayd

y

Axd

x

AAd

If 321321 111111,,,,321

uAuAuAzAyAxAuuuAzyxAA uuuzyx

321321 111111111321321

udAudAudAuAduAduAdzAdyAdxAdAd uuuuuuzyx

BdABAdBAd

BdABAdBAd

CdBACBdACBAdCBAd

CdBACBdACBAdCBAd

then

If are differentiable vector functions of a scalar t.CBA

,,

Table of Contents

87


The Vector Differential Operator Del (, Nabla)

We define the Vector Differential Operator Del (, Nabla) in Cartesian Coordinates as:

zz

yy

xx

111:

This operator has double properties: (a) of a vector, (b) of a differential

Gradient: Nabla operates on a Scalar or Vector Field

zz

yy

xx

zz

yy

xx

111111:

zzz

Ayz

z

Axz

z

A

zyy

Ayy

y

Axy

y

A

zxx

Ayx

x

Axx

x

A

zAyAxAzz

yy

xx

A

zyx

zyx

zyx

zyx

111111

111111

111111

111111:

a scalar

a dyadic

88


The Vector Differential Operator Del (, Nabla) (continue)

We define the Vector Differential Operator in Cartesian Coordinates as:

zz

yy

xx

111:

This operator has double properties: (a) of a vector, (b) of a differential

Divergence: Nabla performs a Scalar Product on a Vector Field

z

A

y

A

x

AzAyAxAz

zy

yx

xA zyx

zyx

111111:

Curl (Rotor): Nabla performs a Vector Product on a Vector Field

zy

A

x

Ay

x

A

z

Ax

z

A

y

A

AAA

zyx

zyx

zAyAxAzz

yy

xx

A

xyzxyz

zyx

zyx

111

111

111111:

Table of Contents

89


Scalar Differential

Let find the differentials of: 321 ,,,, uuuzyx

3

3

2

2

1

1

udu

udu

udu

zdz

ydy

xdx

d

rdzzdyydxxdzz

yy

xx

d

111111

Since zyxuuzyxuuzyxuu ,,,,,,,, 332211

We obtain rduudrduudrduud

332211 ,,

rduu

uu

uu

udu

udu

udu

d

3

3

2

2

1

1

3

3

2

2

1

1

Comparing with we obtainrdd

3

3

2

2

1

13

3

2

2

1

1 uu

uu

uuu

uu

uu

u

Using the Gradient definition: zz

yy

xx

111:

or

3

3

2

2

1

1:u

uu

uu

u

in general curvilinear coordinatesTable of Contents

90


Vector Differential

Let find the differentials of: 321 ,,,, uuuAzyxAA

3

3

2

2

1

1

udu

Aud

u

Aud

u

Azd

z

Ayd

y

Axd

x

AAd

ArdzAyAxAz

zdy

ydx

xd

zdzz

Ay

z

Ax

z

Aydz

y

Ay

y

Ax

y

A

xdzx

Ay

x

Ax

x

AAd

zyx

zyxzyx

zyx

111

111111

111

ArdAu

uu

uu

urdrduu

Au

u

Au

u

AAd

3

3

2

2

1

13

3

2

2

1

1

rduudrduudrduud

332211 ,,

and3

3

2

2

1

1:u

uu

uu

u

ArdAd

Therefore

In Cartesian Coordinates:

In General Curvilinear Coordinates using

Table of Contents

91

Vector AnalysisSOLO

Linearity of operator

Differentiability of operator

BABA


BABA


AAA


AAA


Differential Identities

92

Vector AnalysisSOLO


BAAB

BAAB

BABABA

BA

BA

AAAcbacabcba

2

00

aa

00

baabaa

A

93

Vector AnalysisSOLO


ABBAABBABA

BABABA B

ABBAAB A

ABBABABABAAB

BA

BA

BAABABBABA

BABABA BA

ABABBA AAA

BABABA BBB

94

Vector AnalysisSOLO

Differential Identities Summary

BABA

BABA

AAA

AAA

ABBAABBABA

BAABABBABA

BAABBA

AAA

2

0

0 A

AAAAA

2/

2

Table of Contents

95


Curvilinear Coordinates in a Three Dimensional Space

Let express the cartesiuan coordinates (x, y, z) of any point, in a three dimensional space as a function of three curvilinear coordinates (u1, u2, u3), where:

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

zyxuuzyxuuzyxuu

uuuzuuuyuuuxx

,,,,,,,,

,,,,,,,,

332211

321321321


kzjyixr

3

3

2

2

1

1

3

333

2

222

1

111

3

3

12

2

1

1

3

3

12

2

1

1

3

3

12

2

1

1

udu

rud

u

rud

u

r

udku

zj

u

yi

u

xudk

u

zj

u

yi

u

xudk

u

zj

u

yi

u

x

kudu

zd

u

zud

u

zjud

u

yd

u

yud

u

yiud

u

xd

u

xud

u

x

kzdjydixdrd

or3

3

2

2

1

1

udu

rud

u

rud

u

rrd

96


Curvilinear Coordinates in a Three Dimensional Space (continue – 1)

3

3

2

2

1

1

udu

rud

u

rud

u

rrd

Let define: 3,2,1:1

iu

rr

iu

If and are linear independent (i.e. if and only if αi = 0 i=1,2,3) then they form a base of the space E3.

21, uu rr

3ur 0

3

1

i

ui ir

We have also: 3,2,1,,1 irdzyxuud i

We can write: 3

1

2

1

1

11

1 321,, udurudurudurrdzyxuud uuu

Because du1, du2, du3 are independent increments the precedent equation requires:

001 111

321 ururur uuu

Similarly by multiplying by and we obtain:rd

2u 3u

ji

jiuru

u

r j

i

j

u

j

i 0

1

1

Therefore and are reciprocal systems of vectors.321

,, uuu rrr

321 ,, uuu

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

97



We proved that reciprocal systems of vectors are related by:

and are reciprocal systems of vectors.321

,, uuu rrr

321 ,, uuu

321

21

321

13

321

32

,,,

,,,

,,321

uuu

uu

uuu

uu

uuu

uu

rrr

rru

rrr

rru

rrr

rru

321

21

321

13

321

32

,,,

,,,

,, 321 uuu

uur

uuu

uur

uuu

uur uuu

and 1,,,, 321

321 uuurrr uuu

or

1,,

,,

,,

,, 321

321

333

222

111

333

222

111

zyx

uuuJ

uuu

zyxJ

z

u

y

u

x

u

z

u

y

u

x

u

z

u

y

u

x

u

u

z

u

y

u

x

u

z

u

y

u

x

u

z

u

y

u

x

where is the Jacobian of x,y,z with respect to u1, u2, u3.

321 ,,

,,

uuu

zyxJ Carl Gustav Jacob Jacobi

1804 - 1851

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

98



grrr

u

z

u

y

u

x

u

z

u

y

u

x

u

z

u

y

u

x

uuu

zyxJ uuu

321

,,det:,,

,,

333

222

111

321

If is nonsingular the transformation from x,y,z to u1, u2, u3 is unique.

321 ,,

,,

uuu

zyxJ

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

g

rru

g

rru

g

rru uuuuuu 211332 321 ,,

211332

321,, uugruugruugr uuu

Table of Contents

99


Covariant and Contravariant Components of a Vector in Base .

321,, uuu rrr

Given a vector we have: 321 ,, uuuA

j

u

j

j

u

i

u

i

uuu

urAuAuAuAuA

ruArArArArAA

j

ii

3

3

2

2

1

1

321

321

where:

juj

ii

rAA

uAA

:

: are the contravariant components of A

are the covariant components of A

The Element of Arc ds. The Metric Coefficients gij Riemann and Euler Spaces

Compute: jiuujuiu ududrrudrudrrdrdsdjiji

2

Define:

jiuuuuij grrrrgijji

the metric coefficients

jijijiij ududgududgrdrdsd 2

the element of arc

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

Leonhard Euler1707- 1783 Georg Friedrich Bernhard

Riemann1826 - 1866 A space with the metric defined above is called a

Riemann Space. If gij = δij then the space is called anEuler Space.

100


Covariant and Contravariant Components of a Vector in Base (continue -1).

321,, uuu rrr

or:

If we substitute for we obtain:A

jur

j

ij

j

uu

j

ju ugurruArjii

3

2

1

333231

232221

131211

3

2

1

u

u

u

ggg

ggg

ggg

r

r

r

ugr

u

u

u

j

iju i

Multiplying by gik (where gik gji = δjk) and summing on i and k:

kjk

j

j

ij

ik

u

ik uuuggrgj

Changing k by j we obtain:

3

2

1

333231

232221

131211

3

2

1

u

u

u

u

ijj

r

r

r

ggg

ggg

ggg

u

u

u

rgui

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

101


Covariant and Contravariant Components of a Vector in Base (continue -2).

321,, uuu rrr

Multiplying equation by we get:iu

ijj rgu

iu

ij

u

iijji gruguui

1

jiij uug or

Now: j

ij

j

ij

j

ijui AguAgugArAAj

j

iji AgA

We also found:

321321

,,,,det 2

333231

232221

131211

uuuuuu rrrgrrr

ggg

ggg

ggg

g

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

Table of Contents

102


Coordinate Transformation in Curvilinear Coordinates

from those equations we obtain:

32133

32122

32111

32133

32122

32111

,,

,,

,,

,,

,,

,,

uuuuu

uuuuu

uuuuu

uuuuu

uuuuu

uuuuu

Let and be two general curvilinear coordinates in an E3 space. There exists a unique transformation from one curvilinear coordinates to the other:

321 ,, uuu 321 ,, uuu

k

k

j

k

k

k

j

jj

j

i

j

j

j

i

i udu

uud

u

uudud

u

uud

u

uud

3

1

3

1

k

k

j

j

i

j k

k

k

j

j

i

j

j

j

i

i udu

u

u

uud

u

u

u

uud

u

uud

3

1

3

1

3

1

Because dui and duk are independent:

ki

ki

u

u

u

u i

k

k

j

j

i

0

1

In the same way: k

k

i

i

j

i

i

j

j udu

u

u

udu

u

uud

therefore:

kj

kj

u

u

u

uj

k

k

i

i

j

0

1

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

103


Coordinate Transformation in Curvilinear Coordinates (continue – 1)

If: 321321 ,,,, uuuruuurr

j

i

ij u

u

u

r

u

r

ij u

j

i

u ru

ur

and:

i

j

ji u

u

u

r

u

r

ji u

i

j

u ru

ur

Let be a given vector with respect to two curvilinear coordinates.A

ji u

j

j

ii

u

i rAu

uArAA

i

j

i

ji

u

i

j

i

u

i

ji

u

i

i

j

j

uAu

uurA

u

u

uru

uAurAuAuAA

j

ji

i

j

ij Au

uA

This is a contravariant relation with respect to the reference relation .ij u

j

i

u ru

ur

j

i

j

i Au

uA

This is a covariant relation with respect to the reference relation . ij u

j

i

u ru

ur

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

104


Coordinate Transformation in Curvilinear Coordinates (continue – 2)

Let define: j

i

j

i

i Au

uAuA

11:

By multiplying the last equation by and because then and:

1

i

j

j

i

u

u

u

u

j

i

u

u

j

i

j u

uA

j

j

ij

j

i uu

uuAuA

: j

j

ii uu

uu

From the reference relationmjki u

j

m

uu

i

k

u ru

urr

u

ur

&

mkmkji uu

j

m

i

k

u

j

m

u

i

k

uuij rru

u

u

ur

u

ur

u

urrg

kl

j

m

i

k

ij gu

u

u

ug

This is a two order covariant relation with respect to the reference relation .ij u

j

i

u ru

ur

In the same way m

m

jjk

k

ii uu

uuu

u

uu

&

mk

m

j

k

im

m

jk

k

ijiij uuu

u

u

uu

u

uu

u

uuug

km

m

j

k

iij gu

u

u

ug

This is a two order contravariant relation with respect to the reference relation .ij u

j

i

u ru

ur

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

Table of Contents

105


Covariant Derivative

First we want to find the derivatives and .j

u

u

ri

j

i

u

u

Because are vectors of a base in E3 we can write as a function of this base.

j

u

u

ri

3,2,1iriu

i

i

uijk

ijj

ur

u

r

uu

r

:

ij

kij

kEquivalent notation

Where are the Cristoffel’s Symbols of II kind that we must determine.ijk

Elwin Bruno Cristoffel 1829 - 1900

Because thenk

ji

k ujik

i

u

jiijj

u

uijk r

u

r

u

r

uu

r

uu

rr

ji

kij

k

Let calculate now mjigrrru

rmijkmij

k

uuijk

u

j

u

mkm

i ,: ,

where are the Cristoffel’s Symbols of I kind kmijk

u

j

u

mij gru

rmji

m

i

:,,

Because andjik

ijk mkkm gg kjikij ,,

106


Covariant Derivative (continue – 1)

To find let perform the following calculations:kij ,

ijkjki

k

u

uu

k

u

uu

kk

ij

u

rrr

u

rrr

uu

gj

ij

i

ji ,,

ijkkji

j

u

uu

j

u

uu

jj

ik

u

rrr

u

rrr

uu

gk

ik

i

ki ,,

jikkji

i

u

uu

i

u

uu

ii

jk

u

rrr

u

rrr

uu

gk

jk

j

kj ,,

From those equations we obtain:

k

ij

i

jk

j

ikkij u

g

u

g

u

g

2

1,

k

ij

i

jk

j

ikkm

kij

kmij

m

u

g

u

g

u

ggg

2,

Multiplying the equations by and summing we obtain:kmijm

kij g ,

kmg

The Operator .

107



Now let find . j

i

u

u

Because are vectors of a base in E3 we can write as a function of this base:

3,2,1 iu i

j

i

u

u

Tacking the derivative with respect to uj of the equation we get:k

i

k

u uri

0

j

k

u

k

j

u

u

uru

u

ri

i

or ijkk

mijmk

uijmk

j

u

j

k

u uruu

r

u

ur

m

i

i

kjk

i

j

i

uu

u

:

But we have also ijk

ijk

iuijk

j

k

u uru

ur

ii

Therefore: iij

k

j

k

uu

u

Because we have:mu

imi rgu

iuij

kim

j

k

rgu

u

108



We found that: m

jkim

i

kmjmikjjki

k

ij ggu

g

,,

Let find , where .k

ij

u

g

jiij uug

j

km

iki

km

mjkj

km

ijmi

km

k

jij

k

i

k

ij

gguuuuu

uuu

u

u

u

g

j

km

iki

km

mj

k

ij

ggu

g

We can see that:

0

j

kj

i

ki

j

jk

i

ki

j

km

ik

ij

i

km

mj

ij

m

jkim

iji

kmjm

ij

k

ijij

k

ij

ij

kj

mi

jm

im

ggggggggu

gg

u

gg

This can be proven if we take the derivative with respect to uk of the equation:

1ij

ij ggTable of Contents

109

SOLO

or

Vectors & Tensors in a 3D Space

Covariant Derivative of a Vector .

A

j

ju

i uArAAi

iimi

i

i u

i

mk

i

u

k

i

mi

u

m

ik

i

u

k

i

k

ui

u

k

i

k

rAru

ArAr

u

A

u

rAr

u

A

u

A

ji

mk

i

k

i

iju

i

mk

i

k

i

k

uAu

AgrA

u

A

u

Ai

But we have also

or ji

mk

i

k

i

ij

jm

jkj

k

j

k

uAu

AguA

u

A

u

A

jm

jkj

j

k

j

jm

mj

mkj

j

k

j

k

j

j

j

k

j

k

uAuu

AuAu

u

A

u

uAu

u

A

u

A

Therefore we can write:

i

mk

i

k

i

ij

m

jkj

k

j Au

AgA

u

A

By multiplying the equation by gij and summing we obtain:

m

jkj

k

jiji

mk

i

k

i

Au

AgA

u

A

Table of Contents

110

Vector AnalysisSOLO

Dyadic Identities Summary

CbaCabCba��

CbaCabCba��

CCC��

CCC��

CCC��

2

0 C�

aCCa T ��

TT aCCa��

BCaBaCTT

��

a

Chen-To Tai, "Dyadic Green Function in Electromagnetic Theory",2nd Ed., pg.298

111

SOLO

Vector Integration

Vector Analysis

Ordinary Integration of Vectors Let be a vector depending on the single scalar variablet, with Ax (t), Ay (t), Az (t) continuous in a specific interval, then

ztAytAxtAtA zyx 111

zdttAydttAxdttAdttA zyx 111

If there exists a vector such that then: tStd

dtA

tS

ctSdttStd

ddttA

where is an arbitrary constant.c

The definite integral between t = a and t = b gives

aSbSctSdttStd

ddttA

b

a

b

a

b

a

Table of Contents

112

SOLO

Vector Integration

n

i

iiii

n

i

iiiiin trzyxAtrtrzyxAS11

1 ,,,,

C

1t

ntb

2t

0ta

1it

it

1

2

i

n

itr

Let subdivide C into n parts by n arbitrary pointst1, t2,…,tn, and call a=t0 and b=tn. On each arc joining ti-1 to zi choose a point ξi. Define the sum:

C

b

a

n

i

iiiiznnrdzyxArdzyxAtrzyxAS

i

,,,,,,limlim

10

Properties of Integrals

CCC

rdBrdArdBA

constantrdArdACC

a

b

b

a

rdArdA

b

c

c

a

b

a

rdArdArdA

Vector Analysis

Line Integrals Let be continuous at all points on a curve C of a finite length L, defined by the position vector . tr

zyxA ,,

Let the number of subdivisions n increase in such away that the largest of approaches zero, then the sum approaches a limit that is called the line integral (also Riemann-Stieltjes integral).

itr

Georg Friedrich BernhardRiemann

1826 - 1866

Table of Contents

113

SOLO

Vector Integration

Vector Analysis

Surface Integrals

vur ,

rd

sd

P

O

vdrudrvdrudrsd vuvu

Let be continuous at all points on a surface S of a finite area A, defined by the position vector . vur ,

zyxA ,,

A surface integral over the vector field is defined asA

S

vu

S

sdnsd

S

vdudrrvuxvuyvuxAsdnAsdA

,,,,,ˆˆ

Table of Contents

114

SOLO

Vector Integration

Vector Analysis

Volume Integrals

Let be continuous at all points on a finite volume V, defined by the position vector . 321 ,, uuur

321 ,,,, uuuAzyxA

A volume integral over the vector field is defined asA

V

uuu

V

udududrrruuuAdvA 321321 321,,,,

where

321

333

222

111

,,

,,det,,

321321 uuu

zyxJ

u

z

u

y

u

x

u

z

u

y

u

x

u

z

u

y

u

x

rrrrrr uuuuuu

dr

constu 3

i

j

k

1

1

udu

r

2

2

udu

r

3

3

udu

r

constu 1

constu 2

curveu1

curveu2

curveu3

Table of Contents

115

SOLO

Simply and Multiply Connected Regions

A region R is called simply-connected if any simple closed curve Γ, which lies in Rcan be shrunk to a point without leaving R. A region R that is not simply-connectedis called multiply-connected.

C0

x

y

R C1

C0

x

y

RC1

C2

C3

C

x

y

R

C

x

y

Rsimply-connected

multiply-connected.

Vector Integration

Vector Analysis

Table of Contents

116

SOLO

Green’s Theorem in the Plane

C

R

Let P (x,y) and Q (x,y) be continuous and have continuous partial derivatives in a

region R and on the boundary C.

Green’s Theorem states that:

R

dydxy

P

x

QdyQdxP

C

Vector Integration

Vector Analysis

http://en.wikipedia.org/wiki/George_Green

This Theorem was first published by George Green (1793 – 1841) in 1828 in a paper “An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism”.

117

SOLO

Green’s Theorem in the Plane

Vector Integration

Vector Analysis

Lord Kelvin rediscovered his work four years after his death and gave it wide publicity. Kelvin, James Clerk Maxwell, George Gabriel Stokes and others built on his pioneering work and Green gained a posthumous reputation amongst 19th- and 20th-century mathematicians and scientists. His work has had great influence and nowadays he is remembered principally for Green’s theorem in vector analysis, Green’s tensor (or the Cauchy-Green tensor) in elasticity theory and above all for Green’s functions for solving differential equations.

George Green (1793-1841) was one of the most remarkable of nineteenth century physicists, a self-taught mathematician whose work has contributed greatly to modern physics. He was a pioneer in the application of mathematics to physical problems. He had very little formal education and died without achieving any recognition among other mathematicians.

http://www.historyoftheuniverse.com/george_green/store.htm#Vol_1_paper

George Green1793-1841tomb stone

118

SOLO

Proof of Green’s Theorem in the Plane C

R

P

T

S

Q

a bx

y

xgy 2

xgy 1

Start with a region R and the boundary curve C, definedby S,Q,P,T, where QP and TS are parallel with y axis.

b

a

xgy

Xgy

dyy

Pdxdydx

y

P2

R

By the fundamental lemma of integral calculus:

xgxPxgxPyxPdy

y

yxP xgy

xgy

xgy

Xgy

12 ,,,, 2

1

2

Therefore: b

a

b

a

dxxgxPdxxgxPdydxy

P12 ,,

R

but: a

bSQ

dxxgxPdxxgxP 22 ,, integral along curve SQ

b

aPT

dxxgxPdxxgxP 11 ,, integral along curve PT

If we add to those integrals: 00,, dxsincedxyxPdxyxPQPTS

we obtain:

CTSPTQPSQ

dxyxPdxyxPdxxgxPdxyxPdxxgxPdydxy

P,,,,, 12

R

Assume that PT is defined by the function y = g1 (x) and SQ is defined by the function y = g2 (x), both smooth and

y

P

is continuous in R:

Vector Integration

Vector Analysis

119

SOLO

Proof of Green’s Theorem in the Plane (continue – 1)

C

dxyxPdydxy

P,

R

In the same way:

C

dyyxQdydxx

Q,

R

Therefore we obtain:

R

dydxy

P

x

QdyQdxP

C

The line integral is evaluated by traveling C counterclockwise.

For a general single connected region, as that described in Figure to the right, can be divided in a finite number of sub-regions Ri, each of each are of the type described in the Figure above. Since the adjacent regions boundaries are traveled in opposite directions, there sum is zero, and we obtain again:

R

dydxy

P

x

QdyQdxP

C

C

R4

x

yR

R3

R1

R2

C

R

P

T

S

Q

a bx

y

xgy 2

xgy 1

Vector Integration

Vector Analysis

120

SOLO

Proof of Green’s Theorem in the Plane (continue – 2)

The general multiply-connected regions can be transformed in a simply connected region by infinitesimal slits

C0

x

y

R C1

P0

P1

C0

x

y

RC1

C2

C3

R

dydxy

P

x

QdyQdxPdyQdxP

i CC i0

All line integrals are evaluated by traveling Ci i=0,1,… counterclockwise.

Since the slits boundaries are traveled in opposite directions, there integral sum is zero:

00

1

1

0

P

P

P

P

dyQdxPdyQdxP

We obtain:

Vector Integration

Vector Analysis

Table of Contents

121

SOLO

Stoke’s Theorem

C

R

Let P (x,y) and Q (x,y) be continuous and have continuous

partial derivatives in a region R and on the boundary C.

Green’s Theorem states that:

GEORGE STOCKES1819-1903

A more general theorem was given by Stokes

R

dydxy

P

x

QdyQdxP

C

yzxzxy RRR

dzdyz

Q

y

Rdzdx

x

R

z

Pdydx

y

P

x

QdzRdyQdxP

C

or in vector form: S

dAFdrFC

where: zzyxRyzyxQxzyxPzyxF 1,,1,,1,,,,

zdzydyxdxdr 111

zdydxydzdxxdzdydA 111

GEORGE GREEN1793-1841

zz

yy

xx

111

Vector Integration

Vector Analysis

122

Proof of Stoke’s Theorem

SOLO


Vector Analysis

AB

C

D

v

u duu

dvv

Constantv curves

Constantu curves

vur ,

vuA ,

C

Consider a surface in a 3 dimensional space, defined by two parameters u and v and boundedby a curve Γ. Let choose four points on this surface:

vurA ,: vduurB ,:

dvvurD ,: dvvduurC ,:

Consider also a vector, function of the position: vuA ,

The vector at the four points, A, B, C, D, is given by:

vuAAA ,

vuAdu

u

rvuAvuArdvuAAdvuAvduuABA uu ,,,,,,

vuAdvv

rvuAvuArdvuAAdvuAdvvuADA vv ,,,,,,

vuAdvv

rvuAdu

u

rvuAAdAdvuAdvvduuACA vu ,,,,,

where du, dv are infinitesimals (differentials)

123

Proof of Stoke’s Theorem (continue – 1)

SOLO Vector Analysis

vuAAA ,

vuAdu

u

rvuABA ,,

vuAdvv

rvuADA ,,

vuAdvv

rvuAdu

u

rvuACA ,,,

A

B

C

D

v

u duu

dvv

Constantv curves

Constantu curves

vur ,

vuA ,

C

Let compute the path integral:

dvduu

rAdv

v

r

v

rA

u

r

dvv

rAdv

v

rAdu

u

rAdv

v

rAdu

u

rA

dvv

rAdv

v

rAdu

u

rAdu

u

rAdu

u

rA

drAADA

drDACA

drCABA

drBAAA

drA DACDBCAB

ABCD

2

1

2

1

2

1

2

1

2222

124

Proof of Stoke’s Theorem (continue – 2)


AB

C

D

v

u duu

dvv

Constantv curves

Constantu curves

C

vur ,

vuA ,

Let compute:

dvduu

rA

v

r

v

rA

u

rdrA

ABCD

u

rA

v

r

v

rA

u

r

v

r

u

rAA

u

r

v

r

u

rA

v

r

u

rA

A

Therefore: sdAdvduv

r

u

rAdrA

ABCD

We identify as the vector describing the surface ABCD sinceit is normal to surface and the aria is equal to

dvduv

r

u

rsd

:

dvduv

r

u

r

Let sum over the entire u,v network. Interior line integrals will cancel out in pairsleaving only , and finally we obtain:

C

drA

SC

sdAdrA

Table of Contents

125

Divergence Theorem

This therem is also known as Gauss’ Theorem, Ostrogradsky’s Theorem or Gauss-Ostrogradsky Theorem.

V

A

ds


JOSEPH-LOUIS LAGRANGE

1736-1813

The theorem was first discovered by Lagrange in 1762, than laterrediscovered by Carl Friedrich Gauss in 1813,by George Green in

1825 ,and in 1831 by Michail Vasilievich Ostrogradsky who gavethe first proof.

MIKHAIL VASILIEVICH OSTROGRADSKI

1801-1862

GEORGE GREEN1793-1841tomb stone

http://en.wikipedia.org/Divergence_theorem

JOHANN CARL FRIEDRICH GAUSS

1777-1855

S V

dvAsdA

126

Proof of Divergence Theorem


S V

dvAsdA

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

3

2

2

1

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

321 ,, uuu

1ur

2ur

3ur

S4

S1S2

S5

S3

S6

Fr

321 ,, uuu

dV is any volume, that includes the point (u1,u2,u3) and is closed by the surface S=S1+S2+S3+S4+S5+S6

654321 SSSSSSS

AdsAdsAdsAdsAdsAdsAds

where

32

32

3

3

2

2

41 ududu

r

u

rud

u

rud

u

rdsds

31

13

1

1

3

3

52 ududu

r

u

rud

u

rud

u

rdsds

21

21

2

2

1

1

63 ududu

r

u

rud

u

rud

u

rdsds

321

321

3

3

2

2

1

1

3211

1

321 2,,

2,,

41

udududu

r

u

r

u

A

udu

rud

u

rud

u

AuuuA

ud

u

AuuuAAdsAds

SS

321

132

1

1

3

3

2

2

3212

2

321 2,,

2,,

52

udududu

r

u

r

u

A

udu

rud

u

rud

u

AuuuA

ud

u

AuuuAAdsAds

SS

321

213

2

2

1

1

3

3

3213

3

321 2,,

2,,

63

udududu

r

u

r

u

A

udu

rud

u

rud

u

AuuuA

ud

u

AuuuAAdsAds

SS

127


2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

3

2

2

1

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

321 ,, uuu

1ur

2ur

3ur

S4

S1S2

S5

S3

S6

Fr

321 ,, uuu

654321 SSSSSSS

AdsAdsAdsAdsAdsAdsAds

321

213132321

udududu

r

u

r

u

A

u

r

u

r

u

A

u

r

u

r

u

A

321

321

udududu

r

u

r

u

rvd

Proof of Divergence Theorem (continue – 1)

321

21

321

13

321

32

,,,

,,,

,,321

uuu

uu

uuu

uu

uuu

uu

rrr

rru

rrr

rru

rrr

rru

Using the relations:

VS

vdu

Au

u

Au

u

AuAds

3

3

2

2

1

1

We obtain:

We also obtained the definition of Nabla ( ):3

3

2

2

1

1:u

uu

uu

u

from which: VS

vdAAds

q.e.d.

Table of Contents

128

VECTOR NOTATION CARTESIAN TENSOR NOTATION

Gauss’ Theorem Variations

A analytic in V

A C C const vector .

S V

dvsdGAUSS

2 analytic in V S V k

k dvs

ds

S V

dvAsdAGAUSS

1 S V k

kkk dv

x

AdsA


V

A

ds


129


S V

dvAsdAGAUSS

3

V

dvAA

,A

analytic in V

S V k

k

kk dvx

AdsA

V k

k

k

k dvx

A

xA

B e e e 1 1 2 2 3 3

S V

dvABBAsdABGAUSS

4

S V k

ki

k

ikkki dv

x

AB

x

BAdsAB

A analytic in V

S V

dvAAsdGAUSS

5

S V j

i

i

j

ijji dvx

A

x

AAdsAds


Table of Contents

Gauss’ Theorem Variations (continue)

130

CS

A

ds

dr


Stokes’ Theorem Variations

SC

sdArdAStokes

1

A analytic on S

S

kj

i

i

j

C

ii sdx

A

x

ArdA



SSC

sdAAsdArdAStokes

2

SC

sdrdStokes 3

AsdsdAA const

131


vectorconstCCAA .

SC

sdCArdCA

CsdAnCsdAnnAC

sdACsdACArdC

SS

sdnsd

SSC

ˆˆˆˆ

A d r A d s

C S

SC

sdAnArdStokes

ˆ4

ACACCAconstC

ArdCrdCA

AnnAAn

AnnAAn

A

A

ˆˆˆ

ˆˆˆ AnAnAnAn

ˆˆˆˆ

SSC

sdAnAnAnsdAnArdStokes

ˆˆˆˆ4

Stokes’ Theorem Variations (continue - 1)

a

See H.Lass "Vector and Tensor Analysis", pp.112

132

GAUSS’ AND STOKES’ THEOREMS ARE GENERALIZATIONS OF THEFUNDAMENTAL THEOREM OF CALCULUS

A b A a

d A x

d xd x

a

b

( ) ( )


SC

sdAnArdStokes

ˆ4

Use

with rA

nnnrnrnrn r ˆ2ˆ3ˆˆˆˆ

3

therefore

SC

sdnrdrStokes ˆ2

15

Stokes’ Theorem Variations (continue - 2)

Table of Contents

a

See H.Lass "Vector and Tensor Analysis", pp.112

133

SOLO

GREEN’s IDENTITIES

Start from Gauss’ Theorem that relates the integral of the flux of a union of closed surfaces to it’s divergence.

n

iiSS

1

S

GAUSS

V

dSnFGdvFG 1

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

and must be continuous and twice differentiable in V.G

F

Using the identity we obtain

FGFGFG

S

GAUSS

V

dSnFGdvFGFG 1

First Vector Green Identity

Interchanging and we obtainG

F

S

GAUSS

V

dSnGFdvGFFG 1

By subtracting the second identity from the first we obtain

SV

dSnGFFGdvFGGF 1

Second Vector Green Identity

Vector Analysis


Table of ContentsHarmonic

134

SOLO

Derivation of Nabla ( ) from Gauss’ Theorem

Start from Gauss’ Theorem that relates the integral of the flux of a union of closed surfaces to it’s divergence.

n

iiSS

1

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

Vector Analysis

,,1

lim:

1lim,

1lim,

1lim,

0

0

0

0

SV

SVF

SVF

SVF

dsV

AdsV

trA

AdsV

trA

dsV

tr

SVF

SVF

SVF

AdsV

trA

AdsV

trA

dsV

tr

1lim,

1lim,

1lim,

0

0

0

S

GAUSS

V

S

GAUSS

V

S

GAUSS

V

AdsvdA

AdsvdA

dsvd

5

1

2

Table of Contents

135


The Operator .

,,1

lim:

1lim,

1lim,

1lim,

0

0

0

0

SV

SVF

SVF

SVF

dsV

AdsV

trA

AdsV

trA

dsV

tr

We know that

where V is any volume, that includes the point and is closed by the surface SFr

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

3

2

2

1

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

321 ,, uuu

1ur

2ur

3ur

S4

S1S2

S5

S3

S6

Fr

321 ,, uuu

Let apply those definitions to the infinitesimal volumein the figure, having the point at it’s centerFr

where

321

321

3

3

2

2

1

1

udududu

r

u

r

u

rud

u

rud

u

rud

u

rV

SVF ds

Vtr

1lim,

0

Gradient

V is any volume, that includes the point (u1,u2,u3) and is closed by the surface S=S1+S2+S3+S4+S5+S6

136

SOLO Vectors & Tensors in a 3D SpaceThe Operator (continue – 1)

where

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

321 ,, uuu

1ur

2ur

3ur

S4

S1S2

S5

S3

S6

Fr

321 ,, uuu

654321 SSSSSSS

dsdsdsdsdsdsds

32

32

3

3

2

2

41 ududu

r

u

rud

u

rud

u

rdsds

31

13

1

1

3

3

52 ududu

r

u

rud

u

rud

u

rdsds

21

21

2

2

1

1

63 ududu

r

u

rud

u

rud

u

rdsds

321

321

3

3

2

2

1

1

3211

1

321 2,,

2,,

41

udududu

r

u

r

u

udu

rud

u

rud

uuuu

ud

uuuudsds

SS

321

132

1

1

3

3

2

2

3212

2

321 2,,

2,,

52

udududu

r

u

r

u

udu

rud

u

rud

uuuu

ud

uuuudsds

SS

321

213

2

2

1

1

3

3

3213

3

321 2,,

2,,

63

udududu

r

u

r

u

udu

rud

u

rud

uuuu

ud

uuuudsds

SS

137


or

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

3

2

2

1

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

321 ,, uuu

1ur

2ur

3ur

S4

S1S2

S5

S3

S6

Fr

321 ,, uuu

654321 SSSSSSS

dsdsdsdsdsdsds

321

213132321

udududu

r

u

r

uu

r

u

r

uu

r

u

r

u

321

321

udududu

r

u

r

u

rV

321

321213132

0

1lim

u

r

u

r

u

r

uu

r

u

r

uu

r

u

r

uu

r

u

r

dsV

SV

3

3

2

2

1

1 uu

uu

uu

To the same result we could arrive using:

rduu

uu

uu

rduu

rduu

rduu

udu

udu

udu

rdd

3

3

2

2

1

1

3

3

2

2

1

1

3

3

2

2

1

1

Gradient

138


or

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

3

2

2

1

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

2,

2,

23

32

21

1

udu

udu

udu

321 ,, uuu

1ur

2ur

3ur

S4

S1S2

S5

S3

S6

Fr

321 ,, uuu

321

321213132

0

1lim

u

r

u

r

u

r

u

A

u

r

u

r

u

A

u

r

u

r

u

A

u

r

u

r

AdsV

AS

V

3

3

2

2

1

1 u

Au

u

Au

u

AuA

By the same procedure as before:

Divergence

SV

AdsV

A 1

lim0

or

321

321213132

0

1lim

ur

ur

ur

uA

ur

ur

uA

ur

ur

uA

ur

ur

AdsV

AS

V

3

3

2

2

1

1 u

Au

u

Au

u

AuA

By the same procedure as before:

Rotor

SV

AdsV

A 1

lim0

Summarize3

3

2

2

1

1 uu

uu

uu

139


Let develop those equations in curvilinear coordinates:

Therefore: i

i u

j

ij

j

iji

ugg

We obtain: ii u

i

ij

u

i

ij

i

i

ru

ggru

guu

uuu 1,, 321

where:g

r

r

rr i

i

i

i

u

u

u

u

1

Second Proof:

j

j

jij

i

j

ji

i udu

udgudu

r

u

r

u

r

u

rud

u

rud

u

rud

u

rrdd

3

32

1

1

3

3

2

2

1

1

j

iji

ug

ij

i

j

gu

i

i uu

uu

uu

uuuuu

3

3

2

2

1

1321 ,,Gradient

140


Let develop those equations in curvilinear coordinates: j

ju

i uArAAi

Divergence i

ij

ju

i

u

AuuArAA

i

But j

m

m

ij

i

j

u

mj

mi

i

j

i

uAu

ArA

u

A

u

Aj

Therefore

ij

ij

j

g

ji

m

m

ji

i

j

u

imj

im

i

j

uuAu

AruA

u

AA

or ij

m

m

ji

i

jmi

im

i

i

gAu

AA

u

AA

141


Divergence (continue – 1)j

ju

i uArAAi

ij

m

m

ij

i

jmi

im

i

i

gAu

AA

u

AA

Let compute i

im

konsummationggGggG ki

iki

jkj

ik

ik

ik

Gg

g

m

kiki

m

kiik

m

ki

kim u

ggg

u

gG

u

g

g

g

u

g

k

im

i

mk

m

ikik

mii

im

mjk

ij

i

jk

j

ikkm

kij

kmij

m

u

g

u

g

u

gg

u

g

u

g

u

ggg

22,Start from

Butk

mikigg

k

imkiki

i

mkik

u

gg

u

gg

u

gg

miim

m

ikki

mii

u

gg

2

We found

mmm

ikki

mii

u

g

gu

g

gu

gg

1

2

1

2

i

i

i

ii

iimm

mi

imi

im

i

i

u

Ag

gA

u

g

gu

AA

u

g

gu

AA

u

AA

111

ij

m

m

ij

i

jmi

im

i

i

i

i

gAu

AA

u

A

u

Ag

gA

1

142


Rotor i

ij

ju

i

u

AuuArAA

i

But j

m

m

ij

i

j

u

mj

mi

i

j

i

uAu

ArA

u

A

u

Aj

Therefore ji

m

m

ji

i

j

u

imj

im

i

j

uuAu

AruA

u

AA

j

j

ju

i uArAAi


otherwise

ofnpermutatiocyclicakji

ofnpermutatiocyclicakji

rg

uu kjiu

kjiji

k

0

3,1,2,,1

3,2,1,,1

,,

,,

1,2,3ofnspermutatiocyclicarekj,i,g

r

u

A

u

A

g

rA

u

AA

u

AA

kji

u

j

i

i

j

kji

u

m

m

ij

j

im

m

ji

i

j

k

k

,,

,,

Use the fact that and are reciprocal vectors we haveiujur

143


Laplacian Δ

j

ju

i uArAAi


ij

m

m

ij

ji

j

ji

ii

i

guuu

ugg

ugu

g

g

2

2 11

i

i u

j

ij

j

iji

ugg

Using

ij

m

m

ij

i

jmi

im

i

i

i

i

gAu

AA

u

A

u

Ag

gA

1

ii u

i

ij

u

i

ij

i

i

ru

ggru

guu

uuu 1,, 321

We found

Table of Contents

144


Orthogonal Curvilinear Coordinates in a Three Dimensional Space

Let express the cartesiuan coordinates (x, y, z) of any point, in a three dimensional space as a function of three curvilinear coordinates (u1, u2, u3), where:

dr

constu 3

i

j

k

1

1

111 udu

reudh

2

2

222 udu

reudh

3

3

333 udu

reudh

constu 1

constu 2

curveu1

curveu2

curveu3

zyxuuzyxuuzyxuu

uuuzuuuyuuuxx

,,,,,,,,

,,,,,,,,

332211

321321321


kzjyixr

kzdjydixdrd

For orthogonal coordinates we have:

3332221113

3

2

2

1

1

eudheudheudhudu

rud

u

rud

u

rrd

ji

jiee ji 0

1

2

3

2

3

2

2

2

2

2

1

2

1

2 udhudhudhrdrdsd

3213213

3

2

2

1

1

udududhhhudu

rud

u

rud

u

rVd

33

3

22

2

11

1 /:/:/:u

r

u

re

u

r

u

re

u

r

u

re

145


Orthogonal Curvilinear Coordinates in a Three Dimensional Space(continue – 1)

dr

constu 3

i

j

k

1

1

111 udu

reudh

2

2

222 udu

reudh

3

3

333 udu

reudh

constu 1

constu 2

curveu1

curveu2

curveu3

General Coordinates:

3

3

321

22113

2

2

321

11332

1

1

321

33221

321

21

321

13

321

32

,,,

,,,

,, h

e

hhh

eheh

rrr

rru

h

e

hhh

eheh

rrr

rru

h

e

hhh

eheh

rrr

rru

uuu

uu

uuu

uu

uuu

uu

33

3

22

2

11

1321

,, ehu

rreh

u

rreh

u

rr uuu

i

i

uu

Orthogonal Coordinates:

i

ii

euh

1Gradient:

j

ju

i uri

AAA

321321,, hhhrrrg uuu

33

3

22

2

11

1

uh

e

uh

e

uh

e

146


Orthogonal Curvilinear Coordinates in a Three Dimensional Space(continue – 2)

dr

constu 3

i

j

k

1

1

111 udu

reudh

2

2

222 udu

reudh

3

3

333 udu

reudh

constu 1

constu 2

curveu1

curveu2

curveu3


3

3

321

22113

2

2

321

11332

1

1

321

33221

321

21

321

13

321

32

,,,

,,,

,, h

e

hhh

eheh

rrr

rru

h

e

hhh

eheh

rrr

rru

h

e

hhh

eheh

rrr

rru

uuu

uu

uuu

uu

uuu

uu

33

3

22

2

11

1321

,, ehu

rreh

u

rreh

u

rr uuu

Orthogonal Coordinates:

321 33221133

3

322

2

211

1

1

3

33

2

22

1

11

3

333

2

222

1

111332211

/// uuu rhArhArhAehh

Aeh

h

Aeh

h

A

uhAuhAuhAh

ehA

h

ehA

h

ehAeAeAeA

321

321

AAA

AAA

A

i

i

u

Ag

g

1ADivergence:

3

3

21

2

2

31

1

1

32

321

1

u

Ahh

u

Ahh

u

Ahh

hhhA

j

ju

i uri

AAA


147


Orthogonal Curvilinear Coordinates in a Three Dimensional Space (continue – 3)


3

3

321

22113

2

2

321

11332

1

1

321

33221

321

21

321

13

321

32

,,,

,,,

,, h

e

hhh

eheh

rrr

rru

h

e

hhh

eheh

rrr

rru

h

e

hhh

eheh

rrr

rru

uuu

uu

uuu

uu

uuu

uu

33

3

22

2

11

1321

,, ehu

rreh

u

rreh

u

rr uuu

Orthogonal Coordinates:Rotor:

1,2,3ofnspermutatiocyclicarekj,i,

g

r

u

A

u

A

kji

u

j

i

i

j k

,,

A

j

ju

i uri

AAA

321 33221133

3

322

2

211

1

1

3

33

2

22

1

11

3

333

2

222

1

111332211

/// uuu rhArhArhAehh

Aeh

h

Aeh

h

A

uhAuhAuhAh

ehA

h

ehA

h

ehAeAeAeA

321

321

AAA

AAA

A


332211

321

332211

AhAhAh

uuu

eheheh

A

dr

constu 3

i

j

k

1

1

111 udu

reudh

2

2

222 udu

reudh

3

3

333 udu

reudh

constu 1

constu 2

curveu1

curveu2

curveu3

148


Orthogonal Curvilinear Coordinates in a Three Dimensional Space (continue – 4)


3

3

321

22113

2

2

321

11332

1

1

321

33221

321

21

321

13

321

32

,,,

,,,

,, h

e

hhh

eheh

rrr

rru

h

e

hhh

eheh

rrr

rru

h

e

hhh

eheh

rrr

rru

uuu

uu

uuu

uu

uuu

uu

33

3

22

2

11

1321

,, ehu

rreh

u

rreh

u

rr uuu

Orthogonal Coordinates:Laplacian:

j

ju

i uri

AAA


dr

constu 3

i

j

k

1

1

111 udu

reudh

2

2

222 udu

reudh

3

3

333 udu

reudh

constu 1

constu 2

curveu1

curveu2

curveu3

j

ji

i ugg

ug

1

ji

jih

h

e

h

euug i

j

j

i

ijiji

0

/1 2

33

21

322

31

211

32

1321

1

uh

hh

uuh

hh

uuh

hh

uhhh

Table of Contents

149

Vector AnalysisSOLO

Vector Operations in Various Coordinate Systems

z

y

x

z

y

x

1. Gradient • Cartesian:

zyx zyx 111

z

r

r

z

r

1

• Cylindrical:

zr zr 111

sin

1

1

r

r

rr• Spherical:

111

rr

150

Vector AnalysisSOLO


2. Divergence • Cartesian:

z

A

y

A

x

AA zyx

• Cylindrical: zr zr AAAA 111

• Spherical:

zyx zyx AAAA 111

z

AA

rAr

rrA z

r

11

111

AAAA rr

A

rA

rAr

rrA r sin

1sin

sin

11 2

2

151

Vector AnalysisSOLO


3. Laplacian 2

• Cartesian:

2

2

2

2

2

22

zyx

• Cylindrical:

• Spherical:

2

2

2

2

22

2

2

2

2

2

2

2 1111

zrrrrzrrr

rr

2

2

222

2

2

2

sin

1sin

sin

11

rrr

rrr

152

Vector AnalysisSOLO


4. Curl

• Cartesian:

y

A

x

AA

x

A

z

AA

z

A

y

AA

xy

z

zxy

yzx

• Cylindrical:zr zr AAAA 111

• Spherical:

zyx zyx AAAA 111

111

AAAA rr

zyx zyx AAAA 111

rz

zr

zr

AAr

rrA

r

A

z

AA

z

AA

rA

1

1

zr zr AAAA 111

r

r

r

AAr

rrA

Arr

A

rA

AA

rA

1

sin

1

sinsin

1

111

AAAA rr

153

Vector AnalysisSOLO


5. Scalar Product• Cartesian:

• Cylindrical:

zr zr AAAA 111

• Spherical:

zyx zyx AAAA 111

111

AAAA rr

zyx zyx BBBB 111

zzyyxx BABABABA

zr zr BBBB 111

111

BBBB rr

zzrr BABABABA

BABABABA rr

154

Vector AnalysisSOLO


6. Vector Product• Cartesian:


• Spherical:

zyx zyx AAAA 111

111

AAAA rr

zyx zyx BBBB 111

zyx zyx BABABABA 111

zr zr AAAA 111

111

AAAA rr

xyyxz

zxxzy

yzzyx

BABABA

BABABA

BABABA

zr zr BABABABA 111

rrz

zrrz

zzr

BABABA

BABABA

BABABA

rr

rr

r

BABABA

BABABA

BABABA

111

BABABABA rr

155

Vector AnalysisSOLO


7. Material Derivative• Cartesian:


• Spherical:

zyx zyx AAAA 111

111

AAAA rr

zyx zyx vvvv 111

z

Av

y

Av

x

Av

t

A

tD

AD

z

Av

y

Av

x

Av

t

A

tD

AD

z

Av

y

Av

x

Av

t

A

tD

AD

zz

zy

zx

z

z

y

z

y

y

y

x

y

y

xz

xy

xx

x

x

Av

t

A

tD

AD

zr zr vvvv 111

111

vvvv rr

z

Av

A

r

v

r

Av

t

A

tD

AD

z

Av

A

r

v

r

Av

t

A

tD

AD

z

Av

A

r

v

r

Av

t

A

tD

AD

zz

zzr

z

z

zr

rz

rrr

r

r

A

r

vA

r

v

r

Av

t

A

tD

AD

A

r

vA

r

v

r

Av

t

A

tD

AD

A

r

vA

r

v

r

Av

t

A

tD

AD

r

r

rrrr

r

r

sin

sin

sin

Table of Contents

156

Vector AnalysisSOLO

Applications

Fundamental Theorem of Vector Analysis for a Bounded Region V (Helmholtz’s Theorem)

Reynolds’ Transport Theorem

Poisson’s Non-homogeneous Differential Equation

Kirchhoff’s Solution of the Scalar Helmholtz Non-homogeneous Differential Equation

Table of Contents

Fundamental Theorem of Vector Analysis for a Unbounded Region V (Helmholtz’s Theorem)

Laplace FieldsHarmonic Functions

Rotations

157

ROTATIONS

Rotation of a Rigid Body

SOLO

23r31r

12r1

3

2

P

P

1

2

331r

23r12r

A rigid body in mechanics is defined as a system of mass points subject to theconstraint that the distance between all pair of points remains constant through the motion.

To define a point P in a rigid body it is enough to specify the distance of this point to three non-collinear points. This means that a rigid body is completely definedby three of its non-collinear points. Since each point, in a three dimensional spaceis defined by three coordinates, those three points are defined by 9 coordinates.But the three points are constrained by the three distances between them:

313123231212 && constrconstrconstr

Therefore a rigid body is completely defined by 9 – 3 = 6 degrees of freedom.

This is a part of thePresentation “ROTATIONS”

NOTES ON ROTATIONS

SOLO HERMELIN

INITIAL INTERMEDIATE FINAL

158

ROTATIONS

Rotation of a Rigid Body (continue – 1)

SOLO

We have the following theorems about a rigid body:

Euler’s Theorem (1775) The most general displacement of a rigid body with one point fixed is equivalent toa single rotation about some axis through that point.

Chasles’ Theorem (1839) The most general displacement of a rigid body is a translation plus a rotation.

Leonhard Euler 1707-1783

Michel Chasles 1793-1880

159

ROTATIONS

Rotation of a Rigid Body (continue – 2)

SOLO

Proof of Euler’s Theorem P

'P

OA

'A

B

'BCC

r rr

rr

O – Fixed point in the rigid body

A,B – Two point in the rigid body at equal distance r from O.

rOBOA

__________

A’,B’ – The new position of A,B respectively.

Since the body is rigid rOBOA __________

''

Therefore A,B, A’,B’ are one a spherewith center O.

– plane passing through O such that A and A’ are at the same distance from it. – plane passing through O such that B and B’ are at the same distance from it.

PP’ – Intersection of the planes and The two spherical triangles APB and A’PB’ are equal.

The arcs AA’ and BB’ are equal. That means that rotation around PP’ that moves A to A’ will move B to B’.

q.e.d.

160

ROTATIONS

Mathematical Computation of a Rotation

SOLO

AB

C

O

n

v

1v

We saw that every rotation is defined by three parameters:

• Direction of the rotation axis , defined by by two parameters.n

• The angle of rotation , defines the third parameter. Let rotate the vector around by a large angle , toobtain the new vector

OAv n

OBv1

From the drawing we have:

CBACOAOBv1

vOA

cos1ˆˆ

vnnAC Since direction of is: sinˆˆ&ˆˆ vnnvnn

and it’s length is:

AC

cos1sin v

sinˆ vnCB

Since has the direction and the

absolute valueCB

vn

ˆsinsinv

sinˆcos1ˆˆ1 vnvnnvv

161

ROTATIONS

Computation of the Rotation Matrix

SOLO

We have two frames of coordinates A and B defined by the orthogonal unit vectors and AAA zyx ˆ,ˆ,ˆ BBB zyx ˆ,ˆ,ˆ

The frame B can be reached by rotating the A framearound some direction by an angle . n

We want to find the Rotation Matrixthat describes this rotation from A to B.

,ˆ33 nRC xBA

sinˆˆcos1ˆˆˆˆˆ



AAAB

AAAB

AAAB

znznnxz

ynynnxy

xnxnnxx

Let write those equations in matrix form.

0

0

1

sinˆ

0

0

1

cos1ˆˆ

0

0

1

ˆ AAAAB nnnx

0

0

0

ˆ

xy

xz

yz

A

nn

nn

nn

n 0ˆ ntrace

AxAz

Ay

Bz

By

BxO

n

Rotation Matrix

162

ROTATIONS

Computation of the Rotation Matrix (continue – 1)

SOLO

AxAz

Ay

Bz

By

BxO

n

0

0

1

sinˆ

0

0

1

cos1ˆˆ

0

0

1

ˆ AAAAB nnnx

0

1

0

sinˆ

0

1

0

cos1ˆˆ

0

1

0

ˆ AAAAB nnny

1

0

0

sinˆ

1

0

0

cos1ˆˆ

1

0

0

ˆ AAAAB nnnz

AA

AB

AAAx

AB xCnnnIx ˆ

0

0

1

sinˆcos1ˆˆˆ 33

AA

AB

AAAx

AB yCnnnIy ˆ

0

1

0

sinˆcos1ˆˆˆ 33

AA

AB

AAAx

AB zCnnnIz ˆ

1

0

0

sinˆcos1ˆˆˆ 33

Rotation Matrix (continue – 1)

163

ROTATIONS


SOLO

AxAz

Ay

Bz

By

Bx

O

n

,ˆsinˆcos1ˆˆ 3333 nRnnnICC xAAA

xA

B

A

B

The matrix has the following properties: Anˆ

ATA nn ˆˆ

22

22

22

0

0

0

0

0

0

ˆˆ

yxzyzx

zyzxyx

zxyxyz

xy

xz

yz

xy

xz

yz

AA

nnnnnn

nnnnnn

nnnnnn

nn

nn

nn

nn

nn

nn

nn

Tx

zzyzx

zyyyx

zxyxx

nnI

nnnnn

nnnnn

nnnnn

ˆˆ

000

010

001

33

2

2

2

213ˆˆ AA nntrace

nn

nn

nn

nn

nnnnn

xy

xz

yz

zyxAT ˆˆ000

0

0

0

ˆˆ

AATAATx

AAA nnnnnnnnInnn ˆˆˆˆˆˆˆˆˆˆˆ 22

Tx

AAAAAA nnInnnnnn ˆˆˆˆˆˆˆˆ 33

skew-symmetric


164

ROTATIONS


SOLO

AxAz

Ay

Bz

By

Bx

O

n

BAxx

AAAx

TATATAx

TAB

CnRnR

nnnI

nnnIC

,ˆ,ˆ

sinˆcos1ˆˆ

sinˆcos1ˆˆ

3333

33

33

Note

The last term can be writen in matrix form as

Therefore

In the same way

End Note

In fact is the matrix representation of the vector product: vnn

ˆˆ

vInnvvnn xT

33ˆˆˆˆ

vvnnnnvvnnvnn

ˆˆˆˆˆˆˆˆ

Tx nnInn ˆˆˆˆ 33

nnnnvnvvnnnvnnn ˆˆˆˆˆˆˆˆˆˆˆ

nnnnnnvnnvnnnn ˆˆˆˆˆˆˆˆˆˆˆˆ


165

ROTATIONS


SOLO

AxAz

Ay

Bz

By

Bx

O

n

sin0cos123

sinˆcos1ˆˆ

sinˆcos1ˆˆ

33

33

AAA

x

TATATAx

BA

ntracenntraceItrace

nnnItracetraceC

Therefore cos21BACtrace

Let compute the trace (sum of the diagonal components of a matrix) of B

AC

Also we have

sinˆcos1ˆˆcos

sinˆcos1ˆˆ

sinˆcos1ˆˆ

33

3333

33

ATx

ATxx

TATATAx

BA

nnnI

nnnII

nnnIC

sin

0

0

0

cos1cos

000

010

001

2

2

2

xy

xz

yz

zzyzx

zyyyx

zxyxx

nn

nn

nn

nnnnn

nnnnn

nnnnn


166

ROTATIONS


SOLO

AxAz

Ay

Bz

By

Bx

O

n

Therefore we have

cos1cossincos1sincos1

sincos1cos1cossincos1

sincos1sincos1cos1cos

2

2

2

zxzyyzx

xzyyzyx

yzxzyxx

BA

nnnnnnn

nnnnnnn

nnnnnnn

C

We get

12

1cos B

AtraceC two solutions for

If ; i.e. we obtain0sin ,0

sin2/2,33,2 BA

BAx CCn

sin2/3,11,3 BA

BAy CCn

sin2/1,22,1 BA

BAz CCn


167

ROTATIONS

Consecutive Rotations

SOLO

- Perform first a rotation of the vector , according to the Rotation Matrix to the vector .

v 1133 ,ˆ nR x

1v

- Perform a second a rotation of the vector , according to the Rotation Matrix to the vector .

1v

2233 ,ˆ nR x

2v

vnRv x

11331 ,ˆ

vnRvnRnRvnRv xxxx

,ˆ,ˆ,ˆ,ˆ 3311332233122332

The result of those two consecutive rotation is a rotation defined as:

1133223333 ,ˆ,ˆ,ˆ nRnRnR xxx

Let interchange the order of rotations, first according to the Rotation Matrixand after that according to the Rotation Matrix .

2233 ,ˆ nR x

1133 ,ˆ nR x

The result of those two consecutive rotation is a rotation defined as:

22331133 ,ˆ,ˆ nRnR xx

Since in general, the matrix product is not commutative

2233113311332233 ,ˆ,ˆ,ˆ,ˆ nRnRnRnR xxxx

Therefore, in general, the consecutive rotations are not commutative.


168

ROTATIONSSOLO

INITIALINITIAL INTERMEDIATEINITIAL INTERMEDIATE FINAL

Consecutive Rotations of a DiceRotation Matrix (continue – 7)

169

ROTATIONS

Decomposition of a Vector in Two Different Frames of Coordinates

SOLO

We have two frames of coordinate systems A and B, with the same origin O.

We can reach B from A by performing a rotation.

Let describe the vector in both frames. v

Ax

Az

Ay

Bx

BzBy

v

OxAv

zAv

yAv

xBv

zBv

yBv

BzBByBBxBAzAAyAAxA zvyvxvzvyvxvv

111111

zA

yA

xA

A

v

v

v

v

zB

yB

xB

B

v

v

v

v

&

BBABBABBAA

BBABBABBAA

BBABBABBAA

zzzyyzxxzz

zzyyyyxxyy

zzxyyxxxxx

ˆˆˆˆˆˆˆˆˆˆ

ˆˆˆˆˆˆˆˆˆˆ

ˆˆˆˆˆˆˆ1ˆˆ

zABBABBABBA

yABBABBABBA

xABBABBABBA

vzzzyyzxxz

vzzyyyyxxy

vzzxyyxxxxv

ˆˆˆˆˆˆˆˆˆ

ˆˆˆˆˆˆˆˆˆ

ˆˆˆˆˆˆˆ1ˆ

from which


170

ROTATIONS

Decomposition of a Vector in Two Different Frames of Coordinates (continue – 1)

SOLO

zA

yA

xA

BABABA

BABABA

BABABA

zB

yB

xB

v

v

v

zzzyzx

yzyyyx

xzxyxx

v

v

v

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

Ax

Az

Ay

Bx

BzBy

v

OxAv

zAv

yAv

xBv

zBv

yBv ABA

B vCv

where is the Transformation Matrix (or Direction Cosine Matrix – DCM) from frame A to frame B.

BAC

BABABA

BABABA

BABABAB

A

B

A

zzzyzx

yzyyyx

xzxyxx

CC

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

:

In the same way BAB

BBA

A vCvCv

1

therefore 1 B

AA

B CC


171

ROTATIONS

Decomposition of a Vector in Two Different Frames of Coordinates (continue – 2)

SOLO

Ax

Az

Ay

Bx

BzBy

v

OxAv

zAv

yAv

xBv

zBv

yBv

ATAABA

TBA

TAABA

TABA

BTB vvvCCvvCvCvvv

2

Since the scalar product is independent of the frame of coordinates, we have

1 B

A

TBA

BA

TBA CCICC

100

010

001

3,33,23,1

2,32,22,1

1,31,21,1

3,32,31,3

3,22,21,2

3,12,11,1

BA

BA

BA

BA

BA

BA

BA

BA

BA

BA

BA

BA

BA

BA

BA

BA

BA

BA

BA

TBA

CCC

CCC

CCC

CCC

CCC

CCC

CC

or

3,2,10

3,2,11,,

3

1 jji

ijikjCkiC ij

k

BA

BA

Those are 9 equations in , but by interchanging i with j we get the sameconditions, therefore we have only 6 independent equations.

3,2,1,, jijiC BA

We see that the Rotation Matrix is ortho-normal (having real coefficients and therows/columns are orthogonal to each other and of unit absolute value.


This means that the relation between the two coordinate systems is defined by9 – 6 = 3 independent parameters.

172

ROTATIONS

Differential Equations of the Rotation Matrices

SOLO

We want to develop the differential equation of the Rotation Matrix as a function of the Angular Velocity of the Rotation. Let define by:

-the Rotation Matrix that defines a frame of coordinates B at the time t relative to some frame A.

tC BA

-the Rotation Matrix that defines the frame of coordinates B at the time t+Δt relative to some frame A.

ttC BA

,ˆ33xR -the Rotation Matrix from the frame of coordinates B at the time t to B at time t+Δt relative to some frame A.

tCRttC BAx

BA ,ˆ33

and

2cos

2sinˆ2

2sinˆˆ2

sinˆcos1ˆˆ,ˆ

233

3333

x

xx

I

IR


173

ROTATIONS

Differential Equations of the Rotation Matrices (continue – 1)

SOLO

Let differentiate the Rotation Matrix

tCdt

dIRtC

t

IR

tCt

IR

t

tCtCR

t

tCttC

t

C

dt

dC

BA

xxBA

xx

t

BA

xx

t

BA

BAx

t

BA

BA

t

BA

t

BA

3333

0

3333

0

3333

0

33

0

00

,ˆlim

,ˆlim

,ˆlim

,ˆlim

limlim

ˆ

2cos

2

2sin

ˆ2

2

2sin

ˆˆlim,ˆ

lim 2

2

0

3333

0

xx IR

and

Therefore tC

dt

d

dt

tdC BA

BA ˆ


174

ROTATIONS


SOLO

The final result of the Rotation Matrix differentiation is:

Since defines the unit vector of rotation and the rotation rate from B at time t

to B at time t+Δt, relative to A, then is the angular velocity vector of the frame

B relative to A, at the time t

dt

d

ˆ

dt

d

ˆdt

dBAB

tCtdt

tdC BA

BAB

BA

By changing indixes A and B we obtain

tCtdt

tdC AB

ABA

AB


175

ROTATIONS


SOLO

Let find the relation between and BAB A

AB

For any vector let perform the following computationsv

AAB

BA

BAB

BBAB vCvv

BAB

AAB

BA

ABA

AB

AAB

BA

AAAB

BA vCCvCCCvC

Since this is true for any vector we havev

AB

AAB

BA

BAB CC

Pre-multiplying by and post-multiplying by we get:A

BC BAC

BA

BAB

AB

AAB CC


176

ROTATIONS


SOLO

Let differentiate the equation 33xA

BB

A ICC

to obtain

0 dt

dCC

dt

dCCCC

dt

dCCC

dt

dC ABB

AB

AB

ABB

AA

BB

AB

AB

ABB

AA

B

BA

Post-multiplying by we getA

BC

AB

AAB

AB

BA

BAB

AB

BAB

AB

AB CCCCC

dt

dC

We obtained for the differentiation of the Rotation Matrix

BAB

AB

AB

AAB

AB

ABA

AB ttCtCttCtdt

tdC

Note

We can see that tttt ABBAA

ABA

BA

End Note


177

ROTATIONS


SOLO

Suppose that we have a third frame of coordinates I (for example inertial) and we have the angular velocity vectors of frames A and B relative to I.

We have

BI

BIB

BI C

dt

dC A

IA

IA

AI C

dt

dC

AI

BA

BI CCC

dt

dCCC

dt

dC

dt

dC AIB

AA

I

BA

BI

IA

AI

AIA

BA

IA

BI

BIB

IA

AIB

AI

A

BI

BA CCCCCC

dt

dCCC

dt

dC

dt

dC

or

From which we get:

AIA

BA

BA

BIB

BA CC

dt

dC


178

ROTATIONSSOLO

From the equation

Computation of the Angular Velocity Vector from .AB nRtC xB

A ˆ,33

tCtdt

tdC BA

BAB

BA

we obtain

TBA

BAB

AB tCdt

tdCt

Since the Rotation Matrix is defined also by and

sinˆcos1ˆˆcosˆ, 3333 nnnInRC Tx

BA

tC BA n

we can compute as function of and their derivativesnABtd

d td

ndn

ˆˆ

(this is a long procedure described in the complementary work “Notes on Rotations”, and a simpler derivation will be given later, we give here the final result)

sinˆcos1ˆˆˆ

nnnnAB


179

ROTATIONSSOLO

Computation of and as functions of .ABtd

d td

ndn

ˆˆ

Let pre-multiply the equation by and useTn sinˆcos1ˆˆˆ

nnnnAB

0ˆˆ,0ˆˆ,1ˆˆ

nnnnnn TTT to obtain

ABTTTT

ABT nnnnnnnnn

ˆsinˆˆcos1ˆˆˆˆˆˆ

Let pre-multiply the equation by and use n sinˆcos1ˆˆˆ

nnnnAB

nnInnnnnnn xT ˆˆˆˆˆˆˆ,0ˆˆ 33

to obtain

sinˆˆcos1ˆsinˆˆcos1ˆˆˆˆˆˆ

nnnnnnnnnnn AB

Let pre-multiply the equation by sinˆˆcos1ˆˆ

nnnn AB

n

cos1ˆˆsinˆsinˆˆˆcos1ˆˆˆˆ

nnnnnnnnnn AB


180

ROTATIONS

Computation of and as functions of (continue – 1)

SOLO

ABtd

d td

ndn

ˆˆ

We have two equations:

ABnnnn

ˆsinˆˆcos1ˆ

ABnnnnn

ˆˆcos1ˆˆsinˆ

with two unknowns and

n

nn ˆˆ

From those equations we get:

sinˆˆcos1ˆsincos1ˆ 22ABAB nnnn

or

sinˆˆcos1ˆcos1ˆ2 ABAB nnnn

Finally we obtain:

ABTn ˆ

ABnnnn

2cotˆˆˆ

2

1ˆ


181

ROTATIONS

Quaternions

SOLO

The quaternions method was introduced by Hamilton in 1843. It is based on Euler Theorem (1775) that states:

The most general displacement of a rigid body with one point fixed is equivalent toa single rotation about some axis through that point.

Therefore every rotation is defined by three parameters:

• Direction of the rotation axis , defined by two parameters

• The angle of rotation , defines the third parameter

n

William Rowan Hamilton 1805 - 1865

sinˆcos1ˆˆ1 vnvnnvv

The rotation of around by angle is given by:n v

AB

C

O

n

v

1v

that can be writen

sinˆcos1ˆˆ1 vnvvnnvv

or

sinˆcos1ˆˆcos1 vnvnnvv

182

ROTATIONSQuaternions (continue – 1)

SOLO

Computation of the Rotation Matrix

We found the Rotation Matrixthat describes this rotation from A to B.

,ˆ33 nRC xB

A




AAAB

AAAB

AAAB

znznnxz

ynynnxy

xnxnnxx

AxAz

Ay

Bz

By

BxO

n

AA

AB

AAAx

AB xCnnnIx ˆ

0

0

1

sinˆcos1ˆˆˆ 33

AA

AB

AAAx

AB yCnnnIy ˆ

0

1

0

sinˆcos1ˆˆˆ 33

AA

AB

AAAx

AB zCnnnIz ˆ

1

0

0

sinˆcos1ˆˆˆ 33

or

from which

,ˆsinˆcos1ˆˆ 3333 nRnnnICC xAAA

xA

B

A

B

183


SOLO

Definition of the Quaternions

AxAz

Ay

Bz

By

BxO

n

The quaternions (4 parameters) were defined by Hamilton as a generalization of the complex numbers

32100 , qkqjqiqqq

2/cos0 q

n2/sin

zyx nqnqnq 2/sin&2/sin&2/sin 111

where satisfy the relations: kji

,,

1 kkjjii

kijji

,

ijkkj

,

jkiik

1 kji

i

j

k

the complex conjugate of is defined asq

32100* , qkqjqiqqq

184


SOLO

Product of Quaternions

Product of two quaternions andAq Bq

3210321000 ,, BBBBAAAABBAABA qkqjqiqqkqjqiqqqqq

3210321033221100 AAABBBBABABABABA qkqjqiqqkqjqiqqqqqqqqq

122131132332 BABABABABABA qqqqkqqqqjqqqqi

therefore

BAABBABABABBAABA qqqqqqqq

000000 ,,,

Let use this expression to find

23

22

21

20

222000

*00

* 1ˆˆ2

sin2

cos,,,, qqqqnnqqqqqqqqq

The quaternion product can be writen in matrix form as:

A

A

BxBB

TBB

B

B

AxAA

TAA

BA

q

Iq

qq

Iq

qqq

qq

0

330

00

330

00

1 kjikkjjii

kijji

ijkkj

jkiik

185


SOLO

Rotation Description Using the Quaternions

Let compute the expression:

AAAAAAAA

AAAAA

vvqvqvqvvvqqv

qvvqvqvqqvq

002

000

0000*

,

,,,,0,

A

AAAA

AAAAAAA

vqq

vvqqvv

vvqvqvqvvv

22,0

2,0

,0

02

0

02

0

002

0

Using the relations:

nnq

nnnn

q

n

q

ˆsinˆ2/sin2/cos22

ˆˆcos1ˆˆ2/sin22

1

ˆ2/sin

2/cos

0

2

20

0

and AAAAx

ABA

B vnnnIvCv sinˆcos1ˆˆ33

we obtain

AAABB vqqvqqvqvv

221,0,,0,,0 000*

186


SOLO

Rotation Description Using the Quaternions (continue – 1)

Using the fact that we obtain:

22 033 qIC xB

A

0

0

0

0

0

0

2

0

0

0

2

100

010

001

12

13

23

12

13

23

12

13

23

0

qq

qq

qq

qq

qq

qq

qq

qq

qq

q

22

213231

322

12

321

31212

22

3

1020

1030

2030

2222

2222

2222

022

202

220

100

010

001

qqqqqq

qqqqqq

qqqqqq

qqqq

qqqq

qqqq

2

22

110323120

32102

12

33021

203121302

22

3

2212222

2222122

2222221

qqqqqqqqqq

qqqqqqqqqq

qqqqqqqqqq

123

22

21

20 qqqq

2

32

22

12

010323120

32102

32

22

12

03021

203121302

32

22

12

0

22

22

22

qqqqqqqqqqqq

qqqqqqqqqqqq

qqqqqqqqqqqq

C BA

187


SOLO

Rotation as a Multiplication of Two Matrices

22 033 qIC xB

A

22 0332

0 qIq xT

330332

0 2 xT

x IqIq

33330330 xT

xx IIqIq

For any vector we can write

aaaa

or in matrix notation

Tx

Tx

Tx

TT IIaIa

333333

Therefore we have

33330330 xT

xxBA IIqIqC

Txx IqIq

330330

321

3

2

1

012

103

230

012

103

230

qqq

q

q

q

qqq

qqq

qqq

qqq

qqq

qqq

188


SOLO

Rotation as a Multiplication of Two Matrices (continue – 1)

330

330

012

103

230

321

0123

1032

2301

x

T

xBA

Iq

Iq

qqq

qqq

qqq

qqq

qqqq

qqqq

qqqq

C

330

330

012

103

230

321

0123

1032

2301

x

T

xBA

Iq

Iq

qqq

qqq

qqq

qqq

qqqq

qqqq

qqqq

C

T

x

xBA

Iq

Iq

qqq

qqq

qqq

qqq

qqqq

qqqq

qqqq

C

330

330

321

012

103

230

3012

2103

1230

T

x

xBA

Iq

Iq

qqq

qqq

qqq

qqq

qqqq

qqqq

qqqq

C

330

330

321

012

103

230

3012

2103

1230

189


SOLO

Relation Between Quaternions and Euler Angles

x

Ax

B

x

qvqv

iq

*

2sin

2cos

Rotation around x axis

Ax

1

Ay

1

Az

1Bz

1 By

1

y

Ay

B

y

qvqv

jq

*

2sin

2cos

Ax

1

Ay

1

Az

1Bz

1

Bx

1

Rotation around y axis

z

Az

B

z

qvqv

kq

*

2sin

2cos

Ax

1

Ay

1

Az

1

Bx

1

By

1

Rotation around z axis

190


SOLO

Description of Successive Rotations Using Quaternions

Let describe two consecutive rotations:- First rotation defined by the quaternion

1

111101 ˆ

2sin,

2cos, nqq

- Folowed by the second rotation defined by the quaternion

2

222202 ˆ

2sin,

2cos, nqq

After the first rotation the quaternion of the vector is transferred to 1*

1 qvq

After the second rotation we obtain 21*

2121*

1*

221*

1*

2 qqvqqqqvqqqqvqq

Therefore the quaternion representing those two rotation is:

2121

12

21

212121

21120210212010220110210

ˆˆ2

sin2

sinˆ2

cosˆ2

cos,ˆˆ2

sin2

sin2

cos2

cos

,,,,

nnnnnn

qqqqqqqqqq

210 , qqqq 21

21210 ˆˆ

2sin

2sin

2cos

2cos

2cos nnq

2121

12

21 ˆˆ

2sin

2sinˆ

2cosˆ

2cosˆ

2sin nnnnn

191


SOLO

Description of Successive Rotations Using Quaternions (continue – 1)

210 , qqqq 21

21210 ˆˆ

2sin

2sin

2cos

2cos

2cos nnq

2121

12

21 ˆˆ

2sin

2sinˆ

2cosˆ

2cosˆ

2sin nnnnn

Two consecutive rotations, followed by , are given by:1q 2q

From those equations we can see that:

0ˆˆˆˆˆˆˆˆˆˆ 21212112211221

nnnnnnnnnnifonlyandifqqqq

The rotations are commutative if and only if are collinear.21 ˆ&ˆ nn

In matrix form those two rotations are given by:

First Rotation: 111111331133 sinˆcos1ˆˆcosˆ, nnnInR Txx

Second Rotation: 222222332233 sinˆcos1ˆˆcosˆ, nnnInR Txx

Total Rotation:

sinˆcos1ˆˆcosˆ,ˆ,ˆ, 331133223333 nnnInRnRnR Txxxx

192


SOLO

Description of Successive Rotations Using Quaternions (continue – 2)

Let find the quaternion that describes the Euler Rotations through the

angles respectively. Let write the rotations according to their order

123

,,

2sin

2cos

2sin

2cos

2sin

2cos

ijkqqqq xyz

BA

2sin

2sin

2cos

2sin

2cos

2sin

2cos

2cos

2sin

2cos

kjik

2sin

2sin

2sin

2cos

2cos

2cos

2cos

2sin

2sin

2sin

2cos

2cos

i

2cos

2sin

2cos

2sin

2cos

2sin

j

2sin

2sin

2cos

2cos

2cos

2sin

k

193


SOLO

Differential Equation of the Quaternions

Let define

,0qtq BA - the quaternion that defines the position of B frame

relative to frame A at time t.

tqqttq BA

,00

- the quaternion that defines the position of B frame relative to frame A at time t+Δt.

tBA ntq ˆ

2sin,

2cos

- the quaternion that defines the position of B frame at time t+Δt relative to frame B at time t.

We have the relation: tqtqttq BA

BA

BA

or

,,0,1,,,,,ˆ2

sin,2

cos 00000000* qqqqqqqqttqtqntq B

ABAt

BA

therefore

tnqq ˆ2

sin,12

cos,, 00

tnqq ˆ2

sin,12

cos,, 00

or

194


SOLO

Differential Equation of the Quaternions (continue – 1)

tnqq ˆ2

sin,12

cos,, 00

Let divide both sides by and take the limit .0 tt

tBttBt

ntqnqntt

tqtd

d

tt

qˆ

2

1,0ˆ

2

1,0,ˆ

2

2sin

2

1,

2

12

cos

2

1,lim 0

0

0

But is the instant angular velocity vector of frame B relative to frame A.tnˆ

tB

AB nt ˆ ttn BAB

BABt ,0ˆ,0

So we can write

ttqtqtd

d BAB

BA

BA

2

1

This is the Differential equation of the quaternion that defines the position of B relative to A, at the time t as a function of the angular velocity vector of frame B relative to frame A, .

tq BA

tBAB

195


SOLO


Developing this equation, we get

ttqtqtd

d BAB

BA

BA

2

1

BAB

BAB

BAB

BAB qtq

dt

d

dt

dq

000 ,

2

1,0,

2

1,

from which

BABdt

dq

2

10

BAB

BABq

dt

d

02

1

or in matrix form

t

Iq

q

dt

d BAB

x

T

330

0

2

1

196


SOLO


zBAB

yBAB

xBAB

qqq

qqq

qqq

qqq

q

q

q

q

dt

d

012

103

230

321

3

2

1

0

t

Iq

q

dt

d BAB

x

T

330

0

2

1

BAAB

xBAByBABzBAB

xBABzBAByBAB

yBABzBABxBAB

zBAByBABxBAB

q

q

q

q

q

q

q

q

q

dt

d

2

1

0

0

0

0

2

1

3

2

1

0

3

2

1

0

After rearranging

or zBAByBABxBAB qqq

dt

dq 321

0

2

1

zBAByBABxBAB qqqdt

dq 230

1

2

1

zBAByBABxBAB qqqdt

dq 103

2

2

1

zBAByBABxBAB qqqdt

dq 012

3

2

1

197


SOLO

Pre-multiply the equation

Computation of as a Function of the Quaternion and its Derivatives tB

AB

t

Iq

q

dt

d BAB

x

T

330

0

2

1

by

330 xIq

t

Iq

Iq

q

Iq BAB

x

T

xx

330

330

0

330 2

1

ttIIq

tIq

BBA

BBAx

TTx

T

BBAx

T

2

1

2

12

1

33332

0

332

0

Therefore

0

3302

q

Iqt xB

AB

198


SOLO

Computation of as a Function of the Quaternion and its Derivatives (continue – 1)

But and are related. Differentiating the equation

tBAB

we obtain

0q

120

Tq

3300

0

330 22 xxB

AB Iqq

q

Iqt

0

0332

0330

0

21

2q

qIqIq

q

Tx

xT

From the equation

TT

qqqq

0000

10

we obtain

T

xB

AB qIqq 033

20

0

2

199


SOLO

Computation of as a Function of , and their Derivatives tBAB n

Differentiate the quaternion

nqq ˆ

2sin,

2cos,0

to obtain

nnqq ˆ2

sinˆ2

cos2

,2

sin2

,0

Substitute this in the equation

0

3302

q

Iqt xB

AB

nn

nIn x

ˆ2

sinˆ2

cos2

2sin

2

ˆ2

sin2

cosˆ2

sin2 33

nnnnnnn ˆˆ

2sin2ˆˆ

2cos

2sinˆ

2cos

2sin2ˆ

2cosˆ

2sin 222

nnnnAB ˆˆcos1ˆsinˆ Finally we obtain

We recovered a result obtained before.

200


SOLO

Differential Equation of the Quaternion Between Two Frames A and B Using the AngularVelocities of a Third Frame I

The relations between the components of a vector in the frames A, B and I arev

ABA

IAI

BA

IBI

B vCvCCvCv

Using quaternions the same relations are given by

BA

AI

IAI

BA

BI

IBI

B qqvqqqvqv***

Therefore

BA

AI

BI qqq B

IA

IBA qqq

*

Let perform the following calculationsBA

AI

BA

AI

BI q

dt

dqqq

dt

dq

dt

d

& BIB

BI

BI qq

dt

d

2

1 AIA

AI

AI qq

dt

d

2

1and use

BA

AI

BA

AIA

AI

BIB

BI q

dt

dqqqq

2

1

2

1 BA

AIA

AI

AI

BIB

BI

AI

BA qqqqqq

dt

d

1

**

2

1

2

1

to obtain BA

AIA

BIB

BA

BA qqq

dt

d

2

1

2

1

201


SOLO

Differential Equatio of the Quaternion Between Two Frames A and B Using the AngularVelocities of a Third Frame I (continue – 1)

Using the relations

ABIAIB

and B

AA

IABA

BIA qq

* BA

AIA

BIA

BA qq

we have

0

2

1

2

1

2

1

2

1

2

1 BA

AIA

BIA

BA

BAB

BA

BA

AIA

BIA

BAB

BA

BA qqqqqq

dt

d

from which BA

AAB

BAB

BA

BA qqq

dt

d

2

1

2

1

Since BAAB we get

BA

ABA

BBA

BA

BA

AAB

BAB

BA

BA qqqqq

dt

d

2

1

2

1

2

1

2

1

From we get1*

BA

AB

BA

BA qqqq A

BBA qq

*

Therefore

B

AA

BBA

AB q

dt

dqqq

dt

d0

*BA

BA

AB

AB qq

dt

dqq

dt

d

Table of Contents

202

SOLO

Laplace Fields

Vector Analysis

A vector field is said to be a Laplace Field if rAA

0 rA

In this case we have

and

022

00 2

AAAAA

0 rA

Harmonic Functions

A continuous function φ with continuous first and second partial derivatives is saidto be harmonic if it satisfies Laplace’s Equation 02

Properties of Harmonic Functions

Pierre-Simon Laplace1749-1827

022

SS

dSn

dSn

2

00

21

V

GAUSS

SS

dvdSdSn

Proof:

1 0

S

dSn

Proof:

00

2

0

2

dvdSnn

S

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

2nd Green’s Identity:

203


Harmonic Functions (continue 1)

A continuous function φ with continuous first and second partial derivatives is saidto be harmonic if it satisfies Laplace’s Equation 02 Properties of Harmonic Functions (continue – 1)

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

3 A function φ harmonic in a volume V can be expressed in terms of the functionand its normal derivative on the surface S bounding V.

Proof:Use the solution of the Poisson (Laplace) Equation: 02

S SFSF

F dSrrnnrr

Tr

11

4

where VoutsidendSndSSonF

VinFT

11

2

1

1

204




RS

dSn

1

V

Fr

Sr

FSF rrR

4 If the surface S is a sphere SR of radius R with center at then RS

RF dSR

r

24

1Fr

Proof:

011

SS SF

dSnR

dSnrr

12

11

RrrnRS

SF

Therefore:

RS

R

S SFSF

F dSR

dSrrnnrr

Tr

24

111

4

3

5 If φ is harmonic in a volume V bounded by the surface S and if φ = c = constantat every point on S, then φ = c at every point of V.

Proof:

cdcrr

dScdS

rrnnrrr

S SFS

rr

SFcSF

F

SF

4

2

/10

4

1

4

111

4

1

2

3

205




6 A non-constant function φ harmonic in a region V can have neither a maximumnor a minimum in V.

Proof: For any point inside the region V we can choose an infinitesimal sphere δS centered at this point for which

nn

dvdSdSn

VSS

1000

Since can not be either positive or negative inside the region V, the maximum orminimum of the potential φ can occur only at boundary of the region.

n

S

dSn

1

V

Fr

Sr

SF rrr

SF

206




7 If φ is harmonic in a region V bounded by a surface S and ∂ φ/∂ n = 0 at everypoint of S, then φ = constant at every point of V.

S

dSn

1

V

Fr

Sr

FSF rrr

0

Sn

Proof:

SVV

dSdvdv 2

Use: 0& 2

SV

dSdv 0

2

0

0

2

SV

dSn

dv

Therefore:

VinconstVin .0

1st Green’s Identity:

207




8 If φ1 and φ2 are two solutions of Laplace’s equation in a volume V whose normalderivatives take the same value ∂ φ1/∂ n = ∂ φ2/∂ n on the surface S boundingV, then φ1 and φ2 can differ only by a constant.

S

dSn

1

V

Fr

Sr

FSF rrr

0

Sn

Proof:

Define:

0: 2

2

1

22

21

021

SSSnnn

From it follows that φ is a constant. 7

Table of Contents

208

SOLOFundamental Theorem of Vector Analysis for a Bounded Region V (Helmholtz’s Theorem)

Vector Analysis

Hermann Ludwig Ferdinandvon Helmholtz

1821 - 1894

Let be a continuous vector field with continuous divergence and curl, in a region V bounded by a surface S. Then has a unique representation as sum of a potential field and a solenoidal field , i.e.

rAA

1A

2A

A

FFFFF rArArA 21

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

S FS

S

V FS

SS

V FS

SF dS

rr

rAdv

rr

rAdv

rr

rA

4

1

4

1

4

1:

S FS

S

V FS

SS

V FS

SF dSrr

rAdv

rr

rAdv

rrrA

4

1

4

11

4

1:

S FS

S

V FS

SSF

S FS

S

V FS

SSFF

dSrr

rAdv

rr

rA

dSrr

rAdv

rr

rArA

4

1

4

1

4

1

4

1

Therefore

209

SOLO

Proof of the Fundamental Theorem of Vector Analysis for a Bounded Region V

Vector Analysis

Let use the fact that (see GREEN’s FUNCTION):

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

zzyyxxr SSSS 111

zzyyxxr FFFF 111

where

We can write:

zz

yy

xx SSS

S 111 Sr

We define the operator that accts only on .

The operator acts only on .

zz

yy

xx FFF

F 111 Fr

SF

FS

F rrrr

4

12

1

0

00

dxx

x

x

x

V FS

SF

V FS

FS

V

SFSF dvrr

rAdvrr

rAdvrrrArA

1

4

11

4

1 22

Using the identity we obtain: 2

FFFFFF

V FS

SFF

V FS

SFFF dv

rrrAdv

rr

rArA

1

4

1

4

1

210


Let develop first the divergence expression:

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

S FS

S

V

SS

FS

V FS

SS

V

SS

FS

V FS

SS

V FS

FS

V FS

SF

dSrr

rAdvrA

rr

dvrr

rAdvrA

rr

dvrr

rAdvrr

rAdvrr

rA

4

11

4

1

4

11

4

1

1

4

11

4

11

4

1

Define:

S FS

S

V

SS

FSV FS

SF dSrr

rAdvrA

rrdv

rrrA

4

11

4

11

4

1:

V FS

SFF

V FS

SFFF dv

rrrAdv

rr

rArA

1

4

1

4

1

Proof of the Fundamental Theorem of Vector Analysis for a Bounded Region V(continue – 1)

211


Let develop now the rotor expression:

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

Define:

S FS

S

V FS

SS

V FS

SF dS

rr

rAdv

rr

rAdv

rr

rA

4

1

4

1

4

1:

V FS

S

S

V FS

SS

V FS

SS

V FS

FS

V FS

S

F

dvrr

rAdv

rr

rA

dvrr

rAdvrr

rAdvrr

rA

4

1

4

1

1

4

11

4

1

4

1

but

S FS

S

S FS

S

S FS

S

V FS

S

S

tconsC

V FS

S

S

dSrr

rACdS

rr

rAC

dSrr

rACdv

rr

rACdv

rr

rAC

4

1

4

1

4

1

4

1

4

1

S FS

SGAUSS

V FS

S

S dSrr

rAdv

rr

rA

4

1

4

1 5

V FS

SFF

V FS

SFFF dv

rrrAdv

rr

rArA

1

4

1

4

1

Gauss 5


212


We found:

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

Therefore:

S FS

S

V FS

SS

V FS

SF dS

rr

rAdv

rr

rAdv

rr

rA

4

1

4

1

4

1:

V FS

SFF

V FS

SFFF dv

rrrAdv

rr

rArA

1

4

1

4

1

S FS

S

V

SS

FSV FS

SF dSrr

rAdvrA

rrdv

rrrA

4

11

4

11

4

1:

FFFrA

q.e.d.


Table of Contents

213

SOLO

Fundamental Theorem of Vector Analysis for an Unbounded Region (Helmholtz’s Theorem)

Vector Analysis

For a Bounded region V we found:


1821 - 1894

Let be a continuous vector field with continuous divergence and curl, such that falls off at infinity like 1/r 1+ε while and fall off at infinity like 1/r 2+ε where ε > 0. Then has a unique representation (to within constant vectors) at sum ofa potential field and a solenoidal field , i.e.

rAA

A

A

A

A

rAA

11 rAA

22

0&0

4

1

4

1

UU

Udvrr

rAdv

rr

rArA

FF

V FS

SSF

V FS

SSFF

S FS

S

V FS

SSF

S FS

S

V FS

SSFF

dSrr

rAdv

rr

rA

dSrr

rAdv

rr

rArA

4

1

4

1

4

1

4

1

3/4&4 32 RVRSRrr FS

For an Unbounded region V:The surface integrals are finite only if falls off at infinity like 1/r 1+ε where ε > 0. rA

The volume integrals are finite only if and fall off at infinity like 1/r 2+ε. A

A

214

SOLO

Fundamental Theorem of Vector Analysis for an Unbounded Region (Helmholtz’s Theorem)

Vector Analysis

0,0 11 rArA

Therefore


1821 - 1894

Let be a continuous vector field with continuous divergence and curl, such that falls off at infinity like 1/r 1+ε while and fall off at infinity like 1/r 2+ε where ε > 0. Then has a unique representation (to within constant vectors) at sum ofa potential field and a solenoidal field , i.e.

rAA

A

A

A

A

rAA

11 rAA

22

0&0

4

1

4

1

UU

Udvrr

rAdv

rr

rArA

FF

V FS

SSF

V FS

SSFF

S FS

S

V FS

SSF

S FS

S

V FS

SSFF

dSrr

rAdv

rr

rA

dSrr

rAdv

rr

rArA

4

1

4

1

4

1

4

1

Table of Contents

215

REYNOLDS’ TRANSPORT THEOREM

This is a part of the Presentations

“FLUID DYNAMICS”

v (t)

S(t)

SflowV ,

sd

OSV ,

OSOflowSflow VVV ,,,

OSr ,

md OSV ,

OflowV ,

Or,

-any system of coordinatesOxyz

- any continuous and differentiable functions in

trtr OO ,,, ,,

tandrO,

trO ,,

- flow density at point

and time tOr,

SOLO

- mass flow through the element .mdsdV S , sd

- any control volume, changing shape, bounded by a closed surface S(t)v (t)

- flow velocity, relative to O, at point and time t trV OOflow ,,,

Or,

- position and velocity, relative to O, of an element of surface, part of the control surface S(t).

OSOS Vr ,, ,

- area of the opening i, in the control surface S(t).iopenS

- gradient operator in O frame.O,

- flow relative to the opening i, in the control surface S(t).OSiOflowSi VVV ,,,

- differential of any vector , in O frame.O

td

d

FLUID DYNAMICS FLUID DYNAMICS

MATHEMATICS

SOLO HERMELIN

Updated: 5.03.07

216

Start with LEIBNIZ THEOREM from CALCULUS:

ChangeBoundariesthetodueChange

tb

ta

tb

ta td

tadttaf

td

tbdttbfdx

t

txfdxtxf

td

dLEIBNITZ

)),(()),((

),(),(::

)(

)(

)(

)(

and generalized it for a 3 dimensional vector space on a volume v(t) bounded by thesurface S(t).

Using LEIBNIZ THEOREM followed by GAUSS THEOREM (GAUSS 4):

tv

OSOOOSGAUSS

OpointtoRelativedsofMovement

thetodueChange

tS

OS

tv O

LEIBNITZ

Otv

vdVVt

GAUSSsdVvd

tvd

td

d,,,,)4(

)(

,

This is REYNOLDS’ TRANSPORT THEOREM

OSBORNEREYNOLDS

1842-1912

SOLO

GOTTFRIED WILHELMvon LEIBNIZ

1646-1716

REYNOLDS’ TRANSPORT THEOREM

v (t)

S(t)

SflowV ,

sd

OSV ,


OSr ,

md OSV ,

OflowV ,

Or,

FLUID DYNAMICS

1 .MATHEMATICAL NOTATIONS (CONTINUE)

217

FLUID DYNAMICS


1.11 REYNOLDS’ TRANSPORT THEOREM (CONTINUE)


)(,,,,)4(

,)()()(

tvOSOOOS

OGAUSS

OStStv

O

LEIBNITZ

Otv

vdVVt

GAUSS

sdVvdt

vdtd

d

)(

,

,)4(

,)()()(

tv k

kOS

i

k

i

kOSi

GAUSS

kkOStS

itv

iLEIBNITZ

tvi

vdx

V

xV

t

GAUSS

sdVvdt

vdtd

d

SOLO

v (t)

S(t)

SflowV ,

sd

OSV ,


OSr ,

md OSV ,

OflowV ,

Or,

218

FLUID DYNAMICS




O

OOS td

RduV

,,

CASE 1 (CONTROL VOLUME vF ATTACHED TO THE FLUID)

kkOS uV ,

)(,,,)4(

,)()()(

tvOOO

OGAUSS

OtStv

OOtv

F

FFF

vduut

GAUSS

sduvdt

vdtd

d

)()4(

)()()(

tv k

kI

k

Ik

I

GAUSS

kKtS

Itv

I

tvI

F

FFF

vdx

u

xu

t

GAUSS

sduvdt

vdtd

d

SOLO

vF (t)

SF(t)

sd

OSV ,

0,,, OSOflowS VVV

OSR ,

OR,

md

OSV ,

OflowV ,

219

FLUID DYNAMICS




1&, kkOS uV1&, uV OS

CASE 2 (CONTROL VOLUME vF ATTACHED TO THE FLUID AND )1

)(

,,)(

,)(

)(

tvOO

tSO

tv

F

FFF

vdusduvdtd

d

td

tvd )()()(

)(

tv k

kk

tSk

tv

F

FFF

dvx

udsudv

td

d

td

tvd

td

tvd

tvu F

Ftv

OOF

)(

)(

1lim

0)(,,

td

tvd

tvx

u F

Ftv

k

k

F

)(

)(

1lim

0)(

EULER 1755

SOLO

vF (t)

SF(t)

sd

OSV ,

0,,, OSOflowS VVV

OSR ,

OR,

md

OSV ,

OflowV ,

220

FLUID DYNAMICS




CASE 3 (CONTROL VOLUME vF ATTACHED TO THE FLUID AND )

&, kkOS uV &, uV OS

or, since this is true for any attached volume vF(t)

)(,,

)(,

)( )(

)(0

tvOO

tSO

tv tv

F

FF F

vdut

sduvdt

vdtd

d

td

tmd

)(

)()( )(

)(0

tvk

k

tSkk

tv tv

F

FF F

vduxt

sduvdt

dvtd

d

td

tmd

Because the Control Volume vF is attached to the fluid and they are not sources or sinks ,the mass is constant.

OOOOOO uut

ut ,,,,,,0

k

k

k

k

k x

u

xu

tu

xt

0

SOLO

vF (t)

SF(t)

sd

OSV ,

0,,, OSOflowS VVV

OSR ,

OR,

md

OSV ,

OflowV ,

221

FLUID DYNAMICS




CASE 4 (CONTROL VOLUME WITH FIXED SHAPE C.V. )0,

OSV

Define

.... VC

OOVC

vdt

vdtd

d

.... VC

i

VCi vd

tvd

td

d

r t r t r t, , , i k k i kx t x t x t, , ,

)(,

)()(

tSOS

tvOO

tv

sdV

vdtt

vdtd

d

ktS

kOSi

tvi

i

tvi

sdV

vdtt

vdtd

d

FF

)(,

)()(

We have

but

OOOO ut

ut ,,,, 0

k

k

iik

k

uxt

uxt

0

CASE 5 r t r t r t, , ,

SOLO

222

FLUID DYNAMICS




We have

)(

,,)(

4

.

)(,

)(,,,,,,

)(,

)(,,

)(

tSOOS

tvO

MDG

DerMatGAUSS

tSOS

tvOOOOOO

O

tSOS

tvOO

OOtv

sduVvdtD

D

sdV

vduuut

sdV

vdut

vdtd

d

)(

,)(

4

.

)(,

)(

)(,

)()(

tSkkkOSi

tv

iMDG

DerMatGAUSS

tSkkOSi

tv k

ki

k

ik

k

ik

i

tSkkOSi

tv k

ki

i

tvi

sduVvdtD

D

sdV

vdx

u

xu

xu

t

sdV

vdx

u

tvd

td

d


SOLO

v (t)

S(t)

SflowV ,

sd

OSV ,


OSr ,

md OSV ,

OflowV ,

Or,

223

FLUID DYNAMICS




REYNOLDS 1

)(,,

)(

)(

tSOOS

tvO

Otv

sduVvdtD

D

vdtd

d

)(,

)(

)(

tSkkkOSi

tv

i

tvi

sduVvdtD

D

dvtd

d

REYNOLDS 2

)(

)(,,

)(

tvO

tSOSO

Otv

vdtD

D

sdVuvdtd

d

)(

)(,

)(

tv

i

tSkkOSki

tvi

vdtD

D

sdVuvdtd

d


SOLO

v (t)

S(t)

SflowV ,

sd

OSV ,


OSr ,

md OSV ,

OflowV ,

Or,

224

FLUID DYNAMICS




REYNOLDS 3

CASE 1 (CONTROL VOLUME ATTACHED TO THE FLUID vF(t) )

kkOS uV ,

)()( tv

OOtv FF

vdtD

Dvd

td

d

)()( tv

i

tvi

FF

vdtD

Dvd

td

d

SOLO

O

OOS td

RduV

,,

r t r t r t, , ,

vF (t)

SF(t)

sd

OSV ,

0,,, OSOflowS VVV

OSR ,

OR,

md

OSV ,

OflowV ,

CASE 4 (CONTROL VOLUME WITH FIXED SHAPE C.V. )0,

OSV

REYNOLDS 4

..,

..

..

SCO

OVC

VCO

sduvdtd

d

vdtD

D

....

..

SCkki

VCi

VC

i

sduvdtd

d

vdtD

D

Table of Contents

225

Poisson’s Non-homogeneous Differential EquationSOLO

The Poisson’s Non-homogeneous Differential Equation for the Static Electric Scalar Potential Ve is:

We want to find the Electric Scalar Potential Ve at the point F (field) due to all thesources (S) in the volume V, including its boundaries .

n

iiSS

1

SFSeS rrrV

1

,2

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

F inside V F on the boundary of V

Therefore is the vector from S to F.SF rrr

zzyyxxr 111

zzyyxxr SSSS 111

zzyyxxr FFFF 111

Let define the operator

that acts only on the source coordinate .

zz

yy

xx SSS

S 111

Sr

Note: See development in the Power PointPresentation Table of Contents

226


To find the solution we used the following:

• GREEN’s IDENTITY

S

ee

V

ee dSGVVGdvGVVG 22

• GREEN’s FUNCTION

This Green’s Function is a particlar solution of the following Poisson’sNon-homogeneous Differential Equation:

0;;1

; 2

FSSFSSF

FS rrrrrr

rrG

FSFSS rrrrG

4;2

Siméon Denis Poisson1781-1840


227



S

ee

V

ee dSGVVGdvGVVG 22

Let start from the Gauss’ Identity

SV

dSAdvA


where is any vector field (function of position and time) continuous and differentiable in the volume V. Let define .

A

eVGA

eee VGVGVGA 2

Then

S

e

V

ee

V

e dSVGdvVGVGdvVG 2

If we interchange G with Ve we obtain

S

e

V

ee

V

e dSGVdvGVVGdvGV 2

Subtracting the second equation from the first we obtain

First Green’s Identity

Second Green’s Identity

We have

228

SOLO


Define the Green’s Function is a particlar solution of the following Poisson’sNon-homogeneous Differential Equation:

where δ (x) is the Dirac function

1

0

00

dxx

x

x

x

Let use the Fourier Transformation to write

33

3

exp2

1

exp2

1

exp2

1exp

2

1exp

2

1

dkrrkj

dkdkdkzzkyykxxkj

dkzzjkdkyyjkdkxxjk

zzyyxxrr

SF

zyxSFzSFySFx

zSFzySFyxSFx

SFSFSFSF

zyx

zyx

dkdkdkdk

zkykxkk

3

111

where

FSFSS rrrrG

4;2

Paul Adrien MauriceDirac

1902 - 1984


229

SOLO

• GREEN’s FUNCTION (continue – 1)


Hence

or

SFFS rrkjkgdkrrG

exp; 3

SFSFS rrkjdkrrkjkgdk

exp2

4exp 3

3

32

SFSFS rrkjdkrrkjkgdk

exp2

4exp 3

3

23


Jean Baptiste JosephFourier

1768 - 1830

230

SOLO


Let compute:

Therefore:

Because this is true for all k, we obtain

SFzSFySFxSSSFS zzkyykxxkjrrkj expexp2

SFzSFySFxzyxS zzkyykxxkjzjkyjkxjk exp111

SFzSFySFxSzyx zzkyykxxkjzjkyjkxjk

exp111

SFSFSFS rrkjkrrkjkjkjrrkjkj

expexpexp 2

SFSF rrkjdkrrkjkkgdk

exp2

4exp 3

3

23

22

1

2

1

kkg


231

SOLO


Let use spherical coordinates relative to vector:r

rr

r

r

kk

kk

kk

z

y

x

z

y

x

0

0

cos

sinsin

cossin

dk sindk

dk

dddkkdk sin23

r

xy

z

SFFS rrkjkgdkrrG

exp; 3

22

1

2

1

kkg


232

SOLO


2

3

2

exp

2

1;

k

rkjdkrrG FS

0 0

2

0

222

sincosexp

2

1

dkddkk

jkr

0 0

2coscosexp2

2

1

dkdjkr

00 0

2

expexp2cosexp1dk

kj

jkrjkr

rdk

jkr

jkr

rr

dkk

kr

r

1

2

2sin2

0


2

sin

0

dkk

krwhere we used (see next slide)

SF

FS rrrrrG

11;Therefore

rr

r

r

kk

kk

kk

z

y

x

z

y

x

0

0

cos

sinsin

cossin

dk sindk

dk

dddkkdk sin23

r

xy

z

233

SOLO



0

sindk

k

krLet compute:

x

y

R

A

B

C

D

E

F

G

H

Rx Rx

For this use the integral: 0ABCDEFGHA

zi

dzz

e

Since z = 0 is outside the region of integration

0

BCDEF

ziR xi

GHA

zi

R

xi

ABCDEFGHA

zi

dzz

edx

x

edz

z

edx

x

edz

z

e

00

0000

sin2

sin2

sinlim2limlimlim dk

k

rkidx

x

xidx

x

xidx

x

eedx

x

edx

x

eR

R

R xixi

R

R xi

RR

xi

R

idideideie

edz

z

e i

ii

eii

i

eiez

GHA

zi

00

1

0

0

00limlimlim

012

2

0

/2/2sin

0

sin

00

R

RRReRii

i

eRieRz

BCDEF

zi

eR

dedededeRieR

edz

z

e i

ii

Therefore: 0sin

20

idkk

rkidz

z

e

ABCDEFGHA

zi 2

sin

0

dkk

kr

234



a Green’s Function for the Poisson’s Non-homogeneous Differential Equation

SF

FS rrrrG

1;

Hence

This solution is not unique since we can add any function that satisfies theLaplace’s Equation

0;2 FSS rr

Therefore we have the following Green’s Function

0;;1

; 2

FSSFSSF

FS rrrrrr

rrG

Pierre-Simon Laplace1749-1827

FSFSS rrrrG

4;2

235


Let return to the Poisson’s Non-homogeneous Differential Equation (1812) for the Electric Scalar Potential Ve is:

We want to find the Electric Scalar Potential Ve at the point F (field) due to all the sources (S) in the volume V, includingits boundaries

n

iiSS

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr


SFSeS rrrV

1

,2

Siméon Denis Poisson1781-1840

a

http://maxwell.byu.edu/~spencerr/phys442/node4.htmlHistory

236


Let define the operator that acts only on the source coordinate .Sr

zz

yy

xx SSS

S 111

is the vector from S to F.SF rrr

zzyyxxr 111

zzyyxxr SSSS 111

zzyyxxr FFFF 111

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

0

rr

r

rr

rr

rr

rrrrr

SF

SF

SF

FSSFSS

30 SSSFSS rrrr

01

2 r

r

r

rr

r

dr

dGr

dr

dGrG SS

0031311

34333

2

r

rr

r

r

rr

rr

rr

rrGrG SSSSSS

SFSeS rrrV

1

,2

Since is no defined at r = 0 we define the volume V’ as the volume V minus a smallsphere of radius and surface around the point F, when F is inside V, or asemi-sphere of radius and surface around the point F, when F is on the boundary of V.

rG

00 r

00 r

204 rSF

202 rSF

237


let compute

'' '0

2

''

22 ,1

,V

SFS

V Vin

Se

V

SFS

V

SeeS dVrrrGdvGVdvr

rrGdvGVVG

FF S

SeeS

S

SeeS dSnr

VVr

dSGVVG11

drn

rVV

r SeeSr

2

0

11lim

0

2

0

220

0

2

0

0

0limlimlimlim

drnVdrn

r

nVdnVrdnVr Se

re

reS

reS

r

VoutsideF

SonF

VinF

rVdrV Feerr FS

0

2

4

lim

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

Using the Green’s Identity

S

ee

V

ee dSGVVGdvGVVG 22

238


We obtain

S

e

e

V

SSF

dSndS

nn

S

SeeS

V

SSFFe

dSn

GV

n

VG

TdvrrrG

T

dSGVVGT

dvrrrGT

rV

S

4,

4

4,

4

1

1


VinFT

11

2

1

1

Note If F is outside V from the Green’s Second Identity we obtain

End Note

VoutsidendSndSdSn

GV

n

VGdSGVVGdvGVVG

S

e

e

S

SeeS

V

SeeS

1122

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

SF

FS rrrrG

1;

239

S

De

S

DSeFe dSn

GV

TdSGV

TrV

44


VinFT

11

2

1

1


BOUNDARY CONDITIONS

The General Green Function that is a class of bi-position function, andcontains an arbitrary harmonic function (solution of the Laplace’s Equation)

0;;1

; 2

FSSFSSF

FS rrrrrr

rrG

Let consider the following two simple cases (Dirichlet and Neumann Problems):

1. Dirichlet Problem

Johann Peter Gustav Lejeune Dirichlet

1805-1859

The potential is defined at the boundary S of the volume V.

n

iiFe SSongivenrV

1

In this case

Let choose such that FS rr

; SrrrGG SSFSDirichlet

0;

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

240


BOUNDARY CONDITIONS (continues – 1)

2. Neumann Problem

The potential derivative is defined at the boundary S of the volume V.

In this case

S

SnSSongivennVr

n

V n

ii

SeSF

e

1&11

Let choose such that FS rr

; SrnrrGn

rrGG S

SFSS

S

FSNeumann

01;

;


VinFT

11

2

1

1

S

eN

S

eSNFe dSn

VG

TdSVG

TrV

44

Franz Neumann1798-1895

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

241


Uniqueness of a Laplace Solution that satisfies Dirichlet or Neumann Boundary Conditions

Suppose that we have a solution Ve that satisfies the Laplace Homogeneous Differential Equation:

0,2 FSeS rrV

in the volume V, including its boundaries .n

iiSS

1

Suppose also that Dirichlet or Neumann conditions or a combination of those, are specified. In this case the solution is unique (up to an additive constant).

Proof

Suppose that thee exist two solutions and , and define rVe1 rVe 2

rVrVr ee 21

We have

01

22

1

122

SS r

e

r

e rVrVr

242


Uniqueness of a Laplace Solution that satisfies Dirichlet or Neumann Boundary Conditions

If Dirichlet conditions are satisfied:

Proof (continue)

Let use Green’s First Identity (with G = Φ)

(continue – 1)

n

ii

rVrV

FeFeF SSonrVrVrFeFe

121 0

21

If Neumann conditions are satisfied:

n

ii

rn

Vr

n

V

Fe

Fe

F SSonrn

Vr

n

Vr

n

Fe

Fe

1

21 0

21

SSVV

dSn

dSGdvdv 2

We have

VinconstVVVindSn

dv ee

Dirichlet

NeumannSV

2100

End of Proof

243


In the same way the solution of the Poisson’s Non-homogeneous Differential Equation for the Vector Potential is: FrA

S SF

SS

SFV SF

SSdSndS

nn

S SF

SSSS

SFV SF

SSF

dSrrn

rAn

rA

rr

Tdv

rr

rAT

dSrr

rArArr

Tdv

rr

rATrA

S

11

44

11

44

21

1

2


VinFT

11

2

1

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

Table of Contents

244

ELECTROMAGNETICSSOLO

KIRCHHOFF’s SOLUTION OF THE SCALAR HELMHOLTZ NONHOMOGENEOUS DIFFERENTIAL EQUATION

The Helmholtz Non-homogeneous Differential Equation for the Electric Scalar Potential Ve is:

trtrVtv

trV eee ,1

,1

,2

2

22

We want to find the Electric Scalar Potential Ve at the point F (field) due to all the sources (S) in the volume V, including its boundaries .

n

iiSS

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr


Therefore is the vector from S to F.SF rrr

zzyyxxr 111

zzyyxxr SSSS 111

zzyyxxr FFFF 111

Let define the operator

that acts only on the source coordinate .

zz

yy

xx SSS

S 111

Sr

This is a part of the presentation“Electromagnetics”

SOLO HERMELIN

ELECTROMAGNETICS

245



To find the solution we need to prove the following:


S

ee

V

ee dSGVVGdVGVVG 22


FS

FS

FS rr

v

rrtt

trtrG

'

',;,

This Green’s Function is a particular solution of the following Helmholtz Non-homogeneous Differential Equation:

'4',;,1

',;,2

2

22 ttrrtrtrG

tvtrtrG SFFSFSS

(continue – 1)

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr


246



• Scalar Green’s Identities

S

ee

V

ee dSGVVGdVGVVG 22

(continue – 2)

Let start from the Gauss’ Divergence Theorem

SV

dSAdVA


where is any vector field (function of position and time) continuous and differentiable in the volume V. Let define .

A

eVGA

eee VGVGVGA 2

Then

S

e

Gauss

V

ee

V

e dSVGdVVGVGdVVG 2

If we interchange G with Ve we obtain

S

e

Gauss

V

ee

V

e dSGVdVGVVGdVGV 2

Subtracting the second equation from the first we obtain

First Green’s Identity

Second Green’s Identity

We have


247




Define the Green’s Function as a particular solution of the following Helmholtz Non-homogeneous Differential Equation:

'4',;,1

',;,2

2

22 ttrrtrtrG

tvtrtrG SFFSFSS

(continue – 3)

where δ (x) is the Dirac function

1

0

00

dxx

x

x

x


33

3

exp2

1

exp2

1

exp2

1exp

2

1exp

2

1

dkrrkj

dkdkdkzzkyykxxkj

dkzzjkdkyyjkdkxxjk

zzyyxxrr

SF

zyxSFzSFySFx

zSFzySFyxSFx

SFSFSFSF

zyx

zyx

dkdkdkdk

zkykxkk

3

111

where

Paul Dirac1902-1984

Joseph Fourier 1768-1830

248




In the same way:

(continue – 4)

dttjtt 'exp2

1'

Therefore

'expexp2

1' 3

4ttjrrkjddkttrr SFSF


'expexp,',;, 3 ttjrrkjkgddktrtrG SFFS

Hence

'expexp2

4'expexp,

1 3

4

3

2

2

2

2 ttjrrkjddkttjrrkjkgddktv SFSFS

or

'expexp2

4

'expexp1

exp'exp,

3

4

2

2

2

23

ttjrrkjddk

ttjt

rrkjv

rrkjttjkgddk

SF

SFSFS

249




Let compute:

(continue – 5)

SFSFSFS

SFzSFySFxSzyx

SFzSFySFxzyxS

SFzSFySFxSSSFS

rrkjkrrkjkjkjrrkjkj

zzkyykxxkjzjkyjkxjk

zzkyykxxkjzjkyjkxjk

zzkyykxxkjrrkj

expexpexp

exp111

exp111

expexp

2

2

'exp'exp 2

2

2

ttjttjt

Therefore:

'expexp2

4

'expexp,

3

4

2

223

ttjrrkjddk

ttjrrkjv

kkgddk

SF

SF

Because this is true for all k and ω, we obtain

2

22

3

1

4

1,

vk

kg

250



• GREEN’s FUNCTION (continue – 3) (continue – 6)

'expexp

4

1'expexp,',;,

2

22

33

3 ttjrrkj

vk

ddkttjrrkjkgddktrtrG SFSFFS

We can see that the integral in k has to singular points forv

k

Let consider only the progressive wave, i.e. G = 0 for t > t’.

To find the integral let change ω by ω + jδ where δ is a small negative number

rkj

v

jk

ddktrtrG FS

exp4

1',;,

2

22

33

where and .SF rrr

'tt

251




In the plane ω we close the integration path by the semi-circle withand the singular points on the upper side, for τ > 0 (for t > t’)

rkj

v

jk

ddktrtrG FS

exp4

1',;,

2

22

33

r

'00exp ttdjfUPC

'00exp ttdjfDOWNC

0exp

0exp

exp

DOWN

UP

C

C

djf

djf

djf

jvk jvk Re

Im

252




CC jvkjvk

rkjvdrkj

vj

k

drkj

vj

k

dI

expexpexp

2

2

22

2

22

Let use the Cauchy Integral for a complex function f (z) continuous on a closed path C, in the complex z plane: 0

0

2lim20

zfjzfjdzzz

zfzz

C

We have:

k

kvrkjv

vk

jkv

vk

jkvrkjvj

jvk

rkjvj

jvk

rkjvjI

vkvk

sinexp2

2

exp

2

expexp2

exp2lim

exp2lim

2

2

0,

2

0,

Therefore, we can write:

k

vkrkjdk

v

vk

rkjddktrtrG FS

sinexp

2

exp

4

1',;, 3

2

2

22

33

253




Let use spherical coordinates relative to vector:

k

vkrkjdk

v

vk

rkjddktrtrG FS

sinexp

2

exp

4

1',;, 3

2

2

22

33

r

rr

r

r

kk

kk

kk

z

y

x

z

y

x

0

0

cos

sinsin

cossin

dk sindk

dk

dddkkdk sin23

r

xy

z

254




kvdvr

jkvvr

jkvvr

jkvvr

jkv

r

4

expexpexpexp1

r

v

rrtt

v

rrtt SFSF

''

rvr

vr

dxv

rjx

v

rjx

r

expexp

2

11

kvdvr

jkvvr

jkv

r

vkvd

vr

jkvvr

jkv

r

4

expexp

4

expexp1

dk

j

jvkjvk

j

jkrjkr

r

v

2

expexp

2

expexp

dkvkvkrr

vdkvkvkr

r

v

sinsinsinsin2

0

00 0

sin2

expexp2cosexpsin dkvk

j

jkrjkr

r

vdk

jkr

jkrvkk

v

0 0

2cossincosexp2

2

dkdvkjkrkv

0 0

2

0

22

sinsincosexp

2

dkddkk

vkjkrv

k

vkrkjdk

vtrtrG FS

sinexp

2',;, 3

2

rr

r

r

kk

kk

kk

z

y

x

z

y

x

0

0

cos

sinsin

cossin

dk sindk

dk

dddkkdk sin23

r

xy

z

255




We can see that represents a progressive waveand represents a regressive wave:

v

rrtt

v

rrtt SFSF

''

v

rrtt

v

rrtt SFSF

''

Hence SF

SFSF

FS rr

v

rrtt

v

rrtt

trtrG

''

',;,

We shall consider only the progressive wave and use:

SF

SF

FS rr

v

rrtt

trtrG

'

',;,

Retarded Green Function

The other solution is:

SF

SF

FS rr

v

rrtt

trtrG

'

',;,

Advanced Green Function

256


KIRCHHOFF’s SOLUTION OF THE SCALAR HELMHOLTZ NONHOMOGENEOUS DIFFERENTIAL EQUATION (continue – 12)

Gustav Robert Kirchhoff1824-1887

Let return to the Helmholtz Non-homogeneous Differential Equation for the Electric Scalar Potential Ve is:

trtrVtv

trV eee ,1

,1

,2

2

22

We want to find the Electric Scalar Potential Ve at the point F (field) due to all the sources (S) in the volume V, includingits boundaries

n

iiSS

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr


Hermann von Helmholtz1821-1894

257



Since is no defined at r = 0 we define the volume V’ as the volume V minus a smallsphere of radius and surface around the point F, when F is inside V, or asemi-sphere of radius and surface around the point F, when F is on the boundary of V.

rG

00 r

00 r

204 rSF

202 rSF

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

Let define the operator that acts only on the source coordinate .Sr

zz

yy

xx SSS

S 111

Using the Green’s Identity

FSS

SFSSeSeSSF

V

SFSSeSeSSF dStrtrGtrVtrVtrtrGdVtrtrGtrVtrVtrtrG ',;,',',',;,',;,',',',;,'

22

substitute here

',1

','

1',

2

2

22 trtrV

tvtrV SeSeSeS

'4',;,'

1',;,

2

2

22 ttrrtrtrG

tvtrtrG FSFSFSS

we obtain

FSS

SFSSeeSSSF

V

SFSeS

SF

V

SFSeSeSF

dStrtrGtrVtrVtrtrG

dVttrrtrVtr

trtrG

dVtrtrGt

trVtrVt

trtrGv

',;,',',',;,

'4',',

',;,

',;,'

',','

',;,1

'

'2

2

2

2

2

258



Let integrate the equation between to and choose t such that:1' tt 2' tt 21 /' tvrttt

2

2

1

1

2

1

''4',',

',;,

'',;,'

',','

',;,1

'

'2

2

2

2

2

I

t

t V

SFSeS

SF

I

t

t V

SFSeSeSF

dtdVttrrtrVtr

trtrG

dtdVtrtrGt

trVtrVt

trtrGv

4

2

1

3

2

1

'',;,',',',;,

'',;,',',',;,

I

t

t S

SFSSeeSSSF

I

t

t S

SFSSeeSSSF

dtdStrtrGtrVtrVtrtrG

dtdStrtrGtrVtrVtrtrG

F

259



Integral I1

0

'

',;,',;, 21

SF

SF

SFSF rr

v

rrtt

ttrtrGttrtrG

Since 21 /' tvrttt

then

0',;,'

',','

',;,1

'',;,'

',','

',;,'

1

'',;,'

',','

',;,1

'2

'2

'2

2

2

2

21

2

1

2

1

2

1

V

t

t

SFSeSeSF

V

t

t

SFSeSeSF

t

t V

SFSeSeSF

dVtrtrGt

trVtrVt

trtrGv

dVdttrtrGt

trVtrVt

trtrGtv

dVdttrtrGt

trVtrVt

trtrGv

I

0',;,'

',;,' 21

ttrtrGt

ttrtrGt SFSF

260



Integral I2

2

1

''4',',

',;,'

2

t

t V

SFSeS

SF dtdVttrrtrVtr

trtrGI

2

1

2

1 ''

''',4'',/'

t

t

t

t V

SFSe

V

S

FS

dVdtttrrtrVdtdVtr

rr

vrtt

' /'

0

'' /'

',1'',4

',1

V vrttFS

S

V

SFSe

V vrttFS

S dVrr

trdVttrrtrVdV

rr

tr

The integral is zero since in V’. '

/',V

FSSe dVrrvrttrV

FS rr

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

261



Integral I3

2

1

'',;,',',',;,3

t

t S

SFSSeSeSSF dSdttrtrGtrVtrVtrtrGI

S

t

t SF

SF

SSeSeSSF

SF

dSdtrr

v

rrtt

trVtrVrr

v

rrtt

2

1

'

'

',',

'

S

t

t SFS

SFSeSeS

SF

SF

dSdtrrv

rrtttrVtrV

rr

v

rrtt

2

1

'1

'',',

'

S

t

t SF

SFS

Se dSdtrr

v

rrtt

trV2

1

'

'

',

262



Integral I3 (continue – 1)

But

v

rrvrtt

trrvrtt

rv

rrtt SFS

SFS

rrrSF

S

SF

/''

/''

and

S

t

t

SF

SF

SFSSe

S

t

t SF

SFS

Se dSdtv

rrtt

trrv

rrtrVdSdt

rr

v

rrtt

trV2

1

2

1

'''

','

'

',

S

t

t

SF

SF

SFSSe

S

t

t

SF

SF

SFSSe

partsbyegrating

dSdtv

rrtt

rrv

rrtrV

tdS

v

rrtt

rrv

rrtrV

2

1

2

1

''','

'',

0

int

263




Therefore

S

t

t

SF

Se

SF

SFS

SF

SSe

SF

SeS dSdtv

rrtttrV

trrv

rr

rrtrV

rr

trVI

2

1

''','

1',

',3

S

vrtt

Se

SF

SFS

SF

SSe

SF

SeS dStrVtrrv

rr

rrtrV

rr

trV

/'

','

1',

',

S

vrtt

Se

SF

SF

SF

SSe

SF

Se

dStrVtrrv

rrn

rrtrV

rr

n

trV

/'

','

1',

',

The last equality follows from dSn

UdSnn

n

UdSU SS

11

264



Integral I4

In the same way as for integral I4 we have

2

1

'',;,',',',;,4

t

t S

SFSSeSeSSF

F

dSdttrtrGtrVtrVtrtrGI

FS

vrtt

Se

SF

SFS

SF

SSe

SF

SeS dStrVtrrv

rr

rrtrV

rr

trV

/'

','

1',

',

FS

vrtt

Se

SF

SF

SF

SSe

SF

Se

dStrVtrrv

rrn

rrtrV

rr

n

trV

/'

','

1',

',

265




We use since points inside V normal to and points outside V.

nr

r1

0

0

0

0

r

r

nS

n1

On the sphere or the semi-spherearound the field point F with radius and surface or if the point F is inside V or on the boundary, respectively, we have

FS rrr

002

04 rSF 202 rSF

nr

r

rr

rr

rr

rrrr

SF

SF

SF

FS

SSFSF

10

0

n

rr

r

rrr

rr

rrrr

rr

rrrr SF

SF

SFSF

FS

SFSSFS

F

111111

200

02

022

ndrr

rdrdS

FS

120

0

020

Since we can assume mean values for all field quantities in the integral00 r

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

266




FS

vrtt

Se

SF

SFS

SF

SSe

SF

SeS dStrVtrrv

rr

rrtrV

rr

trVI

/'

4 ','

1',

',

drnnrv

trVt

nr

trVr

trV

vrtt

SeSeSeS

r

2

0

/'0

2

000

111

','

11

',',

lim0

0

0

/'0/'0

0

0/'0',

'lim',lim1',lim

000

dv

rtrV

tdtrVdrntrV

vrtt

ServrttServrttSeSr

trVSonF

VinFtrV

SonFd

VinFd

FeFe ,2

4,

2

0

4

0

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

267



SUMMARIZE

The Kirchhoff’s solution to the Helmholtz Non-homogeneous Differential Equation:

trtrVtv

trV eee ,1

,1

,2

2

22

is

S

v

rrtt

Se

SF

SFS

SF

SSe

SF

SeS

V

v

rrttSF

SFe dStrV

trrv

rr

rrtrV

rr

trVTdV

rr

trTtrV

SFSF

''

','

1',

',

4

',

4,

S

v

rrtt

Se

SF

SF

SF

Se

SF

Se

V

v

rrttSF

SdSndS

nn

dStrVtrrv

rrn

rrntrV

rr

n

trV

TdV

rr

trT

SF

SFS

'

'

ˆ

ˆ

','

1',

',

4

',

4

VoutsidenSonF

VinFT

1

2

1

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

Table of Contents

268

SOLO

References

M.R. Spiegel, “Vector Analysis and an Introduction to Tensor Analysis”, Schaum’s Outline Series, McGraw-Hill, 1959

Vector Analysis

H.Lass, “Vector and Tensor Analysis”, McGraw Hill, 1950

J,N, Reddy & M.L. Rasmussen, “Advanced Engineering Analysis”, John Willey, 1982, Ch.1:”Elements of Vector and Tensor Analysis”

Table of Contents

April 13, 2023 269

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

vector analysis

Science