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Vector and Tensor Algebra Vectors: Scalar Product; Vector Product; Dyadic Product

Lecture 4: Vectors: Vector Product; Dyadic Product

Dr.-Ing. Rene Marklein (University of Kassel) Mathematical Foundations of EFT (MFEFT) WS 2007/2008 73 / 115
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MFEFT - Lecture 4

1 Introduction

2 Vector and Tensor Algebra

Position Vector and Coordinate SystemsCartesian CoordinatesEinsteins Summation ConventionDifferentiation of the Position Vector; Kronecker Symbol (Kronecker Delta)Cylinder Coordinate SystemOrthogonal Curvilinear Coordinate SystemSpherical Coordinate SystemDupin Coordinates

Vectors: Scalar Product; Vector Product; Dyadic ProductScalar ProductVector ProductDyadic ProductComplex Vectors

TensorsDefinition

3 Vector and Tensor Analysis

4 Distributions5 Complex Analysis

6 Special Functions

7 Fourier Transform

8 Laplace Transform

9

ReferencesDr.-Ing. Rene Marklein (University of Kassel) Mathematical Foundations of EFT (MFEFT) WS 2007/2008 74 / 115
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MFEFT - Lecture 4

1 Introduction

2 Vector and Tensor Algebra

Position Vector and Coordinate SystemsCartesian CoordinatesEinsteins Summation ConventionDifferentiation of the Position Vector; Kronecker Symbol (Kronecker Delta)Cylinder Coordinate SystemOrthogonal Curvilinear Coordinate SystemSpherical Coordinate SystemDupin Coordinates

Vectors: Scalar Product; Vector Product; Dyadic ProductScalar ProductVector ProductDyadic ProductComplex Vectors

TensorsDefinition

3 Vector and Tensor Analysis

4 Distributions5 Complex Analysis

6 Special Functions

7 Fourier Transform

8 Laplace Transform

9

ReferencesDr.-Ing. Rene Marklein (University of Kassel) Mathematical Foundations of EFT (MFEFT) WS 2007/2008 75 / 115
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Vector Product

Vector Product

The definition of the vector product isshown in Fig. 8. Two vectors A and Bspan a parallelogram with the surface

F = A B sin . (203)

The vector C with the length equal to F isnormal ( perpendicular) to the surface Fand the orientation ofC is given by theright-hand rule

C = AB . (204)

Thats because the vector product is notcommutative

BA = AB . (205)

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2

2

A

BC

F= |C|

Figure 8: Definition of the Vector Product


V d T Al b V S l P d V P d D di P d
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Vector Product

Vector Product between two Parallel or Antiparallel Vectors

It follows that two vectors are parallel or antiparallel, if the vector product is zero:

A B AB = 0 .

Vector Product between the Unit Vectors of the Cartesian Coordinate System

For example, the vector product between the unit vectors of the Cartesian coordinate system is

exex = 0

eyey = 0 (206)

ezez = 0

and

exey = ez

exez = ey (207)

eyez = ex .


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Vector Product

Vector Product between two Parallel or Antiparallel Vectors

It follows that two vectors are parallel or antiparallel, if the vector product is zero:

A B AB = 0 .

Vector Product between the Unit Vectors of the Cartesian Coordinate System

For example, the vector product between the unit vectors of the Cartesian coordinate system is

exex = 0

eyey = 0 (206)

ezez = 0

and

exey = ez

exez = ey (207)

eyez = ex .


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Vector Product between two Vectors: Components Form in CartesianCoordinates

In components representation ofA and B it follows explicitly by using Eq. (207) and (206)

C = (Axex + Ayey + Azez)(Bxex + Byey + Bzez) (208)

= AxexBxex + AyeyBxex + AzezBxex+ AxexByey + AyeyByey + AzezByey+ AxexBzez + AyeyBzez + AzezBzez (209)

= AxBx exex =0

+AyBx eyex =ez

+AzBx ezex =ey

+ AxBy exey =ez

+AyBy eyey =0

+AzBy ezey =ex

+ AxBz exez =ey

+AyBz eyez =ex

+AzBz ezez =0

(210)

= (AyBz AzBy) =Cx

ex + (AzBx AxBz) =Cy

ey + (AxBy AyBx) =Cz

ez . (211)


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= AxBx exex =0

+AyBx eyex =ez

+AzBx ezex =ey

+ AxBy exey =ez

+AyBy eyey =0

+AzBy ezey =ex

+ AxBz exez =ey

+AyBz eyez =ex

+AzBz ezez =0

(210)

= (AyBz AzBy) =Cx



ez . (211)



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= AxBx exex =0

+AyBx eyex =ez

+AzBx ezex =ey

+ AxBy exey =ez

+AyBy eyey =0

+AzBy ezey =ex

+ AxBz exez =ey

+AyBz eyez =ex

+AzBz ezez =0

(210)

= (AyBz AzBy) =Cx



ez . (211)



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g ; ; y





= AxBx exex =0

+AyBx eyex =ez

+AzBx ezex =ey

+ AxBy exey =ez

+AyBy eyey =0

+AzBy ezey =ex

+ AxBz exez =ey

+AyBz eyez =ex

+AzBz ezez =0

(210)

= (AyBz AzBy) =Cx



ez . (211)



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g y

Sarrus Scheme

Sarrus Scheme

The last result is also obtained by applying the Sarrus Scheme introduced by the Frenchmathematician Pierre Frederic Sarrus (1798-1861):


= det

ex ey ezAx Ay AzBx By Bz

=


. (213)

Computing the determinant of the 3 3 matrix gives:

C =


(214)

= exAyBz + eyAzBx + ezAxBy BxAyez ByAzex BzAxey (215)

= (AyBz AzBy) =Cx



ez . (216)



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Sarrus Scheme

Sarrus Scheme



= det


=


. (213)


C =


(214)


= (AyBz AzBy) =Cx



ez . (216)



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Sarrus Scheme

Sarrus Scheme



= det


=


. (213)


C =


(214)


= (AyBz AzBy) =Cx



ez . (216)



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Sarrus Scheme

Sarrus Scheme



= det


=


. (213)


C =


(214)


= (AyBz AzBy) =Cx



ez . (216)



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Sarrus Scheme

Sarrus Scheme



= det


=


. (213)


C =


(214)


= (AyBz AzBy) =Cx



ez . (216)



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Sarrus Scheme

Sarrus Scheme



= det


=


. (213)


C =


(214)


= (AyBz AzBy) =Cx



ez . (216)



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Levi-Civita Symbol

Levi-Civita Symbol

We can also make use of the so-called Levi-Civita Symbol ijk , i,j,k = 1, 2, 3 introduced by theItalian mathematician T. LeviCivita (1873-1941)

ijk =

0 , if two subscripts are equal

1 , if ijk is a even permutation of 123

1 , if ijk is a odd permutation of 123 .

(217)

Then, the components of the result vector of the vector product of two vectors read

Ci =3

j=1

3k=1

ijkAjBk (218)

and by using the Einstein summation convention:

Ci = ijkAjBk . (219)



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Levi-Civita Symbol

Levi-Civita SymbolWe can also make use of the so-called Levi-Civita Symbol ijk , i,j,k = 1, 2, 3 introduced by theItalian mathematician T. LeviCivita (1873-1941)

ijk =




(217)


Ci =3

j=1

3k=1

ijkAjBk (218)





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Levi-Civita Symbol

Levi-Civita SymbolWe can also make use of the so-called Levi-Civita Symbol ijk , i,j,k = 1, 2, 3 introduced by theItalian mathematician T. LeviCivita (1873-1941)

ijk =




(217)


Ci =3

j=1

3k=1

ijkAjBk (218)





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Multiple Vector and Scalar Products

Vector Triple Product (VTB)

The vector triple product is defined as

A(BC) = B(A C) C(A B) (220)

Say: A cross B cross C = BAC minus CAB.

This is called the Vector Triple Product (VTP), because it involves three terms (vectors) and theresult is a vector. The right-hand side can be shown to be correct by direct evaluation of the

vector product. The VTO appears for example in theMagnetostatic (MS) Case: in the derivation of the vectorial Poisson/Laplace equation forthe magnetic vector potential A(R):

B(R) = A(R) H(R) =1

A(R) (221)

H(R) = 1

A(R)

= 1

A(R) = Je(R) . (222)

Electromagnetic (EM) Case: in the derivation of the vectorial wave equation for the electricfield strength E(R, t):

E(R, t) = 0

t J(R, t) 002

t2E(R, t) (223)Dr.-Ing. Rene Marklein (University of Kassel) Mathematical Foundations of EFT (MFEFT) WS 2007/2008 81 / 115

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A(BC) = B(A C) C(A B) (220)




B(R) = A(R) H(R) =1

A(R) (221)

H(R) = 1

A(R)

= 1

A(R) = Je(R) . (222)


E(R, t) = 0

t J(R, t) 00

2



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A(BC) = B(A C) C(A B) (220)




B(R) = A(R) H(R) =1

A(R) (221)

H(R) = 1

A(R)

= 1

A(R) = Je(R) . (222)


E(R, t) = 0

t J(R, t) 00

2


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A(BC) = B(A C) C(A B) (220)




B(R) = A(R) H(R) =1

A(R) (221)

H(R) = 1

A(R)

= 1

A(R) = Je(R) . (222)


E(R, t) = 0

t J(R, t) 00

2



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A(BC) = B(A C) C(A B) (220)




B(R) = A(R) H(R) =1

A(R) (221)

H(R) = 1

A(R)

= 1

A(R) = Je(R) . (222)


E(R, t) = 0

t J(R, t) 00

2



M l i l V d S l P d

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Proof of the Identity A(BC) = B(A C)C(A B)

BC =

ex ey ezBx By BzCx Cy Cz

(225)

= (ByCz BzCy)ex + (BzCx BxCz)ey + (BxCy ByCx)ez . (226)

A(BC) =

ex ey ezAx Ay Az

(ByCz BzCy) (BzCx BxCz) (BxCy ByCx)

(227)= (Ay(BxCy ByCx) Az(BzCx BxCz))ex

+ (Az(ByCz BzCy) Ax(BxCy ByCx))ey

+ (Ax(BzCx BxCz) Ay(ByCz BzCy))ez (228)

= BxexAyCy AyByCxex AzBzCxex + BxexAzCz

+ ByeyAzCz AzBzCyey AxBxCyey + ByeyAxCx

+ BzezAxCx AxBxCzez AyByCzez + BzezAyCy . (229)



M l i l V d S l P d

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BC =


(225)


A(BC) =

ex ey ezAx Ay Az










M lti l V t d S l P d t

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BC =


(225)


A(BC) =

ex ey ezAx Ay Az










M lti l V t d S l P d t

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BC =


(225)


A(BC) =

ex ey ezAx Ay Az



+ (Az(ByCz BzCy) Ax(BxCy ByCx))ey+ (Ax(BzCx BxCz) Ay(ByCz BzCy))ez (228)







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BC =


(225)


A(BC) =

ex ey ezAx Ay Az



+ (Az(ByCz BzCy) Ax(BxCy ByCx))ey+ (Ax(BzCx BxCz) Ay(ByCz BzCy))ez (228)







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A(BC)


+ByeyAzCz AzBzCyey AxBxCyey + ByeyAxCx

+BzezAxCx AxBxCzez AyByCzez + BzezAyCy (230)

= Bxex (AyCy + AzCz) (AyBy + AzBz) Cxex+ Byey (AzCz + AxCx) (AzBz + AxBx) Cyey+ Bzez (AxCx + AyCy) (AxBx + AyBy) Czez (231)

= Bxex (AyCy + AzCz) +BxexAxCx (AyBy + AzBz) CxexBxexAxCx

+ Byey (AzCz + AxCx) +ByeyAyCy (AzBz + AxBx) CyeyByeyAyCy

+ Bzez (AxCx + AyCy) +BzezAzCz (AxBx + AyBy) CzezBzezAzCz (232)

= Bxex (AxCx + AyCy + AzCz) (BxAx + AyBy + AzBz) Cxex+ Byey (AxCx+AyCy + AzCz) (AxBx+ByAy + AzBz) Cyey+ Bzez (AxCx + AyCy+AzCz) (AxBx + AyBy+BzAz) Czez . (233)




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A(BC)












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A(BC)












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A(BC)












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Proof of the Identity A(BC) = B(AC)C(A

B)

A(BC)

= Bxex (AxCx + AyCy + AzCz) =A C

(BxAx + AyBy + AzBz) =A B

Cxex

+ Byey (AxCx+AyCy + AzCz) =A C

(AxBx+ByAy + AzBz) =A B

Cyey

+ Bzez (AxCx + AyCy+AzCz) =A C

(AxBx + AyBy+BzAz) =A B

Czez (234)

= Bxex

(A C) + Byey

(A C) + Bzez

(A C)

(A B) Cxex (A B) Cyey (A B) Czez (235)

= B (A C) (A B)C . (236)




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B)

A(BC)



Cxex



Cyey



Czez (234)

= Bxex

(A C) + Byey

(A C) + Bzez

(A C)


= B (A C) (A B)C . (236)




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p S


B)

A(BC)



Cxex



Cyey



Czez (234)

= Bxex (A C) + Byey (A C) + Bzez (A C)


= B (A C) (A B)C . (236)




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p

Scalar Triple Product (STP)

The Scalar Triple Product is defined by

A (BC) = [ABC] , (237)

which determines the volume of the parallelepiped represented by the vectors A,B,C. This

scalar value is a so-called Pseudo Scalar [Hafner, 1987], because the sign depends on thehandedness of the involved vector product.

Special Cases of the Scalar Triple Product (STP)

The resulting vector of two vectors is always perpendicular to both vectors, then

A (AB) = 0B (AB) = 0 . (238)




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p

Scalar Triple Product (STP)

The Scalar Triple Product is defined by

A (BC) = [ABC] , (237)

which determines the volume of the parallelepiped represented by the vectors A,B,C. This

scalar value is a so-called Pseudo Scalar [Hafner, 1987], because the sign depends on thehandedness of the involved vector product.

Special Cases of the Scalar Triple Product (STP)

The resulting vector of two vectors is always perpendicular to both vectors, then

A (AB) = 0B (AB) = 0 . (238)



Dyadic Product

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Definition of the Dyadic Product

The Dyadic Product (product without a dot or cross) of two vectors is defined by the formalmultiplication of the vectors in components form

AB = (Axex + Ayey + Azez) (Bxex + Byey + Bzez) . (239)

The multiplication gives the dyadic products of the unit vectors:

AB = AxBxexex + AxByexey + AxBzexez+AyBxeyex + AyByeyey + AyBzeyez

+AzBxezex + AzByezey + AzBzezez . (240)

The dyadic product is not commutative, this means

AB = BA

ex ey = ey exex ez = ez ex

...



Dyadic Product

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AB = BA


...



Dyadic Product

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AB = BA


...



Dyadic Product

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AB = BA


...



Dyadic Product

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AB = BA


...



Dyadic Product

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AB = BA


...



Dyadic Product

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Dyadic Product in Matrix Form

We also can write the components form of the dyadic product in a 33 Matrix. The (cartesian)

components representation of a vector

A = Axex + Ayey + Azez (241)

can be written by keeping the unit vectors in mind in form of a row vector or a column vector

{A} = {Ax Ay Az} (row vector) (242)

{A} =

AxAyAz

(column vector) . (243)

Then, it follows for the dyadic product of two vectors AB by keeping the unit vectors of thefixed coordinate system in mind

AB =

AxAyAz

{Bx By Bz} =

AxBx AxBy AxBzAyBx AyBy AyBzAzBx AzBy AzBz

. (244)

Obviously, the dyadic products exiexj , i,j = 1, 2, 3, in Eq. (240) determine the position of the

AiBj element in the matrix in Eq. (244).



Dyadic Product

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{A} =

AxAyAz



AB =

AxAyAz

{Bx By Bz} =


. (244)





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{A} =

AxAyAz



AB =

AxAyAz

{Bx By Bz} =


. (244)





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{A} =

AxAyAz



AB =

AxAyAz

{Bx By Bz} =


. (244)





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{A} =

AxAyAz



AB =

AxAyAz

{Bx By Bz} =


. (244)





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{A} =

AxAyAz



AB =

AxAyAz

{Bx By Bz} =


. (244)





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The entry AxBz is positioned at the place exez

AB =

AxBx AxBy AxBzAyBx AyBy AyBz

AzBx AzBy AzBz

and the entry AzBx is given by the position ezex

AB =


AzBx AzBy AzBz

.

This proves that the dyadic product is not commutative

AB = BA . (245)

We summarize: In this sense, the dyad has nine components in comparison to the the threecomponents of a vector. In the discussed case AB, these nine components are determined by sixvector components.



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AB =


AzBx AzBy AzBz


AB =


AzBx AzBy AzBz

.


AB = BA . (245)




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AB =


AzBx AzBy AzBz


AB =


AzBx AzBy AzBz

.


AB = BA . (245)




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AB =


AzBx AzBy AzBz


AB =


AzBx AzBy AzBz

.


AB = BA . (245)




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Practical Meaning of the Dyadic Product

A practical meaning of the dyadic product, if according to a matrix-vector multiplication adot product (contraction) is a applied to a vector. We interpret the operation the linearmapping

ABC =

AxBx AxBy AxBz

AyBx AyBy AyBzAzBx AzBy AzBz

Cx

CyCz (246)

or

C AB = {Cx Cy Cz}


(247)

in a meaningful way.


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For example, we find by computing the DyadVector Multiplication in Eq. (246) and

interpretation of the result according to Eq. (243)

AB C =

AxBxCx + AxByCy + AxBzCzAyBxCx + AyByCy + AyBzCz

AzBxCx + AzByCy + AzBzCz

= Ax(BxCx + ByCy + BzCz)

Ay(BxCx + ByCy + BzCz)Az(BxCx + ByCy + BzCz)

= (Axex + Ayey + Azez)(BxCx + ByCy + BzCz) (248)

and respectively by calculation of the Vector-Dyad Multiplication

C AB = (CxAx + CyAy + CzAz)(Bxex + Byey + Bzez) , (249)

which can be written in the coordinate-free form

AB C = A(B C) (250)

C AB = (C A)B . (251)



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AB C =









AB C = A(B C) (250)

C AB = (C A)B . (251)



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AB C =









AB C = A(B C) (250)

C AB = (C A)B . (251)



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The result of the product of a dyadAB

with a vectorC

AB C = A(B C)

is again a vector, but in the direction ofA, where the length is modified by the scalarproduct B Ca).

Similarly, the dot product of a vector C with a dyad AB

C AB = (C A)B

is a vector in direciton ofB, which is stretched/compressed by the scalar product C A.

The contraction of a dyad with a vector determines a rotation ofC in A or ofC in B, where

AB C = C AB . (252)

a)Note that for all vectors C in a plane perpendicular to B is AB C = 0, i. e., the Cs are building the kernel of a(nullspace) of the linear mapping AB C [Burg et al., 1990]; for all vectors in the kernel AB C = D it is impossible to solveC in a unique way.



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with a vectorC

AB C = A(B C)



C AB = (C A)B



AB C = C AB . (252)




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with a vectorC

AB C = A(B C)



C AB = (C A)B



AB C = C AB . (252)




EM Application: Hertzian Dipole Radiation

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The simplest antenna to radiate electromagnetic waves is a Hertzian Dipole. The vector of theelectric field strength E of the radiated electromagnetic field is a function of the observationpoint R and has a different magnitude and direction at every point in space:

E(R, ) = 0 2 p

e() G(0)(R, ) . (253)

The change in direction (rotation) and amplitude (stretching/compression) relative to the

arbitrary but constant directed dipole moment pe is given by a dyad, the so-called DyadicGreens Function:

G(R, ) =

I+

1

k20

ej k0R

4R

= I R R+ jk0R (I 3R R) 1

k20R2 (I 3R R) e

j k0R

4R for R = 0

def= G(0)(R, ) for R = 0 . (254)




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The simplest antenna to radiate electromagnetic waves is a Hertzian Dipole. The vector of theelectric field strength E of the radiated electromagnetic field is a function of the observationpoint R and has a different magnitude and direction at every point in space:

E(R, ) = 0 2 p

e() G(0)(R, ) . (253)

The change in direction (rotation) and amplitude (stretching/compression) relative to the

arbitrary but constant directed dipole moment pe is given by a dyad, the so-called DyadicGreens Function:

G(R, ) =

I+

1

k20

ej k0R

4R

= I R R+ jk0R (I 3R R) 1

k20R2 (I 3R R) e

j k0R

4R for R = 0

def= G(0)(R, ) for R = 0 . (254)



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The Cross Product between a Vector and a Dyad

We can generalize the results obtained for the scalar product to the vector product of a vectorand a dyad and vice versa:

ABC = A(BC) (255)

CAB = (CA)B , (256)

but note that the result is a dyad and these relations are not commutative.



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Dyadic Product with Numbered Coordinates and Summation Convention

If we use the numbered Cartesian coordinates (240) we write:

AB =3

i=1

3j=1

AxiBxjexiexj (257)

and applying the summation convention we find:

AB = Axi Bxj exi exj , (258)

where we have to sum i, j from 1, 2, 3. Note that exi exj = exj exi for i = j. A dyad AB has

components with double indices

Dxixj = Axi Bxj ; (259)

this results in the notation of a so-called 2nd rank tensor D:

D = AB = Axi Bxj exi exj = Dxixj exi exj . (260)

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