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Position Vector and Coordinate Systems Orthogonal Curvilinear Coordinate System

Lecture 3: Position Vector and Coordinate Systems (cont.); Vectors: ScalarProduct; Vector Product; Dyadic Product

Dr.-Ing. Rene Marklein (University of Kassel) Mathematical Foundation of EFT (MFEFT) WS 2007/2008 42 / 77
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Orthogonal Curvilinear Coordinate System

Uniqueness of the Transformation Formulas

The explicit request of a NEW Orthonormal Coordinate System according to

e i e k = ik (110)

transfers this requirement into

ij e xj

kl e xl

= ik (111)

ij kl e x j e x l

= jl= ik (112)

and further to

ij kj = ik on the left-hand sidesum over j from 1 to 3 (113)

with the denition

e x j e x l = jl . (114)


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e i e k = ik (110)


ij e xj

kl e xl

= ik (111)

ij kl e x j e x l

= jl= ik (112)

and further to


with the denition

e x j e x l = jl . (114)


d d h l l d

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e i e k = ik (110)


ij e xj

kl e xl

= ik (111)

ij kl e x j e x l

= jl= ik (112)

and further to


with the denition

e x j e x l = jl . (114)


P iti V t d C di t S t O th l C ili C di t S t

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e i e k = ik (110)


ij e xj

kl e xl

= ik (111)

ij kl e x j e x l

= jl= ik (112)

and further to


with the denition

e x j e x l = jl . (114)


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Orthonormal Relation; Transformation Dyad


The Orthonormal Relation (see Eq. ( 113) )

e x j e x l = jl (115)

says, that the Transposed T

of the Dyada)

is equal to the Inverse , i.e., the dot product T is equal to the unit dyadic I :

T = I (116)

= 1 (117)

is an Orthogonal Dyad .a ) Compared to with the elements ij has T the elements ji .


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e x j e x l = jl (115)


of the Dyada)


T = I (116)

= 1 (117)



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e x j e x l = jl (115)


of the Dyada)


T = I (116)

= 1 (117)




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Via the computation of the products e i e x j we can illustrate the meaning of ij ; it follows

with the application of the orthonormal property of e x j

e i e x j = il e x l

e x j (118)

= il lj (119)= ij (120)= cos (e i , e x j ) . (121)



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y g y






e x j (118)

= il lj (119)= ij (120)= cos (e i , e x j ) . (121)



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y g y






e x j (118)

= il lj (119)= ij (120)= cos (e i , e x j ) . (121)



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e x j (118)

= il lj (119)= ij (120)= cos (e i , e x j ) . (121)



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The ij are

ij = cos (e i , e x j ) (122)

the so-called Direction Cosines of the NEW orthonormal tripod a vectors relative to the OLD Cartesian Coordinate System. The ij are determine the Local Rotation of the NEWorthonormal tripod at every point in space. If this rotation is independent of position, then thecoordinate transformation in Eq. ( 96) is a simple rotation of a cartesian coordinate system.

a Tripod is a word generally used to refer to a three-legged object, generally one used as a platform of some sort, and comesfrom the Greek tripous, meaning three feet.



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Handiness of the Coordinate System

The Handiness of the NEw Coordinate System compared to the OLD one is given by theDeterminant of . With the properties det I = 1 and det T = det it follows

det ( T ) = det det T

= 1 . (123)

and

(det )2 = 1 (124)

and

det = 1 (125)

=+1 the tripod e 1 , e 2 , e 3 is right-handed

1 the tripod e 1 , e 2 , e 3 is left-handed(126)



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= 1 . (123)

and

(det )2 = 1 (124)

and

det = 1 (125)





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= 1 . (123)

and

(det )2 = 1 (124)

and

det = 1 (125)





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= 1 . (123)

and

(det )2 = 1 (124)

and

det = 1 (125)





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If the OLD Coordinate System is not a Cartesian Coordinate System, but also CurvilinearOrthogonal, than we have the Transformation j i , and Eq. ( 109) reads for this general case:

ij =h jh i

j i

=h ih j

i j

. (127)


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If the OLD Coordinate System is not a Cartesian Coordinate System, but also CurvilinearOrthogonal, than we have the Transformation j i , and Eq. ( 109) reads for this general case:

ij =h jh i

j i

=h ih j

i j

. (127)



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Orthogonal Curvilinear Coordinate SystemIllustration of the Meaning of the Metric Coefficients h i

Illustration of the Meaning of the MetricCoefficients h iLike the ij we can illustrate the meaning of themetric coefficients h i . h has been alreadydiscussed. We dene a Line Element ds as the

magnitude of the differential change dR of theposition vector, i. e., a change of R to R + dR(see Fig. 3):

ds 2 = dR dR . (128)

In order the compute ds in the orthogonal

curvilinear coordinates we build the total differentialof R with regard to the dependence of 1 , 2 , 3

dR = R

1d1 +

R

2d2 +

R

3d3 (129)

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. ................

.......................................................................................................................................................

......................................................................................................................................................

...................................................................................................................

O

R

R + dR

dR , |dR | = d s

Figure 3: Denition of the Line Elements



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Illustration of the Meaning of the Metric Coefficients h iWe multiply and make use of the summation convention

ds2

= R

i di

R

j dj (130)= d i h i e i

e j

= ijh j dj (131)

= h 2 j d2j (132)

= h 2 1 d21 + h 2 2 d22 + h 2 3 d23 . (133)



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ds2

= R

i di

R

j dj (130)= d i h i e i

e j

= ijh j dj (131)

= h 2 j d2j (132)

= h 2 1 d21 + h 2 2 d22 + h 2 3 d23 . (133)



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ds2

=

R

i di

R

j dj (130)= d i h i e i

e j

= ijh j dj (131)

= h 2 j d2j (132)

= h2 1 d

21 + h

2 2 d

22 + h

2 3 d

23 . (133)



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ds2

=

R

i di

R

j dj (130)= d i h i e i

e j

= ijh j dj (131)

= h 2 j d2j (132)

= h2 1 d

21 + h

2 2 d

22 + h

2 3 d

23 . (133)



h l l d

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Illustration of the Meaning of the Metric Coefficients h iBecause of the invariance of the Scalar Line Element ds when the Coordinate System ischanging, we can use Eq. ( 133)

ds 2 = h 2 1 d21 + h

2 2 d

22 + h

2 3 d

23

to interpret the Metric Coefficients :In the Cartesian Coordinate System ds read

ds 2 = d x 2 + d y2 + d z 2 [m2 ] (134)

this means ds 2 is given by the Theorem of Pythagoras by adding the squares of the threeMetric Differential Changes dx , dy , dz in the direction of the Coordinate Lines .

In an arbitrary Orthogonal Curvilinear Coordinate System the h i determine the Metric of the Coordinate Lines (in meter).The di can represent as d in cylinder coordinates and d in spherical coordinates aChange in Angle Direction .



O h l C ili C di S

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ds 2 = h 2 1 d21 + h

2 2 d

22 + h

2 3 d

23


ds 2 = d x 2 + d y2 + d z 2 [m2 ] (134)





O th l C ili C di t S t
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ds 2 = h 2 1 d21 + h

2 2 d

22 + h

2 3 d

23


ds 2 = d x 2 + d y2 + d z 2 [m2 ] (134)





Ill t ti f th M i g f th M t i C ffi i t h

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Illustration of the Meaning of the Metric Coefficientsh iDifferential Volume Element

Differential Volume Element

This interpretation of the h i makes clear, that a Differential Volume Element dV in CartesianCoordinate System

dV = d x dy dz (135)

can be generalized to a Differential Volume Element of an arbitrary Curvilinear CoordinateSystem

dV = h 1 d1 h 2 d2 h 3 d3 (136)= h

1h

2h

3d1 d2 d3 (137)



Illustration of the Meaning of the Metric Coefficientsh

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Illustration of the Meaning of the Metric Coefficientsh iDifferential Volume Element

Differential Volume Element

This interpretation of the h i makes clear, that a Differential Volume Element dV in CartesianCoordinate System

dV = d x dy dz (135)

can be generalized to a Differential Volume Element of an arbitrary Curvilinear CoordinateSystem

dV = h 1 d1 h 2 d2 h 3 d3 (136)= h

1h

2h

3d1 d2 d3 (137)


Position Vector and Coordinate Systems Spherical Coordinate System

Spherical Coordinate System

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Spherical Coordinate SystemCoordinates; Unit Vectors, Magnitude of the Position Vector

Spherical Coordinates; Unit Vectors,Magnitude of the Position Vector

Cartesisan Coordinates: R, , in thelimits 0 R < , 0 ,0 < 2

: polar angle; : azimuth angleOrthonormal Unit Vectors: e R , e , e with |e R | = |e | = |e | = 1 a ande R

e

e

b

The straight line from the coordinateorigin O to the (observation) point P is

illustrating the position vector R of P the magnitude of the position vector is

|R | = R = R 2a || stands for the magnitude of the argumentb stands for perpendicular

R eq(q,j)

e (j)j

y

R

z

j

x

O

P R,( )jq,

e R(q,j)

q

Figure 4: Spherical Coordinates of the spatialpoint P and the related position vector




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e

e

b


illustrating the position vectorR

of P the magnitude of the position vector is


R eq(q,j)

e (j)j

y

R

z

j

x

O

P R,( )jq,

e R(q,j)

q





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e

e

b





R eq(q,j)

e (j)j

y

R

z

j

x

O

P R,( )jq,

e R(q,j)

q





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e

e

b





R eq(q,j)

e (j)j

y

R

z

j

x

O

P R,( )jq,

e R(q,j)

q





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Coordinate Transformation FormulasThe Transformation Formulas are following from Fig. 4:

x = R sin cos (138)y = R sin sin (139)z = R cos (140)

Cartesian Position Vector as a Funtion of the Spherical Coordinates

The representation of the Position Vector in the Cartesian Coordinate System as a function of the Spherical Coordinates is:

R= x R sin cos

ex + y

R sin sin e

y + z R cos e

z (141)

= R sin cos e x + R sin sin e y + R cos e z (142)




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Spherical Coordinate SystemMetric Coeffecients; Orthonormal Unit Vectors

Metric Coefficients

The three Metric Coefficients of the Spherical Coordinate System are

h R = 1 (143)h = R (144)h =

Rsin .

(145)

Orthonormal Unit Vectors

The Orthonormal Unit Vectors of the Spherical Coordinate System in form of the VectorDecomposition in the Cartesian Coordinate System as a function of the Spherical Coordinates read

e R (, ) = sin cos e x + sin sin e y + cos e z (146)e

(, ) = cos cos e x + cos sin e y sin e z (147)e () = sin e x + cos e y . (148)




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p yMetric Coeffecients; Orthonormal Unit Vectors

Metric Coefficients

The three Metric Coefficients of the Spherical Coordinate System are

h R = 1 (143)h = R (144)h

= R sin . (145)

Orthonormal Unit Vectors

The Orthonormal Unit Vectors of the Spherical Coordinate System in form of the VectorDecomposition in the Cartesian Coordinate System as a function of the Spherical Coordinates read

e R (, ) = sin cos e x + sin sin e y + cos e z (146)e

(, ) = cos cos e x + cos sin e y sin e z (147)e () = sin e x + cos e y . (148)




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p yPosition Vector in the Spherical Coordinate System

Position Vector in the Spherical Coordinate System

The Position Vector in the Spherical Coordinate System can be found via CoordinateTransformation from the Cartesian Coordinate to the Spherical Coordinate System. The result is:

R = R e R (, ) . (149)

The position vector in the spherical coordinate system has only ONE vector componentR e R (, ) with the scalar vector component R . The dependencies of the angles and arehidden in the unit vector e R (, ) .


For Unit Position Vector it follows then

R =R

R=

R e R (, )R

= e R (, ) . (150)




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R = R e R (, ) . (149)




R =R

R=

R e R (, )R

= e R (, ) . (150)




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R = R e R (, ) . (149)




R =R

R=

R e R (, )R

= e R (, ) . (150)




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R = R e R (, ) . (149)




R =R

R=

R e R (, )R

= e R (, ) . (150)




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R = R e R (, ) . (149)




R =R

R=

R e R (, )R

= e R (, ) . (150)


Position Vector and Coordinate Systems Dupin Coordinates

Dupin Coorindates

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Dupin Coorindates

The (Circular)Cylindrical and SphericalCoodinate System are Special Cases of the so-called general Dupin CoordinateSystem [Tai , 1992], which are veryimportant in the Vector and TensorAnalysis of Surfaces .

The Transition and BoundaryConditions for electromagnetic eldsfrom Maxwells equations are typicallyderived using Dupin Coordinates.Dupin Coordinates are orthogonalcurvilinear coordinates 1 , 2 , 3 with

the unit vectorse

1 ,e

2 ,n

, i. e.e

3 isthe unit normal vector n of the surfacegiven by e 1 and e 2 . The relatedmetric coefficient h 3 is h 3 = 1 . Thecoordinate system is right handed, if n = e 1 e 2 .

Figure 5: Dupin Coordinates of the surface = 1 2 plane with the unit normal vectorn = e

1 e

2

.................................................................................................................................................................................. ................ .......................................................................................................................................................................................................

............................

............................

.....................................

................................................................................................................................................................................O

1

2

n

e 1

e 2



Dupin Coorindates

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Dupin Coorindates



the unit vectorse

1 ,e

2 ,n

, i. e.e



1 e

2

.................................................................................................................................................................................. ................ .......................................................................................................................................................................................................

............................

............................

.....................................

................................................................................................................................................................................O

1

2

n

e 1

e 2



Dupin Coorindates

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Dupin Coorindates



the unit vectorse

1 ,e

2 ,n

, i. e.e



1 e

2

.................................................................................................................................................................................. ................ .......................................................................................................................................................................................................

............................

............................

.....................................

................................................................................................................................................................................O

1

2

n

e 1

e 2


Vectors: Scalar Product; Vector Product; Dyadic Product

MFEFT - Lecture 3

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1 Introduction2 Vector and Tensor Algebra

3 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

4 Vectors: Scalar Product; Vector Product; Dyadic ProductScalar Product

5 Vector and Tensor Analysis6 Distributions7 Complex Analysis8 Special Functions9 Fourier Transform

10 Laplace Transform



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The Scalar, Vector, and Dyadic ProductScalar Product (Dot Product)Example:

A B = C = Scalar! (151)Vector Product (Cross Product)Example:

A B = C = Vector! (152)Dyadic ProductExample:

A B = D = Dyad! (153)




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A B = D = Dyad! (153)




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A B = D = Dyad! (153)


Vectors: Scalar Product; Vector Product; Dyadic Product Scalar Product

Scalar Product

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Scalar Product

Fig. 6 shows a vector A , which is projected to a unitvector e , the result is given by the Scalar Product

A e = A cos , (154)

where determines the enclosed angle between Aand e .

Replacing e by a vector B with the magnitude Byields the general form of Eq. ( 154), theCommutative scalar product A B (say: A dot B ):

A B = B A (155)= A B cos . (156)

Obviously is A B = 0 , if A and B areperpendicular, A

B , to each other; this meansone can dene two orthogonal vectors by avanishing scalar product between both.

...............................................................................................................

..............................................................................................................

...............................................................................................................

......................................................................... ....................... ............................................................................................................................................................................................................................................................................................................................................................................................................................................

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.......................................

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..........................

..........................

.......................................

..........................

......

............. ............. ............. ............. ............

. ............. .......................... ............. ............. ............. .............

............. ............. ............. ............. .............

........................................................................

.......

......................................................................................................................

........................................................................

........................................................................

...................................................................................................................

........................................................................

.....................................................

........................................................................................................................................................................................................................................................ .......................................................................

....................................................................................................

A

e

A

e

Figure 6: Illustration des Skalarprodukts



Scalar Product
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Scalar Product


A e = A cos , (154)



A B = B A (155)= A B cos . (156)



...............................................................................................................

..............................................................................................................

...............................................................................................................

......................................................................... ....................... ............................................................................................................................................................................................................................................................................................................................................................................................................................................

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......

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. ............. .......................... ............. ............. ............. .............

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........

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A

e

A

e




Scalar Product
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Scalar Product


A e = A cos , (154)



A B = B A (155)= A B cos . (156)



...............................................................................................................

..............................................................................................................

...............................................................................................................

......................................................................... ....................... ............................................................................................................................................................................................................................................................................................................................................................................................................................................

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. ............. .......................... ............. ............. ............. ...........

.. ............. .......................... ............. .....

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A

e

A

e




Scalar Product

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Scalar Product: Orthonormal Tripod

The orthonormal tripod of a the cartesian coordinate system is charachterized by

e x i e x j = ij fur i, j = 1 , 2, 3 . (157)

Scalar Product: Scalar Vector Components

Further, we can use the scalar sroduct to determine the scalar vector components of a vector A ,

i. e. in the Cartesian Coordinate System we ndAx = A e xAy = A e y (158)Az = A e z .

We compute for the scalar product in components form of A and B

A B = (A x e x + A y e y + A z e z ) (B x e x + B y e y + B z e z ) (159)

and nd by formal multiplication and the use of Eq. ( 157)

A B = A x Bx + A y By + A z Bz . (160)



Scalar Product

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e x i e x j = ij fur i, j = 1 , 2, 3 . (157)







A B = A x Bx + A y By + A z Bz . (160)



Scalar Product

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e x i e x j = ij fur i, j = 1 , 2, 3 . (157)







A B = A x Bx + A y By + A z Bz . (160)



Scalar Product

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e x i e x j = ij fur i, j = 1 , 2, 3 . (157)







A B = A x Bx + A y By + A z Bz . (160)



Scalar Product

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Enclosed Angle between two General Vectors A and B

The Enclosed Angle between the vector A and B is if A = 0 and B = 0 :

cos =A B

A B

=Ax Bx + A y By + A z Bz

A2x + A 2y + A 2z B2x + B 2y + B 2z. (161)

Magnitude of a General Vector A

The Magnitude A of the Vector A is dened by the scalar product A A :

A = A A = A2x + A 2y + A 2z ; (162)Unit Vector of a General Vector A

Then, the Unit Vector of the Vector A can be computed by

A =A

A A=

A

A=

AxA

e x +AyA

e y +AzA

e z . (163)



Scalar Product
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cos =A B

A B


A2x + A 2y + A 2z B2x + B 2y + B 2z. (161)





A =A

A A=

A

A=

AxA

e x +AyA

e y +AzA

e z . (163)



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cos =A B

A B


A2x + A 2y + A 2z B2x + B 2y + B 2z. (161)





A =A

A A=

A

A=

AxA

e x +AyA

e y +AzA

e z . (163)



Scalar Product
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Scalar Product: More Short-Hand Notations

We are going to cite two other short-hand notation of the scalar product. With Eq. ( 158) innumbered form we nd

Ax i = A e x i fur i = 1 , 2, 3 (164)

and for B we obtain instead Eq. (159 )

A B =3

i =1Ax i Bx i (165)

or applying the summation convention

A B = A x i Bx i . (166)

Obviously, this proves that the scalar product is commutative, i.e.,

A B = B A . (167)



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Ax i = A e x i fur i = 1 , 2, 3 (164)


A B =3

i =1Ax i Bx i (165)


A B = A x i Bx i . (166)


A B = B A . (167)



Scalar Product
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Ax i = A e x i fur i = 1 , 2, 3 (164)


A B =3

i =1Ax i Bx i (165)


A B = A x i Bx i . (166)


A B = B A . (167)



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Ax i = A e x i fur i = 1 , 2, 3 (164)


A B =3

i =1Ax i Bx i (165)


A B = A x i Bx i . (166)


A B = B A . (167)



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Scalar Product: Independence of the Coordinate System

In generalization of Eq. (164 ) we dene for the components of a vector in orthogonal curvilinearcoordinates by a)

A i = A e i , (168)

and obtain by applying the summation convention

A = A i e i

= A i ij e x j= A x j e x j (169)

with

Ax j = ij A i , (170)

by applying the transformation formulas in Eq. (108 ).a ) At the point in space R ( 1 , 2 , 3 ) we project the general position dependent vector A ( 1 , 2 , 3 ) onto the position

dependent unit vectors e i ( 1 , 2 , 3 ) .



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In generalization of Eq. (164 ) we dene for the components of a vector in orthogonal curvilinearcoordinates by a)

A i = A e i , (168)

and obtain by applying the summation convention

A = A i e i= A i ij e x j

= A x j e x j (169)

with

Ax j = ij A i , (170)

by applying the transformation formulas in Eq. (108 ).a ) At the point in space R ( 1 , 2 , 3 ) we project the general position dependent vector A ( 1 , 2 , 3 ) onto the position

dependent unit vectors e i ( 1 , 2 , 3 ) .



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On the other hand, by inserting the cartesian components form A = A x i e x i in

A j = A

=A x i e x i e j

= A x i e x i e j

= A x i e x i e j

= ji

(171)

= ji Ax i . (172)

Number triples, which are transformed from the cartesian to an orthonormal curvilinearcoordinate system with Eq. ( 170) or Eq. (172) are in the mathematical sense (scalar) componentsof vectors. Because of the inverse of is equal to the transpose, Eq. ( 172), A j = ji Ax i canbe derived from Eq. (170 ), Ax

j=

ijA

i, via inversion and vice versa. The vector A as a

directed physical value is independent of the coordinate system (it is koordinatenfrei), simply themathematical representation is coordinate dependent.

The values of the Scalar Product of two Vectors is Independent of the Coordinate System :

A B = A x i Bx i = A i B i . (173)



Scalar Product

S l P d I d d f h C di S
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A j = A

=A x i e x i e j

= A x i e x i e j

= A x i e x i e j

= ji

(171)

= ji Ax i . (172)


j=

ijA




A B = A x i Bx i = A i B i . (173)



Scalar Product

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A j = A

=A x i e x i e j

= A x i e x i e j

= A x i e x i e j

= ji

(171)

= ji Ax i . (172)


j= ij A




A B = A x i Bx i = A i B i . (173)


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