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Cartesian Tensors
Reference: H. Jeffreys Cartesian Tensors

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1 Coordinates and Vectors z = x3 e3 y = x2 e2 e1 x = x1

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Coordinates xi x1 = x Unit vectors ei e1 = e x = i i = 1, 2, 3 e3 = ez = k
(2)

i = 1, 2, 3 x2 = y x3 = z
(1)

e2 = e y = j

General vector (formal definition to follow) denoted by components e.g. u = u i

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Notation The boldface notation for vectors is referred to as dyadic notation The subscript notation is tensor notation. Summation convention Einstein: repeated index means summation:
3

ui vi = ui vi
i=1 3

(3)

u ii = u ii
i=1

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2 Orthogonal Transformations of Coordinates The behaviour of quantities under orthogonal transformations of the coordinate system is the basis of Cartesian tensors. We want to formulate equations in such a way that they are independent of the specific coordinate system.

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x3

x3

x2

x2
x1

x1 General linear transformation
x i = a ij x j

a ij = Transformation Matrix
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Position vector Consider the position vector expressed in terms of coordinates and unit vectors in two related coordinate systems r = xi ei = xi ei
(4)

In view of the transformation from the unprimed to the primed system: r = a ij x j e i = x j ( a ij e i ) Therefore we can write: e j = a ij e i
(6) (5)

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so that we have the two companion transformations: x i = a ij x j Kronecker Delta ij = 1 if i = j = 0 otherwise In matrix form 100 ij = 0 1 0 001
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e i = a ji e j

(7)

(8)

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Substitution property ij T jk ... = T ik ...
(9)

In the summation over j the only term of the sum that makes any contribution is that for which j = i . 2.1 Orthogonal transformation So far, what we have described is valid for any linear transformation. Now impose the condition that both the original and the primed reference frames are orthonormal
e i e j = ij and e i e j = ij

(10)

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Use transformation of the unit vectors:
e i e j = a ki e k a lj e l

= a ki = a ki

a lj e k e l a lj kl

(11)

= a ki a kj NB the last operation is an example of the substitution property of the Kronecker Delta. Since e i e j = ij , then the orthonormal condition on a ij is a ki a kj = ij
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(12)

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In matrix notation: aT a = I aa T = I In tensor notation: ( aa T ) ij = a ik a jk = ij
(15) (13)

We also have as a consequence of the properties of matrices, that
(14)

Any of equations (12), (13), (14) or (15) defines an orthogonal transformation.

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2.2 Reverse transformations
x i = a ij x j a ik x i = a ik a ij x j = kj x j = x k

x k = a ik x i.e. the reverse transformation pose. Similarly, following from

x = a x i i ji j

is simply determined by the trans-

e j = a ij e i we have
e i = a ij e j

(16)

(17)

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Transformations for both coordinates and basis vectors The complementary set of transformations is then x i = a ij x j e i = a ij e j
(18)

2.3 Interpretation of the matrix a ij Since
e i = a ij e j then the a ij are the components of e i wrt the unit vectors in the original system.e.g.

e 1 = a 11 e 1 + a 12 e 2 + a 13 e 3
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2.4 Example: 2D rotation y y e y ey x ex ex It is easiest here to determine the relationship between the unit basis vectors: e x = cos e x + sin e y (20) e y = ­ sin e x + cos e y x In matrix form: ex ex cos sin 0 e y = ­ sin cos 0 e y (21) 0 01e ez z
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Then the transformation equation for the coordinates is: x cos sin 0 x y = ­ sin cos 0 y z 0 01z 3 Scalars, Vectors & Tensors We define these objects by the way in which they transform with respect to orthogonal coordinate transformations. 3.1 Scalar (f): f ( x i ) = f ( xi )
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(22)

(23)

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Example of a scalar is f = r 2 = x i x i . Examples from fluid dynamics are the density and temperature. 3.2 Vector (u): Prototype vector: x i General transformation law:
x i = a ij x j u i = a ij u j

(24)

as the transformation law for a generic vector.

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3.3 Gradient operator Suppose that f is a scalar. The gradient of f is defined by f ( grad f ) i = ( f ) i = xi
(25)

Need to show this is a vector by its transformation properties. Since,
x j = a kj x k

f x j f = x j x xi i

(26)

(27)

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then x j = a kj ki = a ij xi and f f = a ij x j xi

(28)

Hence the gradient operator satisfies our definition of a vector. Scalar Product u v = u i v i = u 1 v 1 + u 2 v 2 + u 3 v 3 is the scalar product of the vectors u i and v i .
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(29)


Exercise: Show that u v is a scalar. 3.4 Tensor Prototype second rank tensor x i x j General definition by transformation of components:
T ij = a ik a jl T kl

(30)

Exercise: Show that u i v j is a second rank tensor if u i and v j are vectors.

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Exercise: u i u i, j = x j is a second rank tensor. (Introduces the comma notation for partial derivatives.) In dyadic form this is written as grad u or u . 3.5 Divergence Exercise: Show that the quantity v i v = div v = xi
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is a scalar. 4 Products and Contractions of Tensors It is easy to form higher order rank tensors, e.g. T ijk = T ij u second rank tensor and u k is tensors by multiplication of lower k is a third rank tensor if T ij is a a vector (first rank tensor). It is

straightforward to show that T ijk has the relevant transformation properties. Similarly, if T ijk is a third rank tensor, then T ijj is a vector. Again the relevant transformation properties are easy to prove.

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5 Differentiation following the motion This involves a common operator occurring in fluid dynamics. Suppose the coordinates of an element of fluid are given as a function of time by xi = xi ( t ) vi
(32)

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The velocities of elements of fluid at all spatial locations within a given region constitute a vector field, i.e. v i = v i ( x j, t ) If we follow the trajectory of an element of fluid, then on a particular trajectory x i = x i ( t ) . The acceleration of an element is then given by: dv i d = v i ( x j ( t ), t ) = fi = dt dt v i v i dx j + = t x jdt v i vi + vj (33) x j t

Exercise: Show that f i is a vector.

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6 The permutation tensor ijk ijk = 0 if any of i, j, k are equal = 1 if i, j, k unequal and in cyclic order = ­ 1 if i, j, k unequal and not in cyclic order e.g. 112 = 0 Is ijk a tensor? In order to show this we have to demonstrate that ijk , when defined the same way in each coordinate system has the correct transformation properties.
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(34)

123 = 1

321 = ­ 1

(35)


Define
ijk = lmn a il a jm a kn = 123 a i 1 a j 2 a k 3 + 312 a i + 213 a i 2 a j 1 a k 3 + 321 a i = ai 1 ( a j 2 ak 3 ­ a j 3 ak 2 ) ­

3 a j 1 a k 2 + 231 a i 2 a j 3 a k 1 3 a j 2 a k 1 + 132 a i 1 a j 3 a k 2

ai 2 ( a j 1 ak 3 ­ a j 3 ak 1 ) + ai 3 ( a j 1 ak 2 ­ a j 2 ak 1 ) ai 3 a j3 ak 3
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=

ai 1 ai 2 a j1 a j2 ak 1 ak 2

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In view of the interpretation of the a ij , the rows of this determinant represent the components of the primed unit vectors in the unprimed system. Hence:
ijk = e i ( e j в e k ) This is zero if any 2 of i, j, k are equal, is +1 for a cyclic permutation of unequal indices and -1 for a non-cyclic permutation of unequal indices. This is just the definition of ijk . Thus ijk transforms as a tensor.

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6.1 Uses of the permutation tensor Cross product Define c i = ijk a j b k then c 1 = 123 c 2 = 231 a 2 b 3 + 132 a 3 a 3 b 1 + 213 a 1 c 3 = 312 a 1 b 2 + 321 a 2 b2 = a2 b3 = a3 b3 ­ a3 b1 ­ a1 b2 b3
(36)

(37)

b1 = a1 b2 ­ a2 b1

These are the components of c = a в b .

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6.2 Triple Product In dyadic notation the triple product of three vectors is: t = u (v в w) In tensor notation this is t = u i ijk v j w k = ijk u i v j w k 6.3 Curl u k ( curl u ) i = ijk x j
(40) (39) (38)

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e.g. u 3 u 2 u 3 u 2 + 132 = ­ ( curl u ) 1 = 123 x2 x3 x2 x3 etc.
(41)

6.4 The tensor iks mps Define T ikmp = iks mps Properties: · If i = k or m = p then T ikmp = 0 .
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(42)


· If i = m we only get a contribution from the terms s i and k i, s . Consequently k = p . Thus iks = ± 1 and mps = iks = ± 1 and the product iks iks = ( ± 1 ) 2 = 1 . · If i = p , similar argument tells us that we must have s i and k = m i . Hence, iks = ± 1 , mps = - 1 iks mps = ­ 1 . + So, i = m, k = p 1 unless i = k 0 i = p, k = m ­ 1 unless i = k 0 These are the components of the tensor im kp ­ ip km . iks mps = im kp ­ ip km
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6.5 Application of iks mps [ curl ( u в v ) ] i = ijk ( klm u l v m ) x j = ijk klm ( ul vm ) x j v m u l = ( il jm ­ im jl ) vm + ul x j x j
(44)

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We then use the substitution property of ij to show that: u i u j v m v i [ curl ( u в v ) ] i = vm ­ vi + ui ­ uj xm x j xm x j u i v i v j u j = vj ­ uj + ui ­ vi x j x j x j x j = ( v u ­ u v + u v ­ v u )i The Laplacean
2 =
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2 + + = --------------2 2 2 xi xi x1 x2 x3

2

2

2

(46)

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7 Tensor Integrals 7.1 Green's Theorem In dyadic form: v ni V S
V



( v ) dV =

S



( v n ) dS

(47)

In tensor form:

V



v i dV = xi

S



v i n i dS = S

(48)

= Flux of v through S

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Extend this to tensors:

V



T ij dV = x j

S



T ij n j dS

(49)

= Flux of T ij through S

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7.2 Stoke's Theorem In dyadic form: ti C S In tensor form:
S



( curl u ) n dS =

C



u t ds

(50)

S



u k ijk n i dS = x j

C



u i t i ds

(51)

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