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Cartesian Tensors
Reference: Jeffreys Cartesian Tensors

1 Coordinates and Vectors
z=x e
3 3

y=x e
2

2

e

1

x=x

1

Coordinates x i, i = 1, 2, 3 Unit vectors: e i, i = 1, 2, 3 General vector (formal definition to follow) denoted by components e.g. u = u i Summation convention (Einstein) repeated index means summation:

Cartesian Tensors

© Geoffrey V. Bicknell


3

3

ui vi =


i=1

ui v

i

u ii =


i=1

u

ii

(1)

2 Orthogonal Transformations of Coordinates
x x
3 3

x

2

x

2

x

1

x

1

x i = a ij x

j

(2) (3)

a ij = Transformation Matrix Position vector r = x i e i = x j e a ji x i e j = x i e xi ( a
ji e j j

i i j

) = xi e

(4)

e i = a ji e

i.e. the transformation of coordinates from the unprimed to the primed frame implies the reverse transformation from the primed to the unprimed frame for the unit vectors.
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Kronecker Delta
ij = 1 if i = j = 0 otherwise (5)

2.1 Orthonormal Condition:
Now impose the condition that the primed reference is orthonormal ei e j = Use the transformation
e i e j = a ki e k a lj e = a ki a lj e k e l l ij

and e i e j =

ij

(6)

= a ki a lj = a ki a
kj

(7)

kl

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 In matrix notation: aT a = I Also have a ik a
jk kj

=

ij

(8)

(9)

= aa T =

ij

(10)

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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 x k = a ik x i x i = a ji x
j k

(11)

i.e. the reverse transformation is simply given by the transpose. Similarly, e i = a ij e
j

(12)

2.3 Interpretation of a
Since

ij

e i = a ij e

j

(13)

then the a ij are the components of e i wrt the unit vectors in the unprimed system.

3 Scalars, Vectors & Tensors
3.1 Scalar (f):
f ( x i ) = f ( xi ) (14)

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:

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x i = a ij x j u i = a ij u

j

(15)

3.2.1 Gradient operator
Suppose that f is a scalar. Gradient defined by ( grad f ) i = ( f ) i = f x (16)
i

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

(17)

(18)

ki

ij

(19)
j

f f = a ij x x i

Hence the gradient operator satisfies our definition of a vector.

3.2.2 Scalar Product
u v = u i v i = u 1 v 1 + u 2 v 2 + u 3 v is the scalar product of the vectors u i and v i . Exercise: Show that u v is a scalar.
3

(20)

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3.3 Tensor
Prototype second rank tensor x i x General definition: T
ij j

= a ik a jl T

kl

(21)

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

Exercise:
u
i, j

=

u x

i j

(22)

is a second rank tensor. (Introduces the comma notation for partial derivatives.) In dyadic form this is written as grad u or u .

3.3.1 Divergence Exercise:
Show that the quantity v i v = div v = xi is a scalar. (23)

4 Products and Contractions of Tensors
It is easy to form higher order tensors by multiplication of lower rank tensors, e.g.T ijk = T ij u k is a third rank tensor if T ij is a second

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rank tensor and u k is 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 tr4ansformation properties are easy to prove.

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 ) (24)

v

i

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 ) The derivative of a function, f ( x i, t ) along the trajectoryof a parcel fluid is given by: df f + ----- = dt t f dx i f f ------ = + vi x x i dt t (25)
i

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Derivative of velocity
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: fi = v i dv i v v i dx j v d = v i ( x j ( t ), t ) = i + = i + vj x j dt dt t x jdt t (26)

Exercise: Show that f i is a vector.

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 (27)

e.g.
112

=0



123

=1



321

= ­1

(28)

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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Define
ijk

= = +

lmn a il a jm a kn 123 a i 1 a j 2 a k 3 213 a i 2 a j 1 a k 3 k3

+ +

312 a i 3 a j 1 a k 2 321 a i 3 a j 2 a k 1

+ +
k3

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

= ai1 ( a j2 a

­ a j3 ak 2 ) ­ ai2 ( a j1 a
k2

­ a j3 ak 2 )

+ ai3 ( a j1 a a = a a
i1 j1 k1

­ a j2 ak 1 )

(29)

a a a

i2 j2

a a a

i3 j3

k2

k3

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

(30)

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 une qual indices. This is just the definition of ijk . Thus ijk transforms as a tensor.

6.1 Uses of the permutation tensor 6.1.1 Cross product
Define c i = ijk a j b then
k

(31)

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c1 = c2 = c3 =

123 a 2 b 3 231 a 3 b 1 312 a 1 b 2

+ + +

132 a 3 b 2 213 a 1 b 3 321 a 2 b 1

= a2 b3 ­ a3 b = a3 b1 ­ a1 b = a1 b2 ­ a2 b

2 3 1

(32)

These are the components of c = a в b .

6.1.2 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

(33)

(34)

6.1.3 Curl
( curl u ) i = e.g. ( curl u ) 1 = etc. u 3 + x2 u 2 u 3 u = ­ x3 x2 x
2 3 ijk

u x

k j

(35)

123

132

(36)

6.1.4 The tensor iks
Define T

mps

ikmp

= iks .

mps

(37)

Properties:
· If i = k or m = p then T
ikmp = 0

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· 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 iks
mps km

(38) . (39)

= im kp ­ ip
mps klm u l v m

km

6.1.5 Application of iks
( curl ( u в v ) ) i =
ijk

( x j
jm

) = ijk

klm

(u v ) x j l m

= ( il

u l v m ­ im jl ) v m + u l x j x j (40)

u i u j v m v i = vm ­ vi + ui ­ uj xm x j xm x j =v
j

u i v i v j u ­ uj + ui ­ vi x j x j x j x

j j i

= (v u ­ u v + u v ­ v u)

7 The Laplacean
2 2 = + + = ------------- 2 2 2 xi xi x1 x2 x3
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2 2 2

(41)

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8 Tensor Integrals
8.1 Green's Theorem

n

i

V S

In dyadic form:


V

( u ) dV =


S

( u n ) dS

(42)

In tensor form:


V

u i dV = xi


S

u i n i dS = Flux of u through S

(43)

Extend this to tensors:


V

T ij dV = x j


S

T ij n j dS = Flux of T ij through S

(44)

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8.2 Stoke's Theorem
n
i

t

C

S

In dyadic form:


S

( curl u ) n dS =


C

u t ds

(45)

In tensor form:


S

u k ijk n i dS = x j


C

u i t i ds

(46)

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