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TENSOR ALGEBRAContinuum Mechanics Course (MMC) - ETSECCPB - UPC
Tensor Algebra
Introduction to Tensors
2
Introduction
SCALAR
VECTOR
MATRIX
?
, , ...σ ε
, , ...v f
, , ...
, ...C
v
3
Concept of Tensor
A TENSOR is an algebraic entity with various components which generalizes the concepts of scalar, vector and matrix.
Many physical quantities are mathematically represented as tensors.
Tensors are independent of any reference system but, by need, are commonly represented in one by means of their “component matrices”.
The components of a tensor will depend on the reference system chosen and will vary with it.
4
Order of a Tensor
The order of a tensor is given by the number of indexes needed to specify without ambiguity a component of a tensor.
Scalar: zero dimension
Vector: 1 dimension
2nd order: 2 dimensions
3rd order: 3 dimensions
4th order …
a,a a
,A A
, AA, AA
3.14 1.20.30.8
vi
0.1 0 1.30 2.4 0.5
1.3 0.5 5.8ijE
5
Cartesian Coordinate System
Given an orthonormal basis formed by three mutually perpendicular unit vectors:
Where:
Note that
1 2 2 3 3 1ˆ ˆ ˆ ˆ ˆ ˆ, , e e e e e e
1 2 3ˆ ˆ ˆ1 , 1 , 1 e e e
1ˆ ˆ
0i j ij
i ji j
e e
if
if
6
Cylindrical Coordinate System
1 2
1 2
3
ˆ ˆ ˆcos sin ˆ ˆ ˆsin cos ˆ ˆ
r
z
θ θθ θ
e e ee e ee e
1
2
3
cos ( , , ) sin
x rr z x r
x z
x
1x
2x
3x
7
Spherical Coordinate System
1 2 3
1 2
1 2 3
ˆ ˆ ˆ ˆsin sin sin cos cos ˆ ˆ ˆcos sin ˆ ˆ ˆ ˆcos sin cos cos sin
r
φ
θ φ θ φ θφ φθ φ θ φ θ
e e e ee e ee e e e
1
2
3
sin cos, , sin sin
cos
x rr x r
x r
x
3x
1x
2x
8
Tensor Algebra
Indicial or (Index) Notation
9
Tensor Bases – VECTOR
A vector can be written as a unique linear combination of the three vector basis for .
In matrix notation:
In index notation:
ˆ ie
1 1 2 2 3 3ˆ ˆ ˆv v v v e e ev
v
1
2
3
vvv
v
ˆvi ii
v e
viiv
tensor as a physical entity
component i of the tensor in the given basis
1v2v
3v
1,2,3i
1,2,3i
10
Tensor Bases – 2nd ORDER TENSOR
A 2nd order tensor can be written as a unique linear combination of the nine dyads for .
Alternatively, this could have been written as:
, 1,2,3i jˆ ˆ ˆ ˆi j i j e e e eA
1 1 1 2 1 3
2 1 2 2 2 3
3 1 3 2 3 3
ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ
11 12 13
21 22 23
31 32 33
A A AA A AA A A
A e e e e e ee e e e e ee e e e e e
1 1 1 2 1 3
2 1 2 2 2 3
3 1 3 2 3 3
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
11 12 13
21 22 23
31 32 33
A A A
A A A
A A A
A e e e e e e
e e e e e e
e e e e e e
11
Tensor Bases – 2nd ORDER TENSOR
In matrix notation:
In index notation:
11 12 13
21 22 23
31 32 33
A A AA A AA A A
A
ˆ ˆAij i jij
A e e
ijijAA
tensor as a physical entity
component ij of the tensor in the given basis
1 1 1 2 1 3
2 1 2 2 2 3
3 1 3 2 3 3
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
11 12 13
21 22 23
31 32 33
A A A
A A A
A A A
A e e e e e e
e e e e e e
e e e e e e
, 1,2,3i j
12
Tensor Bases – 3rd ORDER TENSOR
A 3rd order tensor can be written as a unique linear combination of the 27 tryads for .
Alternatively, this could have been written as:
ˆ ˆ ˆ ˆ ˆ ˆi j k i j k e e e e e eA
, , 1,2,3i j k
1 1 1 1 2 1 1 3 1
2 1 1 2 2 1 2 3 1
3 1 1 3 2 1 3 3 1
1 1 2 1 2 2
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ...
111 121 131
211 221 231
311 321 331
112 122
e e e e e e e e e
e e e e e e e e e
e e e e e e e e e
e e e e e e
A A A A
A A A
A A A
A A
1 1 1 1 2 1 1 3 1
2 1 1 2 2 1 2 3 1
3 1 1 3 2 1 3 3 1
1 1 2 1 2 2
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ...
111 121 131
211 221 231
311 321 331
112 122
e e e e e e e e ee e e e e e e e ee e e e e e e e ee e e e e e
A A A A
A A A
A A A
A A
13
Tensor Bases – 3rd ORDER TENSOR
In matrix notation:
1 1 1 1 2 1 1 3 1
2 1 1 2 2 1 2 3 1
3 1 1 3 2 1 3 3 1
1 1 2 1 2 2
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ...
111 121 131
211 221 231
311 321 331
112 122
e e e e e e e e e
e e e e e e e e e
e e e e e e e e e
e e e e e e
A
A A A
A A A
A A A
A A
113 123 133
213 223 233
313 323 333
A A A
A A A
A A A
112 122 132
212 222 232
312 322 332
A A A
A A A
A A A
111 121 131
211 221 231
311 321 331
A A A
A A A
A A A
A
14
Tensor Bases – 3rd ORDER TENSOR
In index notation:
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
ijk i j kijk
ijk i j k ijk i j k
e e e
e e e e e e
A A
A A
ijkijkA A
1 1 1 1 2 1 1 3 1
2 1 1 2 2 1 2 3 1
3 1 1 3 2 1 3 3 1
1 1 2 1 2 2
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ...
111 121 131
211 221 231
311 321 331
112 122
e e e e e e e e e
e e e e e e e e e
e e e e e e e e e
e e e e e e
A A A A
A A A
A A A
A A
tensor as a physical entity
component ijk of the tensor in the given basis , , 1,2,3i j k
15
A tensor of order n is expressed as:
The number of components in a tensor of order n is 3n.
Higher Order Tensors
1 2 1 2 3, ... ˆ ˆ ˆ ˆ...n ni i i i i i iA A e e e e
1 2, ... 1,2,3ni i i where
16
The Einstein Summation Convention: repeated Roman indices are summed over.
A “MUTE” (or DUMMY) INDEX is an index that does not appear in a monomial after the summation is carried out (it can be arbitrarily changed of “name”).
A “TALKING” INDEX is an index that is not repeated in the same monomial and is transmitted outside of it (it cannot be arbitrarily changed of “name”).
3
1 1 2 2 3 313
1 1 2 2 3 31
i i i ii
ij j ij j i i ij
a b a b a b a b a b
A b A b A b A b A b
REMARKAn index can only appear up to two times in a monomial.
Repeated-index (or Einstein’s) Notation
i is a mute index
i is a talking index and j is a
mute index
17
Rules of this notation:
1. Sum over all repeated indices.
2. Increment all unique indices fully at least once, covering all combinations.
3. Increment repeated indices first.
4. A comma indicates differentiation, with respect to coordinate xi .
5. The number of talking indices indicates the order of the tensor result
Repeated-index (or Einstein’s) Notation
3
,1
i ii i
ii i
u uux x
2 23
, 21
i ii jj
jj j j
u uux x x
3
,1
ij ijij j
jj j
A AA
x x
18
Kronecker Delta δ
The Kronecker delta δij is defined as:
Both i and j may take on any value in Only for the three possible cases where i = j is δij non-zero.
10ij
i ji j
if
if
11 22 33
12 13 21
1 10 ... 0ij
i ji j
if
if
ij ji REMARKFollowing Einsten’s notation: Kronecker delta serves as a replacement operator:
11 22 33 3ii
,ij j i ij jk iku u A A
1,2,3
19
Levi-Civita Epsilon (permutation) ϵ
The Levi-Civita epsilon is defined as:
3 indices 27 possible combinations.
01 123, 231 3121 213,132 321
ijk ijkijk
if there is a repeated index
if or
if or
e
REMARKThe Levi-Civita symbol is also named permutation or alternating symbol.
ijk ikj e e
ijke
20
Relation between δ and ε
1 2 3
1 2 3
1 2 3
deti i i
ijk j j j
k k k
e detip iq ir
ijk pqr jp jq jr
kp kq kr
e e
ijk pqk ip jq iq jp e e
2ijk pjk pie e
6ijk ijk e e
21
Example
Prove the following expression is true:
6ijk ijk e e
22
211 211 212 212 213 213
221 221 222 222 223 223
231 231 232 232 233 233
e e e e e e
e e e e e e
e e e e e e
2i
311 311 312 312 313 313
321 321 322 322 323 323
331 331 332 332 333 333
e e e e e e
e e e e e e
e e e e e e
3i
121 121 122 122 123 123 e e e e e e 2j 131 131 132 132 133 133 e e e e e e 3j
Example - Solution
111 111 112 112 113 113ijk ijk e e e e e e e e1
1
1
1
1
1
6
1i
1j 1k 2k 3k
23
Tensor Algebra
Vector Operations
24
Sum and Subtraction. Parallelogram law.
Scalar multiplication
Vector Operations
a b b a ca b d
1 1 2 2 3 3ˆ ˆ ˆa a a a b e e e
i i i
i i i
c a bd a b
i ib a
25
Scalar or dot product yields a scalar
In index notation:
Norm of a vector
Vector Operations
cos u v u vwhere is the angle
between the vectors u and v
2 ˆ ˆi i j j i j ij i iu u u u u uu u e eu
1 2 1 2i iu uu uu
3
1
ˆ ˆ ˆ ˆv v v v vi
Ti i j j i j i j i j ij i i i i
i
u u u u u
u v e e e e u v
u v ij 0( )1 ( )
i jj i
26
Some properties of the scalar or dot product
Vector Operations
0
000, ,
u v v uu 0u v w u v u wu u u 0u u u 0u v u 0 v 0 u v
Linear operator
27
Vector product (or cross product ) yields another vector
In index notation:
Vector Operations
sin
c a b b ac a b
i iˆ ˆi ijk j k i ijk j kc a b c a b i c e ee e {1,2,3}
2 3 3 2 1 3 1 1 3 2 1 2 2 1 3ˆ ˆ ˆa b a b a b a b a b a b c e e e
where is the angle between the vectors a and b
0
1 2 3
1 2 3
1 2 3
ˆ ˆ ˆdet
symb
a a ab b b
e e e
123 1321 1
2 3 3 2a b a b
e e
1i
231 3 1 213 1 3
1 1
a b a b
e e
2i
312 1 2 321 2 11 1
a b a b
e e
3i
28
Some properties of the vector or cross product
Vector Operations
, , ||
a b a b
u v v uu v 0 u 0 v 0 u vu v w u v u w Linear operator
29
Tensor product (or open or dyadic product) of two vectors:
Also known as the dyad of the vectors u and v, which results in a 2nd
order tensor A.
Deriving the tensor product along an orthonormal basis {êi}:
In matrix notation:
Vector Operations
A u v uv
ˆ ˆ ˆ ˆ ˆ ˆv vi i j j i j i j ij i ju u A A u v e e e e e e
v v v1
T2 1 2 3
3
uuu
u v u v
v ,ij i jij ijA u i j A u v {1,2,3}
v v vv v vv v v
1 1 1 2 1 3 11 12 13
2 1 2 2 2 3 21 22 23
3 1 3 2 3 3 31 32 33
u u u A A Au u u A A Au u u A A A
30
Some properties of the open product:
Vector Operations
u v w u v w u v w v w u
u v v u
u v w u v u w
u v w u v w u v w w u v
u v w x u x v wLinear operator
31
Example
Prove the following property of the tensor product is true:
u v w u v w
32
Example - Solution
v v wk i i i i kkkc u u u v w w
v w v wk i i k i i kik ikc u u u v w u v w
ˆ ˆ ˆv wi j k k k kk ku c e u v w e u v w e
scalar vector
1st order tensor (vector)
1st order tensor (vector)
2nd order tensor (matrix)
u v w u v wvector
c
k-component of vector c
k-component of vector c
33
Example – Solution
u v w u v w
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆv w v w v wi i j j k k i i j k j k i j k i j ku u u u v w e e e e e e e e e
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆv w v w v wi i j j k k i j i j k k i j k i j ku u u u v w e e e e e e e e e
34
Vector Operations
Triple scalar or box product
In index notation:
cos
cos sin
V
a b c
c a b b c a
a b c
a b c
a b c ijk i j kV a b c e 1 2 3
1 2 3
1 2 3
deta a a
V b b bc c c
a b c
base areaheight
35
Vector Operations
Triple vector product
In index notation:
u v w u w v u v w
ˆ ˆ ˆv w v w
ˆ ˆv w v w
ˆ ˆv w v w
u v w e e e
e e
e e
j j klm l m k ijk j klm l m i
ijk lmk j l m i il jm im jl j l m i
m i m i l l i i
u u
u u
u u
e e e
e e
ijk pqk ip jq iq jp e eREMARK
36
Scalar or dot product yields a scalar
Vector or cross product yields another vector
Triple scalar or box product yields a scalar
Triple vector product yields another vector
Summary
c a b b a a b i ijk j kic a b e
cos sinV a b c a b c a b c ijk i j ka b c e
u v w u w v u v w w v v wk k i k k iiu u u v w
cosT u v u v u v vi iu u v
37
Tensor Algebra
Tensor Operations
38
Summation (only for equal order tensors)
Scalar multiplication (scalar times tensor)
Tensor Operations
A B B A C
A C
ij ij ijC A B
ij ijC A
39
Dot product (.) or single index contraction product
Tensor Operations
REMARK2A A A A B B A
Index “j” disappears (index contraction)
i ij jc A b2nd
order1st
order
A b c1st
order
Index “k” disappears (index contraction)
ij i k kjC bA b CA3rd
order1st
order2nd
order
Index “j” disappears (index contraction)
ik i kj jC A B A B C2nd
order2nd
order2nd
order
40
Some properties:
2nd order unit (or identity) tensor
Tensor Operations
1 u u 1 u
[1]ij j i i i
ij ij
1 e e e e
1 0 0
0 1 0
0 0 1
1
A b c A b A c Linear operator
41
Some properties:
2nd Order Tensor Operations
1 A A A 1A B C A B A C
A B C A B C A B C A B B A
42
Example
When does the relation hold true ? n T T n
43
Example - Solution
n T T nvector 2nd order
tensor
vector
c
k i ikc nT
k ki i i kic T n nT
c n T
c T nk kc c ik kiT Tif
ˆ ˆ ˆk k k i ik kkc nT c e n T e e
ˆ ˆ ˆk k k i ki kkc nT c e T n e e
44
Example - Solution
n T T n COMPACT NOTATION
INDEX NOTATION
TT n T T n
1,2,3i ik ki in T T n k
MATRIX NOTATION
Tcc
c
45
Example - Solution
TT n T T n MATRIX NOTATION
Tcc
c
11 12 13 11 12 13 1
1 2 3 21 22 23 21 22 23 2
31 32 33 31 32 33 3
, ,T T T T T T n
n n n T T T T T T nT T T T T T n
1
1 2 3 2
3
cc c c c
c
11 12 13 11 12 13 1
1 2 3 21 22 23 21 22 23 2
31 32 33 31 32 33 3
, ,
TT T T T T T n
n n n T T T T T T nT T T T T T n
1
1 2 3 2
3
Tc
c c c cc
46
Transpose
Trace yields a scalar
Some properties:
2nd Order Tensor Operations
Tjiij AA
11 22 33( )iiTr A A A AA
A ATTr Tr
Tr Tr A B B A Tr Tr Tr A B A B
Tr Tr A A
11 12 13 11 21 31
21 22 23 12 22 32
31 32 33 13 23 33
T
A A A A A AA A A A A AA A A A A A
A A
i j i iTr Tr a b a b a b a b
A ATT
T T T A B B A
u v v uT
A B A BT T T
47
Double index contraction or double (vertical) dot product (:)
Indices contiguous to the double-dot (:) operator get vertically repeated (contraction) and they disappear in the resulting tensor (4 order reduction of the sum of orders).
2nd Order Tensor Operations
Indices “i,j” disappear (double index contraction)
ij ijc A B2nd
order2nd
order
: cA Bzero order
(scalar)
Indices “j,k” disappear (double index contraction)
jk jki i Bc A: B cA3rd
order2nd
order1st
order
Indices “k,l” disappear (double index contraction)
ij ijkl klC B: B CA4th
order2nd
order2nd
order
48
Some properties
2nd Order Tensor Operations
: :T T T TTr Tr Tr Tr A B A B B A A B B A B A
: :
: : :
:
:
1 A A A 1
A B C B A C A C B
A u v u A v
u v w x u w v x
T T
Tr
REMARK: :A B C B A C
49
Double index contraction or double (horizontal) dot product (··)
Indices contiguous to the double-dot (··) operator get horizontally repeated (contraction) and they disappear in the resulting tensor (4 orders reduction of the sum of orders).
2nd Order Tensor Operations
Indices “i,j” disappear (double index contraction)
ij jic A B2nd
order2nd
order
c A B
Indices “j,k” disappear (double index contraction)
jk kji i Bc A B cA3rd
order2nd
order1st
order
Indices “k,l” disappear (double index contraction)
ij ijkl lkC B B CA4th
order2nd
order2nd
order
zero order
(scalar)
50
Norm of a tensor is a non-negative real number defined by
Tensor Operations
REMARK
Unless one of the two tensors is symmetric.
:A B A B Tr 1 A A A 1
1 21 2: 0ij ijA A A A A
Tr Tr A B A B B A B A
51
Example
Prove that:
Tr A B A B
: TTr A B A B
52
A B A BT T Tik ik ij ijikkk ki
c Tr A B A B A B
Example - Solution
k j
A B A Bki ik ij jikk ki ikc Tr A B A B A B
k ii j
53
Determinant yields a scalar
Some properties:
Inverse There exists a unique inverse A-1 of A when A is nonsingular, which satisfies the reciprocal relation:
2nd Order Tensor Operations
11 12 13
21 22 23 1 2 3
31 32 33
1
6det det detA A ijk i j k ijk pqr pi qj rk
A A AA A A A A A A A AA A A
e e e
det det det A B A Bdet detA AT
REMARKThe tensor A is SINGULAR if and only if det A = 0. A is NONSINGULAR if det A ≠ 0.
1 1
1 1 , , {1,2,3}ik kj ik kj ijA A A A i j k
A A 1 A A
3det det A A
54
If A and B are invertible, the following properties apply:
2nd Order Tensor Operations
1 1 1
11
1 1
11
2 1 1
11
1
1det det det
T T T
A B B A
A A
A A
A A A
A A A
A A A
55
Example
Prove that . 1 2 3det A ijk i j kA A A e
56
Example - Solution
11 12 13
21 22 23
31 32 33
11 22 33 21 32 13 31 12 23 13 22 31 23 32 11 33 12 21
det detA A AA A AA A A
A A A A A A A A A A A A A A A A A A
A
57
Example - Solution
111 11 21 31 112 11 21 32 113 11 21 33
121 11 22 31 122 11 22 32 123 11 22 33
131 11 23 31 132 11 23 32 133 11 23 33
211 12 21 31 212 12 21 32 213 12 21 33
221 12 22 31 222 12 22 32
A A A A A A A A A
A A A A A A A A A
A A A A A A A A A
A A A A A A A A A
A A A A A A
e e e
e e e
e e e
e e
e e 223 12 22 33
231 12 23 31 232 12 23 32 233 12 23 33
311 13 21 31 312 13 21 32 313 13 21 33
321 13 22 31 322 13 22 32 323 13 22 33
331 13 23 31 332 13 23 32 333 13 23 33
A A A
A A A A A A A A A
A A A A A A A A A
A A A A A A A A A
A A A A A A A A A
e
e e e
e e e
e e e
e e e
1
1
1
1
1
1
11 22 33 12 23 31 13 21 32 13 22 31 12 21 33 11 23 32A A A A A A A A A A A A A A A A A A
1 2 3ijk i j kA A A e
1i
1j 1k 2k 3k
2i
3i
2j 3j
58
Dot product – contraction of one index:
Summary - Tensor Operations
vi i
ijij ik kj ij
c uA B C
u vC A B
i ij jj j
ij j ii i
u A c
A u d
c u A
d A u
ijk ijk im mjk ijk
ijk ijk ijm mk ijk
A
A
A
A
C B
D B
B C
B D
ijklijkl ijm mkl ijkl
ijkl ijkl ijm mkl ijkl
C
D
A B
B A
A B
B A
CD
ij ij ik kj ijB A D D B A
59
Double dot product – contraction of two indices:
Summary - Tensor Operations
: ij ijc A B A B
:
:
ikm kmj ijij ij
ij ikm kmj ijij
C
D
C
D
A B
B A
A B
B A
:
:ijkl ijkl ijmp mpkl ijkl
ijkl ijkl ijmp mpij ijkl
C A B
D B A
A B CB A D
:
:ijk ijk ijlm lmk ijk
ijk ijk ilm lmjk ijk
AA
C B
D B
A
A
B C
B D
:
:ij ijk kk k
ijk jk ii i
A c
A d
c A
d A
B
B
B
B
60
Transposed double dot product – contraction of two indexes:
Summary - Tensor Operations
ij jic A B A B
ikm mkj ijij ij
ij ikm mkj ijij
C
D
C
D
A B
B A
A B
B A
ijklijkl ijmp pmkl ijkl
ijkl ijkl ijmp pmij ijkl
C A B
D B A
A B CB A D
ijk ijk ijlm mlk ijk
ijk ijk ilm mljk ijk
A
A
C B
D B
B C
B D
AA
ij jik kk k
ijk kj ii i
A c
A d
c A
d A
B
B
B
B
61
Open product – expansion of indexes:
Summary - Tensor Operations
vi j ijij ij
ij kl ijklijkl ijkl
u A
A B
A u v
A BC C
i jk ijkijk ijk
ij k ijkijk ijk
u A
A u
u A
A u
C
D
C
D
ijklmijklm ij klm ijklm
ijklmijklm ijk lm ijklm
A
A
A
A
B
B
B
B
C
D
C
D
vi j ijij ij
ij kl ijklijkl ijkl
u B
B A
B v u
B AD D
62
Tensor Algebra
Differential Operators
63
Differential Operators
A differential operator is a mapping that transforms a field into another field by means of partial derivatives.
The mapping is typically understood to be linear.
Examples: Nabla operator Gradient Divergence Rotation …
, ...v x A x
64
Nabla Operator
The Nabla operator is a differential operator “symbolically” defined as:
In Cartesian coordinates, it can be used as a (symbolic) vector on its own:
.ˆ
symbolic symb
iix
e
x
1
.
2
3
symb
x
x
x
65
Gradient
The gradient (or open product of Nabla) is a differential operator defined as: Gradient of a scalar field Φ(x):
Yields a vector
Gradient of a vector field v(x): Yields a 2nd order tensor
.
{1, 2,3}
ˆ ˆ
symb
i i ii i
i iii
ix x
x
e e
. vv , {1, 2,3}
vˆ ˆ ˆ ˆ
symbj
jij i ji i
ji j i jij
i
i jx x
x
v v
v v v e e e e
ˆ iix
e
vˆ ˆj
i jix
v e e
66
Gradient
Gradient of a 2nd order tensor field A(x): Yields a 3rd order tensor
. AA , , {1,2,3}
Aˆ ˆ ˆ ˆ ˆ ˆ
symbjk
jkijk ijk i jki i
jki j k i j kijk
i
i j kx x
x
A A A
A A A e e e e e e
Aˆ ˆ ˆjk
i j kix
A e e e
67
Divergence
The divergence (or dot product of Nabla) is a differential operator defined as : Divergence of a vector field v(x):
Yields a scalar
Divergence of a 2nd order tensor A(x): Yields a vector
vi
ix
v . vv
symbi
ii ii ix x
v v
Aˆij
jix
A e
. AA {1,2,3}
Aˆ ˆ
symbij
j iji iji i
ijj jj
i
jx x
x
A A
A A e e
68
Divergence
The divergence can only be performed on tensors of order 1 or higher.
If , the vector field is said to be solenoid (or divergence-free).
0 v v x
69
Rotation
The rotation or curl (or vector product of Nabla) is a differential operator defined as: Rotation of a vector field v(x):
Yields a vector
Rotation of a 2nd order tensor A(x): Yields a 2nd order tensor
v ˆv ekijk i
jx
e
A ˆ ˆA e eklijk i l
jx
e
. . vv {1,2,3}
vˆ ˆ
v v
v v e e
symb symbk
i ijk ijk k ijkj kj j
ki i ijk i
j
ix x
x
e e e
e
. AA , , {1, 2,3}
Aˆ ˆ ˆ ˆ
A
A A e e e e
symbkl
il ijk kl ijkj j
klil i l ijk i l
j
i j kx x
x
e e
e
70
Rotation
The rotation can only be performed on tensors of order 1 or higher.
If , the vector field is said to be irrotational (or curl-free).
0 v v x
71
Differential Operators - Summary
scalar fieldΦ(x)
vector field v(x)
2nd order tensor A(x)
GRADIENT
DIVERGENCE
ROTATION AA klil ijk
jx
e vv ki ijk
jx
e
vi
ix
v Aijj
ix
A
i
iix
vij
jij
ix
v
v
Aijk
jkijk
ix
A
A
72
Example
Given the vector determine
1 2 3 1 1 2 2 1 3ˆ ˆ ˆx x x x x x v v x e e e, , . v v v
73
Example - Solution
Divergence:
vi
ix
v
31 22 3 1
1 2 3
v vv vi
i
x x xx x x x
v
1 2 3
1 2
1
x x xx xx
v 1 2 3 1 1 2 2 1 3ˆ ˆ ˆx x x x x x v v x e e e
74
Example - Solution
Divergence:
In matrix notation:
vi
ix
v
1 2 3
1 21 2 3
1
1 2 3 1 2 11 2 3 1 2 1 2 3 1
1 2 3 1 2 3
, ,T
symb symb symbT
symb
x x xx x
x x xx
x x x x x xx x x x x x x x xx x x x x x
v v
1331
11
1 2 3
1 2
1
x x xx xx
v
75
In index notation:
Example - Solution
Rotation:
vv ki ijk
jx
e
3 32 1 1 212 13 21 23 31 32
1 1 2 2 3 3
v
v vv v v v
v ki ijk
j
i i i i i i
x
x x x x x x
e
e e e e e e
1 2 3
1 2
1
x x xx xx
v
76
3 2123 132
2 3
3 1213 231 1 2
1 32 1 3
2 1312 321
1 2
v v
0v v 1
v v
v
x x
x xx x
x x x
x x
e e
e e
e e
Example - Solution
Rotation:
In matrix notation
In compact notation:
32 112 13 21
1 1 2
3 1 223 31 32
2 3 3
vv v
v v v
v i i i i
i i i
x x x
x x x
e e e
e e e
1
1
1
1
1
1
1 2 2 2 1 3 3ˆ ˆ1x x x x x v e e
1 2 3
1 2
1
x x xx xx
v
77
Example - Solution
Rotation: Calculated directly in matrix notation:
1 1 2 3
2 1 2
3 1
vvv
x x xx xx
1 1 2 31
22 1 2 3
31 2 3
3
3 32 1 2 11 2 3
2 3 3 1 1 2
1 2 2 2 1 3 3
ˆ ˆ ˆvv detv
v v v
v vv v v vˆ ˆ ˆ
ˆ ˆ1
symb
x
x x x x
x
x x x x x x
x x x x x
e e e
v
e e e
e e
78
Example - Solution
Gradient:
In matrix notation
In compact notation:
v j
ij ijix
v v
1 1 2 3
2 1 2
3 1
vvv
x x xx xx
1
2 3 2
1 2 3 1 2 1 1 3 12
1 2
3
1, , 0
0 0
symb symb symbT
x x x xx x x x x x x x x
xx x
x
v v v
2 3 1 1 2 1 2 1 3 1 3 2 1 1 2 2 1 2 3 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆx x x x x x x x v e e e e e e e e e e e e
13
31
33
79
Tensor Algebra
Integral Theorems
80
Divergence or Gauss Theorem
Given a field in a volume V with closed boundary surface ∂V and unit outward normal to the boundary n , the Divergence (or Gauss) Theorem states:
Where: represents either a vector field ( v(x) ) or a tensor field ( A(x) ).A
A
V VdV dS
nA A
V VdV dS
nA A
81
Generalized Divergence Theorem
Given a field in a volume V with closed boundary surface ∂V and unit outward normal to the boundary n , the Generalized Divergence Theorem states:
Where: represents either the dot product ( · ), the cross product ( ) or the
tensor product ( ). represents either a scalar field ( ϕ(x) ), a vector field ( v(x) ) or a
tensor field ( A(x) ).
V VdV dS
nA A
V VdV dS
nA A
A
A
82
Curl or Stokes Theorem
Given a vector field u in a surface S with closed boundary surface ∂S and unit outward normal to the boundary n , the Curl (or Stokes) Theorem states:
S S
dS d
u n u r
where the curve of the line integral must have positive orientation, such that dr points counter-clockwise when the unit normal points to the viewer, following the right-hand rule.
83
jn
Example
Use the Generalized Divergence Theorem to show that
where is the position vector of .
i j ijSx n dS V
jnix
V VdS dV
nA A
84
Example - Solution
Applying the Generalized Divergence Theorem:
Applying the definition of gradient of a vector:
V VdS dV
x n x
i jSx n dS S
dS x n
jij
i
xx
x i
ijj
xx
x
i j ijSx n dS V
85
Example - Solution
The Generalized Divergence Theorem in index notation:
Then,
i j ijSx n dS V
ii jS V
j
xx n dS dVx
ii j ij ijS V V
j
xx n dS dV dV Vx
86
Tensor Algebra
References
87
José Mª Goicolea, Mecánica de Medios Continuos: Resumen de Álgebra y Cálculo Tensorial, UPM.
Eduardo W. V. Chaves, Mecánica del Medio Continuo,Vol. 1 Conceptosbásicos, Capítulo 1: Tensores de Mecánica del Medio Continuo, CIMNE, 2007.
L. E. Malvern. Introduction to the mechanics of a continuous medium. Prentice-Hall, Englewood Clis, NJ, 1969.
G. A. Holzapfel. Nonlinear solid mechanics: a continuum approach for engineering. 2000.
References
88