A REVIEW OF VECTORS AND TENSORS - TAMU...

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A REVIEW OF VECTORS AND TENSORS

Much of the material included herein is taken from the instructor’s two books exhibited here(both published by the

Cambridge UniversityPress)

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CONTENTS Physical vectorsMathematical vectors Dot product of vectors Cross product of vectors Plane area as a vector Scalar triple product Components of a vector Index notation Second-order tensorsHigher-order tensors Transformation of tensor

components Invariants of a second-order tensor Eigenvalues of a second-order tensor Del operator (Vector and Tensor calculus) Integral theorems

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In the mathematical description of equations governing acontinuous medium, we derive relations between various quantities that characterize the stress and deformation of the continuum by means of the laws of nature (such as Newton's laws, balance of energy, etc). As a means of expressing a natural law, a coordinate system in a chosen frame of reference is often introduced. The mathematical form of the law thus depends on the chosen coordinate system and may appear different in another type of coordinate system. The laws of nature, however, should be independent of the choice of the coordinate system, and we may seek to represent the law in a manner independent of the particular coordinate system. A way of doing this is provided by vector and tensor analysis.

INTRODUCTION TOVECTOR AND TENSOR ANALYSIS

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When vector notation is used, a particular coordinate system need not be introduced. Consequently, the use of vector notation in formulating natural laws leaves them invariant to coordinate transformations. A study of physical phenomena by means of vector equations often leads to a deeper understanding of the problem in addition to bringing simplicity and versatility into the analysis.

VECTOR AND TENSOR ANALYSIS

In basic engineering courses, the term vector is used often to imply a physical vector that has “magnitude and direction and satisfies the parallelogram law of addition.” In mathematics, vectors are more abstract objects than physical vectors. Like physical vectors, tensors must satisfy the rules of tensor addition and scalar multiplication.

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PHYSICAL VECTORS Physical vector: A directed line segment with an

arrow head.Examples: force, displacement, velocity, weight

Unit vector along a given vector A:

The unit vector,

is that vector which has the same

direction as A but has a magnitude

that is unity.

0Ae ( )A AA

●

●

P

Q

A

eA

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MATHEMATICAL VECTORSRules or Axioms

Vector addition: (i) A + B = B+A (commutative)(ii) (A + B)+C = A+(B+C) (associative)(iii) A+0=A (zero vector)(iv) A+(−A) = 0 (negative vector)

A

B

B

A A+B=B+A

Scalar multiplication of a vector: (i) α(βA)= αβ (A) (associative)(ii) (α + β)A = α A+ βA (distributive w.r.t. scalar addition)(iii) α (A+B)=αA+αB (distributive w.r.t. vector addition)(iv) 1. A=A . 1

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DOT PRODUCT OF VECTORSWork done Magnitude of the force multiplied by the

magnitude of the displacement in the direction of the force:

F

u

F

u

F ucos WD

WD F u cos

F cos

u cos

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Inner product (or scalar product) of two vectors is defined as

A B A B cos cosAB

A = A

B = B

A = A

A A

VECTORS (continued)

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e, sin MF r F M M r F

CROSS PRODUCT OF VECTORSMoment of a force Magnitude of the force

multiplied by the magnitude of the perpendicular distance to the action of the force:

P

F

Or θ

O r

F

θM

eM

r sin sinr

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VECTORS (continued)

Vector product of two vectors is defined as

A B A B e eˆ ˆsin sinAB ABAB

A

BeAB

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The magnitude of the vector C = A × B is equal to the area of the parallelogram formed by the vectors A and B.

In fact, the vector C may be considered to represent both the magnitude and the direction of the product of A and B. Thus, a plane area in space may be looked upon as possessing a direction in addition to a magnitude, the directional character arising out of the need to specify an orientation of the plane area in space. Representation of an area as a vector has many uses in mechanics, as will be seen in the sequel.

PLANE AREA AS A VECTOR

C = A ×B

A

Bê θ S

S nS n

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B

AC

B × C

SCALAR TRIPLE PRODUCTThe product A . (B × C) is a scalar and it is termed the scalar triple product. It can be seen from the figure that the product A . (B × C) , except for the algebraic sign, is the volume of the parallelepiped formed by the vectors A, B, and C.

A . (B × C)

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EXERCISES ON VECTORS0 A B1. If two vectors are such that

what can we conclude?

A B 02. If two vectors are such thatwhat can we conclude?

3. Prove that A B C A B C

0C A B4. If three vectors are such thatwhat can we conclude?

5. The velocity vector in a flow field is .Determine (a) the velocity vector normal to the plane

passing through the point, (b) the angle between, (c) tangential velocity vector on the plane, and

(d) The mass flow rate across the plane through an area if the fluid density is and the

flow is uniform.

2 3v i j ˆ ˆ (m/ s)

3 4n i- kˆvn

v vand n

20 15. mA 3 310 kg/ m

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COMPONENTS OF VECTORS

1 1 2 2 3 3

1 1 2 2 3 3

A e e e e e e

n e e e e e e

ˆ ˆ ˆˆ ˆ ˆ

ˆ ˆ ˆ ˆˆ ˆ ˆ

x x y y z z

x x y y z z

A A AA A An n nn n n

Components of a vector

1e eˆ ˆx

3e eˆ ˆz

2e eˆ ˆy

1x x

2y x

3z x●A

1x x

2y x

3z x

●

1A

2A

3AA

ˆi iA A e

1 1 1 2 1 3

2 2 2 3 3 3

1 1 1 2 3 2 1 3

2 3 1 3 1 2 1 3 2

1 0 01 0 10

e e e e e ee e e e e ee e e e e e e ee e e e e e e e e

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, ,

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AN EXERCISEProblem: Find the vector connecting point P: (2,0,3) to

point Q: (0,2,1).

Solution: The points P and Q are shown in the figure using their coordinates. We have z

x

y

P: (2,0,3)Q: (0,2,1)

●●0 2 2 0 1 3

2 2 2

i j k

i j k

ˆˆ ˆPQ ( ) ( ) ( )

ˆˆ ˆ

2 3 2i k j kˆ ˆˆ ˆP , Q

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Omit the summation sign and understand that a repeated index is to be summed over its range:

1 1 2 2 3 3

3

1

A e e e

e e

ˆ ˆ ˆ

ˆ ˆ (summation convention)i i i ii

A A A

A A

Dummy index

SUMMATION CONVENTION

ˆ ˆ ˆ ˆi i j j A A e e A e e Dummy indices

A B e e

e e

ˆ ˆ

ˆ ˆi i j j

i j i j

i j ij i i

A B

A BA B A B

01

e e , ifˆ ˆ, ifij i j

i ji j

Scalar product

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A B e

e e e ee

ˆsinˆ ˆ ˆ ˆ

ˆ

AB

i i j j i j i j

i j ijk k

ABA B A B

A B

01

1

e e e e e e

, if anytwoindicesarethesame, , andthey permuteˆ ˆ ˆ ˆ ˆ ˆ

inanaturalorder, ,andthey permute

toanaturalorder

ijk i j k i j kif i j k

if i j kopposite

Vector product

SUMMATION CONVENTION (continued)

e e eˆ ˆ ˆi j ijk k

1 2 3

1 2 3

1 2 3

e e eA B

ˆ ˆ ˆA A AB B B

12 ijk i j j k k i

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2 3, ( )i i j j k k i i j j kF A B C G H A B P Q F

Free indices

, andi j k i j k i j kA B C A B F A B C

Incorrect expressions:

Correct expressions:


Contraction of indices:The Kronecker delta modifies (or contracts) the subscripts in the coefficients of an expression in which it appears:

ij

, ,i ij j i j ij i i j j ij ik jkA A A B A B A B

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One must be careful when substituting a quantity with an index into an expression with indices or solving for one quantity with index in terms of the others with indices in an equation. For example, consider the equations

It is correct to write

but it is incorrect to write

which has a totally different meaning

andi i j j k i i kp a b c c d e q

ii

j j

pab c

ij j

i

pb ca

31 2

1 2 3

ij j

i

p pp pb ca a a a


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-


The permutation symbol and the Kronecker delta prove to be very useful in establishing vector identities. Since a vector form of any identity is invariant (i.e., valid in any coordinate system), it suffices to prove it in one coordinate system. The following identity is useful:

ijk imn jm kn jn km

11 12 13

21 22 23 1 2 3 1 2 3

31 32 33

ˆ ˆ ˆ( ) ( ) ( )i i j j k k ijk i j k

s s ss s s s s s s s ss s s

S e e e

The determinant of the matrix of a second-order tensor can be expressed as

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Exercise-3: Simplify the expression ( ) ( ) A B C D

Exercise-2: Prove

Exercise-4: Simplify the expression ( ) A B C

1 2 3

1 2 3

1 2 3

A B C ijk i j k

A A AA B C B B B

C C C

EXERCISES ON INDEX NOTATIONExercise-1: Check which one of the following expressions

are valid:

Exercise-5: Rewrite the expression in vector form

emni i j m n jA B C D

2

3

(a) ( ); (b) ( )(c) ( ); (d)(e) ; (f) ?

m s m r r m s m s s

i j i i i m m

i ij jk ki

a b c d f a b c d fa b c d f x x ra

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SECOND-ORDER TENSORS A second-order tensor is one that has two basis vectors standing

next to each other, and they satisfy the same rules as those of a vector (hence, mathematically, tensors are also called vectors). A second-order tensor and its transpose can be expressed in terms of rectangular Cartesian base vectors as

Second-order identity tensor has the form

S e e e e S e e e eTˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ;ij i j ji j i ji i j ij j iS S S S

I e eˆ ˆij i j

A second-order tensor is symmetric only if

S STij jiS S

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SECOND-ORDER TENSORS

S T e e e e e e e e e e

T S e e e e e e e e e e

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆij i j kl k l ij kl i j k l ij jl i l

ij i j kl k l ij kl i j k l jl ij i l

S T S T S T

T S T S S T

S T T S We note that (where S and T are second-order tensors) because

We also note that (where S and T are second-order tensors and A is a vector)

S T e e e e e e e e e e e

S A e e e e e e e

S A e e e e e e e e

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

ij i j kl k l ij kl i j k l ij kl jkp i p l

ij i j k k ij k i j k ij j i

ij i j k k ij k i j k ij k jkp i p

S T S T S T

S A S A S A

S A S A S A

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CAUCHY STRESS TENSOR Stress tensor is a good example of a second-order tensor. The two

basis vectors represent the direction and the plane on which they act. The Cauchy stress tensor is defined by the Cauchy formula (to be established in the sequel):

ˆ or i j ji ji jt n n σ σt n

σˆ ˆ ˆ ˆ ˆ ˆ,i ij j i i i ij j ij i j t e e t e e e e

t3

t2

t1

1x

2x

3x

1e

3e

2e21

22

233332

31

1112

13

2 21 1 22 2 23 3ˆ ˆ ˆ t e e e

1 11 1 12 2 13 3ˆ ˆ ˆ t e e e

3 31 1 32 2 33 3ˆ ˆ ˆ t e e e

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HIGHER-ORDER TENSORS A nth-order tensor is one that has n basis vectors standing next to

each other, and they satisfy the same rules as those of a vector. A nth-order tensor T can be expressed in terms of rectangular Cartesian base vectors as

T e e e e

e e e

C e e e e

subs base vectors

ˆ ˆ ˆ ˆ ;

ˆ ˆ ˆ

ˆ ˆ ˆ ˆ

n

ijk p i j k p

n

ijk i j k

ijkl i j k l

T

C

ε

The permutation tensor is a third-order tensor

The elasticity tensor is a fourth-order tensor

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TRANSFORMATION OF TENSOR COMPONENTS

A second-order Cartesian tensor S may be represented in barred and unbarred Cartesian coordinate systems as

1 2 3( , , )x x x 1 2 3( , , )x x x

e e e eˆ ˆˆ ˆij i j mn m ns s SThe unit base vectors in the unbarred and barred systems are related by

e e e e e eˆ ˆ ˆˆ ˆ ˆand ,j ij i i ij j ij i j

Thus the components of a second-order tensor transform according to

T[ ] [ ][ ][ ]ij im jn mns s S L S L

The components of a 4th-order tensor transform as

ijkl im jn kp lq mnpqc c

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INVARIANTS OF A SECOND-ORDER TENSOR

When a combination of components of a tensor remains the same in all coordinate systems, they are said to be invariant. There are many invariants of a tensor. For example,

11 22 331( ) iiS S S S is invariant, because

ii mn im in mn mn mm iiS S l l S S S

The determinant of a second –order tensor remains the same in all coordinate systems. The determinant of a second-ordertensor can be expressed as (exercises to the reader)

11 12 131

21 22 23 6

31 32 33

(a) , (b)ir js kt ijk rst rst ijk ir js kt

s s ss s s s s s s s ss s s

S S

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INVARIANTS OF A SECOND-ORDER TENSOR

1 16 6

16

16

16 2( ) is invariant

ir js kt ijk rst mn im rn pq jp sq ab ka tb ijk rst

mn pq ab rn sq tb im jp ka ijk rst

mn pq ab mpa rn sq tb rst

mn pq ab mpa nqb

s s s s l l s l l s l l

s s s l l l l l l

s s s l l l

s s s

S

S

3( ) Thus is invariant

ij ij mn im jn pq ip jq mn pq im ip jn jq

mn pq mp nq mn mn

ij ij

s s s l l s l l s s l l l ls s s s

s s

11 2 321 2 3( ) , ( ) , ( )ii ii jj ij ijI s I s s s s I S

The well-known three invariants are:

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EIGEN VALUES OF A SECOND-ORDER TENSOR

A second-order tensor can be viewed as an operator that changes a vector into another vector (by means of the dot product). Then it is possible that there may exist certain vectors that have only their lengths, not their orientations, changed when operated upon by the tensor: If such vectors exist, then we have

ˆ I 0S n

ˆ ˆ S n n

(1)

Such vectors are called characteristic vectors, or eigenvectors, associated with The parameter is called the eigenvalue, and it characterizes the change in length. Thus, we have

S.

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The vanishing of the determinant, , yields a cubic equation for , called the characteristic equation:

0 IS

3 21 2 3 0I I I

where are the invariants of as defined 1 2 3, , andI I I S

11 32

2 2 2 2 2 2 2 2 2 212 1 11 22 33 12 13 23 21 32

2

312

, ( , |) |,ii ii jj ij ijI I I

I I S S S S S S S

S S S S

S

S

S

S

The eigenvector associated with any particular eigenvalue is calculated using Eq. (1), which gives only two independent relations among the three components . The third equation is provided by

n( )ˆ i

i

1 2 3( ) ( ) ( ), an, di i in n n

2 2 21 2 3 1( ) ( ) ( )( ) ( ) ( )i i in n n

EIGEN VALUES OF A SECOND-ORDER TENSOR

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Problem: Given the matrix of a second-order tensor

Find the eigenvalues and eigenvectors associated with the minimum eigenvalue.

AN EXAMPLE OF AN EIGENVALUE PROBLEM

57 0 240 50 0

24 0 4357 50 43 50 24 24( )( )( ) ( )

57 0 240 50 0

24 0 43[ ]S

Solution: we set

and obtain the eigenvalues The eigenvector is given by

1 2 325 50 75, , .

11 3

3 45 5

( ) ˆ ˆ en e

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THE DEL OPERATOR AND ITS PROPERTIES IN RECTANGULAR CARTESIAN SYSTEM

“Del” operator:1 2 3

1 2 3

e e e eˆ ˆ ˆ ˆiix x x x

“Gradient” operation:

e , where isascalar functionii

FF Fx

“Laplace” operator:2 2 2 2

22 2 21 2 3i ix x x x x

Grad F defines both the direction and magnitude of the maximum rate of increase of F at any point.

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n n

n n

ˆ ˆ, where isaunit vector normal

to the surface constant

ˆ ˆWe also have and

FFn

FF F FF n

“Divergence” operation:

G e e Gˆ ˆ , where isa functionii j j

i i

GG vectorx x

The divergence of a vector function represents the volume density of the outward flux of the vector field.

https://en.wikipedia.org/wiki/Flux

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“Curl” operation:

G e e e

G

ˆ ˆ ˆ ,

where isa function.

ji j j ijk k

i i

GG

x xvector

ε

The curl of a vector function represents its rotation. If the vector field is the velocity of a fluid, curl of the velocity represents the rotation of the fluid at the point.

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CYLINDRICAL COORDINATE SYSTEM

“Del” operator in cylindrical coordinates1e e eˆ ˆ ˆr zr r z

x

z

y

θ

ezˆ

erˆ

e

R 2 2 2r z R

r

y x

z

00

0 0 1

00

0 0 1

e ee ee e

e ee ee e

ˆ ˆcos sinˆ ˆsin cosˆ ˆ

ˆ ˆcos sinˆ ˆsin cosˆ ˆ

r x

y

z z

x r

y

z z

e ee eˆ ˆˆ ˆ,r

r

A e e eˆ ˆ ˆr r z zA A A

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CYLINDRICAL COORDINATE SYSTEM

2 22

2 2

1

1 1

( ) zr ArA A rr r z

r rr r r r z

A

1 1

1

1 1

e e e

e e e e e e e e e e

e e e e

( )ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ

z zr rr z

zr r rr r r r r z z r

zr

A AA A rA Ar z z r r r

AA A A AAr r r r z

AAAr r

A

A

e e e eˆ ˆ ˆ ˆzz z z z

AAz z

Here A is a vector:

Verify these relations to yourself

A e e eˆ ˆ ˆr r z zA A A

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SPHERICAL COORDINATE SYSTEM

“Del” operator1 1e e eˆ ˆ ˆ

sinR R R R

0

e ee e

ee

eee

ˆ ˆsin cos sin sin cosˆ ˆcos cos cos sin sin

ˆsin cosˆ

ˆ sin cos cos cos sinˆ sin sin cos sin cosˆ c

R x

y

z

x

y

z

0

eee

ˆˆ

os sin ˆ

R

e ee e

e ee e

e e e

ˆ ˆˆ ˆ, sin

ˆ ˆˆ ˆ, cos

ˆ ˆ ˆsin cos

R R

R

R

A e e eˆ ˆ ˆR RA A A

x

z

y

θ

eRˆe

e

R

x

y

z

θLine parallel to e

Line parallel to e

R eR 2 2

2 2 2

RˆRR r z

r x y

VECTORS&TENSORS -38

SPHERICAL COORDINATE SYSTEM2

2 22 2 2

1 1 1

2 1 1

sinsin sin

( sin )sin sin

R R

RR R R

AA A AR R R R

A

1 1 1 1

1 1

e e e

e e e e e e e e e e

( )(sin ) ( )ˆ ˆ ˆsin sin

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆsinsin

R RR

R R RR R R R R

A RAA AA RAR R R R R R

AA A AAA AR R R R R

A

A

1 1 1

1

e e e e e e

e e

ˆ ˆ ˆ ˆ ˆ ˆcossin

ˆ ˆsin cossin

R

R

R

A AAA AR R R

AA AR

VECTORS&TENSORS -39

EXERCISES ON VECTOR CALCULUSEstablish the following identities (using rectangular

Cartesian components and index notation):

2

2

12

00

A

( )

( )( )

( )( )( ) ( )

( ) ( ) .

n n

rr

r nrF

F G

r

r0

A A AA B A B B A

A A A A A A

1.

2.3.4.5.

6.7.

8.

VECTORS&TENSORS -40

Exercise: Check appropriate box

Quantity Vector Scalar Nonsense

F

F

F

F

( )

( )

( )

( )

( )

( )

f

f

Fscalar; vectorf

VECTORS&TENSORS -41

INTEGRAL THEOREMSinvolving the del operator

n

n

n

ˆ (Gradient theorem)

ˆ (Divergence theorem)

ˆ (Curl theorem)

d ds

d ds

d ds

A A

A A

n i je ee e1 1 2 2

x y

x x y y

ˆ ˆˆ n nˆ ˆn nˆ ˆn n

x

y nixˆn

jyˆn

e jyˆˆ

e ixˆˆ

ds

VECTORS&TENSORS -Mass-Momenta - 42

n F d F d

THE GRADIENT THEOREM

The gradient of a function F represents the rate of change of F with respect to the coordinate directions.the partial derivative with respect to x, for example, gives the rate of change of F in the x direction.

F normal to the surfaceLevel surfaces

VECTORS&TENSORS -Mass-Momenta - 43

GRADIENT OF A SCALAR FUNCTION

a first- order tensor,that is, aF vector

in rectangular Cartesian system

in cylindrical coordinate system

e e e

e e e

ˆ ˆ ˆ

ˆ ˆ ˆ

x y z

x y x

x y z

F F FFx y z

1

1

e e e

e e e

ˆ ˆ ˆ

ˆ ˆ ˆ

r z

r z

r r z

F F FFr r z

VECTORS&TENSORS -

ˆ d d

n F F

The divergence represents the volume density of the outward flux of a vector field F from an infinitesimal volume around a given point. It is a local measure of its "outgoingness."

d

●

d

dFlux

THE DIVERGENCE THEOREM

VECTORS&TENSORS -

DIVERGENCE OF FIRST-ORDER TENSORS

F a zeroth tensor,that is, a scalar



e e e

F

ˆ ˆ ˆx y z

yx z

x y z

FF Fx y z

1

1

e e e

F

ˆ ˆ ˆ

( )

r z

zr

r r z

FrF F rr r z

VECTORS&TENSORS -

DIVERGENCE OF SECOND-ORDER TENSORS

S afirst- order tensor,that is, a vector

1

1

1

S e e

e

S e

e

ˆ ˆ

ˆ

( ) ˆ

( ) ˆ

yx xy yy zyxx zxx y

yzxz zzz

zrrr rr

zrr

S S S FS Fx y z x y z

SS Fx y z

SrS S r Sr r z

SrS S r Sr r z

e( ) ˆrz z zrz

rS S Srr r z



VECTORS&TENSORS -

n F Fˆ d d

CURL (STOKES’s) THEOREM

The curl of a vector F describes the infinitesimal rotation of F. A physical interpretation is as follows. Suppose the vector field describes the velocity field F = v of a fluid flow, say, in a large tank of liquid, and a small spherical ball is located within the fluid (the center of the ball being fixed at a certain point but free to rotate about an axis perpendicular to the plane of the flow). If the ball has a rough surface, the fluid flowing past the ball will make it rotate. The rotation axis (oriented according to the right hand rule) points in the direction of the curl of the field at the center of the ball, and the angular speed of the rotation is half the magnitude of the curl at this point.

VECTORS&TENSORS -

Curl of a Vector Function in (x,y,z) system

A afirst- order tensor,that is, a vector

A e e e e e e

e e e e e e

e e e e e e

e e e

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ

yx zx x x y x z

yx zy x y y y z

yx zz x z y z z

y yz x z xx y z

AA Ax x x

AA Ay y y

AA Az z z

A AA A A Ay z z x x y

0 eyez

ez 0

0

ex

ex

ey

VECTORS&TENSORS -49

EXERCISES ON INTEGRAL IDENTITIES

21 16 3

2

2

2 2

4 2 2 2 2

1

2

3

4

5

n nˆ ˆ. volume ( )

.

.

.

. ( )

r d d

d dn

d dn

d dn n

d dn n

r

Establish the following identities using the integral theorems:

A REVIEW OF VECTORS AND TENSORS - TAMU...

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