Master in Computational and Applied Physics

Continuum and Fluid MechanicsContinuum and Fluid Mechanics

CHAPTER 1: Tensor CalculusCHAPTER 1: Tensor Calculus

Albert Falqués

Applied Physics DepartmentTechnical University of Catalonia

OUTLINE

1. Scalars, vectors and tensors. Cartesian basis. Rotation of axes.2. Example: stress tensor.3 Matrix algebra Multiplication and contraction3. Matrix algebra. Multiplication and contraction. 4. Isotropic tensors. 5. Algebraic properties of symmetric second order tensors. Eigenvalues and

ieigenvectors. 6. Gradient operator, divergence and curl. Gauss and Stokes theorems.

Vector identities.

BIBLIOGRAPHYBIBLIOGRAPHY

P.K.Kundu. ‘Fluid Mechanics’. Academic Press. 1990. R. Aris. ‘Vectors, tensors and the basic equations of fluid

h i ’ D 1962mechanics’. Dover, 1962. J.M. Massaguer, A. Falqués. ‘Mecánica del continuo.

Geometría y Dinámica’ Edicions UPC 1994Geometría y Dinámica . Edicions UPC, 1994. D.E. Bourne, P.C. Kendall. ‘Análisis vectorial y tensores

cartesianos’ Limusa 1976cartesianos . Limusa, 1976. L.A. Segel. ‘Mathematics applied to continuum mechanics’.

Dover, 1977.Dover, 1977.

1. Scalars, vectors and tensors. Cartesian basis. R t ti f xRotation of axes.

Scalar: quantity which is defined by its magnitude only. Example: temperature, density, kinetic energy

Vector: quantity which is defined by its magnitude and direction Given a Vector: quantity which is defined by its magnitude and direction. Given a coordinate axis, a vector is defined by three components.

Example: force, velocity, acceleration, momentum, torque Tensor: linear or multilinear map between vectors (or vectors and scalars).

Given a coordinate axes system, a tensor is defined by, at least, nine components.p

Example: stress tensor (force per area unit across any section of a body at a point), strain tensor (deformation rate of a body in any direction at a point)

Cartesian axes

x 3x

e3e

ortonormal basis:

cartesian coordinates: 321 ,, xxx

)1,0,0(ˆ,)0,1,0(ˆ,)0,0,1(ˆ 321 eee

1x2x

1e2e

position vector: ii

iexx ˆ3

1

ortonormal basis: )1,0,0(,)0,1,0(,)0,0,1( 321 eee

1 i 1

Rotation of axis3e'ˆ3e

'ˆ

3e

3322221122

3312211111

ˆˆˆ'ˆˆˆˆ'ˆeCeCeCeeCeCeCe

ˆ

2e'1e

'e

3332231133 ˆˆˆ'ˆ eCeCeCe

321ˆ'ˆ3

iC1e 2e 3,2,1,'1

ieCe jj

jii

i = free index; j = dummy indexi ee de ; j du y de

Summation convention over repeated indexes In any product of terms a repeated index is held to be summed over 1,2,3.An index not repeated in any product can take any of the values 1,2,3.

Examples: 3Examples:

33

3,2,1,ˆ'ˆˆ'ˆ3

1

ieCeeCe jj

jiijjii

3,2,1,11,

njRcbaRcbak

kjnkki

knijkikjnkknijki

0

Warning: an index should not be repeated more than twice.Example:

h ld b bb i d 03

a0333222111 aaa should be abbreviated as 01

i

iiia

but not as 0iiia that would mean: 0333222111 aaa

Warning: same free indexes at both sides of an equation or for all the terms in a sum.

??? kjkiij acbaExample: ??? knjkiij acba

Rotation matrix. Properties

232221

131211

CCCCCC

C iii CCC 321 ,, components of ei’ on the basis e1, e2, e3= cosinus of the angles between the new

333231 CCC

cosinus of the angles between the new axis xi’ and the old axes x1’, x2’, x3’

2x'

cos,2/cos2/cos,cos

2212

2111

CCCC

'2x '1x

, 2212

1x

The transpose matrix: verifies: C·CT = 1 (and also CT·C = 1)

322212

312111

CCCCCC

TC

001312111131211 CCCCCC

332313 CCC

100010

332313

322212

333231

232221

CCCCCC

CCCCCC

This is the ‘orthogonality condition’ and implies that: TCC 1

Change of the coordinates of a point.

jjii eCe ˆ'ˆ Given that the basis’ vectors change according to:

How do the coordinates of any point change?

jkkjjkkj

jjkkkkjj

CxxCxx

eCxexexx1''

ˆ''ˆ'ˆ

jkkjjkkj CxxCxx

the coordinates do change with the same matrix Cand since TCC 1

kkjj xCx ' or xCx' T

Orthogonal transformations ….. det C = 1 …..

Formal definition of a vector and a scalar.

A physical quantity b is said to be a scalar if it is invariant under axis rotations, i.e.,

bb' bb '

Three physical quantities (a1, a2, a3) are said to define a vectorif they change as the coordinates of a point under axis rotations, i.e.,

a

kkii aCa '

Tensors arise as linear and multilinear maps between vectors (or vectors and scalars). Given a coordinate system, a tensor is defined by 3n components where n is its order.A scalar can be considered a tensor of order 0 and a vector a tensor of order 1.A formal definition based on the transformation of their components will be given later on.We will first introduce a particular tensor as an example: the stress tensor

2. Stress tensor

F

Force per unit area across a section of a body (force of part B on part A: then the normal is outwards from A):

F

nS

Ff

limSA B S

fS 0lim

f

is a vector that depends on the orientation of the surface, i.e., another vector, nf p , , ,

How is this relation ?2on Newton Law applied to the tetrahedron when S 0 :

n1S

3x

131

332211

SSS

ahSfSfSfSfS

nS2S

2x0

31

33

22

11 ahf

SSf

SSf

SSf

if

external force on iS

3S1xif e te a o ce o iS

so corresponding to a normal ien ˆ

321 SSS 3

32

21

1 fSSf

SSf

SSf

SS Si are the projections of S ii

i nSS

cosnS1S

2S

3S1x

2xif

external force on iSso corresponding to a normal ien ˆ

31force per unit area corresponding to the normals ien ˆ

iif

332211 nnnf therefore, is a linear function of f

n

nf ˆτ

τ stress tensor

nnnf

Matrix expression

Symmetry of the stress tensor !!332211 nnnf

components of

Symmetry of the stress tensor !!ij = ji

131211

321 ,, iii icomponents of

333231

232221

131211

321321 ,,,,nnnfff stress tensor matrix

ij components of the stress tensor 333231 ij

32 components second order tensor33

13 2331 32

3x

meaning of the components

1211 22

13

211x

2x 332211 ,, normal stresses1x stresses...,,,, 23211312 shear

stresses11 > 0 : pulling ; 11 < 0 : pushing (compression)

How do the components of the stress tensor change under axis rotation?

jiji nf jjiijjii nCnfCf ','

',' jijijiji nCnfCf

jikjkjij nCfC ''

jikjkijikjkijikjki nCCnCnCf '''' T1 CC jikjkinjikjknijikjknin nCCnCnCf CC

'''' nCCnf knkinjijkkn nCCnf

Tjiinjkkn CC ' CτCτ T 'or

Formal definition of a tensor.

Nine physical quantities are said to define a second order tensorif they change as the stress tensor under axis rotations, i.e.,

T ijTif they change as the stress tensor under axis rotations, i.e.,

jiinjkkn TCCT '

3n physical quantities are said to define a n-order tensorif they change under axis rotations as

T niiT ...1

if they change under axis rotations as,

nnnn jjijijijii TCCCT ...... 122111...'

nnnn

3. Matrix algebra. Multiplication and contraction.

Matrix algebra can be expressed in several ways:a) Matriciala) Matricialb) Componentsc) Intrinsic or symbolic

Example:

010001

322212

312111

232221

131211

CCCCCC

CCCCCC

a)

100332313333231 CCCCCC

ijjkikCC b) entries of the identity matrix: ij1 if i=j

ijjkikCC b)

c) 1CC T

y ij0 if ij

Tensor multiplication of vectors and tensors.

Tensor product of two vectors

ba,Any pair of vectors, , define a linear map between vectors:

avbvba )()( vba

ba

avbvba )()( v

The components of tensor are simply ba

jiba

Tensor product of vectors and tensorsIn general by multiplying the components of a n-order tensor and those of anIn general, by multiplying the components of a n-order tensor and those of an m–order tensor, an (n+m)-order tensor is defined through its components.

Examples:Examples: From ai and Tij the third order tensor: Qijk=ai Tjk may be defined From two 2nd order tensors, Aij , Bij, the 4th order tensor Pijkl=AijBkl may be defined

Quotient rule.

Let be 3n+3m physical quantities and let be their product mn jjii XA ...... 11

,mnkkB

...1

XAB mnmn jjiijjii XAB ............ 1111

Assume that A and B are tensors

Then, are the components of an m-order tensor mjjX ...1

Namely;

mnmn jjiijjii BXA ............ 1111

tensor tensor X = tensortensor tensor X tensor

Contraction.

Given a n-order tensor, contraction is a procedure to obtain a lower order tensor.Two indices are equated and a summation is performed over these repeated indices.

Examples: From the components of a 2nd order tensor, , the only possible contraction is ijT

332211 TTTTii which is a scalar

From a third order tensor, , three different contractions are possible but all of them give a vector:

332211ii

ijkTgive a vector:

From two 2nd order tensors, , the four order tensor may be defined by multiplying. Four 2nd order tensors may be then obtained

iikkijijijji TcTbTa ,,ijij BA , klijijkl BAQ

y y p y g yby contracting:

iljlij BA BA ikkjij BA TBA jkikij BA BAT kjkiij BA AB

The components of all of them can be computed from standard matrix product

A second contraction may be applied to these 2nd order tensors and a scalar is obtained in two possible ways or They are indicated by AB or by ABTBA BAin two possible ways or . They are indicated by AB or by ABjiij BA ijij BA

4. Isotropic tensors.

3,2,1...' 21 iiiiii iiiTTAn isotropic tensor is one whose components are invariant under axes rotations. i.e.

3,2,1...21...... 2121 niiiiii iiiTT

nn

Isotropic tensor are associated to geometric invariance under rotation.

0-order isotropic tensors: all 0-order tensors are isotropic as they are scalars

1st-order isotropic tensors: there are nonep

2nd-order isotropic tensors: only one, the identity tensor, whose components are given by the Kronecker delta in any basis:

010001

δ

by the Kronecker delta in any basis:

ij1 if i=j

0 if ij

100

Very common use of ij is that in any expression where it appears with index ‘i’ being contracted, it can be dropped out by substituting ‘i’ by ‘j’ in the expression.

0 if ij

co c ed, c be d opped ou by subs u g by j e e p ess o .Examples:

nliiklijknijnjmnimij CBCBAA ,

3rd-order isotropic tensors: only one, the alternating tensor, whose components areare

ijk

1 if i,j,k= even permutation of 123 (i.e., 123 or 231 or 312)

0 if two or three indices are equal

Properties:

-1 if i,j,k= odd permutation of 123 (i.e., 132 or 213 or 321)

Properties:

kijjkiijk even permutations

jikijkkjiijkikjijk ,, odd permutations

kmjnknjmimnijk

The cross product of two vectors, reads

jjj

ikjijk ebaba ˆ

Given any matrix A, det A = ijk A1iA2jA3k

4th-order isotropic tensors: there are three and their linear combinations. So the pmost general is:

)()( jpiqjqipjpiqjqippqij

where , , are arbitrary numbers.

5. Symmetric and antisymmetric second order tensors.Ei l d i tEigenvalues and eigenvectors

A 2nd order tensor B is called symmetric if Bij = Bji antisymmetric if Bij = -Bji

P tiProperties:

Any 2nd order tensor can be represented as the sum of a symmetric part and i ian antisymmetric one: jiijjiijij BBBBB

21

21

t i ti t i

If Aij is antisymmetric and Bij is symmetric, then AijBij = 0

symmetric antisymmetric

A symmetric tensor has only 6 independent components

A ti t i t h di l t d h l 3 An antisymmetric tensor has zero diagonal components and has only 3 independent components. These 3 components are associated with a vector

Properties (continued):p ( )

Every antisymmetric tensor can be associated with a vector and vice versa

R ijkijk R21

0

0

13

23

R

kijkijR

012

Eigenvectors and eigenvaluesg g

Eigenproblem:Given a 2nd order tensor A are there vectors and numbers such thatvGiven a 2 order tensor, A, are there vectors and numbers , such thatv

vv A ?

i t i l

are the solutions of the third order equation (characteristic equation)

0d t δA

v eigenvector , = eigenvalue

for each , is a solution of the eigenvalues and the direction of the eigenvectors are invariant under

0det δA v 0 vδA

the eigenvalues and the direction of the eigenvectors are invariant under axes rotations

the characteristic equation reads: where

0322

13 III

AI 3211 iiAI

1332212

2 21 iiijij AAAI

d AIare invariant under axes rotations

3213 det AI

If A is symmetric:

there are three eigenvalues that are real, 1, 2, 3 (not necessarily distinct)g , 1, 2, 3 ( y )

associated to them, there are three eigenvectors which are mutually orthogonal

if th di t t i t t d t i id ith th i t if the coordinate system is rotated as to coincide with the eigenvectors, matrix A takes a diagonal form:

2

1

0000

A

extremal property: the components A change with the coordinate axes but the

3

2

0000

A

extremal property: the components Aij change with the coordinate axes, but the diagonal elements cannot be larger than the largest and smaller than the smallest

6. Gradient operator, divergence and curl. G d St k th m V t id titiGauss and Stokes theorems. Vector identities

ie

Given a scalar field, (x1,x2,x3), its gradient is defined by

iix

is a vector gives the direction of maximum increase of

n

n

nˆ

|| = magnitude of the derivative of along this direction is perpendicular to the surfaces of (x1,x2,x3) = const. The derivative of along a direction associated to is

nSince is a vector and is a scalar, the operator itself can be considered a vector (quotient rule):

ˆˆ

bl t

iiii

eex

ˆˆ nabla operator

( ill i h f i li i )

(we will omit the arrow for simplicity, )

The gradient of a vector field is a second order tensor:

jiij vv

such that multiplied by a unit vector gives the derivative of such a vectornsuch that multiplied by a unit vector gives the derivative of such a vector along the direction of :

nn

jjii evnvndn

vd ˆˆ

Si il l th di t f t fi ld f d t i +1 d t

dn

Similarly, the gradient of a tensor field of order tensor n is a n+1 order tensor with similar meaning.

The divergence of a vector field is defined as the contraction of its gradient:

v

i

iii x

vvv

(or dot product of and ) v

The divergence of a vector field at a point is associated to its flux going in or out a small surface around this point

dSnvV

vV

ˆ1lim0

V

),,( 321 xxxv

VV

The divergence of a tensor field of n-order may be defined trough a contraction of its gradient and it is a (n-1)-order tensor. There are however several options depending on which contraction is performed. For a 2nd order tensor:

kikiT T jkiijk TT

jjiiT T

The curl of a vector field is defined as the cross product of vector and v

ikjijk evv ˆ

The curl of a vector field is associated with the rotation of the vector field. Example: velocity field of a rigid body:

2 vxv

So, the curl of the velocity field is proportional to the angular velocity of the body

v

Gauss Theorem (divergence theorem)

dAndA ˆ

V V

dAnudVu ˆ

dV

dA

dAnudVuii

i

VdV

V = volume

x Vii

V iV V

V = surface, boundary of V

Al i 2DAlso in 2D: dlndl ˆdl

dldSuu

21Sx2

dS dlnunudSxx SS

2211

2

2

1

1

S

x1

V S surfaceS = curve, boundary of S (in general in n-dimensions, n>1)

Stokes Theorem

dAndA ˆ

A

dludAnuAA

ˆ

Adl

A f

inside

A = surfaceA = curve, boundary of A

sign: choose one of both sides of the surface and define it as the outside, the normal vector pointing from inside to outside.Then the positive direction of is the anticlockwise onedloutside

n

insideThen, the positive direction of is the anticlockwise one looking from the outside

dl

Vector identities

ba , gf ,vector fields scalar fields

f 0 a 0 f

aaa 2

gffggffg 2222

faafaf faafaf

abbaabbaba

bbb baabba

baabbaabba

aaaaaa

21

Laplacian operator (scalar): ii2