SSG 805 Mechanics of Continua - OA Fakoafak.com/wp-content/uploads/2015/03/1.-Vectors-and... · 2015-03-23 · SSG 805 Mechanics of Continua INSTRUCTOR: Fakinlede OA ... Tadmore,

SSG 805 Mechanics of ContinuaINSTRUCTOR: Fakinlede OA [email protected] [email protected]

Department of Systems Engineering, University of Lagos

mailto:[email protected]


Available to beginning graduate students in Engineering

Provides a background to several other courses such as Elasticity, Plasticity, Fluid Mechanics, Heat Transfer, Fracture Mechanics, Rheology, Dynamics, Acoustics, etc. These courses are taught in several of our departments. This present course may be viewed as an advanced introduction to the modern approach to these courses

It is taught so that related graduate courses can build on this modern approach.

Purpose of the Course

[email protected] 12/31/2012Department of Systems Engineering, University of Lagos 2


The slides here are quite extensive. They are meant to assist the serious learner. They are NO SUBSTITUTES for the course text which must be read and followed concurrently.

Preparation by reading ahead is ABSOLUTELYnecessary to follow this course

Assignments are given at the end of each class and they are due (No excuses) exactly five days later.

Late submission carry zero grade.

What you will need



Scope of Instructional Material

Course Schedule:

The read-ahead materials are from Gurtin except the part marked red. There please read Holzapfel. Home work assignments will be drawn from the range of pages in the respective books.

Slide Title Slides Weeks Text Read Pages

Vectors and Linear Spaces 80 2 Gurtin 1-8

Tensor Algebra 110 3 Gurtin 9-37

Tensor Calculus 119 3 Gurtin 39-57

Kinematics: Deformation & Motion 106 3 Gurtin 59-123

Theory of Stress & Heat Flux 43 1 Holz 109-129

Balance Laws 58 2 Gurtin 125-205

The only remedy for late submission is that you fight for the rest of your grade in the final exam if your excuse is considered to be genuine. Ordinarily, the following will hold:


Examination

Evaluation Obtainable

Class Attendance 10

Homework 5

Midterm 45

Exam 40

Total 100

This course was prepared with several textbooks and papers. They will be listed below. However, the main course text is: Gurtin ME, Fried E & Anand L, The Mechanics and Thermodynamics of Continua, Cambridge University Press, www.cambridge.org2010

The course will cover pp1-240 of the book. You can view the course as a way to assist your reading and understanding of this book

The specific pages to be read each week are given ahead of time. It is a waste of time to come to class without the preparation of reading ahead.

This preparation requires going through the slides and the area in the course text that will be covered.

Course Texts


http://www.cambridge.org/

The software for the Course is Mathematica 9 by Wolfram Research. Each student is entitled to a licensed copy. Find out from the LG Laboratory

It your duty to learn to use it. Students will find some examples too laborious to execute by manual computation. It is a good idea to start learning Mathematica ahead of your need of it.

For later courses, commercial FEA Simulations package such as ANSYS, COMSOL or NASTRAN will be needed. Student editions of some of these are available. We have COMSOL in the LG Laboratory

Software


Gurtin, ME, Fried, E & Anand, L, The Mechanics and Thermodynamics of Continua, Cambridge University Press, www.cambridge.org 2010

Bertram, A, Elasticity and Plasticity of Large Deformations, Springer-Verlag Berlin Heidelberg, 2008

Tadmore, E, Miller, R & Elliott, R, Continuum Mechanics and Thermodynamics From Fundamental Concepts to Governing Equations, Cambridge University Press, www.cambridge.org , 2012

Nagahban, M, The Mechanical and ThermodynamicalTheory of Plasticity, CRC Press, Taylor and Francis Group, June 2012

Heinbockel, JH, Introduction to Tensor Calculus and Continuum Mechanics, Trafford, 2003

Texts




Bower, AF, Applied Mechanics of Solids, CRC Press, 2010 Taber, LA, Nonlinear Theory of Elasticity, World Scientific,

2008 Ogden, RW, Nonlinear Elastic Deformations, Dover

Publications, Inc. NY, 1997 Humphrey, JD, Cadiovascular Solid Mechanics: Cells,

Tissues and Organs, Springer-Verlag, NY, 2002 Holzapfel, GA, Nonlinear Solid Mechanics, Wiley NY, 2007 McConnell, AJ, Applications of Tensor Analysis, Dover

Publications, NY 1951 Gibbs, JW “A Method of Geometrical Representation of

the Thermodynamic Properties of Substances by Means of Surfaces,” Transactions of the Connecticut Academy of Arts and Sciences 2, Dec. 1873, pp. 382-404.

Texts


Romano, A, Lancellotta, R, & Marasco A, Continuum Mechanics using Mathematica, Fundamentals, Applications and Scientific Computing, Modeling and Simulation in Science and Technology, Birkhauser, Boston 2006

Reddy, JN, Principles of Continuum Mechanics, Cambridge University Press, www.cambridge.org 2012

Brannon, RM, Functional and Structured Tensor Analysis for Engineers, UNM BOOK DRAFT, 2006, pp 177-184.

Atluri, SN, Alternative Stress and Conjugate Strain Measures, and Mixed Variational Formulations Involving Rigid Rotations, for Computational Analysis of Finitely Deformed Solids with Application to Plates and Shells, Computers and Structures, Vol. 18, No 1, 1984, pp 93-116

Texts



Wang, CC, A New Representation Theorem for Isotropic Functions: An Answer to Professor G. F. Smith's Criticism of my Papers on Representations for Isotropic Functions Part 1. Scalar-Valued Isotropic Functions, Archives of Rational Mechanics, 1969 pp

Dill, EH, Continuum Mechanics, Elasticity, Plasticity, Viscoelasticity, CRC Press, 2007

Bonet J & Wood, RD, Nonlinear Mechanics for Finite Element Analysis, Cambridge University Press, www.cambridge.org 2008

Wenger, J & Haddow, JB, Introduction to Continuum Mechanics & Thermodynamics, Cambridge University Press, www.cambridge.org 2010


Texts



Li, S & Wang G, Introduction to Micromechanics and Nanomechanics, World Scientific, 2008

Wolfram, S The Mathematica Book, 5th Edition Wolgram Media 2003

Trott, M, The Mathematica Guidebook, 4 volumes: Symbolics, Numerics, Graphics &Programming, Springer 2000

Sokolnikoff, IS, Tensor Analysis, Theory and Applications to Geometry and Mechanics of Continua, John Wiley, 1964


Texts

Linear SpacesIntroduction

Continuum Mechanics can be thought of as the grand unifying theory of engineering science.

Many of the courses taught in an engineering curriculum are closely related and can be obtained as special cases of the general framework of continuum mechanics.

This fact is easily lost on most undergraduate and even some graduate students.

Unified Theory


Continuum Mechanics


Continuum Mechanics views matter as continuously distributed in space. The physical quantities we are interested in can be

• Scalars or reals, such as time, energy, power,

• Vectors, for example, position vectors, velocities, or forces,

• Tensors: deformation gradient, strain and stress measures.

Physical Quantities


Since we can also interpret scalars as zeroth-order tensors, and vectors as 1st-order tensors, all continuum mechanical quantities can generally be considered as tensors of different orders.

It is therefore clear that a thorough understanding of Tensors is essential to continuum mechanics. This is NOT always an easy requirement;

The notational representation of tensors is often inconsistent as different authors take the liberty to express themselves in several ways.

Physical Quantities


There are two major divisions common in the literature: Invariant or direct notation and the component notation.

Each has its own advantages and shortcomings. It is possible for a reader that is versatile in one to be handicapped in reading literature written from the other viewpoint. In fact, it has been alleged that

“Continuum Mechanics may appear as a fortress surrounded by the walls of tensor notation” It is our hope that the course helps every serious learner overcome these difficulties

Physical Quantities


The set of real numbers is denoted by R Let R 𝑛 be the set of n-tuples so that when 𝑛 = 2,

R 2 we have the set of pairs of real numbers. For example, such can be used for the 𝑥 and 𝑦 coordinates of points on a Cartesian coordinate system.

Real Numbers & Tuples


A real vector space V is a set of elements (called vectors) such that,1. Addition operation is defined and it is commutative

and associative underV : that is, 𝒖 + 𝐯 ∈V, 𝒖 + 𝐯 =𝐯 + 𝒖, 𝒖 + 𝐯 + 𝒘 = 𝒖 + 𝐯 +𝒘,∀ 𝒖, 𝐯,𝒘 ∈ V. Furthermore,V is closed under addition: That is, given

that 𝒖, 𝐯 ∈ V, then 𝒘 = 𝒖 + 𝐯 = 𝐯 + 𝒖,⇒ 𝒘 ∈ V.

2. V contains a zero element 𝒐 such that 𝒖 + 𝒐 = 𝒖 ∀𝒖 ∈V. For every 𝒖 ∈ V, ∃ − 𝒖: 𝒖 + −𝒖 = 𝒐.

3. Multiplication by a scalar. For 𝛼, 𝛽 ∈ R and 𝒖, 𝒗 ∈ V,𝛼𝒖 ∈ V , 1𝒖 = 𝒖, 𝛼 𝛽𝒖 = 𝛼𝛽 𝒖, 𝛼 + 𝛽 𝒖 = 𝛼𝒖 +𝛽𝒖, 𝛼 𝒖 + 𝐯 = 𝛼𝒖 + 𝛼𝐯.

Vector Space


An Inner-Product (also called a Euclidean Vector) SpaceE is a real vector space that defines the scalar product: for each pair 𝒖, 𝐯 ∈ E, ∃ 𝑙 ∈ R such that, 𝑙 = 𝒖 ⋅ 𝐯 = 𝐯 ⋅ 𝒖 . Further, 𝒖 ⋅ 𝒖 ≥ 0, the zero value occurring only when 𝒖 = 0. It is called “Euclidean” because the laws of Euclidean geometry hold in such a space.

The inner product also called a dot product, is the mapping

" ⋅ ":V × V → Rfrom the product space to the real space.

Euclidean Vector Space


We were previously told that a vector is something that has magnitude and direction. We often represent such objects as directed lines. Do such objects conform to our present definition?

To answer, we only need to see if the three conditions we previously stipulated are met:

Magnitude & Direction


From our definition of the Euclidean space, it is easy to see that,

𝐯 = 𝛼1𝒖1 + 𝛼2𝒖2 +⋯+ 𝛼𝑛𝒖𝑛such that, 𝛼1, 𝛼2, ⋯ , 𝛼𝑛 ∈ R and 𝒖1, 𝒖2, ⋯ , 𝒖𝑛 ∈ E, is also a vector. The subset

𝒖1, 𝒖2, ⋯ , 𝒖𝑛 ⊂ Eis said to be linearly independent or free if for any choice of the subset 𝛼1, 𝛼2, ⋯ , 𝛼𝑛 ⊂ R other than 0,0,⋯ , 0 , 𝐯 ≠ 0.If it is possible to find linearly independent vector systems of order 𝑛 where 𝑛 is a finite integer, but there is no free system of order 𝑛 + 1, then the dimension ofE is 𝑛. In other words,

the dimension of a space is the highest number of linearly independent members it can have.

Dimensionality


Any linearly independent subset ofE is said to form a basis forE in the sense that any other vector inE can be expressed as a linear combination of members of that subset. In particular our familiar Cartesian vectors 𝒊, 𝒋 and 𝒌 is a famous such subset in three dimensional Euclidean space.

From the above definition, it is clear that a basis is not necessarily unique.

Basis


1. Addition operation for a directed line segment is defined by the parallelogram law for addition.

2. V contains a zero element 𝒐 in such a case is simply a point

with zero length..

3. Multiplication by a scalar 𝛼. Has the meaning that

0 < 𝛼 ≤ 1 → Line is shrunk along the same direction by 𝛼

𝛼 > 1 → Elongation by 𝛼

Negative value is same as the above with a change of direction.

Directed Line


Now we have confirmed that our original notion of a vector is accommodated. It is not all that possess magnitude and direction that can be members of a vector space.

A book has a size and a direction but because we cannot define addition, multiplication by a scalar as we have done for the directed line segment, it is not a vector.

Other Vectors


Complex Numbers. The set C of complex numbers is a

real vector space or equivalently, a vector space over R.

2-D Coordinate Space. Another real vector space is the set of all pairs of 𝑥𝑖 ∈ R forms a 2-dimensional vector space overR is the two dimensional coordinate space you have been graphing on! It satisfies each of the requirements:

Other Vectors


𝒙 = {𝑥1, 𝑥2}, 𝑥1, 𝑥2 ∈ R, 𝒚 = {𝑦1, 𝑦2}, 𝑦1, 𝑦2 ∈ R. Addition is easily defined as 𝒙 + 𝒚 = {𝑥1 + 𝑦1, 𝑥2 +𝑦2} clearly 𝒙 + 𝒚 ∈ R 2 since 𝑥1 + 𝑦1, 𝑥2 + 𝑦2 ∈ R. Addition operation creates other members for the vector space – Hence closure exists for the operation.

Multiplication by a scalar: 𝛼𝒙 = {𝛼𝑥1, 𝛼𝑥2}, ∀𝛼 ∈ R. Zero element: 𝒐 = 0,0 . Additive Inverse: −𝒙 =

−𝑥1, −𝑥2 , 𝑥1, 𝑥2 ∈ RType equation here.A standard basis for this : 𝒆1 =1,0 , 𝒆2 = 0,1Type equation here.Any other member can be expressed in terms of this basis.


Set of Pairs

𝒏 −D Coordinate Space. For any positive number 𝑛, we may create 𝑛 −tuples such that, 𝒙 = {𝑥1, 𝑥2, … , 𝑥𝑛}where 𝑥1, 𝑥2, … , 𝑥𝑛 ∈ R are members of R 𝑛 – a real vector space over theR. 𝒚 = {𝑦1, 𝑦2, … , 𝑦𝑛}, 𝑦1, 𝑦2, … , 𝑦𝑛 ∈ R. 𝒙 + 𝒚 = {𝑥1 + 𝑦1, 𝑥2 + 𝑦2, … , 𝑥𝑛 +𝑦𝑛},. 𝒙 + 𝒚 ∈ R 𝑛since 𝑥1 + 𝑦1, 𝑥2 + 𝑦2, … , 𝑥𝑛 + 𝑦𝑛 ∈ R

N-tuples


Addition operation creates other members for the vector space – Hence closure exists for the operation.

Multiplication by a scalar: 𝛼𝒙 = {𝛼𝑥1, 𝛼𝑥2, … , 𝛼𝑥𝑛}, ∀𝛼 ∈ 𝑅.

Zero element 𝒐 = 0,0,… , 0 . Additive Inverse −𝒙 =−𝑥1, −𝑥2, … , −𝑥𝑛 , 𝑥1, 𝑥2, … , 𝑥𝑛 ∈ R

There is also a standard basis which is easily proved to be linearly independent: 𝒆1 = 1,0,… , 0 , 𝒆2 =0,1,… , 0 , … , 𝒆𝑛 = {0,0,… , 0}


N-tuples

Let R𝑚×𝑛 denote the set of matrices with entries that are

real numbers (same thing as saying members of the real

spaceR, Then, R𝑚×𝑛 is a real vector space. Vector

addition is just matrix addition and scalar multiplication is

defined in the obvious way (by multiplying each entry by

the same scalar). The zero vector here is just the zero

matrix. The dimension of this space is 𝑚𝑛. For example, in

R3×3 we can choose basis in the form,

1 0 00 0 00 0 0

,0 1 00 0 00 0 0

, …,0 0 00 0 00 0 1

Matrices


Forming polynomials with a single variable 𝑥 to order 𝑛

when 𝑛 is a real number creates a vector space. It is left as

an exercise to demonstrate that this satisfies all the three

rules of what a vector space is.

The Polynomials


Is a 2D area embedded in a Euclidean Point Space a vector? Does it satisfy the conditions? Discuss.


Area Vector

An Inner-Product (also called a Euclidean Vector) SpaceE is a real vector space that defines the scalar product: for each pair 𝒖, 𝐯 ∈ E, ∃ 𝑙 ∈ R such that, 𝑙 = 𝒖 ⋅ 𝐯 = 𝐯 ⋅ 𝒖 . Further, 𝒖 ⋅ 𝒖 ≥ 0, the zero value occurring only when 𝒖 = 0. It is called “Euclidean” because the laws of Euclidean geometry hold in such a space.

The inner product also called a dot product, is the mapping

" ⋅ ":V × V → Rfrom the product space to the real space.

Euclidean Vector Space


A mapping from a vector space is also called a functional; a term that is more appropriate when we are looking at a function space.

A linear functional 𝐯 ∗:V → R on the vector spaceVis called a covector or a dual vector. For a finite dimensional vector space, the set of all covectorsforms the dual spaceV ∗ ofV . If V is an Inner Product Space, then there is no distinction between the vector space and its dual.

Co-vectors


Magnitude The norm, length or magnitude of 𝒖,denoted 𝒖 is defined as the positive square root of 𝒖 ⋅ 𝒖 = 𝒖 2. When 𝒖 = 1, 𝒖 is said to be a unit vector. When 𝒖 ⋅ 𝐯 = 0, 𝒖 and 𝐯 are said to be orthogonal.

Direction Furthermore, for any two vectors 𝒖 and 𝐯, the angle between them is defined as,

cos−1𝒖 ⋅ 𝐯

𝒖 𝐯

The scalar distance 𝑑 between two vectors 𝒖 and 𝐯𝑑 = 𝒖 − 𝐯

Magnitude & Direction Again


A 3-D Euclidean space is a Normed space because the inner product induces a norm on every member.

It is also a metric space because we can find distances and angles and therefore measure areas and volumes

Furthermore, in this space, we can define the cross product, a mapping from the product space

" × ":V × V → VWhich takes two vectors and produces a vector.

3-D Euclidean Space


Without any further ado, our definition of cross product is exactly the same as what you already know from elementary texts. We simply repeat a few of these for emphasis:

1. The magnitude 𝒂 × 𝒃 = 𝒂 𝒃 sin 𝜃 (0 ≤ 𝜃 ≤ 𝜋)

of the cross product 𝒂 × 𝒃 is the area 𝐴(𝒂, 𝒃)spanned by the vectors 𝒂 and 𝒃. This is the area of the parallelogram defined by these vectors. This area is non-zero only when the two vectors are linearly independent.

2. 𝜃 is the angle between the two vectors.

3. The direction of 𝒂 × 𝒃 is orthogonal to both 𝒂 and 𝒃

Cross Product


Cross Product

The area 𝐴(𝒂, 𝒃) spanned by the vectors 𝒂 and 𝒃. The unit vector 𝒏 in the direction of the cross product can be

obtained from the quotient, 𝒂×𝒃

𝒂×𝒃.


The cross product is bilinear and anti-commutative:

Given 𝛼 ∈ R , ∀𝒂, 𝒃, 𝒄 ∈ V ,𝛼𝒂 + 𝒃 × 𝒄 = 𝛼 𝒂 × 𝒄 + 𝒃 × 𝒄𝒂 × 𝛼𝒃 + 𝒄 = 𝛼 𝒂 × 𝒃 + 𝒂 × 𝒄

So that there is linearity in both arguments.

Furthermore, ∀𝒂, 𝒃 ∈ V𝒂 × 𝒃 = −𝒃 × 𝒂

Cross Product


The trilinear mapping,[ , ,] ∶ V × V × V → R

From the product set V × V × V to real space is defined by:[ u,v,w] ≡ 𝒖 ⋅ 𝒗 × 𝒘 = 𝒖 × 𝒗 ⋅ 𝒘

Has the following properties:

1. [ a,b,c]=[b,c,a]=[c,a,b]=−[b,a,c]=−[c,b,a]=−[a,c,b]

HW: Prove this

1. Vanishes when a, b and c are linearly dependent.

2. It is the volume of the parallelepiped defined by a, b and c

Tripple Products


Tripple product

Parallelepiped defined by u, v and w


We introduce an index notation to facilitate the expression of relationships in indexed objects. Whereas the components of a vector may be three different functions, indexing helps us to have a compact representation instead of using new symbols for each function, we simply index and achieve compactness in notation. As we deal with higher ranked objects, such notational conveniences become even more important. We shall often deal with coordinate transformations.

Summation Convention


When an index occurs twice on the same side of any equation, or term within an equation, it is understood to represent a summation on these repeated indices the summation being over the integer values specified by the range. A repeated index is called a summation index, while an unrepeated index is called a free index. The summation convention requires that one must never allow a summation index to appear more than twice in any given expression.



Consider transformation equations such as,𝑦1 = 𝑎11𝑥1 + 𝑎12𝑥2 + 𝑎13𝑥3𝑦2 = 𝑎21𝑥1 + 𝑎22𝑥2 + 𝑎23𝑥3𝑦3 = 𝑎31𝑥1 + 𝑎32𝑥2 + 𝑎33𝑥3

We may write these equations using the summation symbols as:

𝑦1 =

𝑗=1

𝑛

𝑎1𝑗𝑥𝑗

𝑦2 =

𝑗=1

𝑛

𝑎2𝑗𝑥𝑗

𝑦3 =

𝑗=1

𝑛

𝑎3𝑗𝑥𝑗



In each of these, we can invoke the Einstein summation convention, and write that,

𝑦1 = 𝑎1𝑗𝑥𝑗𝑦2 = 𝑎2𝑗𝑥𝑗𝑦3 = 𝑎3𝑗𝑥𝑗

Finally, we observe that 𝑦1, 𝑦2, and 𝑦3 can be represented as we have been doing by 𝑦𝑖 , 𝑖 = 1,2,3so that the three equations can be written more compactly as,

𝑦𝑖 = 𝑎𝑖𝑗𝑥𝑗 , 𝑖 = 1,2,3



Please note here that while 𝑗 in each equation is a dummy index, 𝑖 is not dummy as it occurs once on the left and in each expression on the right. We therefore cannot arbitrarily alter it on one side without matching that action on the other side. To do so will alter the equation. Again, if we are clear on the range of 𝑖, we may leave it out completely and write,

𝑦𝑖 = 𝑎𝑖𝑗𝑥𝑗to represent compactly, the transformation equations above. It should be obvious there are as many equations as there are free indices.



If 𝑎𝑖𝑗 represents the components of a 3 × 3 matrix 𝐀, we can show that,

𝑎𝑖𝑗𝑎𝑗𝑘 = 𝑏𝑖𝑘Where 𝐁 is the product matrix 𝐀𝐀. To show this, apply summation convention and see that,for 𝑖 = 1, 𝑘 = 1, 𝑎11𝑎11 + 𝑎12𝑎21 + 𝑎13𝑎31 = 𝑏11for 𝑖 = 1, 𝑘 = 2, 𝑎11𝑎12 + 𝑎12𝑎22 + 𝑎13𝑎32 = 𝑏12for 𝑖 = 1, 𝑘 = 3, 𝑎11𝑎13 + 𝑎12𝑎23 + 𝑎13𝑎33 = 𝑏13for 𝑖 = 2, 𝑘 = 1, 𝑎21𝑎11 + 𝑎22𝑎21 + 𝑎23𝑎31 = 𝑏21for 𝑖 = 2, 𝑘 = 2, 𝑎21𝑎12 + 𝑎22𝑎22 + 𝑎23𝑎32 = 𝑏22for 𝑖 = 2, 𝑘 = 3, 𝑎21𝑎13 + 𝑎22𝑎23 + 𝑎23𝑎33 = 𝑏23for 𝑖 = 3, 𝑘 = 1, 𝑎31𝑎11 + 𝑎32𝑎21 + 𝑎33𝑎31 = 𝑏31for 𝑖 = 3, 𝑘 = 2, 𝑎31𝑎12 + 𝑎32𝑎22 + 𝑎33𝑎32 = 𝑏32for 𝑖 = 3, 𝑘 = 3, 𝑎31𝑎13 + 𝑎32𝑎23 + 𝑎33𝑎33 = 𝑏33



The above can easily be verified in matrix notation as,

𝐀𝐀 =

𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

=

𝑏11 𝑏12 𝑏13𝑏21 𝑏22 𝑏23𝑏31 𝑏32 𝑏33

= 𝐁

In this same way, we could have also proved that,𝑎𝑖𝑗𝑎𝑘𝑗 = 𝑏𝑖𝑘

Where 𝐁 is the product matrix 𝐀𝐀T. Note the arrangements could sometimes be counter intuitive.



Consider an orthogonal normal basis 𝐞1, 𝐞2, 𝐞3, 𝑜𝑟 𝐞𝑖, 𝑖 = 1,2,3

The fact that 𝐞1 ⋅ 𝐞2 = 𝐞2 ⋅ 𝐞3 = 𝐞3 ⋅ 𝐞1 = 0, and also that the norm on each, that is, 𝐞1 ⋅ 𝐞1 = 𝐞2 ⋅ 𝐞2 = 𝐞3 ⋅𝐞3 = 1 make the computation of the components on any vector referred to this bases easy to compute.


Vector Components in ONB

Given a known vector 𝐅, (known because its magnitude and direction are given) we can find its components in the ONB in these simple steps:

𝐅 = 𝐹1𝐞1 + 𝐹2𝐞2 + 𝐹3𝐞3Where 𝐹1, 𝐹2 𝑎𝑛𝑑 𝐹3, the scalar quantities to be determined are called the components of 𝐅 in the ONB 𝐞𝑖, 𝑖 = 1,2,3

Take the scalar product of the above equation with 𝐞1, we have,



𝐅 ⋅ 𝐞1 = 𝐹1𝐞1 ⋅ 𝐞1 + 𝐹2𝐞2 ⋅ 𝐞1 + 𝐹3𝐞3 ⋅ 𝐞1= 𝐹1𝐞1 ⋅ 𝐞1 + 𝐹2𝒆2 ⋅ 𝒆1 + 𝐹3𝐞3 ⋅ 𝐞1

The boxed items vanish on account of orthogonality while the first term simply becomes 𝐹1 on account of normality. We therefore find that 𝐹1 = 𝐅 ⋅ 𝐞1. We can repeat this process by taking the dot product with the other basis vectors and find also that, 𝐹2 = 𝐅 ⋅ 𝐞2 and that 𝐹3 = 𝐅 ⋅ 𝐞3.

In index notation, we could write the nine orthonormalityequations as,

𝐞𝑖 ⋅ 𝐞𝑗 = 𝛿𝑖𝑗



Furthermore, the result of calculating the components, that𝐅 = 𝐹1𝐞1 + 𝐹2𝐞2 + 𝐹3𝐞3= 𝐅 ⋅ 𝐞1 𝐞1 + 𝐅 ⋅ 𝐞2 𝐞2 + 𝐅 ⋅ 𝐞3 𝐞3

Can also be written as,

𝐅 = 𝐹𝑖𝐞𝑖 = 𝐅 ⋅ 𝐞𝑗 𝐞𝑗Where the repetition of the indices indicate summation.In the above computation, everything was easy because the orthogonality and normality conditions made some terms vanish and the other terms were products with unity. It is this ease of computation that makes the ONB very attractive.



When we relax the conditions on our basis vectors, these nice properties disappear. Yet, there is an extension of this situation, which, if understood deeply, will make things relatively easy also.



Suppose our basis vectors 𝐠𝑖 , 𝑖 = 1,2,3 are not only not unit in magnitude, but in addition are NOT orthogonal. The only assumption we are making is that 𝐠𝑖 ∈ V , 𝑖 =1,2,3 are linearly independent vectors.

With respect to this basis, we can express vectors 𝐯,𝐰 ∈ V in terms of the basis as,

𝐯 = 𝑣𝑖 𝐠𝑖, 𝐰 = 𝑤𝑖 𝐠𝑖Where each 𝑣𝑖 is called the contravariant component of 𝐯

General Vector Components


Clearly, addition and linearity of the vector space ⇒𝐯 +𝐰 = (𝑣𝑖+𝑤𝑖)𝐠𝑖

Multiplication by scalar rule implies that if 𝛼 ∈ R , ∀𝐯 ∈V ,

𝛼 𝐯 = (𝛼𝑣𝑖)𝐠𝑖

Vector Components


For any basis vectors 𝐠𝑖 ∈ V , 𝑖 = 1,2,3 there is a dual (or reciprocal) basis defined by the reciprocity relationship:

𝐠𝑖 ⋅ 𝐠𝑗 = 𝐠𝑗 ⋅ 𝐠𝑖 = 𝛿𝑗

𝑖

Where 𝛿𝑗𝑖 is the Kronecker delta

𝛿𝑗𝑖 =

0 if 𝑖 ≠ 𝑗1 if 𝑖 = 𝑗

Let the fact that the above equations are actually nine equations each sink. Consider the full meaning:

Reciprocal Basis


Kronecker Delta: 𝛿𝑖𝑗, 𝛿𝑖𝑗 or 𝛿𝑗𝑖 has the following properties:

𝛿11 = 1, 𝛿12 = 0, 𝛿13 = 0𝛿21 = 0, 𝛿22 = 1, 𝛿23 = 0𝛿31 = 0, 𝛿32 = 0, 𝛿33 = 1

As is obvious, these are obtained by allowing the indices toattain all possible values in the range. The Kronecker delta isdefined by the fact that when the indices explicit values areequal, it has the value of unity. Otherwise, it is zero. Theabove nine equations can be written more compactly as,

𝛿𝑗𝑖 =

0 if 𝑖 ≠ 𝑗1 if 𝑖 = 𝑗

Kronecker Delta


For any ∀𝐯 ∈ V ,𝐯 = 𝑣𝑖𝐠𝑖 = 𝑣𝑖𝐠

𝑖

Are two related representations in the reciprocal bases. Taking the inner product of the above equation with the basis vector 𝐠𝑗, we have

𝐯 ⋅ 𝐠𝑗 = 𝑣𝑖𝐠𝑖 ⋅ 𝐠𝑗 = 𝑣𝑖𝐠𝑖 ⋅ 𝐠𝑗

Which gives us the covariant component,

𝐯 ⋅ 𝐠𝑗 = 𝑣𝑖𝑔𝑖𝑗 = 𝑣𝑖𝛿𝑗𝑖 = 𝑣𝑗

The last equality earns the Kronecker delta the epithet of “Substitution symbol”. Work it out

Covariant components


In the same easy manner, we may evaluate the contravariant components of the same vector by taking the dot product of the same equation with the contravariant base vector 𝐠𝑗:

𝐯 ⋅ 𝐠𝑗 = 𝑣𝑖𝐠𝑖 ⋅ 𝐠𝑗 = 𝑣𝑖𝐠

𝑖 ⋅ 𝐠𝑗

So that,

𝐯 ⋅ 𝐠𝑗 = 𝑣𝑖𝛿𝑖𝑗= 𝑣𝑖𝑔

𝑖𝑗 = 𝑣𝑗

Contravariant Components


The nine scalar quantities, 𝑔𝑖𝑗 as well as the nine related quantities 𝑔𝑖𝑗 play important roles in the coordinate

system spanned by these arbitrary reciprocal set of basis vectors as we shall see.

They are called metric coefficients because they metrizethe space defined by these bases.

Metric coefficients


The Levi-Civita Symbol: 𝑒𝑖𝑗𝑘

𝑒111 = 0, 𝑒112 = 0, 𝑒113 = 0, 𝑒121 = 0, 𝑒122 = 0, 𝑒123 =1, 𝑒131 = 0, 𝑒132 = −1, 𝑒133 = 0𝑒211 = 0, 𝑒212 = 0, 𝑒213 = −1, 𝑒221 = 0, 𝑒222 = 0, 𝑒223= 0, 𝑒231 = 1, 𝑒232 = 0, 𝑒233 = 0𝑒311 = 0, 𝑒312 = 1, 𝑒313 = 0, 𝑒321 = −1, 𝑒322 = 0, 𝑒323= 0, 𝑒331 = 0, 𝑒332 = 0, 𝑒333 = 0

Levi Civita Symbol


While the above equations might look arbitrary at first, a closer look shows there is a simple logic to it all. In fact, note that whenever the value of an index is repeated, the symbol has a value of zero. Furthermore, we can see that once the indices are an even arrangement (permutation) of 1,2, and 3, the symbols have the value of 1, When we have an odd arrangement, the value is -1. Again, we desire to avoid writing twenty seven equations to express this simple fact. Hence we use the index notation to define the Levi-Civita symbol as follows:

𝑒𝑖𝑗𝑘 =

1 if 𝑖, 𝑗 and 𝑘 are an even permutation of 1,2 and 3−1 if 𝑖, 𝑗 and 𝑘 are an odd permutation of 1,2 and 30 In all other cases

Levi Civita Symbol


Given that 𝑔 = det 𝑔𝑖𝑗 of the covariant metric coefficients, It is not difficult to prove that

𝐠𝑖 ∙ 𝐠𝑗 × 𝐠𝑘 = 𝜖𝑖𝑗𝑘 ≡ 𝑔𝑒𝑖𝑗𝑘 This relationship immediately implies that,

𝐠𝑖 × 𝐠𝑗 = 𝜖𝑖𝑗𝑘𝐠𝑘 .

The dual of the expression, the equivalent contravariant equivalent also follows from the fact that,

𝐠1 × 𝐠2 ⋅ 𝐠3 = 1/ 𝑔

Cross Product of Basis Vectors


This leads in a similar way to the expression,

𝐠𝑖 × 𝐠𝑗 ⋅ 𝐠𝑘 =𝑒𝑖𝑗𝑘

𝑔= 𝜖𝑖𝑗𝑘

It follows immediately from this that,

𝐠𝑖 × 𝐠𝑗 = 𝜖𝑖𝑗𝑘𝐠𝑘

Cross Product of Basis Vectors


Given that, 𝐠𝟏, 𝐠2 and 𝐠3 are three linearly

independent vectors and satisfy 𝐠𝑖 ⋅ 𝐠𝑗 = 𝛿𝑗𝑖, show

that 𝐠1 =1

𝑉𝐠2 × 𝐠3, 𝐠

2 =𝟏

𝑽𝐠3 × 𝐠1, and 𝐠3 =

1

𝑉𝐠1 ×

𝐠2, where 𝑉 = 𝐠1 ⋅ 𝐠2 × 𝐠3

Exercises


It is clear, for example, that 𝐠1 is perpendicular to 𝐠2 as well as to 𝐠3 (an obvious fact because 𝐠1 ⋅ 𝐠2 = 0 and 𝐠1 ⋅ 𝐠3 = 0), we can say that the vector 𝐠1 must necessarily lie on the cross product 𝐠2 × 𝐠3 of 𝐠2 and 𝐠3. It is therefore correct to write,

𝐠1 =1

𝑉𝐠2 × 𝐠3

Where 𝑉−1is a constant we will now determine. We can do this right away by taking the dot product of both sides of the equation (5) with 𝐠1 we immediately obtain,

𝐠1 ⋅ 𝐠1 = 𝑉−1 𝐠1 ⋅ 𝐠2 × 𝐠3= 1

So that, 𝑉 = 𝐠1 ⋅ 𝐠2 × 𝐠3the volume of the parallelepiped formed by the three vectors 𝐠1, 𝐠2, and 𝐠3 when their origins are made to coincide.


We generalize the result now in terms of natural bases that arise in coordinate transformations from the Cartesian:

In the curvilinear system (𝑢1, 𝑢2, 𝑢3) obtained from the transformation 𝑥𝑖 = 𝑥𝑖(𝑢1, 𝑢2, 𝑢3) from Cartesian

coordinates, let 𝐠𝑖 =𝜕𝐫

𝜕𝑢𝑖and let 𝐠𝑗 be the corresponding dual

basis. Show that 𝑔𝑖𝑗 = 𝐠𝑖 ∙ 𝐠𝑗 =𝜕𝑥𝑘

𝜕𝑢𝑖𝜕𝑥𝑘

𝜕𝑢𝑗. If 𝑉 = 𝐠1 ∙ 𝐠2 × 𝐠3

and 𝑣 = 𝐠1 ⋅ 𝐠2 × 𝐠3, show that 𝑣 𝑉 = 1. Show also that 𝐠𝑖 ∙𝐠𝑗 × 𝐠𝑘 = 𝜖𝑖𝑗𝑘 = 𝑔𝑒𝑖𝑗𝑘.

The position vector 𝐫 𝑥, 𝑦, 𝑧 = 𝑥𝑖 𝑢1, 𝑢2, 𝑢3 𝐞𝑖 where 𝐞𝑖 , 𝑖 =1,2,3 are unit vectors that are orthonormal in the Euclidean space.


Natural Bases

Changing variables, we can write that,

𝐫 𝑥, 𝑦, 𝑧 = 𝑥𝑖(𝑢1, 𝑢2, 𝑢3)𝐞𝑖 = 𝐫 𝑢1, 𝑢2, 𝑢3

So that we have new coordinates 𝑢𝑘 , 𝑘 = 1,2,3. In this new system, the differential of the position vector 𝐫 is,

𝑑𝐫 =𝜕𝐫

𝜕𝑢𝑖𝑑𝑢𝑖 ≡ 𝐠𝑖𝑑𝑢

𝑖

the above equation defining the basis vectors in the new coordinate system. The vectors 𝐠1, 𝐠2 and 𝐠3 are not necessarily unit vectors but they form a basis of the new system provided,

𝑉 = 𝐠1 ∙ 𝐠2 × 𝐠3 ≠ 0


Natural Bases

Clearly, the reciprocal basis vectors are 𝐠1 = 𝑉−1 𝐠2 × 𝐠3𝐠2 = 𝑉−1 𝐠3 × 𝐠1𝐠3 = 𝑉−1 𝐠1 × 𝐠2

(dot the first with 𝐠1 to see) Now we are given that 𝑣 = 𝐠1 ⋅𝐠2 × 𝐠3. Using the above relations, we can write,

𝐠2 × 𝐠3 = 𝑉−1 𝐠3 × 𝐠1 × 𝑉−1 𝐠1 × 𝐠2= 𝑉−2 𝐠3 × 𝐠1⋅ 𝐠2 𝐠1 − 𝐠3 × 𝐠1⋅ 𝐠1= 𝑉−2 𝐠1 × 𝐠2⋅ 𝐠3 𝐠1 = 𝑉−1𝐠1

We can now write,𝑣 = 𝐠1 ⋅ 𝐠2 × 𝐠3 = 𝐠1 ⋅ 𝑉−1𝐠1 = 𝑉−1𝐠1 ⋅ 𝐠1 = 𝑉−1

Showing that, 𝑣 𝑉 = 1 as required.


Dual Bases

If the Jacobian of the transformation 𝜕𝑥𝑘

𝜕𝑢𝑖does not vanish,

the 𝐠𝑖 are independent.

Now,

𝐠𝑖 =𝜕𝐫

𝜕𝑢𝑖=𝜕𝑥𝑘

𝜕𝑢𝑖𝐞𝑘

𝐠1 ∙ 𝐠2 × 𝐠3 =

𝜕𝑥1

𝜕𝑢1𝜕𝑥2

𝜕𝑢1𝜕𝑥3

𝜕𝑢1

𝜕𝑥1

𝜕𝑢2𝜕𝑥2

𝜕𝑢2𝜕𝑥3

𝜕𝑢2

𝜕𝑥1

𝜕𝑢3𝜕𝑥2

𝜕𝑢3𝜕𝑥3

𝜕𝑢3

=𝜕𝑥𝑘

𝜕𝑢𝑖≠ 0.

𝑔𝑖𝑗 = 𝐠𝑖 ∙ 𝐠𝑗 =𝜕𝐫

𝜕𝑢𝑖⋅𝜕𝐫

𝜕𝑢𝑗=

𝜕𝑥𝑘

𝜕𝑢𝑖𝐞𝑘 ∙

𝜕𝑥𝑙

𝜕𝑢𝑗𝐞𝑙

=𝜕𝑥𝑘

𝜕𝑢𝑖𝜕𝑥𝑙

𝜕𝑢𝑗𝐞𝑘 ⋅ 𝐞𝑙 =

𝜕𝑥𝑘

𝜕𝑢𝑖𝜕𝑥𝑙

𝜕𝑢𝑗𝛿𝑘𝑙 =

𝜕𝑥𝑘


𝜕𝑢𝑗


Clearly, the determinant of 𝑔𝑖𝑗

𝑔 ≡ 𝑔𝑖𝑗 =𝜕𝑥𝑘


𝜕𝑢𝑗=

𝜕𝑥𝑘

𝜕𝑢𝑖

2

= 𝑉2

This means, 𝑉 = 𝐠1 ∙ 𝐠2 × 𝐠3 =𝜕𝑥𝑖

𝜕𝑢𝑗= 𝑔. We can therefore write,

𝐠1 ∙ 𝐠2 × 𝐠3 = 𝑒123 𝑔

Swapping indices 2 and 3, we have,𝐠1 ∙ 𝐠3 × 𝐠2 = − 𝑔 = 𝑒132 𝑔 = 𝐠1 × 𝐠3 ⋅ 𝐠2

The second equality coming from the fact that swapping the cross with the dot changes nothing. Lastly, swapping 1 and 3 in the last equation shows that,

𝐠3 × 𝐠1 ⋅ 𝐠2 = − − 𝑔 = 𝑒312 𝑔. These three expression together

imply that,

𝐠𝑖 ∙ 𝐠𝑗 × 𝐠𝑘 = 𝜖𝑖𝑗𝑘 = 𝑔𝑒𝑖𝑗𝑘 as required.


This relationship immediately implies that,

𝐠𝑖 × 𝐠𝑗 = 𝜖𝑖𝑗𝑘𝐠𝑘 as a dot product of this with 𝐠𝛼

recovers the previous. The dual of the expression, the equivalent contravariant equivalent also follows from the fact that .

𝐠1 × 𝐠2 ⋅ 𝐠3 = 1/ 𝑔as it must be since we proved that the two volumes must be inverses. This leads in a similar way to the expression,

𝐠𝑖 × 𝐠𝑗 ⋅ 𝐠𝑘 =𝑒𝑖𝑗𝑘

𝑔= 𝜖𝑖𝑗𝑘

It follows immediately from this that,



Show that 𝐠𝑗 = 𝑔𝑖𝑗𝐠𝑖 = 𝑔𝑗𝑖𝐠𝑖 and establish the relation, 𝑔𝑖𝑗𝑔𝑗𝑘 = 𝛿𝑖

𝑘

First expand 𝐠𝑗 in terms of the 𝐠𝑖s:𝐠𝑗 = 𝛼𝐠1 + 𝛽𝐠2 + 𝛾𝐠3

Dotting with 𝐠1 ⇒ 𝐠𝑗 ⋅ 𝐠1 = 𝛼𝐠1 ⋅ 𝐠1 + 𝛽𝐠2 ⋅ 𝐠

1 + 𝛾𝐠3 ⋅ 𝐠1 = 𝑔𝑗1 = 𝛼. In

the same way we find that 𝛽 = 𝑔𝑗2 and 𝛾 = 𝑔𝑗3so that,𝐠𝑗 = 𝑔𝑗1𝐠1 + 𝑔𝑗2𝐠2 + 𝑔𝑗3𝐠3 = 𝑔𝑗𝑖𝐠𝑖.

Similarly, 𝐠𝑖 = 𝑔𝑖𝛼𝐠𝛼.

Recall the reciprocity relationship: 𝐠𝑖 ⋅ 𝐠𝑘 = 𝛿𝑖

𝑘. Using the above, we can write

𝐠𝑖 ⋅ 𝐠𝑘 = 𝑔𝑖𝛼𝐠

𝛼 ⋅ 𝑔𝑘𝛽𝐠𝛽= 𝑔𝑖𝛼𝑔

𝑘𝛽𝐠𝛼 ⋅ 𝐠𝛽= 𝑔𝑖𝛼𝑔

𝑘𝛽𝛿𝛽𝛼 = 𝛿𝑖

𝑘

which shows that

𝑔𝑖𝛼𝑔𝑘𝛼 = 𝑔𝑖𝑗𝑔

𝑗𝑘 = 𝛿𝑖𝑘

As required. This shows that the tensor 𝑔𝑖𝑗 and 𝑔𝑖𝑗 are inverses of each

other.


Given that the scalar product of the base vectors 𝐠1 ⋅ 𝐠2 × 𝐠3 = 𝑔,

Show that 𝐠𝑖 × 𝐠𝑗 = 𝜖𝑖𝑗𝑘𝐠𝑘 and that 𝐠𝑘 =

1

2𝜖𝑖𝑗𝑘𝐠

𝑖 × 𝐠𝑗

Begin with the reciprocity relationship, note that 𝐠3 ⋅𝐠1 = 0 and 𝐠3 ⋅ 𝐠2 = 0. From these relationships it is immediately obvious that

𝐠3 = 𝛼𝐠1 × 𝐠2

where 𝛼 is a yet to be determined constant. Taking the dot product of the above equation with 𝐠3, again, from reciprocity, we see that,

1 = 𝐠3 ⋅ 𝐠3 = 𝛼𝐠1 × 𝐠2 ⋅ 𝐠3= 𝛼𝐠1 ⋅ 𝐠2 × 𝐠3 = 𝛼 𝑔


So that, 𝛼 =1

𝑔. It is therefore clear that,

𝐠1 × 𝐠2 = 𝑔𝐠3 = 𝑔𝑒123𝐠3 = 𝜖123𝐠

3

This representation is called for, and efficient, because we

may also write conversely that,

𝐠2 × 𝐠1 = − 𝑔𝐠3 = 𝑔𝑒213𝐠3 = 𝜖213𝐠

3.

The same arguments leads us to writing that,

𝐠2 × 𝐠3 = 𝜖231𝐠1 and 𝐠3 × 𝐠1 = 𝜖312𝐠

2. All the

relationships above are compactly written as,



Furthermore, we may contract the contravariant alternating tensor on both sides and obtain,

𝜖𝑖𝑗𝛼𝐠𝑖 × 𝐠𝑗 = 𝜖𝑖𝑗𝛼𝜖𝑖𝑗𝑘𝐠𝑘 = 2𝛿𝑘

𝛼𝐠𝑘 = 2𝐠𝛼

So that,

2𝐠𝛼 = 𝜖𝑖𝑗𝛼𝐠𝑖 × 𝐠𝑗

Or,

𝐠𝑘 =1

2𝜖𝑖𝑗𝑘𝐠𝑖 × 𝐠𝑗

A similar argument is easily constructed to prove that,

𝐠𝑘 =1

2𝜖𝑖𝑗𝑘𝐠

𝑖 × 𝐠𝑗 .


Show that the cross product of vectors 𝒂 and 𝒃 in general coordinates is

𝑎𝑖𝑏𝑗𝜖𝑖𝑗𝑘𝐠𝑘 or 𝜖𝑖𝑗𝑘𝑎𝑖𝑏𝑗𝐠𝑘 where 𝑎𝑖 , 𝑏𝑗 are the respective contravariant

components and 𝑎𝑖 , 𝑏𝑗 the covariant.

Express vectors 𝒂 and 𝒃 as contravariant components: 𝒂 = 𝑎𝑖𝐠𝑖, and 𝒃 =𝑏𝑖𝐠𝑖. Using the above result, we can write that,

𝒂 × 𝒃 = 𝑎𝑖𝐠𝑖 × 𝑏𝑗𝐠𝑗= 𝑎𝑖𝑏𝑗𝐠𝑖 × 𝐠𝑗 = 𝑎𝑖𝑏𝑗𝜖𝑖𝑗𝑘𝐠

𝑘 .

Express vectors 𝒂 and 𝒃 as covariant components: 𝒂 = 𝑎𝑖𝐠𝑖 and 𝒃 = 𝑏𝑖𝐠

𝑖. Again, proceeding as before, we can write,

𝒂 × 𝒃 = 𝑎𝑖𝐠𝑖 × 𝑏𝑗𝐠

𝑗

= 𝜖𝑖𝑗𝑘𝑎𝑖𝑏𝑗𝐠𝑘Express vectors 𝒂 as a contravariant components: 𝒂 = 𝑎𝑖𝐠𝑖 and 𝒃 as covariant components: 𝒃 = 𝑏𝑖𝐠

𝑖

𝒂 × 𝒃 = 𝑎𝑖𝐠𝑖 × 𝑏𝑗𝐠𝑗

= 𝑎𝑖𝑏𝑗 𝐠𝑖 × 𝐠𝑗

= 𝒂 ⋅ 𝒃 𝐠𝑖 × 𝐠𝑗


Given that,

𝛿𝑖𝑗𝑘𝑟𝑠𝑡 ≡ 𝑒𝑟𝑠𝑡𝑒𝑖𝑗𝑘 =

𝛿𝑖𝑟 𝛿𝑗

𝑟 𝛿𝑘𝑟

𝛿𝑖𝑠 𝛿𝑗

𝑠 𝛿𝑘𝑠

𝛿𝑖𝑡 𝛿𝑗

𝑡 𝛿𝑘𝑡

Show that 𝛿𝑖𝑗𝑘𝑟𝑠𝑘 = 𝛿𝑖

𝑟𝛿𝑗𝑠 − 𝛿𝑖

𝑠𝛿𝑗𝑟

Expanding the equation, we have:

𝑒𝑖𝑗𝑘𝑒𝑟𝑠𝑘 = 𝛿𝑖𝑗𝑘

𝑟𝑠𝑘= 𝛿𝑖𝑘𝛿𝑗𝑟 𝛿𝑘

𝑟

𝛿𝑗𝑠 𝛿𝑘

𝑠 − 𝛿𝑗𝑘 𝛿𝑖

𝑟 𝛿𝑘𝑟

𝛿𝑖𝑠 𝛿𝑘

𝑠 + 3𝛿𝑖𝑟 𝛿𝑗

𝑟

𝛿𝑖𝑠 𝛿𝑗

𝑠

= 𝛿𝑖𝑘 𝛿𝑗

𝑟𝛿𝑘𝑠 − 𝛿𝑗

𝑠𝛿𝑘𝑟 − 𝛿𝑗

𝑘 𝛿𝑖𝑟𝛿𝑘

𝑠 − 𝛿𝑖𝑠𝛿𝑘

𝑟

+ 3 𝛿𝑖𝑟𝛿𝑗

𝑠 − 𝛿𝑖𝑠𝛿𝑗

𝑟 = 𝛿𝑗𝑟 𝛿𝑖

𝑠 − 𝛿𝑗𝑠𝛿𝑖

𝑟 − 𝛿𝑖𝑟𝛿𝑗

𝑠

+ 𝛿𝑖𝑠𝛿𝑗

𝑟 +3(𝛿𝑖𝑟 𝛿𝑗


𝑟) = −2(𝛿𝑖𝑟 𝛿𝑗


𝑟) + 3(𝛿𝑖𝑟𝛿𝑗

𝑠

− 𝛿𝑖𝑠𝛿𝑗

𝑟)

= 𝛿𝑖𝑟𝛿𝑗


𝑟


Show that 𝛿𝑖𝑗𝑘𝑟𝑗𝑘

= 2𝛿𝑖𝑟

Contracting one more index, we have:

𝑒𝑖𝑗𝑘𝑒𝑟𝑗𝑘 = 𝛿𝑖𝑗𝑘

𝑟𝑗𝑘= 𝛿𝑖

𝑟𝛿𝑗𝑗− 𝛿𝑖

𝑗𝛿𝑗𝑟 = 3𝛿𝑖

𝑟 − 𝛿𝑖𝑟 = 2𝛿𝑖

𝑟

These results are useful in several situations.


Show that 𝐮 ⋅ 𝐮 × 𝐯 = 0

𝐮 × 𝐯 = ϵ𝑖𝑗𝑘𝑢𝑖𝑣𝑗𝐠𝑘𝐮 ⋅ 𝐮 × 𝐯 = 𝑢𝛼𝐠

𝛼 ⋅ ϵ𝑖𝑗𝑘𝑢𝑖𝑣𝑗𝐠𝑘= 𝑢𝛼 ϵ𝑖𝑗𝑘𝑢𝑖𝑣𝑗 𝛿𝑘

𝛼

= 0On account of the symmetry and antisymmetryin 𝑖 and 𝑘.


Show that 𝐮 × 𝐯 = − 𝐯 × 𝐮

𝐮 × 𝐯 = ϵ𝑖𝑗𝑘𝑢𝑖𝑣𝑗𝐠𝑘= −ϵ𝑗𝑖𝑘𝑢𝑖𝑣𝑗𝐠𝑘 = −ϵ𝑖𝑗𝑘𝑣𝑖𝑢𝑗𝐠𝑘= − 𝐯 × 𝐮


Show that 𝐮 × 𝐯 × 𝐰 = 𝐮 ⋅ 𝐰 𝐯 − 𝐮 ⋅ 𝐯 𝐰.

Let 𝐳 = 𝐯 × 𝐰 = ϵ𝑖𝑗𝑘𝑣𝑖𝑤𝑗𝐠𝑘𝐮 × 𝐯 ×𝐰 = 𝐮 × 𝐳 = 𝜖𝛼𝛽𝛾𝑢

𝛼𝑧𝛽𝐠𝛾

= 𝜖𝛼𝛽𝛾 𝑢𝛼𝑧𝛽𝐠𝛾

= ϵ𝑖𝑗𝛽𝜖𝛾𝛼𝛽𝑢𝛼𝑣𝑖𝑤𝑗𝐠

𝛾

= 𝛿𝛾𝑖𝛿𝛼

𝑗− 𝛿𝛼

𝑖 𝛿𝛾𝑗𝑢𝛼𝑣𝑖𝑤𝑗𝐠

𝛾

= 𝑢𝑗𝑣𝛾𝑤𝑗𝐠𝛾 − 𝑢𝑖𝑣𝑖𝑤𝛾𝐠

𝛾

= 𝐮 ⋅ 𝐰 𝐯 − 𝐮 ⋅ 𝐯 𝐰



1. In the transformation from the 𝑥, 𝑦, 𝑧 system to the 𝑟, 𝜙, 𝑍 coordinate system, the position vector changed from 𝐑 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 to 𝐑 = 𝑟𝒆𝑟 𝜙 +𝑍𝒆𝑧. Show by partial differentiation only, that the basis vectors in respective coordinates are 𝒊, 𝒋, 𝒌

and 𝒆𝑟 , 𝒆𝜙, 𝒆𝑧 respectively where 𝒆𝜙 𝜙 =𝜕𝒆𝑟 𝜙

𝜕𝜙.

2. If the position vector in another system with coordinate variables 𝜌, 𝜙, 𝜃 is 𝐑 = 𝜌𝒆𝜌 𝜙, 𝜃 , use

the same method to find the basis vectors in this system also.

Exercises


3. In Problem 1 above, if the transformation from Cartesian to the other system is given explicitly as 𝑥 𝑟, 𝜙, 𝑍 = 𝑟 cos 𝜙, 𝑦 𝑟, 𝜙, 𝑍 = 𝑟 sin𝜙 and 𝑧 𝑟, 𝜙, 𝑍 = 𝑍, find explicit expression for the basis vectors 𝐠𝑖 , 𝑖 = 1,2,3. Also find the reciprocal basis

vectors 𝐠𝑗 , 𝑗 = 1,2,3. [Hint: 2𝐠𝑖 = 𝜖𝑖𝑗𝑘𝐠𝑗 × 𝐠𝑘]

4. Are these basis vectors orthogonal? Are they normalized?

5. Find the dual bases for the Cartesian system.

6. Find the reciprocal bases for the spherical coordinate systems. Are they orthogonal? Are they normalized?


7. Find the metric tensor for each of the above systems.

8. Find the determinant of the metric tensor and confirm in these cases that 𝐠𝑖 ∙ 𝐠𝑗 × 𝐠𝑘 = 𝜖𝑖𝑗𝑘 =

𝑔𝑒𝑖𝑗𝑘 and that 𝐠𝑖 × 𝐠𝑗 ⋅ 𝐠𝑘 =𝑒𝑖𝑗𝑘

𝑔= 𝜖𝑖𝑗𝑘.

9. Beginning with the equation, 𝐠𝑖 × 𝐠𝑗 = 𝜖𝑖𝑗𝑘𝐠𝑘, take

a contraction with 𝜖𝑖𝑗𝛼 and find the expression for 𝐠𝑘


10. Elliptical Cylindrical Coordinates is defined by the position vector,

𝐑 𝑥, 𝑦, 𝑧 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌= 𝐑 𝜉, 𝜂, 𝑤= 𝒊 cos 𝜂 cosh 𝜉 + 𝒋 sin 𝜂 sinh 𝜉 + 𝒌𝑤

Use Mathematica and show that this system of coordinates is orthogonal. Hint:

𝑅[ξ_, η_,w_] = R 𝜉, 𝜂, 𝑤≔ 𝑖Cos 𝜂 Cosh 𝜉 + 𝑗Sin 𝜂 Sinh 𝜉 + 𝑘𝑤

= i cos 𝜂 cosh 𝜉 + j sin 𝜂 sinh 𝜉 + 𝑘𝑤

g = 𝐷 𝑅 𝜉, 𝜂, 𝑤 , 𝜉, 𝜂, 𝑤 =𝜕 R 𝜉, 𝜂, 𝑤

𝜕 𝜉, 𝜂, 𝑤= {icos 𝜂 sinh 𝜉−𝑖sin(𝜂)cosh(𝜉), 𝑘}

KroneckerProduct[g,g]//MatrixForm


11. Compare the results of

KroneckerProduct[g,g]//MatrixForm

In Q10 to𝐷[{Cos[𝜂]Cosh[𝜉], Sin[𝜂]Sinh[𝜉], 𝑘𝑤}, {{𝜉, 𝜂, 𝑤}}]Transpose[%].%

Explain what the two commands are doing differently.

12. Repeat Q10, 11 for Spherical coordinates

13. Plot Coordinate surfaces for Elliptical Cylindrical coordinates using Mathematica.



Coordinate Surfaces

Documents

SSG 805 Mechanics of Continua - OA Fakoafak.com/wp-content/uploads/2015/03/1.-Vectors-and... · 2015-03-23 · SSG 805 Mechanics of Continua INSTRUCTOR: Fakinlede OA ... Tadmore,