SSG 805 Mechanics of Continua -...

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] 02/17/2016Department 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 to help you prepare for the test every other week.

Assignments are not graded

What you will need


[email protected] 02/17/20164

Scope of Instructional Material

Course Schedule:

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

Slide Title Slides Weeks TextRead

Pages

Vectors and Linear Spaces 105 2

GurtinHeinbo

1-81-60

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

Tests and exams are compulsory. Wrting and pleading after missing your opportunities to do well is a waste of your time and stationery. Ordinarily, the following will hold:


Examination

Evaluation Obtainable

Quiz 10

Tests 50

Midterm 20

Exam 20

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 10 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

Jog, CS, Continuum Mechanics, Foundations & Applications of Mechanics, Vol 1, Cambridge University Press, www.cambridge.org 2015

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 Thermodynamical Theory 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

1. If ∀ 𝐯 ∈V , 𝐚 ⋅ 𝐯 = 𝐛 ⋅ 𝐯, Show that 𝐚 = 𝐛

2. If ∀ 𝐯 ∈V , 𝐚 × 𝐯 = 𝐛 × 𝐯, Show that 𝐚 = 𝐛


Quiz of the day

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


Show that the plane of the board or your sheet of paper is 2-D by proving that once you draw two non-collinear lines, every other vector on that plane can be expressed as a linear combination of these two.

Show that the space of the room is 3-D


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,… , 1}


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


2-D areas embedded in a Euclidean point space are vectors! Show that all the conditions of a vector space are satisfied.


Is Area a 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


From any two vectors 𝒂 and 𝒃 we can form another vector called the cross product 𝐜 ≡ 𝒂 × 𝒃 of 𝒂 and 𝒃.The magnitude 𝐜 ≡ 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.

1. 𝜃 is the angle between the two vectors.

2. 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.



From our knowledge of vectors from secondary school,

given any vector 𝐯 we have always been able to write,

𝐯 = 𝑣𝑥𝐢 + 𝑣𝑦𝐣 + 𝑣𝑧𝐤

where we call the 𝑣𝑥, 𝑣𝑦 , 𝑣𝑧 components of 𝐯 on the

Cartesian bases 𝐢, 𝐣 and 𝐤. There are several issues we

have got accustomed to. Let us re-iterate a few of them.


A Small Notational Change

1. If we change the bases, then the components will change.

2. If we all agree on the bases, we can temporarily forget about them and only talk of the components. In that case we can write the vector as 𝐯 or as 𝑣𝑥 , 𝑣𝑦, 𝑣𝑧 .

3. Given the vector 𝐯 and the Cartesian bases 𝐢, 𝐣 and 𝐤, it is always an easy matter to find the components 𝑣𝑥 , 𝑣𝑦, 𝑣𝑧. In fact, 𝑣𝑥 = 𝐯 ⋅ 𝐢, 𝑣𝑦 = 𝐯 ⋅ 𝐣, 𝑣𝑧 = 𝐯 ⋅ 𝐤.

4. It is therefore correct to write,𝐯 = 𝑣𝑥𝐢 + 𝑣𝑦𝐣 + 𝑣𝑧𝐤 = 𝐯 ⋅ 𝐢 𝐢 + 𝐯 ⋅ 𝐣 𝐣 + 𝐯 ⋅ 𝐤 𝐤

Note that the expressions in the parentheses are all scalars


Old Cartesian System

It is convenient for us, in order to facilitate parsimony and the elegance available from the Einstein summation convention to use the set 𝐞1, 𝐞2 and 𝐞3 in place of 𝐢, 𝐣 and 𝐤for the Cartesian bases vectors, so that the coordinates 𝑥, 𝑦, 𝑧 are understood to be the 1,2,3 coordinates. The previous expressions therefore become,

𝐯 = 𝑣1𝐞1 + 𝑣2𝐞2 + 𝑣3𝐞3 = 𝑣𝑖𝐞𝑖= 𝐯 ⋅ 𝐞1 𝐞1 + 𝐯 ⋅ 𝐞2 𝐞2 + 𝐯 ⋅ 𝐞3 𝐞3= 𝐯 ⋅ 𝐞𝑖 𝐞𝒊

So far, we have not done anything other than simply change the nomenclature from symbolic differentiation to a numeric one. We have not used any other basis than the familiar orthonormal Cartesian system.


Numerically Differentiated Coordinates

We now take a further step. We will recall the fact that we do not need to have as basis, an orthonormal system. The selected basis do not even have to be orthogonal. All, that we really need in a bases set is that the system chosen be free or, equivalently, be linearly independent. We therefore select three vectors, 𝐠1, 𝐠2 and 𝐠3 that are simply linearly independent to represent our vector so that,

𝐯 = 𝑣1𝐠1 + 𝑣𝟐𝐠2 + 𝑣3𝐠3Apart from the fact that we used the superscript rather than the subscript to denote the components in this case, we have not done anything significant other than choosing a set of linearly independent vectors as our bases. There can be a little confusion here when we need to exponentiate. We will take care to say so explicitly unless it is already obvious in the context.


Oblique Bases

First note, again, that the elegance of the Einstein summation convention allows us to write this expression as,

𝐯 = 𝑣1𝐠1 + 𝑣2𝐠2 + 𝑣3𝐠3 = 𝑣𝑖𝐠𝑖and it is precisely this elegance that made us choose to use numbers 1,2,3 instead of using symbols, 𝑥, 𝑦 and 𝑧 to denote our coordinates.

Once the notational changes are clear, it is easy to see that we will now need to find some way to obtain our components in this unfamiliar system when we are no longer able to assume that the basis vectors are unit vectors nor that they are orthogonal. There is an ingenious way to do this such that the pain of tedious calculations of angles and magnitudes is completely avoided! It will be shown shortly:


Working with General Bases

Now that 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

Vector Components


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

(Why is this correct? It comes from the rule of scalar multiplication of vectors!)

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

𝛼 𝐯 = (𝛼𝑣𝑖)𝐠𝑖

Vector Components


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

𝐠𝑖 ⋅ 𝐠𝑗 = 𝐠𝑗 ⋅ 𝐠𝑖 = 𝛿𝑗

𝑖

Where 𝛿𝑗𝑖 is the Kronecker delta

𝛿𝑗𝑖 = ቊ

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 = ቊ

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 =

ቐ1 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, (see over)

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


1. Given that 𝐠1 × 𝐠2 ⋅ 𝐠3 = 𝑔, where 𝑔 is the determinant 𝑔𝑖𝑗 . Show that,

𝐠𝑖 × 𝐠𝑗 ⋅ 𝐠𝑘 = 𝑔𝑒𝑖𝑗𝑘 ≡ 𝜖𝑖𝑗𝑘. Conclude further that 𝐠𝑖 × 𝐠𝑗 = 𝜖𝑖𝑗𝑘𝐠𝑘

Given that 𝐠1 × 𝐠2 ⋅ 𝐠3 = 𝑔, the fact that the triple product obeys the rule,

[ 𝐚, 𝐛, 𝐜] = [𝐛, 𝐜, 𝐚] = [𝐜, 𝐚, 𝐛] = −[𝐛, 𝐚, 𝐜] = −[𝐜, 𝐛, 𝐚] = −[𝐚, 𝐜, 𝐛], combined

with the fact that the triple product vanishes when any two of its vectors are

collinear allow us to write that

𝐠𝑖 × 𝐠𝑗 ⋅ 𝐠𝑘 = 𝑒𝑖𝑗𝑘 𝑔 ≡ 𝜖𝑖𝑗𝑘

By the reciprocity rule, 𝐠𝑖 ⋅ 𝐠𝑗 = 𝛿𝑖

𝑗, we have that, 𝐠1 ⋅ 𝐠

1 = 1, 𝐠1 ⋅ 𝐠2 = 0, 𝐠1 ⋅

𝐠3 = 0. It follows that 𝐠1 must be perpendicular to the plane of 𝐠2and 𝐠3making it parallel to 𝐠2 × 𝐠3 A scalar constant 𝛼 must exist such that, 𝐠1 =

𝛼 𝐠2 × 𝐠3. Dot product of both sides with 𝐠1 shows that 𝛼 = 1/ 𝑔. Therefore,

𝐠2 × 𝐠3= 𝑔𝐠1

𝐠3 × 𝐠1= 𝑔𝐠2

𝐠1 × 𝐠2= 𝑔𝐠3

These three results together can be expressed as, 𝐠𝑖 × 𝐠𝑗 = 𝜖𝑖𝑗𝑘𝐠𝑘 .


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.


Suppose you have a function 𝑓(𝑥, 𝑦, 𝑧) of variables 𝑥, 𝑦 and 𝑧. Let us assume there are some variables 𝑟, 𝜙 and 𝑍 such that, the original variables are themselves functions 𝑥 = 𝑥(𝑟, 𝜙, 𝑍), 𝑦 = 𝑦(𝑟, 𝜙, 𝑍), and 𝑧 = 𝑧(𝑟, 𝜙, 𝑍). A simple example is the polar coordinate transformation: 𝑥 = 𝑟 cos𝜙, 𝑦 = 𝑟 sin𝜙, 𝑧 = 𝑍. We can always get a new function F 𝑟, 𝜙, 𝑍 = 𝑓(𝑥, 𝑦, 𝑧) by doing a coordinate transformation using these equations. It is a well known fact that the transformation equations are invertible provided that the Jacobian of the transformation,

𝜕(𝑥, 𝑦, 𝑧)

𝜕(𝑟, 𝜙, 𝑍)≡

𝜕𝑥

𝜕𝑟

𝜕𝑦

𝜕𝑟

𝜕𝑧

𝜕𝑟𝜕𝑥

𝜕𝜙

𝜕𝑦

𝜕𝜙

𝜕𝑧

𝜕𝜙𝜕𝑥

𝜕𝑍

𝜕𝑦

𝜕𝑍

𝜕𝑧

𝜕𝑍

≠ 0


Jacobian of Transformation

We prefer to use indexed variables. Hence instead of 𝑥, 𝑦 and 𝑧, we prefer 𝑥𝑖 = 𝑥𝑖(𝑢1, 𝑢2, 𝑢3) where 𝑖 = 1,2,3as you can obviously see that instead of 𝑟, 𝜙, 𝑍, we are now talking about 𝑢1, 𝑢2, 𝑢3. As before, we can say that the transformation will have an inverse provided the

Jacobian, 𝜕𝑥𝑘

𝜕𝑢𝑖does not vanish. Therefore to say that

the transformation is invertible ensures that 𝜕𝑥𝑘

𝜕𝑢𝑖≠ 0.

Recall that in Cartesian coordinates, the vector connecting an arbitrary point to the origin, also called a position vector can be written as

𝐫 = 𝑥𝐢 + 𝑦𝐣 + 𝑧𝐤 = 𝑥𝑖𝐞𝑖Or, in order to emphasize the functional dependencies,

𝐫 𝑥, 𝑦, 𝑧 = 𝑥𝑖 𝑢1, 𝑢2, 𝑢3 𝐞𝑖


First notice that once you have a correct expression for your position vector for an arbitrary location, you can, by partial differentiation obtain an alternative representation for your basis vectors. It is elementary, for example to see clearly that,

𝐢 =𝜕𝐫

𝜕𝑥And in general Cartesian coordinates, using index notation,

𝐞𝑖 =𝜕𝐫

𝜕𝑥𝑖, 𝑖 = 1,2,3


Basis Vectors

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, as we shall soon show, defines the natural 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

Let 𝐫 𝑥, 𝑦, 𝑧 = 𝑥𝐢 + 𝑦𝐣 + 𝑧𝐤 = 𝑟 cos 𝜃 𝐢 + 𝑟 sin 𝜃 𝐣 + 𝑧𝐤,𝐫 𝑥, 𝑦, 𝑧 = 𝑥𝑖(𝑟, 𝜃, 𝑧)𝐞𝑖 = 𝐫 𝑟, 𝜃, 𝑧

So that we have new coordinates 𝑟, 𝜃, 𝑧. 𝑥1 𝑟, 𝜃, 𝑧 = 𝑟 cos 𝜃, 𝑥2 𝑟, 𝜃, 𝑧 = 𝑟 sin 𝜃, 𝑥3 𝑟, 𝜃, 𝑧 = 𝑧. In this new system, the differential of the position vector 𝐫 is,

𝑑𝐫 =𝜕𝐫

𝜕𝑟𝑑𝑟 +

𝜕𝐫

𝜕𝜃𝑑𝜃 +

𝜕𝐫

𝜕𝑧𝑑𝑧 ≡ 𝐠𝑖𝑑𝑢

𝑖

= cos 𝜃 𝐢 + sin 𝜃 𝐣 𝑑𝑟 + −𝑟 sin 𝜃𝐢 + 𝑟 cos 𝜃 𝐣 𝑑𝜃 + 𝐤𝑑𝑧

So that 𝐠1 = cos 𝜃 𝐢 + sin 𝜃 𝐣, 𝐠2 =−𝑟 sin 𝜃𝐢 + 𝑟 cos 𝜃 𝐣, and 𝐠3 = 𝐤

defines the natural basis vectors in the new coordinate system. The vectors 𝐠1, 𝐠2 and 𝐠3 are clearly not unit vectors but they form a basis of the new system provided,

𝑉 = 𝐠1 ∙ 𝐠2 × 𝐠3 ≠ 0


Natural Bases (Concrete Example)

Repeat the above example using the spherical coordinate transformations.𝑥 = 𝑥1 = 𝜚 cos 𝜙 sin 𝜃, 𝑦 ≡ 𝑥2 = 𝜚 sin𝜙 sin 𝜃 , 𝑧 = 𝑥3 = 𝜚 cos 𝜃

Again, the position vector 𝐑 = 𝑥𝐢 + 𝑦𝐣 + 𝑧𝐤 = 𝑥𝑖 𝜚, 𝜃, 𝜙 𝐞𝑖 = 𝐑 𝜚, 𝜃, 𝜙

As expected, the natural (tangential) bases of this transformation of coordinates are the vectors,

𝐠1 =𝜕𝐑

𝜕𝜚= cos𝜙 sin 𝜃𝐢 + sin𝜙 sin 𝜃 𝐣 + cos 𝜃 𝐤 ≡ 𝐞𝝔(𝜃, 𝜙)

𝐠2 =𝜕𝐑

𝜕𝜃= 𝜚 cos𝜙 cos 𝜃𝐢 + 𝜚sin𝜙 cos 𝜃 𝐣 − 𝜚 sin 𝜃 𝐤 ≡ 𝜚𝐞𝜽(𝜃, 𝜙)

𝐠3 =𝜕𝐑

𝜕𝜙= −𝜚 sin𝜙 sin 𝜃𝐢 + 𝜚cos 𝜙 sin 𝜃 𝐣 ≡ 𝜚𝐞𝝓(𝜃, 𝜙)


Another Concrete Example

𝐠1 ⋅ 𝐠2 = 𝜚cos2𝜙 cos 𝜃 sin 𝜃 + 𝜚 sin2𝜙 cos 𝜃 sin 𝜃 − 𝜚 cos 𝜃 sin 𝜃 = 0𝐠2 ⋅ 𝐠3 = −𝜚2 cos 𝜙 sin𝜙 cos 𝜃 sin 𝜃 + 𝜚2 cos𝜙 sin𝜙 cos 𝜃 sin 𝜃 = 0𝐠3 ⋅ 𝐠1 = −𝜚 cos𝜙 sin𝜙 cos 𝜃 sin 𝜃 + 𝜚 cos𝜙 sin𝜙 cos 𝜃 sin 𝜃 = 0

From which we see that the natural bases in this case are mutually orthogonal. Show that they are not normalized in the sense that they are not necessarily unit vectors. In fact, 𝐠1 is a unit vector but neither 𝐠2 nor 𝐠3 is.

Furthermore, we also see that for the typical basis vector, we now have, 𝐠𝑖 = 𝐠𝑖 𝜚, 𝜃, 𝜙 unlike the Cartesian bases vectors that are constants everywhere.


Concrete example

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 𝐠2= 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

We now show that if the Jacobian of the transformation 𝜕𝑥𝑘

𝜕𝑢𝑖does not vanish, then 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 (we shall prove later that the determinant of

a product of matrices is the product of the determinants)

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 expressions 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,

𝐠𝑖 × 𝐠𝑗 = 𝜖𝑖𝑗𝑘𝐠𝑘


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.


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


In the Q1.1, we are given that ∀ 𝐯 ∈V , 𝒂 ⋅ 𝐯 = 𝐛 ⋅ 𝐯 this implies,

𝒂 ⋅ 𝐯 − 𝐛 ⋅ 𝐯 = 𝒂 − 𝐛 ⋅ 𝐯 = 0

Define the vector 𝐜 ≡ 𝒂 − 𝐛. The equation becomes,𝐜 ⋅ 𝐯 = 𝐜 𝐯 cos 𝜃 = 0.

Because 𝐯 can be any vector, it does not have to be perpendicular to 𝐜 and we can rule out the trivial case of its being the zero vector. This leaves us with the only choice that 𝐜 = 0. And, the only vector that has zero magnitude is the zero vector. So that,

𝐜 ≡ 𝒂 − 𝐛 = 𝐨, or 𝒂 = 𝐛.


Answer to Quiz 1.1

Starting with 𝐜 ≡ 𝒂 − 𝐛

We could have arrived at the same conclusion from 𝐜 ⋅ 𝐯 = 0

by choosing 𝐯 = 𝐜. We can do this because the equation is valid for every vector 𝐯. It must therefore be valid for a particular vector.

The equation then becomes, 𝐜 ⋅ 𝐜 = 0

which, again is only possible if 𝐜 = 𝐨, the zero vector.

Hence 𝒂 = 𝐛.


Alternative Answer to Quiz 1.1

In the Q1.2, we are given that ∀ 𝐯 ∈V , 𝒂 × 𝐯 = 𝐛 × 𝐯,

Now take a dot product with 𝒂, we have that, 𝒂 ⋅ 𝐛 × 𝐯 = 𝒂 × 𝐛 ⋅ 𝐯 = 0 = 𝐨 ⋅ 𝐯

for all 𝐯 proving from Quiz 1.1 that 𝒂 × 𝐛 = 𝐨. This shows that 𝒂 and 𝐛 are collinear. We can therefore write that 𝐛 =𝛼𝒂

Hence, 𝒂 × 𝐯 = 𝐛 × 𝐯 = 𝛼𝒂 × 𝐯 where 𝛼 is a scalar. So that 𝒂 × 𝐯 1 − 𝛼 = 𝐨 ⇒ 1 = 𝛼

showing that 𝒂 = 𝐛 as was required.


Answer to Quiz 1.2

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



Coordinate Surfaces

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 = ϵ𝑖𝑗𝑘𝑢𝑖𝑢𝑘𝑣𝑗= 0

On 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, 𝒆𝜙 𝜙 = 𝑟𝜕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.


14. Simplify the following by employing the suustitution properties of the Kroneckerdelta 𝑎 𝑒𝑖𝑗𝑘𝛿𝑘𝑛 , 𝑏 𝑒𝑖𝑗𝑘𝛿𝑖𝑠𝛿𝑗𝑚 𝑐 𝑒𝑖𝑗𝑘𝛿𝑖𝑠𝛿𝑗𝑚 𝑑 𝑎𝑖𝑗𝛿𝑖𝑛 𝑒 𝛿𝑖𝑗𝛿𝑗𝑛 𝑓 𝛿𝑖𝑗𝛿𝑗𝑛𝛿𝑛𝑖

15. Show that the moments of inertia 𝑰𝒊𝒋defined by

𝐼11 =ම𝑉

𝑦2 + 𝑧2 𝜌 𝑥, 𝑦, 𝑧 𝑑𝑥𝑑𝑦𝑑𝑧, 𝐼21 = 𝐼12 =ම𝑉

𝑥𝑦 𝜌 𝑥, 𝑦, 𝑧 𝑑𝑥𝑑𝑦𝑑𝑧,

𝐼22 =ම𝑉

𝑧2 + 𝑥2 𝜌 𝑥, 𝑦, 𝑧 𝑑𝑥𝑑𝑦𝑑𝑧, 𝐼32 = 𝐼23 = ම𝑉

𝑦𝑧 𝜌 𝑥, 𝑦, 𝑧 𝑑𝑥𝑑𝑦𝑑𝑧,

𝐼31 = 𝐼13 =ම𝑉

𝑥𝑦 𝜌 𝑥, 𝑦, 𝑧 𝑑𝑥𝑑𝑦𝑑𝑧,

𝐼33 =𝑉𝑥2 + 𝑦2 𝜌 𝑥, 𝑦, 𝑧 𝑑𝑥𝑑𝑦𝑑𝑧

can be represented in the index notation as 𝐼𝑖𝑗 = 𝑉൫𝑥𝑚𝑥𝑚𝛿𝑖𝑗 −


17. Show that the contraction of a symmetric object with an antisymmetric object equals zero. For example given that 𝑎𝑚𝑛, 𝑚, 𝑛 = 1,2,3 is antisymmetric, Show that 𝑎𝑚𝑛𝑥

𝑚𝑥𝑛 = 𝟎

18. Noting that 𝑒𝑖𝑗𝑘𝜎𝑗𝑘 = 0 observe that 𝑒𝑖𝑗𝑘 is perfectly

antisymmetric. What does this tell about 𝜎𝑗𝑘?

19. For any tensor 𝐀, define Sym 𝐀𝑖𝑗=

𝟏

𝟐𝑨𝑖𝑗 + 𝑨𝑗𝑖 .

Show that Sym 𝐀T𝐒𝐀 = 𝐀TSym 𝐒 𝐀


20. Given that, 𝜖𝑟𝑠𝑡𝜖𝑖𝑗𝑘 =

𝑔𝑟𝑖 𝑔𝑟𝑗 𝑔𝑟𝑘𝑔𝑠𝑖 𝑔𝑠𝑗 𝑔𝑠𝑘𝑔𝑡𝑖 𝑔𝑡𝑗 𝑔𝑡𝑘

Find 𝜖𝑖𝑠𝑡𝜖𝑖𝑗𝑘 and 𝜖𝑖𝑗𝑡𝜖𝑖𝑗𝑘.


SSG 805 Mechanics of Continua -...

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