SSG 805 Mechanics of Continua INSTRUCTOR: Fakinlede OA [email protected] [email protected]
ASSISTED BY: Akano T [email protected]
Oyelade AO [email protected]
Department of Systems Engineering, University of Lagos
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
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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 ABSOLUTELY necessary 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
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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:
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Examination
Evaluation Obtainable
Quiz 10
Homework 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.org 2010
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
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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
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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 Thermodynamical Theory of Plasticity, CRC Press, Taylor and Francis Group, June 2012
Heinbockel, JH, Introduction to Tensor Calculus and Continuum Mechanics, Trafford, 2003
Texts
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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 โA Method of Geometrical Representation of
the Thermodynamic Properties of Substances by Means of Surfaces,โ Transactions of the Connecticut Academy of Arts and Sciences 2, Dec. 1873, pp. 382-404.
Texts
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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
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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
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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
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Texts
1. If โ ๐ฏ โV , ๐ โ ๐ฏ = ๐ โ ๐ฏ, Show that ๐ = ๐
2. If โ ๐ฏ โV , ๐ ร ๐ฏ = ๐ ร ๐ฏ, Show that ๐ = ๐
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Quiz of the day
Linear Spaces Introduction
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
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Continuum Mechanics
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Continuum Mechanics views matter as continuously distributed in space. The physical quantities we are interested in can be
โข Scalars or reals, such as time, energy, power,
โข Vectors, for example, position vectors, velocities, or forces,
โข Tensors: deformation gradient, strain and stress measures.
Physical Quantities
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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
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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
โContinuum Mechanics may appear as a fortress surrounded by the walls of tensor notationโ It is our hope that the course helps every serious learner overcome these difficulties
Physical Quantities
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The set of real numbers is denoted by R Let R ๐ be the set of n-tuples so that when ๐ = 2,
R 2 we have the set of pairs of real numbers. For example, such can be used for the ๐ฅ and ๐ฆ coordinates of points on a Cartesian coordinate system.
Real Numbers & Tuples
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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, ๐ + ๐ฏ โV, ๐ + ๐ฏ =๐ฏ + ๐, ๐ + ๐ฏ + ๐ = ๐ + ๐ฏ +๐,โ ๐, ๐ฏ,๐ โ V. Furthermore,V is closed under addition: That is, given
that ๐, ๐ฏ โ V, then ๐ = ๐ + ๐ฏ = ๐ฏ + ๐,โ ๐ โ V.
2. V contains a zero element ๐ such that ๐ + ๐ = ๐ โ ๐ โV. For every ๐ โ V, โ โ ๐: ๐ + โ๐ = ๐.
3. Multiplication by a scalar. For ๐ผ, ๐ฝ โR and ๐, ๐ โ V, ๐ผ๐ โ V , 1๐ = ๐, ๐ผ ๐ฝ๐ = ๐ผ๐ฝ ๐, ๐ผ + ๐ฝ ๐ = ๐ผ๐ +๐ฝ๐, ๐ผ ๐ + ๐ฏ = ๐ผ๐ + ๐ผ๐ฏ.
Vector Space
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An Inner-Product (also called a Euclidean Vector) SpaceE is a real vector space that defines the scalar product: for each pair ๐, ๐ฏ โ E, โ ๐ โ R such that, ๐ = ๐ โ ๐ฏ = ๐ฏ โ ๐ . Further, ๐ โ ๐ โฅ 0, the zero value occurring only when ๐ = 0. It is called โEuclideanโ because the laws of Euclidean geometry hold in such a space.
The inner product also called a dot product, is the mapping
" โ ":V รV โ R from the product space to the real space.
Euclidean Vector Space
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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
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From our definition of the Euclidean space, it is easy to see that,
๐ฏ = ๐ผ1๐1 + ๐ผ2๐2 +โฏ+ ๐ผ๐๐๐
such that, ๐ผ1, ๐ผ2, โฏ , ๐ผ๐ โ R and ๐1, ๐2, โฏ , ๐๐ โ E, is also a vector. The subset
๐1, ๐2, โฏ , ๐๐ โ E is said to be linearly independent or free if for any choice of the subset ๐ผ1, ๐ผ2, โฏ , ๐ผ๐ โR other than 0,0,โฏ , 0 , ๐ฏ โ 0. If it is possible to find linearly independent vector systems of order ๐ where ๐ is a finite integer, but there is no free system of order ๐ + 1, then the dimension ofE is ๐. In other words,
the dimension of a space is the highest number of linearly independent members it can have.
Dimensionality
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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 ๐, ๐ and ๐ 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
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1. Addition operation for a directed line segment is defined by the parallelogram law for addition.
2. V contains a zero element ๐ in such a case is simply a point
with zero length..
3. Multiplication by a scalar ๐ผ. Has the meaning that
0 < ๐ผ โค 1 โ Line is shrunk along the same direction by ๐ผ
๐ผ > 1 โ Elongation by ๐ผ
Negative value is same as the above with a change of direction.
Directed Line
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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
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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 ๐ฅ๐ โ 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
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๐ = *๐ฅ1, ๐ฅ2+, ๐ฅ1, ๐ฅ2 โR, ๐ = *๐ฆ1, ๐ฆ2+, ๐ฆ1, ๐ฆ2 โR. Addition is easily defined as ๐ + ๐ = *๐ฅ1 + ๐ฆ1, ๐ฅ2 +๐ฆ2+ clearly ๐ + ๐ โR 2 since ๐ฅ1 + ๐ฆ1, ๐ฅ2 + ๐ฆ2 โ R. Addition operation creates other members for the vector space โ Hence closure exists for the operation.
Multiplication by a scalar: ๐ผ๐ = *๐ผ๐ฅ1, ๐ผ๐ฅ2+, โ๐ผ โR. Zero element: ๐ = 0,0 . Additive Inverse: โ๐ =โ๐ฅ1, โ๐ฅ2 , ๐ฅ1, ๐ฅ2 โR Type equation here.A standard basis for this : ๐1 = 1,0 , ๐2 = 0,1 Type equation here.Any other member can be expressed in terms of this basis.
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Set of Pairs
๐ โD Coordinate Space. For any positive number ๐, we may create ๐ โtuples such that, ๐ = *๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐+ where ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐ โ R are members of R ๐ โ a real vector space over theR. ๐ = *๐ฆ1, ๐ฆ2, โฆ , ๐ฆ๐+, ๐ฆ1, ๐ฆ2, โฆ , ๐ฆ๐ โR. ๐ + ๐ = *๐ฅ1 + ๐ฆ1, ๐ฅ2 + ๐ฆ2, โฆ , ๐ฅ๐ +๐ฆ๐+,. ๐ + ๐ โ R ๐since ๐ฅ1 + ๐ฆ1, ๐ฅ2 + ๐ฆ2, โฆ , ๐ฅ๐ + ๐ฆ๐ โR
N-tuples
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Addition operation creates other members for the vector space โ Hence closure exists for the operation.
Multiplication by a scalar: ๐ผ๐ = *๐ผ๐ฅ1, ๐ผ๐ฅ2, โฆ , ๐ผ๐ฅ๐+, โ๐ผ โ ๐ .
Zero element ๐ = 0,0,โฆ , 0 . Additive Inverse โ ๐ = โ๐ฅ1, โ๐ฅ2, โฆ , โ๐ฅ๐ , ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐ โR There is also a standard basis which is easily proved to be linearly independent: ๐1 = 1,0,โฆ , 0 , ๐2 =0,1,โฆ , 0 , โฆ , ๐๐ = *0,0,โฆ , 0+
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N-tuples
Let R๐ร๐ denote the set of matrices with entries that are
real numbers (same thing as saying members of the real
spaceR, Then, R๐ร๐ 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 ๐๐. For example, in
R3ร3 we can choose basis in the form,
1 0 00 0 00 0 0
,0 1 00 0 00 0 0
, โฆ, 0 0 00 0 00 0 1
Matrices
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Forming polynomials with a single variable ๐ฅ to order ๐
when ๐ 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
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An Inner-Product (also called a Euclidean Vector) SpaceE is a real vector space that defines the scalar product: for each pair ๐, ๐ฏ โ E, โ ๐ โ R such that, ๐ = ๐ โ ๐ฏ = ๐ฏ โ ๐ . Further, ๐ โ ๐ โฅ 0, the zero value occurring only when ๐ = 0. It is called โEuclideanโ because the laws of Euclidean geometry hold in such a space.
The inner product also called a dot product, is the mapping
" โ ":V รV โ R from the product space to the real space.
Euclidean Vector Space
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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 ๐ฏ โ:V โR on the vector spaceV is called a covector or a dual vector. For a finite dimensional vector space, the set of all covectors forms the dual spaceV โ ofV . If V is an Inner Product Space, then there is no distinction between the vector space and its dual.
Co-vectors
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Magnitude The norm, length or magnitude of ๐, denoted ๐ is defined as the positive square root of ๐ โ ๐ = ๐ 2. When ๐ = 1, ๐ is said to be a unit vector. When ๐ โ ๐ฏ = 0, ๐ and ๐ฏ are said to be orthogonal.
Direction Furthermore, for any two vectors ๐ and ๐ฏ, the angle between them is defined as,
cosโ1๐ โ ๐ฏ
๐ ๐ฏ
The scalar distance ๐ between two vectors ๐ and ๐ฏ ๐ = ๐ โ ๐ฏ
Magnitude & Direction Again
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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
" ร ":V รV โ V Which takes two vectors and produces a vector.
3-D Euclidean Space
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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 ๐ ร ๐ = ๐ ๐ sin ๐ (0 โค ๐ โค ๐)
of the cross product ๐ ร ๐ is the area ๐ด(๐, ๐) spanned by the vectors ๐ and ๐. 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. ๐ is the angle between the two vectors.
3. The direction of ๐ ร ๐ is orthogonal to both ๐ and ๐
Cross Product
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Cross Product
The area ๐ด(๐, ๐) spanned by the vectors ๐ and ๐. The unit vector ๐ in the direction of the cross product can be
obtained from the quotient, ๐ร๐
๐ร๐.
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The cross product is bilinear and anti-commutative:
Given ๐ผ โ R , โ๐, ๐, ๐ โ V , ๐ผ๐ + ๐ ร ๐ = ๐ผ ๐ ร ๐ + ๐ ร ๐ ๐ ร ๐ผ๐ + ๐ = ๐ผ ๐ ร ๐ + ๐ ร ๐
So that there is linearity in both arguments.
Furthermore, โ๐, ๐ โ V ๐ ร ๐ = โ๐ ร ๐
Cross Product
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The trilinear mapping, , , ,- โถ V ร V ร V โR
From the product set V ร V ร V to real space is defined by: , u,v,w- โก ๐ โ ๐ ร ๐ = ๐ ร ๐ โ ๐
Has the following properties:
1. , a,b,c-=,b,c,a-=,c,a,b-=โ,b,a,c-=โ,c,b,a-=โ,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
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Tripple product
Parallelepiped defined by u, v and w
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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
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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.
Summation Convention
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Consider transformation equations such as, ๐ฆ1 = ๐11๐ฅ1 + ๐12๐ฅ2 + ๐13๐ฅ3 ๐ฆ2 = ๐21๐ฅ1 + ๐22๐ฅ2 + ๐23๐ฅ3 ๐ฆ3 = ๐31๐ฅ1 + ๐32๐ฅ2 + ๐33๐ฅ3
We may write these equations using the summation symbols as:
๐ฆ1 = ๐1๐๐ฅ๐
๐
๐=1
๐ฆ2 = ๐2๐๐ฅ๐
๐
๐=1
๐ฆ3 = ๐3๐๐ฅ๐
๐
๐=1
Summation Convention
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In each of these, we can invoke the Einstein summation convention, and write that,
๐ฆ1 = ๐1๐๐ฅ๐ ๐ฆ2 = ๐2๐๐ฅ๐ ๐ฆ3 = ๐3๐๐ฅ๐
Finally, we observe that ๐ฆ1, ๐ฆ2, and ๐ฆ3 can be represented as we have been doing by ๐ฆ๐ , ๐ = 1,2,3 so that the three equations can be written more compactly as,
๐ฆ๐ = ๐๐๐๐ฅ๐ , ๐ = 1,2,3
Summation Convention
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Please note here that while ๐ in each equation is a dummy index, ๐ 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 ๐, we may leave it out completely and write,
๐ฆ๐ = ๐๐๐๐ฅ๐
to represent compactly, the transformation equations above. It should be obvious there are as many equations as there are free indices.
Summation Convention
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If ๐๐๐ represents the components of a 3 ร 3 matrix ๐, we can show that,
๐๐๐๐๐๐ = ๐๐๐
Where ๐ is the product matrix ๐๐. To show this, apply summation convention and see that, for ๐ = 1, ๐ = 1, ๐11๐11 + ๐12๐21 + ๐13๐31 = ๐11 for ๐ = 1, ๐ = 2, ๐11๐12 + ๐12๐22 + ๐13๐32 = ๐12 for ๐ = 1, ๐ = 3, ๐11๐13 + ๐12๐23 + ๐13๐33 = ๐13 for ๐ = 2, ๐ = 1, ๐21๐11 + ๐22๐21 + ๐23๐31 = ๐21 for ๐ = 2, ๐ = 2, ๐21๐12 + ๐22๐22 + ๐23๐32 = ๐22 for ๐ = 2, ๐ = 3, ๐21๐13 + ๐22๐23 + ๐23๐33 = ๐23 for ๐ = 3, ๐ = 1, ๐31๐11 + ๐32๐21 + ๐33๐31 = ๐31 for ๐ = 3, ๐ = 2, ๐31๐12 + ๐32๐22 + ๐33๐32 = ๐32 for ๐ = 3, ๐ = 3, ๐31๐13 + ๐32๐23 + ๐33๐33 = ๐33
Summation Convention
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The above can easily be verified in matrix notation as,
๐๐ =
๐11 ๐12 ๐13๐21 ๐22 ๐23๐31 ๐32 ๐33
๐11 ๐12 ๐13๐21 ๐22 ๐23๐31 ๐32 ๐33
=
๐11 ๐12 ๐13๐21 ๐22 ๐23๐31 ๐32 ๐33
= ๐
In this same way, we could have also proved that, ๐๐๐๐๐๐ = ๐๐๐
Where ๐ is the product matrix ๐๐T. Note the arrangements could sometimes be counter intuitive.
Summation Convention
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Suppose our basis vectors ๐ ๐ , ๐ = 1,2,3 are not only not unit in magnitude, but in addition are NOT orthogonal. The only assumption we are making is that ๐ ๐ โ V , ๐ =1,2,3 are linearly independent vectors.
With respect to this basis, we can express vectors ๐ฏ,๐ฐ โ V in terms of the basis as,
๐ฏ = ๐ฃ๐ ๐ ๐, ๐ฐ = ๐ค๐ ๐ ๐
Where each ๐ฃ๐ is called the contravariant component of ๐ฏ
Vector Components
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Clearly, addition and linearity of the vector space โ ๐ฏ + ๐ฐ = (๐ฃ๐+๐ค๐)๐ ๐
Multiplication by scalar rule implies that if ๐ผ โ R , โ๐ฏ โ V ,
๐ผ ๐ฏ = (๐ผ๐ฃ๐)๐ ๐
Vector Components
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For any basis vectors ๐ ๐ โ V , ๐ = 1,2,3 there is a dual (or reciprocal) basis defined by the reciprocity relationship:
๐ ๐ โ ๐ ๐ = ๐ ๐ โ ๐ ๐ = ๐ฟ๐
๐
Where ๐ฟ๐๐ is the Kronecker delta
๐ฟ๐๐ = 0 if ๐ โ ๐1 if ๐ = ๐
Let the fact that the above equations are actually nine equations each sink. Consider the full meaning:
Reciprocal Basis
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Kronecker Delta: ๐ฟ๐๐, ๐ฟ๐๐ or ๐ฟ๐
๐ has the following properties:
๐ฟ11 = 1, ๐ฟ12 = 0, ๐ฟ13 = 0 ๐ฟ21 = 0, ๐ฟ22 = 1, ๐ฟ23 = 0 ๐ฟ31 = 0, ๐ฟ32 = 0, ๐ฟ33 = 1
As is obvious, these are obtained by allowing the indices to attain all possible values in the range. The Kronecker delta is defined by the fact that when the indices explicit values are equal, it has the value of unity. Otherwise, it is zero. The above nine equations can be written more compactly as,
๐ฟ๐๐ = 0 if ๐ โ ๐1 if ๐ = ๐
Kronecker Delta
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For any โ๐ฏ โ V , ๐ฏ = ๐ฃ๐๐ ๐ = ๐ฃ๐๐
๐
Are two related representations in the reciprocal bases. Taking the inner product of the above equation with the basis vector ๐ ๐, we have
๐ฏ โ ๐ ๐ = ๐ฃ๐๐ ๐ โ ๐ ๐ = ๐ฃ๐๐
๐ โ ๐ ๐
Which gives us the covariant component,
๐ฏ โ ๐ ๐ = ๐ฃ๐๐๐๐ = ๐ฃ๐๐ฟ๐
๐ = ๐ฃ๐
The last equality earns the Kronecker delta the epithet of โSubstitution symbolโ. Work it out
Covariant components
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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 ๐ ๐:
๐ฏ โ ๐ ๐ = ๐ฃ๐๐ ๐ โ ๐ ๐ = ๐ฃ๐๐
๐ โ ๐ ๐
So that,
๐ฏ โ ๐ ๐ = ๐ฃ๐๐ฟ๐๐= ๐ฃ๐๐
๐๐ = ๐ฃ๐
Contravariant Components
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The nine scalar quantities, ๐๐๐ as well as the nine related quantities ๐๐๐ 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 metrize the space defined by these bases.
Metric coefficients
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The Levi-Civita Symbol: ๐๐๐๐
๐111 = 0, ๐112 = 0, ๐113 = 0, ๐121 = 0, ๐122 = 0, ๐123 =1, ๐131 = 0, ๐132 = โ1, ๐133 = 0 ๐211 = 0, ๐212 = 0, ๐213 = โ1, ๐221 = 0, ๐222 = 0, ๐223= 0, ๐231 = 1, ๐232 = 0, ๐233 = 0
๐311 = 0, ๐312 = 1, ๐313 = 0, ๐321 = โ1, ๐322 = 0, ๐323= 0, ๐331 = 0, ๐332 = 0, ๐333 = 0
Levi Civita Symbol
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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:
๐๐๐๐ = 1 if ๐, ๐ and ๐ are an even permutation of 1,2 and 3โ1 if ๐, ๐ and ๐ are an odd permutation of 1,2 and 30 In all other cases
Levi Civita Symbol
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 58
Given that ๐ = det ๐๐๐ of the covariant metric coefficients, It is
not difficult to prove that ๐ ๐ โ ๐ ๐ ร ๐ ๐ = ๐๐๐๐ โก ๐๐๐๐๐
This relationship immediately implies that,
๐ ๐ ร ๐ ๐ = ๐๐๐๐๐ ๐ .
The dual of the expression, the equivalent contravariant equivalent also follows from the fact that,
๐ 1 ร ๐ 2 โ ๐ 3 = 1/ ๐
Cross Product of Basis Vectors
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This leads in a similar way to the expression,
๐ ๐ ร ๐ ๐ โ ๐ ๐ =๐๐๐๐
๐= ๐๐๐๐
It follows immediately from this that, ๐ ๐ ร ๐ ๐ = ๐๐๐๐๐ ๐
Cross Product of Basis Vectors
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Given that, ๐ ๐, ๐ 2 and ๐ 3 are three linearly
independent vectors and satisfy ๐ ๐ โ ๐ ๐ = ๐ฟ๐๐, show
that ๐ 1 =1
๐๐ 2 ร ๐ 3, ๐
2 =๐
๐ฝ๐ 3 ร ๐ 1, and ๐ 3 =
1
๐๐ 1 ร
๐ 2, where ๐ = ๐ 1 โ ๐ 2 ร ๐ 3
Exercises
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It is clear, for example, that ๐ 1 is perpendicular to ๐ 2 as well as to ๐ 3 (an obvious fact because ๐ 1 โ ๐ 2 = 0 and ๐ 1 โ ๐ 3 = 0), we can say that the vector ๐ 1 must necessarily lie on the cross product ๐ 2 ร ๐ 3 of ๐ 2 and ๐ 3. It is therefore correct to write,
๐ 1 =1
๐๐ 2 ร ๐ 3
Where ๐โ1is 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 ๐ 1 we immediately obtain,
๐ 1 โ ๐ 1 = ๐โ1 ๐ 1 โ ๐ 2 ร ๐ 3= 1
So that, ๐ = ๐ 1 โ ๐ 2 ร ๐ 3
the volume of the parallelepiped formed by the three vectors ๐ 1, ๐ 2, and ๐ 3 when their origins are made to coincide.
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 62
Suppose you have a function ๐(๐ฅ, ๐ฆ, ๐ง) of variables ๐ฅ, ๐ฆ and ๐ง. Let us assume there are some variables ๐, ๐ and ๐ such that, the original variables are themselves functions ๐ฅ = ๐ฅ(๐, ๐, ๐), ๐ฆ = ๐ฆ(๐, ๐, ๐), and ๐ง = ๐ง(๐, ๐, ๐). A simple example is the polar coordinate transformation: ๐ฅ = ๐ cos๐, ๐ฆ = ๐ sin๐, ๐ง = ๐. We can always get a new function F ๐, ๐, ๐ = ๐(๐ฅ, ๐ฆ, ๐ง) 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,
๐(๐ฅ, ๐ฆ, ๐ง)
๐(๐, ๐, ๐)โก
๐๐ฅ
๐๐
๐๐ฆ
๐๐
๐๐ง
๐๐๐๐ฅ
๐๐
๐๐ฆ
๐๐
๐๐ง
๐๐๐๐ฅ
๐๐
๐๐ฆ
๐๐
๐๐ง
๐๐
โ 0
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Jacobian of Transformation
We prefer to use indexed variables. Hence instead of ๐ฅ, ๐ฆ and ๐ง, we prefer ๐ฅ๐ = ๐ฅ๐(๐ข1, ๐ข2, ๐ข3) where ๐ = 1,2,3 as you can obviously see that instead of ๐, ๐, ๐, we are now talking about ๐ข1, ๐ข2, ๐ข3. As before, we can say that the transformation will have an inverse provided the
Jacobian, ๐๐ฅ๐
๐๐ข๐ does not vanish. Therefore to say that
the transformation is invertible ensures that ๐๐ฅ๐
๐๐ข๐โ 0.
Recall that in Cartesian coordinates, the vector connecting an arbitrary point to the origin, also called a position vector can be written as
๐ซ = ๐ฅ๐ข + ๐ฆ๐ฃ + ๐ง๐ค = ๐ฅ๐๐๐
Or, in order to emphasize the functional dependencies, ๐ซ ๐ฅ, ๐ฆ, ๐ง = ๐ฅ๐ ๐ข1, ๐ข2, ๐ข3 ๐๐
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 64
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,
๐ข =๐๐ซ
๐๐ฅ
And in general Cartesian coordinates, using index notation,
๐๐ =๐๐ซ
๐๐ฅ๐, ๐ = 1,2,3
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 65
Basis Vectors
We generalize the result now in terms of natural bases that arise in coordinate transformations from the Cartesian:
In the curvilinear system (๐ข1, ๐ข2, ๐ข3) obtained from the transformation ๐ฅ๐ = ๐ฅ๐(๐ข1, ๐ข2, ๐ข3) from Cartesian
coordinates, let ๐ ๐ =๐๐ซ
๐๐ข๐ and let ๐ ๐ be the corresponding dual
basis. Show that ๐๐๐ = ๐ ๐ โ ๐ ๐ =๐๐ฅ๐
๐๐ข๐๐๐ฅ๐
๐๐ข๐. If ๐ = ๐ 1 โ ๐ 2 ร ๐ 3
and ๐ฃ = ๐ 1 โ ๐ 2 ร ๐ 3, show that ๐ฃ ๐ = 1. Show also that ๐ ๐ โ ๐ ๐ ร ๐ ๐ = ๐๐๐๐ = ๐๐๐๐๐.
The position vector ๐ซ ๐ฅ, ๐ฆ, ๐ง = ๐ฅ๐ ๐ข1, ๐ข2, ๐ข3 ๐๐ where ๐๐ , ๐ = 1,2,3 are unit vectors that are orthonormal in the Euclidean space.
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 66
Natural Bases
Changing variables, we can write that,
๐ซ ๐ฅ, ๐ฆ, ๐ง = ๐ฅ๐(๐ข1, ๐ข2, ๐ข3)๐๐ = ๐ซ ๐ข1, ๐ข2, ๐ข3
So that we have new coordinates ๐ข๐ , ๐ = 1,2,3. In this new system, the differential of the position vector ๐ซ is,
๐๐ซ =๐๐ซ
๐๐ข๐ ๐๐ข๐ โก ๐ ๐๐๐ข
๐
the above equation, as we shall soon show, defines the natural basis vectors in the new coordinate system. The vectors ๐ 1, ๐ 2 and ๐ 3 are not necessarily unit vectors but they form a basis of the new system provided,
๐ = ๐ 1 โ ๐ 2 ร ๐ 3 โ 0
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 67
Natural Bases
Clearly, the reciprocal basis vectors are ๐ 1 = ๐โ1 ๐ 2 ร ๐ 3 ๐ 2 = ๐โ1 ๐ 3 ร ๐ 1 ๐ 3 = ๐โ1 ๐ 1 ร ๐ 2
(dot the first with ๐ 1 to see) Now we are given that ๐ฃ = ๐ 1 โ ๐ 2 ร ๐ 3. Using the above relations, we can write, ๐ 2 ร ๐ 3 = ๐โ1 ๐ 3 ร ๐ 1 ร ๐
โ1 ๐ 1 ร ๐ 2 = ๐โ2 ๐ 3 ร ๐ 1โ ๐ 2 ๐ 1 โ ๐ 3 ร ๐ 1โ ๐ 1 ๐ 2 = ๐โ2 ๐ 1 ร ๐ 2โ ๐ 3 ๐ 1 = ๐
โ1๐ 1
We can now write, ๐ฃ = ๐ 1 โ ๐ 2 ร ๐ 3 = ๐ 1 โ ๐โ1๐ 1 = ๐
โ1๐ 1 โ ๐ 1 = ๐โ1
Showing that, ๐ฃ ๐ = 1 as required.
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 68
Dual Bases
We now show that if the Jacobian of the transformation ๐๐ฅ๐
๐๐ข๐ does not vanish, then the ๐ ๐ are independent:
Now,
๐ ๐ =๐๐ซ
๐๐ข๐=๐๐ฅ๐
๐๐ข๐๐๐
๐ 1 โ ๐ 2 ร ๐ 3 =
๐๐ฅ1
๐๐ข1๐๐ฅ2
๐๐ข1๐๐ฅ3
๐๐ข1
๐๐ฅ1
๐๐ข2๐๐ฅ2
๐๐ข2๐๐ฅ3
๐๐ข2
๐๐ฅ1
๐๐ข3๐๐ฅ2
๐๐ข3๐๐ฅ3
๐๐ข3
=๐๐ฅ๐
๐๐ข๐โ 0.
๐๐๐ = ๐ ๐ โ ๐ ๐ =๐๐ซ
๐๐ข๐โ ๐๐ซ
๐๐ข๐=๐๐ฅ๐
๐๐ข๐๐๐ โ
๐๐ฅ๐
๐๐ข๐๐๐
=๐๐ฅ๐
๐๐ข๐๐๐ฅ๐
๐๐ข๐๐๐ โ ๐๐ =
๐๐ฅ๐
๐๐ข๐๐๐ฅ๐
๐๐ข๐๐ฟ๐๐ =๐๐ฅ๐
๐๐ข๐๐๐ฅ๐
๐๐ข๐
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 69
Clearly, the determinant of ๐๐๐ (we shall prove later that the determinant of
a product of matrices is the product of the determinants)
๐ โก ๐๐๐ =๐๐ฅ๐
๐๐ข๐๐๐ฅ๐
๐๐ข๐=๐๐ฅ๐
๐๐ข๐
2
= ๐2
This means, ๐ = ๐ 1 โ ๐ 2 ร ๐ 3 =๐๐ฅ๐
๐๐ข๐= ๐. We can therefore write,
๐ 1 โ ๐ 2 ร ๐ 3 = ๐123 ๐
Swapping indices 2 and 3, we have, ๐ 1 โ ๐ 3 ร ๐ 2 = โ ๐ = ๐132 ๐ = ๐ 1 ร ๐ 3 โ ๐ 2
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,
๐ 3 ร ๐ 1 โ ๐ 2 = โ โ ๐ = ๐312 ๐. These three expressions together
imply that,
๐ ๐ โ ๐ ๐ ร ๐ ๐ = ๐๐๐๐ = ๐๐๐๐๐ as required.
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 70
This relationship immediately implies that,
๐ ๐ ร ๐ ๐ = ๐๐๐๐๐ ๐ as a dot product of this with ๐ ๐ผ
recovers the previous. The dual of the expression, the equivalent contravariant equivalent also follows from the fact that .
๐ 1 ร ๐ 2 โ ๐ 3 = 1/ ๐ as it must be since we proved that the two volumes must be inverses. This leads in a similar way to the expression,
๐ ๐ ร ๐ ๐ โ ๐ ๐ =๐๐๐๐
๐= ๐๐๐๐
It follows immediately from this that,
๐ ๐ ร ๐ ๐ = ๐๐๐๐๐ ๐
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 71
Show that ๐ ๐ = ๐๐๐๐ ๐ = ๐๐๐๐ ๐ and establish the relation, ๐๐๐๐
๐๐ = ๐ฟ๐๐
First expand ๐ ๐ in terms of the ๐ ๐s: ๐ ๐ = ๐ผ๐ 1 + ๐ฝ๐ 2 + ๐พ๐ 3
Dotting with ๐ 1 โ ๐ ๐ โ ๐ 1 = ๐ผ๐ 1 โ ๐ 1 + ๐ฝ๐ 2 โ ๐
1 + ๐พ๐ 3 โ ๐ 1 = ๐๐1 = ๐ผ. In
the same way we find that ๐ฝ = ๐๐2 and ๐พ = ๐๐3so that, ๐ ๐ = ๐๐1๐ 1 + ๐
๐2๐ 2 + ๐๐3๐ 3 = ๐
๐๐๐ ๐ . Similarly, ๐ ๐ = ๐๐๐ผ๐
๐ผ.
Recall the reciprocity relationship: ๐ ๐ โ ๐ ๐ = ๐ฟ๐
๐. Using the above, we can write
๐ ๐ โ ๐ ๐ = ๐๐๐ผ๐
๐ผ โ ๐๐๐ฝ๐ ๐ฝ
= ๐๐๐ผ๐๐๐ฝ๐ ๐ผ โ ๐ ๐ฝ
= ๐๐๐ผ๐๐๐ฝ๐ฟ๐ฝ๐ผ = ๐ฟ๐
๐
which shows that
๐๐๐ผ๐๐๐ผ = ๐๐๐๐
๐๐ = ๐ฟ๐๐
As required. This shows that the tensor ๐๐๐ and ๐๐๐ are inverses of each
other.
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 72
Show that the cross product of vectors ๐ and ๐ in general coordinates is
๐๐๐๐๐๐๐๐๐ ๐ or ๐๐๐๐๐๐๐๐๐ ๐ where ๐๐ , ๐๐ are the respective contravariant
components and ๐๐ , ๐๐ the covariant.
Express vectors ๐ and ๐ as contravariant components: ๐ = ๐๐๐ ๐, and ๐ = ๐๐๐ ๐. Using the above result, we can write that,
๐ ร ๐ = ๐๐๐ ๐ ร ๐๐๐ ๐
= ๐๐๐๐๐ ๐ ร ๐ ๐ = ๐๐๐๐๐๐๐๐๐
๐ .
Express vectors ๐ and ๐ as covariant components: ๐ = ๐๐๐ ๐ and ๐ = ๐๐๐
๐. Again, proceeding as before, we can write,
๐ ร ๐ = ๐๐๐ ๐ ร ๐๐๐
๐
= ๐๐๐๐๐๐๐๐๐ ๐
Express vectors ๐ as a contravariant components: ๐ = ๐๐๐ ๐ and ๐ as covariant components: ๐ = ๐๐๐
๐
๐ ร ๐ = ๐๐๐ ๐ ร ๐๐๐ ๐
= ๐๐๐๐ ๐ ๐ ร ๐ ๐
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 73
The answer to Quiz 1.1 is on page 5 of Gurtin. Once a piece of information has been made available to you in some form, it is your responsibility to take note.
In the Q1.2, we are given that โ ๐ฏ โV , ๐ ร ๐ฏ = ๐ ร ๐ฏ,
Now take a dot product with ๐, we have that, ๐ โ ๐ ร ๐ฏ = ๐ ร ๐ โ ๐ฏ = 0 = ๐จ โ ๐ฏ
for all ๐ฏ proving from Quiz 1.1 that ๐ ร ๐ = ๐จ. This shows that ๐ ร ๐ are collinear. We can therefore write that ๐ = ๐ผ๐
Hence, ๐ ร ๐ฏ = ๐ ร ๐ฏ = ๐ผ๐ ร ๐ฏ where ๐ผ is a scalar. So that ๐ ร ๐ฏ 1 โ ๐ผ = 0 โ 1 = ๐ผ
showing that ๐ = ๐ as was required.
[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 74
Answer to Quiz 1.2
Given that,
๐ฟ๐๐๐๐๐ ๐ก โก ๐๐๐ ๐ก๐๐๐๐ =
๐ฟ๐๐ ๐ฟ๐
๐ ๐ฟ๐๐
๐ฟ๐๐ ๐ฟ๐
๐ ๐ฟ๐๐
๐ฟ๐๐ก ๐ฟ๐
๐ก ๐ฟ๐๐ก
Show that ๐ฟ๐๐๐๐๐ ๐ = ๐ฟ๐
๐๐ฟ๐๐ โ ๐ฟ๐
๐ ๐ฟ๐๐
Expanding the equation, we have:
๐๐๐๐๐๐๐ ๐ = ๐ฟ๐๐๐
๐๐ ๐= ๐ฟ๐๐๐ฟ๐๐ ๐ฟ๐
๐
๐ฟ๐๐ ๐ฟ๐
๐ โ ๐ฟ๐๐ ๐ฟ๐๐ ๐ฟ๐
๐
๐ฟ๐๐ ๐ฟ๐
๐ + 3๐ฟ๐๐ ๐ฟ๐
๐
๐ฟ๐๐ ๐ฟ๐
๐
= ๐ฟ๐๐ ๐ฟ๐๐๐ฟ๐๐ โ ๐ฟ๐
๐ ๐ฟ๐๐ โ ๐ฟ๐
๐ ๐ฟ๐๐๐ฟ๐๐ โ ๐ฟ๐
๐ ๐ฟ๐๐
+ 3 ๐ฟ๐๐๐ฟ๐๐ โ ๐ฟ๐
๐ ๐ฟ๐๐ = ๐ฟ๐
๐ ๐ฟ๐๐ โ ๐ฟ๐
๐ ๐ฟ๐๐ โ ๐ฟ๐
๐๐ฟ๐๐
+ ๐ฟ๐๐ ๐ฟ๐๐ +3(๐ฟ๐
๐ ๐ฟ๐๐
โ ๐ฟ๐๐ ๐ฟ๐๐) = โ2(๐ฟ๐
๐ ๐ฟ๐๐ โ ๐ฟ๐
๐ ๐ฟ๐๐) + 3(๐ฟ๐
๐๐ฟ๐๐ โ ๐ฟ๐
๐ ๐ฟ๐๐)
= ๐ฟ๐๐๐ฟ๐๐ โ ๐ฟ๐
๐ ๐ฟ๐๐
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Coordinate Surfaces
Show that ๐ฟ๐๐๐๐๐๐= 2๐ฟ๐
๐
Contracting one more index, we have:
๐๐๐๐๐๐๐๐ = ๐ฟ๐๐๐
๐๐๐= ๐ฟ๐๐๐ฟ๐๐โ ๐ฟ๐๐๐ฟ๐๐ = 3๐ฟ๐
๐ โ ๐ฟ๐๐ = 2๐ฟ๐
๐
These results are useful in several situations.
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Show that ๐ฎ โ ๐ฎ ร ๐ฏ = 0
๐ฎ ร ๐ฏ = ฯต๐๐๐๐ข๐๐ฃ๐๐ ๐
๐ฎ โ ๐ฎ ร ๐ฏ = ๐ข๐ผ๐ ๐ผ โ ฯต๐๐๐๐ข๐๐ฃ๐๐ ๐
= ๐ข๐ผ ฯต๐๐๐๐ข๐๐ฃ๐ ๐ฟ๐
๐ผ = ฯต๐๐๐๐ข๐๐ฃ๐๐ข๐ = 0
On account of the symmetry and antisymmetry in ๐ and ๐.
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Show that ๐ฎ ร ๐ฏ = โ ๐ฏ ร ๐ฎ
๐ฎ ร ๐ฏ = ฯต๐๐๐๐ข๐๐ฃ๐๐ ๐
= โฯต๐๐๐๐ข๐๐ฃ๐๐ ๐ = โฯต๐๐๐๐ฃ๐๐ข๐๐ ๐
= โ ๐ฏ ร ๐ฎ
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Show that ๐ฎ ร ๐ฏ ร ๐ฐ = ๐ฎ โ ๐ฐ ๐ฏ โ ๐ฎ โ ๐ฏ ๐ฐ.
Let ๐ณ = ๐ฏ ร ๐ฐ = ฯต๐๐๐๐ฃ๐๐ค๐๐ ๐
๐ฎ ร ๐ฏ ร ๐ฐ = ๐ฎ ร ๐ณ = ๐๐ผ๐ฝ๐พ๐ข๐ผ๐ง๐ฝ๐ ๐พ
= ๐๐ผ๐ฝ๐พ ๐ข๐ผ๐ง๐ฝ๐ ๐พ = ๐๐ผ๐ฝ๐พ๐ข
๐ผฯต๐๐๐ฝ๐ฃ๐๐ค๐๐ ๐พ
= ฯต๐๐๐ฝ๐๐พ๐ผ๐ฝ๐ข๐ผ๐ฃ๐๐ค๐๐
๐พ
= ๐ฟ๐พ๐๐ฟ๐ผ๐โ ๐ฟ๐ผ๐ ๐ฟ๐พ๐๐ข๐ผ๐ฃ๐๐ค๐๐
๐พ
= ๐ข๐๐ฃ๐พ๐ค๐๐ ๐พ โ ๐ข๐๐ฃ๐๐ค๐พ๐
๐พ = ๐ฎ โ ๐ฐ ๐ฏ โ ๐ฎ โ ๐ฏ ๐ฐ
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1. In the transformation from the ๐ฅ, ๐ฆ, ๐ง system to the ๐, ๐, ๐ coordinate system, the position vector changed from ๐ = ๐ฅ๐ + ๐ฆ๐ + ๐ง๐ to ๐ = ๐๐๐ ๐ +๐๐๐ง. Show by partial differentiation only, that the basis vectors in respective coordinates are ๐, ๐, ๐
and ๐๐ , ๐๐, ๐๐ง respectively, ๐๐ ๐ = ๐๐๐๐ ๐
๐๐.
2. If the position vector in another system with coordinate variables ๐, ๐, ๐ is ๐ = ๐๐๐ ๐, ๐ , use
the same method to find the basis vectors in this system also.
Exercises
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3. In Problem 1 above, if the transformation from Cartesian to the other system is given explicitly as ๐ฅ ๐, ๐, ๐ = ๐ cos ๐, ๐ฆ ๐, ๐, ๐ = ๐ sin๐ and ๐ง ๐, ๐, ๐ = ๐, find explicit expression for the basis vectors ๐ ๐ , ๐ = 1,2,3. Also find the reciprocal basis
vectors ๐ ๐ , ๐ = 1,2,3. [Hint: 2๐ ๐ = ๐๐๐๐๐ ๐ ร ๐ ๐-
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?
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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 ๐ ๐ โ ๐ ๐ ร ๐ ๐ = ๐๐๐๐ =
๐๐๐๐๐ and that ๐ ๐ ร ๐ ๐ โ ๐ ๐ =๐๐๐๐
๐= ๐๐๐๐.
9. Beginning with the equation, ๐ ๐ ร ๐ ๐ = ๐๐๐๐๐ ๐, take
a contraction with ๐๐๐๐ผ and find the expression for ๐ ๐
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10. Elliptical Cylindrical Coordinates is defined by the position vector, ๐ ๐ฅ, ๐ฆ, ๐ง = ๐ฅ๐ + ๐ฆ๐ + ๐ง๐
= ๐ ๐, ๐, ๐ค = ๐ cos ๐ cosh ๐ + ๐ sin ๐ sinh ๐ + ๐๐ค
Use Mathematica and show that this system of coordinates is orthogonal. Hint: ๐ ,ฮพ_, ฮท_, w_- = R ๐, ๐, ๐ค โ ๐Cos ๐ Cosh ๐ + ๐Sin ๐ Sinh ๐ + ๐๐ค
= i cos ๐ cosh ๐ + j sin ๐ sinh ๐ + ๐๐ค
g = ๐ท ๐ ๐, ๐, ๐ค , ๐, ๐, ๐ค =๐ R ๐, ๐, ๐ค
๐ ๐, ๐, ๐ค
= icos ๐ sinh ๐ + jsin ๐ cosh ๐ , jcos ๐ sinh ๐ โ๐sin(๐)cosh(๐), ๐+
KroneckerProduct[g,g]//MatrixForm
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11. Compare the results of
KroneckerProduct[g,g]//MatrixForm
In Q10 to ๐ท,*Cos,๐-Cosh,๐-, Sin,๐-Sinh,๐-, ๐๐ค+, **๐, ๐, ๐ค++- 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.
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14. Simplify the following by employing the suustitution properties of the Kronecker delta ๐ ๐๐๐๐๐ฟ๐๐ , ๐ ๐๐๐๐๐ฟ๐๐ ๐ฟ๐๐ ๐ ๐๐๐๐๐ฟ๐๐ ๐ฟ๐๐ ๐ ๐๐๐๐ฟ๐๐ ๐ ๐ฟ๐๐๐ฟ๐๐ ๐ ๐ฟ๐๐๐ฟ๐๐๐ฟ๐๐
15. Show that the moments of inertia ๐ฐ๐๐defined by
๐ผ11 = ๐ฆ2 + ๐ง2
๐
๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ฅ๐๐ฆ๐๐ง, ๐ผ21 = ๐ผ12 = ๐ฅ๐ฆ๐
๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ฅ๐๐ฆ๐๐ง,
๐ผ22 = ๐ง2 + ๐ฅ2
๐
๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ฅ๐๐ฆ๐๐ง, ๐ผ32 = ๐ผ23 = ๐ฆ๐ง๐
๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ฅ๐๐ฆ๐๐ง,
๐ผ31 = ๐ผ13 = ๐ฅ๐ฆ๐
๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ฅ๐๐ฆ๐๐ง,
๐ผ33 = ๐ฅ2 + ๐ฆ2
๐๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ฅ๐๐ฆ๐๐ง
can be represented in the index notation as ๐ผ๐๐ = ๐ฅ๐๐ฅ๐๐ฟ๐๐ โ ๐ฅ
๐๐ฅ๐๐
๐ ๐ฅ1, ๐ฅ2, ๐ฅ3 ๐๐ฅ1๐๐ฅ2๐๐ฅ3
where ๐ฅ = ๐ฅ1, ๐ฆ = ๐ฅ2, ๐ง = ๐ฅ3 and ๐ ๐ฅ1, ๐ฅ2, ๐ฅ3 is scalar density of the material 16. Show that the Cylindrical Polar basis vectors,
๐๐ ๐, ๐, ๐ง = cos ๐ ๐ + sin๐ ๐ ๐๐ ๐, ๐, ๐ง = โsin๐ ๐ + cos๐ ๐ ๐๐ง ๐, ๐, ๐ง = ๐
Constitute an orthonormal system. [Hint: Show their magnitudes are unity and they are pairwise orthogonal].
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17. Show that the contraction of a symmetric object with an antisymmetric object equals zero. For example given that ๐๐๐, ๐, ๐ = 1,2,3 is antisymmetric, Show that ๐๐๐๐ฅ
๐๐ฅ๐ = ๐
18. Noting that ๐๐๐๐๐๐๐ = 0 observe that ๐๐๐๐ is perfectly
antisymmetric. What does this tell about ๐๐๐?
19. For any tensor ๐, define Sym ๐๐๐=๐
๐๐จ๐๐ + ๐จ๐๐ .
Show that Sym ๐T๐๐ = ๐TSym ๐ ๐
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20. Given that, ๐๐๐ ๐ก๐๐๐๐ =
๐๐๐ ๐๐๐ ๐๐๐๐๐ ๐ ๐๐ ๐ ๐๐ ๐๐๐ก๐ ๐๐ก๐ ๐๐ก๐
Find ๐๐๐ ๐ก๐๐๐๐ and ๐๐๐๐ก๐๐๐๐.
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