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to Introduction Tensor Calculus and Continuum Mechanics by J.H. Heinbockel Department of Mathematics and Statistics Old Dominion University PREFACE This is an introductory text which presents fundamental concepts from the subject areas of tensor calculus, dierential geometry and continuum mechanics. The material presented is suitable for a two semester course in applied mathematics and is exible enough to be presented to either upper level undergraduate or beginning graduate students majoring in applied mathematics, engineering or physics. The presentation assumes the students have some knowledge from the areas of matrix theory, linear algebra and advanced calculus. Each section includes many illustrative worked examples. At the end of each section there is a large collection of exercises which range in diculty. Many new ideas are presented in the exercises and so the students should be encouraged to read all the exercises. The purpose of preparing these notes is to condense into an introductory text the basic denitions and techniques arising in tensor calculus, dierential geometry and continuum mechanics. In particular, the material is presented to (i) develop a physical understanding of the mathematical concepts associated with tensor calculus and (ii) develop the basic equations of tensor calculus, dierential geometry and continuum mechanics which arise in engineering applications. From these basic equations one can go on to develop more sophisticated models of applied mathematics. The material is presented in an informal manner and uses mathematics which minimizes excessive formalism. The material has been divided into two parts. The rst part deals with an introduction to tensor calculus and dierential geometry which covers such things as the indicial notation, tensor algebra, covariant dierentiation, dual tensors, bilinear and multilinear forms, special tensors, the Riemann Christoel tensor, space curves, surface curves, curvature and fundamental quadratic forms. The second part emphasizes the application of tensor algebra and calculus to a wide variety of applied areas from engineering and physics. The selected applications are from the areas of dynamics, elasticity, uids and electromagnetic theory. The continuum mechanics portion focuses on an introduction of the basic concepts from linear elasticity and uids. The Appendix A contains units of measurements from the Syst`me International dUnit`s along with some selected physical constants. The e e Appendix B contains a listing of Christoel symbols of the second kind associated with various coordinate systems. The Appendix C is a summary of useful vector identities. J.H. Heinbockel, 1996 Copyright c 1996 by J.H. Heinbockel. All rights reserved. Reproduction and distribution of these notes is allowable provided it is for non-prot purposes only. INTRODUCTION TO TENSOR CALCULUS AND CONTINUUM MECHANICS PART 1: INTRODUCTION TO TENSOR CALCULUS 1.1 INDEX NOTATION .............. Exercise 1.1 ..................... 1.2 TENSOR CONCEPTS AND TRANSFORMATIONS Exercise 1.2 . . . . . . . . . . . . . . . . . . . . . . 1.3 SPECIAL TENSORS . . . . . . . . . . . . . . Exercise 1.3 . . . . . . . . . . . . . . . . . . . . . . 1.4 DERIVATIVE OF A TENSOR . . . . . . . . . . Exercise 1.4 . . . . . . . . . . . . . . . . . . . . . . 1.5 DIFFERENTIAL GEOMETRY AND RELATIVITY Exercise 1.5 . . . . . . . . . . . . . . . . . . . . . . PART 2: INTRODUCTION TO CONTINUUM MECHANICS 2.1 TENSOR NOTATION FOR VECTOR QUANTITIES . . . . Exercise 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 DYNAMICS . . . . . . . . . . . . . . . . . . . . . . Exercise 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 BASIC EQUATIONS OF CONTINUUM MECHANICS . . . Exercise 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 CONTINUUM MECHANICS (SOLIDS) ......... Exercise 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 CONTINUUM MECHANICS (FLUIDS) ......... Exercise 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 ELECTRIC AND MAGNETIC FIELDS . . . . . . . . . . Exercise 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . BIBLIOGRAPHY ..................... APPENDIX A UNITS OF MEASUREMENT . . . . . . . APPENDIX B CHRISTOFFEL SYMBOLS OF SECOND KIND APPENDIX C VECTOR IDENTITIES . . . . . . . . . . INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . 171 182 187 206 211 238 243 272 282 317 325 347 352 353 355 362 363 . . . . . . . . . . .. ... .. ... .. ... .. ... .. ... . . . . . . . . . . 1 28 35 54 65 101 108 123 129 162 1 PART 1: INTRODUCTION TO TENSOR CALCULUS A scalar eld describes a one-to-one correspondence between a single scalar number and a point. An ndimensional vector eld is described by a one-to-one correspondence between n-numbers and a point. Let us generalize these concepts by assigning n-squared numbers to a single point or n-cubed numbers to a single point. When these numbers obey certain transformation laws they become examples of tensor elds. In general, scalar elds are referred to as tensor elds of rank or order zero whereas vector elds are called tensor elds of rank or order one. Closely associated with tensor calculus is the indicial or index notation. In section 1 the indicial notation is dened and illustrated. We also dene and investigate scalar, vector and tensor elds when they are subjected to various coordinate transformations. It turns out that tensors have certain properties which are independent of the coordinate system used to describe the tensor. Because of these useful properties, we can use tensors to represent various fundamental laws occurring in physics, engineering, science and mathematics. These representations are extremely useful as they are independent of the coordinate systems considered. 1.1 INDEX NOTATION Two vectors A and B can be expressed in the component form A = A1 e1 + A2 e2 + A3 e3 and B = B1 e1 + B2 e2 + B3 e3 , where e1 , e2 and e3 are orthogonal unit basis vectors. Often when no confusion arises, the vectors A and B are expressed for brevity sake as number triples. For example, we can write A = (A1 , A2 , A3 ) and B = (B1 , B2 , B3 ) where it is understood that only the components of the vectors A and B are given. The unit vectors would be represented e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). A still shorter notation, depicting the vectors A and B is the index or indicial notation. In the index notation, the quantities Ai , i = 1, 2, 3 and Bp , p = 1, 2, 3 represent the components of the vectors A and B. This notation focuses attention only on the components of the vectors and employs a dummy subscript whose range over the integers is specied. The symbol Ai refers to all of the components of the vector A simultaneously. The dummy subscript i can have any of the integer values 1, 2 or 3. For i = 1 we focus attention on the A1 component of the vector A. Setting i = 2 focuses attention on the second component A2 of the vector A and similarly when i = 3 we can focus attention on the third component of A. The subscript i is a dummy subscript and may be replaced by another letter, say p, so long as one species the integer values that this dummy subscript can have. 2 It is also convenient at this time to mention that higher dimensional vectors may be dened as ordered ntuples. For example, the vector X = (X1 , X2 , . . . , XN ) with components Xi , i = 1, 2, . . . , N is called a N dimensional vector. Another notation used to represent this vector is X = X1 e1 + X2 e2 + + XN eN where e1 , e2 , . . . , eN are linearly independent unit base vectors. Note that many of the operations that occur in the use of the index notation apply not only for three dimensional vectors, but also for N dimensional vectors. In future sections it is necessary to dene quantities which can be represented by a letter with subscripts or superscripts attached. Such quantities are referred to as systems. When these quantities obey certain transformation laws they are referred to as tensor systems. For example, quantities like Ak ij eijk ij j i Ai Bj aij . The subscripts or superscripts are referred to as indices or suxes. When such quantities arise, the indices must conform to the following rules: 1. They are lower case Latin or Greek letters. 2. The letters at the end of the alphabet (u, v, w, x, y, z) are never employed as indices. The number of subscripts and superscripts determines the order of the system. A system with one index is a rst order system. A system with two indices is called a second order system. In general, a system with N indices is called a N th order system. A system with no indices is called a scalar or zeroth order system. The type of system depends upon the number of subscripts or superscripts occurring in an expression. m For example, Ai and Bst , (all indices range 1 to N), are of the same type because they have the same jk mn number of subscripts and superscripts. In contrast, the systems Ai and Cp are not of the same type jk because one system has two superscripts and the other system has only one superscript. For certain systems the number of subscripts and superscripts is important. In other systems it is not of importance. The meaning and importance attached to sub- and superscripts will be addressed later in this section. In the use of superscripts one must not confuse powers of a quantity with the superscripts. For example, if we replace the independent variables (x, y, z) by the symbols (x1 , x2 , x3 ), then we are letting y = x2 where x2 is a variable and not x raised to a power. Similarly, the substitution z = x3 is the replacement of z by the variable x3 and this should not be confused with x raised to a power. In order to write a superscript quantity to a power, use parentheses. For example, (x2 )3 is the variable x2 cubed. One of the reasons for introducing the superscript variables is that many equations of mathematics and physics can be made to take on a concise and compact form. There is a range convention associated with the indices. This convention states that whenever there is an expression where the indices occur unrepeated it is to be understood that each of the subscripts or superscripts can take on any of the integer values 1, 2, . . . , N where N is a specied integer. For example, 3 the Kronecker delta symbol ij , dened by ij = 1 if i = j and ij = 0 for i = j, with i, j ranging over the values 1,2,3, represents the 9 quantities 11 = 1 21 = 0 31 = 0 12 = 0 22 = 1 32 = 0 13 = 0 23 = 0 33 = 1. The symbol ij refers to all of the components of the system simultaneously. As another example, consider the equation em en = mn m, n = 1, 2, 3 (1.1.1) the subscripts m, n occur unrepeated on the left side of the equation and hence must also occur on the right hand side of the equation. These indices are called free indices and can take on any of the values 1, 2 or 3 as specied by the range. Since there are three choices for the value for m and three choices for a value of n we nd that equation (1.1.1) represents nine equations simultaneously. These nine equations are e1 e1 = 1 e2 e1 = 0 e3 e1 = 0 Symmetric and Skew-Symmetric Systems A system dened by subscripts and superscripts ranging over a set of values is said to be symmetric in two of its indices if the components are unchanged when the indices are interchanged. For example, the third order system Tijk is symmetric in the indices i and k if Tijk = Tkji for all values of i, j and k. e1 e2 = 0 e2 e2 = 1 e3 e2 = 0 e1 e3 = 0 e2 e3 = 0 e3 e3 = 1. A system dened by subscripts and superscripts is said to be skew-symmetric in two of its indices if the components change sign when the indices are interchanged. For example, the fourth order system Tijkl is skew-symmetric in the indices i and l if Tijkl = Tljki for all values of ijk and l. As another example, consider the third order system aprs , p, r, s = 1, 2, 3 which is completely skewsymmetric in all of its indices. We would then have aprs = apsr = aspr = asrp = arsp = arps . It is left as an exercise to show this completely skew- symmetric systems has 27 elements, 21 of which are zero. The 6 nonzero elements are all related to one another thru the above equations when (p, r, s) = (1, 2, 3). This is expressed as saying that the above system has only one independent component. 4 Summation Convention The summation convention states that whenever there arises an expression where there is an index which 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 specied 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. Because of this rule it is sometimes necessary to replace one dummy summation symbol by some other dummy symbol in order to avoid having three or more indices occurring on the same side of the equation. The index notation is a very powerful notation and can be used to concisely represent many complex equations. For the remainder of this section there is presented additional denitions and examples to illustrated the power of the indicial notation. This notation is then employed to dene tensor components and associated operations with tensors. EXAMPLE 1.1-1 The two equations y1 = a11 x1 + a12 x2 y2 = a21 x1 + a22 x2 can be represented as one equation by introducing a dummy index, say k, and expressing the above equations as yk = ak1 x1 + ak2 x2 , k = 1, 2. The range convention states that k is free to have any one of the values 1 or 2, (k is a free index). This equation can now be written in the form 2 yk = i=1 aki xi = ak1 x1 + ak2 x2 where i is the dummy summation index. When the summation sign is removed and the summation convention is adopted we have yk = aki xi i, k = 1, 2. Since the subscript i repeats itself, the summation convention requires that a summation be performed by letting the summation subscript take on the values specied by the range and then summing the results. The index k which appears only once on the left and only once on the right hand side of the equation is called a free index. It should be noted that both k and i are dummy subscripts and can be replaced by other letters. For example, we can write yn = anm xm n, m = 1, 2 where m is the summation index and n is the free index. Summing on m produces yn = an1 x1 + an2 x2 and letting the free index n take on the values of 1 and 2 we produce the original two equations. 5 EXAMPLE 1.1-2. For yi = aij xj , i, j = 1, 2, 3 and xi = bij zj , i, j = 1, 2, 3 solve for the y variables in terms of the z variables. Solution: In matrix form the given equations can be expressed: y1 a11 y2 = a21 y3 a31 a12 a22 a32 x1 a13 a23 x2 a33 x3 and x1 b11 x2 = b21 x3 b31 b12 b22 b32 z1 b13 b23 z2 . b33 z3 Now solve for the y variables in terms of the z variables and obtain y1 a11 y2 = a21 y3 a31 a12 a22 a32 b11 a13 a23 b21 a33 b31 b12 b22 b32 z1 b13 b23 z2 . b33 z3 The index notation employs indices that are dummy indices and so we can write yn = anm xm , n, m = 1, 2, 3 and xm = bmj zj , m, j = 1, 2, 3. Here we have purposely changed the indices so that when we substitute for xm , from one equation into the other, a summation index does not repeat itself more than twice. Substituting we nd the indicial form of the above matrix equation as yn = anm bmj zj , m, n, j = 1, 2, 3 where n is the free index and m, j are the dummy summation indices. It is left as an exercise to expand both the matrix equation and the indicial equation and verify that they are dierent ways of representing the same thing. EXAMPLE 1.1-3. The dot product of two vectors Aq , q = 1, 2, 3 and Bj , j = 1, 2, 3 can be represented i = 1, 2, 3, A = |A|, B = |B|. Since the with the index notation by the product Ai Bi = AB cos specied, there results subscript i is repeated it is understood to represent a summation index. Summing on i over the range A1 B1 + A2 B2 + A3 B3 = AB cos . Observe that the index notation employs dummy indices. At times these indices are altered in order to conform to the above summation rules, without attention being brought to the change. As in this example, the indices q and j are dummy indices and can be changed to other letters if one desires. Also, in the future, if the range of the indices is not stated it is assumed that the range is over the integer values 1, 2 and 3. To systems containing subscripts and superscripts one can apply certain algebraic operations. We present in an informal way the operations of addition, multiplication and contraction. 6 Addition, Multiplication and Contraction The algebraic operation of addition or subtraction applies to systems of the same type and order. That i is we can add or subtract like components in systems. For example, the sum of Ai and Bjk is again a jk i i system of the same type and is denoted by Cjk = Ai + Bjk , where like components are added. jk The product of two systems is obtained by multiplying each component of the rst system with each component of the second system. Such a product is called an outer product. The order of the resulting product system is the sum of the orders of the two systems involved in forming the product. For example, if Ai is a second order system and B mnl is a third order system, with all indices having the range 1 to N, j imnl = Ai B mnl . The product system represents N 5 then the product system is fth order and is denoted Cj j terms constructed from all possible products of the components from Ai with the components from B mnl . j The operation of contraction occurs when a lower index is set equal to an upper index and the summation imnl convention is invoked. For example, if we have a fth order system Cj and we set i = j and sum, then we form the system jmnl 1mnl 2mnl N = C1 + C2 + + CN mnl . C mnl = Cj Here the symbol C mnl is used to represent the third order system that results when the contraction is performed. Whenever a contraction is performed, the resulting system is always of order 2 less than the original system. Under certain special conditions it is permissible to perform a contraction on two lower case indices. These special conditions will be considered later in the section. The above operations will be more formally dened after we have explained what tensors are. The e-permutation symbol and Kronecker delta Two symbols that are used quite frequently with the indicial notation are the e-permutation symbol and the Kronecker delta. The e-permutation symbol is sometimes referred to as the alternating tensor. The e-permutation symbol, as the name suggests, deals with permutations. A permutation is an arrangement of things. When the order of the arrangement is changed, a new permutation results. A transposition is an interchange of two consecutive terms in an arrangement. As an example, let us change the digits 1 2 3 to 3 2 1 by making a sequence of transpositions. Starting with the digits in the order 1 2 3 we interchange 2 and 3 (rst transposition) to obtain 1 3 2. Next, interchange the digits 1 and 3 ( second transposition) to obtain 3 1 2. Finally, interchange the digits 1 and 2 (third transposition) to achieve 3 2 1. Here the total number of transpositions of 1 2 3 to 3 2 1 is three, an odd number. Other transpositions of 1 2 3 to 3 2 1 can also be written. However, these are also an odd number of transpositions. 7 EXAMPLE 1.1-4. The total number of possible ways of arranging the digits 1 2 3 is six. We have three choices for the rst digit. Having chosen the rst digit, there are only two choices left for the second digit. Hence the remaining number is for the last digit. The product (3)(2)(1) = 3! = 6 is the number of permutations of the digits 1, 2 and 3. These six permutations are 1 2 3 even permutation 1 3 2 odd permutation 3 1 2 even permutation 3 2 1 odd permutation 2 3 1 even permutation 2 1 3 odd permutation. Here a permutation of 1 2 3 is called even or odd depending upon whether there is an even or odd number of transpositions of the digits. A mnemonic device to remember the even and odd permutations of 123 is illustrated in the gure 1.1-1. Note that even permutations of 123 are obtained by selecting any three consecutive numbers from the sequence 123123 and the odd permutations result by selecting any three consecutive numbers from the sequence 321321. Figure 1.1-1. Permutations of 123. In general, the number of permutations of n things taken m at a time is given by the relation P (n, m) = n(n 1)(n 2) (n m + 1). By selecting a subset of m objects from a collection of n objects, m n, without regard to the ordering is called a combination of n objects taken m at a time. For example, combinations of 3 numbers taken from the set {1, 2, 3, 4} are (123), (124), (134), (234). Note that ordering of a combination is not considered. That is, the permutations (123), (132), (231), (213), (312), (321) are considered equal. In general, the number of n n! n where m are the combinations of n objects taken m at a time is given by C(n, m) = = m m!(n m)! binomial coecients which occur in the expansion n (a + b)n = m=0 n anm bm . m 8 The denition of permutations can be used to dene the e-permutation symbol. Denition: (e-Permutation symbol or alternating tensor) The e-permutation symbol is dened if ijk . . . l is an even permutation of the integers 123 . . . n 1 ijk...l if ijk . . . l is an odd permutation of the integers 123 . . . n = eijk...l = 1 e 0 in all other cases EXAMPLE 1.1-5. Find e612453 . Solution: To determine whether 612453 is an even or odd permutation of 123456 we write down the given numbers and below them we write the integers 1 through 6. Like numbers are then connected by a line and we obtain gure 1.1-2. Figure 1.1-2. Permutations of 123456. In gure 1.1-2, there are seven intersections of the lines connecting like numbers. The number of intersections is an odd number and shows that an odd number of transpositions must be performed. These results imply e612453 = 1. Another denition used quite frequently in the representation of mathematical and engineering quantities is the Kronecker delta which we now dene in terms of both subscripts and superscripts. Denition: (Kronecker delta) j ij = i = The Kronecker delta is dened: 1 0 if i equals j if i is dierent from j 9 EXAMPLE 1.1-6. Some examples of the epermutation symbol and Kronecker delta are: e123 = e123 = +1 e213 = e213 = 1 e112 = e 112 1 1 = 1 1 2 = 0 1 3 12 = 0 22 = 1 32 = 0. =0 =0 EXAMPLE 1.1-7. When an index of the Kronecker delta ij is involved in the summation convention, the eect is that of replacing one index with a dierent index. For example, let aij denote the elements of an N N matrix. Here i and j are allowed to range over the integer values 1, 2, . . . , N. Consider the product aij ik where the range of i, j, k is 1, 2, . . . , N. The index i is repeated and therefore it is understood to represent a summation over the range. The index i is called a summation index. The other indices j and k are free indices. They are free to be assigned any values from the range of the indices. They are not involved in any summations and their values, whatever you choose to assign them, are xed. Let us assign a value of j and k to the values of j and k. The underscore is to remind you that these values for j and k are xed and not to be summed. When we perform the summation over the summation index i we assign values to i from the range and then sum over these values. Performing the indicated summation we obtain aij ik = a1j 1k + a2j 2k + + akj kk + + aN j N k . In this summation the Kronecker delta is zero everywhere the subscripts are dierent and equals one where the subscripts are the same. There is only one term in this summation which is nonzero. It is that term where the summation index i was equal to the xed value k This gives the result akj kk = akj where the underscore is to remind you that the quantities have xed values and are not to be summed. Dropping the underscores we write aij ik = akj Here we have substituted the index i by k and so when the Kronecker delta is used in a summation process it is known as a substitution operator. This substitution property of the Kronecker delta can be used to simplify a variety of expressions involving the index notation. Some examples are: Bij js = Bis jk km = jm eijk im jn kp = emnp . Some texts adopt the notation that if indices are capital letters, then no summation is to be performed. For example, aKJ KK = aKJ 10 as KK represents a single term because of the capital letters. Another notation which is used to denote no summation of the indices is to put parenthesis about the indices which are not to be summed. For example, a(k)j (k)(k) = akj , since (k)(k) represents a single term and the parentheses indicate that no summation is to be performed. At any time we may employ either the underscore notation, the capital letter notation or the parenthesis notation to denote that no summation of the indices is to be performed. To avoid confusion altogether, one can write out parenthetical expressions such as (no summation on k). i EXAMPLE 1.1-8. In the Kronecker delta symbol j we set j equal to i and perform a summation. This i operation is called a contraction. There results i , which is to be summed over the range of the index i. Utilizing the range 1, 2, . . . , N we have i 1 2 N i = 1 + 2 + + N i i = 1 + 1 + + 1 i i = N. i In three dimension we have j , i, j = 1, 2, 3 and k 1 2 3 k = 1 + 2 + 3 = 3. In certain circumstances the Kronecker delta can be written with only subscripts. ij , indices so that ii = 3. For example, i, j = 1, 2, 3. We shall nd that these circumstances allow us to perform a contraction on the lower EXAMPLE 1.1-9. The determinant of a matrix A = (aij ) can be represented in the indicial notation. Employing the e-permutation symbol the determinant of an N N matrix is expressed |A| = eij...k a1i a2j aN k where eij...k is an N th order system. In the special case of a 2 2 matrix we write |A| = eij a1i a2j where the summation is over the range 1,2 and the e-permutation symbol is of order 2. In the special case of a 3 3 matrix we have a11 |A| = a21 a31 a12 a22 a32 a13 a23 = eijk ai1 aj2 ak3 = eijk a1i a2j a3k a33 where i, j, k are the summation indices and the summation is over the range 1,2,3. Here eijk denotes the e-permutation symbol of order 3. Note that by interchanging the rows of the 3 3 matrix we can obtain 11 more general results. Consider (p, q, r) as some permutation of the integers (1, 2, 3), and observe that the determinant can be expressed ap1 = aq1 ar1 If (p, q, r) If (p, q, r) If (p, q, r) We can then write eijk api aqj ark = epqr |A|. Each of the above results can be veried by performing the indicated summations. A more formal proof of the above result is given in EXAMPLE 1.1-25, later in this section. ap2 aq2 ar2 ap3 aq3 = eijk api aqj ark . ar3 = |A| = |A| = 0. is an even permutation of (1, 2, 3) then is an odd permutation of (1, 2, 3) then is not a permutation of (1, 2, 3) then EXAMPLE 1.1-10. The expression eijk Bij Ci is meaningless since the index i repeats itself more than twice and the summation convention does not allow this. If you really did want to sum over an index which occurs more than twice, then one must use a summation sign. For example the above expression would be n written i=1 eijk Bij Ci . EXAMPLE 1.1-11. The cross product of the unit vectors ek ei ej = ek 0 e1 , e2 , e3 can be represented in the index notation by if (i, j, k) is an even permutation of (1, 2, 3) if (i, j, k) is an odd permutation of (1, 2, 3) in all other cases This result can be written in the form ei ej = ekij ek . This later result can be veried by summing on the index k and writing out all 9 possible combinations for i and j. EXAMPLE 1.1-12. Given the vectors Ap , p = 1, 2, 3 and Bp , p = 1, 2, 3 the cross product of these two vectors is a vector Cp , p = 1, 2, 3 with components Ci = eijk Aj Bk , i, j, k = 1, 2, 3. (1.1.2) The quantities Ci represent the components of the cross product vector C = A B = C1 e1 + C2 e2 + C3 e3 . The equation (1.1.2), which denes the components of C, is to be summed over each of the indices which repeats itself. We have summing on the index k Ci = eij1 Aj B1 + eij2 Aj B2 + eij3 Aj B3 . (1.1.3) 12 We next sum on the index j which repeats itself in each term of equation (1.1.3). This gives Ci = ei11 A1 B1 + ei21 A2 B1 + ei31 A3 B1 + ei12 A1 B2 + ei22 A2 B2 + ei32 A3 B2 + ei13 A1 B3 + ei23 A2 B3 + ei33 A3 B3 . Now we are left with i being a free index which can have any of the values of 1, 2 or 3. Letting i = 1, then letting i = 2, and nally letting i = 3 produces the cross product components C1 = A2 B3 A3 B2 C2 = A3 B1 A1 B3 C3 = A1 B2 A2 B1 . The cross product can also be expressed in the form A B = eijk Aj Bk ei . This result can be veried by summing over the indices i,j and k. (1.1.4) EXAMPLE 1.1-13. Show eijk = eikj = ejki for i, j, k = 1, 2, 3 Solution: The array i k j represents an odd number of transpositions of the indices i j k and to each transposition there is a sign change of the e-permutation symbol. Similarly, j k i is an even transposition of i j k and so there is no sign change of the e-permutation symbol. The above holds regardless of the numerical values assigned to the indices i, j, k. The e- Identity An identity relating the e-permutation symbol and the Kronecker delta, which is useful in the simplication of tensor expressions, is the e- identity. This identity can be expressed in dierent forms. The subscript form for this identity is eijk eimn = jm kn jn km , i, j, k, m, n = 1, 2, 3 where i is the summation index and j, k, m, n are free indices. A device used to remember the positions of the subscripts is given in the gure 1.1-3. The subscripts on the four Kronecker deltas on the right-hand side of the e- identity then are read (rst)(second)-(outer)(inner). This refers to the positions following the summation index. Thus, j, m are the rst indices after the summation index and k, n are the second indices after the summation index. The indices j, n are outer indices when compared to the inner indices k, m as the indices are viewed as written on the left-hand side of the identity. 13 Figure 1.1-3. Mnemonic device for position of subscripts. Another form of this identity employs both subscripts and superscripts and has the form jk jk eijk eimn = m n n m . (1.1.5) One way of proving this identity is to observe the equation (1.1.5) has the free indices j, k, m, n. Each of these indices can have any of the values of 1, 2 or 3. There are 3 choices we can assign to each of j, k, m or n and this gives a total of 34 = 81 possible equations represented by the identity from equation (1.1.5). By writing out all 81 of these equations we can verify that the identity is true for all possible combinations that can be assigned to the free indices. An alternate proof of the e identity is to consider the determinant 1 1 2 1 3 1 1 2 2 2 3 2 1 3 10 2 3 = 0 1 3 3 00 0 0 = 1. 1 By performing a permutation of the rows of this matrix we can use the permutation symbol and write i 1 j 1 k 1 i 2 j 2 k 2 i 3 j 3 = eijk . k 3 By performing a permutation of the columns, we can write i r j r k r i s j s k s i t j t = eijk erst . k t Now perform a contraction on the indices i and r to obtain i i j i k i i s j s k s i t j t = eijk eist . k t i 1 2 3 Summing on i we have i = 1 + 2 + 3 = 3 and expand the determinant to obtain the desired result jk jk s t t s = eijk eist . 14 Generalized Kronecker delta The generalized Kronecker delta is dened by the (n n) determinant i m j m =. . . k m i n j n . . . k n i p j p .. .. .. . k p ij...k mn...p For example, in three dimensions we can write ijk mnp i m j = m k m i n j n k n i p j p = eijk emnp . k p Performing a contraction on the indices k and p we obtain the fourth order system rs rsp rs rs mn = mnp = ersp emnp = eprs epmn = m n n m . As an exercise one can verify that the denition of the e-permutation symbol can also be dened in terms of the generalized Kronecker delta as 1 ej1 j2 j3 jN = j1 j2 j3 jN . 23 N Additional denitions and results employing the generalized Kronecker delta are found in the exercises. In section 1.3 we shall show that the Kronecker delta and epsilon permutation symbol are numerical tensors which have xed components in every coordinate system. Additional Applications of the Indicial Notation The indicial notation, together with the e identity, can be used to prove various vector identities. EXAMPLE 1.1-14. Solution: Let Show, using the index notation, that A B = B A C = A B = C1 e1 + C2 e2 + C3 e3 = Ci ei D = B A = D1 e1 + D2 e2 + D3 e3 = Di ei . We have shown that the components of the cross products can be represented in the index notation by Ci = eijk Aj Bk and Di = eijk Bj Ak . and let We desire to show that Di = Ci for all values of i. Consider the following manipulations: Let Bj = Bs sj and Ak = Am mk and write Di = eijk Bj Ak = eijk Bs sj Am mk (1.1.6) where all indices have the range 1, 2, 3. In the expression (1.1.6) note that no summation index appears more than twice because if an index appeared more than twice the summation convention would become meaningless. By rearranging terms in equation (1.1.6) we have Di = eijk sj mk Bs Am = eism Bs Am . 15 In this expression the indices s and m are dummy summation indices and can be replaced by any other letters. We replace s by k and m by j to obtain Di = eikj Aj Bk = eijk Aj Bk = Ci . Consequently, we nd that D = C or B A = A B. That is, D = Di ei = Ci ei = C. Note 1. The expressions Ci = eijk Aj Bk and Cm = emnp An Bp with all indices having the range 1, 2, 3, appear to be dierent because dierent letters are used as subscripts. It must be remembered that certain indices are summed according to the summation convention and the other indices are free indices and can take on any values from the assigned range. Thus, after summation, when numerical values are substituted for the indices involved, none of the dummy letters used to represent the components appear in the answer. Note 2. A second important point is that when one is working with expressions involving the index notation, the indices can be changed directly. For example, in the above expression for Di we could have replaced j by k and k by j simultaneously (so that no index repeats itself more than twice) to obtain Di = eijk Bj Ak = eikj Bk Aj = eijk Aj Bk = Ci . Note 3. Be careful in switching back and forth between the vector notation and index notation. Observe that a vector A can be represented A = Ai ei or its components can be represented A ei = Ai , i = 1, 2, 3. Do not set a vector equal to a scalar. That is, do not make the mistake of writing A = Ai as this is a misuse of the equal sign. It is not possible for a vector to equal a scalar because they are two entirely dierent quantities. A vector has both magnitude and direction while a scalar has only magnitude. EXAMPLE 1.1-15. Verify the vector identity A (B C) = B (C A) Solution: Let B C = D = Di ei C A = F = Fi ei where where Di = eijk Bj Ck Fi = eijk Cj Ak and let where all indices have the range 1, 2, 3. To prove the above identity, we have A (B C) = A D = Ai Di = Ai eijk Bj Ck = Bj (eijk Ai Ck ) = Bj (ejki Ck Ai ) 16 since eijk = ejki . We also observe from the expression Fi = eijk Cj Ak that we may obtain, by permuting the symbols, the equivalent expression Fj = ejki Ck Ai . This allows us to write A (B C) = Bj Fj = B F = B (C A) which was to be shown. The quantity A (B C) is called a triple scalar product. The above index representation of the triple scalar product implies that it can be represented as a determinant (See example 1.1-9). We can write A1 A (B C) = B1 C1 A2 B2 C2 A3 B3 = eijk Ai Bj Ck C3 A physical interpretation that can be assigned to this triple scalar product is that its absolute value represents the volume of the parallelepiped formed by the three noncoplaner vectors A, B, C. The absolute value is needed because sometimes the triple scalar product is negative. This physical interpretation can be obtained from an analysis of the gure 1.1-4. Figure 1.1-4. Triple scalar product and volume 17 In gure 1.1-4 observe that: (i) |B C| is the area of the parallelogram P QRS. (ii) the unit vector en = BC |B C| is normal to the plane containing the vectors B and C. (iii) The dot product A en = A BC |B C| =h equals the projection of A on en which represents the height of the parallelepiped. These results demonstrate that A (B C) = |B C| h = (area of base)(height) = volume. EXAMPLE 1.1-16. Verify the vector identity (A B) (C D) = C(D A B) D(C A B) Solution: Let F = A B = Fi ei and E = C D = Ei ei . These vectors have the components Fi = eijk Aj Bk and Em = emnp Cn Dp where all indices have the range 1, 2, 3. The vector G = F E = Gi ei has the components Gq = eqim Fi Em = eqim eijk emnp Aj Bk Cn Dp . From the identity eqim = emqi this can be expressed Gq = (emqi emnp )eijk Aj Bk Cn Dp which is now in a form where we can use the e identity applied to the term in parentheses to produce Gq = (qn ip qp in )eijk Aj Bk Cn Dp . Simplifying this expression we have: Gq = eijk [(Dp ip )(Cn qn )Aj Bk (Dp qp )(Cn in )Aj Bk ] = eijk [Di Cq Aj Bk Dq Ci Aj Bk ] = Cq [Di eijk Aj Bk ] Dq [Ci eijk Aj Bk ] which are the vector components of the vector C(D A B) D(C A B). 18 Transformation Equations Consider two sets of N independent variables which are denoted by the barred and unbarred symbols x and xi with i = 1, . . . , N. The independent variables xi , i = 1, . . . , N can be thought of as dening the coordinates of a point in a N dimensional space. Similarly, the independent barred variables dene a point in some other N dimensional space. These coordinates are assumed to be real quantities and are not complex quantities. Further, we assume that these variables are related by a set of transformation equations. xi = xi (x1 , x2 , . . . , xN ) i = 1, . . . , N. (1.1.7) i It is assumed that these transformation equations are independent. A necessary and sucient condition that these transformation equations be independent is that the Jacobian determinant be dierent from zero, that is xi x = J( ) = x xj x1 x1 x2 x1 xN x1 x1 x2 x2 x2 xN x2 .. . . . . . . . x1 xN x2 xN xN xN . . . = 0. This assumption allows us to obtain a set of inverse relations xi = xi (x1 , x2 , . . . , xN ) i = 1, . . . , N, (1.1.8) where the x s are determined in terms of the x s. Throughout our discussions it is to be understood that the given transformation equations are real and continuous. Further all derivatives that appear in our discussions are assumed to exist and be continuous in the domain of the variables considered. EXAMPLE 1.1-17. The following is an example of a set of transformation equations of the form dened by equations (1.1.7) and (1.1.8) in the case N = 3. Consider the transformation from cylindrical coordinates (r, , z) to spherical coordinates (, , ). From the geometry of the gure 1.1-5 we can nd the transformation equations r = sin = 0 < < 2 0<< r2 + z 2 z = cos with inverse transformation = = r = arctan( ) z Now make the substitutions (x1 , x2 , x3 ) = (r, , z) and (x1 , x2 , x3 ) = (, , ). 19 Figure 1.1-5. Cylindrical and Spherical Coordinates The resulting transformations then have the forms of the equations (1.1.7) and (1.1.8). Calculation of Derivatives We now consider the chain rule applied to the dierentiation of a function of the bar variables. We represent this dierentiation in the indicial notation. Let = (x1 , x2 , . . . , xn ) be a scalar function of the variables xi , i = 1, . . . , N and let these variables be related to the set of variables xi , with i = 1, . . . , N by the transformation equations (1.1.7) and (1.1.8). The partial derivatives of with respect to the variables xi can be expressed in the indicial notation as xj x1 x2 xN = = +2 + + N xi x1 xi x xi xj xi x xi for any xed value of i satisfying 1 i N. equation (1.1.9) partially with respect to xm produces 2 xj 2 = +m i xm x x xj xi xm xj xj . xi (1.1.10) (1.1.9) The second partial derivatives of can also be expressed in the index notation. Dierentiation of This result is nothing more than an application of the general rule for dierentiating a product of two quantities. To evaluate the derivative of the bracketed term in equation (1.1.10) it must be remembered that the quantity inside the brackets is a function of the bar variables. Let = G(x1 , x2 , . . . , xN ) G= xj to emphasize this dependence upon the bar variables, then the derivative of G is G G xk 2 xk = = . (1.1.11) xm xk xm xj xk xm This is just an application of the basic rule from equation (1.1.9) with replaced by G. Hence the derivative from equation (1.1.10) can be expressed 2 xj 2 xj xk 2 = j xi xm + i xm x x xj xk xi xm where i, m are free indices and j, k are dummy summation indices. (1.1.12) 20 Let = (r, ) where r, are polar coordinates related to the Cartesian coordinates 2 and (x, y) by the transformation equations x = r cos y = r sin . Find the partial derivatives x x2 Solution: The partial derivative of with respect to x is found from the relation (1.1.9) and can be written r = + . x r x x (1.1.13) EXAMPLE 1.1-18. The second partial derivative is obtained by dierentiating the rst partial derivative. From the product rule for dierentiation we can write 2 2 r r 2 = + + + . 2 2 2 x r x x x r x x x (1.1.14) To further simplify (1.1.14) it must be remembered that the terms inside the brackets are to be treated as functions of the variables r and and that the derivative of these terms can be evaluated by reapplying the basic rule from equation (1.1.13) with replaced by r and then replaced by . This gives 2 2 2 r r 2 r + = + 2 2 2 x x r x x r r x 2 2 r 2 + + + . 2 x x r x 2 x From the transformation equations we obtain the relations r2 = x2 + y 2 and tan = (1.1.15) y and from x these relations we can calculate all the necessary derivatives needed for the simplication of the equations (1.1.13) and (1.1.15). These derivatives are: r = 2x or x y = 2 or sec2 x x sin2 2r = = sin x2 x r 2r r x = = cos x r y sin = 2 = x r r r r cos x + sin x 2 2 sin cos = = . x2 r2 r2 Therefore, the derivatives from equations (1.1.13) and (1.1.15) can be expressed in the form sin = cos x r r 2 2 sin cos 2 sin 2 cos sin 2 sin2 +2 + = + cos2 2 . 2 2 2 x r r r r r r 2 r2 By letting x1 = r, x2 = , x1 = x, x2 = y and performing the indicated summations in the equations (1.1.9) and (1.1.12) there is produced the same results as above. Vector Identities in Cartesian Coordinates Employing the substitutions x1 = x, x2 = y, x3 = z, where superscript variables are employed and denoting the unit vectors in Cartesian coordinates by e1 , e2 , e3 , we illustrated how various vector operations are written by using the index notation. 21 Gradient. In Cartesian coordinates the gradient of a scalar eld is grad = e1 + e2 + e3 . x y z The index notation focuses attention only on the components of the gradient. In Cartesian coordinates these components are represented using a comma subscript to denote the derivative ej grad = ,j = , xj j = 1, 2, 3. The comma notation will be discussed in section 4. For now we use it to denote derivatives. For example 2 , ,jk = , etc. ,j = j x xj xk Divergence. represented A = div A = A2 A3 A1 + + . x y z In Cartesian coordinates the divergence of a vector eld A is a scalar eld and can be Employing the summation convention and index notation, the divergence in Cartesian coordinates can be represented A = div A = Ai,i = where i is the dummy summation index. Curl. To represent the vector B = curl A = A in Cartesian coordinates, we note that the index notation focuses attention only on the components of this vector. The components Bi , i = 1, 2, 3 of B can be represented Bi = ei curl A = eijk Ak,j , for i, j, k = 1, 2, 3 Ak xj . Ai A1 A2 A3 = + + xi x1 x2 x3 where eijk is the permutation symbol introduced earlier and Ak,j = To verify this representation of the curl A we need only perform the summations indicated by the repeated indices. We have summing on j that Bi = ei1k Ak,1 + ei2k Ak,2 + ei3k Ak,3 . Now summing each term on the repeated index k gives us Bi = ei12 A2,1 + ei13 A3,1 + ei21 A1,2 + ei23 A3,2 + ei31 A1,3 + ei32 A2,3 Here i is a free index which can take on any of the values 1, 2 or 3. Consequently, we have For For For i = 1, i = 2, i = 3, B1 = A3,2 A2,3 = B2 = A1,3 A3,1 B3 = A2,1 A1,2 A3 A2 x2 x3 A1 A3 = x3 x1 A2 A1 = x1 x2 which veries the index notation representation of curl A in Cartesian coordinates. 22 Other Operations. The following examples illustrate how the index notation can be used to represent additional vector operators in Cartesian coordinates. 1. In index notation the components of the vector (B )A are {(B )A} ep = Ap,q Bq p, q = 1, 2, 3 This can be veried by performing the indicated summations. We have by summing on the repeated index q Ap,q Bq = Ap,1 B1 + Ap,2 B2 + Ap,3 B3 . The index p is now a free index which can have any of the values 1, 2 or 3. We have: for p = 1, A1,q Bq = A1,1 B1 + A1,2 B2 + A1,3 B3 A1 A1 A1 = B1 + B2 + B3 x1 x2 x3 A2,q Bq = A2,1 B1 + A2,2 B2 + A2,3 B3 A2 A2 A2 = B+ B+ B3 11 22 x x x3 A3,q Bq = A3,1 B1 + A3,2 B2 + A3,3 B3 A3 A3 A3 = B+ B+ B3 11 22 x x x3 for p = 2, for p = 3, 2. The scalar (B ) has the following form when expressed in the index notation: (B ) = Bi ,i = B1 ,1 + B2 ,2 + B3 ,3 = B1 1 + B2 2 + B3 3 . x x x 3. The components of the vector (B ) is expressed in the index notation by ei (B ) = eijk Bj ,k . This can be veried by performing the indicated summations and is left as an exercise. 4. The scalar (B ) A may be expressed in the index notation. It has the form (B ) A = eijk Bj Ai,k . This can also be veried by performing the indicated summations and is left as an exercise. 5. The vector components of 2 A in the index notation are represented ep 2 A = Ap,qq . The proof of this is left as an exercise. 23 EXAMPLE 1.1-19. In Cartesian coordinates prove the vector identity curl (f A) = (f A) = (f ) A + f ( A). Solution: Let B = curl (f A) and write the components as Bi = eijk (f Ak ),j = eijk [f Ak,j + f,j Ak ] = f eijk Ak,j + eijk f,j Ak . This index form can now be expressed in the vector form B = curl (f A) = f ( A) + (f ) A EXAMPLE 1.1-20. Prove the vector identity (A + B) = A + B Solution: Let A + B = C and write this vector equation in the index notation as Ai + Bi = Ci . We then have C = Ci,i = (Ai + Bi ),i = Ai,i + Bi,i = A + B. EXAMPLE 1.1-21. In Cartesian coordinates prove the vector identity (A )f = A f Solution: In the index notation we write (A )f = Ai f,i = A1 f,1 + A2 f,2 + A3 f,3 f f f = A1 1 + A2 2 + A3 3 = A f. x x x EXAMPLE 1.1-22. In Cartesian coordinates prove the vector identity (A B) = A( B) B( A) + (B )A (A )B Solution: The pth component of the vector (A B) is ep [ (A B)] = epqk [ekji Aj Bi ],q = epqk ekji Aj Bi,q + epqk ekji Aj,q Bi By applying the e identity, the above expression simplies to the desired result. That is, ep [ (A B)] = (pj qi pi qj )Aj Bi,q + (pj qi pi qj )Aj,q Bi = Ap Bi,i Aq Bp,q + Ap,q Bq Aq,q Bp In vector form this is expressed (A B) = A( B) (A )B + (B )A B( A) 24 EXAMPLE 1.1-23. In Cartesian coordinates prove the vector identity ( A) = ( A) 2 A Solution: We have for the ith component of A is given by ei [ A] = eijk Ak,j and consequently the pth component of ( A) is ep [ ( A)] = epqr [erjk Ak,j ],q = epqr erjk Ak,jq . The e identity produces ep [ ( A)] = (pj qk pk qj )Ak,jq = Ak,pk Ap,qq . Expressing this result in vector form we have ( A) = ( A) 2 A. Indicial Form of Integral Theorems The divergence theorem, in both vector and indicial notation, can be written div F d = V S F n d V Fi,i d = S Fi ni d i = 1, 2, 3 (1.1.16) where ni are the direction cosines of the unit exterior normal to the surface, d is a volume element and d is an element of surface area. Note that in using the indicial notation the volume and surface integrals are to be extended over the range specied by the indices. This suggests that the divergence theorem can be applied to vectors in ndimensional spaces. The vector form and indicial notation for the Stokes theorem are ( F ) n d = S C F dr S eijk Fk,j ni d = C Fi dxi i, j, k = 1, 2, 3 (1.1.17) and the Greens theorem in the plane, which is a special case of the Stokes theorem, can be expressed F2 F1 x y dxdy = C F1 dx + F2 dy S e3jk Fk,j dS = C Fi dxi i, j, k = 1, 2 (1.1.18) Other forms of the above integral theorems are d = V S n d obtained from the divergence theorem by letting F = C where C is a constant vector. By replacing F by F C in the divergence theorem one can derive F V d = S F n d. In the divergence theorem make the substitution F = to obtain (2 + () () d = V S () n d. 25 The Greens identity 2 2 d = V S ( ) n d is obtained by rst letting F = in the divergence theorem and then letting F = in the divergence theorem and then subtracting the results. Determinants, Cofactors For A = (aij ), i, j = 1, . . . , n an n n matrix, the determinant of A can be written as det A = |A| = ei1 i2 i3 ...in a1i1 a2i2 a3i3 . . . anin . This gives a summation of the n! permutations of products formed from the elements of the matrix A. The result is a single number called the determinant of A. EXAMPLE 1.1-24. In the case n = 2 we have |A| = a11 a21 a12 = enm a1n a2m a22 = e1m a11 a2m + e2m a12 a2m = e12 a11 a22 + e21 a12 a21 = a11 a22 a12 a21 EXAMPLE 1.1-25. In the case a11 A = a21 a31 n = 3 we can use either of the notations 1 a12 a13 a1 a1 a1 2 3 a22 a23 or A = a2 a2 a2 1 2 3 a32 a33 a3 a3 a3 1 2 3 and represent the determinant of A in any of the forms det A = eijk a1i a2j a3k det A = eijk ai1 aj2 ak3 det A = eijk ai aj ak 123 det A = eijk a1 a2 a3 . ijk These represent row and column expansions of the determinant. An important identity results if we examine the quantity Brst = eijk ai aj ak . It is an easy exercise to rst change the dummy summation indices and rearrange terms in this expression. For example, Brst = eijk ai aj ak = ekji ak aj ai = ekji ai aj ak = eijk ai aj ak = Btsr , rst rst tsr tsr and by considering other permutations of the indices, one can establish that Brst is completely skewsymmetric. In the exercises it is shown that any third order completely skew-symmetric system satises Brst = B123 erst . But B123 = det A and so we arrive at the identity Brst = eijk ai aj ak = |A|erst . rst 26 Other forms of this identity are eijk ar as at = |A|erst ijk and eijk air ajs akt = |A|erst . (1.1.19) Consider the representation of the determinant a1 1 |A| = a2 1 a3 1 a1 2 a2 2 a3 2 a1 3 a2 3 a3 3 by use of the indicial notation. By column expansions, this determinant can be represented |A| = erst ar as at 123 and if one uses row expansions the determinant can be expressed as |A| = eijk a1 a2 a3 . ijk (1.1.21) (1.1.20) Dene Ai as the cofactor of the element am in the determinant |A|. From the equation (1.1.20) the cofactor m i of ar is obtained by deleting this element and we nd 1 A1 = erst as at . r 23 The result (1.1.20) can then be expressed in the form |A| = ar A1 = a1 A1 + a2 A1 + a3 A1 . 1r 11 12 13 (1.1.23) (1.1.22) That is, the determinant |A| is obtained by multiplying each element in the rst column by its corresponding cofactor and summing the result. Observe also that from the equation (1.1.20) we nd the additional cofactors A2 = erst ar at s 13 and A3 = erst ar as . t 12 (1.1.24) Hence, the equation (1.1.20) can also be expressed in one of the forms |A| = as A2 = a1 A2 + a2 A2 + a3 A2 2s 21 22 23 |A| = at A3 = a1 A3 + a2 A3 + a3 A3 3t 31 32 33 The results from equations (1.1.22) and (1.1.24) can be written in a slightly dierent form with the indicial notation. From the notation for a generalized Kronecker delta dened by ijk eijk elmn = lmn , the above cofactors can be written in the form 1 1jk 1 1jk e erst as at = rst as at jk jk 2! 2! 1 2jk 1 2jk s t A2 = e123 esrt as at = e erst as at = rst aj ak r 13 jk 2! 2! 1 3jk 1 3jk 3 123 ts st Ar = e etsr a1 a2 = e erst aj ak = rst as at . jk 2! 2! A1 = e123 erst as at = r 23 27 These cofactors are then combined into the single equation Ai = r 1 ijk s t aa 2! rst j k (1.1.25) which represents the cofactor of ar . When the elements from any row (or column) are multiplied by their i corresponding cofactors, and the results summed, we obtain the value of the determinant. Whenever the elements from any row (or column) are multiplied by the cofactor elements from a dierent row (or column), and the results summed, we get zero. This can be illustrated by considering the summation am Ai = r m 1 ijk s t m 1 mst aj ak ar = eijk emst am as at r jk 2! 2! 1 ijk 1 ijk i = e erjk |A| = rjk |A| = r |A| 2! 2! Here we have used the e identity to obtain ijk ik ik i i i rjk = eijk erjk = ejik ejrk = r k k r = 3r r = 2r which was used to simplify the above result. As an exercise one can show that an alternate form of the above summation of elements by its cofactors is r ar Am = |A|i . mi EXAMPLE 1.1-26. j In N-dimensions the quantity k1 j22...jN is called a generalized Kronecker delta. It 1 k ...kN can be dened in terms of permutation symbols as j ej1 j2 ...jN ek1 k2 ...kN = k1 j22...jN 1 k ...kN (1.1.26) Observe that j k1 j22...jN ek1 k2 ...kN = (N !) ej1 j2 ...jN 1 k ...kN This follows because ek1 k2 ...kN is skew-symmetric in all pairs of its superscripts. The left-hand side denotes a summation of N ! terms. The rst term in the summation has superscripts j1 j2 . . . jN and all other terms have superscripts which are some permutation of this ordering with minus signs associated with those terms having an odd permutation. Because ej1 j2 ...jN is completely skew-symmetric we nd that all terms in the summation have the value +ej1 j2 ...jN . We thus obtain N ! of these terms. 28 EXERCISE 1.1 1. Simplify each of the following by employing the summation property of the Kronecker delta. Perform sums on the summation indices only if your are unsure of the result. (a) eijk kn (b) eijk is jm 2. (c) eijk is jm kn (d) aij in (e) ij jn (f ) ij jn ni Simplify and perform the indicated summations over the range 1, 2, 3 (a) ii (c) eijk Ai Aj Ak (d) eijk eijk (e) eijk jk (f ) Ai Bj ji Bm An mn (b) ij ij 3. Express each of the following in index notation. Be careful of the notation you use. Note that A = Ai is an incorrect notation because a vector can not equal a scalar. The notation A ei = Ai should be used to express the ith component of a vector. (a) A (B C) (b) A (B C) 4. 5. 6. Show the e permutation symbol satises: (a) (c) B(A C) (d) B(A C) C(A B) (b) eijk = ejik = eikj = ekji eijk = ejki = ekij Use index notation to verify the vector identity A (B C) = B(A C) C(A B) Let yi = aij xj and xm = aim zi where the range of the indices is 1, 2 (a) Solve for yi in terms of zi using the indicial notation and check your result to be sure that no index repeats itself more than twice. (b) Perform the indicated summations and write out expressions for y1 , y2 in terms of z1 , z2 (c) Express the above equations in matrix form. Expand the matrix equations and check the solution obtained in part (b). 7. 8. Use the e identity to simplify (a) Prove the following vector identities: (a) eijk ejik (b) eijk ejki A (B C) = B (C A) = C (A B) triple scalar product (b) (A B) C = B(A C) A(B C) 9. Prove the following vector identities: (a) (A B) (C D) = (A C)(B D) (A D)(B C) (b) A (B C) + B (C A) + C (A B) = 0 (c) (A B) (C D) = B(A C D) A(B C D) 29 10. For A = (1, 1, 0) and B = (4, 3, 2) nd using the index notation, (a) Ci = eijk Aj Bk , i = 1, 2, 3 (b) Ai Bi (c) What do the results in (a) and (b) represent? dy1 = a11 y1 + a12 y2 dt dy2 = a21 y1 + a22 y2 dt 11. Represent the dierential equations and using the index notation. 12. Let = (r, ) where r, are polar coordinates related to Cartesian coordinates (x, y) by the transforand y = r sin . 2 , and (a) Find the partial derivatives y y 2 (b) Combine the result in part (a) with the result from EXAMPLE 1.1-18 to calculate the Laplacian 2 = in polar coordinates. 13. (Index notation) Let a11 = 3, a12 = 4, a21 = 5, a22 = 6. 2 2 + x2 y 2 mation equations x = r cos Calculate the quantity C = aij aij , i, j = 1, 2. 14. Show the moments of inertia Iij dened by I11 = R (y 2 + z 2 )(x, y, z) d (x + z )(x, y, z) d R 2 2 I23 = I32 = R yz(x, y, z) d xy(x, y, z) d R I22 = I33 = R I12 = I21 = I13 = I31 = R (x2 + y 2 )(x, y, z) d xz(x, y, z) d, can be represented in the index notation as Iij = R xm xm ij xi xj d, where is the density, x1 = x, x2 = y, x3 = z and d = dxdydz is an element of volume. 15. Determine if the following relation is true or false. Justify your answer. ei ( ej ek ) = ( ei ej ) ek = eijk , Hint: Let em = (1m , 2m , 3m ). 16. Without substituting values for i, l = 1, 2, 3 calculate all nine terms of the given quantities (a) 17. i i B il = (j Ak + k Aj )ejkl i, j, k = 1, 2, 3. (b) m k Ail = (i B k + i B m )emlk Let Amn xm y n = 0 for arbitrary xi and y i , i = 1, 2, 3, and show that Aij = 0 for all values of i, j. 30 18. (a) For amn , m, n = 1, 2, 3 skew-symmetric, show that amn xm xn = 0. (b) Let amn xm xn = 0, symmetric. 19. Let A and B denote 3 3 matrices with elements aij and bij respectively. Show that if C = AB is a Hint: Use the result from example 1.1-9. 20. (a) Let u1 , u2 , u3 be functions of the variables s1 , s2 , s3 . Further, assume that s1 , s2 , s3 are in turn each um (u1 , u2 , u3 ) = denote the Jacobian of the u s with functions of the variables x1 , x2 , x3 . Let xn (x1 , x2 , x3 ) respect to the x s. Show that ui sj ui sj ui = = . xm sj xm sj xm (b) Note that xi xj xi i x = = m and show that J( x )J( x ) = 1, where J( x ) is the Jacobian determinant x x j xm x xm of the transformation (1.1.7). A third order system a = am =a mn m, n = 1, 2, 3 for all values of xi , i = 1, 2, 3 and show that amn must be skew- matrix product, then det(C) = det(A) det(B). 21. a with , m, n = 1, 2, 3 is said to be symmetric in two of its subscripts if the mn components are unaltered when these subscripts are interchanged. When a mn n nm is completely symmetric then = amn = anm = an m. Whenever this third order system is completely symmetric, (ii) =m=n (iii) = m = n. then: (i) How many components are there? (ii) How many of these components are distinct? Hint: Consider the three cases (i) = m = n 22. A third order system b mn with , m, n = 1, 2, 3 is said to be skew-symmetric in two of its subscripts = bmn = bnm = bn = b nm . if the components change sign when the subscripts are interchanged. A completely skew-symmetric third order system satises b mn = bm n m (i) How many components does a completely skew-symmetric system have? (ii) How many of these components are zero? (iii) How many components can be dierent from zero? (iv) Show that there is one distinct component b123 and that b mn =e mn b123 . Hint: Consider the three cases (i) = m = n 23. (ii) =m=n (iii) = m = n. Let i, j, k = 1, 2, 3 and assume that eijk jk = 0 for all values of i. What does this equation tell you about the values ij , i, j = 1, 2, 3? 24. Assume that Amn and Bmn are symmetric for m, n = 1, 2, 3. Let Amn xm xn = Bmn xm xn for arbitrary values of xi , i = 1, 2, 3, and show that Aij = Bij for all values of i and j. 25. Assume Bmn is symmetric and Bmn xm xn = 0 for arbitrary values of xi , i = 1, 2, 3, show that Bij = 0. 31 26. (Generalized Kronecker delta) i m j m =. . . k m i n j n . . . k n Dene the generalized Kronecker delta as the n n determinant i p j p . .. .. . k p ij...k mn...p r where s is the Kronecker delta. (a) Show 123 eijk = ijk ijk eijk = 123 ij mn = eij emn rs rsp mn = mnp (b) Show (c) (d) Show Dene (summation on p) and show rs rs rs mn = m n n m Note that by combining the above result with the result from part (c) we obtain the two dimensional form of the e identity r rn (e) Dene m = 1 mn 2 rs rs ers emn = m n n m . (summation on n) and show rst r pst = 2p (f ) Show rst rst = 3! 27. Let Ai r denote the cofactor of ar i a1 1 in the determinant a2 1 a3 1 a1 2 a2 2 a3 2 a1 3 a2 as given by equation (1.1.25). 3 a3 3 (a) Show erst Ai = eijk as at r jk (b) Show erst Ar = eijk aj ak i st 28. (a) Show that if Aijk = Ajik , i, j, k = 1, 2, 3 there is a total of 27 elements, but only 18 are distinct. (b) Show that for i, j, k = 1, 2, . . . , N there are N 3 elements, but only N 2 (N + 1)/2 are distinct. 29. 30. (a) (b) (c) (d) 31. For For For For A = (aij ), i, j = 1, 2, 3, A = (ai ), i, j = 1, 2, 3, j A = (ai ), i, j = 1, 2, 3, j i I = (j ), i, j = 1, 2, 3, Let aij = Bi Bj for i, j = 1, 2, 3 where B1 , B2 , B3 are arbitrary constants. Calculate det(aij ) = |A|. |A| = eijk ai1 aj2 ak3 . |A| = eijk ai aj ak . 123 |A| = eijk a1 a2 a3 . ijk |I| = 1. show show show show Let |A| = eijk ai1 aj2 ak3 and dene Aim as the cofactor of aim . Show the determinant can be expressed in any of the forms: (a) |A| = Ai1 ai1 (b) |A| = Aj2 aj2 (c) |A| = Ak3 ak3 where Ai1 = eijk aj2 ak3 where Ai2 = ejik aj1 ak3 where Ai3 = ejki aj1 ak2 32 32. Show the results in problem 31 can be written in the forms: 1 e1st eijk ajs akt , 2! Ai2 = 1 e2st eijk ajs akt , 2! Ai3 = 1 e3st eijk ajs akt , 2! or Aim = 1 emst eijk ajs akt 2! Ai1 = 33. Use the results in problems 31 and 32 to prove that apm Aim = |A|ip . 1 Let (aij ) = 1 2 Let 21 0 3 and calculate C = aij aij , i, j = 1, 2, 3. 32 a111 = 1, a211 = 1, a112 = 3, a212 = 5, a121 = 4, a221 = 2, a122 = 2 a222 = 2 34. 35. and calculate the quantity C = aijk aijk , i, j, k = 1, 2. 36. Let a1111 = 2, a1211 = 5, a2111 = 1, a2211 = 2, a1112 = 1, a1212 = 2, a2112 = 0, a2212 = 1, a1121 = 3, a1221 = 4, a2121 = 2, a2221 = 2, a1122 = 1 a1222 = 2 a2122 = 1 a2222 = 2 and calculate the quantity C = aijkl aijkl , i, j, k, l = 1, 2. 37. Simplify the expressions: (a) (Aijkl + Ajkli + Aklij + Alijk )xi xj xk xl (b) (Pijk + Pjki + Pkij )xi xj xk (c) (d) xi xj aij 2 xi xj 2 xm xi r ami t s x x x xs xt xr 38. Let g denote the determinant of the matrix having the components gij , i, j = 1, 2, 3. Show that g1r (a) g erst = g2r g3r g1s g2s g3s g1t g2t g3t i n j n k n i p j p k p gir (b) g erst eijk = gjr gkr gis gjs gks git gjt gkt 39. 40. Show that e ijk emnp = ijk mnp i m j = m k m Show that eijk emnp Amnp = Aijk Aikj + Akij Ajik + Ajki Akji Hint: Use the results from problem 39. 41. Show that (a) (b) eij eij = 2! eijk eijk = 3! (c) (d) eijkl eijkl = 4! Guess at the result ei1 i2 ...in ei1 i2 ...in 33 42. 43. Determine if the following statement is true or false. Justify your answer. eijk Ai Bj Ck = eijk Aj Bk Ci . Let aij , i, j = 1, 2 denote the components of a 2 2 matrix A, which are functions of time t. a a12 to verify that these representations are the same. (a) Expand both |A| = eij ai1 aj2 and |A| = 11 a21 a22 (b) Verify the equivalence of the derivative relations d|A| dai1 daj2 = eij aj2 + eij ai1 dt dt dt and d|A| = dt da11 dt a21 da12 dt a22 + a11 da21 dt a12 da22 dt (c) Let aij , i, j = 1, 2, 3 denote the components of a 3 3 matrix A, which are functions of time t. Develop appropriate relations, expand them and verify, similar to parts (a) and (b) above, the representation of a determinant and its derivative. 44. For f = f (x1 , x2 , x3 ) and = (f ) dierentiable scalar functions, use the indicial notation to nd a formula to calculate grad . Use the indicial notation to prove (a) = 0 (b) A = 0 45. 46. 47. Amn 48. If Aij is symmetric and Bij is skew-symmetric, i, j = 1, 2, 3, then calculate C = Aij Bij . Assume Aij = Aij (x1 , x2 , x3 ) and Aij = Aij (x1 , x2 , x3 ) for i, j = 1, 2, 3 are related by the expression xi xj Amn = Aij m n . Calculate the derivative . x x xk Prove that if any two rows (or two columns) of a matrix are interchanged, then the value of the determinant of the matrix is multiplied by minus one. Construct your proof using 3 3 matrices. 49. Prove that if two rows (or columns) of a matrix are proportional, then the value of the determinant of the matrix is zero. Construct your proof using 3 3 matrices. 50. Prove that if a row (or column) of a matrix is altered by adding some constant multiple of some other row (or column), then the value of the determinant of the matrix remains unchanged. Construct your proof using 3 3 matrices. 51. 52. Simplify the expression = eijk e mn Ai Ajm Akn . Let Aijk denote a third order system where i, j, k = 1, 2. (a) How many components does this system have? (b) Let Aijk be skew-symmetric in the last pair of indices, how many independent components does the system have? 53. Let Aijk denote a third order system where i, j, k = 1, 2, 3. (a) How many components does this system have? (b) In addition let Aijk = Ajik and Aikj = Aijk and determine the number of distinct nonzero components for Aijk . 34 54. Show that every second order system Tij can be expressed as the sum of a symmetric system Aij and skew-symmetric system Bij . Find Aij and Bij in terms of the components of Tij . 55. Consider the system Aijk , i, j, k = 1, 2, 3, 4. (a) How many components does this system have? (b) Assume Aijk is skew-symmetric in the last pair of indices, how many independent components does this system have? (c) Assume that in addition to being skew-symmetric in the last pair of indices, Aijk + Ajki + Akij = 0 is satised for all values of i, j, and k, then how many independent components does the system have? 56. (a) Write the equation of a line r = r0 + t A in indicial form. (b) Write the equation of the plane n (r r0 ) = 0 in indicial form. (c) Write the equation of a general line in scalar form. (d) Write the equation of a plane in scalar form. (e) Find the equation of the line dened by the intersection of the planes 2x + 3y + 6z = 12 and 6x + 3y + z = 6. (f) Find the equation of the plane through the points (5, 3, 2), (3, 1, 5), (1, 3, 3). Find also the normal to this plane. 57. The angle 0 between two skew lines in space is dened as the angle between their direction vectors when these vectors are placed at the origin. Show that for two lines with direction numbers ai and bi i = 1, 2, 3, the cosine of the angle between these lines satises ai b i cos = ai ai b i b i Let aij = aji for i, j = 1, 2, . . . , N and prove that for N odd det(aij ) = 0. 2 (b) Let = Aij xi xj where Aij = Aji and calculate (a) xm xm xk Given an arbitrary nonzero vector Uk , k = 1, 2, 3, dene the matrix elements aij = eijk Uk , where eijk 58. 59. 60. is the e-permutation symbol. Determine if aij is symmetric or skew-symmetric. Suppose Uk is dened by the above equation for arbitrary nonzero aij , then solve for Uk in terms of the aij . 61. If Aij = Ai Bj = 0 for all i, j values and Aij = Aji for i, j = 1, 2, . . . , N , show that Aij = Bi Bj where is a constant. State what is. 62. Assume that Aijkm , with i, j, k, m = 1, 2, 3, is completely skew-symmetric. How many independent components does this quantity have? 63. Consider Rijkm , i, j, k, m = 1, 2, 3, 4. (a) How many components does this quantity have? (b) If Rijkm = Rijmk = Rjikm then how many independent components does Rijkm have? (c) If in addition Rijkm = Rkmij determine the number of independent components. Let xi = aij xj , i, j = 1, 2, 3 denote a change of variables from a barred system of coordinates to an unbarred system of coordinates and assume that Ai = aij Aj where aij are constants, Ai is a function of the i A . xj variables and Aj is a function of the xj variables. Calculate xm 64. 35 1.2 TENSOR CONCEPTS AND TRANSFORMATIONS For e1 , e2 , e3 independent orthogonal unit vectors (base vectors), we may write any vector A as A = A1 e1 + A2 e2 + A3 e3 where (A1 , A2 , A3 ) are the coordinates of A relative to the base vectors chosen. These components are the projection of A onto the base vectors and A = (A e1 ) e1 + (A e2 ) e2 + (A e3 ) e3 . Select any three independent orthogonal vectors, (E1 , E2 , E3 ), not necessarily of unit length, we can then write e1 = E1 |E1 | , e2 = E2 |E2 | , e3 = E3 |E3 | , and consequently, the vector A can be expressed as A= Here we say that A E(i) E(i) E(i) , i = 1, 2, 3 A E1 E1 E1 E1 + A E2 E2 E2 E2 + A E3 E3 E3 E3 . are the components of A relative to the chosen base vectors E1 , E2 , E3 . Recall that the parenthesis about the subscript i denotes that there is no summation on this subscript. It is then treated as a free subscript which can have any of the values 1, 2 or 3. Reciprocal Basis Consider a set of any three independent vectors (E1 , E2 , E3 ) which are not necessarily orthogonal, nor of unit length. In order to represent the vector A in terms of these vectors we must nd components (A1 , A2 , A3 ) such that A = A1 E1 + A2 E2 + A3 E3 . This can be done by taking appropriate projections and obtaining three equations and three unknowns from which the components are determined. A much easier way to nd the components (A1 , A2 , A3 ) is to construct a reciprocal basis (E 1 , E 2 , E 3 ). Recall that two bases (E1 , E2 , E3 ) and (E 1 , E 2 , E 3 ) are said to be reciprocal if they satisfy the condition j Ei E j = i = 1 Note that E2 E 1 = 2 = 0 1 0 if i = j . if i = j 1 and E3 E 1 = 3 = 0 so that the vector E 1 is perpendicular to both the vectors E2 and E3 . (i.e. A vector from one basis is orthogonal to two of the vectors from the other basis.) We can therefore write E 1 = V 1 E2 E3 where V is a constant to be determined. By taking the dot product of both sides of this equation with the vector E1 we nd that V = E1 (E2 E3 ) is the volume of the parallelepiped formed by the three vectors E1 , E2 , E3 when their origins are made to coincide. In a 36 similar manner it can be demonstrated that for (E1 , E2 , E3 ) a given set of basis vectors, then the reciprocal basis vectors are determined from the relations 1 1 1 E 1 = E2 E3 , E 2 = E3 E1 , E 3 = E1 E2 , V V V where V = E1 (E2 E3 ) = 0 is a triple scalar product and represents the volume of the parallelepiped having the basis vectors for its sides. Let (E1 , E2 , E3 ) and (E 1 , E 2 , E 3 ) denote a system of reciprocal bases. We can represent any vector A with respect to either of these bases. If we select the basis (E1 , E2 , E3 ) and represent A in the form A = A1 E1 + A2 E2 + A3 E3 , (1.2.1) then the components (A1 , A2 , A3 ) of A relative to the basis vectors (E1 , E2 , E3 ) are called the contravariant components of A. These components can be determined from the equations A E 1 = A1 , A E 2 = A2 , A E 3 = A3 . Similarly, if we choose the reciprocal basis (E 1 , E 2 , E 3 ) and represent A in the form A = A1 E 1 + A2 E 2 + A3 E 3 , (1.2.2) then the components (A1 , A2 , A3 ) relative to the basis (E 1 , E 2 , E 3 ) are called the covariant components of A. These components can be determined from the relations A E1 = A1 , A E2 = A2 , A E3 = A3 . The contravariant and covariant components are dierent ways of representing the same vector with respect to a set of reciprocal basis vectors. There is a simple relationship between these components which we now develop. We introduce the notation Ei Ej = gij = gji , and E i E j = g ij = g ji (1.2.3) where gij are called the metric components of the space and g ij are called the conjugate metric components of the space. We can then write A E1 = A1 (E 1 E1 ) + A2 (E 2 E1 ) + A3 (E 3 E1 ) = A1 A E1 = A1 (E1 E1 ) + A2 (E2 E1 ) + A3 (E3 E1 ) = A1 or A1 = A1 g11 + A2 g12 + A3 g13 . In a similar manner, by considering the dot products A E2 and A E3 one can establish the results A2 = A1 g21 + A2 g22 + A3 g23 A3 = A1 g31 + A2 g32 + A3 g33 . (1.2.4) These results can be expressed with the index notation as Ai = gik Ak . Forming the dot products A E 1 , A E2, A E 3 it can be veried that Ai = g ik Ak . (1.2.7) (1.2.6) The equations (1.2.6) and (1.2.7) are relations which exist between the contravariant and covariant components of the vector A. Similarly, if for some value j we have E j = E1 + E2 + E3 , then one can show that E j = g ij Ei . This is left as an exercise. 37 Coordinate Transformations Consider a coordinate transformation from a set of coordinates (x, y, z) to (u, v, w) dened by a set of transformation equations x = x(u, v, w) y = y(u, v, w) z = z(u, v, w) It is assumed that these transformations are single valued, continuous and possess the inverse transformation u = u(x, y, z) v = v(x, y, z) w = w(x, y, z). These transformation equations dene a set of coordinate surfaces and coordinate curves. The coordinate surfaces are dened by the equations u(x, y, z) = c1 v(x, y, z) = c2 w(x, y, z) = c3 where c1 , c2 , c3 are constants. These surfaces intersect in the coordinate curves r(u, c2 , c3 ), where r(u, v, w) = x(u, v, w) e1 + y(u, v, w) e2 + z(u, v, w) e3 . The general situation is illustrated in the gure 1.2-1. Consider the vectors E 1 = grad u = u, E 2 = grad v = v, E 3 = grad w = w (1.2.12) r(c1 , v, c3 ), r(c1 , c2 , w), (1.2.11) (1.2.10) (1.2.9) (1.2.8) evaluated at the common point of intersection (c1 , c2 , c3 ) of the coordinate surfaces. The system of vectors (E 1 , E 2 , E 3 ) can be selected as a system of basis vectors which are normal to the coordinate surfaces. Similarly, the vectors E1 = r , u E2 = r , v E3 = r w (1.2.13) when evaluated at the common point of intersection (c1 , c2 , c3 ) forms a system of vectors (E1 , E2 , E3 ) which we can select as a basis. This basis is a set of tangent vectors to the coordinate curves. It is now demonstrated that the normal basis (E 1 , E 2 , E 3 ) and the tangential basis (E1 , E2 , E3 ) are a set of reciprocal bases. Recall that r = x e1 + y e2 + z e3 denotes the position vector of a variable point. By substitution for x, y, z from (1.2.8) there results r = r(u, v, w) = x(u, v, w) e1 + y(u, v, w) e2 + z(u, v, w) e3 . (1.2.14) 38 Figure 1.2-1. Coordinate curves and coordinate surfaces. A small change in r is denoted dr = dx e1 + dy e2 + dz e3 = where r r r du + dv + dw u v w (1.2.15) r x y z = e1 + e2 + e3 u u u u x r y z (1.2.16) = e1 + e2 + e3 v v v v x y z r = e1 + e2 + e3 . w w w w In terms of the u, v, w coordinates, this change can be thought of as moving along the diagonal of a paralr r r du, dv, and dw. lelepiped having the vector sides u v w Assume u = u(x, y, z) is dened by equation (1.2.9) and dierentiate this relation to obtain du = u u u dx + dy + dz. x y z (1.2.17) The equation (1.2.15) enables us to represent this dierential in the form: du = grad u dr du = grad u du = r r r du + dv + dw u v w r r r grad u du + grad u dv + grad u u v w (1.2.18) dw. By comparing like terms in this last equation we nd that E 1 E1 = 1, E 1 E2 = 0, E 1 E3 = 0. (1.2.19) and w = w(x, y, z) it Similarly, from the other equations in equation (1.2.9) which dene v = v(x, y, z), can be demonstrated that dv = grad v r u du + grad v r v dv + grad v r w dw (1.2.20) 39 and dw = grad w r u du + grad w r v dv + grad w r w dw. (1.2.21) By comparing like terms in equations (1.2.20) and (1.2.21) we nd E 2 E1 = 0, E 3 E1 = 0, E 2 E2 = 1, E 3 E2 = 0, E 2 E3 = 0 E 3 E3 = 1. (1.2.22) The equations (1.2.22) and (1.2.19) show us that the basis vectors dened by equations (1.2.12) and (1.2.13) are reciprocal. Introducing the notation (x1 , x2 , x3 ) = (u, v, w) (y 1 , y 2 , y 3 ) = (x, y, z) (1.2.23) where the x s denote the generalized coordinates and the y s denote the rectangular Cartesian coordinates, the above equations can be expressed in a more concise form with the index notation. For example, if xi = xi (x, y, z) = xi (y 1 , y 2 , y 3 ), and y i = y i (u, v, w) = y i (x1 , x2 , x3 ), i = 1, 2, 3 (1.2.24) then the reciprocal basis vectors can be represented E i = grad xi , and Ei = r , xi i = 1, 2, 3. (1.2.26) i = 1, 2, 3 (1.2.25) We now show that these basis vectors are reciprocal. Observe that r = r(x1 , x2 , x3 ) with dr = and consequently dxi = grad xi dr = grad xi r i dxm = E i Em dxm = m dxm , xm i = 1, 2, 3 (1.2.28) r dxm xm (1.2.27) Comparing like terms in this last equation establishes the result that i E i Em = m , i, m = 1, 2, 3 (1.2.29) which demonstrates that the basis vectors are reciprocal. 40 Scalars, Vectors and Tensors Tensors are quantities which obey certain transformation laws. That is, scalars, vectors, matrices and higher order arrays can be thought of as components of a tensor quantity. We shall be interested in nding how these components are represented in various coordinate systems. We desire knowledge of these transformation laws in order that we can represent various physical laws in a form which is independent of the coordinate system chosen. Before dening dierent types of tensors let us examine what we mean by a coordinate transformation. Coordinate transformations of the type found in equations (1.2.8) and (1.2.9) can be generalized to higher dimensions. Let xi , i = 1, 2, . . . , N denote N variables. These quantities can be thought of as representing a variable point (x1 , x2 , . . . , xN ) in an N dimensional space VN . Another set of N quantities, call them barred quantities, xi , i = 1, 2, . . . , N, can be used to represent a variable point (x1 , x2 , . . . , xN ) in an N dimensional space V N . When the x s are related to the x s by equations of the form xi = xi (x1 , x2 , . . . , xN ), i = 1, 2, . . . , N (1.2.30) then a transformation is said to exist between the coordinates xi and xi , i = 1, 2, . . . , N. Whenever the relations (1.2.30) are functionally independent, single valued and possess partial derivatives such that the Jacobian of the transformation x =J J x x1 , x2 , . . . , xN x1 , x2 , . . . , xN x1 x1 x1 x2 xN x2 x1 xN xN xN ... ... ... = xN x1 . . . . . . . . . (1.2.31) is dierent from zero, then there exists an inverse transformation xi = xi (x1 , x2 , . . . , xN ), i = 1, 2, . . . , N. (1.2.32) For brevity the transformation equations (1.2.30) and (1.2.32) are sometimes expressed by the notation xi = xi (x), i = 1, . . . , N and xi = xi (x), i = 1, . . . , N. (1.2.33) Consider a sequence of transformations from x to x and then from x to x coordinates. For simplicity let x = y and x = z. If we denote by T1 , T2 and T3 the transformations T1 : T2 : y i = y i (x1 , . . . , xN ) i = 1, . . . , N z = z (y , . . . , y ) i = 1, . . . , N i i 1 N or T1 x = y or T2 y = z Then the transformation T3 obtained by substituting T1 into T2 is called the product of two successive transformations and is written T3 : z i = z i (y 1 (x1 , . . . , xN ), . . . , y N (x1 , . . . , xN )) i = 1, . . . , N or T3 x = T2 T1 x = z. This product transformation is denoted symbolically by T3 = T2 T1 . The Jacobian of the product transformation is equal to the product of Jacobians associated with the product transformation and J3 = J2 J1 . 41 Transformations Form a Group A group G is a nonempty set of elements together with a law, for combining the elements. The combined elements are denoted by a product. Thus, if a and b are elements in G then no matter how you dene the law for combining elements, the product combination is denoted ab. The set G and combining law forms a group if the following properties are satised: (i) For all a, b G, then ab G. This is called the closure property. (ii) There exists an identity element I such that for all a G we have Ia = aI = a. (iii) There exists an inverse element. That is, for all a G there exists an inverse element a1 such that a a1 = a1 a = I. (iv) The associative law holds under the combining law and a(bc) = (ab)c for all a, b, c G. For example, the set of elements G = {1, 1, i, i}, where i2 = 1 together with the combining law of ordinary multiplication, forms a group. This can be seen from the multiplication table. 1 -1 -i i 1 1 -1 -i i -1 -1 1 i -i i i -i 1 -1 -i -i i -1 1 The set of all coordinate transformations of the form found in equation (1.2.30), with Jacobian dierent from zero, forms a group because: (i) The product transformation, which consists of two successive transformations, belongs to the set of transformations. (closure) (ii) The identity transformation exists in the special case that x and x are the same coordinates. (iii) The inverse transformation exists because the Jacobian of each individual transformation is dierent from zero. (iv) The associative law is satised in that the transformations satisfy the property T3 (T2 T1 ) = (T3 T2 )T1 . When the given transformation equations contain a parameter the combining law is often times represented as a product of symbolic operators. For example, we denote by T a transformation of coordinates 1 having a parameter . The inverse transformation can be denoted by T and one can write T x = x or 1 x = T x. We let T denote the same transformation, but with a parameter , then the transitive property is expressed symbolically by T T = T where the product T T represents the result of performing two successive transformations. The rst coordinate transformation uses the given transformation equations and uses the parameter in these equations. This transformation is then followed by another coordinate transformation using the same set of transformation equations, but this time the parameter value is . The above symbolic product is used to demonstrate that the result of applying two successive transformations produces a result which is equivalent to performing a single transformation of coordinates having the parameter value . Usually some relationship can then be established between the parameter values , and . 42 Figure 1.2-2. Cylindrical coordinates. In this symbolic notation, we let T denote the identity transformation. That is, using the parameter value of in the given set of transformation equations produces the identity transformation. The inverse transformation can then be expressed in the form of nding the parameter value such that T T = T . Cartesian Coordinates At times it is convenient to introduce an orthogonal Cartesian coordinate system having coordinates y, i i = 1, 2, . . . , N. This space is denoted EN and represents an N-dimensional Euclidean space. Whenever the generalized independent coordinates xi , i = 1, . . . , N are functions of the y s, and these equations are functionally independent, then there exists independent transformation equations y i = y i (x1 , x2 , . . . , xN ), i = 1, 2, . . . , N, (1.2.34) with Jacobian dierent from zero. Similarly, if there is some other set of generalized coordinates, say a barred system xi , i = 1, . . . , N where the x s are independent functions of the y s, then there will exist another set of independent transformation equations y i = y i (x1 , x2 , . . . , xN ), i = 1, 2, . . . , N, (1.2.35) with Jacobian dierent from zero. The transformations found in the equations (1.2.34) and (1.2.35) imply that there exists relations between the x s and x s of the form (1.2.30) with inverse transformations of the form (1.2.32). It should be remembered that the concepts and ideas developed in this section can be applied to a space VN of any nite dimension. Two dimensional surfaces (N = 2) and three dimensional spaces (N = 3) will occupy most of our applications. In relativity, one must consider spaces where N = 4. EXAMPLE 1.2-1. (cylindrical coordinates (r, , z)) Consider the transformation x = x(r, , z) = r cos y = y(r, , z) = r sin z = z(r, , z) = z from rectangular coordinates (x, y, z) to cylindrical coordinates (r, , z), illustrated in the gure 1.2-2. By letting y 1 = x, y 2 = y, y3 = z x1 = r, x2 = , x3 = z the above set of equations are examples of the transformation equations (1.2.8) with u = r, v = , w = z as the generalized coordinates. 43 EXAMPLE 1.2.2. (Spherical Coordinates) (, , ) Consider the transformation x = x(, , ) = sin cos y = y(, , ) = sin sin z = z(, , ) = cos from rectangular coordinates (x, y, z) to spherical coordinates (, , ). By letting y 1 = x, y 2 = y, y 3 = z x1 = , x2 = , x3 = the above set of equations has the form found in equation (1.2.8) with u = , v = , w = the generalized coordinates. One could place bars over the x s in this example in order to distinguish these coordinates from the x s of the previous example. The spherical coordinates (, , ) are illustrated in the gure 1.2-3. Figure 1.2-3. Spherical coordinates. Scalar Functions and Invariance We are now at a point where we can begin to dene what tensor quantities are. The rst denition is for a scalar invariant or tensor of order zero. 44 Denition: ( Absolute scalar eld) Assume there exists a coordinate transformation of the type (1.2.30) with Jacobian J dierent from zero. Let the scalar function f = f (x1 , x2 , . . . , xN ) (1.2.36) be a function of the coordinates xi , i = 1, . . . , N in a space VN . Whenever there exists a function f = f (x1 , x2 , . . . , xN ) (1.2.37) which is a function of the coordinates xi , i = 1, . . . , N such that f = J W f, then f is called a tensor of rank or order zero of weight W in the space VN . Whenever W = 0, the scalar f is called the component of an absolute scalar eld and is referred to as an absolute tensor of rank or order zero. That is, an absolute scalar eld is an invariant object in the space VN with respect to the group of coordinate transformations. It has a single component in each coordinate system. For any scalar function of the type dened by equation (1.2.36), we can substitute the transformation equations (1.2.30) and obtain f = f (x1 , . . . , xN ) = f (x1 (x), . . . , xN (x)) = f (x1 , . . . , xN ). (1.2.38) Vector Transformation, Contravariant Components In VN consider a curve C dened by the set of parametric equations C: xi = xi (t), i = 1, . . . , N where t is a parameter. The tangent vector to the curve C is the vector T= dx1 dx2 dxN , ,..., dt dt dt . In index notation, which focuses attention on the components, this tangent vector is denoted Ti = dxi , dt i = 1, . . . , N. For a coordinate transformation of the type dened by equation (1.2.30) with its inverse transformation dened by equation (1.2.32), the curve C is represented in the barred space by xi = xi (x1 (t), x2 (t), . . . , xN (t)) = xi (t), i = 1, . . . , N, with t unchanged. The tangent to the curve in the barred system of coordinates is represented by dxi xi dxj = , dt xj dt i = 1, . . . , N. (1.2.39) 45 Letting T , i = 1, . . . , N denote the components of this tangent vector in the barred system of coordinates, the equation (1.2.39) can then be expressed in the form T= i i xi j T, xj i, j = 1, . . . , N. (1.2.40) This equation is said to dene the transformation law associated with an absolute contravariant tensor of rank or order one. In the case N = 3 the matrix form of this transformation is represented 1 x 1 1 T x2 2 T = x1 x 3 x3 T 1 x x1 x2 x2 x2 x3 x2 x1 x3 x2 x3 x3 x3 T1 2 T T3 (1.2.41) A more general denition is Denition: (Contravariant tensor) Whenever N quantities Ai in a coordinate system (x1 , . . . , xN ) are related to N quantities A in a coordinate system (x1 , . . . , xN ) such that the Jacobian J is dierent from zero, then if the transformation law A = JW i i xi j A xj is satised, these quantities are called the components of a relative tensor of rank or order one with weight W . Whenever W = 0 these quantities are called the components of an absolute tensor of rank or order one. We see that the above transformation law satises the group properties. EXAMPLE 1.2-3. (Transitive Property of Contravariant Transformation) Show that successive contravariant transformations is also a contravariant transformation. Solution: Consider the transformation of a vector from an unbarred to a barred system of coordinates. A vector or absolute tensor of rank one Ai = Ai (x), i = 1, . . . , N will transform like the equation (1.2.40) and A (x) = i xi j A (x). xj (1.2.42) Another transformation from x x coordinates will produce the components i A (x) = x j A (x) xj i (1.2.43) Here we have used the notation Aj (x) to emphasize the dependence of the components Aj upon the x coordinates. Changing indices and substituting equation (1.2.42) into (1.2.43) we nd i A (x) = x xj m A (x). xj xm i (1.2.44) 46 From the fact that x x xj = , xm xj xm the equation (1.2.44) simplies to i i i A (x) = x m A (x) xm i (1.2.45) and hence this transformation is also contravariant. We express this by saying that the above are transitive with respect to the group of coordinate transformations. Note that from the chain rule one can write xm x1 xm x2 xm x3 xm xm xj m 1 xn + 2 xn + 3 xn = xn = n . j xn = x x x x Do not make the mistake of writing xm x2 xm 2 xn = xn x or xm x3 xm 3 xn = xn x as these expressions are incorrect. Note that there are no summations in these terms, whereas there is a summation index in the representation of the chain rule. Vector Transformation, Covariant Components Consider a scalar invariant A(x) = A(x) which is a shorthand notation for the equation A(x1 , x2 , . . . , xn ) = A(x1 , x2 , . . . , xn ) involving the coordinate transformation of equation (1.2.30). By the chain rule we dierentiate this invariant and nd that the components of the gradient must satisfy A A xj . i = xj x xi Let Aj = A xj and Ai = A , xi (1.2.46) then equation (1.2.46) can be expressed as the transformation law Ai = Aj xj . xi (1.2.47) This is the transformation law for an absolute covariant tensor of rank or order one. A more general denition is 47 Denition: (Covariant tensor) 1 N Whenever N quantities Ai in a coordinate system (x , . . . , x ) are related to N quantities Ai in a coordinate system (x1 , . . . , xN ), with Jacobian J dierent from zero, such that the transformation law Ai = J W xj Aj xi (1.2.48) is satised, then these quantities are called the components of a relative covariant tensor of rank or order one having a weight of W . Whenever W = 0, these quantities are called the components of an absolute covariant tensor of rank or order one. Again we note that the above transformation satises the group properties. Absolute tensors of rank or order one are referred to as vectors while absolute tensors of rank or order zero are referred to as scalars. EXAMPLE 1.2-4. (Transitive Property of Covariant Transformation) Consider a sequence of transformation laws of the type dened by the equation (1.2.47) xx xx xj xi xm Ak (x) = Am (x) k x Ai (x) = Aj (x) We can therefore express the transformation of the components associated with the coordinate transformation x x and Ak (x) = Aj (x) xj xm xm k = Aj (x) xj k , x x which demonstrates the transitive property of a covariant transformation. Higher Order Tensors We have shown that rst order tensors are quantities which obey certain transformation laws. Higher order tensors are dened in a similar manner and also satisfy the group properties. We assume that we are given transformations of the type illustrated in equations (1.2.30) and (1.2.32) which are single valued and continuous with Jacobian J dierent from zero. Further, the quantities xi and xi , i = 1, . . . , n represent the coordinates in any two coordinate systems. The following transformation laws dene second order and third order tensors. 48 Denition: (Second order contravariant tensor) Whenever N-squared quantities Aij in a coordinate system (x1 , . . . , xN ) are related to N-squared quantities A system (x , . . . , x ) such that the transformation law A mn 1 N mn in a coordinate (x) = Aij (x)J W xm xn xi xj (1.2.49) is satised, then these quantities are called components of a relative contravariant tensor of rank or order two with weight W . Whenever W = 0 these quantities are called the components of an absolute contravariant tensor of rank or order two. Denition: (Second order covariant tensor) 1 N Whenever N-squared quantities Aij in a coordinate system (x , . . . , x ) are related to N-squared quantities Amn in a coordinate system (x1 , . . . , xN ) such that the transformation law Amn (x) = Aij (x)J W xi xj xm xn (1.2.50) is satised, then these quantities are called components of a relative covariant tensor of rank or order two with weight W . Whenever W = 0 these quantities are called the components of an absolute covariant tensor of rank or order two. Denition: (Second order mixed tensor) Ai j 1 N Whenever N-squared quantities m in a coordinate system (x , . . . , x ) are related to N-squared quantities An in a coordinate system (x1 , . . . , xN ) such that the transformation law An (x) = Ai (x)J W j m xm xj xi xn (1.2.51) is satised, then these quantities are called components of a relative mixed tensor of rank or order two with weight W . Whenever W = 0 these quantities are called the components of an absolute mixed tensor of rank or order two. It is contravariant of order one and covariant of order one. Higher order tensors are dened in a similar manner. For example, if we can nd N-cubed quantities Am np such that Ajk (x) = A (x)J W i xi x x (1.2.52) x xj xk then this is a relative mixed tensor of order three with weight W . It is contravariant of order one and covariant of order two. 49 General Denition In general a mixed tensor of rank or order (m + n) 1 i2 ...i Tji1 j2 ...jm n (1.2.53) is contravariant of order m and covariant of order n if it obeys the transformation law T j1 j2 ...jn = J where J i1 i2 ...im x x W a Tb11ba2 ...am 2 ...bn xi1 xi2 xim xb1 xb2 xbn am jn xa1 xa2 x xj1 xj2 x (1.2.54) x (x1 , x2 , . . . , xN ) x = = x x (x1 , x2 , . . . , xN ) is the Jacobian of the transformation. When W = 0 the tensor is called an absolute tensor, otherwise it is called a relative tensor of weight W. Here superscripts are used to denote contravariant components and subscripts are used to denote covariant components. Thus, if we are given the tensor components in one coordinate system, then the components in any other coordinate system are determined by the transformation law of equation (1.2.54). Throughout the remainder of this text one should treat all tensors as absolute tensors unless specied otherwise. Dyads and Polyads Note that vectors can be represented in bold face type with the notation A = Ai Ei This notation can also be generalized to tensor quantities. Higher order tensors can also be denoted by bold face type. For example the tensor components Tij and Bijk can be represented in terms of the basis vectors Ei , i = 1, . . . , N by using a notation which is similar to that for the representation of vectors. For example, T = Tij Ei Ej B = Bijk Ei Ej Ek . Here T denotes a tensor with components Tij and B denotes a tensor with components Bijk . The quantities Ei Ej are called unit dyads and Ei Ej Ek are called unit triads. There is no multiplication sign between the basis vectors. This notation is called a polyad notation. A further generalization of this notation is the representation of an arbitrary tensor using the basis and reciprocal basis vectors in bold type. For example, a mixed tensor would have the polyadic representation ij...k T = Tlm...n Ei Ej . . . Ek El Em . . . En . A dyadic is formed by the outer or direct product of two vectors. For example, the outer product of the vectors a = a1 E 1 + a2 E 2 + a3 E 3 and b = b1 E1 + b2 E2 + b3 E3 50 gives the dyad ab =a1 b1 E1 E1 + a1 b2 E1 E2 + a1 b3 E1 E3 a2 b 1 E 2 E 1 + a2 b 2 E 2 E 2 + a2 b 3 E 2 E 3 a3 b 1 E 3 E 1 + a3 b 2 E 3 E 2 + a3 b 3 E 3 E 3 . In general, a dyad can be represented A = Aij Ei Ej i, j = 1, . . . , N where the summation convention is in eect for the repeated indices. The coecients Aij are called the coecients of the dyad. When the coecients are written as an N N array it is called a matrix. Every second order tensor can be written as a linear combination of dyads. The dyads form a basis for the second order tensors. As the example above illustrates, the nine dyads {E1 E1 , E1 E2 , . . . , E3 E3 }, associated with the outer products of three dimensional base vectors, constitute a basis for the second order tensor A = ab having the components Aij = ai bj with i, j = 1, 2, 3. Similarly, a triad has the form T = Tijk Ei Ej Ek Sum on repeated indices where i, j, k have the range 1, 2, . . . , N. The set of outer or direct products { Ei Ej Ek }, with i, j, k = 1, . . . , N i constitutes a basis for all third order tensors. Tensor components with mixed suxes like Cjk are associated with triad basis of the form i C = Cjk Ei Ej Ek where i, j, k have the range 1, 2, . . . N. Dyads are associated with the outer product of two vectors, while triads, tetrads,... are associated with higher-order outer products. These higher-order outer or direct products are referred to as polyads. The polyad notation is a generalization of the vector notation. The subject of how polyad components transform between coordinate systems is the subject of tensor calculus. In Cartesian coordinates we have Ei = Ei = ei and a dyadic with components called dyads is written A = Aij ei ej or A =A11 e1 e1 + A12 e1 e2 + A13 e1 e3 A21 e2 e1 + A22 e2 e2 + A23 e2 e3 A31 e3 e1 + A32 e3 e2 + A33 e3 e3 where the terms ei ej are called unit dyads. Note that a dyadic has nine components as compared with a vector which has only three components. The conjugate dyadic Ac is dened by a transposition of the unit vectors in A, to obtain Ac =A11 e1 e1 + A12 e2 e1 + A13 e3 e1 A21 e1 e2 + A22 e2 e2 + A23 e3 e2 A31 e1 e3 + A32 e2 e3 + A33 e3 e3 51 If a dyadic equals its conjugate A = Ac , then Aij = Aji and the dyadic is called symmetric. If a dyadic equals the negative of its conjugate A = Ac , then Aij = Aji and the dyadic is called skew-symmetric. A special dyadic called the identical dyadic or idemfactor is dened by J = e1 e1 + e2 e2 + e3 e3 . This dyadic has the property that pre or post dot product multiplication of J with a vector V produces the same vector V . For example, V J = (V1 e1 + V2 e2 + V3 e3 ) J = V1 e1 e1 e1 + V2 e2 e2 e2 + V3 e3 e3 e3 = V and J V = J (V1 e1 + V2 e2 + V3 e3 ) = V1 e1 e1 e1 + V2 e2 e2 e2 + V3 e3 e3 e3 = V A dyadic operation often used in physics and chemistry is the double dot product A : B where A and B are both dyadics. Here both dyadics are expanded using the distributive law of multiplication, and then each unit dyad pair ei ej : em en are combined according to the rule ei ej : em en = ( ei em )( ej en ). For example, if A = Aij ei ej and B = Bij ei ej , then the double dot product A : B is calculated as follows. A : B = (Aij ei ej ) : (Bmn em en ) = Aij Bmn ( ei ej : em en ) = Aij Bmn ( ei em )( ej en ) = Aij Bmn im jn = Amj Bmj = A11 B11 + A12 B12 + A13 B13 + A21 B21 + A22 B22 + A23 B23 + A31 B31 + A32 B32 + A33 B33 When operating with dyads, triads and polyads, there is a denite order to the way vectors and polyad components are represented. For example, for A = Ai ei and B = Bi ei vectors with outer product AB = Am Bn em en = there is produced the dyadic with components Am Bn . In comparison, the outer product B A = Bm An em en = produces the dyadic with components Bm An . That is = AB =A1 B1 e1 e1 + A1 B2 e1 e2 + A1 B3 e1 e3 A2 B1 e2 e1 + A2 B2 e2 e2 + A2 B3 e2 e3 A3 B1 e3 e1 + A3 B2 e3 e2 + A3 B3 e3 e3 and = B A =B1 A1 e1 e1 + B1 A2 e1 e2 + B1 A3 e1 e3 B2 A1 e2 e1 + B2 A2 e2 e2 + B2 A3 e2 e3 B3 A1 e3 e1 + B3 A2 e3 e2 + B3 A3 e3 e3 are dierent dyadics. The scalar dot product of a dyad with a vector C is dened for both pre and post multiplication as C = AB C =A(B C) C = C AB =(C A)B These products are, in general, not equal. 52 Operations Using Tensors The following are some important tensor operations which are used to derive special equations and to prove various identities. Addition and Subtraction Tensors of the same type and weight can be added or subtracted. For example, two third order mixed i tensors, when added, produce another third order mixed tensor. Let Ai and Bjk denote two third order jk mixed tensors. Their sum is denoted i i Cjk = Ai + Bjk . jk That is, like components are added. The sum is also a mixed tensor as we now verify. By hypothesis Ai jk i and Bjk are third order mixed tensors and hence must obey the transformation laws Ajk = Am np B jk i i i i i xi xn xp xm xj xk i n p m x x x = Bnp m . x xj xk We let C jk = Ajk + B jk denote the sum in the transformed coordinates. Then the addition of the above transformation equations produces i n p xi xn xp m x x x = Cnp m . xm xj xk x xj xk Consequently, the sum transforms as a mixed third order tensor. m C jk = Ajk + B jk = Am + Bnp np i i i Multiplication (Outer Product) The product of two tensors is also a tensor. The rank or order of the resulting tensor is the sum of the ranks of the tensors occurring in the multiplication. As an example, let Ai denote a mixed third order jk l tensor and let Bm denote a mixed second order tensor. The outer product of these two tensors is the fth order tensor il l Cjkm = Ai Bm , i, j, k, l, m = 1, 2, . . . , N. jk i l Here all indices are free indices as i, j, k, l, m take on any of the integer values 1, 2, . . . , N. Let Ajk and B m denote the components of the given tensors in the barred system of coordinates. We dene C jkm as the il l outer product of these components. Observe that Cjkm is a tensor for by hypothesis Ai and Bm are tensors jk il and hence obey the transformation laws x xj xk xi x x m l x x B = Bm l . x x The outer product of these components produces A = Ai jk l C = A B = Ai Bm jk (1.2.55) il which demonstrates that Cjkm x xj xk x xm xi x x xl x (1.2.56) x xj xk x xm il = Cjkm i x x x xl x transforms as a mixed fth order absolute tensor. Other outer products are analyzed in a similar way. 53 Contraction The operation of contraction on any mixed tensor of rank m is performed when an upper index is set equal to a lower index and the summation convention is invoked. When the summation is performed over the repeated indices the resulting quantity is also a tensor of rank or order (m 2). For example, let Ai , i, j, k = 1, 2, . . . , N denote a mixed tensor and perform a contraction by setting j equal to i. We obtain jk Ai = A1 + A2 + + AN k = Ak ik 1k 2k N where k is a free index. To show that Ak is a tensor, we let transformed components of Ai . jk By hypothesis i i Aik (1.2.57) = Ak denote the contraction on the Ai jk is a mixed tensor and hence the components must satisfy the transformation law xi xn xp . xm xj xk Now execute a contraction by setting j equal to i and perform a summation over the repeated index. We Ajk = Am np nd xi xn xp xn xp = Am np k m xi x xm xk x (1.2.58) p p xp m n x n x = Anp m k = Anp k = Ap k . x x x Hence, the contraction produces a tensor of rank two less than the original tensor. Contractions on other Aik = Ak = Am np i mixed tensors can be analyzed in a similar manner. New tensors can be constructed from old tensors by performing a contraction on an upper and lower index. This process can be repeated as long as there is an upper and lower index upon which to perform the contraction. Each time a contraction is performed the rank of the resulting tensor is two less than the rank of the original tensor. Multiplication (Inner Product) The inner product of two tensors is obtained by: (i) rst taking the outer product of the given tensors and (ii) performing a contraction on two of the indices. EXAMPLE 1.2-5. (Inner product) Let Ai and Bj denote the components of two rst order tensors (vectors). The outer product of these tensors is i Cj = Ai Bj , i, j = 1, 2, . . . , N. The inner product of these tensors is the scalar C = Ai Bi = A1 B1 + A2 B2 + + AN BN . Note that in some situations the inner product is performed by employing only subscript indices. For example, the above inner product is sometimes expressed as C = Ai Bi = A1 B1 + A2 B2 + AN BN . This notation is discussed later when Cartesian tensors are considered. 54 Quotient Law qs s Assume Br and Cp are arbitrary absolute tensors. Further assume we have a quantity A(ijk) which we think might be a third order mixed tensor Ai . By showing that the equation jk qs s Ar Br = Cp qp is satised, then it follows that Ar must be a tensor. This is an example of the quotient law. Obviously, qp this result can be generalized to apply to tensors of any order or rank. To prove the above assertion we shall show from the above equation that Ai is a tensor. Let xi and xi denote a barred and unbarred system of jk coordinates which are related by transformations of the form dened by equation (1.2.30). In the barred system, we assume that Aqp B r = C p r qs s (1.2.59) ij l where by hypothesis Bk and Cm are arbitrary absolute tensors and therefore must satisfy the transformation equations ij B r = Bk xq xs xk xi xj xr s m s l x x C p = Cm l . x xp qs We substitute for B r and C p in the equation (1.2.59) and obtain the equation ij Aqp Bk r qs s xq xs xk xi xj xr = xs xm xl xp s m ql x x = Ar Br . qm xl xp l Cm Since the summation indices are dummy indices they can be replaced by other symbols. We change l to j, q to i and r to k and write the above equation as xs xj Use inner multiplication by xn xs Aqp r xq xk xm Ak im xi xr xp ij Bk = 0. and simplify this equation to the form n j Aqp r r xq xk xm ij Bk = 0 Ak im xi xr xp xq xk xm in Ak Bk = 0. im xi xr xp or Aqp in Because Bk is an arbitrary tensor, the quantity inside the brackets is zero and therefore Aqp r xq xk xm Ak = 0. im i xr x xp xi xl xj xk This equation is simplied by inner multiplication by ql j r Aqp Ak im r to obtain or Ajp l xm xi xl =0 xp xj xk xm xi xl = Ak im xp xj xk which is the transformation law for a third order mixed tensor. 55 EXERCISE 1.2 1. Consider the transformation equations representing a rotation of axes through an angle . T : x1 x2 = x1 cos x2 sin = x1 sin + x2 cos Treat as a parameter and show this set of transformations constitutes a group by nding the value of which: (i) gives the identity transformation. (ii) gives the inverse transformation. (iii) show the transformation is transitive in that a transformation with = 1 followed by a transformation with = 2 is equivalent to the transformation using = 1 + 2 . 2. Show the transformation T : x1 x2 = x1 1 = x2 forms a group with as a parameter. Find the value of such that: (i) the identity transformation exists. (ii) the inverse transformation exists. (iii) the transitive property is satised. 3. Show the given transformation forms a group with parameter . T : 4. x1 x2 = = x1 1x1 x2 1x1 Consider the Lorentz transformation from relativity theory having the velocity parameter V, c is the 1 x 2 x x3 4 x = x1 V x4 1 V2 c 2 speed of light and x4 = t is time. TV : = x2 = x3 = x4 Vcx 2 c 1 2 1 V2 Show this set of transformations constitutes a group, by establishing: (i) V = 0 gives the identity transformation T0 . (ii) TV2 TV1 = T0 requires that V2 = V1 . (iii) TV2 TV1 = TV3 requires that V3 = 5. V1 + V2 . V 1 + V1 2 2 c For (E1 , E2 , E3 ) an arbitrary independent basis, (a) Verify that E1 = 1 E2 E3 , V E2 = 1 E3 E1 , V E3 = 1 E1 E2 V is a reciprocal basis, where V = E1 (E2 E3 ) (b) Show that E j = g ij Ei . 56 Figure 1.2-4. Cylindrical coordinates (r, , z). 6. For the cylindrical coordinates (r, , z) illustrated in the gure 1.2-4. coordinates. Also write out the inverse transformation. (b) Determine the following basis vectors in cylindrical coordinates and represent your results in terms of cylindrical coordinates. (i) The tangential basis E1 , E2 , E3 . (ii)The normal basis E 1 , E 2 , E 3 . (iii) er , e , ez where er , e , ez are normalized vectors in the directions of the tangential basis. (c) A vector A = Ax e1 + Ay e2 + Az e3 can be represented in any of the forms: A = A1 E1 + A2 E2 + A3 E3 A = A1 E 1 + A2 E 2 + A3 E 3 A = Ar er + A e + Az ez depending upon the basis vectors selected . In terms of the components Ax , Ay , Az (i) Solve for the contravariant components A1 , A2 , A3 . (ii) Solve for the covariant components A1 , A2 , A3 . (iii) Solve for the components Ar , A , Az . Express all results in cylindrical coordinates. (Note the components Ar , A , Az are referred to as physical components. Physical components are considered in more detail in a later section.) (a) Write out the transformation equations from rectangular (x, y, z) coordinates to cylindrical (r, , z) 57 Figure 1.2-5. Spherical coordinates (, , ). 7. For the spherical coordinates (, , ) illustrated in the gure 1.2-5. ordinates. Also write out the equations which describe the inverse transformation. (b) Determine the following basis vectors in spherical coordinates (i) The tangential basis E1 , E2 , E3 . (ii) The normal basis E 1 , E 2 , E 3 . (iii) e , e , e which are normalized vectors in the directions of the tangential basis. Express all results in terms of spherical coordinates. (c) A vector A = Ax e1 + Ay e2 + Az e3 can be represented in any of the forms: A = A1 E1 + A2 E2 + A3 E3 A = A1 E 1 + A2 E 2 + A3 E 3 A = A e + A e + A e depending upon the basis vectors selected . Calculate, in terms of the coordinates (, , ) and the components Ax , Ay , Az (i) The contravariant components A1 , A2 , A3 . (ii) The covariant components A1 , A2 , A3 . (iii) The components A , A , A which are called physical components. 8. Work the problems 6,7 and then let (x1 , x2 , x3 ) = (r, , z) denote the coordinates in the cylindrical (a) Write out the transformation equations from rectangular (x, y, z) coordinates to spherical (, , ) co- system and let (x1 , x2 , x3 ) = (, , ) denote the coordinates in the spherical system. (a) Write the transformation equations x x from cylindrical to spherical coordinates. Also nd the inverse transformations. ( Hint: See the gures 1.2-4 and 1.2-5.) (b) Use the results from part (a) and the results from problems 6,7 to verify that Ai = Aj xj xi for i = 1, 2, 3. (i.e. Substitute Aj from problem 6 to get Ai given in problem 7.) 58 (c) Use the results from part (a) and the results from problems 6,7 to verify that A = Aj i xi xj for i = 1, 2, 3. (i.e. Substitute Aj from problem 6 to get Ai given by problem 7.) 9. Pick two arbitrary noncolinear vectors in the x, y plane, say V1 = 5 e1 + e2 and V2 = e1 + 5 e2 and let V3 = e3 be a unit vector perpendicular to both V1 and V2 . The vectors V1 and V2 can be thought of as dening an oblique coordinate system, as illustrated in the gure 1.2-6. (a) Find the reciprocal basis (V 1 , V 2 , V 3 ). (b) Let r = x e1 + y e2 + z e3 = V1 + V2 + V3 and show that y 5x 24 24 5y x = + 24 24 =z x = 5 + y = + 5 z= (d) For = 0 constant, show the coordinate lines are described by = constant and sketch some of these coordinate lines. (See gure 1.2-6.) (e) Find the metrics gij and conjugate metrices g ij associated with the (, , ) space. and = constant, = (c) Show Figure 1.2-6. Oblique coordinates. 59 10. Consider the transformation equations x = x(u, v, w) y = y(u, v, w) z = z(u, v, w) substituted into the position vector r = x e1 + y e2 + z e3 . Dene the basis vectors (E1 , E2 , E3 ) = with the reciprocal basis E1 = where V = E1 (E2 E3 ). Let v = E 1 (E 2 E 3 ) and show that v V = 1. 11. Given the coordinate transformation x = u 2v y = u v z=z 1 E2 E3 , V E2 = 1 E3 E1 , V E3 = 1 E1 E2 . V r r r , , u v w (a) Find and illustrate graphically some of the coordinate curves. (b) For r = r(u, v, z) a position vector, dene the basis vectors E1 = r , u E2 = r , v E3 = r . z Calculate these vectors and then calculate the reciprocal basis E 1 , E 2 , E 3 . (c) With respect to the basis vectors in (b) nd the contravariant components Ai associated with the vector A = 1 e1 + 2 e2 + 3 e3 where (1 , 2 , 3 ) are constants. (d) Find the covariant components Ai associated with the vector A given in part (c). (e) Calculate the metric tensor gij and conjugate metric tensor g ij . k (f) From the results (e), verify that gij g jk = i (g) Use the results from (c)(d) and (e) to verify that Ai = gik Ak (h) Use the results from (c)(d) and (e) to verify that Ai = g ik Ak (i) Find the projection of the vector A on unit vectors in the directions E1 , E2 , E3 . (j) Find the projection of the vector A on unit vectors the directions E 1 , E 2 , E 3 . 60 12. For r = y i ei where y i = y i (x1 , x2 , x3 ), i = 1, 2, 3 we have by denition Ej = and consequently gij = Ei Ej = 13. y m y m , xi xj and g ij = E i E j = xi xj , y m y m i, j, m = 1, . . . , 3 r y i ei . From this relation show that = xj xj Em = xm ej y j Consider the set of all coordinate transformations of the form y i = ai xj + bi j where ai and bi are constants and the determinant of ai is dierent from zero. Show this set of transformaj j tions forms a group. 14. For i , i constants and t a parameter, xi = i + t i ,i = 1, 2, 3 is the parametric representation of dr dt a straight line. Find the parametric equation of the line which passes through the two points (1, 2, 3) and (14, 7, 3). What does the vector 15. represent? A surface can be represented using two parameters u, v by introducing the parametric equations xi = xi (u, v), i = 1, 2, 3, a < u < b and c < v < d. The parameters u, v are called the curvilinear coordinates of a point on the surface. A point on the surface can be represented by the position vector r = r(u, v) = x1 (u, v) e1 + x2 (u, v) e2 + x3 (u, v) e3 . The vectors and r v r u are tangent vectors to the coordinate surface curves r(u, c2 ) and r(c1 , v) respectively. An element of r v dv. surface area dS on the surface is dened as the area of the elemental parallelogram having the vector sides r u du and Show that dS = | r r | dudv = u v g11 g22 (g12 )2 dudv where g11 = r r u u g12 = r r u v g22 = r r . v v Hint: (A B) (A B) = |A B|2 See Exercise 1.1, problem 9(c). 16. (a) Use the results from problem 15 and nd the element of surface area of the circular cone x = u sin cos v a constant (b) Find the surface area of the above cone. y = u sin sin v 0ub z = u cos 0 v 2 61 17. The equation of a plane is dened in terms of two parameters u and v and has the form xi = i u + i v + i i = 1, 2, 3, where i i and i are constants. Find the equation of the plane which passes through the points (1, 2, 3), (14, 7, 3) and (5, 5, 5). What does this problem have to do with the position vector r(u, v), the vectors r r u , v and r(0, 0)? Hint: See problem 15. Determine the points of intersection of the curve x1 = t, x2 = (t)2 , x3 = (t)3 with the plane 8 x1 5 x2 + x3 4 = 0. 18. 19. Verify the relations V eijk E k = Ei Ej and v 1 eijk Ek = E i E j where v = E 1 (E 2 E 3 ) and V = E1 (E2 E3 ).. 20. Let xi and xi , i = 1, 2, 3 be related by the linear transformation xi = ci xj , where ci are constants j j n such that the determinant c = det(ci ) is dierent from zero. Let m denote the cofactor of cm divided by n j the determinant c. j i i (a) Show that ci k = j cj = k . j k i (b) Show the inverse transformation can be expressed xi = j xj . (c) Show that if Ai is a contravariant vector, then its transformed components are Ap = cp Aq . q (d) Show that if Ai is a covariant vector, then its transformed components are Ai = p Ap . i 21. Show that the outer product of two contravariant vectors Ai and B i , i = 1, 2, 3 results in a second order contravariant tensor. Show that for the position vector r = y i (x1 , x2 , x3 ) ei the element of arc length squared is y m y m ds2 = dr dr = gij dxi dxj where gij = Ei Ej = . xi xj 22. 23. 24. p m k i For Ai , Bn and Ctq absolute tensors, show that if Ai Bn = Cjn then Ajk B n = C jn . jk jk i k i Let Aij denote an absolute covariant tensor of order 2. Show that the determinant A = det(Aij ) is (A) is an invariant of weight 1. an invariant of weight 2 and 25. Let B ij denote an absolute contravariant tensor of order 2. Show that the determinant B = det(B ij ) is an invariant of weight 2 and B is an invariant of weight 1. 26. (a) Write out the contravariant components of the following vectors (i) E1 (ii) E2 (iii) E3 where Ei = r xi for i = 1, 2, 3. (b) Write out the covariant components of the following vectors (i) E 1 (ii) E 2 (ii) E 3 where E i = grad xi , for i = 1, 2, 3. 62 27. 28. Let Aij and Aij denote absolute second order tensors. Show that = Aij Aij is a scalar invariant. Assume that aij , i, j = 1, 2, 3, 4 is a skew-symmetric second order absolute tensor. (a) Show that bijk = ajk aki aij + + i j x x xk is a third order tensor. (b) Show bijk is skew-symmetric in all pairs of indices and (c) determine the number of independent components this tensor has. 29. Show the linear forms A1 x + B1 y + C1 and A2 x + B2 y + C2 , with respect to the group of rotations and translations x = x cos y sin + h and y = x sin + y cos + k, have the forms A1 x + B 1 y + C 1 and A2 x + B 2 y + C 2 . Also show that the quantities A1 B2 A2 B1 and A1 A2 + B1 B2 are invariants. 30. Show that the curvature of a curve y = f (x) is = y (1 + y 2 )3/2 and that this curvature remains dy dx dy dx dx dx . invariant under the group of rotations given in the problem 1. Hint: Calculate 31. = Show that when the equation of a curve is given in the parametric form x = x(t), y = y(t), then x yx y the curvature is = 2 and remains invariant under the change of parameter t = t(t), where (x + y 2 )3/2 x = dx , etc. dt 32. Let Aij denote a third order mixed tensor. (a) Show that the contraction Aij is a rst order i k contravariant tensor. (b) Show that contraction of i and j produces Aii which is not a tensor. This shows k that in general, the process of contraction does not always apply to indices at the same level. 33. Let = (x1 , x2 , . . . , xN ) denote an absolute scalar invariant. (a) Is the quantity xi xj 2 xi a tensor? (b) Is the quantity 34. a tensor? Consider the second order absolute tensor aij , i, j = 1, 2 where a11 = 1, a12 = 2, a21 = 3 and a22 = 4. Find the components of aij under the transformation of coordinates x1 = x1 + x2 and x2 = x1 x2 . 35. Let Ai , Bi denote the components of two covariant absolute tensors of order one. Show that Cij = Ai Bj is an absolute second order covariant tensor. 36. Let Ai denote the components of an absolute contravariant tensor of order one and let Bi denote the i components of an absolute covariant tensor of order one, show that Cj = Ai Bj transforms as an absolute mixed tensor of order two. 37. (a) Show the sum and dierence of two tensors of the same kind is also a tensor of this kind. (b) Show that the outer product of two tensors is a tensor. Do parts (a) (b) in the special case where one tensor Ai j is a relative tensor of weight 4 and the other tensor Bk is a relative tensor of weight 3. What is the weight ij j of the outer product tensor Tk = Ai Bk in this special case? 38. j Let Aij denote the components of a mixed tensor of weight M . Form the contraction Bm = Aij im km j and determine how Bm transforms. What is its weight? 39. Let Ai denote the components of an absolute mixed tensor of order two. Show that the scalar j contraction S = Ai is an invariant. i 63 40. Let Ai = Ai (x1 , x2 , . . . , xN ) denote the components of an absolute contravariant tensor. Form the Ai xj i and determine if Bj transforms like a tensor. i quantity Bj = Ai Aj Let Ai denote the components of a covariant vector. (a) Show that aij = are the j x xi ajk aki aij + + = 0. components of a second order tensor. (b) Show that xk xi xj 42. Show that xi = K eijk Aj Bk , with K = 0 and arbitrary, is a general solution of the system of equations 41. Ai xi = 0, Bi xi = 0, i = 1, 2, 3. Give a geometric interpretation of this result in terms of vectors. 43. Given the vector A = y e1 + z e2 + x e3 where e1 , e2 , e3 denote a set of unit basis vectors which dene a set of orthogonal x, y, z axes. Let E1 = 3 e1 + 4 e2 , E2 = 4 e1 + 7 e2 and E3 = e3 denote a set of basis vectors which dene a set of u, v, w axes. (a) Find the coordinate transformation between these two sets of axes. (b) Find a set of reciprocal vectors E 1 , E 3 , E 3 . (c) Calculate the covariant components of A. (d) Calculate the contravariant components of A. 44. Let A = Aij ei ej denote a dyadic. Show that A : Ac = A11 A11 + A12 A21 + A13 A31 + A21 A12 + A22 A22 + A23 A32 + A31 A13 + A32 A23 + A23 A33 45. Let A = Ai ei , B = Bi ei , C = Ci ei , D = Di ei denote vectors and let = AB, = C D denote dyadics which are the outer products involving the above vectors. Show that the double dot product satises : = AB : C D = (A C)(B D) 46. Show that if aij is a symmetric tensor in one coordinate system, then it is symmetric in all coordinate systems. 47. 48. Write the transformation laws for the given tensors. (a) Show that if Ai = Aj Ak ij (b) Aij k (c) Aijk m and unbarred systems. 49. xj xj i , then Ai = Aj xi . Note that this is equivalent to interchanging the bar x (a) Show that under the linear homogeneous transformation x1 =a1 x1 + a2 x2 1 1 x2 =a1 x1 + a2 x2 2 2 the quadratic form Q(x1 , x2 ) = g11 (x1 )2 + 2g12 x1 x2 + g22 (x2 )2 becomes Q(x1 , x2 ) = g11 (x1 )2 + 2g12 x1 x2 + g 22 (x2 )2 where g ij = g11 aj ai + g12 (ai aj + aj ai ) + g22 ai aj . 12 22 11 12 (b) Show F = g11 g22 (g12 )2 is a relative invariant of weight 2 of the quadratic form Q(x1 , x2 ) with respect to the group of linear homogeneous transformations. i.e. Show that F = 2 F where F = g 11 g22 (g12 )2 and = (a1 a2 a2 a1 ). 12 12 64 50. Let ai and bi for i = 1, . . . , n denote arbitrary vectors and form the dyadic = a1 b1 + a2 b2 + + an bn . By denition the rst scalar invariant of is 1 = a1 b1 + a2 b2 + + an bn where a dot product operator has been placed between the vectors. The rst vector invariant of is dened = a1 b1 + a2 b2 + + an bn where a vector cross product operator has been placed between the vectors. (a) Show that the rst scalar and vector invariant of = e1 e2 + e2 e3 + e3 e3 are respectively 1 and e1 + e3 . (b) From the vector f = f1 e1 + f2 e2 + f3 e3 one can form the dyadic f having the matrix components f1 f2 f3 f = x f1 y f1 z x f2 y f2 z x f3 y f3 z . Show that this dyadic has the rst scalar and vector invariants given by f2 f3 f1 + + f = x y z f1 f2 f3 f2 f3 f1 e1 + e2 + f = y z z x x y 51. Let denote the dyadic given in problem 50. The dyadic 2 dened by 1 2 = ai aj bi bj 2 i,j e3 is called the Gibbs second dyadic of , where the summation is taken over all permutations of i and j. When i = j the dyad vanishes. Note that the permutations i, j and j, i give the same dyad and so occurs twice in the nal sum. The factor 1/2 removes this doubling. Associated with the Gibbs dyad 2 are the scalar invariants 2 = 3 = Show that the dyad = as + tq + cu has the rst scalar invariant 1 = a s + b t + c u the rst vector invariant = a s + b t + c u Gibbs second dyad 2 = b ct u + c au s + a bs t 1 2 1 6 (ai aj ) (bi bj ) i,j (ai aj ak )(bi bj bk ) i,j,k second scalar of 2 = (b c) (t u) + (c a) (u s) + (a b) (s t) third scalar of 3 = (a b c)(s t u) 65 52. (Spherical Trigonometry) Construct a spherical triangle ABC on the surface of a unit sphere with sides and angles less than 180 degrees. Denote by a,b c the unit vectors from the origin of the sphere to the vertices A,B and C. Make the construction such that a (b c) is positive with a, b, c forming a right-handed system. Let , , denote the angles between these unit vectors such that a b = cos c a = cos b c = cos . (1) The great circles through the vertices A,B,C then make up the sides of the spherical triangle where side is opposite vertex A, side is opposite vertex B and side is opposite the vertex C. The angles A,B and C between the various planes formed by the vectors a, b and c are called the interior dihedral angles of the spherical triangle. Note that the cross products a b = sin c b c = sin a c a = sin b (2) dene unit vectors a, b and c perpendicular to the planes determined by the unit vectors a, b and c. The dot products a b = cos b c = cos c a = cos (3) dene the angles , and which are called the exterior dihedral angles at the vertices A,B and C and are such that =A =B = C. (4) (a) Using appropriate scaling, show that the vectors a, b, c and a, b, c form a reciprocal set. (b) Show that a (b c) = sin a a = sin b b = sin c c (c) Show that a (b c) = sin a a = sin b b = sin c c (d) Using parts (b) and (c) show that sin sin sin = = sin sin sin (e) Use the results from equation (4) to derive the law of sines for spherical triangles sin sin sin = = sin A sin B sin C (f) Using the equations (2) show that sin sin b c = (c a) (a b) = (c a)(a b) b c and hence show that cos = cos cos sin sin cos . In a similar manner show also that cos = cos cos sin sin cos . (g) Using part (f) derive the law of cosines for spherical triangles cos = cos cos + sin sin cos A cos A = cos B cos C + sin B sin C cos A cyclic permutation of the symbols produces similar results involving the other angles and sides of the spherical triangle. 65 1.3 SPECIAL TENSORS Knowing how tensors are dened and recognizing a tensor when it pops up in front of you are two dierent things. Some quantities, which are tensors, frequently arise in applied problems and you should learn to recognize these special tensors when they occur. In this section some important tensor quantities are dened. We also consider how these special tensors can in turn be used to dene other tensors. Metric Tensor Dene y i , i = 1, . . . , N as independent coordinates in an N dimensional orthogonal Cartesian coordinate system. The distance squared between two points y i expression ds2 = dy m dy m = (dy 1 )2 + (dy 2 )2 + + (dy N )2 . (1.3.1) and y i + dy i , i = 1, . . . , N is dened by the Assume that the coordinates y i are related to a set of independent generalized coordinates xi , i = 1, . . . , N by a set of transformation equations y i = y i (x1 , x2 , . . . , xN ), i = 1, . . . , N. (1.3.2) To emphasize that each y i depends upon the x coordinates we sometimes use the notation y i = y i (x), for i = 1, . . . , N. The dierential of each coordinate can be written as dy m = y m j dx , xj m = 1, . . . , N, (1.3.3) and consequently in the x-generalized coordinates the distance squared, found from the equation (1.3.1), becomes a quadratic form. Substituting equation (1.3.3) into equation (1.3.1) we nd ds2 = where gij = y m y m i j dx dx = gij dxi dxj xi xj y m y m , xi xj (1.3.4) i, j = 1, . . . , N (1.3.5) are called the metrices of the space dened by the coordinates xi , i = 1, . . . , N. Here the gij are functions of the x coordinates and is sometimes written as gij = gij (x). Further, the metrices gij are symmetric in the indices i and j so that gij = gji for all values of i and j over the range of the indices. If we transform to another coordinate system, say xi , i = 1, . . . , N , then the element of arc length squared is expressed in terms of the barred coordinates and ds2 = g ij dxi dxj , where gij = g ij (x) is a function of the barred coordinates. The following example demonstrates that these metrices are second order covariant tensors. 66 EXAMPLE 1.3-1. Show the metric components gij are covariant tensors of the second order. Solution: In a coordinate system xi , i = 1, . . . , N the element of arc length squared is ds2 = gij dxi dxj (1.3.6) while in a coordinate system xi , i = 1, . . . , N the element of arc length squared is represented in the form ds2 = g mn dxm dxn . The element of arc length squared is to be an invariant and so we require that gmn dxm dxn = gij dxi dxj (1.3.8) (1.3.7) Here it is assumed that there exists a coordinate transformation of the form dened by equation (1.2.30) together with an inverse transformation, as in equation (1.2.32), which relates the barred and unbarred coordinates. In general, if xi = xi (x), then for i = 1, . . . , N we have dxi = xi dxm xm and dxj = xj dxn xn (1.3.9) Substituting these dierentials in equation (1.3.8) gives us the result g mn dxm dxn = gij xi xj dxm dxn xm xn or g mn gij xi xj xm xn dxm dxn = 0 For arbitrary changes in dxm this equation implies that g mn = gij as a second order absolute covariant tensor. xi xj and consequently gij transforms xm xn EXAMPLE 1.3-2. (Curvilinear coordinates) Consider a set of general transformation equations from rectangular coordinates (x, y, z) to curvilinear coordinates (u, v, w). These transformation equations and the corresponding inverse transformations are represented x = x(u, v, w) y = y(u, v, w) z = z(u, v, w). Here y 1 = x, y 2 = y, y 3 = z u = u(x, y, z) v = v(x, y, z) w = w(x, y, z) (1.3.10) and x1 = u, x2 = v, x3 = w are the Cartesian and generalized coordinates and N = 3. The intersection of the coordinate surfaces u = c1 ,v = c2 and w = c3 dene coordinate curves of the curvilinear coordinate system. The substitution of the given transformation equations (1.3.10) into the position vector r = x e1 + y e2 + z e3 produces the position vector which is a function of the generalized coordinates and r = r(u, v, w) = x(u, v, w) e1 + y(u, v, w) e2 + z(u, v, w) e3 67 and consequently dr = r r r du + dv + dw, where u v w x r y z = e1 + e2 + e3 u u u u x r y z E2 = = e1 + e2 + e3 v v v v x r y z E3 = = e1 + e2 + e3 . w w w w E1 = (1.3.11) are tangent vectors to the coordinate curves. The element of arc length in the curvilinear coordinates is ds2 = dr dr = r r r r r r dudu + dudv + dudw u u u v u w r r r r r r dvdu + dvdv + dvdw + v u v v v w r r r r r r + dwdu + dwdv + dwdw. w u w v w w (1.3.12) Utilizing the summation convention, the above can be expressed in the index notation. Dene the quantities g11 = g21 g31 and let x1 = u, x2 = v, r r u u r r = v u r r = w u g12 = g22 g32 r r u v r r = v v r r = w v g13 = g23 g33 r r u w r r = v w r r = w w x3 = w. Then the above element of arc length can be expressed as ds2 = Ei Ej dxi dxj = gij dxi dxj , i, j = 1, 2, 3 where gij = Ei Ej = r r y m y m j= , xi x xi xj i, j free indices (1.3.13) are called the metric components of the curvilinear coordinate system. The metric components may be thought of as the elements of a symmetric matrix, since gij = gji . In the rectangular coordinate system x, y, z, the element of arc length squared is ds2 = dx2 + dy 2 + dz 2 . In this space the metric components are 10 gij = 0 1 00 0 0. 1 68 EXAMPLE 1.3-3. (Cylindrical coordinates (r, , z)) The transformation equations from rectangular coordinates to cylindrical coordinates can be expressed as x = r cos , y = r sin , z = z. Here y 1 = x, y 2 = y, y 3 = z and x1 = r, x2 = , x3 = z, and the position vector can be expressed r = r(r, , z) = r cos e1 + r sin e2 + z e3 . The derivatives of this position vector are calculated and we nd E1 = r = cos e1 + sin e2 , r E2 = r = r sin e1 + r cos e2 , E3 = r = e3 . z From the results in equation (1.3.13), the metric components of this space are 10 gij = 0 r2 00 0 0. 1 We note that since gij = 0 when i = j, the coordinate system is orthogonal. Given a set of transformations of the form found in equation (1.3.10), one can readily determine the metric components associated with the generalized coordinates. For future reference we list several dierent coordinate systems together with their metric components. Each of the listed coordinate systems are orthogonal and so gij = 0 for i = j. The metric components of these orthogonal systems have the form h2 1 0 gij = 0 and the element of arc length squared is ds2 = h2 (dx1 )2 + h2 (dx2 )2 + h2 (dx3 )2 . 1 2 3 0 h2 2 0 0 0 h2 3 1. Cartesian coordinates (x, y, z) x=x y=y z=z h1 = 1 h2 = 1 h3 = 1 The coordinate curves are formed by the intersection of the coordinate surfaces x =Constant, y =Constant and z =Constant. 69 Figure 1.3-1. Cylindrical coordinates. 2. Cylindrical coordinates (r, , z) x = r cos y = r sin z=z r0 0 2 <z < h1 = 1 h2 = r h3 = 1 The coordinate curves, illustrated in the gure 1.3-1, are formed by the intersection of the coordinate surfaces x2 + y 2 = r2 , y/x = tan Cylinders Planes Planes. z = Constant 3. Spherical coordinates (, , ) x = sin cos y = sin sin z = cos 0 0 0 2 h1 = 1 h2 = h3 = sin The coordinate curves, illustrated in the gure 1.3-2, are formed by the intersection of the coordinate surfaces x2 + y 2 + z 2 = 2 Spheres Cones x2 + y 2 = tan2 z 2 y = x tan Planes. 4. Parabolic cylindrical coordinates (, , z) x = 1 y = ( 2 2 ) 2 z=z < < <z < 0 h1 = h2 = h3 = 1 2 + 2 2 + 2 70 Figure 1.3-2. Spherical coordinates. The coordinate curves, illustrated in the gure 1.3-3, are formed by the intersection of the coordinate surfaces x2 = 2 2 (y x2 = 2 2 (y + 2 ) 2 Parabolic cylinders 2 ) Parabolic cylinders 2 z = Constant Planes. Figure 1.3-3. Parabolic cylindrical coordinates in plane z = 0. 5. Parabolic coordinates (, , ) x = cos y = sin 1 z = ( 2 2 ) 2 0 0 0 < < 2 h1 = h2 = h3 = 2 + 2 2 + 2 71 The coordinate curves, illustrated in the gure 1.3-4, are formed by the intersection of the coordinate surfaces x2 + y 2 = 2 2 (z x2 + y 2 = 2 2 (z + 2 ) 2 Paraboloids 2 ) Paraboloids 2 y = x tan Planes. Figure 1.3-4. Parabolic coordinates, = /4. 6. Elliptic cylindrical coordinates (, , z) x = cosh cos y = sinh sin z=z 0 0 2 <z < h1 = h2 = h3 = 1 sinh2 + sin2 sinh2 + sin2 The coordinate curves, illustrated in the gure 1.3-5, are formed by the intersection of the coordinate surfaces y2 x2 + =1 cosh2 sinh2 y2 x2 =1 cos2 sin2 Elliptic cylinders Hyperbolic cylinders Planes. z = Constant 72 Figure 1.3-5. Elliptic cylindrical coordinates in the plane z = 0. 7. Elliptic coordinates (, , ) 2 2 2 1 2 2 1 2 (1 2 )( 2 1) x= y= z = (1 2 )( 2 1) cos 1< 1 1 0 < 2 h1 = h2 = h3 = (1 2 )( 2 1) sin The coordinate curves, illustrated in the gure 1.3-6, are formed by the intersection of the coordinate surfaces x2 y2 z2 +2 + 2 =1 Prolate ellipsoid 2 1 1 z2 x2 y2 =1 Two-sheeted hyperboloid 2 1 2 1 2 y = x tan Planes 8. Bipolar coordinates (u, v, z) x= a sinh v , 0 u < 2 cosh v cos u a sin u , < v < y= cosh v cos u z=z <z < h2 = h2 1 2 h2 = 2 a2 (cosh v cos u)2 h2 = 1 3 73 Figure 1.3-6. Elliptic coordinates = /4. Figure 1.3-7. Bipolar coordinates. The coordinate curves, illustrated in the gure 1.3-7, are formed by the intersection of the coordinate surfaces (x a coth v)2 + y 2 = a2 Cylinders sinh2 v a2 x2 + (y a cot u)2 = Cylinders sin2 u z = Constant Planes. 74 9. Conical coordinates (u, v, w) x= uvw , b 2 > v 2 > a2 > w 2 , ab u (v 2 a2 )(w2 a2 ) y= a a2 b 2 2 b2 )(w2 b2 ) u (v z= b b 2 a2 u0 h2 = 1 1 h2 = 2 u2 (v 2 w2 ) a2 )(b2 v 2 ) u2 (v 2 w2 ) h2 = 3 2 a2 )(w2 b2 ) (w (v 2 The coordinate curves, illustrated in the gure 1.3-8, are formed by the intersection of the coordinate surfaces 2 2 x2 + y 2 + z 2 = u2 2 Spheres y z x +2 +2 =0, Cones v2 v a2 v b2 x2 y2 z2 +2 +2 = 0, Cones. w2 w a2 w b2 Figure 1.3-8. Conical coordinates. 10. Prolate spheroidal coordinates (u, v, ) x = a sinh u sin v cos , y = a sinh u sin v sin , z = a cosh u cos v, u0 0v h2 = h2 1 2 h2 = a2 (sinh2 u + sin2 v) 2 h2 = a2 sinh2 u sin2 v 3 0 < 2 The coordinate curves, illustrated in the gure 1.3-9, are formed by the intersection of the coordinate surfaces x2 y2 z2 + + = 1, 2 2 (a sinh u) (a sinh u) (a cosh u)2 z2 x2 y2 = 1, 2 2 (a cos v) (a sin v) (a sin v)2 Prolate ellipsoids Two-sheeted hyperboloid Planes. y = x tan , 75 Figure 1.3-9. Prolate spheroidal coordinates 11. Oblate spheroidal coordinates (, , ) x = a cosh cos cos , 0 y = a cosh cos sin , 2 2 z = a sinh sin , 0 2 h2 = h2 1 2 h2 = a2 (sinh2 + sin2 ) 2 h2 = a2 cosh2 cos2 3 The coordinate curves, illustrated in the gure 1.3-10, are formed by the intersection of the coordinate surfaces x2 y2 z2 + + = 1, 2 2 (a cosh ) (a cosh ) (a sinh )2 x2 y2 z2 + = 1, 2 2 (a cos ) (a cos ) (a sin )2 Oblate ellipsoids One-sheet hyperboloids Planes. y = x tan , 12. Toroidal coordinates (u, v, ) x= a sinh v cos , cosh v cos u a sinh v sin , y= cosh v cos u a sin u , z= cosh v cos u 0 u < 2 < v < 0 < 2 h2 = h2 1 2 h2 = 2 h2 = 3 a2 (cosh v cos u)2 a2 sinh2 v (cosh v cos u)2 The coordinate curves, illustrated in the gure 1.3-11, are formed by the intersection of the coordinate surfaces a cos u 2 a2 , = sin u sin2 u 2 cosh v a2 x2 + y 2 a + z2 = , sinh v sinh2 v y = x tan , x2 + y 2 + z Spheres Torus planes 76 Figure 1.3-10. Oblate spheroidal coordinates Figure 1.3-11. Toroidal coordinates i EXAMPLE 1.3-4. Show the Kronecker delta j is a mixed second order tensor. Solution: Assume we have a coordinate transformation xi = xi (x), i = 1, . . . , N of the form (1.2.30) and i possessing an inverse transformation of the form (1.2.32). Let j and j denote the Kronecker delta in the i barred and unbarred system of coordinates. By denition the Kronecker delta is dened i j = j = i 0, 1, if if i=j i=j . 77 Employing the chain rule we write xm xi xm xk i xm n= n= x xi x xi xn k By hypothesis, the xi , i = 1, . . . , N are independent coordinates and therefore we have simplies to i n = k m xm xn m (1.3.14) = n and (1.3.14) xm xk . xi xn Therefore, the Kronecker delta transforms as a mixed second order tensor. Conjugate Metric Tensor Let g denote the determinant of the matrix having the metric tensor gij , i, j = 1, . . . , N as its elements. In our study of cofactor elements of a matrix we have shown that j cof (g1j )g1k + cof (g2j )g2k + . . . + cof (gN j )gN k = gk . (1.3.15) We can use this fact to nd the elements in the inverse matrix associated with the matrix having the components gij . The elements of this inverse matrix are g ij = 1 cof (gij ) g (1.3.16) and are called the conjugate metric components. We examine the summation g ij gik and nd: g ij gik = g 1j g1k + g 2j g2k + . . . + g N j gN k 1 = [cof (g1j )g1k + cof (g2j )g2k + . . . + cof (gN j )gN k ] g 1 j j gk = k = g The equation j g ij gik = k (1.3.17) is an example where we can use the quotient law to show g ij is a second order contravariant tensor. Because of the symmetry of g ij and gij the equation (1.3.17) can be represented in other forms. EXAMPLE 1.3-5. Let Ai and Ai denote respectively the covariant and contravariant components of a vector A. Show these components are related by the equations Ai = gij Aj A = g Aj where gij and g ij are the metric and conjugate metric components of the space. k jk (1.3.18) (1.3.19) 78 Solution: We multiply the equation (1.3.18) by g im (inner product) and use equation (1.3.17) to simplify m the results. This produces the equation g im Ai = g im gij Aj = j Aj = Am . Changing indices produces the result given in equation (1.3.19). Conversely, if we start with equation (1.3.19) and multiply by gkm (inner j product) we obtain gkm Ak = gkm g jk Aj = m Aj = Am which is another form of the equation (1.3.18) with the indices changed. Notice the consequences of what the equations (1.3.18) and (1.3.19) imply when we are in an orthogonal Cartesian coordinate system where 1 gij = 0 0 In this special case, we have A1 = g11 A1 + g12 A2 + g13 A3 = A1 A2 = g21 A1 + g22 A2 + g23 A3 = A2 A3 = g31 A1 + g32 A2 + g33 A3 = A3 . These equations tell us that in a Cartesian coordinate system the contravariant and covariant components are identically the same. 00 1 0 01 and g ij 1 = 0 0 00 1 0. 01 EXAMPLE 1.3-6. We have previously shown that if Ai is a covariant tensor of rank 1 its components in a barred system of coordinates are Ai = Aj xj . xi (1.3.20) Solve for the Aj in terms of the Aj . (i.e. nd the inverse transformation). Solution: Multiply equation (1.3.20) by xi xm (inner product) and obtain (1.3.21) Ai xi xj xi = Aj i m . xm x x xj xi xj j = = m since xj and xm are assumed to be independent xm xi xm coordinates. This reduces equation (1.3.21) to the form In the above product we have Ai which is the desired inverse transformation. This result can be obtained in another way. Examine the transformation equation (1.3.20) and ask the question, When we have two coordinate systems, say a barred and an unbarred system, does it matter which system we call the barred system? With some thought it should be obvious that it doesnt matter which system you label as the barred system. Therefore, we can interchange the barred and unbarred symbols in xj equation (1.3.20) and obtain the result Ai = Aj i which is the same form as equation (1.3.22), but with x a dierent set of indices. xi j = Aj m = Am xm (1.3.22) 79 Associated Tensors Associated tensors can be constructed by taking the inner product of known tensors with either the metric or conjugate metric tensor. Denition: (Associated tensor) Any tensor constructed by multiplying (inner product) a given tensor with the metric or conjugate metric tensor is called an associated tensor. Associated tensors are dierent ways of representing a tensor. The multiplication of a tensor by the metric or conjugate metric tensor has the eect of lowering or raising indices. For example the covariant and contravariant components of a vector are dierent representations of the same vector in dierent forms. These forms are associated with one another by way of the metric and conjugate metric tensor and g ij Ai = Aj gij Aj = Ai . EXAMPLE 1.3-7. The following are some examples of associated tensors. Aj = g ij Ai Am = g mi Aijk .jk A.nm = g mk g nj Aijk i.. Aj = gij Ai Ai.k = gmj Aijk m Amjk = gim Ai .jk Sometimes dotsare used as indices in order to represent the location of the index that was raised or lowered. If a tensor is symmetric, the position of the index is immaterial and so a dot is not needed. For example, if Amn is a symmetric tensor, then it is easy to show that An and A.n are equal and therefore can be written .m m as An without confusion. m Higher order tensors are similarly related. For example, if we nd a fourth order covariant tensor Tijkm we can then construct the fourth order contravariant tensor T pqrs from the relation T pqrs = g pi g qj g rk g sm Tijkm . This fourth order tensor can also be expressed as a mixed tensor. Some mixed tensors associated with the given fourth order covariant tensor are: p T.jkm = g pi Tijkm , pq p T..km = g qj T.jkm . 80 Riemann Space VN A Riemannian space VN is said to exist if the element of arc length squared has the form ds2 = gij dxi dxj (1.3.23) where the metrices gij = gij (x1 , x2 , . . . , xN ) are continuous functions of the coordinates and are dierent from constants. In the special case gij = ij the Riemannian space VN reduces to a Euclidean space EN . The element of arc length squared dened by equation (1.3.23) is called the Riemannian metric and any geometry which results by using this metric is called a Riemannian geometry. A space VN is called at if it is possible to nd a coordinate transformation where the element of arclength squared is ds2 = where each i i2 i (dx ) is either +1 or 1. A space which is not at is called curved. Geometry in VN Given two vectors A = Ai Ei and B = B j Ej , then their dot product can be represented A B = Ai B j Ei Ej = gij Ai B j = Aj B j = Ai Bi = g ij Aj Bi = |A||B| cos . (1.3.24) Consequently, in an N dimensional Riemannian space VN the dot or inner product of two vectors A and B is dened: gij Ai B j = Aj B j = Ai Bi = g ij Aj Bi = AB cos . (1.3.25) In this denition A is the magnitude of the vector Ai , the quantity B is the magnitude of the vector Bi and is the angle between the vectors when their origins are made to coincide. In the special case that = 90 we have gij Ai B j = 0 as the condition that must be satised in order that the given vectors Ai and B i are orthogonal to one another. Consider also the special case of equation (1.3.25) when Ai = B i and = 0. In this case the equations (1.3.25) inform us that g in An Ai = Ai Ai = gin Ai An = (A)2 . (1.3.26) From this equation one can determine the magnitude of the vector Ai . The magnitudes A and B can be written A = (gin Ai An ) 2 1 and B = (gpq B p B q ) 2 and so we can express equation (1.3.24) in the form cos = gij Ai B j (gmn Am An ) 2 (gpq B p B q ) 2 1 1 1 . (1.3.27) An import application of the above concepts arises in the dynamics of rigid body motion. Note that if a vector Ai has constant magnitude and the magnitude of dA dt i dAi dt is dierent from zero, then the vectors Ai and j must be orthogonal to one another due to the fact that gij Ai dA = 0. As an example, consider the unit dt vectors e1 , e2 and e3 on a rotating system of Cartesian axes. We have for constants ci , i = 1, 6 that d e1 = c1 e 2 + c2 e 3 dt d e2 = c3 e 3 + c4 e 1 dt d e3 = c5 e 1 + c6 e 2 dt because the derivative of any ei (i xed) constant vector must lie in a plane containing the vectors ej and ek , (j = i , k = i and j = k), since any vector in this plane must be perpendicular to ei . 81 The above denition of a dot product in VN can be used to dene unit vectors in VN . Denition: (Unit vector) i Whenever the magnitude of a vec- tor A is unity, the vector is called a unit vector. In this case we have gij Ai Aj = 1. (1.3.28) EXAMPLE 1.3-8. (Unit vectors) In VN the element of arc length squared is expressed ds2 = gij dxi dxj which can be expressed in the dxi dxi dxj . This equation states that the vector , i = 1, . . . , N is a unit vector. One application form 1 = gij ds ds ds of this equation is to consider a particle moving along a curve in VN which is described by the parametric equations xi = xi (t), for i = 1, . . . , N. The vector V i = particle. By chain rule dierentiation we have Vi = where V = ds dt dxi dt , i = 1, . . . , N represents a velocity vector of the dxi ds dxi dxi = =V , dt ds dt ds dxi ds (1.3.29) is the scalar speed of the particle and is a unit tangent vector to the curve. The equation (1.3.29) shows that the velocity is directed along the tangent to the curve and has a magnitude V. That is ds dt 2 = (V )2 = gij V i V j . EXAMPLE 1.3-9. (Curvilinear coordinates) Find an expression for the cosine of the angles between the coordinate curves associated with the transformation equations x = x(u, v, w), y = y(u, v, w), z = z(u, v, w). 82 Figure 1.3-12. Angles between curvilinear coordinates. Solution: Let y 1 = x, y 2 = y, y 3 = z and x1 = u, x2 = v, x3 = w denote the Cartesian and curvilinear coordinates respectively. With reference to the gure 1.3-12 we can interpret the intersection of the surfaces v = c2 and w = c3 as the curve r = r(u, c2 , c3 ) which is a function of the parameter u. By moving only along r du and consequently this curve we have dr = u r r ds2 = dr dr = dudu = g11 (dx1 )2 , u u or 1= This equation shows that the vector be represented by vector tr = (2) tr (1) = r 1 1 . g11 dx1 ds dr dr = g11 ds ds 1 g11 dx1 ds 2 . = is a unit vector along this curve. This tangent vector can The curve which is dened by the intersection of the surfaces u = c1 and w = c3 has the unit tangent r 1 2 . g22 Similarly, the curve which is dened as the intersection of the surfaces u = c1 and r 1 3 . g33 v = c2 has the unit tangent vector tr = (3) The cosine of the angle 12 , which is the angle between the unit vectors tr and tr , is obtained from the result of equation (1.3.25). We nd (1) (2) g12 1 p1 q cos 12 = gpq tp tq = gpq 1 2 = . (1) (2) g11 g22 g11 g22 For 13 the angle between the directions ti and ti we nd (1) (3) g13 cos 13 = . g11 g33 Finally, for 23 the angle between the directions ti and ti we nd (2) (3) g23 cos 23 = . g22 g33 When 13 = 12 = 23 = 90 , we have g12 = g13 = g23 = 0 and the coordinate curves which make up the curvilinear coordinate system are orthogonal to one another. In an orthogonal coordinate system we adopt the notation g11 = (h1 )2 , g22 = (h2 )2 , g33 = (h3 )2 and gij = 0, i = j. 83 Epsilon Permutation Symbol Associated with the epermutation symbols there are the epsilon permutation symbols dened by the relations ijk = geijk and ijk 1 = eijk g (1.3.30) where g is the determinant of the metrices gij . It can be demonstrated that the eijk permutation symbol is a relative tensor of weight 1 whereas the ijk permutation symbol is an absolute tensor. Similarly, the eijk permutation symbol is a relative tensor of ijk weight +1 and the corresponding permutation symbol is an absolute tensor. permutation symbol is an EXAMPLE 1.3-10. ( permutation symbol) Show that eijk is a relative tensor of weight 1 and the corresponding absolute tensor. Solution: Examine the Jacobian J and make the substitution ai = j x = x x1 x1 x2 x1 x3 x1 x1 x2 x2 x2 x3 x2 x1 x3 x2 x3 x3 x3 ijk xi , i, j = 1, 2, 3. xj From the denition of a determinant we may write x eijk ai aj ak = J( )emnp . mnp x (1.3.31) By denition, emnp = emnp in all coordinate systems and hence equation (1.3.31) can be expressed in the form x J( ) x 1 eijk xi xj xk = emnp xm xn xp (1.3.32) which demonstrates that eijk transforms as a relative tensor of weight 1. We have previously shown the metric tensor gij is a second order covariant tensor and transforms xm xn . Taking the determinant of this result we nd according to the rule g ij = gmn xi xj g = |gij | = |gmn | xm xi 2 x = g J( ) x 2 (1.3.33) where g is the determinant of (gij ) and g is the determinant of (g ij ). This result demonstrates that g is a scalar invariant of weight +2. Taking the square root of this result we nd that g= x gJ( ). x (1.3.34) Consequently, we call g a scalar invariant of weight +1. Now multiply both sides of equation (1.3.32) by g and use (1.3.34) to verify the relation xi xj xk g eijk m n p = g emnp . x x x This equation demonstrates that the quantity ijk = g eijk transforms like an absolute tensor. (1.3.35) 84 Figure 1.3-14. Translation followed by rotation of axes In a similar manner one can show eijk is a relative tensor of weight +1 and tensor. This is left as an exercise. Another exercise found at the end of this section is to show that a generalization of the e identity is the epsilon identity g ij Cartesian Tensors Consider the motion of a rigid rod in two dimensions. No matter how complicated the movement of the rod is we can describe the motion as a translation followed by a rotation. Consider the rigid rod AB illustrated in the gure 1.3-13. ipt jrs ijk 1 eijk g = is an absolute = gpr gts gps gtr . (1.3.36) Figure 1.3-13. Motion of rigid rod In this gure there is a before and after picture of the rods position. By moving the point B to B we have a translation. This is then followed by a rotation holding B xed. 85 Figure 1.3-15. Rotation of axes A similar situation exists in three dimensions. Consider two sets of Cartesian axes, say a barred and unbarred system as illustrated in the gure 1.3-14. Let us translate the origin 0 to 0 and then rotate the (x, y, z) axes until they coincide with the (x, y, z) axes. We consider rst the rotation of axes when the origins 0 and 0 coincide as the translational distance can be represented by a vector bk , k = 1, 2, 3. When the origin 0 is translated to 0 we have the situation illustrated in the gure 1.3-15, where the barred axes can be thought of as a transformation due to rotation. Let r = x e1 + y e2 + z e3 (1.3.37) denote the position vector of a variable point P with coordinates (x, y, z) with respect to the origin 0 and the unit vectors e1 , e2 , e3 . This same point, when referenced with respect to the origin 0 and the unit vectors e1 , e2 , e3 , has the representation r = x e1 + y e2 + z e3 . (1.3.38) By considering the projections of r upon the barred and unbarred axes we can construct the transformation equations relating the barred and unbarred axes. We calculate the projections of r onto the x, y and z axes and nd: r e1 = x = x( e1 e1 ) + y( e2 e1 ) + z( e3 e1 ) r e2 = y = x( e1 e2 ) + y( e2 e2 ) + z( e3 e2 ) r e3 = z = x( e1 e3 ) + y( e2 e3 ) + z( e3 e3 ). We also calculate the projection of r onto the x, y, z axes and nd: r e1 = x = x( e1 e1 ) + y( e2 e1 ) + z( e3 e1 ) r e2 = y = x( e1 e2 ) + y( e2 e2 ) + z( e3 e2 ) r e3 = z = x( e1 e3 ) + y( e2 e3 ) + z( e3 e3 ). By introducing the notation (y1 , y2 , y3 ) = (x, y, z) (y 1 , y2 , y3 ) = (x, y, z) and dening ij as the angle j , we can represent the above transformation equations in a more concise between the unit vectors ei and e (1.3.40) (1.3.39) 86 form. We observe that the direction cosines can be written as 11 21 31 = e1 e1 = cos 11 = e2 e1 = cos 21 = e3 e1 = cos 31 12 22 32 = e1 e2 = cos 12 = e2 e2 = cos 22 = e3 e2 = cos 32 13 23 33 = e1 e3 = cos 13 = e2 e3 = cos 23 = e3 e3 = cos 33 (1.3.41) which enables us to write the equations (1.3.39) and (1.3.40) in the form yi = ij y j and yi = ji yj . (1.3.42) Using the index notation we represent the unit vectors as: er = where pr pr ep or ep = pr er (1.3.43) are the direction cosines. In both the barred and unbarred system the unit vectors are orthogonal and consequently we must have the dot products er ep = rp and em en = mn (1.3.44) where ij is the Kronecker delta. Substituting equation (1.3.43) into equation (1.3.44) we nd the direction cosines ij must satisfy the relations: er es = and er es = pr ep ms em = en = pr ms ep em = em en = pr ms pm = = mr ms = rs = rs . rm em sn rm sn rm sn mn rm sm The relations mr ms = rs and rm sm = rs , (1.3.45) with summation index m, are important relations which are satised by the direction cosines associated with a rotation of axes. Combining the rotation and translation equations we nd yi = ij y j rotation + bi translation . (1.3.46) We multiply this equation by ik and make use of the relations (1.3.45) to nd the inverse transformation yk = ik (yi bi ). (1.3.47) These transformations are called linear or ane transformations. Consider the xi axes as xed, while the xi axes are rotating with respect to the xi axes where both sets of axes have a common origin. Let A = Ai ei denote a vector xed in and rotating with the xi axes. We dA dA denote by and the derivatives of A with respect to the xed (f) and rotating (r) axes. We can dt f dt r 87 write, with respect to the xed axes, that d ei d ei dA dAi ei + Ai . Note that is the derivative of a = dt f dt dt dt vector with constant magnitude. Therefore there exists constants i , i = 1, . . . , 6 such that d e1 = 3 e2 2 e3 dt d e2 = 1 e3 4 e1 dt d e3 = 5 e1 6 e2 dt d e2 dt i.e. see page 80. From the dot product e1 e2 = 0 we obtain by dierentiation e1 + d e1 dt e2 = 0 which implies 4 = 3 . Similarly, from the dot products e1 e3 and e2 e3 we obtain by dierentiation the additional relations 5 = 2 and 6 = 1 . The derivative of A with respect to the xed axes can now be represented dA dt = f dAi dA ei + (2 A3 3 A2 ) e1 + (3 A1 1 A3 ) e2 + (1 A2 2 A1 ) e3 = dt dt +A r where = i ei is called an angular velocity vector of the rotating system. The term A represents the dA dAi velocity of the rotating system relative to the xed system and = ei represents the derivative with dt r dt respect to the rotating system. Employing the special transformation equations (1.3.46) let us examine how tensor quantities transform when subjected to a translation and rotation of axes. These are our special transformation laws for Cartesian tensors. We examine only the transformation laws for rst and second order Cartesian tensor as higher order transformation laws are easily discerned. We have previously shown that in general the rst and second order tensor quantities satisfy the transformation laws: yj y i y i A = Aj i yj y y mn A = Aij m n yi yj yi yj Amn = Aij y m yn y yj m An = Ai m j yi y n Ai = Aj dierentiate the equations (1.3.46) and (1.3.47) and nd yi = yk ij (1.3.48) (1.3.49) (1.3.50) (1.3.51) (1.3.52) For the special case of Cartesian tensors we assume that yi and y i , i = 1, 2, 3 are linearly independent. We yj = yk ij jk = ik , and y k = ym ik yi = ym ik im = mk . Substituting these derivatives into the transformation equations (1.3.48) through (1.3.52) we produce the transformation equations Ai = Aj A = Aj A mn i ji ji im jn im jn im jn . =A ij Amn = Aij m An = Ai j 88 Figure 1.3-16. Transformation to curvilinear coordinates These are the transformation laws when moving from one orthogonal system to another. In this case the direction cosines im are constants and satisfy the relations given in equation (1.3.45). The transformation laws for higher ordered tensors are similar in nature to those given above. In the unbarred system (y1 , y2 , y3 ) the metric tensor and conjugate metric tensor are: gij = ij and g ij = ij where ij is the Kronecker delta. In the barred system of coordinates, which is also orthogonal, we have g ij = From the orthogonality relations (1.3.45) we nd g ij = We examine the associated tensors Ai = g ij Aj Aij = g im g jn Amn Ai = g im Amn n Ai = gij Aj Amn = gmi gnj Aij Ai = gnj Aij n mi mj ym ym . y i yj = ij and gij = ij . and nd that the contravariant and covariant components are identical to one another. This holds also in the barred system of coordinates. Also note that these special circumstances allow the representation of contractions using subscript quantities only. This type of a contraction is not allowed for general tensors. It is left as an exercise to try a contraction on a general tensor using only subscripts to see what happens. Note that such a contraction does not produce a tensor. These special situations are considered in the exercises. Physical Components We have previously shown an arbitrary vector A can be represented in many forms depending upon the coordinate system and basis vectors selected. For example, consider the gure 1.3-16 which illustrates a Cartesian coordinate system and a curvilinear coordinate system. 89 Figure 1.3-17. Physical components In the Cartesian coordinate system we can represent a vector A as A = Ax e1 + Ay e2 + Az e3 where ( e1 , e2 , e3 ) are the basis vectors. Consider a coordinate transformation to a more general coordinate system, say (x1 , x2 , x3 ). The vector A can be represented with contravariant components as A = A1 E1 + A2 E2 + A3 E3 (1.3.53) with respect to the tangential basis vectors (E1 , E2 , E3 ). Alternatively, the same vector A can be represented in the form A = A1 E 1 + A2 E 2 + A3 E 3 (1.3.54) having covariant components with respect to the gradient basis vectors (E 1 , E 2 , E 3 ). These equations are just dierent ways of representing the same vector. In the above representations the basis vectors need not be orthogonal and they need not be unit vectors. In general, the physical dimensions of the components Ai and Aj are not the same. The physical components of the vector A in a direction is dened as the projection of A upon a unit vector in the desired direction. For example, the physical component of A in the direction E1 is A E1 |E1 | = A1 |E1 | = projection of A on E1 . (1.3.58) Similarly, the physical component of A in the direction E 1 is A E1 |E 1 | = A1 |E 1 | = projection of A on E 1 . (1.3.59) EXAMPLE 1.3-11. (Physical components) Let , , denote nonzero positive constants such that the product relation = 1 is satised. Consider the nonorthogonal basis vectors E1 = e1 , illustrated in the gure 1.3-17. E2 = e1 + e2 , E3 = e3 90 It is readily veried that the reciprocal basis is E 1 = e1 e2 , E 2 = e2 , E 3 = e3 . Consider the problem of representing the vector A = Ax e1 + Ay e2 in the contravariant vector form A = A1 E1 + A2 E2 This vector has the contravariant components A1 = A E 1 = Ax Ay and A2 = A E 2 = Ay . or tensor form Ai , i = 1, 2. Alternatively, this same vector can be represented as the covariant vector A = A1 E 1 + A2 E 2 which has the tensor form Ai , i = 1, 2. The covariant components are found from the relations A1 = A E1 = Ax A2 = A E2 = Ax + Ay . The physical components of A in the directions E 1 and E 2 are found to be: A A E1 |E 1 | E2 |E 2 | = = A1 |E 1 | A2 |E 2 | = = Ax Ay 2 + 2 = A(1) Ay = Ay = A(2). Note that these same results are obtained from the dot product relations using either form of the vector A. For example, we can write A and A E1 |E 1 | E2 |E 2 | = = A1 (E 1 E 1 ) + A2 (E 2 E 1 ) |E 1 | A1 (E 1 E 2 ) + A2 (E 2 E 2 ) |E 2 | = A(1) = A(2). In general, the physical components of a vector A in a direction of a unit vector i is the generalized dot product in VN . This dot product is an invariant and can be expressed gij Ai j = Ai i = Ai i = projection of A in direction of i 91 Physical Components For Orthogonal Coordinates In orthogonal coordinates observe the element of arc length squared in V3 is ds2 = gij dxi dxj = (h1 )2 (dx1 )2 + (h2 )2 (dx2 )2 + (h3 )2 (dx3 )2 where 0 0 . (h3 )2 (h1 )2 0 gij = 0 0 (h2 )2 0 (1.3.60) In this case the curvilinear coordinates are orthogonal and h2 = g(i)(i) (i) i not summed and gij = 0, i = j. At an arbitrary point in this coordinate system we take i , i = 1, 2, 3 as a unit vector in the direction of the coordinate x1 . We then obtain 1 = This is a unit vector since 1 = gij i j = g11 1 1 = h2 (1 )2 1 or 1 = 1 h1 . dx1 , ds 2 = 0, 3 = 0. Here the curvilinear coordinate system is orthogonal and in this case the physical component of a vector Ai , in the direction xi , is the projection of Ai on i in V3 . The projection in the x1 direction is determined from A(1) = gij Ai j = g11 A1 1 = h2 A1 1 1 = h1 A1 . h1 Similarly, we choose unit vectors i and i , i = 1, 2, 3 in the x2 and x3 directions. These unit vectors can be represented 1 =0, 1 =0, 2 = 1 dx2 = , ds h2 2 =0, 3 =0 3 = 1 dx3 = ds h3 and the physical components of the vector Ai in these directions are calculated as A(2) = h2 A2 and A(3) = h3 A3 . In summary, we can say that in an orthogonal coordinate system the physical components of a contravariant tensor of order one can be determined from the equations A(i) = h(i) A(i) = g(i)(i) A(i) , i = 1, 2 or 3 no summation on i, which is a short hand notation for the physical components (h1 A1 , h2 A2 , h3 A3 ). In an orthogonal coordinate system the nonzero conjugate metric components are g (i)(i) = 1 , g(i)(i) i = 1, 2, or 3 no summation on i. 92 These components are needed to calculate the physical components associated with a covariant tensor of order one. For example, in the x1 direction, we have the covariant components 1 = g11 1 = h2 1 1 = h1 , h1 2 = 0, 3 = 0 and consequently the projection in V3 can be represented gij Ai j = gij Ai g jm m = Aj g jm m = A1 1 g 11 = A1 h1 In a similar manner we calculate the relations A(2) = A2 h2 and A(3) = A3 h3 1 A1 = = A(1). h2 h1 1 for the other physical components in the directions x2 and x3 . These physical components can be represented in the short hand notation A(i) = A(i) A(i) = , h(i) g(i)(i) i = 1, 2 or 3 no summation on i. In an orthogonal coordinate system the physical components associated with both the contravariant and covariant components are the same. To show this we note that when Ai gij = Aj is summed on i we obtain A1 g1j + A2 g2j + A3 g3j = Aj . Since gij = 0 for i = j this equation reduces to A(i) g(i)(i) = A(i) , Another form for this equation is A(i) A(i) = A(i) g(i)(i) = g(i)(i) i not summed, i not summed. which demonstrates that the physical components associated with the contravariant and covariant components are identical. NOTATION The physical components are sometimes expressed by symbols with subscripts which represent the coordinate curve along which the projection is taken. For example, let H i denote the contravariant components of a rst order tensor. The following are some examples of the representation of the physical components of H i in various coordinate systems: orthogonal coordinates general rectangular cylindrical spherical general coordinate system (x1 , x2 , x3 ) (x, y, z) (r, , z) (, , ) (u, v, w) tensor components Hi H H H H i i i i physical components H(1), H(2), H(3) Hx , Hy , Hz Hr , H , Hz H , H , H Hu , Hv , Hw 93 Higher Order Tensors The physical components associated with higher ordered tensors are dened by projections in VN just like the case with rst order tensors. For an nth ordered tensor Tij...k we can select n unit vectors i , i , . . . , i and form the inner product (projection) Tij...k i j . . . k . When projecting the tensor components onto the coordinate curves, there are N choices for each of the unit vectors. This produces N n physical components. The above inner product represents the physical component of the tensor Tij...k along the directions of the unit vectors i , i , . . . , i . The selected unit vectors may or may not be orthogonal. In the cases where the selected unit vectors are all orthogonal to one another, the calculation of the physical components is greatly simplied. By relabeling the unit vectors i , i , . . . , i where (m), (n), ..., (p) represent one of (m) (n) (p) the N directions, the physical components of a general nth order tensor is represented T (m n . . . p) = Tij...k i j . . . k (m) (n) (p) EXAMPLE 1.3-12. (Physical components) In an orthogonal curvilinear coordinate system V3 with metric gij , i, j = 1, 2, 3, nd the physical components of (i) the second order tensor Aij . (ii) the second order tensor Aij . (iii) the second order tensor Ai . j Solution: The physical components of Amn , m, n = 1, 2, 3 along the directions of two unit vectors i and i is dened as the inner product in V3 . These physical components can be expressed A(ij) = Amn m n (i) (j) i, j = 1, 2, 3, where the subscripts (i) and (j) represent one of the coordinate directions. Dropping the subscripts (i) and (j), we make the observation that in an orthogonal curvilinear coordinate system there are three choices for the direction of the unit vector i and also three choices for the direction of the unit vector i . These three choices represent the directions along the x1 , x2 or x3 coordinate curves which emanate from a point of the curvilinear coordinate system. This produces a total of nine possible physical components associated with the tensor Amn . For example, we can obtain the components of the unit vector i , i = 1, 2, 3 in the x1 direction directly from an examination of the element of arc length squared ds2 = (h1 )2 (dx1 )2 + (h2 )2 (dx2 )2 + (h3 )2 (dx3 )2 . By setting dx2 = dx3 = 0, we nd 1 dx1 = = 1 , ds h1 2 = 0, 3 = 0. This is the vector i , i = 1, 2, 3. Similarly, if we choose to select the unit vector i , i = 1, 2, 3 in the x2 (1) direction, we set dx1 = dx3 = 0 in the element of arc length squared and nd the components 1 = 0, 2 = 1 dx2 = , ds h2 3 = 0. 94 This is the vector i , i = 1, 2, 3. Finally, if we select i , i = 1, 2, 3 in the x3 direction, we set dx1 = dx2 = 0 (2) in the element of arc length squared and determine the unit vector 1 = 0, 2 = 0, 3 = 1 dx3 = . ds h3 This is the vector i , i = 1, 2, 3. Similarly, the unit vector i can be selected as one of the above three (3) directions. Examining all nine possible combinations for selecting the unit vectors, we calculate the physical components in an orthogonal coordinate system as: A11 h1 h1 A21 A(21) = h1 h2 A31 A(31) = h3 h1 A(11) = A12 h1 h2 A22 A(22) = h2 h2 A32 A(32) = h3 h2 A(12) = A13 h1 h3 A23 A(23) = h2 h3 A33 A(33) = h3 h3 A(13) = These results can be written in the more compact form A(ij) = For mixed tensors we have Ai = g im Amj = g i1 A1j + g i2 A2j + g i3 A3j . j (1.3.62) A(i)(j) h(i) h(j) no summation on i or j . (1.3.61) From the fact g ij = 0 for i = j, together with the physical components from equation (1.3.61), the equation (1.3.62) reduces to A(j) = g (i)(i) A(i)(j) = (i) 1 h(i) h(j) A(ij) no summation on i and i, j = 1, 2 or 3. h2 (i) This can also be written in the form A(ij) = A(j) (i) h(i) h(j) no summation on i or j. (1.3.63) Hence, the physical components associated with the mixed tensor Ai in an orthogonal coordinate system j can be expressed as A(11) = A1 1 h2 h1 h3 A(31) = A3 1 h1 A(21) = A2 1 A(12) = A1 2 A(22) = A2 2 A(32) = A3 2 h3 h2 h1 h2 h1 h3 h2 A(23) = A2 3 h3 A(33) = A3 . 3 A(13) = A1 3 For second order contravariant tensors we may write Aij gjm = Ai = Ai1 g1m + Ai2 g2m + Ai3 g3m . m 95 We use the fact gij = 0 for i = j together with the physical components from equation (1.3.63) to reduce the above equation to the form A(m) = A(i)(m) g(m)(m) we have h(m) A(im) = A(i)(m) h2 (m) h(i) or A(im) = A(i)(m) h(i) h(m) . no summation i, m = 1, 2, 3 (1.3.64) (i) no summation on m . In terms of physical components Examining the results from equation (1.3.64) we nd that the physical components associated with the contravariant tensor Aij , in an orthogonal coordinate system, can be written as: A(11) = A11 h1 h1 A(21) = A21 h2 h1 A(31) = A31 h3 h1 A(12) = A12 h1 h2 A(22) = A22 h2 h2 A(32) = A32 h3 h2 A(13) = A13 h1 h3 A(23) = A23 h2 h3 A(33) = A33 h3 h3 . Physical Components in General In an orthogonal curvilinear coordinate system, the physical components associated with the nth order tensor Tij...kl along the curvilinear coordinate directions can be represented: T (ij . . . kl) = T(i)(j)...(k)(l) h(i) h(j) . . . h(k) h(l) no summations. These physical components can be related to the various tensors associated with Tij...kl . For example, in ij...m an orthogonal coordinate system, the physical components associated with the mixed tensor Tn...kl can be expressed as: T (ij . . . m n . . . kl) = T(n)...(k)(l) (i)(j)...(m) h(i) h(j) . . . h(m) h(n) . . . h(k) h(l) no summations. (1.3.65) EXAMPLE 1.3-13. (Physical components) Let xi = xi (t), i = 1, 2, 3 denote the position vector of a particle which moves as a function of time t. Assume there exists a coordinate transformation xi = xi (x), for i = 1, 2, 3, of the form given by equations (1.2.33). The position of the particle when referenced with respect to the barred system of coordinates can be found by substitution. The generalized velocity of the particle in the unbarred system is a vector with components vi = dxi , i = 1, 2, 3. dt The generalized velocity components of the same particle in the barred system is obtained from the chain rule. We nd this velocity is represented by vi = dxi xi dxj xi j v. = = j dt dt x xj This equation implies that the contravariant quantities (v 1 , v 2 , v 3 ) = ( dx1 dx2 dx3 , , ) dt dt dt 96 are tensor quantities. These quantities are called the components of the generalized velocity. The coordinates x1 , x2 , x3 are generalized coordinates. This means we can select any set of three independent variables for the representation of the motion. The variables selected might not have the same dimensions. For example, in cylindrical coordinates we let (x1 = r, x2 = , x3 = z). Here x1 and x3 have dimensions of distance but x2 has dimensions of angular displacement. The generalized velocities are v1 = dr dx1 = , dt dt v2 = d dx2 = , dt dt v3 = dz dx3 = . dt dt Here v 1 and v 3 have units of length divided by time while v 2 has the units of angular velocity or angular change divided by time. Clearly, these dimensions are not all the same. Let us examine the physical components of the generalized velocities. We nd in cylindrical coordinates h1 = 1, h2 = r, h3 = 1 and the physical components of the velocity have the forms: vr = v(1) = v 1 h1 = dr , dt v = v(2) = v 2 h2 = r d , dt vz = v(3) = v 3 h3 = dz . dt Now the physical components of the velocity all have the same units of length divided by time. Additional examples of the use of physical components are considered later. For the time being, just remember that when tensor equations are derived, the equations are valid in any generalized coordinate system. In particular, we are interested in the representation of physical laws which are to be invariant and independent of the coordinate system used to represent these laws. Once a tensor equation is derived, we can chose any type of generalized coordinates and expand the tensor equations. Before using any expanded tensor equations we must replace all the tensor components by their corresponding physical components in order that the equations are dimensionally homogeneous. It is these expanded equations, expressed in terms of the physical components, which are used to solve applied problems. Tensors and Multilinear Forms Tensors can be thought of as being created by multilinear forms dened on some vector space V. Let us dene on a vector space V a linear form, a bilinear form and a general multilinear form. We can then illustrate how tensors are created from these forms. Denition: (Linear form) Let V denote a vector space which contains vectors x, x1 , x2 , . . . . A linear form in x is a scalar function (x) having a single vector argument x which satises the linearity properties: (i) (ii) (x1 + x2 ) = (x1 ) + (x2 ) (x1 ) = (x1 ) (1.3.66) for all arbitrary vectors x1 , x2 in V and all real numbers . 97 An example of a linear form is the dot product relation (x) = A x where A is a constant vector and x is an arbitrary vector belonging to the vector space V. Note that a linear form in x can be expressed in terms of the components of the vector x and the base vectors ( e1 , e2 , e3 ) used to represent x. To show this, we write the vector x in the component form x = xi ei = x1 e1 + x2 e2 + x3 e3 , where xi , i = 1, 2, 3 are the components of x with respect to the basis vectors ( e1 , e2 , e3 ). By the linearity property of we can write (x) = (xi ei ) = (x1 e1 + x2 e2 + x3 e3 ) = (x1 e1 ) + (x2 e2 ) + (x3 e3 ) = x1 ( e1 ) + x2 ( e2 ) + x3 ( e3 ) = xi ( ei ) Thus we can write (x) = xi ( ei ) and by dening the quantity ( ei ) = ai as a tensor we obtain (x) = xi ai . Note that if we change basis from ( e1 , e2 , e3 ) to (E1 , E2 , E3 ) then the components of x also must change. Letting xi denote the components of x with respect to the new basis, we would have x = xi Ei and (x) = (xi Ei ) = xi (Ei ). (1.3.67) The linear form denes a new tensor ai = (Ei ) so that (x) = xi ai . Whenever there is a denite relation between the basis vectors ( e1 , e2 , e3 ) and (E1 , E2 , E3 ), say, Ei = xj ej , xi then there exists a denite relation between the tensors ai and ai . This relation is ai = (Ei ) = ( xj xj xj aj . i ej ) = i ( ej ) = x x xi This is the transformation law for an absolute covariant tensor of rank or order one. The above idea is now extended to higher order tensors. Denition: ( Bilinear form) the linearity properties: A bilinear form in x and y is a scalar function (x, y) with two vector arguments, which satises (i) (x1 + x2 , y1 ) = (x1 , y1 ) + (x2 , y1 ) (ii) (x1 , y1 + y2 ) = (x1 , y1 ) + (x1 , y2 ) (iii) (x1 , y1 ) = (x1 , y1 ) (iv) (x1 , y1 ) = (x1 , y1 ) for arbitrary vectors x1 , x2 , y1 , y2 in the vector space V and for all real numbers . (1.3.68) 98 Note in the denition of a bilinear form that the scalar function is linear in both the arguments x and y. An example of a bilinear form is the dot product relation (x, y) = x y where both x and y belong to the same vector space V. The denition of a bilinear form suggests how multilinear forms can be dened. (1.3.69) Denition: (Multilinear forms) A multilinear form of degree M or a M degree linear form in the vector arguments x1 , x2 , . . . , xM is a scalar function (x1 , x2 , . . . , xM ) of M vector arguments which satises the property that it is a linear form in each of its arguments. That is, must satisfy for each j = 1, 2, . . . , M the properties: (i) (x1 , . . . , xj1 + xj2 , . . . xM ) = (x1 , . . . , xj1 , . . . , xM ) + (x1 , . . . , xj2 , . . . , xM ) (ii) (x1 , . . . , xj , . . . , xM ) = (x1 , . . . , xj , . . . , xM ) (1.3.70) for all arbitrary vectors x1 , . . . , xM in the vector space V and all real numbers . An example of a third degree multilinear form or trilinear form is the triple scalar product (x, y, z) = x (y z). (1.3.71) Note that multilinear forms are independent of the coordinate system selected and depend only upon the vector arguments. In a three dimensional vector space we select the basis vectors ( e1 , e2 , e3 ) and represent all vectors with respect to this basis set. For example, if x, y, z are three vectors we can represent these vectors in the component forms x = xi ei , y = y j ej , z = z k ek (1.3.72) where we have employed the summation convention on the repeated indices i, j and k. Substituting equations (1.3.72) into equation (1.3.71) we obtain (xi ei , y j ej , z k ek ) = xi y j z k ( ei , ej , ek ), since is linear in all its arguments. By dening the tensor quantity ( ei , ej , ek ) = eijk (1.3.74) (1.3.73) 99 (See exercise 1.1, problem 15) the trilinear form, given by equation (1.3.71), with vectors from equations (1.3.72), can be expressed as (x, y, z) = eijk xi y j z k , i, j, k = 1, 2, 3. (1.3.75) The coecients eijk of the trilinear form is called a third order tensor. It is the familiar permutation symbol considered earlier. In a multilinear form of degree M , (x, y, . . . , z), the M arguments can be represented in a component form with respect to a set of basis vectors ( e1 , e2 , e3 ). Let these vectors have components xi , y i , z i , i = 1, 2, 3 with respect to the selected basis vectors. We then can write x = xi ei , y = y j ej , z = z k ek . Substituting these vectors into the M degree multilinear form produces (xi ei , y j ej , . . . , z k ek ) = xi y j z k ( ei , ej , . . . , ek ). Consequently, the multilinear form denes a set of coecients aij...k = ( ei , ej , . . . , ek ) (1.3.77) (1.3.76) which are referred to as the components of a tensor of order M. The tensor is thus created by the multilinear form and has M indices if is of degree M. Note that if we change to a dierent set of basis vectors, say, (E1 , E2 , E3 ) the multilinear form denes a new tensor aij...k = (Ei , Ej , . . . , Ek ). (1.3.78) This new tensor has a bar over it to distinguish it from the previous tensor. A denite relation exists between the new and old basis vectors and consequently there exists a denite relation between the components of the barred and unbarred tensors components. Recall that if we are given a set of transformation equations y i = y i (x1 , x2 , x3 ), i = 1, 2, 3, (1.3.79) from rectangular to generalized curvilinear coordinates, we can express the basis vectors in the new system by the equations y j ej , i = 1, 2, 3. (1.3.80) xi For example, see equations (1.3.11) with y 1 = x, y 2 = y, y 3 = z, x1 = u, x2 = v, x3 = w. Substituting Ei = equations (1.3.80) into equations (1.3.78) we obtain y y y e, e , . . . , k e ). i xj x x By the linearity property of , this equation is expressible in the form aij...k = ( y y y . . . k ( e , e , . . . , e ) i xj x x y y y aij...k = . . . k a... xi xj x This is the familiar transformation law for a covariant tensor of degree M. By selecting reciprocal basis aij...k = vectors the corresponding transformation laws for contravariant vectors can be determined. The above examples illustrate that tensors can be considered as quantities derivable from multilinear forms dened on some vector space. 100 Dual Tensors The e-permutation symbol is often used to generate new tensors from given tensors. For Ti1 i2 ...im a skew-symmetric tensor, we dene the tensor 1 j1 j2 ...jnm i1 i2 ...im T j1 j2 ...jnm = e Ti1 i2 ...im m! mn (1.3.81) as the dual tensor associated with Ti1 i2 ...im . Note that the e-permutation symbol or alternating tensor has a weight of +1 and consequently the dual tensor will have a higher weight than the original tensor. The e-permutation symbol has the following properties ei1 i2 ...iN ei1 i2 ...iN = N ! i ei1 i2 ...iN ej1 j2 ...jN = j1 i2 ...iN 1 j2 ...jN j ek1 k2 ...km i1 i2 ...iN m ej1 j2 ...jm i1 i2 ...iN m = (N m)!k1 j22...jm 1 k ...km j k1 j22...jm Tj1 j2 ...jm = m!Tk1 k2 ...km . 1 k ...km (1.3.82) Using the above properties we can solve for the skew-symmetric tensor in terms of the dual tensor. We nd Ti1 i2 ...im = 1 T j1 j2 ...jnm . ei i ...i j j ...j (n m)! 1 2 m 1 2 nm (1.3.83) For example, if Aij i, j = 1, 2, 3 is a skew-symmetric tensor, we may associate with it the dual tensor Vi = 1 ijk e Ajk , 2! which is a rst order tensor or vector. Note that Aij has the components 0 A12 A13 A12 0 A23 A13 A23 0 and consequently, the components of the vector V are (V 1 , V 2 , V 3 ) = (A23 , A31 , A12 ). (1.3.84) (1.3.85) Note that the vector components have a cyclic order to the indices which comes from the cyclic properties of the e-permutation symbol. As another example, consider the fourth order skew-symmetric tensor Aijkl , i, j, k, l = 1, . . . , n. We can associate with this tensor any of the dual tensor quantities 1 ijkl e Aijkl 4! 1 V i = eijklm Ajklm 4! 1 V ij = eijklmn Aklmn 4! 1 V ijk = eijklmnp Almnp 4! 1 V ijkl = eijklmnpr Amnpr 4! V= Applications of dual tensors can be found in section 2.2. (1.3.86) 101 EXERCISE 1.3 1. (a) From the transformation law for the second order tensor g ij = gab solve for the gab in terms of gij . xa xb xi xj (b) Show that if gij is symmetric in one coordinate system it is symmetric in all coordinate systems. x (c) Let g = det(gij ) and g = det(gij ) and show that g = gJ 2 ( x ) and consequently g = gJ( ). This x x shows that g is a scalar invariant of weight 2 and g is a scalar invariant of weight 1. 2. For gij = 3. y m y m xi xj show that g ij = xi xj y m y m Show that in a curvilinear coordinate system which is orthogonal we have: (a) (b) (c) g = det(gij ) = g11 g22 g33 gmn = g mn = 0 for m = n 1 gN N = for N = 1, 2, 3 (no summation on N) gN N y i xj 2 4. Show that g = det(gij ) = = J 2 , where J is the Jacobian. r |, u h2 = hv = | r |, v h3 = hw = | r | and construct the unit w 5. Dene the quantities h1 = hu = | eu = vectors 1 r 1 r 1 r , ev = , ew = . h1 u h2 v h3 w (a) Assume the coordinate system is orthogonal and show that g11 = h2 = 1 g22 = h2 = 2 g33 = h2 = 3 x u x v x w 2 + 2 y u y v y w 2 + 2 z u z v z w 2 , 2 + 2 + 2 , 2 + + . (b) Show that dr can be expressed in the form dr = h1 eu du + h2 ev dv + h3 ew dw. (c) Show that the volume of the elemental parallelepiped having dr as diagonal can be represented d = Hint: A1 |A (B C)| = B1 C1 A2 B2 C2 A3 B3 C3 (x, y, z) dudvdw. g dudvdw = J dudvdw = (u, v, w) 102 Figure 1.3-18 Oblique cylindrical coordinates. 6. For the change dr given in problem 5, show the elemental parallelepiped with diagonal dr has: 2 g22 g33 g23 dvdw in the u =constant surface. 2 g33 g11 g13 dudw in the v =constant surface. 2 g11 g22 g12 dudv in the w =constant surface. (a) the element of area dS1 = (b) The element of area dS2 = (c) the element of area dS3 = (d) What do the above elements of area reduce to in the special case the curvilinear coordinates are orthogonal? Hint: |A B| = = 7. (A B) (A B) (A A)(B B) (A B)(A B) ij xj . In Cartesian coordinates you are given the ane transformation. xi = x1 = 1 (5x1 14x2 + 2x3 ), 15 1 x2 = (2x1 + x2 + 2x3 ), 3 x3 = where 1 (10x1 + 2x2 11x3 ) 15 (a) Show the transformation is orthogonal. (b) A vector A(x1 , x2 , x3 ) in the unbarred system has the components A1 = (x1 )2 , A2 = (x2 )2 A3 = (x3 )2 . Find the components of this vector in the barred system of coordinates. 8. 9. 10. 11. 12. Calculate the metric and conjugate metric tensors in cylindrical coordinates (r, , z). Calculate the metric and conjugate metric tensors in spherical coordinates (, , ). Calculate the metric and conjugate metric tensors in parabolic cylindrical coordinates (, , z). Calculate the metric and conjugate metric components in elliptic cylindrical coordinates (, , z). Calculate the metric and conjugate metric components for the oblique cylindrical coordinates (r, , ), y = r sin + cos , z = sin and is a parameter 2. illustrated in gure 1.3-18, where x = r cos , 0< Note: When = 2 cylindrical coordinates result. 103 13. Calculate the metric and conjugate metric tensor associated with the toroidal surface coordinates (, ) illustrated in the gure 1.3-19, where x = (a + b cos ) cos y = (a + b cos ) sin z = b sin a>b>0 0 < < 2 0 < < 2 Figure 1.3-19. Toroidal surface coordinates 14. Calculate the metric and conjugate metric tensor associated with the spherical surface coordinates (, ), illustrated in the gure 1.3-20, where x = a sin cos y = a sin sin z = a cos 15. a>0 is constant 0 < < 2 0<< 2 Consider gij , i, j = 1, 2 g22 g12 g11 , g 12 = g 21 = , g 22 = where = g11 g22 g12 g21 . (a) Show that g 11 = ik k (b) Use the results in part (a) and verify that gij g = j , i, j, k = 1, 2. 16. Let Ax , Ay , Az denote the constant components of a vector in Cartesian coordinates. Using the transformation laws (1.2.42) and (1.2.47) to nd the contravariant and covariant components of this vector upon changing to (a) cylindrical coordinates (r, , z). (b) spherical coordinates (, , ) and (c) Parabolic cylindrical coordinates. 17. Find the relationship which exists between the given associated tensors. (a) Apqk r. (b) Ap .mrs and Apq rs and Apq ..rs (c) (d) Ai.j. .l.m Amnk and A.s.p r.t. and Aij ..k 104 Figure 1.3-20. Spherical surface coordinates 18. Given the fourth order tensor Cikmp = ik mp + (im kp + ip km ) + (im kp ip km ) where , ij xj and are scalars and ij is the Kronecker delta. Show that under an orthogonal transformation of rotation of axes with xi = where rs is = mr mi = ri the components of the above tensor are unaltered. Any tensor whose components are unaltered under an orthogonal transformation is called an isotropic tensor. Another way of stating this problem is to say Show Cikmp is an isotropic tensor. 19. Assume Aijl is a third order covariant tensor and B pqmn is a fourth order contravariant tensor. Prove that Aikl B klmn is a mixed tensor of order three, with one covariant and two contravariant indices. 20. Assume that Tmnrs is an absolute tensor. Show that if Tijkl + Tijlk = 0 in the coordinate system xr then T ijkl + T ijlk = 0 in any other coordinate system xr . 21. Show that ijk rst gir = gjr gkr gis gjs gks git gjt gkt Hint: See problem 38, Exercise 1.1 22. 23. 24. rmn Determine if the tensor equation Prove the epsilon identity g ij mnp mij + mnj mpi = mni mpj is true or false. Justify your answer. ipt jrs = gpr gts gps gtr . Hint: See problem 38, Exercise 1.1 1 Let Ars denote a skew-symmetric contravariant tensor and let cr = rmn Amn where 2 = germn . Show that cr are the components of a covariant tensor. Write out all the components. Let Ars denote a skew-symmetric covariant tensor and let cr = rmn 1 rmn Amn where 2 Show that cr are the components of a contravariant tensor. Write out all the components. 25. 1 = ermn . g 105 26. qs s qs s Let Apq Br = Cpr where Br is a relative tensor of weight 1 and Cpr is a relative tensor of weight 2 . Prove that Apq is a relative tensor of weight (2 1 ). 27. 28. 29. (a) Show eijk is a relative tensor of weight +1. (b) Show 30. ijk When Ai is an absolute tensor prove that j When Ai is an absolute tensor prove that j i gAj is a relative tensor of weight +1. 1 Ai gj is a relative tensor of weight 1. = 1 eijk g is an absolute tensor. Hint: See example 1.1-25. The equation of a surface can be represented by an equation of the form (x1 , x2 , x3 ) = constant. Show that a unit normal vector to the surface can be represented by the vector ni = g ij xj 1 (g mn xm xn ) 2 . 31. 32. Assume that g ij = gij with a nonzero constant. Find and calculate g ij in terms of g ij . Determine if the following tensor equation is true. Justify your answer. r rjk Ai + r irk Aj + r ijr Ak = r ijk Ar . Hint: See problem 21, Exercise 1.1. 33. 34. 35. 36. Show that for Ci and C i associated tensors, and C i = Prove that Show ijk ijk ijk Aj Bk , then Ci = jk ijk A B and ijk are associated tensors. Hint: Consider the determinant of gij . Ai Bj Ck = ijk ijk A B C . Let Tji , i, j = 1, 2, 3 denote a second order mixed tensor. Show that the given quantities are scalar (i) I1 = Tii 1 i (Tii )2 Tm Tim (ii) I2 = 2 (iii) I3 = det|Tji | invariants. 37. (a) Assume Aij and B ij , i, j = 1, 2, 3 are absolute contravariant tensors, and determine if the inner product C ik = Aij B jk is an absolute tensor? xj xj = nm is satised, and determine whether the inner product in (b) Assume that the condition xn xm part (a) is a tensor? (c) Consider only transformations which are a rotation and translation of axes y i = y j yj direction cosines for the rotation of axes. Show that = nm yn ym ij yj + bi , where ij are 106 38. For Aijk a Cartesian tensor, determine if a contraction on the indices i and j is allowed. That (summation on i) is a tensor. Hint: See part(c) of the previous is, determine if the quantity Ak = Aiik , problem. 39. 40. jk jk Prove the e- identity eijk eimn = m n n m . Consider the vector Vk , k = 1, 2, 3 and dene the matrix (aij ) having the elements aij = eijk Vk , where eijk is the epermutation symbol. (a) Solve for Vi in terms of amn by multiplying both sides of the given equation by eijl and note the e identity allows us to simplify the result. (b) Sum the given expression on k and then assign values to the free indices (i,j=1,2,3) and compare your results with part (a). (c) Is aij symmetric, skew-symmetric, or neither? 41. It can be shown that the continuity equation of uid dynamics can be expressed in the tensor form 1 = 0, ( g V r) + r g x t where is the density of the uid, t is time, V r , with r = 1, 2, 3 are the velocity components and g = |gij | is the determinant of the metric tensor. Employing the summation convention and replacing the tensor components of velocity by their physical components, express the continuity equation in (a) Cartesian coordinates (x, y, z) with physical components Vx , Vy , Vz . (b) Cylindrical coordinates (r, , z) with physical components Vr , V , Vz . (c) Spherical coordinates (, , ) with physical components V , V , V . 42. Let x1 , x2 , x3 denote a set of skewed coordinates with respect to the Cartesian coordinates y 1 , y 2 , y 3 . Assume that E1 , E2 , E3 are unit vectors in the directions of the x1 , x2 and x3 axes respectively. If the unit vectors satisfy the relations E1 E1 = 1 E2 E2 = 1 E3 E3 = 1 E1 E2 = cos 12 E1 E3 = cos 13 E2 E3 = cos 23 , then calculate the metrices gij and conjugate metrices g ij . 43. Let Aij , i, j = 1, 2, 3, 4 denote the skew-symmetric second rank tensor 0 a b d a 0 Aij = b d 0 c e f c e , f 0 where a, b, c, d, e, f are complex constants. Calculate the components of the dual tensor V ij = 1 ijkl e Akl . 2 107 In Cartesian coordinates the vorticity tensor at a point in a uid medium is dened 1 Vj Vi j ij = 2 xi x where Vi are the velocity components of the uid at the point. The vorticity vector at a point in a uid 1 medium in Cartesian coordinates is dened by i = eijk jk . Show that these tensors are dual tensors. 2 Write out the relation between each of the components of the dual tensors 1 T ij = eijkl Tkl i, j, k, l = 1, 2, 3, 4 2 and show that if ijkl is an even permutation of 1234, then T ij = Tkl . 46. Consider the general ane transformation xi = aij xj where (x1 , x2 , x3 ) = (x, y, z) with inverse 45. 44. transformation xi = bij xj . Determine (a) the image of the plane Ax + By + Cz + D = 0 under this transformation and (b) the image of a second degree conic section Ax2 + 2Bxy + Cy 2 + Dx + Ey + F = 0. 47. Using a multilinear form of degree M, derive the transformation law for a contravariant vector of g gij = gg ij k . xk x degree M. 48. 49. Let g denote the determinant of gij and show that We have shown that for a rotation of xyz axes with respect to a set of xed xy z axes, the derivative dA dt dA dt of a vector A with respect to an observer on the barred axes is given by = f + A. r Introduce the operators Df A = Dr A = (a) Show that Df A = (Dr + )A. (b) Consider the special case that the vector A is the position vector r. Show that Df r = (Dr + )r produces V f dA dt dA dt = derivative in xed system f = derivative in rotating system r =V r + r where V f represents the velocity of a particle relative to the xed system and V r represents the velocity of a particle with respect to the rotating system of coordinates. =a f r (c) Show that a + ( r) where a f represents the acceleration of a particle relative to the xed system and a r represents the acceleration of a particle with respect to the rotating system. (d) Show in the special case is a constant that a f = 2 V + ( r) where V is the velocity of the particle relative to the rotating system. The term 2 V is referred to as the Coriolis acceleration and the term ( r) is referred to as the centripetal acceleration. 108 1.4 DERIVATIVE OF A TENSOR In this section we develop some additional operations associated with tensors. Historically, one of the basic problems of the tensor calculus was to try and nd a tensor quantity which is a function of the metric gij 2 gij tensor gij and some of its derivatives , , . . . . A solution of this problem is the fourth order xm xm xn Riemann Christoel tensor Rijkl to be developed shortly. In order to understand how this tensor was arrived at, we must rst develop some preliminary relationships involving Christoel symbols. Christoel Symbols Let us consider the metric tensor gij which we know satises the transformation law g = gab Dene the quantity (, , ) = g 2 xa xb xa 2 xb gab xc xa xb + gab + gab = x xc x x x x x x x x x 1 [(, , ) + (, , ) (, , )] to obtain the result 2 (1.4.1) xa xb . x x and form the combination of terms g g 1 g xb 2 xa gbc gca xa xb xc 1 gab + + + gab . = 2 x x 2 xc xa xb x x x x x x x In this equation the combination of derivatives occurring inside the brackets is called a Christoel symbol of the rst kind and is dened by the notation [ac, b] = [ca, b] = 1 gab gbc gac + . 2 xc xa xb (1.4.2) The equation (1.4.1) denes the transformation for a Christoel symbol of the rst kind and can be expressed as [ , ] = [ac, b] xa xb xc 2 xa xb + gab . x x x x x x (1.4.3) Observe that the Christoel symbol of the rst kind [ac, b] does not transform like a tensor. However, it is symmetric in the indices a and c. At this time it is convenient to use the equation (1.4.3) to develop an expression for the second derivative term which occurs in that equation as this second derivative term arises in some of our future considerations. x de To solve for this second derivative we can multiply equation (1.4.3) by g and simplify the result to the xd form xa xc x 2 xe = g de[ac, d] + [ , ] d g de . (1.4.4) x x x x x The transformation g de = g xd xe allows us to express the equation (1.4.4) in the form x x (1.4.5) 2 xe xa xc xe de [ , ] . = g [ac, d] +g x x x x x 109 Dene the Christoel symbol of the second kind as i jk = i kj = g i [jk, ] = 1 i g 2 gk gj gjk + xj xk x . (1.4.6) This Christoel symbol of the second kind is symmetric in the indices j and k and from equation (1.4.5) we see that it satises the transformation law xe = x e ac xa xc 2 xe . + x x x x (1.4.7) Observe that the Christoel symbol of the second kind does not transform like a tensor quantity. We can use the relation dened by equation (1.4.7) to express the second derivative of the transformation equations in terms of the Christoel symbols of the second kind. At times it will be convenient to represent the Christoel symbols with a subscript to indicate the metric from which they are calculated. Thus, an alternative notation for i jk i jk is the notation . g EXAMPLE 1.4-1. (Christoel symbols) Solve for the Christoel symbol of the rst kind in terms of the Christoel symbol of the second kind. Solution: By the denition from equation (1.4.6) we have i jk We multiply this equation by gi and nd gi and so [jk, ] = gi i jk = g1 1 jk + + gN N jk . i jk = [jk, ] = [jk, ] = g i [jk, ]. EXAMPLE 1.4-2. (Christoel symbols of rst kind) Derive formulas to nd the Christoel symbols of the rst kind in a generalized orthogonal coordinate system with metric coecients gij = 0 where i is not summed. Solution: In an orthogonal coordinate system where gij = 0 for i = j we observe that [ab, c] = 1 2 gac gbc gab + b a x x xc . (1.4.8) for i=j and g(i)(i) = h2 , (i) i = 1, 2, 3 Here there are 33 = 27 quantities to calculate. We consider the following cases: 110 CASE I Let a = b = c = i, then the equation (1.4.8) simplies to [ab, c] = [ii, i] = 1 gii 2 xi (no summation on i). (1.4.9) From this equation we can calculate any of the Christoel symbols [11, 1], [22, 2], or [33, 3]. CASE II Let a = b = i = c, then the equation (1.4.8) simplies to the form [ab, c] = [ii, c] = 1 gii 2 xc (no summation on i and i = c). (1.4.10) since, gic = 0 for i = c. This equation shows how we may calculate any of the six Christoel symbols [11, 2], [11, 3], [22, 1], [22, 3], [33, 1], [33, 2]. CASE III Let a = c = i = b, and noting that gib = 0 for i = b, it can be veried that the equation (1.4.8) simplies to the form [ab, c] = [ib, i] = [bi, i] = 1 gii 2 xb (no summation on i and i = b). (1.4.11) From this equation we can calculate any of the twelve Christoel symbols [12, 1] = [21, 1] [32, 3] = [23, 3] [13, 1] = [31, 1] [31, 3] = [13, 3] [21, 2] = [12, 2] [23, 2] = [32, 2] CASE IV Let a = b = c and show that the equation (1.4.8) reduces to [ab, c] = 0, This represents the six Christoel symbols [12, 3] = [21, 3] = [23, 1] = [32, 1] = [31, 2] = [13, 2] = 0. From the Cases I,II,III,IV all twenty seven Christoel symbols of the rst kind can be determined. In practice, only the nonzero Christoel symbols are listed. (a = b = c.) EXAMPLE 1.4-3. (Christoel symbols of the rst kind)Find the nonzero Christoel symbols of the rst kind in cylindrical coordinates. Solution: From the results of example 1.4-2 we nd that for x1 = r, g11 = 1, g22 = (x1 )2 = r2 , x2 = , x3 = z and g33 = 1 the nonzero Christoel symbols of the rst kind in cylindrical coordinates are: 1 g22 = x1 = r 2 x1 1 g22 = x1 = r. [21, 2] = [12, 2] = 2 x1 [22, 1] = 111 EXAMPLE 1.4-4. (Christoel symbols of the second kind) Find formulas for the calculation of the Christoel symbols of the second kind in a generalized orthogonal coordinate system with metric coecients gij = 0 where i is not summed. Solution: By denition we have i jk = g im [jk, m] = g i1 [jk, 1] + g i2 [jk, 2] + g i3 [jk, 3] (1.4.12) for i=j and g(i)(i) = h2 , (i) i = 1, 2, 3 By hypothesis the coordinate system is orthogonal and so g ij = 0 for i=j and g ii = 1 gii i not summed. The only nonzero term in the equation (1.4.12) occurs when m = i and consequently i jk = g ii [jk, i] = [jk, i] gii no summation on i. (1.4.13) We can now consider the four cases considered in the example 1.4-2. CASE I Let j = k = i and show i ii = [ii, i] 1 gii 1 = = ln gii gii 2gii xi 2 xi no summation on i. (1.4.14) CASE II Let k = j = i and show i jj = 1 gjj [jj, i] = gii 2gii xi no summation on i or j. (1.4.15) CASE III Let i = j = k and verify that j jk = j kj = [jk, j] 1 gjj 1 = = ln gjj gjj 2gjj xk 2 xk no summation on i or j. (1.4.16) CASE IV For the case i = j = k we nd i jk = [jk, i] = 0, gii i=j=k no summation on i. The above cases represent all 27 terms. 112 EXAMPLE 1.4-5. (Notation) In the case of cylindrical coordinates we can use the above relations and nd the nonzero Christoel symbols of the second kind: 1 22 2 12 = = 1 g22 = x1 = r 2g11 x1 2 21 = 1 1 1 g22 = 1= 2g22 x1 x r Note 1: The notation for the above Christoel symbols are based upon the assumption that x1 = r, x2 = and x3 = z. However, in tensor calculus the choice of the coordinates can be arbitrary. We could just as well have dened x1 = z, x2 = r and x3 = . In this latter case, the numbering system of the Christoel symbols changes. To avoid confusion, an alternate method of writing the Christoel symbols is to use coordinates in place of the integers 1,2 and 3. For example, in cylindrical coordinates we can write r = r = 1 r and r = r. If we dene x1 = r, x2 = , x3 = z, then the nonzero Christoel symbols are written as 2 12 = 2 21 = 1 r and 1 22 = r. In contrast, if we dene x1 = z, x2 = r, x3 = , then the nonzero Christoel symbols are written 3 23 = 3 32 = 1 r and 2 33 = r. Note 2: Some textbooks use the notation a,bc for Christoel symbols of the rst kind and d = g da a,bc for bc Christoel symbols of the second kind. This notation is not used in these notes since the notation suggests that the Christoel symbols are third order tensors, which is not true. The Christoel symbols of the rst and second kind are not tensors. This fact is clearly illustrated by the transformation equations (1.4.3) and (1.4.7). Covariant Dierentiation Let Ai denote a covariant tensor of rank 1 which obeys the transformation law A = Ai Dierentiate this relation with respect to x and show A 2 xi Ai xj xi = Ai + . xj x x x x x (1.4.18) xi . x (1.4.17) Now use the relation from equation (1.4.7) to eliminate the second derivative term from (1.4.18) and express it in the form A = Ai x xi x i jk xj xk x x + Ai xj xi . xj x x (1.4.19) 113 Employing the equation (1.4.17), with replaced by , the equation (1.4.19) is expressible in the form A A x or alternatively A A x Dene the quantity Aj,k = Aj Ai xk i jk (1.4.22) = Aj Ai xk i jk xj xk . x x (1.4.21) = Aj xj xk Ai xk x x i jk xj xk x x (1.4.20) as the covariant derivative of Aj with respect to xk . The equation (1.4.21) demonstrates that the covariant derivative of a covariant tensor produces a second order tensor which satises the transformation law A, = Aj,k xj xk . x x (1.4.23) Other notations frequently used to denote the covariant derivative are: Aj,k = Aj;k = Aj/k = k Aj = Aj |k . (1.4.24) In the special case where gij are constants the Christoel symbols of the second kind are zero, and conseAj . That is, under the special circumstances where the quently the covariant derivative reduces to Aj,k = xk Christoel symbols of the second kind are zero, the covariant derivative reduces to an ordinary derivative. Covariant Derivative of Contravariant Tensor A contravariant tensor Ai obeys the transformation law A = A form Ai = A i xi which can be expressed in the x (1.4.24) xi x by interchanging the barred and unbarred quantities. We write the transformation law in the form of equation (1.4.24) in order to make use of the second derivative relation from the previously derived equation (1.4.7). Dierentiate equation (1.4.24) with respect to xj to obtain the relation 2i A x xi x Ai x =A + . xj x x xj x xj x (1.4.25) Changing the indices in equation (1.4.25) and substituting for the second derivative term, using the relation from equation (1.4.7), produces the equation Ai =A j x xi x i mk xm xk x x A x xi x + . xj x xj x (1.4.26) Applying the relation found in equation (1.4.24), with i replaced by m, together with the relation x xk k = j , xj x 114 we simplify equation (1.4.26) to the form Ai + xj Dene the quantity Ai ,j = Ai + xj i Am mj (1.4.28) i mj Am = A + x i x x A . j x x (1.4.27) as the covariant derivative of the contravariant tensor Ai . The equation (1.4.27) demonstrates that a covariant derivative of a contravariant tensor will transform like a mixed second order tensor and Ai ,j = A x , j xi . x x (1.4.29) Again it should be observed that for the condition where gij are constants we have Ai ,j = Ai and the xj covariant derivative of a contravariant tensor reduces to an ordinary derivative in this special case. In a similar manner the covariant derivative of second rank tensors can be derived. We nd these derivatives have the forms: Aij,k = Ai ,k j Aij Aj Ai k ik jk x Ai i j = + A Ai j k jk xk Aij i + Aj k xk + Ai j . k (1.4.30) Aij ,k = In general, the covariant derivative of a mixed tensor Aij...k lm...p of rank n has the form Aij...k = lm...p,q Aij...k lm...p xq + Aj...k lm...p Aij...k m...p i q lq + Ai...k lm...p Aij...k l...p j q mq + + Aij... lm...p Aij...k lm... k q pq (1.4.31) and this derivative is a tensor of rank n + 1. Note the pattern of the + signs for the contravariant indices and the signs for the covariant indices. Observe that the covariant derivative of an nth order tensor produces an n+ 1st order tensor, the indices of these higher order tensors can also be raised and lowered by multiplication by the metric or conjugate metric tensor. For example we can write g im Ajk |m = Ajk |i and g im Ajk |m = Ajk |i 115 Rules for Covariant Dierentiation The rules for covariant dierentiation are the same as for ordinary dierentiation. That is: (i) The covariant derivative of a sum is the sum of the covariant derivatives. (ii) The covariant derivative of a product of tensors is the rst times the covariant derivative of the second plus the second times the covariant derivative of the rst. (iii) Higher derivatives are dened as derivatives of derivatives. Be careful in calculating higher order derivatives as in general Ai,jk = Ai,kj . EXAMPLE 1.4-6. (Covariant dierentiation) Solution: The covariant derivative of Ai is Ai,j = Calculate the second covariant derivative Ai,jk . Ai A . ij xj By denition, the second covariant derivative is the covariant derivative of a covariant derivative and hence Ai,jk = (Ai,j ) ,k = xk Ai A ij xj Am,j m ik Ai,m m . jk Simplifying this expression one obtains Ai,jk = 2 Ai A xj xk xk Am A mj xj ij A m ik xk ij Ai A im xm m . jk Rearranging terms, the second covariant derivative can be expressed in the form Ai,jk = 2 Ai A xj xk xk xk ij ij Am xj m jk m ik m ik Ai xm mj m jk . A im (1.4.32) 116 Riemann Christoel Tensor Utilizing the equation (1.4.32), it is left as an exercise to show that Ai,jk Ai,kj = A Rijk where Rijk = xj ik xk ij + m ik mj m ij mk (1.4.33) is called the Riemann Christoel tensor. The covariant form of this tensor is i Rhjkl = gih Rjkl . (1.4.34) It is an easy exercise to show that this covariant form can be expressed in either of the forms Rinjk = or Rijkl s s [nk, i] k [nj, i] + [ik, s] [ij, s] j nj nk x x 2 gil 1 2 gjl 2 gik 2 gjk = i k j l + i l + g ([jk, ][il, ] [jl, ][ik, ]) . 2 xj xk x x x x x x From these forms we nd that the Riemann Christoel tensor is skew symmetric in the rst two indices and the last two indices as well as being symmetric in the interchange of the rst pair and last pairs of indices and consequently Rjikl = Rijkl Rijlk = Rijkl Rklij = Rijkl . In a two dimensional space there are only four components of the Riemann Christoel tensor to consider. These four components are either +R1212 or R1212 since they are all related by R1212 = R2112 = R2121 = R1221 . In a Cartesian coordinate system Rhijk = 0. The Riemann Christoel tensor is important because it occurs in dierential geometry and relativity which are two areas of interest to be considered later. Additional properties of this tensor are found in the exercises of section 1.5. 117 Physical Interpretation of Covariant Dierentiation In a system of generalized coordinates (x1 , x2 , x3 ) we can construct the basis vectors (E1 , E2 , E3 ). These basis vectors change with position. That is, each basis vector is a function of the coordinates at which they are evaluated. We can emphasize this dependence by writing Ei = Ei (x1 , x2 , x3 ) = r xi i = 1, 2, 3. Associated with these basis vectors we have the reciprocal basis vectors E i = E i (x1 , x2 , x3 ), i = 1, 2, 3 which are also functions of position. A vector A can be represented in terms of contravariant components as A = A1 E1 + A2 E2 + A3 E3 = Aj Ej or it can be represented in terms of covariant components as A = A1 E 1 + A2 E 2 + A3 E 3 = Aj E j . A change in the vector A is represented as dA = where from equation (1.4.35) we nd Ej Aj A Ej = Aj k + xk x xk or alternatively from equation (1.4.36) we may write Ej Aj j A E. = Aj k + k x x xk We dene the covariant derivative of the covariant components as Ai,k = Ej A Ai Ei = + Aj k Ei . k k x x x (1.4.39) (1.4.38) (1.4.37) A dxk xk (1.4.36) (1.4.35) The covariant derivative of the contravariant components are dened by the relation Ai ,k = Introduce the notation Ej = xk We then have Ei Ej = xk m Em E i = jk m i = jk m i jk (1.4.42) m Em jk and j Ej Em. = k mk x (1.4.41) A Ai Ej Ei = + Aj k E i . k x xk x (1.4.40) 118 and Ei Ej j j j = E m Ei = m = . xk mk mk i ik (1.4.43) Then equations (1.4.39) and (1.4.40) become Ai,k = Ai ,k = Ai xk Ai + xk j Aj ik i Aj , jk which is consistent with our earlier denitions from equations (1.4.22) and (1.4.28). Here the rst term of the covariant derivative represents the rate of change of the tensor eld as we move along a coordinate curve. The second term in the covariant derivative represents the change in the local basis vectors as we move along the coordinate curves. This is the physical interpretation associated with the Christoel symbols of the second kind. We make the observation that the derivatives of the basis vectors in equations (1.4.39) and (1.4.40) are related since j Ei E j = i and consequently Ej Ei (Ei E j ) = Ei + k Ej = 0 k x xk x Ej Ei or Ei = E j k k x x Hence we can express equation (1.4.39) in the form Ai,k = Ai Ei Aj E j k . k x x (1.4.44) We write the rst equation in (1.4.41) in the form Ej = xk and consequently (1.4.46) Ej i i and Em =[jk, i]E Em = [jk, i]m = [jk, m]. xk These results also reduce the equations (1.4.40) and (1.4.44) to our previous forms for the covariant derivatives. The equations (1.4.41) are representations of the vectors Ei xk m gim E i = [jk, i]E i jk i m = jk i m jk (1.4.45) i Ej Ei E m = Em = k jk x and Ej xk in terms of the basis vectors and reciprocal basis vectors of the space. The covariant derivative relations then take into account how these vectors change with position and aect changes in the tensor eld. The Christoel symbols in equations (1.4.46) are symmetric in the indices j and k since Ej = xk xk r xj = xj r xk = Ek . xj (1.4.47) 119 The equations (1.4.46) and (1.4.47) enable us to write [jk, m] =Em = Ej 1 Ej Ek = + Em Em xk 2 xk xj 1 Em Em Ek Em Ej + j Em Ek Ej k k 2 x x x xj 1 Ek Ej Em Ej + j Em Ek Ej m Ek 2 xk x x xm = = 1 Em Ej + j Em Ek m Ej Ek k 2 x x x gmk gjk 1 gmj + m = [kj, m] = 2 xk xj x which again agrees with our previous result. For future reference we make the observation that if the vector A is represented in the form A = Aj Ej , involving contravariant components, then we may write dA = A dxk = xk = = Ej Aj E + Aj k kj x x dxk dxk (1.4.48) Aj i Ei E + Aj kj jk x Aj + xk j Am Ej dxk = Aj,k dxk Ej . mk Similarly, if the vector A is represented in the form A = Aj E j involving covariant components it is left as an exercise to show that dA = Aj,k dxk E j Riccis Theorem Riccis theorem states that the covariant derivative of the metric tensor vanishes and gik,l = 0. Proof: We have gik,l = gik,l gik,l m m gik gim gmk kl il xl gik = [kl, i] [il, k] xl gik 1 gik gil gkl gkl gil 1 gik = + k + k = 0. l l i l i x 2 x x x 2 x x x (1.4.49) Because of Riccis theorem the components of the metric tensor can be regarded as constants during covariant dierentiation. i EXAMPLE 1.4-7. (Covariant dierentiation) Show that j,k = 0. Solution i j,k = i j i + j k xk i jk = i jk i jk = 0. 120 EXAMPLE 1.4-8. (Covariant dierentiation) Show that g ij,k = 0. k Solution: Since gij g jk = i we take the covariant derivative of this expression and nd k (gij g jk ),l = i,l = 0 gij g jk,l + gij,l g jk = 0. But gij,l = 0 by Riccis theorem and hence gij g jk,l = 0. We multiply this expression by g im and obtain m g im gij g jk,l = j g jk,l = g mk = 0 ,l which demonstrates that the covariant derivative of the conjugate metric tensor is also zero. EXAMPLE 1.4-9. (Covariant dierentiation) Some additional examples of covariant dierentiation are: (i) (gil Al ),k = gil Al ,k = Ai,k (ii) (gim gjn Aij ) ,k = gim gjn Aij = Amn,k ,k Intrinsic or Absolute Dierentiation The intrinsic or absolute derivative of a covariant vector Ai taken along a curve xi = xi (t), i = 1, . . . , N is dened as the inner product of the covariant derivative with the tangent vector to the curve. The intrinsic derivative is represented dxj Ai = Ai,j t dt Ai Ai = A ij t xj dAi Ai = A ij t dt dxj dt dxj . dt (1.4.50) Similarly, the absolute or intrinsic derivative of a contravariant tensor Ai is represented Ai dAi dxj = Ai ,j = + t dt dt i dxj Ak . jk dt The intrinsic or absolute derivative is used to dierentiate sums and products in the same manner as used in ordinary dierentiation. Also if the coordinate system is Cartesian the intrinsic derivative becomes an ordinary derivative. The intrinsic derivative of higher order tensors is similarly dened as an inner product of the covariant derivative with the tangent vector to the given curve. For example, Aij dxp klm = Aij klm,p t dt is the intrinsic derivative of the fth order mixed tensor Aij . klm 121 EXAMPLE 1.4-10. (Generalized velocity and acceleration) Let t denote time and let xi = xi (t) for i = 1, . . . , N , denote the position vector of a particle in the generalized coordinates (x1 , . . . , xN ). From the transformation equations (1.2.30), the position vector of the same particle in the barred system of coordinates, (x1 , x2 , . . . , xN ), is xi = xi (x1 (t), x2 (t), . . . , xN (t)) = xi (t), The generalized velocity is v i = dxi dt , i = 1, . . . , N. i = 1, . . . , N. The quantity v i transforms as a tensor since by denition vi = xi dxj xi j dxi = = v. dt xj dt xj (1.4.51) Let us now nd an expression for the generalized acceleration. Write equation (1.4.51) in the form vj = v i and dierentiate with respect to time to obtain dv i xj dv j 2 xj dxk = vi i k + dt dt xi x x dt The equation (1.4.53) demonstrates that dv i dt xj xi (1.4.52) (1.4.53) does not transform like a tensor. From the equation (1.4.7) previously derived, we change indices and write equation (1.4.53) in the form dv j dxk = vi dt dt Rearranging terms we nd v j dxk + xk dt j ac xa i v xi v j + xk xc dxk xk dt = xj v i dxk + xi xk dt xj dxk vi ik x dt or ik xj x j xa xc xj dv i . + i k a c x x xi dt j dxk v = va + ak dt xk v xj v j . = t t x dxk xj vi ik dt x The above equation illustrates that the intrinsic derivative of the velocity is a tensor quantity. This derivative is called the generalized acceleration and is denoted fi = dv i dxj v i = v i,j = + t dt dt i d2 xi vm vn = + mn dt2 i dxm dxn , m n dt dt i = 1, . . . , N (1.4.54) To summarize, we have shown that if xi = xi (t), i i = 1, . . . , N is the generalized position vector, then dx , i = 1, . . . , N is the generalized velocity, and dt i dxj v = v i,j , i = 1, . . . , N is the generalized acceleration. fi = t dt vi = 122 Parallel Vector Fields Let y i = y i (t), i = 1, 2, 3 denote a space curve C in a Cartesian coordinate system and let Y i dene a constant vector in this system. Construct at each point of the curve C the vector Y i . This produces a eld of parallel vectors along the curve C. What happens to the curve and the eld of parallel vectors when we transform to an arbitrary coordinate system using the transformation equations y i = y i (x1 , x2 , x3 ), with inverse transformation xi = xi (y 1 , y 2 , y 3 ), i = 1, 2, 3? i = 1, 2, 3 The space curve C in the new coordinates is obtained directly from the transformation equations and can be written xi = xi (y 1 (t), y 2 (t), y 3 (t)) = xi (t), i = 1, 2, 3. The eld of parallel vectors Y i become X i in the new coordinates where Y i = Xj y i . xj (1.4.55) Since the components of Y i are constants, their derivatives will be zero and consequently we obtain by dierentiating the equation (1.4.55), with respect to the parameter t, that the eld of parallel vectors X i must satisfy the dierential equation dY i 2 y i dxm dX j y i = = 0. + Xj j m j dt x x x dt dt (1.4.56) Changing symbols in the equation (1.4.7) and setting the Christoel symbol to zero in the Cartesian system of coordinates, we represent equation (1.4.7) in the form 2yi = xj xm y i j m x and consequently, the equation (1.4.56) can be reduced to the form dX j X j = + t dt j dxm Xk = 0. km dt (1.4.57) The equation (1.4.57) is the dierential equation which must be satised by a parallel eld of vectors X i along an arbitrary curve xi (t). 123 EXERCISE 1.4 1. 1 Find the nonzero Christoel symbols of the rst and second kind in cylindrical coordinates 2 (x , x , x3 ) = (r, , z), where x = r cos , 2. 1 y = r sin , z = z. Find the nonzero Christoel symbols of the rst and second kind in spherical coordinates 2 (x , x , x3 ) = (, , ), where x = sin cos , 3. y = sin sin , z = cos . Find the nonzero Christoel symbols of the rst and second kind in parabolic cylindrical coordinates 1 (x , x , x3 ) = (, , z), where x = , y = ( 2 2 ), z = z. 2 1 2 Find the nonzero Christoel symbols of the rst and second kind in parabolic coordinates 1 (x , x , x3 ) = (, , ), where x = cos , y = sin , z = ( 2 2 ). 2 1 2 4. 5. 1 Find the nonzero Christoel symbols of the rst and second kind in elliptic cylindrical coordinates 2 (x , x , x3 ) = (, , z), where x = cosh cos , 6. 1 y = sinh sin , z = z. Find the nonzero Christoel symbols of the rst and second kind for the oblique cylindrical coordinates y = r sin + cos , z = sin with 0 < < 2 (x , x2 , x3 ) = (r, , ), where x = r cos , and constant. Hint: See gure 1.3-18 and exercise 1.3, problem 12. 7. 8. r = g ri [st, i] and solve for the Christoel symbol of the rst kind in terms of the Christoel st symbol of the second kind. n (b) Assume [st, i] = gni and solve for the Christoel symbol of the second kind in terms of the st Christoel symbol of the rst kind. (a) Let 9. (a) Write down the transformation law satised by the fourth order tensor (b) Show that ijk,m = 0 in all coordinate systems. (c) Show that ( g),k = 0. ijk ,m ijk,m . Show [ij, k] + [kj, i] = gik . xj 10. 11. 12. Show = 0. Calculate the second covariant derivative Ai ,kj . The gradient of a scalar eld (x1 , x2 , x3 ) is the vector grad = E i . xi (a) Find the physical components associated with the covariant components ,i Ai ,i d = (b) Show the directional derivative of in a direction Ai is . dA (gmn Am An )1/2 124 13. (a) Show g is a relative scalar of weight +1. (b) Use the results from problem 9(c) and problem 44, Exercise 1.4, to show that g m g = 0. ( g),k = xk km m 1 g (c) Show that = ln( g) = . k km x 2g xk 14. Use the result from problem 9(b) to show Hint: Expand the covariant derivative multiplication with 15. erst g rst,p m 1 g = ln( g) = . km xk 2g xk and then substitute rst = gerst . Simplify by inner and note the Exercise 1.1, problem 26. Calculate the covariant derivative Ai,m and then contract on m and i to show that 1i gA . Ai,i = g xi 1 ij Show gg + g xj 16. 17. i pq g pq = 0. Hint: See problem 14. Prove that the covariant derivative of a sum equals the sum of the covariant derivatives. Hint: Assume Ci = Ai + Bi and write out the covariant derivative for Ci,j . 18. i Let Cj = Ai Bj and prove that the covariant derivative of a product equals the rst term times the covariant derivative of the second term plus the second term times the covariant derivative of the rst term. x x and take an ordinary derivative of both sides Start with the transformation law Aij = A i x xj k with respect to x and hence derive the relation for Aij,k given in (1.4.30). 19. xi xj and take an ordinary derivative of both sides x x with respect to xk and hence derive the relation for Aij given in (1.4.30). ,k 20. Start with the transformation law Aij = A 21. Find the covariant derivatives of (a) Aijk (b) Aij k (c) Ai jk (d) Aijk 22. Find the intrinsic derivative along the curve xi = xi (t), (a) Aijk (b) Aij k (c) Ai jk i = 1, . . . , N for (d) Aijk 23. (a) Assume A = Ai Ei and show that dA = Ai ,k dxk Ei . (b) Assume A = Ai E i and show that dA = Ai,k dxk E i . 125 24. (parallel vector eld) Imagine a vector eld Ai = Ai (x1 , x2 , x3 ) which is a function of position. Assume that at all points along a curve xi = xi (t), i = 1, 2, 3 the vector eld points in the same direction, we would then have a parallel vector eld or homogeneous vector eld. Assume A is a constant, then dA = 25. A xk dxk = 0. Show that for a parallel vector eld the condition Ai,k = 0 must be satised. [ik, n] = gn j xj x ik + ([nj, ] + [j, n]) . ik Show that 26. 27. Show Ar,s As,r = Ar As . xs xr In cylindrical coordinates you are given the contravariant vector components A1 = r A2 = cos A3 = z sin (a) Find the physical components Ar , A , and Az . Arr Ar A Az Arz Az Azz . (b) Denote the physical components of Ai,j , i, j = 1, 2, 3, by Ar Azr Find these physical components. 28. Find the covariant form of the contravariant tensor C i = Akj . , ijk Ak,j . Express your answer in terms of 29. 1 In Cartesian coordinates let x denote the magnitude of the position vector xi . Show that (a) x ,j = xj x 1 1 1 2 ij 3xi xj (b) x ,ij = ij 3 xi xj (c) x ,ii = . (d) LetU = , x = 0, and show that U ,ij = + and 3 x x x x x x5 U ,ii = 0. 30. Consider a two dimensional space with element of arc length squared ds2 = g11 (du1 )2 + g22 (du2 )2 where u1 , u2 are surface coordinates. (a) Find formulas to calculate the Christoel symbols of the rst kind. (b) Find formulas to calculate the Christoel symbols of the second kind. 31. Find the metric tensor and Christoel symbols of the rst and second kind associated with the and metric gij = g11 0 0 g22 two dimensional space describing points on a cylinder of radius a. Let u1 = and u2 = z denote surface coordinates where x = a cos = a cos u1 y = a sin = a sin u1 z = z = u2 126 32. Find the metric tensor and Christoel symbols of the rst and second kind associated with the two dimensional space describing points on a sphere of radius a. Let u1 = and u2 = denote surface coordinates where x = a sin cos = a sin u1 cos u2 y = a sin sin = a sin u1 sin u2 z = a cos = a cos u1 33. u1 = , Find the metric tensor and Christoel symbols of the rst and second kind associated with the u2 = . illustrated in the gure 1.3-19. The points on the surface of the torus are given in terms two dimensional space describing points on a torus having the parameters a and b and surface coordinates of the surface coordinates by the equations x = (a + b cos ) cos y = (a + b cos ) sin z = b sin Prove that eijk am bj ck ui,m + eijk ai bm ck uj,m + eijk ai bj cm uk = ur eijk ai bj ck . Hint: See Exercise 1.3, ,m ,r 34. problem 32 and Exercise 1.1, problem 21. 35. 36. 37. Calculate the second covariant derivative Ai,jk . 1 ij Show that ij = g + mn ,j g xj i mn Find the contravariant, covariant and physical components of velocity and acceleration in (a) Cartesian coordinates and (b) cylindrical coordinates. 38. Find the contravariant, covariant and physical components of velocity and acceleration in spherical coordinates. 39. In spherical coordinates (, , ) show that the acceleration components can be represented in terms of the velocity components as f = v 2 2 v + v , f = v + 2 v v v , tan f = v + v v v v + tan Hint: Calculate v , v , v . 40. The divergence of a vector Ai is Ai,i . That is, perform a contraction on the covariant derivative Ai,j to obtain Ai,i . Calculate the divergence in (a) Cartesian coordinates (b) cylindrical coordinates and (c) spherical coordinates. 41. If S is a scalar invariant of weight one and Ai is a third order relative tensor of weight W , show jk that S W Ai is an absolute tensor. jk 127 Let Y i ,i = 1, 2, 3 denote the components of a eld of parallel vectors along the curve C dened by yi the equations y i = y i (t), i = 1, 2, 3 in a space with metric tensor gij , i, j = 1, 2, 3. Assume that Y i and d dt are unit vectors such that at each point of the curve C we have 42. gij Y i dj y = cos = Constant. dt (i.e. The eld of parallel vectors makes a constant angle with the tangent to each point of the curve C.) i i i i123 Show that if Y and y (t) undergo a transformation x = x ( , y , y ), i = 1, 2, 3 then the transformed y m i xm vector X = Y j makes a constant angle with the tangent vector to the transformed curve C given by y y xi = xi (1 (t), y 2 (t), y 3 (t)). 43. Let J denote the Jacobian determinant | xi |. Dierentiate J with respect to xm and show that xj J xp r =J J . m p xm rm x Hint: See Exercise 1.1, problem 27 and (1.4.7). 44. Assume that is a relative scalar of weight W so that = J W . Dierentiate this relation with respect to xk . Use the result from problem 43 to obtain the transformation law: W xk = JW W k xm r xm . mr xk The quantity inside the brackets is called the covariant derivative of a relative scalar of weight W. The covariant derivative of a relative scalar of weight W is dened as ,k = W xk r kr and this denition has an extra term involving the weight. It can be shown that similar results hold for relative tensors of weight W. For example, the covariant derivative of rst and second order relative tensors of weight W have the forms i T ,k = Tji ,k T i + xk Tji = + xk i Tm W km i Tj k r Ti kr r Ti kr j Ti W jk When the weight term is zero these covariant derivatives reduce to the results given in our previous denitions. 45. Let dxi dt = v i denote a generalized velocity and dene the scalar function of kinetic energy T of a T= 1 1 m gij v i v j = m gij xi xj . 2 2 T T particle with mass m as Show that the intrinsic derivative of T is the same as an ordinary derivative of T. (i.e. Show that = dT dt .) 128 46. Verify the relations gij = gmj gni xk g in = g mn g ij xk 47. g nm xk gjm xk 1 ijk a tensor? Justify Assume that B ijk is an absolute tensor. Is the quantity T jk = gB g xi your answer. If your answer is no, explain your answer and determine if there any conditions you can impose upon B ijk such that the above quantity will be a tensor? 48. The e-permutation symbol can be used to dene various vector products. Let Ai , Bi , Ci , Di (a) In two dimensions R =eij Ai Bj Ri =eij Aj (b) In three dimensions S =eijk Ai Bj Ck Si =eijk Bj Ck Sij =eijk Ck (c) In four dimensions T =eijkm Ai Bj Ck Dm Ti =eijkm Bj Ck Dm Tij =eijkm Ck Dm Tijk =eikm Dm a scalar determinant. a scalar determinant. a vector cross product. a scalar determinant. a vector (rotation). i = 1, . . . , N denote vectors, then expand and verify the following products: a skew-symmetric matrix 4-dimensional cross product. skew-symmetric matrix. skew-symmetric tensor. with similar products in higher dimensions. 49. Expand the curl operator for: (a) Two dimensions B = eij Aj,i (b) Three dimensions Bi = eijk Ak,j (c) Four dimensions Bij = eijkm Am,k 129 1.5 DIFFERENTIAL GEOMETRY AND RELATIVITY In this section we will examine some fundamental properties of curves and surfaces. In particular, at each point of a space curve we can construct a moving coordinate system consisting of a tangent vector, a normal vector and a binormal vector which is perpendicular to both the tangent and normal vectors. How these vectors change as we move along the space curve brings up the subjects of curvature and torsion associated with a space curve. The curvature is a measure of how the tangent vector to the curve is changing and the torsion is a measure of the twisting of the curve out of a plane. We will nd that straight lines have zero curvature and plane curves have zero torsion. In a similar fashion, associated with every smooth surface there are two coordinate surface curves and a normal surface vector through each point on the surface. The coordinate surface curves have tangent vectors which together with the normal surface vectors create a set of basis vectors. These vectors can be used to dene such things as a two dimensional surface metric and a second order curvature tensor. The coordinate curves have tangent vectors which together with the surface normal form a coordinate system at each point of the surface. How these surface vectors change brings into consideration two dierent curvatures. A normal curvature and a tangential curvature (geodesic curvature). How these curvatures are related to the curvature tensor and to the Riemann Christoel tensor, introduced in the last section, as well as other interesting relationships between the various surface vectors and curvatures, is the subject area of dierential geometry. Also presented in this section is a brief introduction to relativity where again the Riemann Christoel tensor will occur. Properties of this important tensor are developed in the exercises of this section. Space Curves and Curvature For xi = xi (s),i = 1, 2, 3, a 3-dimensional space curve in a Riemannian space Vn with metric tensor gij , and arc length parameter s, the vector T i = i dxi ds represents a tangent vector to the curve at a point P on dxi dxj = 1. ds ds the curve. The vector T is a unit vector because gij T i T j = gij (1.5.1) Dierentiate intrinsically, with respect to arc length, the relation (1.5.1) and verify that gij T i which implies that gij T j Hence, the vector T i s T j T i j + gij T = 0, s s T i = 0. s T i s (1.5.2) (1.5.3) is perpendicular to the tangent vector T i . Dene the unit normal vector N i to the and write (1.5.4) space curve to be in the same direction as the vector Ni = 1 T i s where is a scale factor, called the curvature, and is selected such that gij N i N j = 1 which implies gij T i T j = 2 . s s (1.5.5) 130 The reciprocal of curvature is called the radius of curvature. The curvature measures the rate of change of the tangent vector to the curve as the arc length varies. By dierentiating intrinsically, with respect to arc length s, the relation gij T i N j = 0 we nd that gij T i N j T i j + gij N = 0. s s (1.5.6) Consequently, the curvature can be determined from the relation gij T i N j T i j = gij N = gij N i N j = s s (1.5.7) which denes the sign of the curvature. In a similar fashion we dierentiate the relation (1.5.5) and nd that gij N i This later equation indicates that the vector N j s N j = 0. s (1.5.8) is perpendicular to the unit normal N i . The equation (1.5.3) indicates that T i is also perpendicular to N i and hence any linear combination of these vectors will also be perpendicular to N i . The unit binormal vector is dened by selecting the linear combination N j + T j s and then scaling it into a unit vector by dening Bj = 1 N j + T j s (1.5.10) (1.5.9) where is a scalar called the torsion. The sign of is selected such that the vectors T i , N i and B i form a right handed system with satisfying gij B i B j = 1. (1.5.11) ijk T i N j B k = 1 and the magnitude of is selected such that B i is a unit vector The triad of vectors T i , N i , B i at a point on the curve form three planes. The plane containing T i and B i is called the rectifying plane. The plane containing N i and B i is called the normal plane. The plane containing T i and N i is called the osculating plane. The reciprocal of the torsion is called the radius of torsion. The torsion measures the rate of change of the osculating plane. The vectors T i , N i and B i form a right-handed orthogonal system at a point on the space curve and satisfy the relation Bi = ijk T j Nk . (1.5.12) By using the equation (1.5.10) it can be shown that B i is perpendicular to both the vectors T i and N i since gij B i T j = 0 and gij B i N j = 0. B i s It is left as an exercise to show that the binormal vector B i satises the relation relations T i = N i s N i = B i T i s B i = N i s = N i . The three (1.5.13) 131 are known as the Frenet-Serret formulas of dierential geometry. Surfaces and Curvature Let us examine surfaces in a Cartesian frame of reference and then later we can generalize our results to other coordinate systems. A surface in Euclidean 3-dimensional space can be dened in several dierent ways. Explicitly, z = f (x, y), implicitly, F (x, y, z) = 0 or parametrically by dening a set of parametric equations of the form x = x(u, v), y = y(u, v), z = z(u, v) which contain two independent parameters u, v called surface coordinates. For example, the equations x = a sin cos , y = a sin sin , z = a cos are the parametric equations which dene a spherical surface of radius a with parameters u = and v = . See for example gure 1.3-20 in section 1.3. By eliminating the parameters u, v one can derive the implicit form of the surface and by solving for z one obtains the explicit form of the surface. Using the parametric form of a surface we can dene the position vector to a point on the surface which is then represented in terms of the parameters u, v as r = r(u, v) = x(u, v) e1 + y(u, v) e2 + z(u, v) e3 . (1.5.14) The coordinates (u, v) are called the curvilinear coordinates of a point on the surface. The functions x(u, v), y(u, v), z(u, v) are assumed to be real and dierentiable such that r(u, c2 ) and r(c1 , v) r u r v = 0. The curves (1.5.15) with c1 , c2 constants, then dene two surface curves called coordinate curves, which intersect at the surface coordinates (c1 , c2 ). The family of curves dened by equations (1.5.15) with equally spaced constant values ci , ci + ci , ci + 2ci , . . . dene a surface coordinate grid system. The vectors r u r and v evaluated at the surface coordinates (c1 , c2 ) on the surface, are tangent vectors to the coordinate curves through the point and are basis vectors for any vector lying in the surface. Letting (x, y, z) = (y 1 , y 2 , y 3 ) and (u, v) = (u1 , u2 ) and utilizing the summation convention, we can write the position vector in the form r = r(u1 , u2 ) = y i (u1 , u2 ) ei . (1.5.16) The tangent vectors to the coordinate curves at a point P can then be represented as the basis vectors E = r y i ei , = u u = 1, 2 (1.5.17) where the partial derivatives are to be evaluated at the point P where the coordinate curves on the surface intersect. From these basis vectors we construct a unit normal vector to the surface at the point P by calculating the cross product of the tangent vector ru = n = n(u, v) = r u and rv = = r v . A unit normal is then (1.5.18) E1 E2 |E1 E2 | ru rv |ru rv | 132 and is such that the vectors E1 , E2 and n form a right-handed system of coordinates. If we transform from one set of curvilinear coordinates (u, v) to another set (, v ), which are determined u by a set of transformation laws u = u(, v ), u the equation of the surface becomes u u u u r = r(, v ) = x(u(, v ), v(, v )) e1 + y(u(, v ), v(, v )) e2 + z(u(, v ), v(, v )) e3 u u u and the tangent vectors to the new coordinate curves are r u r v r = + u u u v u and r r u r v = + . v u v v v v = v(, v ), u Using the indicial notation this result can be represented as y i y i u = . u u u This is the transformation law connecting the two systems of basis vectors on the surface. A curve on the surface is dened by a relation f (u, v) = 0 between the curvilinear coordinates. Another way to represent a curve on the surface is to represent it in a parametric form where u = u(t) and v = v(t), where t is a parameter. The vector r du r dv dr = + dt u dt v dt is tangent to the curve on the surface. An element of arc length with respect to the surface coordinates is represented by ds2 = dr dr = where a = r u r r du du = a du du u u (1.5.19) r u with , = 1, 2 denes a surface metric. This element of arc length on the surface is 1 EG F 2 2 (E du + F dv)2 + dv E E often written as the quadratic form A = ds2 = E(du)2 + 2F du dv + G(dv)2 = (1.5.20) and called the rst fundamental form of the surface. Observe that for ds2 to be positive denite the quantities E and EG F 2 must be positive. The surface metric associated with the two dimensional surface is dened by a = E E = r r y i y i = , u u u u , = 1, 2 (1.5.21) with conjugate metric tensor a dened such that a a = . Here the surface is embedded in a three dimensional space with metric gij and a is the two dimensional surface metric. In the equation (1.5.20) the quantities E, F, G are functions of the surface coordinates u, v and are determined from the relations E =a11 = F =a12 G =a22 y i y i r r = u u u1 u1 y i y i r r = = u v u1 u2 y i y i r r = = v v u2 u2 (1.5.22) 133 Here and throughout the remainder of this section, we adopt the convention that Greek letters have the range 1,2, while Latin letters have the range 1,2,3. Construct at a general point P on the surface the unit normal vector n at this point. Also construct a plane which contains this unit surface normal vector n. Observe that there are an innite number of planes which contain this unit surface normal. For now, select one of these planes, then later on we will consider all such planes. Let r = r(s) denote the position vector dening a curve C which is the intersection of the selected plane with the surface, where s is the arc length along the curve, which is measured from some xed point on the curve. Let us nd the curvature of this curve of intersection. The vector T = dr ds , evaluated at the point P, is a unit tangent vector to the curve C and lies in the tangent plane to the surface at the point P. Here we are using ordinary dierentiation rather than intrinsic dierentiation because we are in a Cartesian system of coordinates. Dierentiating the relation T T = 1, with respect to arc length s we nd that T K= dT ds dT ds = 0 which implies that the vector dT ds is perpendicular to the tangent vector T . Since the coordinate system is Cartesian we can treat the curve of intersection C as a space curve, then the vector , evaluated at point P, is dened as the curvature vector with curvature |K| = and radius of curvature R = 1/. A unit normal N to the space curve is taken in the same direction as dT so that the ds dT . Consider the geometry of gure 1.5-1 curvature will always be positive. We can then write K = N = ds and dene on the surface a unit vector u = n T which is perpendicular to both the surface tangent vector T and the surface normal vector n, such that the vectors T i ,ui and ni forms a right-handed system. Figure 1.5-1 Surface curve with tangent plane and a normal plane. 134 The direction of u in relation to T is in the same sense as the surface tangents E1 and E2 . Note that the vector dT ds is perpendicular to the tangent vector T and lies in the plane which contains the vectors n and u. We can therefore write the curvature vector K in the component form K= dT = (n) n + (g) u = Kn + Kg ds (1.5.23) where (n) is called the normal curvature and (g) is called the geodesic curvature. The subscripts are not indices. These curvatures can be calculated as follows. From the orthogonality condition n T = 0 we obtain dn dT +T = 0. Consequently, the normal by dierentiation with respect to arc length s the result n ds ds curvature is determined from the dot product relation n K = (n) = T dr dn dn = . ds ds ds (1.5.24) By taking the dot product of u with equation (1.5.23) we nd that the geodesic curvature is determined from the triple scalar product relation (g) = u Normal Curvature The equation (1.5.24) can be expressed in terms of a quadratic form by writing (n) ds2 = dr dn. (1.5.26) dT dT = (n T ) . ds ds (1.5.25) The unit normal to the surface n and position vector r are functions of the surface coordinates u, v with dr = We dene the quadratic form B = dr dn = 2 r r du + dv u v and dn = n n du + dv. u v (1.5.27) r r du + dv u v 2 n n du + dv u v (1.5.28) B = e(du) + 2f du dv + g(dv) = b du du where e= and b r n , u u 2f = r n n r + u v u v , g= r n v v (1.5.29) , = 1, 2 is called the curvature tensor and a b = b is an associated curvature tensor. The quadratic form of equation (1.5.28) is called the second fundamental form of the surface. Alternative methods for calculating the coecients of this quadratic form result from the following considerations. The unit surface normal is perpendicular to the tangent vectors to the coordinate curves at the point P and therefore we have the orthogonality relationships r n=0 u and r n = 0. v (1.5.30) 135 Observe that by dierentiating the relations in equation (1.5.30), with respect to both u and v, one can derive the results 2r r n = b11 n= 2 u u u r n n r 2r n= = = b21 = b12 f= uv u v u v 2r r n = b22 g= n= 2 v v v and consequently the curvature tensor can be expressed as e= b = r n . u u (1.5.31) (1.5.32) The quadratic forms from equations (1.5.20) and (1.5.28) enable us to represent the normal curvature in the form of a ratio of quadratic forms. We nd from equation (1.5.26) that the normal curvature in the direction du dv is (n) = e(du)2 + 2f du dv + g(dv)2 B = . A E(du)2 + 2F du dv + G(dv)2 (1.5.33) If we write the unit tangent vector to the curve in the form T = of the unit surface normal with respect to arc length as expressed in the form (n) = T dn = ds r n u u dn ds = dr r du ds = u ds and express the derivative n du u ds , then the normal curvature can be du du ds ds b du du b du du = = . ds2 a du du Observe that the curvature tensor is a second order symmetric tensor. (1.5.34) In the previous discussions, the plane containing the unit normal vector was arbitrary. Let us now consider all such planes that pass through this unit surface normal. As we vary the plane containing the unit surface normal n at P we get dierent curves of intersection with the surface. Each curve has a curvature associated with it. By examining all such planes we can nd the maximum and minimum normal curvatures associated with the surface. We write equation (1.5.33) in the form (n) = where = dv du . e + 2f + g2 E + 2F + G2 (1.5.35) From the theory of proportions we can also write this equation in the form (n) = f + g e + f (e + f ) + (f + g) = = . (E + F ) + (F + G) F + G E + F (1.5.36) Consequently, the curvature will satisfy the dierential equations (e E)du + (f F )dv = 0 and (f F )du + (g G)dv = 0. d(n) d (1.5.37) The maximum and minimum curvatures occur in those directions where = 0. Calculating the deriva- tive of (n) with respect to and setting the derivative to zero we obtain a quadratic equation in (F g Gf )2 + (Eg Ge) + (Ef F e) = 0, (F g Gf ) = 0. 136 This equation has two roots 1 and 2 which satisfy 1 + 2 = Eg Ge F g Gf and 1 2 = Ef F e , F g Gf (1.5.38) where F g Gf = 0. The curvatures (1) ,(2) corresponding to the roots 1 and 2 are called the principal curvatures at the point P. Several quantities of interest that are related to (1) and (2) are: (1) the principal radii of curvature Ri = 1/i ,i = 1, 2; (2) H = two directions on the surface dr1 r r = + 1 du u v If these directions are orthogonal we will have dr1 dr2 r r r r =( + 1 )( + 2 ) = 0. du du u v u v This requires that G1 2 + F (1 + 2 ) + E = 0. (1.5.39) and r r dr2 = + 2 . du u v 1 2 ((1) + (2) ) called the mean curvature and K = (1) (2) called the total curvature or Gaussian curvature of the surface. Observe that the roots 1 and 2 determine It is left as an exercise to verify that this is indeed the case and so the directions determined by the principal curvatures must be orthogonal. In the case where F g Gf = 0 we have that F = 0 and f = 0 because the coordinate curves are orthogonal and G must be positive. In this special case there are still two directions determined by the dierential equations (1.5.37) with dv = 0, du arbitrary, and du = 0, dv arbitrary. From the dierential equations (1.5.37) we nd these directions correspond to (1) = We let = du ds e E and (2) = g . G denote a unit vector on the surface satisfying a = 1. Then the equation (1.5.34) can be written as (n) = b or we can write (b (n) a ) = 0. The maximum and minimum normal curvature occurs in those directions where (b (n) a ) = 0 and so (n) must be a root of the determinant equation |b (n) a | = 0 or |a b (n) | = b1 (n) 1 b2 1 b2 2 b b1 2 = 2 b a (n) + = 0. (n) (n) a (1.5.40) This is a quadratic equation in (n) of the form 2 ((1) + (2) )(n) + (1) (2) = 0. In other words the (n) principal curvatures (1) and (2) are the eigenvalues of the matrix with elements b = a b . Observe that from the determinant equation in (n) we can directly nd the total curvature or Gaussian curvature which is an invariant given by K = (1) (2) = |b | = |a b | = b/a. The mean curvature is also an invariant obtained from H = 1 2 ((1) + (2) ) = 1 b , 2a where a = a11 a22 a12 a21 and b = b11 b22 b12 b21 are the determinants formed from the surface metric tensor and curvature tensor components. 137 The equations of Gauss, Weingarten and Codazzi At each point on a space curve we can construct a unit tangent T , a unit normal N and unit binormal B. The derivatives of these vectors, with respect to arc length, can also be represented as linear combinations of the base vectors T , N , B. See for example the Frenet-Serret formulas from equations (1.5.13). In a similar fashion the surface vectors ru , rv , n form a basis and the derivatives of these basis vectors with respect to the surface coordinates u, v can also be expressed as linear combinations of the basis vectors ru , rv , n. For example, the derivatives ruu , ruv , rvv can be expressed as linear combinations of ru , rv , n. We can write ruu = c1 ru + c2 rv + c3 n ruv = c4 ru + c5 rv + c6 n rvv = c7 ru + c8 rv + c9 n where c1 , . . . , c9 are constants to be determined. It is an easy exercise (see exercise 1.5, problem 8) to show that these equations can be written in the indicial notation as 2r = u u r + b n. u (1.5.42) (1.5.41) These equations are known as the Gauss equations. In a similar fashion the derivatives of the normal vector can be represented as linear combinations of the surface basis vectors. If we write n = c1 ru + c2 rv u n = c3 ru + c4 rv v r n n = c + c 1 2 u u v n n r = c + c 3 4 v u v or (1.5.43) where c1 , . . . , c4 and c , . . . , c are constants. These equations are known as the Weingarten equations. It 1 4 is easily demonstrated (see exercise 1.5, problem 9) that the Weingarten equations can be written in the indicial form r n = b u u is the mixed second order form of the curvature tensor. (1.5.44) where b = a b i The equations of Gauss produce a system of partial dierential equations dening the surface coordinates x as a function of the curvilinear coordinates u and v. The equations are not independent as certain compatibility conditions must be satised. In particular, it is required that the mixed partial derivatives must satisfy 3r u u u We calculate 3r = u u u 2r + u u = 3r u u u . u b r n + b + n u u u and use the equations of Gauss and Weingarten to express this derivative in the form r 3r b = b b + b + n. u + u u u u u 138 Forming the dierence 3r u u u equal to zero produces the Codazzi equations b b + b b = 0. u u (1.5.45) 3r u u u r u =0 must be zero. Setting the coecient of n we nd that the coecients of the independent vectors n and These equations are sometimes referred to as the Mainardi-Codazzi equations. Equating to zero the coecient of r u we nd that R = b b b b or changing indices we have the covariant form a R = R = b b b b , where R = u u + (1.5.46) (1.5.47) is the mixed Riemann curvature tensor. EXAMPLE 1.5-1 Show that the Gaussian or total curvature K = (1) (2) depends only upon the metric a and is R1212 where a = det(a ). K= a Solution: Utilizing the two-dimensional alternating tensor e and the property of determinants we can write e K = e b b and to obtain e e K = e e b b = 2K 2K = e e (a bu ) a b 2K = e a e b b . Using ae = where from page 137, K = |b | = |a b |. Now multiply by e and then contract on But e a a = ae so that we have 2K = b b . Interchanging indices we can write 2K = b b and 2K = b b . Adding these last two results we nd that 4K = both sides by (b b b b ) = R . Now multiply to obtain 4K = R . From exercise 1.5, problem 16, the Riemann curvature tensor Rijkl is skew symmetric in the (i, j), (k, l) as well as being symmetric in the (ij), (kl) pair of indices. Consequently, R = 4R and hence R = K and we have the special case b R1212 . A much simpler way to obtain this result is to observe K = where K ae12 ae12 = R1212 or K = a a (bottom of page 137) and note from equation (1.5.46) that R1212 = b11 b22 b12 b21 = b. Note that on a surface ds2 = a du du where a are the metrices for the surface. This metric is a u u and by taking determinants we nd tensor and satises a = a u u a = a u u = aJ 2 u u 139 where J is the Jacobian of the surface coordinate transformation. Here the curvature tensor for the surface R has only one independent component since R1212 = R2121 = R1221 = R2112 (See exercises 20,21). u u u u u u u u one can sum over the repeated indices and show that R1212 = R1212 J 2 and consequently R From the transformation law = R R1212 R1212 = =K a a which shows that the Gaussian curvature is a scalar invariant in V2 . Geodesic Curvature For C an arbitrary curve on a given surface the curvature vector K, associated with this curve, is the vector sum of the normal curvature (n) n and geodesic curvature (g) u and lies in a plane which is perpendicular to the tangent vector to the given curve on the surface. The geodesic curvature (g) is obtained from the equation (1.5.25) and can be represented (g) = u K = u Substituting into this expression the vectors T= dr du dv = ru + rv ds ds ds dT dT = (n T ) = ds ds T dT ds n. dT = K = ruu (u )2 + 2ruv u v + rvv (v )2 + ru u + rv v , ds where = d ds , and by utilizing the results from problem 10 of the exercises following this section, we nd that the geodesic curvature can be represented as (g) = 2 11 2 22 (u )3 + 2 2 1 12 2 12 2 1 11 1 22 (u )2 v + (1.5.48) (v ) + (u v u v ) 3 u (v ) EG F 2. This equation indicates that the geodesic curvature is only a function of the surface metrices E, F, G and the derivatives u , v , u , v . When the geodesic curvature is zero the curve is called a geodesic curve. Such curves are often times, but not always, the lines of shortest distance between two points on a surface. For example, the great circle on a sphere which passes through two given points on the sphere is a geodesic curve. If you erase that part of the circle which represents the shortest distance between two points on the circle you are left with a geodesic curve connecting the two points, however, the path is not the shortest distance between the two points. For plane curves we let u = x and v = y so that the geodesic curvature reduces to kg = u v u v = d ds 140 where is the angle between the tangent T to the curve and the unit vector e1 . Geodesics are curves on the surface where the geodesic curvature is zero. Since kg = 0 along a geodesic surface curve, then at every point on this surface curve the normal N to the curve will be in the same direction as the normal n to the surface. In this case, we have ru n = 0 and rv n = 0 which reduces to dT ru = 0 ds since the vectors n and dT ds and dT rv = 0, ds (1.5.49) have the same direction. In particular, we may write T= dr r du r dv = + = ru u + rv v ds u ds v ds dT = ruu (u )2 + 2ruv u v + rvv (v )2 + ru u + rv v ds Consequently, the equations (1.5.49) become dT ru = (ruu ru ) (u )2 + 2(ruv ru ) u v + (rvv ru ) (v )2 + Eu + F v = 0 ds . dT 2 2 rv = (ruu rv ) (u ) + 2(ruv rv ) u v + (rvv rv ) (v ) + F u + Gv = 0. ds Utilizing the results from exercise 1.5,(See problems 4,5 and 6), we can eliminate v (1.5.50) to obtain d2 u + ds2 and eliminating u 1 11 du ds 2 (1.5.50) from the equations +2 1 12 du dv + ds ds 1 22 dv ds 2 =0 from the equations (1.5.50) produces the equation d2 v + ds2 2 11 du ds 2 +2 2 12 du dv + ds ds 2 22 dv ds 2 = 0. In tensor form, these last two equations are written d2 u + ds2 du du = 0, ds ds , , = 1, 2 (1.5.51) a where u = u1 and v = u2 . The equations (1.5.51) are the dierential equations dening a geodesic curve on a surface. We will nd that these same type of equations arise in considering the shortest distance between two points in a generalized coordinate system. See for example problem 18 in exercise 2.2. 141 Tensor Derivatives Let u = u (t) denote the parametric equations of a curve on the surface dened by the parametric equations xi = xi (u1 , u2 ). We can then represent the surface curve in the spatial geometry since the surface curve can be represented in the spatial coordinates through the representation xi = xi (u1 (t), u2 (t)) = xi (t). Recall that for xi = xi (t) a given curve C , the intrinsic derivative of a vector eld Ai along C is dened as the inner product of the covariant derivative of the vector eld with the tangent vector to the curve. This intrinsic derivative is written Ai Ai dxj = Ai,j = + t dt xj or dAi Ai = + t dt i jk i jk Ak g dxj dt Ak g dxj dt where the subscript g indicates that the Christoel symbol is formed from the spatial metric gij . If A is a surface vector dened along the curve C, the intrinsic derivative is represented A A du = A = + , t dt u or A dA = + t dt A a du dt A a du dt where the subscript a denotes that the Christoel is formed from the surface metric a . Similarly, the formulas for the intrinsic derivative of a covariant spatial vector Ai or covariant surface vector A are given by dAi Ai = t dt and A dA = t dt Consider a mixed tensor i i T k ij Ak g dxj dt du . dt A a which is contravariant with respect to a transformation of space coordinates i x and covariant with respect to a transformation of surface coordinates u . For T dened over the surface i curve C, which can also be viewed as a space curve C, dene the scalar invariant = (t) = T Ai B where Ai is a parallel vector eld along the curve C when it is viewed as a space curve and B is also a parallel vector eld along the curve C when it is viewed as a surface curve. Recall that these parallel vector elds must satisfy the dierential equations dAi Ai = t dt k ij Ak g dxj = 0 and dt B dB = + t dt B a du = 0. dt (1.5.52) The scalar invariant is a function of the parameter t of the space curve since both the tensor and the parallel vector elds are to be evaluated along the curve C. By dierentiating the function with respect to the parameter t there results d dT i dB i dAi i = Ai B + T B + T Ai . dt dt dt dt (1.5.53) 142 But the vectors Ai and B are parallel vector elds and must satisfy the relations given by equations (1.5.52). This implies that equation (1.5.53) can be written in the form i d dT = + dt dt i kj k T g dxj dt i T a du Ai B . dt (1.5.54) The quantity inside the brackets of equation (1.5.54) is dened as the intrinsic tensor derivative with respect to the parameter t along the curve C. This intrinsic tensor derivative is written i i T dT = + dt dt i kj k T g dxj dt i T a du . dt (1.5.55) The spatial representation of the curve C is related to the surface representation of the curve C through the dening equations. Therefore, we can express the equation (1.5.55) in the form i i T T = + dt u i kj k T g xj u i T a du dt (1.5.56) i The quantity inside the brackets is a mixed tensor which is dened as the tensor derivative of T with i respect to the surface coordinates u . The tensor derivative of the mixed tensor T with respect to the surface coordinates u is written i T, = i T + u i kj k T g xj u i T . a i...j In general, given a mixed tensor T... which is contravariant with respect to transformations of the space coordinates and covariant with respect to transformations of the surface coordinates, then we can dene the scalar eld along the surface curve C as i...j (t) = T... Ai Aj B B (1.5.57) where Ai , . . . , Aj and B , . . . , B are parallel vector elds along the curve C. The intrinsic tensor derivative is then derived by dierentiating the equation (1.5.57) with respect to the parameter t. Tensor derivatives of the metric tensors gij , a and the alternating tensors ijk , and their associated tensors are all zero. Hence, they can be treated as constants during the tensor dierentiation process. Generalizations In a Riemannian space Vn with metric gij and curvilinear coordinates xi , i = 1, 2, 3, the equations of a surface can be written in the parametric form xi = xi (u1 , u2 ) where u , = 1, 2 are called the curvilinear coordinates of the surface. Since dxi = xi du u (1.5.58) then a small change du on the surface results in change dxi in the space coordinates. Hence an element of arc length on the surface can be represented in terms of the curvilinear coordinates of the surface. This same element of arc length can also be represented in terms of the curvilinear coordinates of the space. Thus, an element of arc length squared in terms of the surface coordinates is represented ds2 = a du du (1.5.59) 143 where a is the metric of the surface. This same element when viewed as a spatial element is represented ds2 = gij dxi dxj . By equating the equations (1.5.59) and (1.5.60) we nd that gij dxi dxj = gij xi xj du du = a du du . u u (1.5.61) (1.5.60) The equation (1.5.61) shows that the surface metric is related to the spatial metric and can be calculated xi xj from the relation a = gij . This equation reduces to the equation (1.5.21) in the special case of u u Cartesian coordinates. In the surface coordinates we dene the quadratic form A = a du du as the rst fundamental form of the surface. The tangent vector to the coordinate curves dening the surface are given by xi u and can be viewed as either a covariant surface vector or a contravariant spatial vector. We dene xi = xi , u i = 1, 2, 3, = 1, 2. (1.5.62) this vector as Any vector which is a linear combination of the tangent vectors to the coordinate curves is called a surface vector. A surface vector A can also be viewed as a spatial vector Ai . The relation between the spatial representation and surface representation is Ai = A xi . The surface representation A , = 1, 2 and the spatial representation Ai , i = 1, 2, 3 dene the same direction and magnitude since gij Ai Aj = gij A xi A xj = gij xi xj A A = a A A . Consider any two surface vectors A and B and their spatial representations Ai and B i where Ai = A xi and B i = B xi . (1.5.63) These vectors are tangent to the surface and so a unit normal vector to the surface can be dened from the cross product relation ni AB sin = jk ijk A B (1.5.64) where A, B are the magnitudes of Ai , B i and is the angle between the vectors when their origins are made to coincide. Substituting equations (1.5.63) into the equation (1.5.64) we nd ni AB sin = In terms of the surface metric we have AB sin = form (ni which for arbitrary surface vectors implies ni j k ijk A x B x . (1.5.65) A B so that equation (1.5.65) can be written in the =0 (1.5.66) jk ijk x x )A B = jk ijk x x or ni = 1 2 jk ijk x x . (1.5.67) The equation (1.5.67) denes a unit normal vector to the surface in terms of the tangent vectors to the coordinate curves. This unit normal vector is related to the covariant derivative of the surface tangents as 144 is now demonstrated. By using the results from equation (1.5.50), the tensor derivative of equation (1.5.59), with respect to the surface coordinates, produces xi = , 2 xi + u u i pq xp xq g xi a (1.5.68) where the subscripts on the Christoel symbols refer to the metric from which they are calculated. Also the tensor derivative of the equation (1.5.57) produces the result gij xi xj + gij xi xj = a, = 0. , , Interchanging the indices , , cyclically in the equation (1.5.69) one can verify that gij xi xj = 0. , (1.5.70) (1.5.69) The equation (1.5.70) indicates that in terms of the space coordinates, the vector xi is perpendicular to , the surface tangent vector xi and so must have the same direction as the unit surface normal ni . Therefore, there must exist a second order tensor b such that b ni = xi . , By using the relation gij ni nj = 1 we can transform equation (1.5.71) to the form b = gij nj xi = , 1 2 i jk ijk x, x x . (1.5.71) (1.5.72) The second order symmetric tensor b is called the curvature tensor and the quadratic form B = b du du is called the second fundamental form of the surface. Consider also the tensor derivative with respect to the surface coordinates of the unit normal vector to the surface. This derivative is ni = , ni + u i jk nj xk . g (1.5.73) (1.5.74) Taking the tensor derivative of gij ni nj = 1 with respect to the surface coordinates produces the result gij ni nj, = 0 which shows that the vector nj, is perpendicular to ni and must lie in the tangent plane to the surface. It can therefore be expressed as a linear combination of the surface tangent vectors xi and written in the form ni, = xi (1.5.75) where the coecients can be written in terms of the surface metric components a and the curvature components b as follows. The unit vector ni is normal to the surface so that gij ni xj = 0. (1.5.76) 145 The tensor derivative of this equation with respect to the surface coordinates gives gij ni xj + gij ni xj = 0. , (1.5.77) Substitute into equation (1.5.77) the relations from equations (1.5.57), (1.5.71) and (1.5.75) and show that b = a . Solving the equation (1.5.78) for the coecients we nd = a b . (1.5.78) (1.5.79) Now substituting equation (1.5.79) into the equation (1.5.75) produces the Weingarten formula ni, = a b xi . (1.5.80) This is a relation for the derivative of the unit normal in terms of the surface metric, curvature tensor and surface tangents. A third fundamental form of the surface is given by the quadratic form C = c du du where c is dened as the symmetric surface tensor c = gij ni, nj, . By using the Weingarten formula in the equation (1.5.81) one can verify that c = a b b . (1.5.83) (1.5.82) (1.5.81) Geodesic Coordinates In a Cartesian coordinate system the metric tensor gij is a constant and consequently the Christoel symbols are zero at all points of the space. This is because the Christoel symbols are dependent upon the derivatives of the metric tensor which is constant. If the space VN is not Cartesian then the Christoel symbols do not vanish at all points of the space. However, it is possible to nd a coordinate system where the Christoel symbols will all vanish at a given point P of the space. Such coordinates are called geodesic coordinates of the point P. Consider a two dimensional surface with surface coordinates u and surface metric a . If we transform to some other two dimensional coordinate system, say u with metric a , where the two coordinates are related by transformation equations of the form u u = u ( 1 , u 2 ), = 1, 2, (1.5.84) 146 then from the transformation equation (1.4.7) we can write, after changing symbols, u = a u u u 2 u + . u u a u u (1.5.85) This is a relationship between the Christoel symbols in the two coordinate systems. If a point P , then for that particular point the equation (1.5.85) reduces to 2 u = u u u u a u u vanishes at a (1.5.86) where all terms are evaluated at the point P. Conversely, if the equation (1.5.86) is satised at the point P, then the Christoel symbol tion must be zero at this point. Consider the special coordinate transformaa u = u + u 0 1 2 u u a (1.5.87) where u are the surface coordinates of the point P. The point P in the new coordinates is given by 0 u = 0. We now dierentiate the relation (1.5.87) to see if it satises the equation (1.5.86). We calculate the derivatives 1 u = u 2 u a 1 2 u a u =0 (1.5.88) and 2 u = u u a u =0 (1.5.89) where these derivative are evaluated at u = 0. We nd the derivative equations (1.5.88) and (1.5.89) do satisfy the equation (1.5.86) locally at the point P. Hence, the Christoel symbols will all be zero at this particular point. The new coordinates can then be called geodesic coordinates. Riemann Christoel Tensor Consider the Riemann Christoel tensor dened by the equation (1.4.33). Various properties of this tensor are derived in the exercises at the end of this section. We will be particularly interested in the Riemann Christoel tensor in a two dimensional space with metric a and coordinates u . We nd the Riemann Christoel tensor has the form R. = u u + (1.5.90) where the Christoel symbols are evaluated with respect to the surface metric. The above tensor has the associated tensor R = a R. (1.5.91) which is skew-symmetric in the indices (, ) and (, ) such that R = R and R = R . (1.5.92) The two dimensional alternating tensor is used to dene the constant K= 1 4 R (1.5.93) 147 (see example 1.5-1) which is an invariant of the surface and called the Gaussian curvature or total curvature. In the exercises following this section it is shown that the Riemann Christoel tensor of the surface can be expressed in terms of the total curvature and the alternating tensors as R = K . (1.5.94) Consider the second tensor derivative of xr which is given by xr , = xr , + u r mn xr xn , g xr , a xr , a (1.5.95) which can be shown to satisfy the relation r r xr , x, = R . x . (1.5.96) Using the relation (1.5.96) we can now derive some interesting properties relating to the tensors a , b , c , R , the mean curvature H and the total curvature K. Consider the tensor derivative of the equation (1.5.71) which can be written i i xi , = b, n + b n , (1.5.97) where b, = b u b a b . a (1.5.98) By using the Weingarten formula, given in equation (1.5.80), the equation (1.5.97) can be expressed in the form i i xi , = b, n b a b x (1.5.99) and by using the equations (1.5.98) and (1.5.99) it can be established that r r r xr , x, = (b, b, )n a (b b b b )x . (1.5.100) Now by equating the results from the equations (1.5.96) and (1.5.100) we arrive at the relation r r r R . x = (b, b, )n a (b b b b )x . (1.5.101) Multiplying the equation (1.5.101) by nr and using the results from the equation (1.5.76) there results the Codazzi equations b, b, = 0. (1.5.102) Multiplying the equation (1.5.101) by grm xm and simplifying one can derive the Gauss equations of the surface R = b b b b . By using the Gauss equations (1.5.103) the equation (1.5.94) can be written as K (1.5.103) = b b b b . (1.5.104) 148 Another form of equation (1.5.104) is obtained by using the equation (1.5.83) together with the relation a = a . It is left as an exercise to verify the resulting form Ka = c a b b . (1.5.106) Dene the quantity H= 1 a b 2 (1.5.107) as the mean curvature of the surface, then the equation (1.5.106) can be written in the form c 2H b + K a = 0. By multiplying the equation (1.5.108) by du du and summing, we nd C 2H B + K A = 0 is a relation connecting the rst, second and third fundamental forms. EXAMPLE 1.5-2 In a two dimensional space the Riemann Christoel tensor has only one nonzero independent component R1212 . ( See Exercise 1.5, problem number 21.) Consequently, the equation (1.5.104) can be written in the form K ae12 ae12 = b22 b11 b21 b12 and solving for the Gaussian curvature K we nd K= R1212 b b22 b11 b12 b21 . == a11 a22 a12 a21 a a (1.5.110) (1.5.109) (1.5.108) Surface Curvature For a surface curve u = u (s), = 1, 2 lying upon a surface xi = xi (u1 , u2 ),i = 1, 2, 3, we have a two du is a unit tangent vector to dimensional space embedded in a three dimensional space. Thus, if t = ds du du the surface curve then a = a t t = 1. This same vector can be represented as the unit tangent ds ds dxi dxj dxi vector to the space curve xi = xi (u1 (s), u2 (s)) with T i = . That is we will have gij = gij T i T j = 1. ds ds ds The surface vector t and the space vector T i are related by Ti = xi du = xi t . u ds t s (1.5.111) The surface vector t is a unit vector so that a t t = 1. If we dierentiate this equation intrinsically with respect to the parameter s, we nd that a t t = 0. This shows that the surface vector s is perpendicular to the surface vector t . Let u denote a unit normal vector in the surface plane which is orthogonal to the tangent vector t . The direction of u is selected such that such that t u = 1. Therefore, there exists a scalar (g) (1.5.112) t = (g) u s 149 where (g) is called the geodesic curvature of the curve. In a similar manner it can be shown that is a surface vector orthogonal to t . Let u s u s = t where is a scalar constant to be determined. By dierentiating the relation a t u = 0 intrinsically and simplifying we nd that = (g) and therefore u = (g) t . s (1.5.113) The equations (1.5.112) and (1.5.113) are sometimes referred to as the Frenet-Serret formula for a curve relative to a surface. Taking the intrinsic derivative of equation (1.5.111), with respect to the parameter s, we nd that t du T i = xi + xi t. , s s ds (1.5.114) Treating the curve as a space curve we use the Frenet formulas (1.5.13). If we treat the curve as a surface curve, then we use the Frenet formulas (1.5.112) and (1.5.113). In this way the equation (1.5.114) can be written in the form N i = xi (g) u + xi t t . , By using the results from equation (1.5.71) in equation (1.5.115) we obtain N i = (g) ui + b ni t t (1.5.116) (1.5.115) where ui is the space vector counterpart of the surface vector u . Let denote the angle between the surface normal ni and the principal normal N i , then we have that cos = ni N i . Hence, by multiplying the equation (1.5.116) by ni we obtain cos = b t t . (1.5.117) Consequently, for all curves on the surface with the same tangent vector t , the quantity cos will remain constant. This result is known as Meusniers theorem. Note also that cos = (n) is the normal component of the curvature and sin = (g) is the geodesic component of the curvature. Therefore, we write the equation (1.5.117) as (n) = b t t (1.5.118) which represents the normal curvature of the surface in the direction t . The equation (1.5.118) can also be written in the form (n) = b which is a ratio of quadratic forms. The surface directions for which (n) has a maximum or minimum value is determined from the equation (1.5.119) which is written as (b (n) a ) = 0. The direction giving a maximum or minimum value to (n) must then satisfy (b (n) a ) = 0 (1.5.121) (1.5.120) B du du = ds ds A (1.5.119) 150 so that (n) must be a root of the determinant equation det(b (n) a ) = 0. The expanded form of equation (1.5.122) can be written as 2 a b (n) + (n) b =0 a (1.5.123) (1.5.122) where a = a11 a22 a12 a21 and b = b11 b22 b12 b21 . Using the denition given in equation (1.5.107) and using the result from equation (1.5.110), the equation (1.5.123) can be expressed in the form 2 2H (n) + K = 0. (n) The roots (1) and (2) of the equation (1.5.124) then satisfy the relations H= and K = (1) (2) . (1.5.126) 1 ((1) + (2) ) 2 (1.5.125) (1.5.124) Here H is the mean value of the principal curvatures and K is the Gaussian or total curvature which is the product of the principal curvatures. It is readily veried that H= Eg 2f F + eG 2(EG F 2 ) and K = eg f 2 EG F 2 are invariants obtained from the surface metric and curvature tensor. Relativity Sir Isaac Newton and Albert Einstein viewed the world dierently when it came to describing gravity and the motion of the planets. In this brief introduction to relativity we will compare the Newtonian equations with the relativistic equations in describing planetary motion. We begin with an examination of Newtonian systems. Newtons viewpoint of planetary motion is a multiple bodied problem, but for simplicity we consider only a two body problem, say the sun and some planet where the motion takes place in a plane. Newtons law of gravitation states that two masses m and M are attracted toward each other with a force of magnitude GmM 2 , where G is a constant, is the distance between the masses, m is the mass of the planet and M is the mass of the sun. One can construct an x, y plane containing the two masses with the origin located at the center of mass of the sun. Let e = cos e1 + sin e2 denote a unit vector at the origin of this coordinate system and pointing in the direction of the mass m. The vector force of attraction of mass M on mass m is given by the relation F= GmM e . 2 (1.5.127) 151 Figure 1.5-2. Parabolic and elliptic conic sections The equation of motion of mass m with respect to mass M is obtained from Newtons second law. Let = e denote the position vector of mass m with respect to the origin. Newtons second law can then be written in any of the forms F= GmM GmM d2 dV e = m 2 = m = 2 dt dt 3 (1.5.128) and from this equation we can show that the motion of the mass m can be described as a conic section. Recall that a conic section is dened as a locus of points p(x, y) such that the distance of p from a xed point (or points), called a focus (foci), is proportional to the distance of the point p from a xed line, called a directrix, that does not contain the xed point. The constant of proportionality is called the eccentricity and is denoted by the symbol . For hyperbola results; and if = 1 a parabola results; for 0 1 an ellipse results; for FP PD >1a = 0 the conic section is a circle. = where F P = and ratio we solve for and obtain the polar representation for the conic section = p 1 + cos (1.5.129) With reference to gure 1.5-2, a conic section is dened in terms of the ratio P D = 2q cos . From the 152 where p = 2q and the angle is known as the true anomaly associated with the orbit. The quantity p is called the semi-parameter of the conic section. (Note that when = of the above equation is = p 1 + cos( 0 ) or u = 1 = A[1 + cos( 0 )], (1.5.130) 2, then = p.) A more general form where 0 is an arbitrary starting anomaly. An additional symbol a, known as the semi-major axes of an elliptical orbit can be introduced where q, p, , a are related by p 1+ = q = a(1 ) or p = a(1 2 ). (1.5.131) To show that the equation (1.5.128) produces a conic section for the motion of mass m with respect to mass M we will show that one form of the solution of equation (1.5.128) is given by the equation (1.5.129). To verify this we use the following vector identities: e =0 d dt d d2 = 2 dt dt d e e =0 dt d e d e e . = dt dt (1.5.132) e From the equation (1.5.128) we nd that d dt d dt = GM d2 = 2 e = 0 2 dt (1.5.133) so that an integration of equation (1.5.133) produces d = h = constant. dt (1.5.134) The quantity H = mV = m d is the angular momentum of the mass m so that the quantity h dt represents the angular momentum per unit mass. The equation (1.5.134) tells us that h is a constant for our two body system. Note that because h is constant we have dV GM d d V h = h = 2 e dt dt dt d d e GM + e )] = 2 e [ e ( dt dt GM d e 2 d e = 2 e ( e ) = GM dt dt and consequently an integration produces V h = GM e + C 153 where C is a vector constant of integration. The triple scalar product formula gives us (V h) = h ( or h2 = GM + C cos where is the angle between the vectors C and . From the equation (1.5.135) we nd that = where p = h2 /GM and p 1 + cos (1.5.136) <1 (1.5.135) d ) = h2 = GM e + C dt = C/GM. This result is known as Keplers rst law and implies that when the mass m describes an elliptical orbit with the sun at one focus. We present now an alternate derivation of equation (1.5.130) for later use. From the equation (1.5.128) we have 2 d d d2 2= dt dt dt d d dt dt = 2 GM d d GM = 3 ( ) . 3 dt dt d dt , (1.5.137) d dt , sin d dt Consider the equation (1.5.137) in spherical coordinates , , . The tensor velocity components are V 1 = V2 = d dt , V3 = d dt and the physical components of velocity are given by V = V = d , V = dt so that the velocity can be written V= d d d d = e + e + sin e . dt dt dt dt (1.5.138) Substituting equation (1.5.138) into equation (1.5.137) gives the result d dt d dt 2 + 2 d dt 2 + 2 sin2 d dt 2 = d GM d 2 2GM d ( ) = 2 = 2GM 3 dt dt dt 1 which can be integrated directly to give d dt 2 + 2 d dt 2 + 2 sin2 d dt 2 = 2GM E 2 (1.5.139) constant so that where E is a constant of integration. In the special case of a planar orbit we set = the equation (1.5.139) reduces to d dt d d d dt 2 + 2 2 d dt d dt 2 = 2 2GM E (1.5.140) + 2 2GM E. = Also for this special case of planar motion we have | By eliminating d dt d d | = 2 = h. dt dt (1.5.141) from the equation (1.5.140) we obtain the result d d 2 + 2 = 2GM 3 E 2 4 . h2 h (1.5.142) 154 Figure 1.5-3. Relative motion of two inertial systems. The substitution = 1 u can be used to represent the equation (1.5.142) in the form du d 2 + u2 2GM E u+ 2 =0 2 h h (1.5.143) which is a form we will return to later in this section. Note that we can separate the variables in equations (1.5.142) or (1.5.143). The results can then be integrate to produce the equation (1.5.130). Newton also considered the relative motion of two inertial systems, say S and S. Consider two such systems as depicted in the gure 1.5-3 where the S system is moving in the xdirection with speed v relative to the system S. For a Newtonian system, if at time t = 0 we have clocks in both systems which coincide, than at time t a point P (x, y, z) in the S system can be described by the transformation equations x =x + vt y =y or z =z t =t z =z t =t. x =x vt y =y (1.5.144) These are the transformation equation of Newtons relativity sometimes referred to as a Galilean transformation. Before Einstein the principle of relativity required that velocities be additive and obey Galileos velocity addition rule VP/R = VP/Q + VQ/R . (1.5.145) 155 That is, the velocity of P with respect to R equals the velocity of P with respect to Q plus the velocity of Q with respect to R. For example, a person (P ) running north at 3 km/hr on a train (Q) moving north at 60 km/hr with respect to the ground (R) has a velocity of 63 km/hr with respect to the ground. What happens when (P ) is a light wave moving on a train (Q) which is moving with velocity V relative to the ground? Are the velocities still additive? This type of question led to the famous Michelson-Morley experiment which has been labeled as the starting point for relativity. Einsteins answer to the above question was NO and required that VP/R = VP/Q = c =speed of light be a universal constant. In contrast to the Newtonian equations, Einstein considered the motion of light from the origins 0 and 0 of the systems S and S. If the S system moves with velocity v relative to the S system and at time t = 0 a light signal is sent from the S system to the S system, then this light signal will move out in a spherical wave front and lie on the sphere x2 + y 2 + z 2 = c2 t2 (1.5.146) where c is the speed of light. Conversely, if a light signal is sent out from the S system at time t = 0, it will lie on the spherical wave front x2 + y 2 + z 2 = c2 t . 2 (1.5.147) Observe that the Newtonian equations (1.5.144) do not satisfy the equations (1.5.146) and (1.5.147) identically. If y = y and z = z then the space variables (x, x) and time variables (t, t) must somehow be related. Einstein suggested the following transformation equations between these variables x = (x vt) and x = (x + vt) where is a constant to be determined. The dierentials of equations (1.5.148) produce dx = (dx vdt) from which we obtain the ratios dx (dx v dt) = dx (dx + v dt) When or 1 (1 + dx ) dt (1.5.148) and dx = (dx + vdt) (1.5.149) v = (1 v dx dt ). (1.5.150) dx dx = = c, the speed of light, the equation (1.5.150) requires that dt dt 2 = (1 v 2 1 ) c2 or = (1 v 2 1/2 ) . c2 (1.5.151) From the equations (1.5.148) we eliminate x and nd t = (t v x). c2 (1.5.152) We can now replace the Newtonian equations (1.5.144) by the relativistic transformation equations x =(x + vt) y =y z =z t =(t + v x) c2 or x =(x vt) y =y z =z t =(t v x) c2 (1.5.153) 156 where is given by equation (1.5.151). These equations are also known as the Lorentz transformation. v Note that for v << c, then 2 0, 1 , then the equations (1.5.153) closely approximate the equations c (1.5.144). The equations (1.5.153) also satisfy the equations (1.5.146) and (1.5.147) identically as can be readily veried by substitution. Further, by using chain rule dierentiation we obtain from the relations (1.5.148) that dx = dt dx dt +v dx dt . (1.5.154) 1+ v cc The equation (1.5.154) is the Einstein relative velocity addition rule which replaces the previous Newtonian rule given by equation (1.5.145). We can rewrite equation (1.5.154) in the notation of equation (1.5.145) as VP/R = VP/Q + VQ/R 1+ VP/Q VQ/R c c . (1.5.155) Observe that when VP/Q << c and VQ/R << c then equation (1.5.155) approximates closely the equation (1.5.145). Also as VP/Q and VQ/R approach the speed of light we have lim VP/Q + VQ/R 1+ VP/Q VQ/R c c VP/Q C VQ/R C =c (1.5.156) which agrees with Einsteins hypothesis that the speed of light is an invariant. Let us return now to the viewpoint of what gravitation is. Einstein thought of space and time as being related and viewed the motion of the planets as being that of geodesic paths in a space-time continuum. Recall the equations of geodesics are given by d2 xi + ds2 i jk dxj dxk = 0, ds ds (1.5.157) where s is arc length. These equations are to be associated with a 4-dimensional space-time metric gij where the indices i, j take on the values 1, 2, 3, 4 and the xi are generalized coordinates. Einstein asked the question, Can one introduce a space-time metric gij such that the equations (1.5.157) can somehow reproduce the law of gravitational attraction d2 dt2 + GM 3 = 0? Then the motion of the planets can be viewed as optimized motion in a space-time continuum where the metrices of the space simulate the law of gravitational attraction. Einstein thought that this motion should be related to the curvature of the space which can be obtained from the Riemann-Christoel tensor Rijkl . The metric we desire gij , i, j = 1, 2, 3, 4 i has 16 components. The conjugate metric tensor g ij is dened such that g ij gjk = k and an element of arc length squared is given by ds2 = gij dxi dxj . Einstein thought that the metrices should come from the Riemann-Christoel curvature tensor which, for n = 4 has 256 components, but only 20 of these are linearly independent. This seems like a large number of equations from which to obtain the law of gravitational attraction and so Einstein considered the contracted tensor Gij = Rtijt = xj n in xn n ij + m in n mj m ij n mn . (1.5.158) Spherical coordinates (, , ) suggests a metric similar to ds2 = (d)2 2 (d)2 2 sin2 (d)2 + c2 (dt)2 157 where g11 = 1, g22 = 2 , g33 = 2 sin2 , g44 = c2 and gij = 0 for i = j. The negative signs are introduced so that ds 2 dt = c2 v 2 is positive when v < c and the velocity is not greater than c. However, this metric will not work since the curvature tensor vanishes. The spherical symmetry of the problem suggest that g11 and g44 change while g22 and g33 remain xed. Let (x1 , x2 , x3 , x4 ) = (, , , t) and assume g11 = eu , g22 = 2 , g33 = 2 sin2 , g44 = ev (1.5.159) where u and v are unknown functions of to be determined. This gives the conjugate metric tensor g 11 = eu , g 22 = 1 , 2 g 33 = 1 , sin2 g 44 = ev (1.5.160) 2 and g ij = 0 for i = j. This choice of a metric produces ds2 = eu (d)2 2 (d)2 2 sin2 (d)2 + ev (dt)2 together with the nonzero Christoel symbols 1 11 1 22 1 33 1 44 = 1 du 2 d 3 13 3 23 3 31 3 32 = = = = 1 cos sin 1 cos sin 4 14 4 41 = = 1 dv 2 d 1 dv . 2 d (1.5.161) = eu = eu sin2 dv 1 = evu 2 dr 2 12 2 21 2 33 = 1 1 = = sin cos (1.5.162) The equation (1.5.158) is used to calculate the nonzero Gij and we nd that G11 = 1 d2 v 1 + 2 2 d 4 dv 1 du 1 du dv d 4 d d d 1 dv 1 du 1+ eu 2 d 2 d 1 du 1 dv eu sin2 1+ 2 d 2 d 1 1 d2 v 1 du dv + 2 d2 4 d d 4 dv d 2 2 G22 =eu G33 =eu (1.5.163) G44 = evu + 1 dv d and Gij = 0 for i = j. The assumption that Gij = 0 for all i, j leads to the dierential equations d2 v 1 + d2 2 2 du 1 du dv =0 2 d d d 1 du 1 dv eu =0 1+ 2 d 2 d dv d 2 dv d 2 (1.5.164) d2 v 1 + 2 d 2 2 dv 1 du dv + =0. 2 d d d 158 Subtracting the rst equation from the third equation gives du dv + =0 d d The second equation in (1.5.164) then becomes du = 1 eu d (1.5.166) or u + v = c1 = constant. (1.5.165) Separate the variables in equation (1.5.166) and integrate to obtain the result eu = 1 2 1 c (1.5.167) where c2 is a constant of integration and consequently ev = ec1 u = ec1 1 c2 . (1.5.168) The constant c1 is selected such that g44 approaches c2 as increases without bound. This produces the metrices g11 = 1 , 2 1 c g22 = 2 , g33 = 2 sin2 , g44 = c2 (1 c2 ) (1.5.169) where c2 is a constant still to be determined. The metrices given by equation (1.5.169) are now used to expand the equations (1.5.157) representing the geodesics in this four dimensional space. The dierential equations representing the geodesics are found to be 1 du d2 + ds2 2 d d ds 2 eu d ds d ds 2 eu sin2 =0 d ds 2 + 1 vu dv e 2 d dt ds 2 =0 (1.5.170) (1.5.171) (1.5.172) (1.5.173) d2 2 d d + sin cos ds2 ds ds 2 d2 2 d d cos d d + +2 =0 ds2 ds ds sin ds ds d2 t dv dt d + = 0. ds2 d ds ds The equation (1.5.171) is identically satised if we examine planar orbits where = dierential equation d ds 2 d ds =0 or 2 d = c4 , ds 2 is a constant. This value of also simplies the equations (1.5.170) and (1.5.172). The equation (1.5.172) becomes an exact (1.5.174) and the equation (1.5.173) also becomes an exact dierential d ds dt v e ds =0 or dt v e = c5 , ds (1.5.175) where c4 and c5 are constants of integration. This leaves the equation (1.5.170) which determines . Substituting the results from equations (1.5.174) and (1.5.175), together with the relation (1.5.161), the equation (1.5.170) reduces to d2 c2 c2 c2 c2 c2 4 + 2+ (1 ) 4 = 0. ds2 2 24 3 (1.5.176) 159 By the chain rule we have d2 d2 =2 2 ds d d ds 2 + d d2 d2 c2 4 = + 2 d ds d2 4 d d 2 2c2 4 5 and so equation (1.5.176) can be written in the form d2 2 2 d The substitution = 1 u d d 2 + c2 c2 2 c2 2 + 2 1 2 c4 = 0. (1.5.177) reduces the equation (1.5.177) to the form d2 u c2 3 + u 2 = c2 u 2 . d2 2c4 2 (1.5.178) du Multiply the equation (1.5.178) by 2 d and integrate with respect to to obtain du d 2 + u2 c2 u = c2 u 3 + c6 . c2 4 (1.5.179) where c6 is a constant of integration. To determine the constant c6 we write the equation (1.5.161) in the special case = 2 and use the substitutions from the equations (1.5.174) and (1.5.175) to obtain eu d ds d d 2 = eu 2 d d d ds c2 2 = 1 2 d ds 2 + ev dt ds 2 or + 1 2 + 1 c2 c2 5 c2 4 = 0. c2 4 (1.5.180) The substitution = 1 u reduces the equation (1.5.180) to the form du d 2 + u 2 c2 u 3 + 1 c2 c2 5 2 c2 u c2 c2 = 0. c4 4 4 (1.5.181) Now comparing the equations (1.5.181) and (1.5.179) we select c6 = so that the equation (1.5.179) takes on the form du d 2 c2 5 1 c2 1 c2 4 + u2 c2 c2 u+ 1 5 c2 c2 4 1 = c2 u 3 c2 4 (1.5.182) Now we can compare our relativistic equation (1.5.182) with our Newtonian equation (1.5.143). In order that the two equations almost agree we select the constants c2 , c4 , c5 so that c2 2GM 2= c4 h2 and 1 c5 E 2 = 2. 2 c4 h c2 (1.5.183) The equations (1.5.183) are only two equations in three unknowns and so we use the additional equation lim 2 d d ds = lim 2 =h dt ds dt (1.5.184) 160 which is obtained from equation (1.5.141). Substituting equations (1.5.174) and (1.5.175) into equation (1.5.184), rearranging terms and taking the limit we nd that c4 c2 = h. c5 From equations (1.5.183) and (1.5.185) we obtain the results that c2 = 5 c2 , E 1 + c2 c2 = 2GM c2 1 1 + E/c2 , c4 = h c 1 + E/c2 (1.5.186) (1.5.185) These values substituted into equation (1.5.181) produce the dierential equation du d Let = c2 c2 4 2 + u2 2GM E 2GM u+ 2 = 2 h h c2 1 1 + E/c2 u3 . (1.5.187) = 2GM h2 and = c2 = 2GM 1 c2 ( 1+E/c2 ) then the dierential equation (1.5.178) can be written as (1.5.188) 3 d2 u + u = u2 . d2 2 2 We know the solution to equation (1.5.143) is given by u= 1 = A(1 + cos( 0 )) (1.5.189) and so we assume a solution to equation (1.5.188) of this same general form. We know that A is small and so we make the assumption that the solution of equation (1.5.188) given by equation (1.5.189) is such that 0 is approximately constant and varies slowly as a function of A. Observe that if 0 = 0 (A), then and d 0 d2 2 d0 d = 0 A = 0 A , where primes denote dierentiation with respect to the argument of the function. (i.e. 2 A for this problem.) The derivatives of equation (1.5.189) produce du = A sin( 0 )(1 0 A) d d2 u = A3 sin( 0 )0 A cos( 0 )(1 2A0 + A2 (0 )2 ) d2 = A cos( 0 ) + 2 A2 0 cos( 0 ) + O(A3 ). Substituting these derivatives into the dierential equation (1.5.188) produces the equations 2 A2 0 cos( 0 ) + A 3 = A2 + 2 A2 cos( 0 ) + 2 2 2 A2 cos2 ( 0 ) + O(A3 ). Now A is small so that terms O(A3 ) can be neglected. Equating the constant terms and the coecient of the cos( 0 ) terms we obtain the equations A 3 2 = A 2 2 2 A2 0 = 3 A2 + 3 2 2 A2 cos( 0 ). Treating 0 as essentially constant, the above system has the approximate solutions A 2 0 3 3 A + A sin( 0 ) 2 4 (1.5.190) 161 The solutions given by equations (1.5.190) tells us that 0 varies slowly with time. For at a rate given by d0 dt . less than 1, the elliptical motion is aected by this change in 0 . It causes the semi-major axis of the ellipse to slowly rotate Using the following values for the planet Mercury G =6.67(108) dyne cm2 /g2 M =1.99(1033 ) g a =5.78(1012 ) cm =0.206 c =3(1010 ) cm/sec 2GM 2 = 2.95(105) cm c h GM a(1 2 ) = 2.71(1019) cm2 /sec d dt GM a3 1/2 (1.5.191) sec1 Keplers third law we calculate the slow rate of rotation of the semi-major axis to be approximately d0 d0 d 3 d = A 3 dt d dt 2 dt GM ch 2 GM a3 1/2 =6.628(1014) rad/sec =43.01 seconds of arc per century. (1.5.192) This slow variation in Mercurys semi-major axis has been observed and measured and is in agreement with the above value. Newtonian mechanics could not account for the changes in Mercurys semi-major axis, but Einsteins theory of relativity does give this prediction. The resulting solution of equation (1.5.188) can be viewed as being caused by the curvature of the space-time continuum. The contracted curvature tensor Gij set equal to zero is just one of many conditions that can be assumed in order to arrive at a metric for the space-time continuum. Any assumption on the value of Gij relates to imposing some kind of curvature on the space. Within the large expanse of our universe only our imaginations limit us as to how space, time and matter interact. You can also imagine the existence of other tensor metrics in higher dimensional spaces where the geodesics within the space-time continuum give rise to the motion of other physical quantities. This short introduction to relativity is concluded with a quote from the NASA News@hg.nasa.gov news release, spring 1998, Release:98-51. An international team of NASA and university researchers has found the rst direct evidence of a phenomenon predicted 80 years ago using Einsteins theory of general relativity that the Earth is dragging space and time around itself as it rotates.The news release explains that the eect is known as frame dragging and goes on to say Frame dragging is like what happens if a bowling ball spins in a thick uid such as molasses. As the ball spins, it pulls the molasses around itself. Anything stuck in the molasses will also move around the ball. Similarly, as the Earth rotates it pulls space-time in its vicinity around itself. This will shift the orbits of satellites near the Earth.This research is reported in the journal Science. 162 EXERCISE 1.5 1. Let = T s N and = N s B. Assume in turn that each of the intrinsic derivatives of T , N , B are some linear combination of T , N , B and hence derive the Frenet-Serret formulas of dierential geometry. 2. Determine the given surfaces. Describe and sketch the curvilinear coordinates upon each surface. (b) r(u, v) = u cos v e1 + u sin v e2 (c) r(u, v) = 2uv 2 2u2 v e1 + 2 e2 . u2 + v 2 u + v2 (a) r(u, v) = u e1 + v e2 3. Determine the given surfaces and describe the curvilinear coordinates upon the surface. Use some graphics package to plot the surface and illustrate the coordinate curves on the surface. Find element of area dS in terms of u and v. a, b, c constants 0 u, v 2 (a) r(u, v) = a sin u cos v e1 + b sin u sin v e2 + c cos u e3 u u u (b) r(u, v) = (4 + v sin ) cos u e1 + (4 + v sin ) sin u e2 + v cos e3 1 v 1, 0 u 2 2 2 2 (c) r(u, v) = au cos v e1 + bu sin v e2 + cu e3 (d) r(u, v) = u cos v e1 + u sin v e2 + v e3 constant a, b constant (e) r(u, v) = a cos v e1 + b sin v e2 + u e3 (f ) r(u, v) = u cos v e1 + u sin v e2 + u e3 2 EF . Assume that the surface is FG described by equations of the form y i = y i (u, v) and that any point on the surface is given by the position 4. Consider a two dimensional space with metric tensor (a ) = vector r = r(u, v) = y i ei . Show that the metrices E, F, G are functions of the parameters u, v and are given by E = ru ru , 5. F = ru rv , G = rv rv where ru = r u and rv = r . v For the metric given in problem 4 show that the Christoel symbols of the rst kind are given by [1 1, 1] = ru ruu [1 1, 2] = rv ruu [1 2, 1] = [2 1, 1] = ru ruv [1 2, 2] = [2 1, 2] = rv ruv [2 2, 1] = ru rvv [2 2, 2] = rv rvv which can be represented [ , ] = 6. 2r r , , , = 1, 2. u u u Show that the results in problem 5 can also be written in the form [1 1, 1] = 1 Eu 2 [1 2, 1] = [2 1, 1] = 1 [2 2, 1] = Fv Gu 2 1 [2 2, 2] = Gv 2 1 Ev 2 1 1 [1 1, 2] = Fu Ev [1 2, 2] = [2 1, 2] = Gu 2 2 where the subscripts indicate partial dierentiation. 7. For the metric given in problem 4, show that the Christoel symbols of the second kind can be expressed in the form = a [ , ], , , = 1, 2 and produce the results 1 11 1 22 = = GEu 2F Fu + F Ev 2(EG F 2 ) 2GFv GGu F Gv 2(EG F 2 ) 1 12 2 12 = = 1 21 2 21 = = GEv F Gu 2(EG F 2 ) EGu F Ev 2(EG F 2 ) 2 11 2 22 = = 2EFu EEv F Eu 2(EG F 2 ) EGv 2F Fv + F Gu 2(EG F 2 ) where the subscripts indicate partial dierentiation. 163 8. Derive the Gauss equations by assuming that ruu = c1 ru + c2 rv + c3 n, ruv = c4 ru + c5 rv + c6 n, rvv = c7 ru + c8 rv + c9 n where c1 , . . . , c9 are constants determined by taking dot products of the above vectors with the vectors ru , rv , 1 2 1 2 and n. Show that c1 = , c2 = , c3 = e, c4 = , c5 = , c6 = f, 11 11 12 12 2 1 2 r r c7 = , c8 = , c9 = g Show the Gauss equations can be written = + b n. u 22 22 u u 9. Derive the Weingarten equations nu = c1 ru + c2 rv nv = c3 ru + c4 rv and show f F eG EG F 2 eF f E c2 = EG F 2 gF f G EG F 2 f F gE c4 = EG F 2 c3 = and ru = c nu + c nv 1 2 rv = c nu + c nv 3 4 f F gE eg f 2 f E eF c = 2 eg f 2 c = 1 f G gF eg f 2 f F eG c = 4 eg f 2 c = 3 c1 = The constants in the above equations are determined in a manner similar to that suggested in problem 8. Show that the Weingarten equations can be written in the form n r = b . u u ru rv , the results from exercise 1.1, problem 9(a), and the results from problem 5, Using n = EG F 2 verify that 10. (ru ruu ) n = (ru ruv ) n = (rv ruu ) n = (ru rvv ) n = 2 11 2 12 1 11 2 22 EG F 2 EG F2 F2 (rv ruv ) n = (rv rvv ) n = (ru rv ) n = EG F2 1 21 1 22 EG F 2 EG F 2 EG F 2 EG and then derive the formula for the geodesic curvature given by equation (1.5.48). dT dT = (T ) n and a ] , ] = . Hint:(n T ) ds ds 164 11. Verify the equation (1.5.39) which shows that the normal curvature directions are orthogonal. i.e. Verify that R = 4R . verify that G1 2 + F (1 + 2 ) + E = 0. 12. 13. 14. Find the rst fundamental form and unit normal to the surface dened by z = f (x, y). Verify Ai,jk Ai,kj = A R.ijk where R.ijk = xj ik xk ij + n ik nj n ij . nk which is sometimes written Rinjk = For Rijkl = gi R.jkl show xj xk + s nj s nk [nj, k] [nk, i] 15. [ij, s] [ik, s] Rinjk = which is sometimes written s [nk, i] k [nj, i] + [ik, s] nj xj x [ij, s] s nk R.ijk = xj ij xk n ik n ij nj + ik nk 16. Show Rijkl = 1 2 2 gjl 2 gik 2 gjk 2 gil i k j l+ i l xj xk x x x x x x + g ([jk, ][il, ] [jl, ][ik, ]) . 17. Use the results from problem 15 to show (i) Rjikl = Rijkl , (ii) Rijlk = Rijkl , (iii) Rklij = Rijkl Hence, the tensor Rijkl is skew-symmetric in the indices i, j and k, l. Also the tensor Rijkl is symmetric with respect to the (ij) and (kl) pair of indices. 18. Verify the following cyclic properties of the Riemann Christoel symbol: (i) Rnijk + Rnjki + Rnkij = 0 (ii) Rinjk + Rjnki + Rknij = 0 (iii) Rijnk + Rjkni + Rkinj = 0 (iv) 19. Riijk , Rikjn + Rkjin + Rjikn = 0 rst index xed second index xed third index xed fourth index xed By employing the results from the previous problems, show all components of the form: Rinjj , Riijj , Riiii , (no summation on i or j) must be zero. 165 20. Rijkm , Find the number of independent components associated with the Riemann Christoel tensor i, j, k, m = 1, 2, . . . , N. There are N 4 components to examine in an N dimensional space. Many of these components are zero and many of the nonzero components are related to one another by symmetries or the cyclic properties. Verify the following cases: CASE I We examine components of the form Rinin , that Rinin = Rnini , leaves M1 = 1 2 N (N i=n with no summation of i or n. The rst index can be chosen in N ways and therefore with i = n the second index can be chosen in N 1 ways. Observe (no summation on i or n) and so one half of the total combinations are repeated. This 1) components of the form Rinin . The quantity M1 can also be thought of as the i = n = j where there is no summation on number of distinct pairs of indices (i, n). CASE II We next examine components of the form Rinji , the index i. We have previously shown that the rst pair of indices can be chosen in M1 ways. Therefore, the third index can be selected in N 2 ways and consequently there are M2 = 1 N (N 1)(N 2) distinct 2 components of the form Rinji with i = n = j. CASE III Next examine components of the form Rinjk where i = n = j = k. From CASE I the rst pairs of indices (i, n) can be chosen in M1 ways. Taking into account symmetries, it can be shown that the second pair of indices can be chosen in 1 (N 2)(N 3) ways. This implies that there are 1 N (N 1)(N 2)(N 3) 2 4 ways of choosing the indices i, n, j and k with i = n = j = k. By symmetry the pairs (i, n) and (j, k) can be interchanged and therefore only one half of these combinations are distinct. This leaves 1 N (N 1)(N 2)(N 3) 8 distinct pairs of indices. Also from the cyclic relations we nd that only two thirds of the above components are distinct. This produces M3 = N (N 1)(N 2)(N 3) 12 distinct components of the form Rinjk with i = n = j = k. Adding the above components from each case we nd there are M4 = M1 + M2 + M3 = distinct and independent components. Verify the entries in the following table: Dimension of space N Number of components N 4 N 2 (N 2 1) 12 1 1 0 2 16 1 3 81 6 4 256 20 5 625 50 M4 = Independent components of Rijkm Note 1: A one dimensional space can not be curved and all one dimensional spaces are Euclidean. (i.e. if we have an element of arc length squared given by ds2 = f (x)(dx)2 , we can make the coordinate transformation f (x)dx = du and reduce the arc length squared to the form ds2 = du2 .) Note 2: In a two dimensional space, the indices can only take on the values 1 and 2. In this special case there are 16 possible components. It can be shown that the only nonvanishing components are: R1212 = R1221 = R2112 = R2121 . 166 For these nonvanishing components only one independent component exists. By convention, the component R1212 is selected as the single independent component and all other nonzero components are expressed in terms of this component. Find the nonvanishing independent components Rijkl for i, j, k, l = 1, 2, 3, 4 and show that R1212 R1313 R2323 R1414 R2424 R3434 R1231 R1421 R1341 R2132 R2142 R2342 R3213 R3243 R3143 R4124 R4314 R4234 R1324 R1432 can be selected as the twenty independent components. 21. (a) For N = 2 show R1212 is the only nonzero independent component and R1212 = R2121 = R1221 = R2112 . 2 (b) Show that on the surface of a sphere of radius r0 we have R1212 = r0 sin2 . 22. Show for N = 2 that R1212 = R1212 J 2 = R1212 x x 2 23. s i i i Dene Rij = R.ijs as the Ricci tensor and Gi = Rj 1 j R as the Einstein tensor, where Rj = g ik Rkj j 2 i and R = Ri . Show that (a) Rjk = g ab Rjabk 2 log g (b) Rij = xi xj i Rijk = 0 b ij log g a b x x a ij + b ia a jb (c) 24. By employing the results from the previous problem show that in the case N = 2 we have R22 R12 R1212 R11 = = = g11 g22 g12 g where g is the determinant of gij . 25. Consider the case N = 2 where we have g12 = g21 = 0 and show that (a) (b) R12 = R21 = 0 R11 g22 = R22 g11 = R1221 (c) (d) 2R1221 g11 g22 1 Rij = Rgij , 2 R= where R = g ij Rij i i The scalar invariant R is known as the Einstein curvature of the surface and the tensor Gi = Rj 1 j R is j 2 known as the Einstein tensor. 26. For N = 3 show that R1212 , R1313 , R2323 , R1213 , R2123 , R3132 are independent components of the Riemann Christoel tensor. 167 27. For N = 2 and a = a11 0 0 a22 show that 1 a22 a u1 u2 1 a11 a u2 K= 1 R1212 = a 2 a u1 a11 a21 a12 a22 + . 28. For N = 2 and a = 1 K= 2a show that a12 a11 2 a12 1 a22 1 a11 a12 a11 + 1a 2 1 2 1 2 u a u u a u a u a11 a u1 11 a u . Check your results by setting a12 = a21 = 0 and comparing this answer with that given in the problem 27. 29. 30. (a) Use the fact that for n = 2 we have R1212 = R2121 = R2112 = R1221 together with e , e the two dimensional alternating tensors to show that the equation (1.5.110) can be written as R = K Write out the Frenet-Serret formulas (1.5.112)(1.5.113) for surface curves in terms of Christoel symbols of the second kind. where = ae and 1 = e a are the corresponding epsilon tensors. (b) Show that from the result in part (a) we obtain 31. 32. 33. 1 R 4 Hint: See equations (1.3.82),(1.5.93) and (1.5.94). Verify the result given by the equation (1.5.100). Show that a c = 4H 2 2K. Find equations for the principal curvatures associated with the surface x = u, y = v, z = f (u, v). = K. 34. Geodesics on a sphere Let (, ) denote the surface coordinates of the sphere of radius dened by the parametric equations x = sin cos , y = sin sin , z = cos . (1) Consider also a plane which passes through the origin with normal having the direction numbers (n1 , n2 , n3 ). This plane is represented by n1 x + n2 y + n3 z = 0 and intersects the sphere in a great circle which is described by the relation n1 sin cos + n2 sin sin + n3 cos = 0. (2) This is an implicit relation between the surface coordinates , which describes the great circle lying on the sphere. We can write this later equation in the form n1 cos + n2 sin = n3 tan (3) 168 and in the special case where n1 = cos , n2 = sin ,n3 = tan is expressible in the form cos( ) = tan tan or = cos1 tan tan . (4) The above equation denes an explicit relationship between the surface coordinates which denes a great circle on the sphere. The arc length squared relation satised by the surface coordinates together with the equation obtained by dierentiating equation (4) with respect to arc length s gives the relations sin2 d = ds tan 1 tan2 tan2 d ds (5) (6) ds2 = 2 d2 + 2 sin2 d2 The above equations (1)-(6) are needed to consider the following problem. (a) Show that the dierential equations dening the geodesics on the surface of a sphere (equations (1.5.51)) are d d2 sin cos =0 ds2 ds d2 d d =0 + 2 cot ds2 ds ds (b) Multiply equation (8) by sin2 and integrate to obtain sin2 where c1 is a constant of integration. (c) Multiply equation (7) by d ds 2 (7) (8) d = c1 ds (9) and use the result of equation (9) to show that an integration produces d ds 2 = c2 1 + c2 2 sin2 (10) where c2 is a constant of integration. 2 (d) Use the equations (5)(6) to show that c2 = 1/ and c1 = (e) Show that equations (9) and (10) imply that tan d = d tan2 and making the substitution u = tan tan sin . sec2 1 tan2 tan2 this equation can be integrated to obtain the equation (4). We can now expand the equation (4) and express the results in terms of x, y, z to obtain the equation (3). This produces a plane which intersects the sphere in a great circle. Consequently, the geodesics on a sphere are great circles. 169 35. 36. Find the dierential equations dening the geodesics on the surface of a cylinder. Find the dierential equations dening the geodesics on the surface of a torus. (See problem 13, Exercise 1.3) 37. Find the dierential equations dening the geodesics on the surface of revolution x = r cos , y = r sin , z = f (r). Note the curve z = f (x) gives a prole of the surface. The curves r = Constant are the parallels, while the curves = Constant are the meridians of the surface and ds2 = (1 + f 2 ) dr2 + r2 d2 . 38. Find the unit normal and tangent plane to an arbitrary point on the right circular cone x = u sin cos , y = u sin sin , z = u cos . This is a surface of revolution with r = u sin and f (r) = r cot with constant. Let s denote arc length and assume the position vector r(s) is analytic about a point s0 . Show that h2 h3 the Taylor series r(s) = r(s0 ) + hr (s0 ) + r (s0 ) + r (s0 ) + about the point s0 , with h = s s0 is 2! 3! given by r(s) = r(s0 ) + hT + 1 h2 N + 1 h3 (2 T + N + B) + which is obtained by dierentiating 2 6 the Frenet formulas. 40. (a) Show that the circular helix dened by x = a cos t, (i.e. T e3 = cos = constant.) (b) Show also that N e3 = 0 and consequently e3 is parallel to the rectifying plane, which implies that e3 = T cos + B sin . (c) Dierentiate the result in part (b) and show that / = tan is a constant. Consider a space curve xi = xi (s) in Cartesian coordinates. dT = xi xi (a) Show that = ds (b) Show that = 42. (a) Find the direction cosines of a normal to a surface z = f (x, y). (b) Find the direction cosines of a normal to a surface F (x, y, z) = 0. (c) Find the direction cosines of a normal to a surface x = x(u, v), y = y(u, v), z = z(u, v). 43. by K= Show that for a smooth surface z = f (x, y) the Gaussian curvature at a point on the surface is given 2 fxx fyy fxy . 2 2 (fx + fy + 1)2 39. y = a sin t, z = bt with a, b constants, has the property that any tangent to the curve makes a constant angle with the line dening the z-axis. 41. 1 eijk xi xj xk . Hint: Consider r r r 2 170 44. Show that for a smooth surface z = f (x, y) the mean curvature at a point on the surface is given by H= 2 2 (1 + fy )fxx 2fx fy fxy + (1 + fx )fyy . 2 2 2(fx + fy + 1)3/2 45. 46. 47. Express the Frenet-Serret formulas (1.5.13) in terms of Christoel symbols of the second kind. Verify the relation (1.5.106). In Vn assume that Rij = gij and show that = R n where R = g ij Rij . This result is known as Einsteins gravitational equation at points where matter is present. It is analogous to the Poisson equation 2 V = from the Newtonian theory of gravitation. 48. 49. In Vn assume that Rijkl = K(gik gjl gil gjk ) and show that R = Kn(1 n). (Hint: See problem 23.) Assume gij = 0 for i = j and verify the following. (a) Rhijk = 0 for h = i = j = k gii log ghh gii log gkk 2 gii (b) Rhiik = gii for h, i, k unequal. xh k x xh xk xk xh n gii ghh 1 gii 1 ghh (c) Rhiih = gii ghh h +i + where h = i. x ghh xh x gii xi xm xm m=1 m=h m=i 50. Consider a surface of revolution where x = r cos , y = r sin and z = f (r) is a given function of r. d ds . (a) Show in this V2 we have ds2 = (1 + (f )2 )dr2 + r2 d2 where = (b) Show the geodesic equations in this V2 are dr ff d2 r + 2 2 ds 1 + (f ) ds d2 2 d dr =0 + ds2 r ds ds (c) Solve the second equation in part (b) to obtain d = 2 r 1 + (f )2 d ds 2 =0 a d = 2 . Substitute this result for ds in part (a) to show ds r a 1 + (f )2 dr which theoretically can be integrated. r r 2 a2 171 PART 2: INTRODUCTION TO CONTINUUM MECHANICS In the following sections we develop some applications of tensor calculus in the areas of dynamics, elasticity, uids and electricity and magnetism. We begin by rst developing generalized expressions for the vector operations of gradient, divergence, and curl. Also generalized expressions for other vector operators are considered in order that tensor equations can be converted to vector equations. We construct a table to aid in the translating of generalized tensor equations to vector form and vice versa. The basic equations of continuum mechanics are developed in the later sections. These equations are developed in both Cartesian and generalized tensor form and then converted to vector form. 2.1 TENSOR NOTATION FOR SCALAR AND VECTOR QUANTITIES We consider the tensor representation of some vector expressions. Our goal is to develop the ability to convert vector equations to tensor form as well as being able to represent tensor equations in vector form. In this section the basic equations of continuum mechanics are represented using both a vector notation and the indicial notation which focuses attention on the tensor components. In order to move back and forth between these notations, the representation of vector quantities in tensor form is now considered. Gradient For = (x1 , x2 , . . . , xN ) a scalar function of the coordinates xi , i = 1, . . . , N , the gradient of is dened as the covariant vector ,i = The contravariant form of the gradient is g im ,m . (2.1.2) , xi i = 1, . . . , N. (2.1.1) Note, if C i = g im ,m , i = 1, 2, 3 are the tensor components of the gradient then in an orthogonal coordinate system we will have C 1 = g 11 ,1 , C 2 = g 22 ,2 , C 3 = g 33 ,3 . We note that in an orthogonal coordinate system that g ii = 1/h2 , (no sum on i), i = 1, 2, 3 and hence i replacing the tensor components by their equivalent physical components there results the equations 1 C(1) = 2 1, h1 h1 x C(2) 1 = 2 2, h2 h2 x C(3) 1 = 2 3. h3 h3 x Simplifying, we nd the physical components of the gradient are C(1) = 1 , h1 x1 C(2) = 1 , h2 x2 C(3) = 1 . h3 x3 These results are only valid when the coordinate system is orthogonal and gij = 0 for i = j and gii = h2 , i with i = 1, 2, 3, and where i is not summed. 172 Divergence The divergence of a contravariant tensor Ar is obtained by taking the covariant derivative with respect to xk and then performing a contraction. This produces div Ar = Ar,r . (2.1.3) Still another form for the divergence is obtained by simplifying the expression (2.1.3). The covariant derivative can be represented Ar,k = Ar + xk r Am . mk Upon contracting the indices r and k and using the result from Exercise 1.4, problem 13, we obtain Ar,r = Ar,r Ar,r Ar 1 ( g) m + A xr g xm 1 Ar r g = g r +A g x xr 1r = ( gA ) . g xr (2.1.4) EXAMPLE 2.1-1. (Divergence) Find the representation of the divergence of a vector Ar in spherical coordinates (, , ). Solution: x1 = , In spherical coordinates we have x2 = , x3 = with g22 = h2 = 2 , 2 gij = 0 for i=j and g11 = h2 = 1, 1 g33 = h2 = 2 sin2 . 3 The determinant of gij is g = |gij | = 4 sin2 and g = 2 sin . Employing the relation (2.1.4) we nd 1 1 div Ar = ( gA ) + 2 ( gA2 ) + 3 ( gA3 ) . 1 g x x x In terms of the physical components this equation becomes A(1) 1 A(2) A(3) (g (g (g )+ )+ ). div Ar = g h1 h2 h3 By using the notation A(1) = A , A(2) = A , A(3) = A for the physical components, the divergence can be expressed in either of the forms: div Ar = 2 1 2 A A 2 ( sin A ) + ( sin ) + ( sin ) 2 sin sin 1 2 1 1 A div Ar = 2 ( A ) + (sin A ) + . sin sin or 173 Curl The contravariant components of the vector C = curl A are represented Ci = In expanded form this representation becomes: 1 C1 = g 1 C2 = g 1 C3 = g A3 A2 2 x x3 A3 A1 3 x x1 A2 A1 1 x x2 . (2.1.6) ijk Ak,j . (2.1.5) EXAMPLE 2.1-2. (Curl) Find the representation for the components of curl A in spherical coordinates (, , ). Solution: In spherical coordinates we have :x1 = , g11 = h2 = 1, 1 x2 = , x3 = with gij = 0 for i = j and g22 = h2 = 2 , 2 g33 = h2 = 2 sin2 . 3 The determinant of gij is g = |gij | = 4 sin2 with g = 2 sin . The relations (2.1.6) are tensor equations representing the components of the vector curl A. To nd the components of curl A in spherical components we write the equations (2.1.6) in terms of their physical components. These equations take on the form: C(1) 1 (h3 A(3)) (h2 A(2)) = h1 g C(2) 1 (h1 A(1)) (h3 A(3)) = h2 g C(3) 1 (h2 A(2)) (h1 A(1)) . = h3 g We employ the notations C(1) = C , C(2) = C , C(3) = C , A(1) = A , A(2) = A , A(3) = A (2.1.7) to denote the physical components, and nd the components of the vector curl A, in spherical coordinates, are expressible in the form: 1 ( sin A ) (A ) 2 sin 1 (A ) ( sin A ) C = sin 1 C = (A ) (A ) . C = (2.1.8) 174 Laplacian The Laplacian 2 U has the contravariant form 2 U = g ij U,ij = (g ij U,i ),j = Expanding this expression produces the equations: xj 2 U = xj 2 U = U xi U g ij i x g ij j U xi m j 1 g ij U g + g xj xi + g im g ij U xi . + g ij g ij U xi . ,j (2.1.9) 1 2 U = gj g x 1 2 U = g xj U g xi xj (2.1.10) gg ij U xi In orthogonal coordinates we have g ij = 0 for i = j and g11 = h2 , 1 g22 = h2 , 2 g33 = h2 3 and so (2.1.10) when expanded reduces to the form 2 U = 1 h1 h2 h3 x1 h2 h3 U h1 x1 + x2 h1 h3 U h2 x2 + x3 h1 h2 U h3 x3 . (2.1.11) This representation is only valid in an orthogonal system of coordinates. EXAMPLE 2.1-3. (Laplacian) has the form 2 U = This simplies to 2 U = 1 2U 1 2U 2U 2 U cot U +2 +2 2 + +2 . 2 2 sin 2 (2.1.13) 2 1 sin 2 sin U + sin U + 1 U sin . (2.1.12) Find the Laplacian in spherical coordinates. Solution: Utilizing the results given in the previous example we nd the Laplacian in spherical coordinates The table 1 gives the vector and tensor representation for various quantities of interest. 175 VECTOR GENERAL TENSOR Ai or Ai Ai Bi = gij Ai B j = Ai B i Ai Bi = g ij Ai Bj 1 C i = eijk Aj Bk g g im ,m 1r g mn Am,n = Ar,r = ( gA ) g xr Ci = ijk CARTESIAN TENSOR A AB C =AB = grad A = div A A = C = curl A 2 U C = (A )B C = A( B) C = 2 A A A A Ai Ai Bi Ci = eijk Aj Bk xi Ai xi Ak xj ,i = Ai,i = Ak,j ij U gg xi Ci = eijk xi 1 g mn U ,mn = g xj U xi Bi xm C i = Am B i,m C i = Ai B j,j C i = g jm Ai ,mj or Ci = g jm Ai,mj Ci = Ci = Am Ci = Ai xm Bm xm Ai xm Ai ,i 2 Ar xi xr g im Ai ,m g im Ar ,r ijk g jm kst ,m At,s ,m 2 Aj 2 Ai xj xi xj xj Table 1 Vector and tensor representations. 176 EXAMPLE 2.1-4. (Maxwells equations) In the study of electrodynamics there arises the following vectors and scalars: E =Electric force vector, [E] = Newton/coulomb B =Magnetic force vector, [B] = Weber/m2 D =Displacement vector, [D] = coulomb/m2 H =Auxilary magnetic force vector, [H] = ampere/m J =Free current density, [J] = ampere/m2 =free charge density, [ ] = coulomb/m3 The above quantities arise in the representation of the following laws: Faradays Law This law states the line integral of the electromagnetic force around a loop is proportional to the rate of ux of magnetic induction through the loop. This gives rise to the rst electromagnetic eld equation: E = Amperes Law B t or ijk Ek,j = B i . t (2.1.15) This law states the line integral of the magnetic force vector around a closed loop is proportional to the sum of the current through the loop and the rate of ux of the displacement vector through the loop. This produces the second electromagnetic eld equation: H =J + D t or ijk Hk,j = J i + Di . t (2.1.16) Gausss Law for Electricity This law states that the ux of the electric force vector through a closed surface is proportional to the total charge enclosed by the surface. This results in the third electromagnetic eld equation: D = or 1i gD = . g xi (2.1.17) Gausss Law for Magnetism This law states the magnetic ux through any closed volume is zero. This produces the fourth electromagnetic eld equation: B = 0 or 1i gB = 0. g xi (2.1.18) The four electromagnetic eld equations are referred to as Maxwells equations. These equations arise in the study of electrodynamics and can be represented in other forms. These other forms will depend upon such things as the material assumptions and units of measurements used. Note that the tensor equations (2.1.15) through (2.1.18) are representations of Maxwells equations in a form which is independent of the coordinate system chosen. In applications, the tensor quantities must be expressed in terms of their physical components. In a general orthogonal curvilinear coordinate system we will have g11 = h2 , 1 This produces the result g22 = h2 , 2 g33 = h2 , 3 and gij = 0 for i = j. g = h1 h2 h3 . Further, if we represent the physical components of Di , Bi , Ei , Hi by D(i), B(i), E(i), and H(i) 177 the Maxwell equations can be represented by the equations in table 2. The tables 3, 4 and 5 are the representation of Maxwells equations in rectangular, cylindrical, and spherical coordinates. These latter tables are special cases associated with the more general table 2. 1 (h3 E(3)) h1 h2 h3 x2 1 (h1 E(1)) h1 h2 h3 x3 1 (h2 E(2)) h1 h2 h3 x1 1 (h3 H(3)) h1 h2 h3 x2 1 (h1 H(1)) h1 h2 h3 x3 1 (h2 H(2)) h1 h2 h3 x1 1 h1 h2 h3 x1 1 h1 h2 h3 x1 D(1) h1 B(1) h1 x2 x2 1 B(1) (h2 E(2)) = 3 x h1 t 1 B(2) (h3 E(3)) = 1 x h2 t 1 B(3) (h1 E(1)) = 2 x h3 t J(1) 1 D(1) (h2 H(2)) = + 3 x h1 h1 t J(2) 1 D(2) (h3 H(3)) = + 1 x h2 h2 t J(3) 1 D(3) (h1 H(1)) = + x2 h3 h3 t D(2) h2 B(2) h2 x3 x3 D(3) h3 B(3) h3 h1 h2 h3 + h1 h2 h3 + h1 h2 h3 = h1 h2 h3 + h1 h2 h3 + h1 h2 h3 =0 Table 2 Maxwells equations in generalized orthogonal coordinates. Note that all the tensor components have been replaced by their physical components. 178 Ez Ey Bx = y z t Ex Ez By = z x t Ex Bz Ey = x y t Hy Hz = Jx + y z Hx Hz = Jy + z x Hx Hy = Jz + x y Dx t Dy t Dz t Dy Dz Dx + + = x y z By Bz Bx + + =0 x y z Here we have introduced the notations: Dx = D(1) Dy = D(2) Dz = D(3) with x1 = x, x2 = y, Bx = B(1) By = B(2) Bz = B(3) Hx = H(1) Hy = H(2) Hz = H(3) Jx = J(1) Jy = J(2) Jz = J(3) Ex = E(1) Ey = E(2) Ez = E(3) x3 = z, h1 = h2 = h3 = 1 Table 3 Maxwells equations Cartesian coordinates E Br 1 Ez = r z t Ez B Er = z r t Bz 1 1 Er (rE ) = r r r t Dz 1 D 1 (rDr ) + + = r r r z Here we have introduced the notations: Dr = D(1) D = D(2) Dz = D(3) with x1 = r, x2 = , Br = B(1) B = B(2) Bz = B(3) h1 = 1, H Dr 1 Hz = Jr + r z t Hz Hr D = J + z r t 1 1 Hr Dz (rH ) = Jz + r r r t 1 Bz 1 B (rBr ) + + =0 r r r z Hr = H(1) H = H(2) Hz = H(3) h2 = r, Jr = J(1) J = J(2) Jz = J(3) Er = E(1) E = E(2) Ez = E(3) x3 = z, h3 = 1. Table 4 Maxwells equations in cylindrical coordinates. 179 1 E B (sin E ) = sin t 1 1 E B (E ) = sin t 1 B 1 E (E ) = t 1 H D (sin H ) = J + sin t 1 1 H D (H ) = J + sin t 1 1 H D (H ) = J + t 1 2 1 1 D ( D ) + (sin D ) + = 2 sin sin 1 2 1 1 B ( B ) + (sin B ) + =0 2 sin sin Here we have introduced the notations: D = D(1) D = D(2) D = D(3) with x1 = , x2 = , B = B(1) B = B(2) B = B(3) h1 = 1, H = H(1) H = H(2) H = H(3) h2 = , J = J(1) J = J(2) J = J(3) E = E(1) E = E(2) E = E(3) x3 = , h3 = sin Table 5 Maxwells equations spherical coordinates. Eigenvalues and Eigenvectors of Symmetric Tensors Consider the equation Tij Aj = Ai , i, j = 1, 2, 3, (2.1.19) where Tij = Tji is symmetric, Ai are the components of a vector and is a scalar. Any nonzero solution Ai of equation (2.1.19) is called an eigenvector of the tensor Tij and the associated scalar is called an eigenvalue. When expanded these equations have the form (T11 )A1 + T12 A2 + T13 A3 = 0 T23 A3 = 0 T21 A1 + (T22 )A2 + T31 A1 + T32 A2 + (T33 )A3 = 0. The condition for equation (2.1.19) to have a nonzero solution Ai is that the characteristic equation should be zero. This equation is found from the determinant equation T11 T21 T31 T12 T13 T22 T23 = 0, T32 T33 f () = (2.1.20) 180 which when expanded is a cubic equation of the form f () = 3 + I1 2 I2 + I3 = 0, where I1 , I2 and I3 are invariants dened by the relations I1 = Tii 1 1 I2 = Tii Tjj Tij Tij 2 2 I3 = eijk Ti1 Tj2 Tk3 . When Tij is subjected to an orthogonal transformation, where Tmn = Tij im jn im jn , (2.1.21) (2.1.22) then (Tmn mn ) = Tij ij and det (Tmn mn ) = det Tij ij . Hence, the eigenvalues of a second order tensor remain invariant under an orthogonal transformation. If Tij is real and symmetric then the eigenvalues of Tij will be real, and the eigenvectors corresponding to distinct eigenvalues will be orthogonal. Proof: To show a quantity is real we show that the conjugate of the quantity equals the given quantity. If (2.1.19) is satised, we multiply by the conjugate Ai and obtain Ai Tij Aj = Ai Ai . (2.1.25) The right hand side of this equation has the inner product Ai Ai which is real. It remains to show the left hand side of equation (2.1.25) is also real. Consider the conjugate of this left hand side and write Ai Tij Aj = Ai T ij Aj = Ai Tji Aj = Ai Tij Aj . Consequently, the left hand side of equation (2.1.25) is real and the eigenvalue can be represented as the ratio of two real quantities. Assume that (1) and (2) are two distinct eigenvalues which produce the unit eigenvectors L1 and L2 with components i1 and i2 , i = 1, 2, 3 respectively. We then have Tij j1 = (1) i1 and Tij j2 = (2) i2 . (2.1.26) Consider the products (1) (2) i1 i2 i1 i2 = Tij = j1 i2 , (2.1.27) j1 Tji i2 . i1 Tij j2 = and subtract these equations. We nd that [(1) (2) ] i1 i2 = 0. i1 i2 (2.1.28) must be zero. Therefore, By hypothesis, (1) is dierent from (2) and consequently the inner product the eigenvectors corresponding to distinct eigenvalues are orthogonal. 181 Therefore, associated with distinct eigenvalues (i) , i = 1, 2, 3 there are unit eigenvectors L(i) = with components im , m i1 e1 + i2 e2 + i3 e3 = 1, 2, 3 which are direction cosines and satisfy in im = mn and ij jm = im . (2.1.23) The unit eigenvectors satisfy the relations Tij j1 = (1) i1 Tij j2 = (2) i2 Tij j3 = (3) i3 and can be written as the single equation Tij jm = (m) im , m = 1, 2, or 3 m not summed. Consider the transformation xi = which represents a rotation of axes, where linear transformation where the ij ij xj or xm = mj xj ij are the direction cosines from the eigenvectors of Tij . This is a satisfy equation (2.1.23). Such a transformation is called an orthogonal transformation. In the new x coordinate system, called principal axes, we have T mn = Tij xi xj = Tij xm xn im jn = (n) in im = (n) mn (no sum on n). (2.1.24) This equation shows that in the barred coordinate system there are the components (1) = 0 0 0 (2) 0 0 0 . (3) T mn That is, along the principal axes the tensor components Tij are transformed to the components T ij where T ij = 0 for i = j. The elements T (i)(i) , i not summed, represent the eigenvalues of the transformation (2.1.19). 182 EXERCISE 2.1 1. In cylindrical coordinates (r, , z) with f = f (r, , z) nd the gradient of f. 2. In cylindrical coordinates (r, , z) with A = A(r, , z) nd div A. 3. In cylindrical coordinates (r, , z) for A = A(r, , z) nd curl A. 4. In cylindrical coordinates (r, , z) for f = f (r, , z) nd 2 f. 5. In spherical coordinates (, , ) with f = f (, , ) nd the gradient of f. 6. In spherical coordinates (, , ) with A = A(, , ) nd div A. 7. In spherical coordinates (, , ) for A = A(, , ) nd curl A. 8. In spherical coordinates (, , ) for f = f (, , ) nd 2 f. 9. Let r = x e1 + y e2 + z e3 denote the position vector of a variable point (x, y, z) in Cartesian coordinates. Let r = |r| denote the distance of this point from the origin. Find in terms of r and r: (a) grad (r) (b) grad (rm ) (c) 1 grad ( ) r (d) grad (ln r) (e) grad () where = (r) is an arbitrary function of r. 10. Let r = x e1 +y e2 +z e3 denote the position vector of a variable point (x, y, z) in Cartesian coordinates. Let r = |r| denote the distance of this point from the origin. Find: (a) div (r) (b) div (rm r) (c) div (r3 r) (d) div ( r) where = (r) is an arbitrary function or r. 11. Let r = x e1 + y e2 + z e3 denote the position vector of a variable point (x, y, z) in Cartesian curl r (b) curl ( r) coordinates. Let r = |r| denote the distance of this point from the origin. Find: (a) where = (r) is an arbitrary function of r. 12. Expand and simplify the representation for curl (curl A). 13. Show that the curl of the gradient is zero in generalized coordinates. 14. Write out the physical components associated with the gradient of = (x1 , x2 , x3 ). 15. Show that 1 im 1i g im Ai,m = gg Am = Ai,i = gA . i g x g xi 183 16. Let r = (r r)1/2 = 17. x2 + y 2 + z 2 ) and calculate (a) 2 (r) 1 2 (vi,j (b) 2 (1/r) (c) 2 (r2 ) (d) 2 (1/r2 ) Given the tensor equations Dij = + vj,i ), i, j = 1, 2, 3. Let v(1), v(2), v(3) denote the physical components of v1 , v2 , v3 and let D(ij) denote the physical components associated with Dij . Assume the coordinate system (x1 , x2 , x3 ) is orthogonal with metric coecients g(i)(i) = h2 , i = 1, 2, 3 and gij = 0 i for i = j. (a) Find expressions for the physical components D(11), D(22) and D(33) in terms of the physical compoV (j) hi 1 V (i) + no sum on i. nents v(i), i = 1, 2, 3. Answer: D(ii) = i hi x hi hj xj j=i (b) Find expressions for the physical components D(12), D(13) and D(23) in terms of the physical compoV (i) V (j) hj 1 hi + nents v(i), i = 1, 2, 3. Answer: D(ij) = 2 hj xj hi hi xi hj 18. Write out the tensor equations in problem 17 in Cartesian coordinates. 19. Write out the tensor equations in problem 17 in cylindrical coordinates. 20. Write out the tensor equations in problem 17 in spherical coordinates. 21. Express the vector equation ( + 2) 2 + F = 0 in tensor form. 22. Write out the equations in problem 21 for a generalized orthogonal coordinate system in terms of physical components. 23. Write out the equations in problem 22 for cylindrical coordinates. 24. Write out the equations in problem 22 for spherical coordinates. 25. Use equation (2.1.4) to represent the divergence in parabolic cylindrical coordinates (, , z). 26. Use equation (2.1.4) to represent the divergence in parabolic coordinates (, , ). 27. Use equation (2.1.4) to represent the divergence in elliptic cylindrical coordinates (, , z). Change the given equations from a vector notation to a tensor notation. 28. 29. 30. 31. 32. B = v A + ( v) A d dA dB dC [A (B C)] = (B C) + A ( C) + A (B ) dt dt dt dt v dv = + (v )v dt t 1 H = curl E c t dB (B )v + B( v) = 0 dt 184 Change the given equations from a tensor notation to a vector notation. 33. 34. 35. 36. ijk Bk,j + F i = 0 gij jkl Bl,k + Fi = 0 + ( vi ), i = 0 t vi P 2 vi vi + vm m ) = i + m m + Fi ( t x x x x 37. The moment of inertia of an area or second moment of area is dened by Iij = A (ym ym ij yi yj ) dA where dA is an element of area. Calculate the moment of inertia Iij , i, j = 1, 2 for the triangle illustrated in 1 1 3 24 b2 h2 12 bh the gure 2.1-1 and show that Iij = . 122 13 24 b h 12 b h Figure 2.1-1 Moments of inertia for a triangle 38. Use the results from problem 37 and rotate the axes in gure 2.1-1 through an angle to a barred system of coordinates. (a) Show that in the barred system of coordinates I 11 = I 12 = I 21 I 22 I11 + I22 I11 I22 + cos 2 + I12 sin 2 2 2 I11 I22 = sin 2 + I12 cos 2 2 I11 + I22 I11 I22 = cos 2 I12 sin 2 2 2 (b) For what value of will I 11 have a maximum value? (c) Show that when I 11 is a maximum, we will have I 22 a minimum and I 12 = I 21 = 0. 185 Figure 2.1-2 Mohrs circle 39. Otto Mohr1 gave the following physical interpretation to the results obtained in problem 38: Plot the points A(I11 , I12 ) and B(I22 , I12 ) as illustrated in the gure 2.1-2 Draw the line AB and calculate the point C where this line intersects the I axes. Show the point C has the coordinates I11 + I22 , 0) 2 Calculate the radius of the circle with center at the point C and with diagonal AB and show this ( radius is r= I11 I22 2 2 2 + I12 Show the maximum and minimum values of I occur where the constructed circle intersects the I axes. I11 + I22 I11 + I22 Show that Imax = I 11 = +r Imin = I 22 = r. 2 2 I11 I12 40. Show directly that the eigenvalues of the symmetric matrix Iij = are 1 = Imax and I21 I22 2 = Imin where Imax and Imin are given in problem 39. 41. Find the principal axes and moments of inertia for the triangle given in problem 37 and summarize your results from problems 37,38,39, and 40. 42. Verify for orthogonal coordinates the relations 3 A e(i) = k=1 e(i)jk (h(k) A(k)) h(i) h1 h2 h3 xj h2 e 2 h3 e 3 or A = 1 h1 h2 h3 h1 e 1 h1 A(1) x1 x2 h2 A(2) x3 h3 A(3) . 43. Verify for orthogonal coordinates the relation 3 ( A) e(i) = m=1 1 e(i)jr ersm h(i) h1 h2 h3 xj h2 (r) (h(m) A(m)) h1 h2 h3 xs Christian Otto Mohr (1835-1918) German civil engineer. 186 44. Verify for orthogonal coordinates the relation 1 (h2 h3 A(1)) (h1 h3 A(2)) (h1 h2 A(3)) 1 A e(i) = + + h(i) x(i) h1 h2 h3 x1 x2 x3 45. Verify the relation 3 (A )B e(i) = k=1 A(k) B(i) + h(k) xk k=i B(k) hk h(i) A(i) h(i) hk A(k) xk x(i) 46. The Gauss divergence theorem is written F 2 F 3 F 1 + + x y z V d = S n1 F 1 + n2 F 2 + n3 F 3 d where V is the volume within a simple closed surface S. Here it is assumed that F i = F i (x, y, z) are continuous functions with continuous rst order derivatives throughout V and ni are the direction cosines of the outward normal to S, d is an element of volume and d is an element of surface area. (a) Show that in a Cartesian coordinate system i F,i = F 2 F 3 F 1 + + x y z i F,i d = V S and that the tensor form of this theorem is (b) Write the vector form of theorem. this (c) Show that if we dene u v , vr = xr xr = g im (uvi,m + um vi ) ur = F i ni d. and Fr = grm F m = uvr i then F,i = g im Fi,m (d) Show that another form of the Gauss divergence theorem is g im um vi d = V S uvm nm d V ug im vi,m d 12 2 1. 11 21 1 0. 01 10 1 1. 11 Write out the above equation in Cartesian coordinates. 47. Show 48. Show 49. Show 1 Find the eigenvalues and eigenvectors associated with the matrix A = 1 2 that the eigenvectors are orthogonal. 1 Find the eigenvalues and eigenvectors associated with the matrix A = 2 1 that the eigenvectors are orthogonal. 1 Find the eigenvalues and eigenvectors associated with the matrix A = 1 0 that the eigenvectors are orthogonal. 50. The harmonic and biharmonic functions or potential functions occur in the mathematical modeling of many physical problems. Any solution of Laplaces equation 2 = 0 is called a harmonic function and any solution of the biharmonic equation 4 = 0 is called a biharmonic function. (a) Expand the Laplace equation in Cartesian, cylindrical and spherical coordinates. (b) Expand the biharmonic equation in two dimensional Cartesian and polar coordinates. Hint: Consider 4 = 2 (2 ). In Cartesian coordinates 2 = ,ii and 4 = ,iijj . 187 2.2 DYNAMICS Dynamics is concerned with studying the motion of particles and rigid bodies. By studying the motion of a single hypothetical particle, one can discern the motion of a system of particles. This in turn leads to the study of the motion of individual points in a continuous deformable medium. Particle Movement The trajectory of a particle in a generalized coordinate system is described by the parametric equations xi = xi (t), i = 1, . . . , N (2.2.1) where t is a time parameter. If the coordinates are changed to a barred system by introducing a coordinate transformation xi = xi (x1 , x2 , . . . , xN ), i = 1, . . . , N then the trajectory of the particle in the barred system of coordinates is xi = xi (x1 (t), x2 (t), . . . , xN (t)), i = 1, . . . , N. (2.2.2) The generalized velocity of the particle in the unbarred system is dened by vi = dxi , dt i = 1, . . . , N. (2.2.3) By the chain rule dierentiation of the transformation equations (2.2.2) one can verify that the velocity in the barred system is vr = xr dxj xr j dxr = = v, j dt dt x xj r = 1, . . . , N. (2.2.4) Consequently, the generalized velocity v i is a rst order contravariant tensor. The speed of the particle is obtained from the magnitude of the velocity and is v 2 = gij v i v j . The generalized acceleration f i of the particle is dened as the intrinsic derivative of the generalized velocity. The generalized acceleration has the form fi = n dv i v i i dx = v,n = + t dt dt i d2 xi vm vn = + mn dt2 i dxm dxn m n dt dt (2.2.5) and the magnitude of the acceleration is f 2 = gij f i f j . 188 Figure 2.2-1 Tangent, normal and binormal to point P on curve. Frenet-Serret Formulas The parametric equations (2.2.1) describe a curve in our generalized space. With reference to the gure 2.2-1 we wish to dene at each point P of the curve the following orthogonal unit vectors: T i = unit tangent vector at each point P. N i = unit normal vector at each point P. B i = unit binormal vector at each point P. These vectors dene the osculating, normal and rectifying planes illustrated in the gure 2.2-1. In the generalized coordinates the arc length squared is ds2 = gij dxi dxj . Dene T i = dxi ds as the tangent vector to the parametric curve dened by equation (2.2.1). This vector is a unit tangent vector because if we write the element of arc length squared in the form 1 = gij dxi dxj = gij T i T j , ds ds (2.2.6) we obtain the generalized dot product for T i . This generalized dot product implies that the tangent vector is a unit vector. Dierentiating the equation (2.2.6) intrinsically with respect to arc length s along the curve produces gmn which simplies to gmn T n T m n T n T + gmn T m = 0, s s T m = 0. s (2.2.7) 189 The equation (2.2.7) is a statement that the vector vector is dened as Ni = 1 T i s T m s is orthogonal to the vector T m . The unit normal Ni = 1 Ti , s (2.2.8) or where is a scalar called the curvature and is chosen such that the magnitude of N i is unity. The reciprocal of the curvature is R = as the arc length varies. The equation (2.2.7) can be expressed in the form gij T i N j = 0. Taking the intrinsic derivative of equation (2.2.9) with respect to the arc length s produces gij T i or gij T i N j T i j + gij N =0 s s (2.2.9) 1 , which is called the radius of curvature. The curvature of a straight line is zero while the curvature of a circle is a constant. The curvature measures the rate of change of the tangent vector N j T i j = gij N = gij N i N j = . s s (2.2.10) The generalized dot product can be written gij T i T j = 1, and consequently we can express equation (2.2.10) in the form gij T i Consequently, the vector N j + T j s with respect to the arc length s to show that gij N i This in turn can be expressed in the form gij N i N j + T j s = 0. N j = 0. s (2.2.12) N j = gij T i T j s or gij T i N j + T j s = 0. (2.2.11) is orthogonal to T i . In a similar manner, we can use the relation gij N i N j = 1 and dierentiate intrinsically This form of the equation implies that the vector represented in equation (2.2.12) is also orthogonal to the unit normal N i . We dene the unit binormal vector as Bi = 1 N i + T i s or Bi = 1 Ni + Ti s (2.2.13) where is a scalar called the torsion. The torsion is chosen such that the binormal vector is a unit vector. The torsion measures the rate of change of the osculating plane and consequently, the torsion is a measure 190 of the twisting of the curve out of a plane. The value = 0 corresponds to a plane curve. The vectors T i , N i , B i , i = 1, 2, 3 satisfy the cross product relation Bi = ijk T j Nk . If we dierentiate this relation intrinsically with respect to arc length s we nd B i = s = ijk ijk Tj Tj Nk + Nk s s (2.2.14) ikj [Tj ( Bk Tk ) + Nj Nk ] Tj Bk = Bk Tj = N i . = ijk The relations (2.2.8),(2.2.13) and (2.2.14) are now summarized and written T i = N i s N i = B i T i s B i = N i . s These equations are known as the Frenet-Serret formulas of dierential geometry. Velocity and Acceleration Chain rule dierentiation of the generalized velocity is expressible in the form vi = where v = ds dt (2.2.15) dxi ds dxi = = T i v, dt ds dt (2.2.16) is the speed of the particle and is the magnitude of v i . The vector T i is the unit tangent vector to the trajectory curve at the time t. The equation (2.2.16) is a statement of the fact that the velocity of a particle is always in the direction of the tangent vector to the curve and has the speed v. By chain rule dierentiation, the generalized acceleration is expressible in the form fr = dv r v r T r = T +v t dt t dv r T r ds T +v = dt s dt dv r T + v 2 N r . = dt (2.2.17) The equation (2.2.17) states that the acceleration lies in the osculating plane. Further, the equation (2.2.17) indicates that the tangential component of the acceleration is eration is v . 2 dv dt , while the normal component of the accel- 191 Work and Potential Energy Dene M as the constant mass of the particle as it moves along the curve dened by equation (2.2.1). Also let Qr denote the components of a force vector (in appropriate units of measurements) which acts upon the particle. Newtons second law of motion can then be expressed in the form Qr = M f r or Qr = M fr . (2.2.18) The work done W in moving a particle from a point P0 to a point P1 along a curve xr = xr (t), r = 1, 2, 3, with parameter t, is represented by a summation of the tangential components of the forces acting along the path and is dened as the line integral P1 W= P0 Qr dxr ds = ds P1 P0 Qr dxr = t1 Qr t0 dxr dt = dt t1 t0 Qr v r dt (2.2.19) where Qr = grs Qs is the covariant form of the force vector, t is the time parameter and s is arc length along the curve. Conservative Systems If the force vector is conservative it means that the force is derivable from a scalar potential function V = V (x1 , x2 , . . . , xN ) such that Qr = V ,r = V , xr r = 1, . . . , N. (2.2.20) In this case the equation (2.2.19) can be integrated and we nd that to within an additive constant we will have V = W. The potential function V is called the potential energy of the particle and the work done becomes the change in potential energy between the starting and end points and is independent of the path connecting the points. Lagranges Equations of Motion The kinetic energy T of the particle is dened as one half the mass times the velocity squared and can be expressed in any of the forms T= 1 M 2 ds dt 2 = 1 1 1 M v 2 = M gmn v m v n = M gmn xm xn , 2 2 2 (2.2.21) where the dot notation denotes dierentiation with respect to time. It is an easy exercise to calculate the derivatives d dt T = M grmxm xr T grm n m xx = M grm xm + r x xn T 1 gmn m n =M x x, r x 2 xr (2.2.22) and thereby verify the relation d dt T xr T = M fr = Qr , xr r = 1, . . . , N. (2.2.23) 192 This equation is called the Lagranges form of the equations of motion. EXAMPLE 2.2-1. (Equations of motion in spherical coordinates) the equations of motion in spherical coordinates. Solution: Let x1 = , x2 = , x3 = then the element of arc length squared in spherical coordinates has the form ds2 = (d)2 + 2 (d)2 + 2 sin2 (d)2 . The element of arc length squared can be used to construct the kinetic energy. For example, T= 1 M 2 ds dt 2 Find the Lagranges form of = 1 M ()2 + 2 ()2 + 2 sin2 ()2 . 2 The Lagrange form of the equations of motion of a particle are found from the relations (2.2.23) and are calculated to be: M f1 = Q1 = d dt d M f2 = Q2 = dt d M f3 = Q3 = dt T T T T = M ()2 sin2 ()2 d T =M 2 2 sin cos ()2 dt d T =M 2 sin2 . dt In terms of physical components we have Q = M ()2 sin2 ()2 Md 2 2 sin cos ()2 dt d M 2 sin2 . Q = sin dt Q = Euler-Lagrange Equations of Motion Starting with the Lagranges form of the equations of motion from equation (2.2.23), we assume that the external force Qr is derivable from a potential function V as specied by the equation (2.2.20). That is, we assume the system is conservative and express the equations of motion in the form d dt T xr V T = r = Qr , xr x r = 1, . . . , N (2.2.24) The Lagrangian is dened by the equation L = T V = T (x1 , . . . , xN , x1 , . . . , xN ) V (x1 , . . . , xN ) = L(xi , xi ). (2.2.25) Employing the dening equation (2.2.25), it is readily veried that the equations of motion are expressible in the form d dt L xr L = 0, xr r = 1, . . . , N, (2.2.26) which are called the Euler-Lagrange form for the equations of motion. 193 Figure 2.2-2 Simply pulley system EXAMPLE 2.2-2. (Simple pulley system) Find the equation of motion for the simply pulley system illustrated in the gure 2.2-2. Solution: The given system has only one degree of freedom, say y1 . It is assumed that y1 + y2 = = a constant. The kinetic energy of the system is T= 1 (m1 + m2 )y1 . 2 2 Let y1 increase by an amount dy1 and show the work done by gravity can be expressed as dW = m1 g dy1 + m2 g dy2 dW = m1 g dy1 m2 g dy1 dW = (m1 m2 )g dy1 = Q1 dy1 . Here Q1 = (m1 m2 )g is the external force acting on the system where g is the acceleration of gravity. The Lagrange equation of motion is d dt or y (m1 + m2 )1 = (m1 m2 )g. Initial conditions must be applied to y1 and y1 before this equation can be solved. T y1 T = Q1 y1 194 EXAMPLE 2.2-3. (Simple pendulum) Find the equation of motion for the pendulum system illustrated in the gure 2.2-3. Solution: Choose the angle illustrated in the gure 2.2-3 as the generalized coordinate. If the pendulum is moved from a vertical position through an angle , we observe that the mass m moves up a distance h = cos . The work done in moving this mass a vertical distance h is W = mgh = mg (1 cos ), since the force is mg in this coordinate system. In moving the pendulum through an angle , the arc length s swept out by the mass m is s = . This implies that the kinetic energy can be expressed T= 1 m 2 ds dt 2 = 1 m 2 2 = 1 2 2 m () . 2 Figure 2.2-3 Simple pendulum system The Lagrangian of the system is L=T V = and from this we nd the equation of motion d dt L L =0 or d m 2 mg ( sin ) = 0. dt 1 2 2 m () mg (1 cos ) 2 This in turn simplies to the equation g + sin = 0. This equation together with a set of initial conditions for and represents the nonlinear dierential equation which describes the motion of a pendulum without damping. 195 EXAMPLE 2.2-4. (Compound pendulum) illustrated in the gure 2.2-4. Solution: Choose for the generalized coordinates the angles x1 = 1 and x2 = 2 illustrated in the gure 2.2-4. To nd the potential function V for this system we consider the work done as the masses m1 and m2 are moved. Consider independent motions of the angles 1 and 2 . Imagine the compound pendulum initially in the vertical position as illustrated in the gure 2.2-4(a). Now let m1 be displaced due to a change in 1 and obtain the gure 2.2-4(b). The work done to achieve this position is W1 = (m1 + m2 )gh1 = (m1 + m2 )gL1 (1 cos 1 ). Starting from the position in gure 2.2-4(b) we now let 2 undergo a displacement and achieve the conguration in the gure 2.2-4(c). Find the equations of motion for the compound pendulum Figure 2.2-4 Compound pendulum The work done due to the displacement 2 can be represented W2 = m2 gh2 = m2 gL2 (1 cos 2 ). Since the potential energy V satises V = W to within an additive constant, we can write V = W = W1 W2 = (m1 + m2 )gL1 cos 1 m2 gL2 cos 2 + constant, where the constant term in the potential energy has been neglected since it does not contribute anything to the equations of motion. (i.e. the derivative of a constant is zero.) The kinetic energy term for this system can be represented 1 ds1 ds2 1 + m2 m1 2 dt 2 dt 1 1 T = m1 (x2 + y1 ) + m2 (x2 + y2 ), 1 2 2 2 2 2 T= 2 2 (2.2.27) 196 where (x1 , y1 ) = (L1 sin 1 , L1 cos 1 ) (x2 , y2 ) = (L1 sin 1 + L2 sin 2 , L1 cos 1 L2 cos 2 ) (2.2.27) and simplifying produces the kinetic energy expression T= 1 1 2 2 (m1 + m2 )L2 1 + m2 L1 L2 1 2 cos(1 2 ) + m2 L2 2 . 1 2 2 2 (2.2.29) (2.2.28) are the coordinates of the masses m1 and m2 respectively. Substituting the equations (2.2.28) into equation Writing the Lagrangian as L = T V , the equations describing the motion of the compound pendulum are obtained from the Lagrangian equations d dt L 1 L =0 1 and d dt L 2 L = 0. 2 Calculating the necessary derivatives, substituting them into the Lagrangian equations of motion and then simplifying we derive the equations of motion L 1 1 + m2 m2 L2 2 cos(1 2 ) + L2 (2 )2 sin(1 2 ) + g sin 1 = 0 m1 + m2 m1 + m2 L1 1 cos(1 2 ) + L2 2 L1 (1 )2 sin(1 2 ) + g sin 2 = 0. These equations are a set of coupled, second order nonlinear ordinary dierential equations. These equations are subject to initial conditions being imposed upon the angular displacements (1 , 2 ) and the angular velocities (1 , 2 ). Alternative Derivation of Lagranges Equations of Motion Let c denote a given curve represented in the parametric form xi = xi (t), i = 1, . . . , N, t0 t t 1 and let P0 , P1 denote two points on this curve corresponding to the parameter values t0 and t1 respectively. Let c denote another curve which also passes through the two points P0 and P1 as illustrated in the gure 2.2-5. The curve c is represented in the parametric form xi = xi (t) = xi (t) + i (t), i = 1, . . . , N, t0 t t1 in terms of a parameter . In this representation the function i (t) must satisfy the end conditions i (t0 ) = 0 and i (t1 ) = 0 i = 1, . . . , N since the curve c is assumed to pass through the end points P0 and P1 . Consider the line integral t1 I( ) = t0 L(t, xi + i , xi + i ) dt, (2.2.30) 197 Figure 2.2-5. Motion along curves c and c where i L = T V = L(t, xi , x ) is the Lagrangian evaluated along the curve c. We ask the question, What conditions must be satised by the curve c in order that the integral I( ) have an extremum value when a minimum value when we will have dI( ) d Employing the denition dI d = lim =0 0 is zero?If the integral I( ) has will be zero at this value and is zero it follows that its derivative with respect to = 0. =0 I( ) I(0) = I (0) = 0 we expand the Lagrangian in equation (2.2.30) in a series about the point = 0. Substituting the expansion L(t, xi + i , xi + i ) = L(t, xi , xi ) + into equation (2.2.30) we calculate the derivative I (0) = lim I( ) I(0) t1 L i L + i i + i x x 2 [ ] + 0 = lim 0 t0 L i L (t) + i i (t) dt + [ ] + = 0, i x x since is approaching zero. Analysis of this equation where we have neglected higher order powers of informs us that the integral I has a minimum value at t1 = 0 provided that the integral (2.2.31) I = t0 L i L (t) + i i (t) dt = 0 i x x 198 is satised. Integrating the second term of this integral by parts we nd t1 I = t0 L i L i dt + (t) i x xi t1 t0 t1 t0 d dt L xi i (t) dt = 0. (2.2.32) The end condition on i (t) makes the middle term in equation (2.2.32) vanish and we are left with the integral t1 I = t0 i (t) L d xi dt L xi dt = 0, (2.2.33) which must equal zero for all i (t). Since i (t) is arbitrary, the only way the integral in equation (2.2.33) can be zero for all i (t) is for the term inside the brackets to vanish. This produces the result that the integral of the Lagrangian is an extremum when the Euler-Lagrange equations d dt L xi L = 0, xi i = 1, . . . , N (2.2.34) are satised. This is a necessary condition for the integral I( ) to have a minimum value. In general, any line integral of the form t1 I= t0 (t, xi , xi ) dt (2.2.35) has an extremum value if the curve c dened by xi = xi (t), i = 1, . . . , N satises the Euler-Lagrange equations d dt xi = 0, xi i = 1, . . . , N. (2.2.36) The above derivation is a special case of (2.2.36) when = L. Note that the equations of motion equations (2.2.34) are just another form of the equations (2.2.24). Note also that T = t t 1 mgij v i v j 2 = mgij v i f j = mfi v i = mfi x i V and if we assume that the force Qi is derivable from a potential function V , then mfi = Qi = i , so x V i V T i i = mfi x = Qi x = i x = or (T + V ) = 0 or T + V = h = constant called the energy that t x t t constant of the system. Action Integral The equations of motion (2.2.34) or (2.2.24) are interpreted as describing geodesics in a space whose line-element is ds2 = 2m(h V )gjk dxj dxk where V is the potential function for the force system and T + V = h is the energy constant of the motion. The integral of ds along a curve C between two points P1 and P2 is called an action integral and is A = 2m P2 P1 dxj dxk (h V )gjk d d 1/2 d 199 where is a parameter used to describe the curve C. The principle of stationary action states that of all curves through the points P1 and P2 the one which makes the action an extremum is the curve specied by Newtons second law. The extremum is usually a minimum. To show this let = 2m (h V )gjk dxk d dxj dxk d d 1/2 in equation (2.2.36). Using the notation x k = we nd that 2m (h V )gik x k = i x gjk j k 2m V 2m (h V ) = xx gjk x j x k . xi 2 xi 2 xi The equation (2.2.36) which describe the extremum trajectories are found to be gjk j k 2m V d 2m 2m (h V )gik x k (h V ) xx + gjk x j x k = 0. dt 2 xi xi By changing variables from to t where satisfy the equation m d dt gik dxk dt V m gjk dxj dxk + i =0 i dt dt 2 x x dt d = m 2(hV ) we nd that the trajectory for an extremum must which are the same equations as (2.2.24). (i.e. See also the equations (2.2.22).) Dynamics of Rigid Body Motion Let us derive the equations of motion of a rigid body which is rotating due to external forces acting upon it. We neglect any translational motion of the body since this type of motion can be discerned using our knowledge of particle dynamics. The derivation of the equations of motion is restricted to Cartesian tensors and rotational motion. Consider a system of N particles rotating with angular velocity i , i = 1, 2, 3, about a line L through the center of mass of the system. Let V () denote the velocity of the th particle which has mass m() and position xi , i = 1, 2, 3 with respect to an origin on the line L. Without loss of generality we can assume that the origin of the coordinate system is also at the center of mass of the system of particles, as this choice of an origin simplies the derivation. The velocity components for each particle is obtained by taking cross products and we can write V () = r () () or Vi () = eijk j xk . () (2.2.37) The kinetic energy of the system of particles is written as the sum of the kinetic energies of each individual particle and is 1 1 () () () T= m() Vi Vi = m() eijk j xk eimn m x() . n 2 =1 2 =1 N N (2.2.38) 200 Employing the e identity the equation (2.2.38) can be simplied to the form T= 1 () () () m() m m xk xk n k xk x() . n 2 =1 N Dene the second moments and products of inertia by the equation N Iij = =1 m() xk xk ij xi xj () () () () (2.2.39) and write the kinetic energy in the form T= 1 Iij i j . 2 (2.2.40) Similarly, the angular momentum of the system of particles can also be represented in terms of the second moments and products of inertia. The angular momentum of a system of particles is dened as a summation of the moments of the linear momentum of each individual particle and is N N Hi = =1 m() eijk xj vk () () = =1 m() eijk xj ekmn m x() . n () (2.2.41) The e identity simplies the equation (2.2.41) to the form N Hi = j =1 m() x() x() ij xj xi n n () () = j Iji . (2.2.42) The equations of motion of a rigid body is obtained by applying Newtons second law of motion to the system of N particles. The equation of motion of the th particle is written m() xi () = Fi () . (2.2.43) Summing equation (2.2.43) over all particles gives the result N N m() xi =1 () = =1 Fi () . (2.2.44) This represents the translational equations of motion of the rigid body. The equation (2.2.44) represents the rate of change of linear momentum being equal to the total external force acting upon the system. Taking the cross product of equation (2.2.43) with the position vector xj () () produces m() xt erst x() = erst x() Ft s s and summing over all particles we nd the equation N () m() erst x() xt s =1 N () = =1 erst x() Ft s () . (2.2.45) 201 The equations (2.2.44) and (2.2.45) represent the conservation of linear and angular momentum and can be written in the forms d dt and d dt By denition we have Gr = to the origin, and Mr = N () m() erst x() xt s =1 () N N N m() x() r =1 = =1 () Fr (2.2.46) = =1 erst x() Ft s () . Fr () (2.2.47) the total force m() xr representing the linear momentum, Fr = () () m() erst xs xt acting on the system of particles, Hr = () () erst xs Ft is the angular momentum of the system relative is the total moment of the system relative to the origin. We can therefore express the equations (2.2.46) and (2.2.47) in the form dGr = Fr dt and dHr = Mr . dt (2.2.49) (2.2.48) The equation (2.2.49) expresses the fact that the rate of change of angular momentum is equal to the moment of the external forces about the origin. These equations show that the motion of a system of particles can be studied by considering the motion of the center of mass of the system (translational motion) and simultaneously considering the motion of points about the center of mass (rotational motion). We now develop some relations in order to express the equations (2.2.49) in an alternate form. Toward this purpose we consider rst the concepts of relative motion and angular velocity. Relative Motion and Angular Velocity Consider two dierent reference frames denoted by S and S. Both reference frames are Cartesian coordinates with axes xi and xi , i = 1, 2, 3, respectively. The reference frame S is xed in space and is called an inertial reference frame or space-xed reference system of axes. The reference frame S is xed to and rotates with the rigid body and is called a body-xed system of axes. Again, for convenience, it is assumed that the origins of both reference systems are xed at the center of mass of the rigid body. Further, we let the system S have the basis vectors ei , i = 1, 2, 3, while the reference system S has the basis vectors ei , i = 1, 2, 3. The transformation equations between the two sets of reference axes are the ane transformations xi = where ij ji xj and xi = ij xj ij (2.2.50) are the cosines of the = ij (t) are direction cosines which are functions of time t (i.e. the angles between the barred and unbarred axes where the barred axes are rotating relative to the space-xed unbarred axes.) The direction cosines satisfy the relations ij ik = jk and ij kj = ik . (2.2.51) 202 EXAMPLE 2.2-5. (Euler angles , , ) Consider the following sequence of transformations which are used in celestial mechanics. First a rotation about the x3 axis taking the xi axes to the yi axes y1 cos sin 0 x1 y2 = sin cos 0 x2 0 0 1 y3 x3 where the rotation angle is called the longitude of the ascending node. Second, a rotation about the y1 axis taking the yi axes to the yi axes 1 0 0 y1 y1 y2 = 0 cos sin y2 y3 y3 0 sin cos where the rotation angle is called the angle of inclination of the orbital plane. Finally, a rotation about the y3 axis taking the yi axes to the xi axes cos sin 0 y1 x1 x2 = sin cos 0 y2 0 0 1 x3 y3 where the rotation angle is called the argument of perigee. The Euler angle is the angle x3 0x3 , the angle x is the angle x1 0y1 and is the angle y1 01 . These angles are illustrated in the gure 2.2-6. Note also that the rotation vectors associated with these transformations are vectors of magnitude , , in the directions indicated in the gure 2.2-6. Figure 2.2-6. Euler angles. By combining the above transformations there results the transformation equations (2.2.50) cos cos cos sin sin cos sin + cos cos sin sin sin x1 x1 x2 = sin cos cos sin cos sin sin + cos cos cos cos sin x2 . sin sin sin cos cos x3 x3 It is left as an exercise to verify that the transformation matrix is orthogonal and the components satisfy the relations (2.2.51). ji 203 Consider the velocity of a point which is rotating with the rigid body. Denote by vi = vi (S), for i = 1, 2, 3, the velocity components relative to the S reference frame and by v i = v i (S), i = 1, 2, 3 the velocity components of the same point relative to the body-xed axes. In terms of the basis vectors we can write V = v1 (S) e1 + v2 (S) e2 + v3 (S) e3 = as the velocity in the S reference frame. Similarly, we write V = v 1 (S)e1 + v 2 (S)e2 + v 3 (S)e3 = dxi ei dt (2.2.53) dxi ei dt (2.2.52) as the velocity components relative to the body-xed reference frame. There are occasions when it is desirable to represent V in the S frame of reference and V in the S frame of reference. In these instances we can write V = v1 (S)e1 + v2 (S)e2 + v3 (S)e3 and V = v 1 (S) e1 + v 2 (S) e2 + v 3 (S) e3 . (2.2.55) (2.2.54) Here we have adopted the notation that vi (S) are the velocity components relative to the S reference frame and vi (S) are the same velocity components relative to the S reference frame. Similarly, v i (S) denotes the velocity components relative to the S reference frame, while v i (S) denotes the same velocity components relative to the S reference frame. Here both V and V are vectors and so their components are rst order tensors and satisfy the transformation laws v i (S) = ji vj (S) = ji xj and vi (S) = ij v j (S) = ij xj . (2.2.56) The equations (2.2.56) dene the relative velocity components as functions of time t. By dierentiating the equations (2.2.50) we obtain dxi = v i (S) = dt and dxi = vi (S) = dt Multiply the equation (2.2.57) by mi ji xj + ji xj (2.2.57) ij xj + ij xj . im (2.2.58) and derive the relations (2.2.59) and multiply the equation (2.2.58) by vm (S) = vm (S) + mi ji xj and v m (S) = v m (S) + im ij xj . (2.2.60) The equations (2.2.59) and (2.2.60) describe the transformation laws of the velocity components upon changing from the S to the S reference frame. These equations can be expressed in terms of the angular velocity by making certain substitutions which are now dened. The rst order angular velocity vector i is related to the second order skew-symmetric angular velocity tensor ij by the dening equation mn = eimn i . (2.2.61) 204 The equation (2.2.61) implies that i and ij are dual tensors and i = 1 eijk jk . 2 Also the velocity of a point which is rotating about the origin relative to the S frame of reference is vi (S) = eijk j xk which can also be written in the form vm (S) = mk xk . Since the barred axes rotate with the rigid body, then a particle in the barred reference frame will have vm (S) = 0, since the coordinates of a point in the rigid body will be constants with respect to this reference frame. Consequently, we write equation (2.2.59) in the form 0 = vm (S) + mi ji xj which implies that vm (S) = mi ji xj = mk xk or mj = mj (S, S) = mi ji . This equation is interpreted as describing the angular velocity tensor of S relative to S. Since ij is a tensor, it can be represented in the barred system by mn (S, S) = = im jn ij (S, S) im jn is js jn js = ms = (2.2.62) jn jm By dierentiating the equations (2.2.51) it is an easy exercise to show that ij is skew-symmetric. The second order angular velocity tensor can be used to write the equations (2.2.59) and (2.2.60) in the forms vm (S) = vm (S) + mj (S, S)xj v m (S) = v m (S) + jm (S, S)xj The above relations are now employed to derive the celebrated Eulers equations of motion of a rigid body. Eulers Equations of Motion We desire to nd the equations of motion of a rigid body which is subjected to external forces. These equations are the formulas (2.2.49), and we now proceed to write these equations in a slightly dierent form. Similar to the introduction of the angular velocity tensor, given in equation (2.2.61), we now introduce the following tensors 1. The fourth order moment of inertia tensor Imnst which is related to the second order moment of inertia tensor Iij by the equations Imnst = 1 ejmn eist Iij 2 or Iij = 1 Ipqrs eipq ejrs 2 (2.2.64) (2.2.63) 2. The second order angular momentum tensor Hjk which is related to the angular momentum vector Hi by the equation Hi = 1 eijk Hjk 2 or Hjk = eijk Hi (2.2.65) 3. The second order moment tensor Mjk which is related to the moment Mi by the relation Mi = 1 eijk Mjk 2 or Mjk = eijk Mi . (2.2.66) 205 Now if we multiply equation (2.2.49) by erjk , then it can be written in the form dHij = Mij . (2.2.67) dt Similarly, if we multiply the equation (2.2.42) by eimn , then it can be expressed in the alternate form Hmn = eimn j Iji = Imnst st and because of this relation the equation (2.2.67) can be expressed as d (Iijst st ) = Mij . (2.2.68) dt We write this equation in the barred system of coordinates where I pqrs will be a constant and consequently its derivative will be zero. We employ the transformation equations Iijst = ij = M pq = and then multiply the equation (2.2.68) by ip jq ip jq ip jq sr tk I pqrk si tj st ip jq Mij and simplify to obtain d i j I rk rk = M pq . dt Expand all terms in this equation and take note that the derivative of the I rk is zero. The expanded equation then simplies to d rk + (u pv q + p u qv ) I rk rk uv = M pq . (2.2.69) dt Substitute into equation (2.2.69) the relations from equations (2.2.61),(2.2.64) and (2.2.66), and then multiply I pqrk by empq and simplify to obtain the Eulers equations of motion d i etmj I ij i t = M m . (2.2.70) dt Dropping the bar notation and performing the indicated summations over the range 1,2,3 we nd the I im Euler equations have the form d1 d2 d3 + I21 + I31 dt dt dt d1 d2 d3 + I22 + I32 I12 dt dt dt d1 d2 d3 + I23 + I33 I13 dt dt dt In the special case where I11 + (I13 1 + I23 2 + I33 3 ) 2 (I12 1 + I22 2 + I32 3 ) 3 = M1 + (I11 1 + I21 2 + I31 3 ) 3 (I13 1 + I23 2 + I33 3 ) 1 = M2 + (I12 1 + I22 2 + I32 3 ) 1 (I11 1 + I21 2 + I31 3 ) 2 = M3 . the barred axes are principal axes, then Iij = 0 for i = j and the Eulers (2.2.71) equations reduces to the system of nonlinear dierential equations d1 + (I33 I22 )2 3 = M1 dt d2 (2.2.72) + (I11 I33 )3 1 = M2 I22 dt d3 + (I22 I11 )1 2 = M3 . I33 dt In the case of constant coecients and constant moments the solutions of the above dierential equations I11 can be expressed in terms of Jacobi elliptic functions. 206 EXERCISE 2.2 1. Find a set of parametric equations for the straight line which passes through the points P1 (1, 1, 1) and P2 (2, 3, 4). Find the unit tangent vector to any point on this line. 2. Consider the space curve x = 1 2 sin2 t, y = 1 t 1 sin 2t, z = sin t where t is a parameter. Find the unit 2 4 vectors T i , B i , N i , i = 1, 2, 3 at the point where t = . 3. A claim has been made that the space curve x = t, y = t2 , z = t3 intersects the plane 11x-6y+z=6 in three distinct points. Determine if this claim is true or false. Justify your answer and nd the three points of intersection if they exist. 4. Find a set of parametric equations xi = xi (s1 , s2 ), i = 1, 2, 3 for the plane which passes through the points P1 (3, 0, 0), P2 (0, 4, 0) and P3 (0, 0, 5). Find a unit normal to this plane. 2 t nd the equation of the tangent plane to the curve at the point where t = /4. Find the equation of the tangent line to the curve at the point where t = /4. 5. For the helix x = sin t y = cos t z = 6. Verify the derivative T = M grm xm . xr d dt T xr = M grm xm + grm n m xx . xn 7. Verify the derivative 8. Verify the derivative T 1 gmn m n =M x x. xr 2 xr 9. Use the results from problems 6,7 and 8 to derive the Lagranges form for the equations of motion dened by equation (2.2.23). 10. Express the generalized velocity and acceleration in cylindrical coordinates (x1 , x2 , x3 ) = (r, , z) and show dr dx1 = dt dt d dx2 2 = V= dt dt 3 dz dx = V3 = dt dt V1 = f1 = d d2 r V 1 = 2 r t dt dt d2 2 dr d V 2 = 2+ f2 = t dt r dt dt d2 z V 3 3 =2 f= t dt Find the physical components of velocity and acceleration in cylindrical coordinates and show dr dt d V =r dt dz Vz = dt Vr = d d2 r r dt2 dt 2 d dr d f =r 2 + 2 dt dt dt d2 z fz = 2 dt fr = 2 2 207 11. Express the generalized velocity and acceleration in spherical coordinates (x2 , x2 , x3 ) = (, , ) and show d dx1 = V= dt dt d dx2 = V2 = dt dt d dx3 = V3 = dt dt 1 f1 = f2 = d2 V 1 = 2 t dt d dt 2 sin2 2 d dt 2 d d2 V 2 2 d d = 2 sin cos + t dt dt dt dt d2 2 d d d d V 3 = 2+ + 2 cot f3 = t dt dt dt dt dt Find the physical components of velocity and acceleration in spherical coordinates and show d V = dt d V = dt V = sin d dt f = d2 dt2 d dt 2 sin2 2 d dt 2 d2 d d d sin cos +2 2 dt dt dt dt 2 d d d d d + 2 cos f = sin 2 + 2 sin dt dt dt dt dt f = 12. Expand equation (2.2.39) and write out all the components of the moment of inertia tensor Iij . 13. For the density of a continuous material and d an element of volume inside a region R where the material is situated, we write d as an element of mass inside R. Find an equation which describes the center of mass of the region R. 14. Use the equation (2.2.68) to derive the equation (2.2.69). 15. Drop the bar notation and expand the equation (2.2.70) and derive the equations (2.2.71). 16. Verify the Euler transformation, given in example 2.2-5, is orthogonal. 17. For the pulley and mass system illustrated in the gure 2.2-7 let a = the radius of each pulley. 1 2 = the length of the upper chord. = the length of the lower chord. Neglect the weight of the pulley and nd the equations of motion for the pulley mass system. 208 Figure 2.2-7. Pulley and mass system 18. Let = ds dt , where s is the arc length between two points on a curve in generalized coordinates. gmn xm xn dt and show the integral I, dened by (a) Write the arc length in general coordinates as ds = equation (2.2.35), represents the distance between two points on a curve. (b) Using the Euler-Lagrange equations (2.2.36) show that the shortest distance between two points in a d2 s i 2 xj xk = xi dt ds generalized space is the curve dened by the equations: xi + jk dt i dxj dxk d2 xi = 0, for (c) Show in the special case t = s the equations in part (b) reduce to + 2 j k ds ds ds i = 1, . . . , N. An examination of equation (1.5.51) shows that the above curves are geodesic curves. (d) Show that the shortest distance between two points in a plane is a straight line. (e) Consider two points on the surface of a cylinder of radius a. Let u1 = and u2 = z denote surface coordinates in the two dimensional space dened by the surface of the cylinder. Show that the shortest distance between the points where = 0, z = 0 and = , z = H is L = a2 2 + H 2 . 19. For T = 1 mgij v i v j the kinetic energy of a particle and V the potential energy of the particle show 2 that T + V = constant. Hint: V mfi = Qi = xi , i = 1, 2, 3 and dxi dt = xi = v i , i = 1, 2, 3. 20. Dene H = T + V as the sum of the kinetic energy and potential energy of a particle. The quantity H = H(xr , pr ) is called the Hamiltonian of the particle and it is expressed in terms of: the particle position xi and the particle momentum pi = mvi = mgij xj . Here xr and pr are treated as independent variables. (a) Show that the particle momentum is a covariant tensor of rank 1. (b) Express the kinetic energy T in terms of the particle momentum. T . (c) Show that pi = xi 209 Figure 2.2-8. Compound pendulum (d) Show that dpi dxi H H and = = i . These are a set of dierential equations describing the dt pi dt x position change and momentum change of the particle and are known as Hamiltons equations of motion for a particle. 21. Bi = Let ijk T i s = N i and B i s N i s = B i T i and calculate the intrinsic derivative of the cross product Tj Nk and nd in terms of the unit normal vector. 22. For T the kinetic energy of a particle and V the potential energy of a particle, dene the Lagrangian 1 L = L(xi , xi ) = T V = M gij xi xj V as a function of the independent variables xi , xi . Dene the 2 1 ij g pi pj + V, as a function of the independent variables xi , pi , Hamiltonian H = H(xi , pi ) = T + V = 2M where pi is the momentum vector of the particle and M is the mass of the particle. T (a) Show that pi = . xi L H = i (b) Show that i x x 23. When the Euler angles, gure 2.2-6, are applied to the motion of rotating objects, is the angle of nutation, is the angle of precession and is the angle of spin. Take projections and show that the time derivative of the Euler angles are related to the angular velocity vector components x , y , z by the relations x = cos + sin sin y = sin + sin cos z = + cos where x , y , z are the angular velocity components along the x1 , x2 , x3 axes. 24. Find the equations of motion for the compound pendulum illustrated in the gure 2.2-8. 210 GM m r denote the inverse square law force of attraction between the earth and sun, with r3 G a universal constant, M the mass of the sun, m the mass of the earth and r a unit vector from origin r 25. Let F = at the center of the sun pointing toward the earth. (a) Write down Newtons second law, in both vector d (r v) = 0 and and tensor form, which describes the motion of the earth about the sun. (b) Show that dt consequently r v = r dr = h = a constant. dt 26. Construct a set of axes xed and attached to an airplane. Let the x axis be a longitudinal axis running from the rear to the front of the plane along its center line. Let the y axis run between the wing tips and let the z axis form a right-handed system of coordinates. The y axis is called a lateral axis and the z axis is called a normal axis. Dene pitch as any angular motion about the lateral axis. Dene roll as any angular motion about the longitudinal axis. Dene yaw as any angular motion about the normal axis. Consider two sets of axes. One set is the x, y, z axes attached to and moving with the aircraft. The other set of axes is denoted X, Y, Z and is xed in space ( an inertial set of axes). Describe the pitch, roll and yaw of an aircraft with respect to the inertial set of axes. Show the transformation is orthogonal. Hint: Consider pitch with respect to the xed axes, then consider roll with respect to the pitch axes and nally consider yaw with respect to the roll axes. This produces three separate transformation matrices which can then be combined to describe the motions of pitch, roll and yaw of an aircraft. 27. In Cartesian coordinates let Fi = Fi (x1 , x2 , x3 ) denote a force eld and let xi = xi (t) denote a curve 2 dxi d1 dxi C. (a) Show Newtons second law implies that along the curve C m = Fi (x1 , x2 , x3 ) dt 2 dt dt (no summation on i) and hence d dt 1 m 2 dx1 dt 2 + dx2 dt 2 + dx3 dt 2 = dx1 dx2 dx3 d1 mv 2 = F1 + F2 + F3 dt 2 dt dt dt (b) Consider two points on the curve C, say point A, xi (tA ) and point B, xi (tB ) and show that the work done in moving from A to B in the force eld Fi is 1 mv 2 2 tB B = tA A Fi dx1 + F2 dx2 + F3 dx3 where the right hand side is a line integral along the path C from A to B. (c) Show that if the force eld is derivable from a potential function U (x1 , x2 , x3 ) by taking the gradient, then the work done is independent of the path C and depends only upon the end points A and B. 28. Find the Lagrangian equations of motion of a spherical pendulum which consists of a bob of mass m suspended at the end of a wire of length , which is free to swing in any direction subject to the constraint that the wire length is constant. Neglect the weight of the wire and show that for the wire attached to the origin of a right handed x, y, z coordinate system, with the z axis downward, the angle between the wire and the z axis and the angle of rotation of the bob from the y axis, that there results the equations of 2 d d d g d2 motion sin cos + sin = 0 sin2 =0 and dt dt dt2 dt 211 29. In Cartesian coordinates show the Frenet formulas can be written dT = T, ds dN = N, ds dB =B ds where is the Darboux vector and is dened = T + B. 30. Consider the following two cases for rigid body rotation. Case 1: Rigid body rotation about a xed line which is called the xed axis of rotation. Select a point 0 on this xed axis and denote by e a unit vector from 0 in the direction of the xed line and denote by eR a unit vector which is perpendicular to the xed axis of rotation. The position vector of a general point in the rigid body can then be represented by a position vector from the point 0 given by r = h e + r0 eR where h, r0 and e are all constants and the vector eR is xed in and rotating with the rigid body. d the scalar angular change with respect to time of the vector eR as it rotates about Denote by = dt d d e where e is dened as the the xed line and dene the vector angular velocity as = ( e) = dt dt vector angle of rotation. d eR = e eR . Show that d dr d eR d eR d Show that V = = r0 = r0 = (r0 eR ) = (h e + r0 eR ) = r. dt dt d dt Case 2: Rigid body rotation about a xed point 0. Construct at point 0 the unit vector e1 which is d e1 xed in and rotating with the rigid body. From pages 80,87 we know that must be perpendicular dt d e1 to e1 and so we can dene the vector e2 as a unit vector which is in the direction of such that dt d e1 = e2 for some constant . We can then dene the unit vector e3 from e3 = e1 e2 . dt d e3 , which must be perpendicular to e3 , is also perpendicular to e1 . Show that dt d e3 d e3 Show that can be written as = e2 for some constant . dt dt d e2 = ( e3 e1 ) e2 From e2 = e3 e1 show that dt d e1 d e2 d e3 = e1 , = e2 , = e3 Dene = e3 e1 and show that dt dt dt Let r = x e1 + y e2 + z e3 denote an arbitrary point within the rigid body with respect to the point 0. dr Show that = r. dt Note that in Case 2 the direction of is not xed as the unit vectors e3 and e1 are constantly changing. In this case the direction is called an instantaneous axis of rotation and , which also can change in magnitude and direction, is called the instantaneous angular velocity. (a) (b) (a) (b) (c) (d) (e) 211 2.3 BASIC EQUATIONS OF CONTINUUM MECHANICS Continuum mechanics is the study of how materials behave when subjected to external inuences. External inuences which aect the properties of a substance are such things as forces, temperature, chemical reactions, and electric phenomena. Examples of forces are gravitational forces, electromagnetic forces, and mechanical forces. Solids deform under external forces and so deformations are studied. Fluids move under external forces and so the velocity of the uid is studied. A material is considered to be a continuous media which is a collection of material points interconnected by internal forces (forces between the atoms making up the material). We concentrate upon the macroscopic properties rather than the microscopic properties of the material. We treat the material as a body which is homogeneous and continuous in its makeup. In this introduction we will only consider solid media and liquid media. In general, most of the ideas and concepts developed in this section can be applied to any type of material which is assumed to be a collection of material points held together by some kind of internal forces. An elastic material is one which deforms under applied forces in such a way that it will return to its original unloaded state when the applied forces are removed. When a linear relation exists between the applied forces and material displacements, then the material is called a linear elastic material. In contrast, a plastic material is one which deforms under applied forces in such a way that it does not return to its original state after removal of the applied forces. Plastic materials will always exhibit some permanent deformation after removal of the applied forces. An elastic material is called homogeneous if it has the same properties throughout. An isotropic material has the same properties, at a point, in all directions about the point. In this introduction we develop the basic mathematical equations which describe how a continuum behaves when subjected to external forces. We shall discover that there exists a set of basic equations associated with all continuous material media. These basic equations are developed for linear elastic materials and applied to solids and uids in later sections. Introduction to Elasticity Take a rubber band, which has a rectangular cross section, and mark on it a parallelepiped having a length , a width w and a height h, as illustrated in the gure 2.3-1. Now apply a force F to both ends of the parallelepiped cross section on the rubber band and examine what happens to the parallelepiped. You will see that: 1. 2. 3. increases by an amount . w decreases by an amount w. h decreases by an amount h. There are many materials which behave in a manner very similar to the rubber band. Most materials, when subjected to tension forces will break if the change is only one or two percent of the original length. The above example introduces us to several concepts which arise in the study of materials when they are subjected to external forces. The rst concept is that of strain which is dened as strain = change in length , original length (dimensionless). 212 Figure 2.3-1. Section of a rubber band When the force F is applied to our rubber band example there arises the strains , w , w h . h The second concept introduced by our simple example is stress. Stress is dened as a force per unit area. In particular, Force force , with dimension of . Area over which force acts unit area We will be interested in studying stress and strain in homogeneous, isotropic materials which are in equilibstress = rium with respect to the force system acting on the material. Hookes Law For linear elastic materials, where the forces are all one dimensional, the stress and strains are related by Hookes law which has two parts. The Hookes law, part one, states that stress is proportional to strain in the stretch direction, where the Youngs modulus E is the proportionality constant. This is written Hookes law part 1 F =E A . (2.3.1) A graph of stress vs strain is a straight line with slope E in the linear elastic range of the material. The Hookes law, part two, involves the fact that there is a strain contraction perpendicular to the stretch direction. The strain contraction is the same for both the width and height and is proportional to the strain in the stretch direction. The proportionality constant being the Poissons ratio . Hookes law part 2 h w = = , w h 0<< 1 . 2 (2.3.2) The proportionality constants E and depend upon the material being considered. The constant is called the Poissons ratio and it is always a positive number which is less than one half. Some representative values for E and are as follows. Various types of steel Various types of aluminium 28 (10)6 psi E 30 (10)6 psi 9.0 (10)6 psi E 11.0 (10)6 psi 0.26 0.31 0.3 0.35 213 Figure 2.3-2. Typical Stress-strain curve. Consider a typical stress-strain curve, such as the one illustrated in the gure 2.3-2, which is obtained by placing a material in the shape of a rod or wire in a machine capable of performing tensile straining at a low rate. The engineering stress is the tensile force F divided by the original cross sectional area A0 . Note that during a tensile straining the cross sectional area A of the sample is continually changing and getting smaller so that the actual stress will be larger than the engineering stress. Observe in the gure 2.3-2 that the stress-strain relation remains linear up to a point labeled the proportional limit. For stress-strain points in this linear region the Hookes law holds and the material will return to its original shape when the loading is removed. For points beyond the proportional limit, but less than the yield point, the material no longer obeys Hookes law. In this nonlinear region the material still returns to its original shape when the loading is removed. The region beyond the yield point is called the plastic region. At the yield point and beyond, there is a great deal of material deformation while the loading undergoes only small changes. For points in this plastic region, the material undergoes a permanent deformation and does not return to its original shape when the loading is removed. In the plastic region there usually occurs deformation due to slipping of atomic planes within the material. In this introductory section we will restrict our discussions of material stress-strain properties to the linear region. EXAMPLE 2.3-1. (One dimensional elasticity) Consider a circular rod with cross sectional area A which is subjected to an external force F applied to both ends. The gure 2.3-3 illustrates what happens to the rod after the tension force F is applied. Consider two neighboring points P and Q on the rod, where P is at the point x and Q is at the point x + x. When the force F is applied to the rod it is stretched and P moves to P and Q moves to Q . We assume that when F is applied to the rod there is a displacement function u = u(x, t) which describes how each point in the rod moves as a function of time t. If we know the displacement function u = u(x, t) we would then be able to calculate the following distances in terms of the displacement function P P = u(x, t), 0P = x + u(x, t), QQ = u(x + x, t) 0Q = x + x + u(x + x, t). 214 Figure 2.3-3. One dimensional rod subjected to tension force The strain associated with the distance e= = = x = P Q is P Q PQ (0Q 0P ) (0Q 0P ) = PQ PQ [x + x + u(x + x, t) (x + u(x, t))] [(x + x) x] e= x u(x + x, t) u(x, t) e= . x Use the Hookes law part(i) and write F u(x + x, t) u(x, t) =E . A x Taking the limit as x approaches zero we nd that F u(x, t) =E . A x Hence, the stress is proportional to the spatial derivative of the displacement function. Normal and Shearing Stresses Let us consider a more general situation in which we have some material which can be described as having a surface area S which encloses a volume V. Assume that the density of the material is and the material is homogeneous and isotropic. Further assume that the material is subjected to the forces b and t (n) where b is a body force per unit mass [f orce/mass], and t (n) is a surface traction per unit area [f orce/area]. The superscript (n) on the vector is to remind you that we will only be interested in the normal component of the surface forces. We will neglect body couples, surface couples, and concentrated forces or couples that act at a single point. If the forces described above are everywhere continuous we can calculate the resultant force F and resultant moment M acting on the material by constructing various surface and volume integrals which sum the forces acting upon the material. In particular, the resultant force F acting on our material can be described by the surface and volume integrals: F= S t (n) dS + V b d (2.3.3) 215 Figure 2.3-4. Stress vectors acting upon an element of volume which is a summation of all the body forces and surface tractions acting upon our material. Here density of the material, dS is an element of surface area, and d is an element of volume. The resultant moment M about the origin is similarly expressed as M= S is the r t (n) dS + V (r b) d. (2.3.4) The global motion of the material is governed by the Euler equations of motion. The time rate of change of linear momentum equals the resultant force or d dt v d = F = V S t (n) dS + V b d. (2.3.5) This is a statement concerning the conservation of linear momentum. The time rate of change of angular momentum equals the resultant moment or d dt r v d = M = V S r t (n) dS + V (r b) d. (2.3.6) This is a statement concerning conservation of angular momentum. The Stress Tensor Dene the stress vectors t 1 = 11 e1 + 12 e2 + 13 e3 t 2 = 21 e1 + 22 e2 + 23 e3 t 3 = 31 e1 + 32 e2 + 33 e3 , where ij , i, j = 1, 2, 3 is the stress tensor acting at each point of the material. The index i indicates the coordinate surface xi = a constant, upon which t i acts. The second index j denotes the direction associated with the components of t i . (2.3.7) 216 Figure 2.3-5. Stress distribution at a point For i = 1, 2, 3 we adopt the convention of sketching the components of t i in the positive directions if the exterior normal to the surface xi = constant also points in the positive direction. This gives rise to the gure 2.3-4 which illustrates the stress vectors acting upon an element of volume in rectangular Cartesian coordinates. The components 11 , 22 , 33 are called normal stresses while the components ij , i = j are called shearing stresses. The equations (2.3.7) can be written in the more compact form using the indicial notation as t i = ij ej , i, j = 1, 2, 3. (2.3.8) If we know the stress distribution at three orthogonal interfaces at a point P in a solid body, we can then determine the stress at the point P with respect to any plane passing through the point P. With reference to the gure 2.3-5, consider an arbitrary plane passing through the point P which lies within the material body being considered. Construct the elemental tetrahedron with orthogonal axes parallel to the x1 = x, x2 = y and x3 = z axes. In this gure we have the following surface tractions: t 1 t 2 t 3 t (n) on the surface 0BC on the surface 0AC on the surface 0AB on the surface ABC The superscript parenthesis n is to remind you that this surface traction depends upon the orientation of the plane ABC which is determined by a unit normal vector having the direction cosines n1 , n2 and n3 . 217 Let S1 = the surface area 0BC S2 = the surface area 0AC S3 = the surface area 0AB S = the surface area ABC . These surface areas are related by the relations S1 = n1 S, S2 = n2 S, S3 = n3 S (2.3.9) which can be thought of as projections of S upon the planes xi =constant for i = 1, 2, 3. Cauchy Stress Law Let tj (n) denote the components of the surface traction on the surface ABC. That is, we let t (n) = t1 (n) e1 + t2 (n) e2 + t3 (n) e3 = tj (n) ej . (2.3.10) It will be demonstrated that the components tj (n) of the surface traction forces t (n) associated with a plane through P and having the unit normal with direction cosines n1 , n2 and n3 , must satisfy the relations tj (n) = ni ij , This relation is known as the Cauchy stress law. Proof: Sum the forces acting on the elemental tetrahedron in the gure 2.3-5. If the body is in equilibrium, then the sum of these forces must equal zero or (t S1 ) + (t 2 S2 ) + (t 3 S3 ) + t (n) S = 0. 1 i, j = 1, 2, 3. (2.3.11) (2.3.12) The relations in the equations (2.3.9) are used to simplify the sum of forces in the equation (2.3.12). It is readily veried that the sum of forces simplies to t (n) = n1 t 1 + n2 t 2 + n3 t 3 = ni t i . Substituting in the relations from equation (2.3.8) we nd t (n) = tj (n) ej = ni ij ej , or in component form tj (n) = ni ij which is the Cauchy stress law. (2.3.15) i, j = 1, 2, 3 (2.3.14) (2.3.13) 218 Conservation of Linear Momentum Let R denote a region in space where there exists a material volume with density i having surface tractions and body forces acting upon it. Let v denote the velocity of the material volume and use Newtons second law to set the time rate of change of linear momentum equal to the forces acting upon the volume as in (2.3.5). We nd t v j d = R S ij ni dS + R bj d. Here d is an element of volume, dS is an element of surface area, bj are body forces per unit mass, and ij are the stresses. Employing the Gauss divergence theorem, the surface integral term is replaced by a volume integral and Newtons second law is expressed in the form f j bj ij ,i d = 0, R (2.3.16) where f j is the acceleration from equation (1.4.54). Since R is an arbitrary region, the equation (2.3.16) implies that ij ,i + bj = f j . (2.3.17) This equation arises from a balance of linear momentum and represents the equations of motion for material in a continuum. If there is no velocity term, then equation (2.3.17) reduces to an equilibrium equation which can be written ij ,i + bj = 0. This equation can also be written in the covariant form g si ms,i + bm = 0, which reduces to ij,j + bi = 0 in Cartesian coordinates. The equation (2.3.18) is an equilibrium equation and is one of our fundamental equations describing a continuum. Conservation of Angular Momentum The conservation of angular momentum equation (2.3.6) has the Cartesian tensors representation d dt eijk xj vk d = R S (2.3.18) eijk xj pk np dS + R eijk xj bk d. (2.3.19) Employing the Gauss divergence theorem, the surface integral term is replaced by a volume integral to obtain eijk R d (xj vk ) eijk dt xj bk + (xj pk ) xp d = 0. (2.3.20) Since equation (2.3.20) must hold for all arbitrary volumes R we conclude that eijk d (xj vk ) = eijk dt xj bk + xj pk + jk xp 219 Figure 2.3-6. Shearing parallel to the y axis which can be rewritten in the form eijk jk + xj ( pk dvk ) vj vk = 0. + bk xp dt (2.3.21) In the equation (2.3.21) the middle term is zero because of the equation (2.3.17). Also the last term in (2.3.21) is zero because eijk vj vk represents the cross product of a vector with itself. The equation (2.3.21) therefore reduces to eijk jk = 0, (2.3.22) which implies (see exercise 1.1, problem 22) that ij = ji for all i and j. Thus, the conservation of angular momentum requires that the stress tensor be symmetric. Consequently, there are only 6 independent stress components to be determined. This is another fundamental law for a continuum. Strain in Two Dimensions Consider the matrix equation x y = 1 0 1 x y (2.3.23) which can be used to transform points (x, y) to points (x, y). When this transformation is applied to the unit square illustrated in the gure 2.3-6(a) we obtain the geometry illustrated in the gure 2.3-6(b) which represents a shearing parallel to the y axis. If is very small, we can use the approximation tan and then this transformation can be thought of as a rotation of the element P1 P2 through an angle to the position P1 P2 when the barred axes are placed atop the unbarred axes. Similarly, the matrix equation x y = 1 0 1 x y (2.3.24) can be used to represent a shearing of the unit square parallel to the x axis as illustrated in the gure 2.3-7(b). 220 Figure 2.3-7. Shearing parallel to the x axis Figure 2.3-8. Shearing parallel to x and y axes Again, if is very small, we may use the approximation tan and interpret as an angular rotation of the element P1 P4 to the position P1 P4 . Now let us multiply the matrices given in equations (2.3.23) and (2.3.24). Note that the order of multiplication is important as can be seen by an examination of the products x y x y = = 1 1 0 0 1 1 1 01 1 0 1 x y x y = = 1 1 + 1 x y x y . 1 + (2.3.25) In equation (2.3.25) we will assume that the product is very, very small and can be neglected. Then the order of matrix multiplication will be immaterial and the transformation equation (2.3.25) will reduce to x y = 1 1 x y . (2.3.26) Applying this transformation to our unit square we obtain the simultaneous shearing parallel to both the x and y axes as illustrated in the gure 2.3-8. This transformation can then be interpreted as the superposition of the two shearing elements depicted in the gure 2.3-9. For comparison, we consider also the transformation equation x y = 10 1 x y (2.3.27) 221 Figure 2.3-9. Superposition of shearing elements Figure 2.3-10. Rotation of element P1 P2 where is very small. Applying this transformation to the unit square previously considered we obtain the results illustrated in the gure 2.3-10. Note the dierence in the direction of shearing associated with the transformation equations (2.3.27) and (2.3.23) illustrated in the gures 2.3-6 and 2.3-10. If the matrices appearing in the equations (2.3.24) and (2.3.27) are multiplied and we neglect product terms because is assumed to be very small, we obtain the matrix equation x y = 1 1 x y = 1 0 0 1 identity x y + 0 0 rotation x . y (2.3.28) This can be interpreted as a superposition of the transformation equations (2.3.24) and (2.3.27) which represents a rotation of the unit square as illustrated in the gure 2.3-11. The matrix on the right-hand side of equation (2.3.28) is referred to as a rotation matrix. The ideas illustrated by the above simple transformations will appear again when we consider the transformation of an arbitrary small element in a continuum when it under goes a strain. In particular, we will be interested in extracting the rigid body rotation from a deformed element and treating this rotation separately from the strain displacement. 222 Figure 2.3-11. Rotation of unit square Transformation of an Arbitrary Element In two dimensions, we consider a rectangular element ABCD as illustrated in the gure 2.3-12. Let the points ABCD have the coordinates A(x, y), and denote by u = u(x, y), v = v(x, y) B(x + x, y), C(x, y + y), D(x + x, y + y) (2.3.29) the displacement eld associated with each of the points in the material continuum when it undergoes a deformation. Assume that the deformation of the element ABCD in gure 2.3-12 can be represented by the matrix equation x y = b11 b21 b12 b22 x y (2.3.30) where the coecients bij , i, j = 1, 2, 3 are to be determined. Let us dene u = u(x, y) as the horizontal displacement of the point (x, y) and v = v(x, y) as the vertical displacement of the same point. We can now express the displacement of each of the points A, B, C and D in terms of the displacement eld u = u(x, y) and v = v(x, y). Consider rst the displacement of the point A to A . Here the coordinates (x, y) deform to the new coordinates x = x + u, y = y + v. That is, the coecients bij must be chosen such that the equation x+u y+v = b11 b21 b12 b22 x y (2.3.31) is satised. We next examine the displacement of the point B to B . This displacement is described by the coordinates (x + x, y) transforming to (x, y), where x = x + x + u(x + x, y), y = y + v(x + x, y). (2.3.32) 223 Figure 2.3-12. Displacement of element ABCD to A B C D Expanding u and v in (2.3.32) in Taylor series about the point (x, y) we nd x = x + x + u + y =y+v+ u x + h.o.t. x v x + h.o.t., x (2.3.33) where h.o.t. denotes higher order terms which have been neglected. The equations (2.3.33) require that the coecients bij satisfy the matrix equation x + u + x + u x x v y + v + x x = b11 b21 b12 b22 x + x y . (2.3.34) 224 The displacement of the point C to C is described by the coordinates (x, y + y) transforming to (x, y) where x = x + u(x, y + y), y = y + y + v(x, y + y). (2.3.35) Again we expand the displacement eld components u and v in a Taylor series about the point (x, y) and nd u y + h.o.t. y v y + h.o.t. y = y + y + v + y x = x+u+ (2.3.36) This equation implies that the coecients bij must be chosen such that x + u + u y y y + v + y + v y y = b11 b21 b12 b22 x y + y . (2.3.37) Finally, it can be veried that the point D with coordinates (x + x, y + y) moves to the point D with coordinates x = x + x + u(x + x, y + y), y = y + y + v(x + x, y + y). (2.3.38) Expanding u and v in a Taylor series about the point (x, y) we nd the coecients bij must be chosen to satisfy the matrix equation x + x + u + y + y + v + u x x + v x x + u y y v y y = b11 b21 b12 b22 x + x y + y . (2.3.39) The equations (2.3.31),(2.3.34),(2.3.37) and (2.3.39) give rise to the simultaneous equations b11 x + b12 y = x + u b21 x + b22 y = y + v b11 (x + x) + b12 y = x + u + x + b21 (x + x) + b22 y = y + v + u x x v x x u y b11 x + b12 (y + y) = x + u + y v b21 x + b22 (y + y) = y + v + y + y y u u b11 (x + x) + b12 (y + y) = x + x + u + x + y x y v v b21 (x + x) + b22 (y + y) = y + y + v + x + y. x y It is readily veried that the system of equations (2.3.40) has the solution b11 = 1 + b21 v = x u x b12 = b22 u y (2.3.40) v . =1+ y (2.3.41) 225 Figure 2.3-13. Change in 45 line Hence the transformation equation (2.3.30) can be written as x y = 1+ u x v x 1 u y + v y x y . (2.3.42) A physical interpretation associated with this transformation is obtained by writing it in the form: x y where 1 u v + 2 y x v e12 e22 = y are the elements of a symmetric matrix called the strain matrix and e11 = u x u 1 v + = 2 x y e21 = 11 = 0 21 = 1 2 u v x y 12 = 1 2 u v y x (2.3.44) = 1 0 0 1 identity x y + e11 e21 e12 e22 x y + 11 21 12 22 x , y (2.3.43) strain matrix rotation matrix (2.3.45) 22 = 0 are the elements of a skew symmetric matrix called the rotation matrix. The strain per unit length in the x-direction associated with the point A in the gure 2.3-12 is x and the strain per unit length of the point A in the y direction is e22 = y + v y y e11 = x + u x x x = u x (2.3.46) v . (2.3.47) y y These are the terms along the main diagonal in the strain matrix. The geometry of the gure 2.3-12 implies = that tan = v x x , x + u x x u y y . v y + y y y and tan = (2.3.48) For small derivatives associated with the displacements u and v it is assumed that the angles and are small and the equations (2.3.48) therefore reduce to the approximate equations u v tan = . (2.3.49) x y For a physical interpretation of these terms we consider the deformation of a small rectangular element which tan = undergoes a shearing as illustrated in the gure 2.3-13. 226 Figure 2.3-14. Displacement eld due to state of strain The quantity + = u v + y x = 2e12 = 2e21 (2.3.50) as representing a change from a 45 angle due to the deformation. The quantities e21 , e12 are called the shear strains and the quantity 12 = 2e12 is called the shear angle. In the equation (2.3.45), the quantities 21 = 12 are the elements of the rigid body rotation matrix and are interpreted as angles associated with a rotation. The situation is analogous to the transformations and gures for the deformation of the unit square which was considered earlier. Strain in Three Dimensions The development of strain in three dimensions is approached from two dierent viewpoints. The rst approach considers the derivation using Cartesian tensors and the second approach considers the derivation of strain using generalized tensors. Cartesian Tensor Derivation of Strain. Consider a material which is subjected to external forces such that all the points in the material undergo a deformation. Let (y1 , y2 , y3 ) denote a set of orthogonal Cartesian coordinates, xed in space, which is used to describe the deformations within the material. Further, let ui = ui (y1 , y2 , y3 ), i = 1, 2, 3 denote a displacement eld which describes the displacement of each point within the material. With reference to the gure 2.3-14 let P and Q denote two neighboring points within the material while it is in an unstrained state. These points move to the points P and Q when the material is in a state of strain. We let yi , i = 1, 2, 3 represent the position vector to the general point P in the material, which is in an unstrained state, and denote by yi + ui , i = 1, 2, 3 the position vector of the point P when the material is in a state of strain. (2.3.51) is the change from a ninety degree angle due to the deformation and hence we can write 1 ( + ) = e12 = e21 2 227 For Q a neighboring point of P which moves to Q when the material is in a state of strain, we have from the gure 2.3-14 the following vectors: position of P : yi , i = 1, 2, 3 i = 1, 2, 3 (2.3.52) i = 1, 2, 3 position of P : yi + ui (y1 , y2 , y3 ), position of Q : yi + yi , i = 1, 2, 3 position of Q : yi + yi + ui (y1 + y1 , y2 + y2 , y3 + y3 ), Employing our earlier one dimensional denition of strain, we dene the strain associated with the point P L L0 in the direction P Q as e = , where L0 = P Q and L = P Q . To calculate the strain we need to rst L0 calculate the distances L0 and L. The quantities L2 and L2 are easily calculated by considering dot products 0 of vectors. For example, we have L2 = yi yi , and the distance L = P Q is the magnitude of the vector 0 yi + yi + ui (y1 + y1 , y2 + y2 , y3 + y3 ) (yi + ui (y1 , y2 , y3 )), i = 1, 2, 3. Expanding the quantity ui (y1 + y1 , y2 + y2 , y3 + y3 ) in a Taylor series about the point P and neglecting higher order terms of the expansion we nd that L2 = (yi + ui ui ym )(yi + yn ). ym yn Expanding the terms in this expression produces the equation L2 = yi yi + ui ui ui ui yi yn + ym yi + ym yn . yn ym ym yn Note that L and L0 are very small and so we express the dierence L2 L2 in terms of the strain e. We can 0 write L2 L2 = (L + L0 )(L L0 ) = (L L0 + 2L0 )(L L0 ) = (e + 2)eL2 . 0 0 Now for e very small, and e2 negligible, the above equation produces the approximation eL2 0 The quantities emn = 1 um un ur ur + + 2 yn ym ym yn = jk = ij ik . 1 um L2 L2 un ur ur 0 = + + ym yn . 2 2 yn ym ym yn (2.3.53) is called the Green strain tensor or Lagrangian strain tensor. To show that eij is indeed a tensor, we consider the transformation yi = ij y j +bi , where ji ki Note that from the derivative relation yi y j = ij and the transformation equations ui = 1 ui uj ur ur + + 2 yj yi y i yj 1 ( 2 yn ik uk ) ij uj , i = 1, 2, 3 we can express the strain in the barred system of coordinates. Performing the necessary calculations produces eij = = = yn + ( yj ym jk uk ) ym + ( yi yk ki tj rs us ) yk ( y i yt rm um ) yt y j um uk 1 + jk mi + rs rp im nj 2 yn ym 1 um un us us = + + im nj 2 yn ym ym yn im nj .Consequently, us up yk yt or eij = emn the strain eij transforms like a second order Cartesian tensor. 228 Lagrangian and Eulerian Systems Let xi denote the initial position of a material particle in a continuum. Assume that at a later time the particle has moved to another point whose coordinates are xi . Both sets of coordinates are referred to the same coordinate system. When the nal position can be expressed as a function of the initial position and time we can write xi = xi (x1 , x2 , x3 , t). Whenever the changes of any physical quantity is represented in terms of its initial position and time, the representation is referred to as a Lagrangian or material representation of the quantity. This can be thought of as a transformation of the coordinates. When the Jacobian J( x ) of this x transformation is dierent from zero, the above set of equations have a unique inverse xi = xi (x1 , x2 , x3 , t), where the position of the particle is now expressed in terms of its instantaneous position and time. Such a representation is referred to as an Eulerian or spatial description of the motion. Let (x1 , x2 , x3 ) denote the initial position of a particle whose motion is described by xi = xi (x1 , x2 , x3 , t), then ui = xi xi denotes the displacement vector which can by represented in a Lagrangian or Eulerian form. For example, if x1 = 2(x1 x2 )(et 1) + (x2 x1 )(et 1) + x1 x2 = (x1 x2 )(et 1) + (x2 x1 )(et 1) + x2 x3 = x3 then the displacement vector can be represented in the Lagrangian form u1 = 2(x1 x2 )(et 1) + (x2 x1 )(et 1) u2 = (x1 x2 )(et 1) + (x2 x1 )(et 1) u3 = 0 or the Eulerian form u1 = x1 (2x2 x1 )(1 et ) (x1 x2 )(e2t et ) x1 et u2 = x2 (2x2 x1 )(1 et ) (x2 x1 )(e2t et ) x2 et u3 = 0. Note that in the Lagrangian system the displacements are expressed in terms of the initial position and time, while in the Eulerian system the independent variables are the position coordinates and time. Euler equations describe, as a function of time, how such things as density, pressure, and uid velocity change at a xed point in the medium. In contrast, the Lagrangian viewpoint follows the time history of a moving individual uid particle as it moves through the medium. 229 General Tensor Derivation of Strain. With reference to the gure 2.3-15 consider the deformation of a point P within a continuum. Let (y , y 2 , y 3 ) denote a Cartesian coordinate system which is xed in space. We can introduce a coordinate transformation y i = y i (x1 , x2 , x3 ), i = 1, 2, 3 and represent all points within the continuum with respect to a set of generalized coordinates (x1 , x2 , x3 ). Let P denote a general point in the continuum while it is in an unstrained state and assume that this point gets transformed to a point P when the continuum experiences external forces. If P moves to P , then all points Q which are near P will move to points Q near P . We can imagine that in the unstrained state all the points of the continuum are referenced with respect to the set of generalized coordinates (x1 , x2 , x3 ). After the strain occurs, we can imagine that it will be convenient to represent all points of the continuum with respect to a new barred system of coordinates (x1 , x2 , x3 ). We call the original set of coordinates the Lagrangian system of coordinates and the new set of barred coordinates the Eulerian coordinates. The Eulerian coordinates are assumed to be described by a set of coordinate transformation equations xi = xi (x1 , x2 , x3 ), x = x (x , x , x ), i i 1 2 3 1 i = 1, 2, 3 with inverse transformations i = 1, 2, 3, which are assumed to exist. The barred and unbarred coordinates can be related to a xed set of Cartesian coordinates y i , i = 1, 2, 3, and we may assume that there exists transformation equations y i = y i (x1 , x2 , x3 ), i = 1, 2, 3 and y i = y i (x1 , x2 , x3 ), i = 1, 2, 3 which relate the barred and unbarred coordinates to the Cartesian axes. In the discussion that follows be sure to note whether there is a bar over a symbol, as we will be jumping back and forth between the Lagrangian and Eulerian reference frames. Figure 2.3-15. Strain in generalized coordinates r which produce xi the metrices gij = Ei Ej . Similarly, in the Eulerian system of barred coordinates we have the basis vectors r Ei = which produces the metrices Gij = E i E j . These basis vectors are illustrated in the gure 2.3-15. xi In the Lagrangian system of unbarred coordinates we have the basis vectors Ei = 230 We assume that an element of arc length squared ds2 in the unstrained state is deformed to the element of arc length squared ds2 in the strained state. An element of arc length squared can be expressed in terms of the barred or unbarred coordinates. For example, in the Lagrangian system, let dr = P Q so that L2 = dr dr = ds2 = gij dxi dxj , 0 (2.3.54) where gij are the metrices in the Lagrangian coordinate system. This same element of arc length squared can be expressed in the barred system by L2 = ds2 = gij dxi dxj , 0 where g ij = gmn xm xn . xi xj (2.3.55) Similarly, in the Eulerian system of coordinates the deformed arc length squared is L2 = dr dr = ds2 = Gij dxi dxj , (2.3.56) where Gij are the metrices in the Eulerian system of coordinates. This same element of arc length squared can be expressed in the Lagrangian system by the relation L2 = ds2 = Gij dxi dxj , In the Lagrangian system we have ds2 ds2 = (Gij gij )dxi dxj = 2eij dxi dxj where eij = 1 (Gij gij ) 2 (2.3.58) where Gij = Gmn xm xn . xi xj (2.3.57) is called the Green strain tensor or Lagrangian strain tensor. Alternatively, in the Eulerian system of coordinates we may write ds2 ds2 = Gij g ij dxi dxj = 2eij dxi dxj where eij = 1 Gij gij 2 (2.3.59) is called the Almansi strain tensor or Eulerian strain tensor. 231 Note also in the gure 2.3-15 there is the displacement vector u. This vector can be represented in any of the following forms: u = ui Ei u = ui E i u = ui E i u = ui E i contravariant, Lagrangian basis covariant, Lagrangian reciprocal basis contravariant, Eulerian basis covariant, Eulerian reciprocal basis. By vector addition we have r + u = r and consequently dr + du = dr. In the Lagrangian frame of reference at the point P we represent u in the contravariant form u = ui Ei and write dr in the form dr = dxi Ei . By use of the equation (1.4.48) we can express du in the form du = ui,k dxk Ei . These substitutions produce the representation dr = (dxi + ui,k dxk )Ei in the Lagrangian coordinate system. We can then express ds2 in the Lagrangian system. We nd dr dr = ds2 = (dxi + ui,k dxk )Ei (dxj + uj,m dxm )Ej = (dxi dxj + uj,m dxm dxi + ui,k dxk dxj + ui,k uj,m dxk dxm )gij and consequently from the relation (2.3.58) we derive the representation eij = 1 ui,j + uj,i + um,i um . ,j 2 (2.3.60) This is the representation of the Lagrangian strain tensor in any system of coordinates. The strain tensor eij is symmetric. We will restrict our study to small deformations and neglect the product terms in equation (2.3.60). Under these conditions the equation (2.3.60) reduces to eij = 1 (ui,j + uj,i ). 2 If instead, we chose to represent the displacement u with respect to the Eulerian basis, then we can write u = ui E i These relations imply that dr = dr du = (dxi ui,k dxk )E i . This representation of dr in the Eulerian frame of reference can be used to calculate the strain eij from the relation ds2 ds2 . It is left as an exercise to show that there results eij = 1 ui,j + uj,i um,i um . ,j 2 (2.3.61) with du = ui,k dxk E i . The equation (2.3.61) is the representation of the Eulerian strain tensor in any system of coordinates. Under conditions of small deformations both the equations (2.3.60) and (2.3.61) reduce to the linearized Lagrangian and Eulerian strain tensor eij = 1 (ui,j + uj,i ). In the case of large deformations the equations (2.3.60) and 2 (2.3.61) describe the strains. In the case of linear elasticity, where the deformations are very small, the product terms in equations (2.3.60) and (2.3.61) are neglected and the Lagrangian and Eulerian strains reduce to their linearized forms eij = 1 [ui,j + uj,i ] 2 eij = 1 [ui,j + uj,i ] . 2 (2.3.62) 232 Figure 2.3-16. Displacement due to strain Compressible and Incompressible Material With reference to gure 2.3-16, let xi , i = 1, 2, 3 denote the position vector of an arbitrary point P in a continuum before there is a state of strain. Let Q be a neighboring point of P with position vector xi + dxi , i = 1, 2, 3. Also in the gure 2.3-16 there is the displacement vector u. Here it is assumed that u = u(x1 , x2 , x3 ) denotes the displacement eld when the continuum is in a state of strain. The gure 2.3-16 illustrates that in a state of strain P moves to P and Q moves to Q . Let us nd a relationship between the distance P Q before the strain and the distance P Q when the continuum is in a state of strain. For E1 , E2 , E3 basis functions constructed at P we have previously shown that if u(x1 , x2 , x3 ) = ui Ei then du = ui,j dxj Ei . Now for u + du the displacement of the point Q we may use vector addition and write P Q + u + du = u + P Q . (2.3.63) Let P Q = dxi Ei = ai Ei denote an arbitrary small change in the continuum. This arbitrary displacement gets deformed to P Q = Ai Ei due to the state of strain in the continuum. Employing the equation (2.3.63) we write dxi + ui,j dxj = ai + ui,j aj = Ai which can be written in the form ai = Ai ai = ui,j aj where dxi = ai , i = 1, 2, 3 (2.3.64) denotes an arbitrary small change. The tensor ui,j and the associated tensor ui,j = git ut,j are in general not symmetric tensors. However, we know we can express ui,j as the sum of a symmetric (eij ) and skewsymmetric(ij ) tensor. We therefore write ui,j = eij + ij where eij = 1 1 (ui,j + uj,i ) = (gim um ,j + gjm um ,i ) 2 2 and ij = 1 1 (ui,j uj,i ) = (gim um ,j gjm um ,i ) . 2 2 or ui,j = eij + ij , The deformation of a small quantity ai can therefore be represented by a pure strain Ai ai = ei as followed s i by a rotation Ai ai = s as . 233 Consider now a small element of volume inside a material medium. With reference to the gure 2.317(a) we let a, b, c denote three small arbitrary independent vectors constructed at a general point P within the material before any external forces are applied. We imagine a, b, c as representing the sides of a small parallelepiped before any deformation has occurred. When the material is placed in a state of strain the point P will move to P and the vectors a, b, c will become deformed to the vectors A, B, C as illustrated in the gure 2.3-17(b). The vectors A, B, C represent the sides of the parallelepiped after the deformation. Figure 2.3-17. Deformation of a parallelepiped Let V denote the volume of the parallelepiped with sides a, b, c at P before the strain and let V denote the volume of the deformed parallelepiped after the strain, when it then has sides A, B, C at the point P . We dene the ratio of the change in volume due to the strain divided by the original volume as the dilatation at the point P. The dilatation is thus expressed as = V V = dilatation. V (2.3.65) Since ui , i = 1, 2, 3 represents the displacement eld due to the strain, we use the result from equation (2.3.64) and represent the displaced vectors A, B, C in the form Ai = ai + ui,j aj B i = bi + ui,j bj C =c + i i (2.3.66) ui,j cj where a, b, c are arbitrary small vectors emanating from the point P in the unstrained state. The element of volume V, before the strain, is calculated from the triple scalar product relation V = a (b c) = eijk ai bj ck . The element of volume V , which occurs due to the strain, is calculated from the triple scalar product V = A (B C) = eijk Ai B j C k . 234 Substituting the relations from the equations (2.3.66) into the triple scalar product gives V = eijk (ai + ui,m am )(bj + uj,n bn )(ck + uk cp ). ,p Expanding the triple scalar product and employing the result from Exercise 1.4, problem 34, we nd the simplied result gives us the dilatation = V V = ur = div (u). ,r V (2.3.67) That is, the dilatation is the divergence of the displacement eld. If the divergence of the displacement eld is zero, there is no volume change and the material is said to be incompressible. If the divergence of the displacement eld is dierent from zero, the material is said to be compressible. Note that the strain eij is expressible in terms of the displacement eld by the relation eij = 1 (ui,j + uj,i ), 2 and consequently g mn emn = ur . ,r (2.3.68) Hence, for an orthogonal system of coordinates the dilatation can be expressed in terms of the strain elements along the main diagonal. Conservation of Mass Consider the material in an arbitrary region R of a continuum. Let = (x, y, z, t) denote the density is gm/cm3 in the cgs system of the material within the region. Assume that the dimension of the density of units. We shall assume that the region R is bounded by a closed surface S with exterior unit normal n dened everywhere on the surface. Further, we let v = v(x, y, z, t) denote a velocity eld associated with all points within the continuum. The velocity eld has units of cm/sec in the cgs system of units. Neglecting sources and sinks, the law of conservation of mass examines all the material entering and leaving a region R. Enclosed within R is the material mass m where m = R d with dimensions of gm in the cgs system of units. Here d denotes an element of volume inside the region R. The change of mass with time is obtained by dierentiating the above relation. Dierentiating the mass produces the equation m = t and has the dimensions of gm/sec. Consider also the surface integral I= S R d t (2.3.69) v n d (2.3.70) where d is an element of surface area on the surface S which encloses R and n is the exterior unit normal vector to the surface S. The dimensions of the integral I is determined by examining the dimensions of each term in the integrand of I. We nd that [I] = gm gm cm cm2 = 3 cm sec sec and so the dimension of I is the same as the dimensions for the change of mass within the region R. The surface integral I is the ux rate of material crossing the surface of R and represents the change of mass 235 entering the region if v n is negative and the change of mass leaving the region if v n is positive, as n is always an exterior unit normal vector. Equating the relations from equations (2.3.69) and (2.3.70) we obtain a mathematical statement for mass conservation m = t d = t v n d. S (2.3.71) R The equation (2.3.71) implies that the rate at which the mass contained in R increases must equal the rate at which the mass ows into R through the surface S. The negative sign changes the direction of the exterior normal so that we consider ow of material into the region. Employing the Gauss divergence theorem, the surface integral in equation (2.3.71) can be replaced by a volume integral and the law of conservation of mass is then expressible in the form + div ( v) d = 0. t (2.3.72) R Since the region R is an arbitrary volume we conclude that the term inside the brackets must equal zero. This gives us the continuity equation + div ( v) = 0 t tation of continuity of mass ow. Equivalent forms of the continuity equation are: + v grad + div v = 0 t vi + vi i + =0 t x xi vi D + =0 Dt xi dxi D = + = + vi is called the material derivative of the density . Note that the Dt t xi dt t xi material derivative contains the expression xi vi which is known as the convective or advection term. If the where density = (x, y, z, t) is a constant we have D dx dy dz dxi = + + + = +i =0 Dt t x dt y dt z dt t x dt (2.3.74) (2.3.73) which represents the mass conservation law in terms of velocity components. This is the Eulerian represen- and hence the continuity equation reduces to div (v) = 0. Thus, if div (v) is zero, then the material is incompressible. EXAMPLE 2.3-2. (Continuity Equation) Find the Lagrangian representation of mass conservation. Solution: Let (X, Y, Z) denote the initial position of a uid particle and denote the density of the uid by (X, Y, Z, t) so that (X, Y, Z, 0) denotes the density at the time t = 0. Consider a simple closed region in our continuum and denote this region by R(0) at time t = 0 and by R(t) at some later time t. That is, all the points in R(0) move in a one-to-one fashion to points in R(t). Initially the mass of material in R(0) is m(0) = R(0) (X, Y, Z, 0) d (0) where d (0) = dXdY dZ is an element of volume in R(0). We have after a 236 time t has elapsed the mass of material in the region R(t) given by m(t) = R(t) (X, Y, Z, t) d (t) where x,y,z X,Y,Z d (t) = dxdydz is a deformed element of volume related to the d (0) by d (t) = J mass conservation we require that m(t) = m(0) for all t. This implies that (X, Y, Z, t)J = (X, Y, Z, 0) d (0) where J is the Jacobian of the Eulerian (x, y, z) variables with respect to the Lagrangian (X, Y, Z) representation. For (2.3.75) for all time, since the initial region R(0) is arbitrary. The right hand side of equation (2.3.75) is independent of time and so d ( (X, Y, Z, t)J) = 0. dt (2.3.76) This is the Lagrangian form of the continuity equation which expresses mass conservation. Using the result dJ = Jdiv V , (see problem 28, Exercise 2.3), the equation (2.3.76) can be expanded and written in the that dt form D + div V = 0 (2.3.77) Dt where D Dt is from equation (2.3.74). The form of the continuity equation (2.3.77) is one of the Eulerian forms previously developed. In the Eulerian coordinates the continuity equation is written system the continuity equation is written d( J) dt t + div ( v) = 0, while in the Lagrangian = 0. Note that the velocity carries the Lagrangian axes and the density change grad . This is reective of the advection term v grad . Thus, in order for mass to be conserved it need not remain stationary. The mass can ow and the density can change. The material derivative is a transport rule depicting the relation between the Eulerian and Lagrangian viewpoints. In general, from a Lagrangian viewpoint, any quantity Q(x, y, z, t) which is a function of both position and time is seen as being transported by the uid velocity (v1 , v2 , v3 ) to Q(x + v1 dt, y + v2 dt, z + v3 dt, t + dt). Then the time derivative of Q contains both Q t and the advection term v Q. In terms of mass ow, the Eulerian viewpoint sees ow into and out of a xed volume in space, as depicted by the equation (2.3.71), In contrast, the Lagrangian viewpoint sees the same volume moving with the uid and consequently D Dt d = 0, R(t) where R(t) represents the volume moving with the uid. Both viewpoints produce the same continuity equation reecting the conservation of mass. Summary of Basic Equations Let us summarize the basic equations which are valid for all types of a continuum. We have derived: Conservation of mass (continuity equation) + ( v i ),i = 0 t 237 Conservation of linear momentum sometimes called the Cauchy equation of motion. ij ,i + bj = f j , Conservation of angular momentum ij = ji Strain tensor for linear elasticity eij = 1 (ui,j + uj,i ). 2 j = 1, 2, 3. If we assume that the continuum is in equilibrium, and there is no motion, then the velocity and acceleration terms above will be zero. The continuity equation then implies that the density is a constant. The conservation of angular momentum equation requires that the stress tensor be symmetric and we need nd only six stresses. The remaining equations reduce to a set of nine equations in the fteen unknowns: 3 displacements u1 , u2 , u3 6 strains e11 , e12 , e13 , e22 , e23 , e33 6 stresses 11 , 12 , 13 , 22 , 23 , 33 Consequently, we still need additional information if we desire to determine these unknowns. Note that the above equations do not involve any equations describing the material properties of the continuum. We would expect solid materials to act dierently from liquid material when subjected to external forces. An equation or equations which describe the material properties are called constitutive equations. In the following sections we will investigate constitutive equations for solids and liquids. We will restrict our study to linear elastic materials over a range where there is a linear relationship between the stress and strain. We will not consider plastic or viscoelastic materials. Viscoelastic materials have the property that the stress is not only a function of strain but also a function of the rates of change of the stresses and strains and consequently properties of these materials are time dependent. 238 EXERCISE 2.3 1. Assume an orthogonal coordinate system with metric tensor gij = 0 for i = j and g(i)(i) = h2 (no i 1 1 (ur,s + us,r ) = grt ut,s + gst ut,r 2 2 eij hi hj summation on i). Use the denition of strain ers = and show that in terms of the physical components e(ij) = no summation on i or j no summation on i u(i) = hi ui there results the equations: t ut + um no summation on i mi xi ut ut 2eij = git j + gjt i , i=j x x eii = git e(ii) = 2e(ij) = xi u(i) hi + 1 2h2 i + u(m) h2 hm xm i m=1 u(j) hj , 3 no summation on i hi hj xj u(i) hi hj hi xi no summation on i or j, i = j. 2. Use the results from problem 1 to write out all components of the strain tensor in Cartesian coordinates. Use the notation u(1) = u,u(2) = v,u(3) = w and e(11) = exx , to verify the relations: exx = eyy ezz u x v = y w = z exy = exz ezy 1 2 1 = 2 1 = 2 v u + x y u w + z x w v + y z e(22) = eyy , e(33) = ezz , e(12) = exy , e(13) = exz , e(23) = eyz 3. Use the results from problem 1 to write out all components of the strain tensor in cylindrical coordinates. Use the notation u(1) = ur , u(2) = u , u(3) = uz and e(11) = err , to verify the relations: err = e ezz ur r ur 1 u + = r r uz = z er = erz ez 1 2 1 = 2 1 = 2 1 ur u u + r r r uz ur + r z u 1 uz + z r e(22) = e , e(33) = ezz , e(12) = er , e(13) = erz , e(23) = ez 239 4. Use the results from problem 1 to write out all components of the strain tensor in spherical coordinates. Use the notation u(1) = u ,u(2) = u ,u(3) = u and e(11) = e , to verify the relations e = e e u u 1 u + = u 1 u u + + cot = sin e = e e 1 2 1 = 2 1 = 2 1 u u u + 1 u u u + sin 1 u u 1 u cot + sin e(22) = e , e(33) = e , e(12) = e , e(13) = e , e(23) = e 5. Expand equation (2.3.67) and nd the dilatation in terms of the physical components of an orthogonal system and verify that = (h2 h3 u(1)) (h1 h3 u(2)) (h1 h2 u(3)) 1 + + h1 h2 h3 x1 x2 x3 6. Verify that the dilatation in Cartesian coordinates is = exx + eyy + ezz = u v w + + . x y z 7. Verify that the dilatation in cylindrical coordinates is = err + e + ezz = 1 u 1 ur uz + + ur + . r r r z 8. Verify that the dilatation in spherical coordinates is = e + e + e = 1 u 2 u cot u 1 u + + u + + . sin 9. Show that in an orthogonal set of coordinates the rotation tensor ij can be written in terms of physical components in the form (ij) = Hint: See problem 1. 10. Use the result from problem 9 to verify that in Cartesian coordinates yx = xz zy 1 2 1 = 2 1 = 2 v u x y u w z x w v y z 1 2hi hj (hi u(i)) (hj u(j)) , xj xi no summations 240 11. Use the results from problem 9 to verify that in cylindrical coordinates r = rz z 1 2r 1 = 2 1 = 2 (ru ) ur r ur uz z r 1 uz u r z 12. Use the results from problem 9 to verify that in spherical coordinates = 1 (u ) u 2 1 u (u ) 1 = 2 sin (u sin ) u 1 = 2 sin 13. The conditions for static equilibrium in a linear elastic material are determined from the conservation law j i ,j + bi = 0, i, j = 1, 2, 3, is the density i where j are the stress tensor components, bi are the external body forces per unit mass and of the material. Assume an orthogonal coordinate system and verify the following results. (a) Show that 1j j i ,j = ( gi ) [ij, m] mj g xj j (ij) = i (b) Use the substitutions hj hi no summation on i or j bi no summation on i hi (ij) = ij hi hj no summation on i or j b(i) = and express the equilibrium equations in terms of physical components and verify the relations 3 j=1 1 g xj ghi (ij) hj 1 2 3 j=1 (jj) (h2 ) j + hi b(i) = 0, h2 xi j where there is no summation on i. 14. Use the results from problem 13 and verify that the equilibrium equations in Cartesian coordinates can be expressed xx xy xz + + + bx = 0 x y z yy yz yx + + + by = 0 x y z zy zz zx + + + bz = 0 x y z 241 15. Use the results from problem 13 and verify that the equilibrium equations in cylindrical coordinates can be expressed 1 r rz 1 rr + + + (rr ) + br = 0 r r z r 1 z 2 r + + + r + b = 0 r r z r 1 z zz 1 zr + + + zr + bz = 0 r r z r 16. Use the results from problem 13 and verify that the equilibrium equations in spherical coordinates can be expressed 1 1 1 + + + (2 + cot ) + b = 0 sin 1 1 1 + + + (3 + [ ] cot ) + b = 0 sin 1 1 1 + + + (3 + 2 cot ) + b = 0 sin 17. Derive the result for the Lagrangian strain dened by the equation (2.3.60). 18. Derive the result for the Eulerian strain dened by equation (2.3.61). 19. The equation ai = ui,j aj , describes the deformation in an elastic solid subjected to forces. The quantity ai denotes the dierence vector Ai ai between the undeformed and deformed states. (a) Let |a| denote the magnitude of the vector ai and show that the strain e in the direction ai can be represented e= |a| = eij |a| ai |a| aj |a| = eij i j , where i is a unit vector in the direction ai . (b) Show that for 1 = 1, 2 = 0, 3 = 0 there results e = e11 , with similar results applying to vectors i in the y and z directions. Hint: Consider the magnitude squared |a|2 = gij ai aj . 20. At the point (1, 2, 3) of an elastic solid construct the small vector a = ( 2 e1 + 3 results. u = (xy e1 + yz e2 + xz e3 ) 102 Calculate the deformed vector A after the displacement eld has been imposed. 21. For the displacement eld u = (x2 + yz) e1 + (xy + z 2 ) e2 + xyz e3 (a) Calculate the strain matrix at the point (1, 2, 3). (b) Calculate the rotation matrix at the point (1, 2, 3). 2 3 e2 + 1 3 e3 ), where > 0 is a small positive quantity. The solid is subjected to forces such that the following displacement eld 242 22. Show that for an orthogonal coordinate system the ith component of the convective operator can be written [(V ) A]i = V (m) A(i) + hm xm m=1 3 3 m=1 m=i A(m) hm hi V (i) hm hi V (m) i xm x 23. Consider a parallelepiped with dimensions , w, h which has a uniform pressure P applied to each face. Show that the volume strain can be expressed as V w h 3P (1 2) = + + = . V w h E The quantity k = E/3(1 2) is called the bulk modulus of elasticity. 24. Show in Cartesian coordinates the continuity equation is ( u) ( v) ( w) + + + = 0, t x y z where (u, v, w) are the velocity components. 25. Show in cylindrical coordinates the continuity equation is 1 (r Vr ) 1 ( V ) ( Vz ) + + + =0 t r r r z where Vr , V , Vz are the velocity components. 26. Show in spherical coordinates the continuity equation is 1 (2 V ) 1 ( V sin ) 1 ( V ) +2 + + =0 t sin sin where V , V , V are the velocity components. 27. (a) Apply a stress yy to both ends of a square element in a x, y continuum. Illustrate and label all changes that occur due to this stress. (b) Apply a stress xx to both ends of a square element in a x, y continuum. Illustrate and label all changes that occur due to this stress. (c) Use superposition of your results in parts (a) and (b) and explain each term in the relations exx = yy xx E E and eyy = xx yy . E E satises dy , dt dJ = J div V where dt dz . dt 28. Show that the time derivative of the Jacobian J = J V1 V2 V3 + + x y z x, y, z X, Y, Z dx , dt V2 = div V = and V1 = V3 = Hint: Let (x, y, z) = (x1 , x2 , x3 ) and (X, Y, Z) = (X1 , X2 , X3 ), then note that eijk V1 x2 x3 V1 xm x2 x3 x1 x2 x3 V1 = eijk = eijk , Xi Xj Xk xm Xi Xj Xk Xi Xj Xk x1 etc. 243 2.4 CONTINUUM MECHANICS (SOLIDS) In this introduction to continuum mechanics we consider the basic equations describing the physical eects created by external forces acting upon solids and uids. In addition to the basic equations that are applicable to all continua, there are equations which are constructed to take into account material characteristics. These equations are called constitutive equations. For example, in the study of solids the constitutive equations for a linear elastic material is a set of relations between stress and strain. In the study of uids, the constitutive equations consists of a set of relations between stress and rate of strain. Constitutive equations are usually constructed from some basic axioms. The resulting equations have unknown material parameters which can be determined from experimental investigations. One of the basic axioms, used in the study of elastic solids, is that of material invariance. This axiom requires that certain symmetry conditions of solids are to remain invariant under a set of orthogonal transformations and translations. This axiom is employed in the next section to simplify the constitutive equations for elasticity. We begin our study of continuum mechanics by investigating the development of constitutive equations for linear elastic solids. Generalized Hookes Law If the continuum material is a linear elastic material, we introduce the generalized Hookes law in Cartesian coordinates ij = cijkl ekl , i, j, k, l = 1, 2, 3. (2.4.1) The Hookes law is a statement that the stress is proportional to the gradient of the deformation occurring in the material. These equations assume a linear relationship exists between the components of the stress tensor and strain tensor and we say stress is a linear function of strain. Such relations are referred to as a set of constitutive equations. Constitutive equations serve to describe the material properties of the medium when it is subjected to external forces. Constitutive Equations The equations (2.4.1) are constitutive equations which are applicable for materials exhibiting small deformations when subjected to external forces. The 81 constants cijkl are called the elastic stiness of the material. The above relations can also be expressed in the form eij = sijkl kl , i, j, k, l = 1, 2, 3 (2.4.2) where sijkl are constants called the elastic compliance of the material. Since the stress ij and strain eij have been shown to be tensors we can conclude that both the elastic stiness cijkl and elastic compliance sijkl are fourth order tensors. Due to the symmetry of the stress and strain tensors we nd that the elastic stiness and elastic compliance tensor must satisfy the relations cijkl = cjikl = cijlk = cjilk sijkl = sjikl = sijlk = sjilk and consequently only 36 of the 81 constants are actually independent. If all 36 of the material (crystal) constants are independent the material is called triclinic and there are no material symmetries. (2.4.3) 244 Restrictions on Elastic Constants due to Symmetry The equations (2.4.1) and (2.4.2) can be replaced by an equivalent set of equations which are easier to analyze. This is accomplished by dening the quantities e1 , 1 , where e1 e4 e5 and 1 4 5 the forms i = cij ej or ei = sij j where i, j = 1, . . . , 6 (2.4.4) e4 e2 e6 4 2 6 e2 , 2 , e3 , 3 , e4 , 4 , e5 , 5 , e12 e22 e32 12 22 32 e6 6 e13 e23 e33 13 23 . 33 e5 e11 e6 = e21 e3 e31 5 11 6 = 21 3 31 Then the generalized Hookes law from the equations (2.4.1) and (2.4.2) can be represented in either of where cij are constants related to the elastic stiness and sij are constants related to the elastic compliance. These constants satisfy the relation smi cij = mj Here eij = and similarly ij = i , 1+i+j , i = j = 1, 2, 3 i = j, and i = 1, or, 2. where i, m, j = 1, . . . , 6 (2.4.5) ei , e1+i+j , i = j = 1, 2, 3 i = j, and i = 1, or, 2 These relations show that the constants cij are related to the elastic stiness coecients cpqrs by the relations cm1 = cij11 cm2 = cij22 cm3 = cij33 where m= i, 1 + i + j, cm4 = 2cij12 cm5 = 2cij13 cm6 = 2cij23 if i = j = 1, 2, or 3 if i = j and i = 1 or 2. A similar type relation holds for the constants sij and spqrs . The above relations can be veried by expanding the equations (2.4.1) and (2.4.2) and comparing like terms with the expanded form of the equation (2.4.4). 245 The generalized Hookes law can now be expressed in a form where the 36 independent constants can be examined in more detail under special material symmetries. We will examine the form s11 s12 s13 s14 s15 s16 1 e1 e2 s21 s22 s23 s24 s25 s26 2 e3 s31 s32 s33 s34 s35 s36 3 = . e4 s41 s42 s43 s44 s45 s46 4 e5 s51 s52 s53 s54 s55 s56 5 e6 s61 s62 s63 s64 s65 s66 6 Alternatively, in the arguments that follow, one can examine 1 c11 c12 c13 c14 c15 2 c21 c22 c23 c24 c25 3 c31 c32 c33 c34 c35 = 4 c41 c42 c43 c44 c45 5 c51 c52 c53 c54 c55 6 c61 c62 c63 c64 c65 Material Symmetries A material (crystal) with one plane of symmetry is called an aelotropic material. If we let the x1 x2 plane be a plane of symmetry then the equations (2.4.6) must remain invariant under the coordinate transformation 10 x1 x2 = 0 1 x3 00 0 x1 0 x2 x3 1 the equivalent form e1 c16 c26 e2 c36 e3 . c46 e4 c56 e5 c66 e6 (2.4.6) (2.4.7) which represents an inversion of the x3 axis. That is, if the x1 -x2 plane is a plane of symmetry we should be able to replace x3 by x3 and the equations (2.4.6) should remain unchanged. This is equivalent to saying that a transformation of the type from equation (2.4.7) changes the Hookes law to the form ei = sij j where the sij remain unaltered because it is the same material. Employing the transformation equations x1 = x1 , x2 = x2 , x3 = x3 (2.4.8) we examine the stress and strain transformation equations ij = pq xp xq xi xj and eij = epq xp xq . xi xj (2.4.9) If we expand both of the equations (2.4.9) and substitute in the nonzero derivatives x1 = 1, x1 we obtain the relations 11 = 11 22 = 22 33 = 33 21 = 21 31 = 31 23 = 23 e11 = e11 e22 = e22 e33 = e33 e21 = e21 e31 = e31 e23 = e23 . (2.4.11) x2 = 1, x2 x3 = 1, x3 (2.4.10) 246 We conclude that if the material undergoes a strain, with the x1 -x2 plane as a plane of symmetry then e5 and e6 change sign upon reversal of the x3 axis and e1 , e2 , e3 , e4 remain unchanged. Similarly, we nd 5 and 6 change sign while 1 , 2 , 3 , 4 remain unchanged. The equation (2.4.6) then becomes s11 e1 e2 s21 e3 s31 = e4 s41 e5 s51 e6 s61 s12 s22 s32 s42 s52 s62 s13 s23 s33 s43 s53 s63 s14 s24 s34 s44 s54 s64 s15 s25 s35 s45 s55 s65 1 s16 s26 2 s36 3 . s46 4 s56 5 s66 6 (2.4.12) If the stress strain relation for the new orientation of the x3 axis is to have the same form as the old orientation, then the equations (2.4.6) and (2.4.12) must give the same results. Comparison of these equations we nd that s15 = s16 = 0 s25 = s26 = 0 s35 = s36 = 0 s45 = s46 = 0 s51 = s52 = s53 = s54 = 0 s61 = s62 = s63 = s64 = 0. In summary, from an examination of the equations (2.4.6) and (2.4.12) we nd that for an aelotropic material (crystal), with one plane of symmetry, the 36 constants sij reduce to 20 constants and the generalized Hookes law (constitutive equation) has the form e1 s11 e2 s21 e3 s31 = e4 s41 e5 0 e6 0 s12 s22 s32 s42 0 0 s13 s23 s33 s43 0 0 s14 s24 s34 s44 0 0 0 0 0 0 s55 s65 0 1 0 2 0 3 . 0 4 s56 5 s66 6 (2.4.13) (2.4.14) Alternatively, the Hookes law can be represented in the form c11 1 2 c21 3 c31 = 4 c41 5 0 6 0 c12 c22 c32 c42 0 0 c13 c23 c33 c43 0 0 c14 c24 c34 c44 0 0 0 0 0 0 c55 c65 e1 0 0 e2 0 e3 . 0 e4 c56 e5 c66 e6 247 Additional Symmetries If the material (crystal) is such that there is an additional plane of symmetry, say the x2 -x3 plane, then reversal of the x1 axis should leave the equations (2.4.14) unaltered. If there are two planes of symmetry then there will automatically be a third plane of symmetry. Such a material (crystal) is called orthotropic. Introducing the additional transformation x1 = x1 , x2 = x2 , x3 = x3 which represents the reversal of the x1 axes, the expanded form of equations (2.4.9) are used to calculate the eect of such a transformation upon the stress and strain tensor. We nd 1 , 2 , 3 , 6 , e1 , e2 , e3 , e6 remain unchanged while 4 , 5 , e4 , e5 change sign. The equation (2.4.14) then becomes e1 s11 e2 s21 e3 s31 = e4 s41 0 e5 e6 0 s12 s22 s32 s42 0 0 s13 s23 s33 s43 0 0 s14 s24 s34 s44 0 0 0 0 0 0 s55 s65 0 1 0 2 0 3 . 0 4 s56 5 s66 6 (2.4.15) Note that if the constitutive equations (2.4.14) and (2.4.15) are to produce the same results upon reversal of the x1 axes, then we require that the following coecients be equated to zero: s14 = s24 = s34 = 0 s41 = s42 = s43 = 0 s56 = s65 = 0. This then produces the constitutive equation s11 e1 e2 s21 e3 s31 = e4 0 e5 0 0 e6 or its equivalent form c11 1 2 c21 3 c31 = 4 0 0 5 0 6 s12 s22 s32 0 0 0 c12 c22 c32 0 0 0 s13 s23 s33 0 0 0 c13 c23 c33 0 0 0 0 0 0 s44 0 0 0 0 0 c44 0 0 0 0 0 0 s55 0 0 0 0 0 c55 0 1 0 0 2 0 3 0 4 0 5 s66 6 e1 0 0 e2 0 e3 0 e4 0 e5 c66 e6 (2.4.16) and the original 36 constants have been reduced to 12 constants. This is the constitutive equation for orthotropic material (crystals). 248 Axis of Symmetry If in addition to three planes of symmetry there is an axis of symmetry then the material (crystal) is termed hexagonal. Assume that the x1 axis is an axis of symmetry and consider the eect of the transformation x1 = x1 , x2 = x3 x3 = x2 upon the constitutive equations. It is left as an exercise to verify that the constitutive equations reduce to the form where there are 7 independent constants having either of the forms s11 e1 e2 s21 e3 s21 = e4 0 0 e5 0 e6 or 1 c11 2 c21 3 c21 = 4 0 0 5 6 0 s12 s22 s23 0 0 0 c12 c22 c23 0 0 0 s12 s23 s22 0 0 0 c12 c23 c22 0 0 0 0 0 0 s44 0 0 0 0 0 c44 0 0 0 0 0 0 s44 0 0 0 0 0 c44 0 1 0 0 2 0 3 0 4 0 5 s66 6 0 e1 0 e2 0 e3 . 0 e4 0 e5 e6 c66 Finally, if the material is completely symmetric, the x2 axis is also an axis of symmetry and we can consider the eect of the transformation x1 = x3 , x2 = x2 , x3 = x1 upon the constitutive equations. It can be veried that these transformations reduce the Hookes law constitutive equation to the form e1 s11 e2 s12 e3 s12 = e4 0 0 e5 e6 0 s12 s11 s12 0 0 0 s12 s12 s11 0 0 0 0 0 0 s44 0 0 0 0 0 0 s44 0 0 1 0 2 0 3 . 0 4 0 5 6 s44 (2.4.17) Materials (crystals) with atomic arrangements that exhibit the above symmetries are called isotropic materials. An equivalent form of (2.4.17) is the relation 1 c11 2 c12 3 c12 = 4 0 0 5 6 0 c12 c11 c12 0 0 0 c12 c12 c11 0 0 0 0 0 0 c44 0 0 0 0 0 0 c44 0 0 e1 0 e2 0 e3 . 0 e4 0 e5 e6 c44 The gure 2.4-1 lists values for the elastic stiness associated with some metals which are isotropic1 1 Additional constants are given in International Tables of Selected Constants, Metals: Thermal and Mechanical Data, Vol. 16, Edited by S. Allard, Pergamon Press, 1969. 249 Metal Na Pb Cu Ni Cr Mo W c11 0.074 0.495 1.684 2.508 3.500 4.630 5.233 c12 0.062 0.423 1.214 1.500 0.678 1.610 2.045 c44 0.042 0.149 0.754 1.235 1.008 1.090 1.607 Figure 2.4-1. Elastic stiness coecients for some metals which are cubic. Constants are given in units of 1012 dynes/cm2 Under these conditions the stress strain constitutive relations can be written as 1 = 11 = (c11 c12 )e11 + c12 (e11 + e22 + e33 ) 2 = 22 = (c11 c12 )e22 + c12 (e11 + e22 + e33 ) 3 = 33 = (c11 c12 )e33 + c12 (e11 + e22 + e33 ) 4 = 12 = c44 e12 5 = 13 = c44 e13 6 = 23 = c44 e23 . Isotropic Material Materials (crystals) which are elastically the same in all directions are called isotropic. We have shown that for a cubic material which exhibits symmetry with respect to all axes and planes, the constitutive stress-strain relation reduces to the form found in equation (2.4.17). Dene the quantities s11 = 1 , E s12 = , E s44 = 1 2 (2.4.18) where E is the Youngs Modulus of elasticity, is the Poissons ratio, and is the shear or rigidity modulus. For isotropic materials the three constants E, , are not independent as the following example demonstrates. EXAMPLE 2.4-1. (Elastic constants) For an isotropic material, consider a cross section of material in the x1 -x2 plane which is subjected to pure shearing so that 4 = 12 is the only nonzero stress as illustrated in the gure 2.4-2. For the above conditions, the equation (2.4.17) reduces to the single equation e4 = e12 = s44 4 = s44 12 or = 12 12 and so the shear modulus is the ratio of the shear stress to the shear angle. Now rotate the axes through a 45 degree angle to a barred system of coordinates where x1 = x1 cos x2 sin x2 = x1 sin + x2 cos 250 Figure 2.4-2. Element subjected to pure shearing where = 4. Expanding the transformation equations (2.4.9) we nd that 1 = 11 = cos sin 12 + sin cos 21 = 12 = 4 2 = 22 = sin cos 12 sin cos 21 = 12 = 4 , and similarly e1 = e11 = e4 , In the barred system, the Hookes law becomes e1 = s11 1 + s12 2 or e2 = e22 = e4 . e4 = s11 4 s12 4 = s44 4 . Hence, the constants s11 , s12 , s44 are related by the relation s11 s12 = s44 or 1 1 + = . E E 2 (2.4.19) This is an important relation connecting the elastic constants associated with isotropic materials. The above transformation can also be applied to triclinic, aelotropic, orthotropic, and hexagonal materials to nd relationships between the elastic constants. Observe also that some texts postulate the existence of a strain energy function U which has the property that ij = U eij . In this case the strain energy function, in the single index notation, is written U = cij ei ej where cij and consequently sij are symmetric. In this case the previous discussed symmetries give the following results for the nonzero elastic compliances sij : 13 nonzero constants instead of 20 for aelotropic material, 9 nonzero constants instead of 12 for orthotropic material, and 6 nonzero constants instead of 7 for hexagonal material. This is because of the additional property that sij = sji be symmetric. 251 The previous discussion has shown that for an isotropic material the generalized Hookes law (constitutive equations) have the form e11 = e22 = e33 = e21 = e12 = e32 = e23 = e31 = e13 = 1 [11 (22 + 33 )] E 1 [22 (33 + 11 )] E 1 [33 (11 + 22 )] E , 1+ 12 E 1+ 23 E 1+ 13 E (2.4.20) where equation (2.4.19) holds. These equations can be expressed in the indicial notation and have the form eij = 1+ ij kk ij , E E (2.4.21) where kk = 11 + 22 + 33 is a stress invariant and ij is the Kronecker delta. We can solve for the stress in terms of the strain by performing a contraction on i and j in equation (2.4.21). This gives the dilatation eii = 1+ 3 1 2 ii kk = kk . E E E Note that from the result in equation (2.4.21) we are now able to solve for the stress in terms of the strain. We nd 1+ ij ekk ij E 1 2 E E eij = ij ekk ij 1+ (1 + )(1 2) E E or ij = eij + ekk ij . 1+ (1 + )(1 2) eij = The tensor equation (2.4.22) represents the six scalar equations 11 = 22 33 E [(1 )e11 + (e22 + e33 )] (1 + )(1 2) E [(1 )e22 + (e33 + e11 )] = (1 + )(1 2) E [(1 )e33 + (e22 + e11 )] = (1 + )(1 2) 12 = 13 23 E e12 1+ E e13 = 1+ E e23 . = 1+ (2.4.22) 252 Alternative Approach to Constitutive Equations The constitutive equation dened by Hookes generalized law for isotropic materials can be approached from another point of view. Consider the generalized Hookes law ij = cijkl ekl , i, j, k, l = 1, 2, 3. If we transform to a barred system of coordinates, we will have the new Hookes law ij = cijkl ekl , For an isotropic material we require that cijkl = cijkl . Tensors whose components are the same in all coordinate systems are called isotropic tensors. We have previously shown in Exercise 1.3, problem 18, that cpqrs = pq rs + (pr qs + ps qr ) + (pr qs ps qr ) is an isotropic tensor when we consider ane type transformations. If we further require the symmetry conditions found in equations (2.4.3) be satised, we nd that = 0 and consequently the generalized Hookes law must have the form pq = cpqrs ers = [pq rs + (pr qs + ps qr )] ers pq = pq err + (epq + eqp ) or pq = 2epq + err pq , (2.4.23) i, j, k, l = 1, 2, 3. where err = e11 + e22 + e33 = is the dilatation. The constants and are called Lames constants. Comparing the equation (2.4.22) with equation (2.4.23) we nd that the constants and satisfy the relations = E 2(1 + ) = E . (1 + )(1 2) (2.4.24) In addition to the constants E, , , , it is sometimes convenient to introduce the constant k, called the bulk modulus of elasticity, (Exercise 2.3, problem 23), dened by k= E . 3(1 2) (2.4.25) The stress-strain constitutive equation (2.4.23) was derived using Cartesian tensors. To generalize the equation (2.4.23) we consider a transformation from a Cartesian coordinate system y i , i = 1, 2, 3 to a general coordinate system xi , i = 1, 2, 3. We employ the relations g ij = and mn = ij y i y j , xm xn y m y m , xi xj g ij = xi xj y m y m xi xj y r y q emn = eij y i y j , xm xn or erq = eij 253 and convert equation (2.4.23) to a more generalized form. Multiply equation (2.4.23) by the result mn = which can be simplied to the form mn = g mn eij g ij + (emn + enm ) . Dropping the bar notation, we have mn = gmn g ij eij + (emn + enm ) . The contravariant form of this equation is sr = g sr g ij eij + (g ms g nr + g ns g mr ) emn . Employing the equations (2.4.24) the above result can also be expressed in the form rs = E 2(1 + ) g ms g nr + g ns g mr + 2 g sr g mn emn . 1 2 (2.4.26) y q y q err + (emn + enm ) , xm xn y p y q and verify xm xn This is a more general form for the stress-strain constitutive equations which is valid in all coordinate systems. Multiplying by gsk and employing the use of associative tensors, one can verify i j = E 1+ ei + j em i 1 2 m j or i i j = 2ei + em j , j m are alternate forms for the equation (2.4.26). As an exercise, solve for the strains in terms of the stresses and show that i mi Eei = (1 + )j m j . j EXAMPLE 2.4-2. (Hookes law) Let us construct a simple example to test the results we have developed so far. Consider the tension in a cylindrical bar illustrated in the gure 2.4-3. Figure 2.4-3. Stress in a cylindrical bar 254 Assume that F ij = 0 0 A 0 0 0 0 0 0 where F is the constant applied force and A is the cross sectional area of the cylinder. Consequently, the generalized Hookes law (2.4.21) produces the nonzero strains e11 = e22 e33 From these equations we obtain: The rst part of Hookes law 11 = Ee11 or The second part of Hookes law e22 e33 lateral contraction = = = = Poissons ratio. longitudinal extension e11 e11 This example demonstrates that the generalized Hookes law for homogeneous and isotropic materials reduces to our previous one dimensional result given in (2.3.1) and (2.3.2). F = Ee11 . A 1+ 11 11 (11 + 22 + 33 ) = E E E 11 = E 11 = E Basic Equations of Elasticity Assuming the density is constant, the basic equations of elasticity reduce to the equations representing conservation of linear momentum and angular momentum together with the strain-displacement relations and constitutive equations. In these equations the body forces are assumed known. These basic equations produce 15 equations in 15 unknowns and are a formidable set of equations to solve. Methods for solving these simultaneous equations are: 1) Express the linear momentum equations in terms of the displacements ui and obtain a system of partial dierential equations. Solve the system of partial dierential equations for the displacements ui and then calculate the corresponding strains. The strains can be used to calculate the stresses from the constitutive equations. 2) Solve for the stresses and from the stresses calculate the strains and from the strains calculate the displacements. This converse problem requires some additional considerations which will be addressed shortly. 255 Basic Equations of Linear Elasticity Conservation of linear momentum. ij + bj = f j ,i j = 1, 2, 3. (2.4.27(a)) where ij is the stress tensor, bj is the body force per unit mass and f j is the acceleration. If there is no motion, then f j = 0 and these equations reduce to the equilibrium equations ij + bj = 0 ,i Conservation of angular momentum. Strain tensor. eij = j = 1, 2, 3. ij = ji (2.4.28) (2.4.27(b)) 1 (ui,j + uj,i ) 2 where ui denotes the displacement eld. Constitutive equation. For a linear elastic isotropic material we have i j = Ei E e+ ek i 1 + j (1 + )(1 2) k j i, j = 1, 2, 3 (2.4.29(a)) or its equivalent form i i j = 2ei + er j j r i, j = 1, 2, 3, (2.4.29(b)) where er is the dilatation. This produces 15 equations for the 15 unknowns r u1 , u2 , u3 , 11 , 12 , 13 , 22 , 23 , 33 , e11 , e12 , e13 , e22 , e23 , e33 , which represents 3 displacements, 6 strains and 6 stresses. In the above equations it is assumed that the body forces are known. Naviers Equations The equations (2.4.27) through (2.4.29) can be combined and written as one set of equations. The resulting equations are known as Naviers equations for the displacements ui over the range i = 1, 2, 3. To derive the Naviers equations in Cartesian coordinates, we write the equations (2.4.27),(2.4.28) and (2.4.29) in Cartesian coordinates. We then calculate ij,j in terms of the displacements ui and substitute the results into the momentum equation (2.4.27(a)). Dierentiation of the constitutive equations (2.4.29(b)) produces ij,j = 2eij,j + ekk,j ij . (2.4.30) 256 A contraction of the strain produces the dilatation err = 1 (ur,r + ur,r ) = ur,r 2 (2.4.31) From the dilatation we calculate the covariant derivative ekk,j = uk,kj . Employing the strain relation from equation (2.4.28), we calculate the covariant derivative eij,j = 1 (ui,jj + uj,ij ). 2 (2.4.33) (2.4.32) These results allow us to express the covariant derivative of the stress in terms of the displacement eld. We nd ij,j = [ui,jj + uj,ij ] + ij uk,kj or ij,j = ( + )uk,ki + ui,jj . (2.4.34) Substituting equation (2.4.34) into the linear momentum equation produces the Navier equations: ( + )uk,ki + ui,jj + bi = fi , In vector form these equations can be expressed ( + ) ( u) + 2 u + b = f , (2.4.36) i = 1, 2, 3. (2.4.35) where u is the displacement vector, b is the body force per unit mass and f is the acceleration. In Cartesian coordinates these equations have the form: ( + ) for i = 1, 2, 3, where 2 ui = 2 ui 2 ui 2 ui + + . x1 2 x2 2 x3 2 2 u1 2 u2 2 u3 + + x1 xi x2 xi x3 xi + 2 ui + bi = 2 ui , t2 The Navier equations must be satised by a set of functions ui = ui (x1 , x2 , x3 ) which represent the displacement at each point inside some prescribed region R. Knowing the displacement eld we can calculate the strain eld directly using the equation (2.4.28). Knowledge of the strain eld enables us to construct the corresponding stress eld from the constitutive equations. In the absence of body forces, such as gravity, the solution to equation (2.4.36) can be represented in the form u = u (1) + u (2) , where u (1) satises div u (1) = u (1) = 0 and the vector u (2) satises curl u (2) = u (2) = 0. The vector eld u (1) is called a solenoidal eld, while the vector eld u (2) is called an irrotational eld. Substituting u into the equation (2.4.36) and setting b = 0, we nd in Cartesian coordinates that 2 u (1) 2 u (2) + 2 t t2 = ( + ) u (2) + 2 u (1) + 2 u (2) . (2.4.37) 257 The vector eld u (1) can be eliminated from equation (2.4.37) by taking the divergence of both sides of the equation. This produces 2 u (2) = ( + )2 ( u (2) ) + 2 u (2) . t2 The displacement eld is assumed to be continuous and so we can interchange the order of the operators 2 and and write This last equation implies that 2 u (2) = ( + 2)2 u(2) t2 and consequently, u (2) is a vector wave which moves with the speed eld u (2) 2 u (2) ( + 2)2 u (2) t2 = 0. ( + 2)/ . Similarly, when the vector is eliminated from the equation (2.4.37), by taking the curl of both sides, we nd the vector u (1) also satises a wave equation having the form 2 u (1) = 2 u (1) . t2 This later wave moves with the speed a shearing wave. The exercises 30 through 38 enable us to write the Naviers equations in Cartesian, cylindrical or spherical coordinates. In particular, we have for cartesian coordinates 2w 2u 2u 2u 2u 2v 2u + ) + ( 2 + 2 + 2 ) + bx = + 2 x xy xz x y z t2 2 2 2 2 2 2 v v u w v v 2v + 2+ ) + ( 2 + 2 + 2 ) + by = ( + )( xy y yz x y z t2 2 2 2 2 2 2 v w u w w w 2w + + ( + )( ) + ( 2 + + ) + bz = xz yz z 2 x y 2 z 2 t2 ( + )( and in cylindrical coordinates uz 1 1 u (rur ) + + r r r r z 2 2 2 1 ur 1 ur ur ur 2 u ur +2 ) + br + 2 2 ( 2 + r r r r 2 z 2 r r uz 1 1 1 u (rur ) + + ( + ) r r r r z 2 2 2 1 u u 1 u u 2 ur u +2 2 ) + b + +2 ( 2 + r r r r 2 z 2 r r uz 1 1 u (rur ) + + ( + ) z r r r z 1 2 uz 1 uz 2 uz 2 uz +2 + ) + bz ( 2 + r r r r 2 z 2 ( + ) + = + = + = 2 uz t2 2 u t2 2 ur t2 / . The vector u (2) is a compressive wave, while the wave u (1) is 258 and in spherical coordinates 1 2 1 1 u ( u ) + (u sin ) + 2 sin sin 2u cot 2 2 u 2 u ) + b 2 (2 u 2 u 2 2 sin 1 2 1 1 1 u ( u ) + (u sin ) + ( + ) 2 sin sin u 2 u 2 cos u 2 22 ) + b (2 u + 2 sin sin2 1 2 1 1 1 u ( u ) + (u sin ) + ( + ) sin 2 sin sin 2 cos u 1 2 u +2 2 ) + b (2 u 2 2 u + 2 sin sin sin ( + ) where 2 is determined from either equation (2.1.12) or (2.1.13). Boundary Conditions In elasticity the body forces per unit mass (bi , i = 1, 2, 3) are assumed known. In addition one of the following type of boundary conditions is usually prescribed: The displacements ui , is desired. The stresses (surface tractions) are prescribed on the boundary of the region R over which a solution is desired. The displacements ui , i = 1, 2, 3 are given over one portion of the boundary and stresses (surface tractions) are specied over the remaining portion of the boundary. This type of boundary condition is known as a mixed boundary condition. General Solution of Naviers Equations There has been derived a general solution to the Naviers equations. It is known as the Papkovich-Neuber solution. In the case of a solid in equilibrium one must solve the equilibrium equations ( + ) ( u) + 2 u + b = 0 or 1 1 ( u) + b = 0 ( = ) 2 u + 1 2 2 i = 1, 2, 3 are prescribed on the boundary of the region R over which a solution + = + = + = 2 u t2 2 u t2 2 u t2 (2.4.38) 259 THEOREM is u = grad ( + r ) 4(1 ) where and are continuous solutions of the equations 2 = rb 4(1 ) and 2 = b 4(1 ) (2.4.40) (2.4.39) A general elastostatic solution of the equation (2.4.38) in terms of harmonic potentials , with r = x e1 + y e2 + z e3 a position vector to a general point (x, y, z) within the continuum. Proof: First we write equation (2.4.38) in the tensor form ui,kk + 1 (uj,j ) ,i + bi = 0 1 2 (2.4.41) Now our problem is to show that equation (2.4.39), in tensor form, ui = ,i + (xj j ),i 4(1 )i is a solution of equation (2.4.41). Toward this purpose, we dierentiate equation (2.4.42) ui,k = ,ik + (xj j ),ik 4(1 )i,k and then contract on i and k giving ui,i = ,ii + (xj j ),ii 4(1 )i,i . Employing the identity (xj j ),ii = 2i,i + xi i,kk the equation (2.4.44) becomes ui,i = ,ii + 2i,i + xi i,kk 4(1 )i,i . By dierentiating equation (2.4.43) we establish that ui,kk = ,ikk + (xj j ),ikk 4(1 )i,kk = (,kk ),i + ((xj j ),kk ),i 4(1 )i,kk = [,kk + 2j,j + xj j,kk ],i 4(1 )i,kk . We use the hypothesis ,kk = xj Fj 4(1 ) Fj , 4(1 ) (2.4.46) (2.4.45) (2.4.44) (2.4.43) (2.4.42) and j,kk = and simplify the equation (2.4.46) to the form ui,kk = 2j,ji 4(1 )i,kk . Also by dierentiating (2.4.45) one can establish that uj,ji = (,jj ),i + 2j,ji + (xj j,kk ),i 4(1 )j,ji = xj Fj 4(1 ) + 2j,ji + ,i (2.4.47) xj Fj 4(1 ) 4(1 )j,ji ,i (2.4.48) = 2(1 2)j,ji . 260 Finally, from the equations (2.4.47) and (2.4.48) we obtain the desired result that ui,kk + 1 Fi uj,ji + = 0. 1 2 Consequently, the equation (2.4.39) is a solution of equation (2.4.38). As a special case of the above theorem, note that when the body forces are zero, the equations (2.4.40) become 2 = 0 and 2 = 0. In this case, we nd that equation (2.4.39) is a solution of equation (2.4.38) provided and each component of are harmonic functions. The Papkovich-Neuber potentials are used together with complex variable theory to solve various two-dimensional elastostatic problems of elasticity. Note also that the Papkovich-Neuber potentials are not unique as dierent combinations of and can produce the same value for u. Compatibility Equations If we know or can derive the displacement eld ui , i = 1, 2, 3 we can then calculate the components of the strain tensor eij = 1 (ui,j + uj,i ). 2 (2.4.49) Knowing the strain components, the stress is found using the constitutive relations. Consider the converse problem where the strain tensor is given or implied due to the assigned stress eld and we are asked to determine the displacement eld ui , i = 1, 2, 3. Is this a realistic request? Is it even possible to solve for three displacements given six strain components? It turns out that certain mathematical restrictions must be placed upon the strain components in order that the inverse problem have a solution. These mathematical restrictions are known as compatibility equations. That is, we cannot arbitrarily assign six strain components eij and expect to nd a displacement eld ui , i = 1, 2, 3 with three components which satises the strain relation as given in equation (2.4.49). EXAMPLE 2.4-3. Suppose we are given the two partial dierential equations, u =x+y x and u = x3 . y Can we solve for u = u(x, y)? The answer to this question is no, because the given equations are inconsistent. The inconsistency is illustrated if we calculate the mixed second derivatives from each equation. We 2u 2u = 1 and from the second equation we calculate = 3x2 . These nd from the rst equation that xy yx mixed second partial derivatives are unequal for all x dierent from 3/3. In general, if we have two rst u u order partial dierential equations = f (x, y) and = g(x, y), then for consistency (integrability of x y the equations) we require that the mixed partial derivatives f 2u g 2u = = = xy y yx x be equal to one another for all x and y values over the domain for which the solution is desired. This is an example of a compatibility equation. 261 A similar situation occurs in two dimensions for a material in a state of strain where ezz = ezx = ezy = 0, called plane strain. In this case, are we allowed to arbitrarily assign values to the strains exx , eyy and exy and from these strains determine the displacement eld u = u(x, y) and v = v(x, y) in the x and ydirections? Let us try to answer this question. Assume a state of plane strain where ezz = ezx = ezy = 0. Further, let us assign 3 arbitrary functional values f, g, h such that exx = u = f (x, y), x exy = 1 2 u v + y x = g(x, y), eyy = v = h(x, y). y We must now decide whether these equations are consistent. That is, will we be able to solve for the displacement eld u = u(x, y) and v = v(x, y)? To answer this question, let us derive a compatibility equation (integrability condition). From the given equations we can calculate the following partial derivatives 2 exx 3u 2f = = 2 2 y xy y 2 2 eyy 3v 2h = = 2 2 x yx x2 2 3 u exy 3v 2g = . 2 + =2 xy xy 2 yx2 xy This last equation gives us the compatibility equation 2 2 exx 2 eyy 2 exy = + xy y 2 x2 or the functions g, f, h must satisfy the relation 2 2f 2h 2g = + 2. xy y 2 x Cartesian Derivation of Compatibility Equations If the displacement eld ui , i = 1, 2, 3 is known we can derive the strain and rotation tensors eij = 1 (ui,j + uj,i ) 2 and ij = 1 (ui,j uj,i ). 2 (2.4.50) Now work backwards. Assume the strain and rotation tensors are given and ask the question, Is it possible to solve for the displacement eld ui , i = 1, 2, 3? If we view the equation (2.4.50) as a system of equations with unknowns eij , ij and ui and if by some means we can eliminate the unknowns ij and ui then we will be left with equations which must be satised by the strains eij . These equations are known as the compatibility equations and they represent conditions which the strain components must satisfy in order that a displacement function exist and the equations (2.4.37) are satised. Let us see if we can operate upon the equations (2.4.50) to eliminate the quantities ui and ij and hence derive the compatibility equations. Addition of the equations (2.4.50) produces ui,j = ui = eij + ij . xj (2.4.51) 262 Dierentiate this expression with respect to xk and verify the result 2 ui eij ij = + . xj xk xk xk and (2.4.52) We further assume that the displacement eld is continuous so that the mixed partial derivatives are equal 2 ui 2 ui = . xj xk xk xj 2 ui eik ik = + . xk xj xj xj result eij eik ik ij = (2.4.55) xk xj xj xk ik ij jk satises = , the equation (2.4.55) simplies to the xj xk xi (2.4.56) (2.4.53) Interchanging j and k in equation (2.4.52) gives us (2.4.54) Equating the second derivatives from equations (2.4.54) and (2.4.52) and rearranging terms produces the Making the observation that ij form The term involving jk eik jk eij = . xk xj xi can be eliminated by using the mixed partial derivative relation 2 jk 2 jk = . (2.4.57) xi xm xm xi To derive the compatibility equations we dierentiate equation (2.4.56) with respect to xm and then interchanging the indices i and m and substitute the results into equation (2.4.57). This will produce the compatibility equations 2 eij 2 emk 2 eik 2 emj + = 0. (2.4.58) xm xk xi xj xm xj xi xk This is a set of 81 partial dierential equations which must be satised by the strain components. Fortunately, due to symmetry considerations only 6 of these 81 equations are distinct. These 6 distinct equations are known as the St. Venants compatibility equations and can be written as 2 e11 2 e12 2 e23 2 e31 = + 2 x2 x3 x1 x3 x1 x1 x2 2 2 2 e22 e23 e31 2 e12 = + 2 x1 x3 x2 x1 x2 x2 x3 2 e33 2 e31 2 e12 2 e23 = + x1 x2 x3 x2 x3 2 x3 x1 2 2 2 e12 e11 e22 2 = + x1 x2 x2 2 x1 2 2 2 e23 e22 2 e33 2 = + x2 x3 x3 2 x2 2 2 2 e31 e33 2 e11 2 = + . 2 x3 x1 x1 x3 2 Observe that the fourth compatibility equation is the same as that derived in the example 2.4-3. These compatibility equations can also be expressed in the indicial form eij,km + emk,ji eik,jm emj,ki = 0. (2.4.60) (2.4.59) 263 Compatibility Equations in Terms of Stress In the generalized Hookes law, equation (2.4.29), we can solve for the strain in terms of stress. This in turn will give rise to a representation of the compatibility equations in terms of stress. The resulting equations are known as the Beltrami-Michell equations. Utilizing the strain-stress relation eij = 1+ ij kk ij E E we substitute for the strain in the equations (2.4.60) and rearrange terms to produce the result ij,km + mk,ji ik,jm mj,ki = [ij nn,km + mk nn,ji ik nn,jm mj nn,ki ] . 1+ (2.4.61) Now only 6 of these 81 equations are linearly independent. It can be shown that the 6 linearly independent equations are equivalent to the equations obtained by setting k = m and summing over the repeated indices. We then obtain the equations ij,mm + mm,ij (im,m ),j (mj,m ),i = [ij nn,mm + nn,ij ] . 1+ Employing the equilibrium equation ij,i + bj = 0 the above result can be written in the form ij,mm + or 2 ij + 1 kk,ij ij nn,mm = ( bi ),j ( bj ),i . 1+ 1+ 1 kk,ij ij nn,mm = ( bi ),j ( bj ),i 1+ 1+ This result can be further simplied by observing that a contraction on the indices k and i in equation (2.4.61) followed by a contraction on the indices m and j produces the result ij,ij = 1 nn,jj . 1+ Consequently, the Beltrami-Michell equations can be written in the form 2 ij + 1 pp,ij = ij ( bk ) ,k ( bi ) ,j ( bj ) ,i . 1+ 1 (2.4.62) Their derivation is left as an exercise. The Beltrami-Michell equations together with the linear momentum (equilibrium) equations ij,i + bj = 0 represent 9 equations in six unknown stresses. This combinations of equations is dicult to handle. An easier combination of equations in terms of stress functions will be developed shortly. The Navier equations with boundary conditions are dicult to solve in general. Let us take the momentum equations (2.4.27(a)), the strain relations (2.4.28) and constitutive equations (Hookes law) (2.4.29) and make simplifying assumptions so that a more tractable systems results. 264 Plane Strain The plane strain assumption usually is applied in situations where there is a cylindrical shaped body whose axis is parallel to the z axis and loads are applied along the zdirection. In any x-y plane we assume that the surface tractions and body forces are independent of z. We set all strains with a subscript z equal to zero. Further, all solutions for the stresses, strains and displacements are assumed to be only functions of x and y and independent of z. Note that in plane strain the stress zz is dierent from zero. In Cartesian coordinates the strain tensor is expressible in terms of its physical components which can be represented in the matrix form e11 e21 e31 e12 e22 e32 e13 exx e23 = eyx e33 ezx exy eyy ezy exz eyz . ezz If we assume that all strains which contain a subscript z are zero and the remaining strain components are functions of only x and y, we obtain a state of plane strain. For a state of plane strain, the stress components are obtained from the constitutive equations. The condition of plane strain reduces the constitutive equations to the form: exx = eyy = 0= exy = eyx = ezy = eyz = ezx = exz = where xx , yy , 1 [xx (yy + zz )] E 1 [yy (zz + xx )] E 1 [zz (xx + yy )] E 1+ xy E 1+ yz = 0 E 1+ xz = 0 E xy , xz , xx = yy zz xy xz E [(1 )exx + eyy ] (1 + )(1 2) E [(1 )eyy + exx ] = (1 + )(1 2) E [(eyy + exx )] = (1 + )(1 2) E exy = 1+ =0 (2.4.63) yz = 0 zz , yz are the physical components of the stress. The above constitutive equations imply that for a state of plane strain we will have zz = (xx + yy ) 1+ exx = [(1 )xx yy ] E 1+ [(1 )yy xx ] eyy = E 1+ xy . exy = E Also under these conditions the compatibility equations reduce to 2 exx 2 eyy 2 exy . + =2 2 2 y x xy 265 Plane Stress An assumption of plane stress is usually applied to thin at plates. The plate thinness is assumed to be in the zdirection and loads are applied perpendicular to z. Under these conditions all stress components with a subscript z are assumed to be zero. The remaining stress components are then treated as functions of x and y. In Cartesian coordinates the stress tensor is expressible in terms of its physical components and can be represented by the matrix 11 21 31 12 22 32 xx 13 23 = yx 33 zx xy yy zy xz yz . zz If we assume that all the stresses with a subscript z are zero and the remaining stresses are only functions of x and y we obtain a state of plane stress. The constitutive equations simplify if we assume a state of plane stress. These simplied equations are exx = eyy ezz exy exz 1 xx yy E E 1 = yy xx E E = (xx + yy ) E 1+ xy = E =0 xx = yy zz xy yz E [exx + eyy ] 1 2 E = [eyy + exx ] 1 2 = 0 = (1 )ezz + (exx + eyy ) E exy = 1+ =0 (2.4.64) eyz = 0. xz = 0 For a state of plane stress the compatibility equations reduce to 2 exx 2 eyy 2 exy + =2 2 2 y x xy and the three additional equations 2 ezz = 0, x2 2 ezz = 0, y 2 2 ezz = 0. xy (2.4.65) These three additional equations complicate the plane stress problem. Airy Stress Function In Cartesian coordinates we examine the equilibrium equations (2.4.25(b)) under the conditions of plane strain. In terms of physical components we nd that these equations reduce to xx xy + + bx = 0, x y yy yx + + by = 0, x y zz = 0. z or bi = V ,i The last equation is satised since zz is a function of x and y. If we further assume that the body forces are conservative and derivable from a potential function V by the operation b = grad V we can express the above equilibrium equations in the form: xx xy V + =0 x y x yx yy V + =0 x y y (2.4.66) 266 We will consider these equations together with the compatibility equations (2.4.65). The equations (2.4.66) will be automatically satised if we introduce a scalar function = (x, y) and assume that the stresses are derivable from this function and the potential function V according to the rules: xx = 2 +V y 2 xy = 2 xy yy = 2 + V. x2 (2.4.67) The function = (x, y) is called the Airy stress function after the English astronomer and mathematician Sir George Airy (18011892). Since the equations (2.4.67) satisfy the equilibrium equations we need only consider the compatibility equation(s). For a state of plane strain we substitute the relations (2.4.63) into the compatibility equation (2.4.65) and write the compatibility equation in terms of stresses. We then substitute the relations (2.4.67) and express the compatibility equation in terms of the Airy stress function . These substitutions are left as exercises. After all these substitutions the compatibility equation, for a state of plane strain, reduces to the form 4 4 4 1 2 +2 2 2 + 4 + 4 x x y y 1 2V 2V + 2 x y 2 = 0. (2.4.68) In the special case where there are no body forces we have V = 0 and equation (2.4.68) is further simplied to the biharmonic equation. 4 4 4 + 2 2 2 + 4 = 0. x4 x y y In polar coordinates the biharmonic equation is written 4 = 4 = 2 (2 ) = 2 1 2 1 +2 2 + 2 r r r r 2 1 1 2 +2 2 + 2 r r r r = 0. (2.4.69) For conditions of plane stress, we can again introduce an Airy stress function using the equations (2.4.67). However, an exact solution of the plane stress problem which satises all the compatibility equations is dicult to obtain. By removing the assumptions that xx , yy , xy are independent of z, and neglecting body forces, it can be shown that for symmetrically distributed external loads the stress function can be represented in the form = z 2 2 2(1 + ) (2.4.70) where is a solution of the biharmonic equation 4 = 0. Observe that if z is very small, (the condition of a thin plate), then equation (2.4.70) gives the approximation . Under these conditions, we obtain the approximate solution by using only the compatibility equation (2.4.65) together with the stress function dened by equations (2.4.67) with V = 0. Note that the solution we obtain from equation (2.4.69) does not satisfy all the compatibility equations, however, it does give an excellent rst approximation to the solution in the case where the plate is very thin. In general, for plane strain or plane stress problems, the equation (2.4.68) or (2.4.69) must be solved for the Airy stress function which is dened over some region R. In addition to specifying a region of the x, y plane, there are certain boundary conditions which must be satised. The boundary conditions specied for the stress will translate through the equations (2.4.67) to boundary conditions being specied for . In the special case where there are no body forces, both the problems for plane stress and plane strain are governed by the biharmonic dierential equation with appropriate boundary conditions. 267 EXAMPLE 2.4-4 Assume there exist a state of plane strain with zero body forces. For F11 , F12 , F22 constants, consider the function dened by = (x, y) = 1 F22 x2 2F12 xy + F11 y 2 . 2 This function is an Airy stress function because it satises the biharmonic equation 4 = 0. The resulting stress eld is xx = 2 = F11 y 2 yy = 2 = F22 x2 xy = 2 = F12 . xy This example, corresponds to stresses on an innite at plate and illustrates a situation where all the stress components are constants for all values of x and y. In this case, we have zz = (F11 +F22 ). The corresponding strain eld is obtained from the constitutive equations. We nd these strains are exx = 1+ [(1 )F11 F22 ] E eyy = 1+ [(1 )F22 F11 ] E exy = 1+ F12 . E The displacement eld is found to be u = u(x, y) = 1+ [(1 )F11 F22 ] x + E 1+ [(1 )F22 F11 ] y + v = v(x, y) = E 1+ E 1+ E F12 y + c1 y + c2 F12 x c1 x + c3 , with c1 , c2 , c3 constants, and is obtained by integrating the strain displacement equations given in Exercise 2.3, problem 2. EXAMPLE 2.4-5. A special case from the previous example is obtained by setting F22 = F12 = 0. This is the situation of an innite plate with only tension in the xdirection. In this special case we have = 1 F11 y 2 . Changing to polar coordinates we write 2 = (r, ) = F11 2 F11 2 2 r sin = r (1 cos 2). 2 4 The Exercise 2.4, problem 20, suggests we utilize the Airy equations in polar coordinates and calculate the stresses rr = r 1 2 1 F11 + 2 2 = F11 cos2 = (1 + cos 2) r r r 2 2 F11 (1 cos 2) = = F11 sin2 = 2 r 2 F11 1 1 2 = sin 2. =2 r r r 2 268 EXAMPLE 2.4-6. We now consider an innite plate with a circular hole x2 + y 2 = a2 which is traction free. Assume the plate has boundary conditions at innity dened by xx = F11 , the stress eld. Solution: The traction boundary condition at r = a is ti = mi nm or t1 = 11 n1 + 12 n2 and t2 = 12 n1 + 22 n2 . yy = 0, xy = 0. Find For polar coordinates we have n1 = nr = 1, n2 = n = 0 and so the traction free boundary conditions at the surface of the hole are written rr |r=a = 0 and r |r=a = 0. The results from the previous example are used as the boundary conditions at innity. Our problem is now to solve for the Airy stress function = (r, ) which is a solution of the biharmonic equation. The previous example 2.4-5 and the form of the boundary conditions at innity suggests that we assume a solution to the biharmonic equation which has the form = (r, ) = f1 (r) + f2 (r) cos 2, where f1 , f2 are unknown functions to be determined. Substituting the assumed solution into the biharmonic equation produces the equation d2 1d + dr2 r dr 1 f1 + f1 r + d2 4 1d + dr2 r dr r2 1 f2 f2 + f2 4 2 r r cos 2 = 0. We therefore require that f1 , f2 be chosen to satisfy the equations d2 1d + 2 dr r dr or r 4 f1 (iv) 1 f1 + f1 r =0 d2 4 1d 2 + 2 dr r dr r r 4 f2 (iv) 1 f2 f2 + f2 4 2 r r =0 + 2r3 f1 r2 f1 + rf1 = 0 + 2r3 f2 9r2 f2 + 9rf2 = 0 These equations are Cauchy type equations. Their solutions are obtained by assuming a solution of the form f1 = r and f2 = rm and then solving for the constants and m. We nd the general solutions of the above equations are f1 = c1 r2 ln r + c2 r2 + c3 ln r + c4 and f2 = c5 r2 + c6 r4 + c7 + c8 . r2 The constants ci , i = 1, . . . , 8 are now determined from the boundary conditions. The constant c4 can be arbitrary since the derivative of a constant is zero. The remaining constants are determined from the stress conditions. Using the results from Exercise 2.4, problem 20, we calculate the stresses rr = c1 (1 + 2 ln r) + 2c2 + r c3 c7 c8 2c5 + 6 4 + 4 2 cos 2 2 r r r c3 c7 2 = c1 (3 + 2 ln r) + 2c2 2 + 2c5 + 12c6 r + 6 4 cos 2 r r c7 c8 = 2c5 + 6c6 r2 6 4 2 2 sin 2. r r 269 The stresses are to remain bounded for all values of r and consequently we require c1 and c6 to be zero to avoid innite stresses for large values of r. The stress rr |r=a = 0 requires that 2c2 + The stress r |r=a = 0 requires that 2c5 6 c3 =0 a2 and 2c5 + 6 c7 c8 + 4 2 = 0. 4 a a c7 c8 2 2 = 0. 4 a a In the limit as r we require that the stresses must satisfy the boundary conditions from the previous F11 F11 example 2.4-5. This leads to the equations 2c2 = and 2c5 = . Solving the above system of equations 2 2 produces the Airy stress function = (r, ) = F11 2 a2 F11 + r F11 ln r + c4 + 4 4 2 F11 a2 F11 2 F11 a4 r 2 4 4r2 cos 2 and the corresponding stress eld is rr = r F11 a2 F11 a2 a4 1 2 + 1 + 3 4 4 2 cos 2 2 r 2 r r F11 a2 a4 = 1 3 4 + 2 2 sin 2 2 r r F11 F11 a2 a4 = 1+ 2 1 + 3 4 cos 2. 2 r 2 r There is a maximum stress = 3F11 at = /2, 3/2 and a minimum stress = F11 at = 0, . The eect of the circular hole has been to magnify the applied stress. The factor of 3 is known as a stress concentration factor. In general, sharp corners and unusually shaped boundaries produce much higher stress concentration factors than rounded boundaries. EXAMPLE 2.4-7. Consider an innite cylindrical tube, with inner radius R1 and the outer radius R0 , which is subjected to an internal pressure P1 and an external pressure P0 as illustrated in the gure 2.4-7. Find the stress and displacement elds. Solution: Let ur , u , uz denote the displacement eld. We assume that u = 0 and uz = 0 since the cylindrical surface r equal to a constant does not move in the or z directions. The displacement ur = ur (r) is assumed to depend only upon the radial distance r. Under these conditions the Navier equations become ( + 2) This equation has the solution ur = c1 err = dur , dr d dr 1d (rur ) r dr = 0. c2 r + and the strain components are found from the relations 2 r e = ur , r ezz = er = erz = ez = 0. The stresses are determined from Hookes law (the constitutive equations) and we write ij = ij + 2eij , 270 where = ur 1 ur + = (rur ) r r r r is the dilatation. These stresses are found to be rr = ( + )c1 2 c2 r2 = ( + )c1 + 2 c2 r2 zz = c1 r = rz = z = 0. We now apply the boundary conditions rr |r=R1 nr = ( + )c1 2 2 c2 = +P1 R1 and rr |r=R0 nr = ( + )c1 2 2 c2 = P0 . R0 Solving for the constants c1 and c2 we nd c1 = 2 2 R1 P1 R0 P0 2 R2 ) , ( + )(R0 1 c2 = 22 R1 R0 (P1 P0 ) . 2 2 2(R0 R1 ) This produces the displacement eld ur = 2 R1 P1 2 R2 ) 2(R0 1 r R2 +0 + r 2 R0 P0 2 R2 ) 2(R0 1 r R2 +1 + r , u = 0, uz = 0, and stress elds rr = zz 2 2 R1 P1 R0 R2 P0 1 2 20 2 2 R2 R0 r R0 R1 1 2 2 R P1 R0 R2 P0 = 21 2 1+ 2 20 2 R0 R1 r R0 R1 2 2 R1 P1 R0 P0 = 2 2 + R0 R1 2 R1 2 r 2 R1 1+ 2 r 1 rz = z = r = 0 EXAMPLE 2.4-8. By making simplifying assumptions the Navier equations can be reduced to a more tractable form. For example, we can reduce the Navier equations to a one dimensional problem by making the following assumptions 1. Cartesian coordinates x1 = x, 2. u1 = u1 (x, t), u2 = u3 = 0. x2 = y, x3 = z 3. There are no body forces. u1 (x, 0) =0 t 5. Boundary conditions of the displacement type u1 (0, t) = f (t), 4. Initial conditions of u1 (x, 0) = 0 and where f (t) is a specied function. These assumptions reduce the Navier equations to the single one dimensional wave equation 2 u1 2 u1 = 2 , 2 t x2 The solution of this equation is u1 (x, t) = f (t x/), 0, x t . x > t 2 = + 2 . 271 The solution represents a longitudinal elastic wave propagating in the xdirection with speed . The stress wave associated with this displacement is determined from the constitutive equations. We nd xx = ( + )exx = ( + ) This produces the stress wave xx = (+) f (t x/), 0, x t x > t . u1 . x Here there is a discontinuity in the stress wave front at x = t. Summary of Basic Equations of Elasticity The equilibrium equations for a continuum have been shown to have the form ij + bi = 0, where ,j bi are the body forces per unit mass and ij is the stress tensor. In addition to the above equations we have the constitutive equations ij = ekk ij + 2eij which is a generalized Hookes law relating stress to strain for a linear elastic isotropic material. The strain tensor is related to the displacement eld ui by 1 the strain equations eij = (ui,j + uj,i ) . These equations can be combined to obtain the Navier equations 2 ui,jj + ( + )uj,ji + bi = 0. The above equations must be satised at all interior points of the material body. A boundary value problem results when conditions on the displacement of the boundary are specied. That is, the Navier equations must be solved subject to the prescribed displacement boundary conditions. If conditions on the stress at the boundary are specied, then these prescribed stresses are called surface tractions and must satisfy the relations ti (n) = ij nj , where ni is a unit outward normal vector to the boundary. For surface tractions, we need to use the compatibility equations combined with the constitutive equations and equilibrium equations. This gives rise to the Beltrami-Michell equations of compatibility ij,kk + 1 kk,ij + (bi,j + bj,i ) + bk,k = 0. 1+ 1 Here we must solve for the stress components throughout the continuum where the above equations hold subject to the surface traction boundary conditions. Note that if an elasticity problem is formed in terms of the displacement functions, then the compatibility equations can be ignored. For mixed boundary value problems we must solve a system of equations consisting of the equilibrium equations, constitutive equations, and strain displacement equations. We must solve these equations subject to conditions where the displacements ui are prescribed on some portion(s) of the boundary and stresses are prescribed on the remaining portion(s) of the boundary. Mixed boundary value problems are more dicult to solve. For elastodynamic problems, the equilibrium equations are replaced by equations of motion. In this case we need a set of initial conditions as well as boundary conditions before attempting to solve our basic system of equations. 272 EXERCISE 2.4 1. Verify the generalized Hookes law constitutive equations for hexagonal materials. In the following problems the Youngs modulus E, Poissons ratio , the shear modulus or modulus of rigidity (sometimes denoted by G in Engineering texts), Lames constant and the bulk modulus of elasticity k are assumed to satisfy the equations (2.4.19), (2.4.24) and (2.4.25). Show that these relations imply the additional relations given in the problems 2 through 6. 2. E= (3 + 2) + (1 + )(1 2) E= E= 9k(k ) 3k 9k + 3k E= E = 2(1 + ) E = 3(1 2)k E 2 2 = 3k = 2(1 + ) 3(1 2) (1 + ) k= 3 3. 3k E 6k = 2( + ) = (E + )2 + 82 (E + ) 4 3k 2 = 2( + 3k) = E 3(1 2) E k= 3(3 E) 4. (E + )2 + 82 + (E + 3) 6 2 + 3 k= 3 k= 5. 3(k ) 2 (1 2) = 2 = 3k(1 2) 2(1 + ) 3Ek = 9k E k= k= = = (E + )2 + 82 + (E 3) 4 E = 2(1 + ) E (1 + )(1 2) 2 = 1 2 6. 3k 1+ (2 E) = E 3 = = 3k 2 3 3k(3k E) = 9k E = 7. The previous exercises 2 through 6 imply that the generalized Hookes law ij = 2eij + ij ekk is expressible in a variety of forms. From the set of constants (,,,E,k) we can select any two constants and then express Hookes law in terms of these constants. (a) Express the above Hookes law in terms of the constants E and . (b) Express the above Hookes law in terms of the constants k and E. (c) Express the above Hookes law in terms of physical components. Hint: The quantity ekk is an invariant hence all you need to know is how second order tensors are represented in terms of physical components. See also problems 10,11,12. 273 8. Verify the equations dening the stress for plane strain in Cartesian coordinates are xx = yy zz xy yz = xz E [(1 )exx + eyy ] (1 + )(1 2) E [(1 )eyy + exx ] = (1 + )(1 2) E [exx + eyy ] = (1 + )(1 2) E exy = 1+ =0 9. Verify the equations dening the stress for plane strain in polar coordinates are rr = zz r rz = z E [(1 )err + e ] (1 + )(1 2) E [(1 )e + err ] = (1 + )(1 2) E [err + e ] = (1 + )(1 2) E er = 1+ =0 10. Write out the independent components of Hookes generalized law for strain in terms of stress, and stress in terms of strain, in Cartesian coordinates. Express your results using the parameters and E. (Assume a linear elastic, homogeneous, isotropic material.) 11. Write out the independent components of Hookes generalized law for strain in terms of stress, and stress in terms of strain, in cylindrical coordinates. Express your results using the parameters and E. (Assume a linear elastic, homogeneous, isotropic material.) 12. Write out the independent components of Hookes generalized law for strain in terms of stress, and stress in terms of strain in spherical coordinates. Express your results using the parameters and E. (Assume a linear elastic, homogeneous, isotropic material.) 13. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane strain in Cartesian coordinates. Verify the equilibrium equations are xx xy + + bx = 0 x y yy yx + + by = 0 x y zz + bz = 0 z Hint: See problem 14, Exercise 2.3. 274 14 . For a linear elastic, homogeneous, isotropic material assume there exists a state of plane strain in polar coordinates. Verify the equilibrium equations are 1 r 1 rr + + (rr ) + br = 0 r r r 1 2 r + + r + b = 0 r r r zz + bz = 0 z Hint: See problem 15, Exercise 2.3. 15. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane stress in Cartesian coordinates. Verify the equilibrium equations are xx xy + + bx = 0 x y yy yx + + by = 0 x y 16. Determine the compatibility equations in terms of the Airy stress function when there exists a state of plane stress. Assume the body forces are derivable from a potential function V. 17. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane stress in polar coordinates. Verify the equilibrium equations are rr 1 r 1 + + (rr ) + br = 0 r r r 1 2 r + + r + b = 0 r r r 275 18. Figure 2.4-4 illustrates the state of equilibrium on an element in polar coordinates assumed to be of unit length in the z-direction. Verify the stresses given in the gure and then sum the forces in the r and directions to derive the same equilibrium laws developed in the previous exercise. Figure 2.4-4. Polar element in equilibrium. Hint: Resolve the stresses into components in the r and directions. Use the results that sin d 2 cos d 2 d 2 and 1 for small values of d. Sum forces and then divide by rdr d and take the limit as dr 0 and d 0. 19. Express each of the physical components of plane stress in polar coordinates, rr , , and r in terms of the physical components of stress in Cartesian coordinates xx , yy , xy . Hint: Consider the xa xb . transformation law ij = ab i x xj 20. Use the results from problem 19 and assume the stresses are derivable from the relations xx = V + 2 , y 2 xy = 2 , xy yy = V + 2 x2 where V is a potential function and is the Airy stress function. Show that upon changing to polar coordinates the Airy equations for stress become rr = V + 1 2 1 + 2 2, r r r r = 1 1 2 , r2 r r = V + 2 . r2 21. Verify that the Airy stress equations in polar coordinates, given in problem 20, satisfy the equilibrium equations in polar coordinates derived in problem 17. 276 22. In Cartesian coordinates show that the traction boundary conditions, equations (2.3.11), can be written in terms of the constants and as T1 = n1 ekk + 2n1 T2 = n2 ekk + n1 T3 = n3 ekk + n1 u1 u1 u1 u2 u3 + n2 + 1 + n3 +1 x1 x2 x x3 x u2 u2 u2 u1 u3 + 2 + 2n2 2 + n3 +2 x1 x x x3 x u3 u3 u3 u1 u2 + 3 + n2 + 3 + 2n3 3 x1 x x2 x x where (n1 , n2 , n3 ) are the direction cosines of the unit normal to the surface, u1 , u2 , u3 are the components of the displacements and T1 , T2 , T3 are the surface tractions. 23. Consider an innite plane subject to tension in the xdirection only. Assume a state of plane strain and let xx = T with xy = yy = 0. Find the strain components exx , eyy and exy . Also nd the displacement eld u = u(x, y) and v = v(x, y). 24. Consider an innite plane subject to tension in the y-direction only. Assume a state of plane strain and let yy = T with xx = xy = 0. Find the strain components exx , eyy and exy . Also nd the displacement eld u = u(x, y) and v = v(x, y). 25. Consider an innite plane subject to tension in both the x and y directions. Assume a state of plane strain and let xx = T , yy = T and xy = 0. Find the strain components exx , eyy and exy . Also nd the displacement eld u = u(x, y) and v = v(x, y). 26. An innite cylindrical rod of radius R0 has an external pressure P0 as illustrated in gure 2.5-5. Find the stress and displacement elds. Figure 2.4-5. External pressure on a rod. 277 Figure 2.4-6. Internal pressure on circular hole. Figure 2.4-7. Tube with internal and external pressure. 27. An innite plane has a circular hole of radius R1 with an internal pressure P1 as illustrated in the gure 2.4-6. Find the stress and displacement elds. 28. A tube of inner radius R1 and outer radius R0 has an internal pressure of P1 and an external pressure of P0 as illustrated in the gure 2.4-7. Verify the stress and displacement elds derived in example 2.4-7. 29. Use Cartesian tensors and combine the equations of equilibrium ij,j + bi = 0, Hookes law ij = 1 ekk ij + 2eij and the strain tensor eij = (ui,j + uj,i ) and derive the Navier equations of equilibrium 2 ij,j + bi = ( + ) where = e11 + e22 + e33 is the dilatation. 30. Show the Navier equations in problem 29 can be written in the tensor form ui,jj + ( + )uj,ji + bi = 0 or the vector form 2 u + ( + ) ( u) + b = 0. 2 ui + k k + bi = 0, xi x x 278 31. Show that in an orthogonal coordinate system the components of ( u) can be expressed in terms of physical components by the relation [ ( u)]i = 1 hi xi 1 (h2 h3 u(1)) (h1 h3 u(2)) (h1 h2 u(3)) + + h1 h2 h3 x1 x2 x3 32. Show that in orthogonal coordinates the components of 2 u can be written 2 u i = g jk ui,jk = Ai and in terms of physical components one can write 3 3 m 1 2 (hi u(i)) hi A(i) = 2 ij h2 xj xj m=1 j=1 j 3 (hm u(m)) xj m=1 m ip p jj 3 3 m jj m jp m=1 hm u(m) xj m ij 3 p=1 p=1 (hi u(i)) xm p ij 33. Use the results in problem 32 to show in Cartesian coordinates the physical components of [2 u]i = Ai can be represented 2 u e1 = A(1) = 2u 2u 2u + 2+ 2 x2 y z 2 2 v v 2v 2 u e2 = A(2) = + 2+ 2 x2 y z 2 2 w w 2w 2 u e3 = A(3) = + + x2 y 2 z 2 where (u, v, w) are the components of the displacement vector u. 34. Use the results in problem 32 to show in cylindrical coordinates the physical components of [2 u]i = Ai can be represented 1 2 u u2 2r r r 1 2 ur 2 2 2 u u e = A(2) = u + 2 r r 2 u ez = A(3) = 2 uz 2 u er = A(1) = 2 ur where ur , u , uz are the physical components of u and 2 = 1 2 2 2 1 + 2 2+ + r2 r r r z 2 35. Use the results in problem 32 to show in spherical coordinates the physical components of [2 u]i = Ai 2 cot 2 2 u 2 u u2 u 2 2 2 sin 1 2 u 2 cos u 2 u e = A(2) = 2 u + 2 2 u 2 2 sin sin 1 2 u 2 cos u +2 2 2 u e = A(3) = 2 u 2 2 u + 2 sin sin sin 2 u e = A(1) = 2 u where u , u , u are the physical components of u and where 2 = 2 1 2 cot 1 2 2 + 2 2+ 2 +2 2 + 2 sin 2 can be represented 279 36. Combine the results from problems 30,31,32 and 33 and write the Navier equations of equilibrium in Cartesian coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical components and then use these results, together with the results from Exercise 2.3, problems 2 and 14, to derive the Navier equations. 37. Combine the results from problems 30,31,32 and 34 and write the Navier equations of equilibrium in cylindrical coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical components and then use these results, together with the results from Exercise 2.3, problems 3 and 15, to derive the Navier equations. 38. Combine the results from problems 30,31,32 and 35 and write the Navier equations of equilibrium in spherical coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical components and then use these results, together with the results from Exercise 2.3, problems 4 and 16, to derive the Navier equations. 39. Assume b = grad V and let denote the Airy stress function dened by 2 xx =V + 2 y 2 yy =V + x2 2 xy = xy (a) Show that for conditions of plane strain the equilibrium equations in two dimensions are satised by the above denitions. (b) Express the compatibility equation 2 exx 2 eyy 2 exy + =2 y 2 x2 xy in terms of and V and show that 1 2 2 V = 0. 4 + 1 40. Consider the case where the body forces are conservative and derivable from a scalar potential function such that bi = V,i . Show that under conditions of plane strain in rectangular Cartesian coordinates the 1 2 V compatibility equation e11,22 + e22,11 = 2e12,12 can be reduced to the form 2 ii = , i = 1, 2 1 involving the stresses and the potential. Hint: Dierentiate the equilibrium equations. i i 41. Use the relation j = 2ei + em j and solve for the strain in terms of the stress. j m 42. Derive the equation (2.4.26) from the equation (2.4.23). 43. In two dimensions assume that the body forces are derivable from a potential function V and im jn bi = g ij V ,j . Also assume that the stress is derivable from the Airy stress function and the potential function by employing the relations ij = pq um,n + g ij V i, j, m, n = 1, 2 where um = ,m and is the two dimensional epsilon permutation symbol and all indices have the range 1,2. im jn (a) Show that (b) Show that (m ) ,nj = 0. ij ,j = bi . (c) Verify the stress laws for cylindrical and Cartesian coordinates given in problem 20 by using the above expression for ij . Hint: Expand the contravariant derivative and convert all terms to physical components. Also recall that ij = 1 eij . g 280 44. Consider a material with body forces per unit volume F i , i = 1, 2, 3 and surface tractions denoted by r = rj nj , where nj is a unit surface normal. Further, let ui denote a small displacement vector associated with a small variation in the strain eij . (a) Show the work done during a small variation in strain is W = WB + WS where WB = V F i ui d is a volume integral representing the work done by the body forces and WS = integral representing the work done by the surface forces. 1 2 1 2 S r ur dS is a surface (b) Using the Gauss divergence theorem show that the work done can be represented as W = cijmn [emn eij ] d V or W = ij eij d. V The scalar quantity 1 ij eij is called the strain energy density or strain energy per unit volume. 2 Hint: Interchange subscripts, add terms and calculate 2W = V ij [ui,j + uj,i ] d. 45. Consider a spherical shell subjected to an internal pressure pi and external pressure po . Let a denote the inner radius and b the outer radius of the spherical shell. Find the displacement and stress elds in spherical coordinates (, , ). Hint: Assume symmetry in the and directions and let the physical components of displacements satisfy the relations u = u (), 46. u = u = 0. (a) Verify the average normal stress is proportional to the dilatation, where the proportionality E 1i 12 3 ei i constant is the bulk modulus of elasticity. i.e. Show that 1 i = 3 = kei where k is the bulk modulus i of elasticity. (b) Dene the quantities of strain deviation and stress deviation in terms of the average normal stress i s = 1 i and average cubic dilatation e = 1 ei as follows 3 3i i i = ei ej j j i i si = j sj j strain deviator stress deviator Show that zero results when a contraction is performed on the stress and strain deviators. (The above denitions are used to split the strain tensor into two parts. One part represents pure dilatation and the other part represents pure distortion.) (c) Show that (1 2)s = Ee or s = (3 + 2)e (d) Express Hookes law in terms of the strain and stress deviator and show i i E(i + ej ) = (1 + )si + (1 2)sj j j which simplies to si = 2i . j j 47. Show the strain energy density (problem 44) can be written in terms of the stress and strain deviators (problem 46) and W= and from Hookes law W= 3 2 ((3 + 2)e2 + V 1 2 ij eij d = V 1 2 (3se + sij ij ) d V 2 ij ij ) d. 3 281 48. Find the stress rr ,r and in an innite plate with a small circular hole, which is traction free, when the plate is subjected to a pure shearing force F12 . Determine the maximum stress. 49. Show that in terms of E and C1111 = E(1 ) (1 + )(1 2) C1122 = E (1 + )(1 2) C1212 = E 2(1 + ) 50. Show that in Cartesian coordinates the quantity S = xx yy + yy zz + zz xx (xy )2 (yz )2 (xz )2 is a stress invariant. Hint: First verify that in tensor form S = 1 (ii jj ij ij ). 2 51. Show that in Cartesian coordinates for a state of plane strain where the displacements are given by u = u(x, y),v = v(x, y) and w = 0, the stress components must satisfy the equations xy xx + + bx =0 x y yy yx + + by =0 x y 2 (xx + yy ) = 1 bx by + x y 52. Show that in Cartesian coordinates for a state of plane stress where xx = xx (x, y), yy = yy (x, y), xy = xy (x, y) and xz = yz = zz = 0 the stress components must satisfy xx xy + + bx =0 x y yy yx + + by =0 x y 2 (xx + yy ) = ( + 1) bx by + x y 282 2.5 CONTINUUM MECHANICS (FLUIDS) Let us consider a uid medium and use Cartesian tensors to derive the mathematical equations that describe how a uid behaves. A uid continuum, like a solid continuum, is characterized by equations describing: 1. Conservation of linear momentum ij,j + bi = vi 2. Conservation of angular momentum ij = ji . 3. Conservation of mass (continuity equation) vi + vi + =0 t xi xi In the above equations vi , i = 1, 2, 3 is a velocity eld, have dimensions [vj ] = cm/sec2 , [bj ] = dynes/g, [ij ] = dyne/cm2 , [ ] = g/cm3 . (2.5.3) or D + V = 0. Dt (2.5.2) (2.5.1) is the density of the uid, ij is the stress tensor and bj is an external force per unit mass. In the cgs system of units of measurement, the above quantities The displacement eld ui , i = 1, 2, 3 can be represented in terms of the velocity eld vi , i = 1, 2, 3, by the relation ui = 0 t vi dt. (2.5.4) The strain tensor components of the medium can then be represented in terms of the velocity eld as eij = where 1 (vi,j + vj,i ) 2 is called the rate of deformation tensor , velocity strain tensor, or rate of strain tensor. Dij = (2.5.6) 1 (ui,j + uj,i ) = 2 t 0 1 (vi,j + vj,i ) dt = 2 t Dij dt, 0 (2.5.5) Note the dierence in the equations describing a solid continuum compared with those for a uid continuum. In describing a solid continuum we were primarily interested in calculating the displacement eld ui , i = 1, 2, 3 when the continuum was subjected to external forces. In describing a uid medium, we calculate the velocity eld vi , i = 1, 2, 3 when the continuum is subjected to external forces. We therefore replace the strain tensor relations by the velocity strain tensor relations in all future considerations concerning the study of uid motion. Constitutive Equations for Fluids In addition to the above basic equations, we will need a set of constitutive equations which describe the material properties of the uid. Toward this purpose consider an arbitrary point within the uid medium and pass an imaginary plane through the point. The orientation of the plane is determined by a unit normal ni , i = 1, 2, 3 to the planar surface. For a uid at rest we wish to determine the stress vector ti on the plane element passing through the selected point P. We desire to express element depends upon the orientation of the plane through the point. (n) ti (n) acting in terms of the stress tensor ij . The superscript (n) on the stress vector is to remind you that the stress acting on the planar 283 We make the assumption that ti (n) is colinear with the normal vector to the surface passing through the selected point. It is also assumed that for uid elements at rest, there are no shear forces acting on the planar element through an arbitrary point and therefore the stress tensor ij should be independent of the orientation of the plane. That is, we desire for the stress vector ij to be an isotropic tensor. This requires ij to have a specic form. To nd this specic form we let ij denote the stress components in a general coordinate system xi , i = 1, 2, 3 and let ij denote the components of stress in a barred coordinate system xi , i = 1, 2, 3. Since ij is a tensor, it must satisfy the transformation law mn = ij xi xj , xm xn i, j, m, n = 1, 2, 3. (2.5.7) We desire for the stress tensor ij to be an invariant under an arbitrary rotation of axes. Consider therefore the special coordinate transformations illustrated in the gures 2.5-1(a) and (b). Figure 2.5-1. Coordinate transformations due to rotations For the transformation equations given in gure 2.5-1(a), the stress tensor in the barred system of coordinates is 11 = 22 12 = 23 13 = 21 21 = 32 22 = 33 23 = 31 31 = 12 32 = 13 33 = 11 . (2.5.8) If ij is to be isotropic, we desire that 11 = 11 , 22 = 22 and 33 = 33 . If the equations (2.5.8) are to produce these results, we require that 11 , 22 and 33 must be equal. We denote these common values by (p). In particular, the equations (2.5.8) show that if 11 = 11 , 22 = 22 and 33 = 33 , then we must require that 11 = 22 = 33 = p. If 12 = 12 and 23 = 23 , then we also require that 12 = 23 = 31 . We note that if 13 = 13 and 32 = 32 , then we require that 21 = 32 = 13 . If the equations (2.5.7) are expanded using the transformation given in gure 2.5-1(b), we obtain the additional requirements that 11 = 22 12 = 21 13 = 23 21 = 12 22 = 11 23 = 13 31 = 32 32 = 31 33 = 33 . (2.5.9) 284 Analysis of these equations implies that if ij is to be isotropic, then 21 = 21 = 12 = 21 or 21 = 0 which implies 12 = 23 = 31 = 21 = 32 = 13 = 0. (2.5.10) The above analysis demonstrates that if the stress tensor ij is to be isotropic, it must have the form ij = pij . Use the traction condition (2.3.11), and express the stress vector as tj (n) (2.5.11) = ij ni = pnj . (2.5.12) This equation is interpreted as representing the stress vector at a point on a surface with outward unit normal ni , where p is the pressure (hydrostatic pressure) stress magnitude assumed to be positive. The negative sign in equation (2.5.12) denotes a compressive stress. Imagine a submerged object in a uid medium. We further imagine the object to be covered with unit normal vectors emanating from each point on its surface. The equation (2.5.12) shows that the hydrostatic pressure always acts on the object in a compressive manner. A force results from the stress vector acting on the object. The direction of the force is opposite to the direction of the unit outward normal vectors. It is a compressive force at each point on the surface of the object. The above considerations were for a uid at rest (hydrostatics). For a uid in motion (hydrodynamics) a dierent set of assumptions must be made. Hydrodynamical experiments show that the shear stress components are not zero and so we assume a stress tensor having the form ij = pij + ij , i, j = 1, 2, 3, (2.5.13) where ij is called the viscous stress tensor. Note that all real uids are both viscous and compressible. Denition: (Viscous/inviscid uid) If the viscous stress tensor ij is zero for all i, j, then the uid is called an inviscid, nonviscous, ideal or perfect uid. The uid is called viscous when ij is dierent from zero. In these notes it is assumed that the equation (2.5.13) represents the basic form for constitutive equations describing uid motion. 285 Figure 2.5-2. Viscosity experiment. Viscosity Most uids are characterized by the fact that they cannot resist shearing stresses. That is, if you put a shearing stress on the uid, the uid gives way and ows. Consider the experiment illustrated in the gure 2.5-2 which illustrates a uid moving between two parallel plane surfaces. Let S denote the distance between the two planes. Now keep the lower surface xed or stationary and move the upper surface parallel to the lower surface with a constant velocity V0 . If you measure the force F required to maintain the constant velocity of the upper surface, you discover that the force F varies directly as the area A of the surface and the ratio V0 /S. This is expressed in the form F V0 = . A S (2.5.14) The constant is a proportionality constant called the coecient of viscosity. The viscosity usually depends upon temperature, but throughout our discussions we will assume the temperature is constant. A dimensional analysis of the equation (2.5.14) implies that the basic dimension of the viscosity is [ ] = M L1 T 1 . For example, [ ] = gm/(cm sec) in the cgs system of units. The viscosity is usually measured in units of centipoise where one centipoise represents one-hundredth of a poise, where the unit of 1 poise= 1 gram per centimeter per second. The result of the above experiment shows that the stress is proportional to the change in velocity with change in distance or gradient of the velocity. Linear Viscous Fluids The above experiment with viscosity suggest that the viscous stress tensor ij is dependent upon both the gradient of the uid velocity and the density of the uid. In Cartesian coordinates, the simplest model suggested by the above experiment is that the viscous stress tensor ij is proportional to the velocity gradient vi,j and so we write ik = cikmp vm,p , where cikmp is a proportionality constant which is dependent upon the uid density. The viscous stress tensor must be independent of any reference frame, and hence we assume that the proportionality constants cikmp can be represented by an isotropic tensor. Recall that an isotropic tensor has the basic form cikmp = ik mp + (im kp + ip km ) + (im kp ip km ) (2.5.16) (2.5.15) 286 where , and are constants. Examining the results from equations (2.5.11) and (2.5.13) we nd that if the viscous stress is symmetric, then ij = ji . This requires be chosen as zero. Consequently, the viscous stress tensor reduces to the form ik = ik vp,p + (vk,i + vi,k ). (2.5.17) The coecient is called the rst coecient of viscosity and the coecient is called the second coecient of viscosity. Sometimes it is convenient to dene 2 = + 3 (2.5.18) as another second coecient of viscosity, or bulk coecient of viscosity. The condition of zero bulk viscosity is known as Stokes hypothesis. Many uids problems assume the Stokes hypothesis. This requires that the bulk coecient be zero or very small. Under these circumstances the second coecient of viscosity 2 is related to the rst coecient of viscosity by the relation = 3 . In the study of shock waves and acoustic waves the Stokes hypothesis is not applicable. There are many tables and empirical formulas where the viscosity of dierent types of uids or gases can be obtained. For example, in the study of the kinetic theory of gases the viscosity can be calculated C1 gT 3/2 from the Sutherland formula = where C1 , C2 are constants for a specic gas. These constants T + C2 can be found in certain tables. The quantity g is the gravitational constant and T is the temperature in degrees Rankine (o R = 460 + o F ). Many other empirical formulas like the above exist. Also many graphs and tabular values of viscosity can be found. The table 5.1 lists the approximate values of the viscosity of some selected uids and gases. Viscosity of selected uids and gases gram in units of cmsec = Poise at Atmospheric Pressure. 20 C 0.01002 0.012 0.199 0.0157 1.94(104 ) 1.74(104 ) 60 C 0.00469 0.00592 0.0495 0.013 100 C 0.00284 Table 5.1 Substance Water Alcohol Ethyl Alcohol Glycol Mercury Air Helium Nitrogen 0 C 0.01798 0.01773 0.017 1.708(104) 1.86(104) 1.658(104) 1.92(104) 0.0199 0.0100 2.175(104) 2.28(104) 2.09(104) The viscous stress tensor given in equation (2.5.17) may also be expressed in terms of the rate of deformation tensor dened by equation (2.5.6). This representation is ij = ij Dkk + 2 Dij , (2.5.19) where 2Dij = vi,j + vj,i and Dkk = D11 + D22 + D33 = v1,1 + v2,2 + v3,3 = vi,i = is the rate of change of the dilatation considered earlier. In Cartesian form, with velocity components u, v, w, the viscous stress 287 tensor components are xx =( + 2 ) yy zz u + x v =( + 2 ) + y w + =( + 2 ) z v w + y z u w + x z u v + x y yx = xy = zx = xz = zy = yz = u v + y x w u + x z w v + z y In cylindrical form, with velocity components vr , v , vz , the viscous stess tensor components are vr + V r 1 v vr =2 + V + r r vz + V zz =2 z 1 1 v vz V = (rvr ) + + r r r z rr =2 r = r = rz = zr = z = z = v v 1 vr + r r r vr vz + z r v 1 vz + r z where In spherical coordinates, with velocity components v , v , v , the viscous stress tensor components have the form v + V 1 v v =2 + V + v v cot 1 v + V + + =2 sin 1 1 1 v (sin v ) + V = 2 2 v + sin sin =2 = = = = = = 1 v v + 1 vr v + sin v 1 v sin + sin sin where Note that the viscous stress tensor is a linear function of the rate of deformation tensor Dij . Such a uid is called a Newtonian uid. In cases where the viscous stress tensor is a nonlinear function of Dij the uid is called non-Newtonian. Denition: (Newtonian Fluid) If the viscous stress tensor ij is expressible as a linear function of the rate of deformation tensor Dij , the uid is called a Newtonian uid. Otherwise, the uid is called a non-Newtonian uid. Important note: Do not assume an arbitrary form for the constitutive equations unless there is experimental evidence to support your assumption. A constitutive equation is a very important step in the modeling processes as it describes the material you are working with. One cannot arbitrarily assign a form to the viscous stress and expect the mathematical equations to describe the correct uid behavior. The form of the viscous stress is an important part of the modeling process and by assigning dierent forms to the viscous stress tensor then various types of materials can be modeled. We restrict our study in these notes to Newtonian uids. In Cartesian coordinates the rate of deformation-stress constitutive equations for a Newtonian uid can be written as ij = pij + ij Dkk + 2 Dij (2.5.20) 288 which can also be written in the alternative form ij = pij + ij vk,k + (vi,j + vj,i ) involving the gradient of the velocity. Upon transforming from a Cartesian coordinate system y i , i = 1, 2, 3 to a more general system of coordinates xi , i = 1, 2, 3, we write mn = ij y i y j . xm xn (2.5.22) (2.5.21) Now using the divergence from equation (2.1.3) and substituting equation (2.5.21) into equation (2.5.22) we obtain a more general expression for the constitutive equation. Performing the indicated substitutions there results mn = pij + ij v k + (vi,j + vj,i ) ,k mn y i y j xm xn k = pgmn + gmn v ,k + (v m,n + v n,m ). Dropping the bar notation, the stress-velocity strain relationships in the general coordinates xi , i = 1, 2, 3, is mn = pgmn + gmn g ik vi,k + (vm,n + vn,m ). (2.5.23) Summary The basic equations which describe the motion of a Newtonian uid are : Continuity equation (Conservation of mass) + t vi = 0, or D + V = 0 Dt 3 equations (2.5.25) 1 equation. (2.5.24) ,i Conservation of linear momentum ij + bi = v i , ,j or in vector form where DV = b + = b p + Dt = 3 i=1 3 j=1 (pij + ij ) ei ej and ij = 3 i=1 3 j=1 ij ei ej are second order tensors. Conser vation of angular momentum = , ji (Reduces the set of equations (2.5.23) to 6 equations.) Rate of deformation tensor (Velocity strain tensor) Dij = Constitutive equations mn = pgmn + gmn g ik vi,k + (vm,n + vn,m ), 6 equations. (2.5.27) 1 (vi,j + vj,i ) , 2 6 equations. (2.5.26) 289 In the cgs system of units the above quantities have the following units of measurements in Cartesian coordinates vi ij is the velocity eld , i = 1, 2, 3, is the stress tensor, i, j = 1, 2, 3, is the uid density bi is the external body forces per unit mass Dij , is the rate of deformation tensor p is the pressure are coecients of viscosity [vi ] = cm/sec [ij ] = dyne/cm2 [ ] = gm/cm3 [bi ] = dyne/gm [Dij ] = sec1 [p] = dyne/cm2 [ ] = [ ] = Poise where 1 Poise = 1gm/cm sec If we assume the external body forces per unit mass are known, then the equations (2.5.24), (2.5.25), (2.5.26), and (2.5.27) represent 16 equations in the 16 unknowns , v1 , v2 , v3 , 11 , 12 , 13 , 22 , 23 , 33 , D11 , D12 , D13 , D22 , D23 , D33 . Navier-Stokes-Duhem Equations of Fluid Motion Substituting the stress tensor from equation (2.5.27) into the linear momentum equation (2.5.25), and assuming that the viscosity coecients are constants, we obtain the Navier-Stokes-Duhem equations for uid motion. In Cartesian coordinates these equations can be represented in any of the equivalent forms vi = bi p,j ij + ( + )vk,ki + vi,jj vi + vj vi,j = bi + (pij + ij ) ,j t vi + ( vi vj + pij ij ) ,j = bi t Dv = b p + ( + ) ( v) + 2 v Dt where (2.5.28) v Dv = + (v ) v is the material derivative, substantial derivative or convective derivative. This Dt t derivative is represented as vi = vi dxj vi vi vi vi +j = + j vj = + vi,j v j . t x dt t x t (2.5.29) In the vector form of equations (2.5.28), the terms on the right-hand side of the equation represent force terms. The term b represents external body forces per unit volume. If these forces are derivable from a potential function , then the external forces are conservative and can be represented in the form . The term p is the gradient of the pressure and represents a force per unit volume due to hydrostatic pressure. The above statement is veried in the exercises that follow this section. The remaining terms can be written fviscous = ( + ) ( v) + 2 v (2.5.30) 290 and are given the physical interpretation of an internal force per unit volume. These internal forces arise from the shearing stresses in the moving uid. If fviscous is zero the vector equation in (2.5.28) is called Eulers equation. If the viscosity coecients are nonconstant, then the Navier-Stokes equations can be written in the Cartesian form [ vi vi vi vk vj + vj ] = bi + + + pij + ij t xj xj xk xj xi vi vk p vj = bi + + + j xi xi xk x xj xi which can also be written in terms of the bulk coecient of viscosity = + 2 as 3 [ vi p vi + vj ] = bi + t xj xi xi p = bi + xi xi vi vk vj 2 + ( ) + j 3 xk x xj xi vi vj 2 vk vk + ij + j xk x xj xi 3 xk These equations form the basics of viscous ow theory. In the case of orthogonal coordinates, where g(i)(i) = h2 (no summation) and gij = 0 for i = j, general i expressions for the Navier-Stokes equations in terms of the physical components v(1), v(2), v(3) are: Navier-Stokes-Duhem equations for compressible uid in terms of physical components: (i = j = k) v(2) v(i) v(3) v(i) v(i) v(1) v(i) + + + t h1 x1 h2 x2 h3 x3 v(j) hi hj v(j) hj hi v(i) xi xj + + v(k) hi hk v(i) hi hk v(k) xk xi v(j) hj + hi hj xj = v(i) hi hi hj 1 p 1 b(i) + V hi hi xi hi xi + + hi hk 2 hi hk xj hi hk xk v(i) hi + hk hi xi hi hj v(k) hk hj hi xi 2 hi xk hi hj v(k) hj v(i) hj 1 v(j) + + hj xj hj hk xk hi hj xi xi 2 hj hk 1 v(i) v(j) hi v(k) hi + + hi xi hi hj hj hi hk xk v(i) hi + hk hi xi v(k) hk 1 v(k) v(i) hk v(k) hk + + hk xk hi hk xi hk hj xi hi hk hj hi xi v(j) hj + hi hj xj hk 1 + xi hi hj hk v(i) hi + xk hi hj hi hk xk (2.5.31) where v is found in equation (2.1.4). In the above equation, cyclic values are assigned to i, j and k. That is, for the x1 components assign the values i = 1, j = 2, k = 3; for the x2 components assign the values i = 2, j = 3, k = 1; and for the x3 components assign the values i = 3, j = 1, k = 2. The tables 5.2, 5.3 and 5.4 show the expanded form of the Navier-Stokes equations in Cartesian, cylindrical and spherical coordinates respectively. 291 DVx p Vx 2 = bx + + V Dt x x x DVy p = by + Dt y x DVz p = bz + Dt z x Vy Vx + x y Vz Vx + x z + y Vx Vy + y x + z z Vx Vz + z x Vy Vz + z y + Vy 2 + V y y y Vz Vy + y z + + + Vz 2 + V z z where D ( ) ( ) ( ) ( ) () = + Vx + Vy + Vz Dt t x y z Vx Vy Vz + + x y z and V = (2.5.31a) Table 5.2 Navier-Stokes equations for compressible uids in Cartesian coordinates. V2 DVr Dt r = br + p Vr 2 + + V r r r Vr Vz + z r + 2 r + 1 r 1 Vr V V + r r r z Vr 1 V Vr r r r 1 V Vr + r r + V DV Vr V + Dt r = b 1 Vr 1 p 1 V V + 2 + + r r r r r r 1 Vz V V V 2 1 Vr + + + + z r z r r r r Vr Vz + z r + 1 r DVz p 1 r = bz + Dt z r r 1 Vz V + r z + Vz 2 + V z z where D ( ) ( ) ( ) V ( ) () = + Vr + + Vz Dt t r r z 1 (rVr ) 1 V Vz + + r r r z and V = (2.5.31b) Table 5.3 Navier-Stokes equations for compressible uids in cylindrical coordinates. 292 Observe that for incompressible ow D Dt = 0 which implies V = 0. Therefore, the assumptions of constant viscosity and incompressibility of the ow will simplify the above equations. If on the other hand the viscosity is temperature dependent and the ow is compressible, then one should add to the above equations the continuity equation, an energy equation and an equation of state. The energy equation comes from the rst law of thermodynamics applied to a control volume within the uid and will be considered in the sections ahead. The equation of state is a relation between thermodynamic variables which is added so that the number of equations equals the number of unknowns. Such a system of equations is known as a closed system. An example of an equation of state is the ideal gas law where pressure p is related to gas density and temperature T by the relation p = RT where R is the universal gas constant. 2 2 V + V DV Dt = b p V 2 + + V + 1 V + V V V 1 + sin sin 2 V 4V 2 V 2V cot V + cot + 4 sin + 2 V cot DV V V + Dt V + cot V = b 1 p + V + V 1 2 V + V + V V sin 1 V + + sin sin sin 1 V 1 V V cot + 2 cot +3 sin + DV Dt + V V + V V cot = b 1 p + sin V + sin V + 1 V V V 1 sin + sin 1 V 1 2 + + V + V cot + V sin sin V V 3 V sin + + 2 cot + 3 sin sin V + sin + 1 V sin where V ( ) D ( ) ( ) V ( ) () = + V + + Dt t sin 1 (2 V ) 1 V sin 1 V + + 2 sin sin and V = (2.5.31c) Table 5.4 Navier-Stokes equations for compressible uids in spherical coordinates. 293 We now consider various special cases of the Navier-Stokes-Duhem equations. Special Case 1: Assume that b is a conservative force such that b = . Also assume that the viscous force terms are zero. Consider steady ow ( v = 0) and show that equation (2.5.28) reduces to the equation t (v ) v = Employing the vector identity 1 (v ) v = ( v) v + (v v), 2 (2.5.33) 1 p is constant. (2.5.32) we take the dot product of equation (2.5.32) with the vector v. Noting that v [( v) v] = 0 we obtain v p 1 + + v 2 = 0. 2 (2.5.34) This equation shows that for steady ow we will have p 1 + + v 2 = constant 2 (2.5.35) along a streamline. This result is known as Bernoullis theorem. In the special case where = gh is a v2 p + gh = constant. This equation is known as force due to gravity, the equation (2.5.35) reduces to + 2 Bernoullis equation. It is a conservation of energy statement which has many applications in uids. Special Case 2: Assume that b = is conservative and dene the quantity by = v = curl v = 1 2 (2.5.36) as the vorticity vector associated with the uid ow and observe that its magnitude is equivalent to twice the angular velocity of a uid particle. Then using the identity from equation (2.5.33) we can write the Navier-Stokes-Duhem equations in terms of the vorticity vector. We obtain the hydrodynamic equations v 1 1 1 + v + v 2 = p + fviscous , t 2 (2.5.37) where fviscous is dened by equation (2.5.30). In the special case of nonviscous ow this further reduces to the Euler equation 1 1 v + v + v 2 = p . t 2 If the density is a function of the pressure only it is customary to introduce the function p P= c dp so that P = dP 1 p = p dp then the Euler equation becomes v 1 + v = (P + + v 2 ). t 2 Some examples of vorticies are smoke rings, hurricanes, tornadoes, and some sun spots. You can create a vortex by letting water stand in a sink and then remove the plug. Watch the water and you will see that a rotation or vortex begins to occur. Vortices are associated with circulating motion. 294 Pick an arbitrary simple closed curve C and place it in the uid ow and dene the line integral K= C v et ds, where ds is an element of arc length along the curve C, v is the vector eld dening the velocity, and et is a unit tangent vector to the curve C. The integral K is called the circulation of the uid around the closed curve C. The circulation is the summation of the tangential components of the velocity eld along the curve C. The local vorticity at a point is dened as the limit lim Area0 Circulation around C = circulation per unit area. Area inside C By Stokes theorem, if curl v = 0, then the uid is called irrotational and the circulation is zero. Otherwise the uid is rotational and possesses vorticity. If we are only interested in the velocity eld we can eliminate the pressure by taking the curl of both sides of the equation (2.5.37). If we further assume that the uid is incompressible we obtain the special equations v = 0 = curl v 2 + ( v) = t Incompressible uid, is constant. (2.5.38) Denition of vorticity vector. Results because curl of gradient is zero. Note that when is identically zero, we have irrotational motion and the above equations reduce to the Cauchy-Riemann equations. Note also that if the term ( v) is neglected, then the last equation in equation (2.5.38) reduces to a diusion equation. This suggests that the vorticity diuses through the uid once it is created. Vorticity can be caused by a rigid rotation or by shear ow. For example, in cylindrical coordinates let V = r e , with r, constants, denote a rotational motion, then curl V = V = 2 ez , which shows the vorticity is twice the rotation vector. Shear can also produce vorticity. For example, consider the velocity eld V = y e1 with y 0. Observe that this type of ow produces shear because |V | increases as y increases. For this ow eld we have curl V = V = e3 . The right-hand rule tells us that if an imaginary paddle wheel is placed in the ow it would rotate clockwise because of the shear eects. Scaled Variables In the Navier-Stokes-Duhem equations for uid ow we make the assumption that the external body forces are derivable from a potential function and write b = [dyne/gm] We also want to write the Navier-Stokes equations in terms of scaled variables v= v v0 p p= p0 = 0 = t t= , gL x x= L y= y L z z= L which can be referred to as the barred system of dimensionless variables. Dimensionless variables are introduced by scaling each variable associated with a set of equations by an appropriate constant term called a characteristic constant associated with that variable. Usually the characteristic constants are chosen from various parameters used in the formulation of the set of equations. The characteristic constants assigned to each variable are not unique and so problems can be scaled in a variety of ways. The characteristic constants 295 assigned to each variable are scales, of the appropriate dimension, which act as reference quantities which reect the order of magnitude changes expected of that variable over a certain range or area of interest associated with the problem. An inappropriate magnitude selected for a characteristic constant can result in a scaling where signicant information concerning the problem can be lost. This is analogous to selecting an inappropriate mesh size in a numerical method. The numerical method might give you an answer but details of the answer might be lost. In the above scaling of the variables occurring in the Navier-Stokes equations we let v0 denote some characteristic speed, p0 a characteristic pressure, 0 a characteristic density, L a characteristic length, g the acceleration of gravity and a characteristic time (for example = L/v0 ), then the barred variables v, p, ,, t, x, y and z are dimensionless. Dene the barred gradient operator by = e1 + e2 + e3 x y z where all derivatives are with respect to the barred variables. The above change of variables reduces the Navier-Stokes-Duhem equations v + (v ) v = p + ( + ) ( v) + 2 v, t (2.5.39) to the form 0 v0 v + t 2 0 v0 L v v = 0g p0 p L v0 L2 v. 2 ( + ) v0 v + + L2 Now if each term in the equation (2.5.40) is divided by the coecient S v + t v v = 1 Ep + F +1 (2.5.40) 2 0 v0 /L, we obtain the equation (2.5.41) 1 12 v + v R R which has the dimensionless coecients E= p0 2 = Euler number 0 v0 v2 F = 0 = Froude number, g is acceleration of gravity gL R= S= 0 V0 L = Reynolds number L = Strouhal number. v0 Dropping the bars over the symbols, we write the dimensionless equation using the above coecients. The scaled equation is found to have the form S 1 v + (v )v = Ep + t F +1 1 1 ( v) + 2 v R R (2.5.42) 296 Boundary Conditions Fluids problems can be classied as internal ows or external ows. An example of an internal ow problem is that of uid moving through a converging-diverging nozzle. An example of an external ow problem is uid ow around the boundary of an aircraft. For both types of problems there is some sort of boundary which inuences how the uid behaves. In these types of problems the uid is assumed to adhere to a boundary. Let rb denote the position vector to a point on a boundary associated with a moving uid, and let r denote the position vector to a general point in the uid. Dene v(r) as the velocity of the uid at the point r and dene v(rb ) as the known velocity of the boundary. The boundary might be moving within the uid or it could be xed in which case the velocity at all points on the boundary is zero. We dene the boundary condition associated with a moving uid as an adherence boundary condition. Denition: (Adherence Boundary Condition) An adherence boundary condition associated with a uid in motion is dened as the limit lim v(r) = v(rb ) where rb is the position rrb vector to a point on the boundary. Sometimes, when no nite boundaries are present, it is necessary to impose conditions on the components of the velocity far from the origin. Such conditions are referred to as boundary conditions at innity. Summary and Additional Considerations Throughout the development of the basic equations of continuum mechanics we have neglected thermodynamical and electromagnetic eects. The inclusion of thermodynamics and electromagnetic elds adds additional terms to the basic equations of a continua. These basic equations describing a continuum are: Conservation of mass The conservation of mass is a statement that the total mass of a body is unchanged during its motion. This is represented by the continuity equation + ( v k ),k = 0 or t where is the mass density and v k is the velocity. D + V =0 Dt Conservation of linear momentum The conservation of linear momentum requires that the time rate of change of linear momentum equal the resultant of all forces acting on the body. In symbols, we write D Dt where Dv i Dt n v i d = V S i F(s) ni dS + V i F(b) d + =1 i F() (2.5.43) = v i t + v i xk i i v k is the material derivative, F(s) are the surface forces per unit area, F(b) are the i body forces per unit mass and F() represents isolated external forces. Here S represents the surface and V represents the volume of the control volume. The right-hand side of this conservation law represents the resultant force coming from the consideration of all surface forces and body forces acting on a control volume. 297 Surface forces acting upon the control volume involve such things as pressures and viscous forces, while body forces are due to such things as gravitational, magnetic and electric elds. Conservation of angular momentum The conservation of angular momentum states that the time rate of change of angular momentum (moment of linear momentum) must equal the total moment of all forces and couples acting upon the body. In symbols, D Dt n V eijk x v d = jk S eijk x j k F(s) dS + V eijk x j k F(b) d + =1 k i (eijk xj F() + M() ) () (2.5.44) i k where M() represents concentrated couples and F() represents isolated forces. Conservation of energy The conservation of energy law requires that the time rate of change of kinetic energy plus internal energies is equal to the sum of the rate of work from all forces and couples plus a summation of all external energies that enter or leave a control volume per unit of time. The energy equation results from the rst law of thermodynamics and can be written D (E + K) = W + Qh Dt (2.5.45) where E is the internal energy, K is the kinetic energy, W is the rate of work associated with surface and body forces, and Qh is the input heat rate from surface and internal eects. Let e denote the internal specic energy density within a control volume, then E = V e d represents the total internal energy of the control volume. The kinetic energy of the control volume is expressed as 1 gij v i v j d where v i is the velocity, is the density and d is a volume element. The energy (rate K= 2V of work) associated with the body and surface forces is represented n W= S i gij F(s) v j dS + V i gij F(b) v j d + i i (gij F() v j + gij M() j ) =1 i i where j is the angular velocity of the point xi , F() are isolated forces, and M() are isolated couples. () Two external energy sources due to thermal sources are heat ow q i and rate of internal heat production per unit volume. The conservation of energy can thus be represented D Dt 1 (e + gij v i v j ) d = 2 i (gij F(s) v j qi ni ) dS + n i ( gij F(b) v j + Q t V S V Q ) d t (2.5.46) + i i (gij F() v j + gij M() j + U() ) =1 where U() represents all other energies resulting from thermal, mechanical, electric, magnetic or chemical sources which inux the control volume and D/Dt is the material derivative. In equation (2.5.46) the left hand side is the material derivative of an integral of the total energy et = (e + 1 gij v i v j ) over the control volume. Material derivatives are not like ordinary derivatives and so 2 298 we cannot interchange the order of dierentiation and integration in this term. Here we must use the result that D Dt V et d = V et + (et V ) t d. To prove this result we consider a more general problem. Let A denote the amount of some quantity per unit mass. The quantity A can be a scalar, vector or tensor. The total amount of this quantity inside the control volume is A = V A d and therefore the rate of change of this quantity is A = t D ( A) d = t Dt A d AV n dS, V V S which represents the rate of change of material within the control volume plus the inux into the control volume. The minus sign is because n is always a unit outward normal. By converting the surface integral to a volume integral, by the Gauss divergence theorem, and rearranging terms we nd that D Dt A d = ( A) + ( AV ) d. t V V i i In equation (2.5.46) we neglect all isolated external forces and substitute F(s) = ij nj , F(b) = bi where ij = pij + ij . We then replace all surface integrals by volume integrals and nd that the conservation of energy can be represented in the form et Q + (et V ) = ( V ) q + b V + t t 2 2 2 where et = e + (v1 + v2 + v3 )/2 is the total energy and = 3 i=1 3 3 j=1 (2.5.47) ij ei ej is the second order stress tensor. Here V 3 3 = pV + j=1 1j vj e1 + j=1 2j vj e2 + j=1 3j vj e3 = pV + V and ij = (vi,j + vj,i ) + ij vk,k is the viscous stress tensor. Using the identities et D(et / ) = + (et V ) Dt t and D(et / ) De D(V 2 /2) = + Dt Dt Dt together with the momentum equation (2.5.25) dotted with V as DV V = b V p V + ( ) V Dt the energy equation (2.5.47) can then be represented in the form Q De + p( V ) = q + + Dt t where is the dissipation function and can be represented = (ij vi ) ,j vi ij,j = ( V ) ( ) V . As an exercise it can be shown that the dissipation function can also be represented as = 2 Dij Dij + 2 where is the dilatation. The heat ow vector is determined from the Fourier law of heat conduction in (2.5.48) 299 terms of the temperature T as q = T , where is the thermal conductivity. Consequently, the energy equation can be written as Q De + p( V ) = + + (kT ). Dt t In Cartesian coordinates (x, y, z) we use D = + Vx + Vy + Vz Dt t x y z Vx Vy Vz V = + + x y z T T (T ) = + + x x y y z In cylindrical coordinates (r, , z) V D = + Vr + + Vz Dt t r r z Vz 1 1 V (rVr ) + 2 + V = r r r z 1 T 1 T (T ) = r +2 r r r r and in spherical coordinates (, , ) V V D = + V + Dt t sin 1 1 1 V V = 2 (V ) + (V sin ) + sin sin 1 T 1 T 1 (T ) = 2 2 +2 sin +2 2 sin sin (2.5.49) T z + z T z T The combination of terms h = e + p/ is known as enthalpy and at times is used to express the energy equation in the form D p Q Dh = + q + . Dt Dt t The derivation of this equation is left as an exercise. Conservative Systems Let Q denote some physical quantity per unit volume. Here Q can be either a scalar, vector or tensor eld. Place within this eld an imaginary simple closed surface S which encloses a volume V. The total amount of Q within the surface is given by to time is t V Q d and the rate of change of this amount with respect Q d. The total amount of Q within S changes due to sources (or sinks) within the volume J n d and by transport processes. Transport processes introduce a quantity J, called current, which represents a ow per unit area across the surface S. The inward ux of material into the volume is denoted volume so that S ( is a unit outward normal.) The sources (or sinks) SQ denotes a generation (or loss) of material per unit n V SQ d denotes addition (or loss) of material to the volume. For a xed volume we then Q d = t J n d + S V have the material balance SQ d. V 300 Using the divergence theorem of Gauss one can derive the general conservation law Q + J = SQ t (2.5.50) The continuity equation and energy equations are examples of a scalar conservation law in the special case where SQ = 0. In Cartesian coordinates, we can represent the continuity equation by letting Q= and J = V = (Vx e1 + Vy e2 + Vz e3 ) (2.5.51) The energy equation conservation law is represented by selecting Q = et and neglecting the rate of internal heat energy we let 3 J = (et + p)v1 i=1 3 vi xi + qx e1 + vi yi + qy e2 + i=1 3 (et + p)v2 (et + p)v3 i=1 (2.5.52) vi zi + qz e3 . In a general orthogonal system of coordinates (x1 , x2 , x3 ) the equation (2.5.50) is written ((h1 h2 h3 Q)) + ((h2 h3 J1 )) + ((h1 h3 J2 )) + ((h1 h2 J3 )) = 0, t x1 x2 x3 where h1 , h2 , h3 are scale factors obtained from the transformation equations to the general orthogonal coordinates. The momentum equations are examples of a vector conservation law having the form a + (T ) = b t where a is a vector and T is a second order symmetric tensor T = we let a = (Vx e1 + Vy e2 + Vz e3 ) and Tij = 3 3 (2.5.53) Tjk ej ek . In Cartesian coordinates k=1 j=1 vi vj + pij ij . In general coordinates (x1 , x2 , x3 ) the momentum equations result by selecting a = V and Tij = vi vj + pij ij . In a general orthogonal system the conservation law (2.5.53) has the general form (h2 h3 T e1 ) + (h1 h3 T e2 ) + (h1 h2 T e3 ) = b. ((h1 h2 h3 a)) + t x1 x2 x3 can be expressed in the strong conservative form U E F G + + + =0 t x y z where Vx U = Vy Vz et (2.5.55) (2.5.54) Neglecting body forces and internal heat production, the continuity, momentum and energy equations (2.5.56) 301 Vx 2 Vx + p xx E= Vx Vy xy Vx Vz xz (et + p)Vx Vx xx Vy xy Vz xz + qx Vy Vx Vy xy F = Vy2 + p yy Vy Vz yz (et + p)Vy Vx yx Vy yy Vz yz + qy Vz Vx Vz xz G= Vy Vz yz 2 Vz + p zz (et + p)Vz Vx zx Vy zy Vz zz + qz where the shear stresses are ij = (Vi,j + Vj,i ) + ij Vk,k for i, j, k = 1, 2, 3. Computational Coordinates To transform the conservative system (2.5.55) from a physical (x, y, z) domain to a computational (, , ) domain requires that a general change of variables take place. Consider the following general transformation of the independent variables = (x, y, z) = (x, y, z) = (x, y, z) (2.5.60) (2.5.57) (2.5.58) (2.5.59) with Jacobian dierent from zero. The chain rule for changing variables in equation (2.5.55) requires the operators ( ) ( ) ( ) ( ) = x + x + x x ( ) ( ) ( ) ( ) = y + y + y y ( ) ( ) ( ) ( ) = z + z + z z The partial derivatives in these equations occur in the dierential expressions d =x dx + y dy + z dz d =x dx + y dy + z dz d =x dx + y dy + z dz or d x d = x x d y y y z dx z dy z dz (2.5.61) (2.5.62) In a similar mannaer from the inverse transformation equations x = x(, , ) we can write the dierentials dx =x d + x d + x d dy =y d + y d + y d dz =z d + z d + z d or dx x dy = y z dz x y z d x y d z d y = y(, , ) z = z(, , ) (2.5.63) (2.5.64) 302 The transformations (2.5.62) and (2.5.64) are inverses of each other and so we can write x x x y y y x z z = y z z x y z 1 x y z (x z x z ) x y x y y z y z =J (y z y z ) x z x z (x y x y ) y z y z (x z x z ) x y x y (2.5.65) By comparing like elements in equation (2.5.65) we obtain the relations x =J(y z y z ) y = J(x z x z ) z =J(x y x y ) x = J(y z y z ) y =J(x z z z ) z = J(x y x y ) x =J(y z y z ) y = J(x z x z ) z =J(x y x y ) (2.5.66) The equations (2.5.55) can now be written in terms of the new variables (, , ) as U E E E F F F G G G + x + x + x + y + y + y + z + z + z = 0 t Now divide each term by the Jacobian J and write the equation (2.5.67) in the form t U J + + + Ex + F y + Gz J Ex + F y + Gz J Ex + F y + Gz J x x + + J J y y + + J J z z + + J J (2.5.67) E F G x J y J z J (2.5.68) =0 Using the relations given in equation (2.5.66) one can show that the curly bracketed terms above are all zero and so the transformed equations (2.5.55) can also be written in the conservative form E F G U + + + =0 t where U J Ex + F y + Gz E= J Ex + F y + Gz F= J Ex + F y + Gz G= J (2.5.69) U= (2.5.70) 303 Fourier law of heat conduction The Fourier law of heat conduction can be written qi = T,i for isotropic material and qi = ij T,j for anisotropic material. The Prandtl number is a nondimensional constant dened as P r = the heat ow terms can be represented in Cartesian coordinates as qx = cp T P r x qy = cp T P r y qz = cp T P r z R 1 cp so that Now one can employ the equation of state relations P = e( 1), cp = above equations in the alternate forms qx = P r( 1) x P qy = P P r( 1) y P , cp T = RT 1 and write the qz = P r( 1) z P The speed of sound is given by a = in the above equations. = RT and so one can substitute a2 in place of the ratio P Equilibrium and Nonequilibrium Thermodynamics High temperature gas ows require special considerations. In particular, the specic heat for monotonic and diatomic gases are dierent and are in general a function of temperature. The energy of a gas can be written as e = et + er + ev + ee + en where et represents translational energy, er is rotational energy, ev is vibrational energy, ee is electronic energy, and en is nuclear energy. The gases follow a Boltzmann distribution for each degree of freedom and consequently at very high temperatures the rotational, translational and vibrational degrees of freedom can each have their own temperature. Under these conditions the gas is said to be in a state of nonequilibrium. In such a situation one needs additional energy equations. The energy equation developed in these notes is for equilibrium thermodynamics where the rotational, translational and vibrational temperatures are the same. Equation of state It is assumed that an equation of state such as the universal gas law or perfect gas law pV = nRT holds which relates pressure p [N/m2 ], volume V [m3 ], amount of gas n [mol],and temperature T [K] where R [J/mol K] is the universal molar gas constant. If the ideal gas law is represented in the form p = RT where [Kg/m3 ] is the gas density, then the universal gas constant must be expressed in units of [J/Kg K] (See Appendix A). Many gases deviate from this ideal behavior. In order to account for the intermolecular forces associated with high density gases, an empirical equation of state of the form M1 p = RT + n=1 n n+r1 + e1 2 2 M2 cn n+r2 n=1 involving constants M1 , M2 , n , cn , r1 , r2 , 1 , 2 is often used. For a perfect gas the relations e = cv T = cp cv cv = R 1 cp = R 1 h = cp T hold, where R is the universal gas constant, cv is the specic heat at constant volume, cp is the specic heat at constant pressure, is the ratio of specic heats and h is the enthalpy. For cv and cp constants the relations p = ( 1) e and RT = ( 1)e can be veried. 304 EXAMPLE 2.5-1. (One-dimensional uid ow) Construct an x-axis running along the center line of a long cylinder with cross sectional area A. Consider the motion of a gas driven by a piston and moving with velocity v1 = u in the x-direction. From an Eulerian point of view we imagine a control volume xed within the cylinder and assume zero body forces. We require the following equations be satised. + div( V ) = 0 which in one-dimension reduces to + ( u) = 0. Conservation of mass t t x p ( u) + u2 + = 0. Conservation of momentum, equation (2.5.28) reduces to t x x Conservation of energy, equation (2.5.48) in the absence of heat ow and internal heat production, e e u becomes in one dimension +u +p = 0. Using the conservation of mass relation this t x x u equation can be written in the form ( e) + ( eu) + p = 0. t x x In contrast, from a Lagrangian point of view we let the control volume move with the ow and consider advection terms. This gives the following three equations which can then be compared with the above Eulerian equations of motion. D u d + = 0. Conservation of mass ( J) = 0 which in one-dimension is equivalent to dt Dt x Du p + = 0. Conservation of momentum, equation (2.5.25) in one-dimension Dt x u De +p = 0. In the above equations Conservation of energy, equation (2.5.48) in one-dimension Dt x D( ) Dt = t ( ) + u x ( ). The Lagrangian viewpoint gives three equations in the three unknowns , u, e. In both the Eulerian and Lagrangian equations the pressure p represents the total pressure p = pg + pv where pg is the gas pressure and pv is the viscous pressure which causes loss of kinetic energy. The gas pressure is a function of , e and is determined from the ideal gas law pg = RT = (cp cv )T = ( cp 1)cv T or v pg = ( 1)e. Some kind of assumption is usually made to represent the viscous pressure pv as a function of e, u. The above equations are then subjected to boundary and initial conditions and are usually solved numerically. c Entropy inequality Energy transfer is not always reversible. Many energy transfer processes are irreversible. The second law of thermodynamics allows energy transfer to be reversible only in special circumstances. In general, the second law of thermodynamics can be written as an entropy inequality, known as the Clausius-Duhem inequality. This inequality states that the time rate of change of the total entropy is greater than or equal to the total entropy change occurring across the surface and within the body of a control volume. The ClausiusDuhem inequality places restrictions on the constitutive equations. This inequality can be expressed in the form D Dt s d V n S si ni dS + b d + V =1 B() Rate of entropy increase Entropy input rate into control volume where s is the specic entropy density, si is an entropy ux, b is an entropy source and B() are isolated entropy sources. Irreversible processes are characterized by the use of the inequality sign while for reversible 305 Figure 2.5-3. Interaction of various elds. processes the equality sign holds. The Clausius-Duhem inequality is assumed to hold for all independent thermodynamical processes. If in addition there are electric and magnetic elds to consider, then these elds place additional forces upon the material continuum and we must add all forces and moments due to these eects. In particular we must add the following equations Gausss law for magnetism B = 0 1 i ( gB ) = 0. g xi 1 i ( gD ) = g xi Ek,j = B i . t Di . t Gausss law for electricity D = e e. Faradays law E = B t ijk Amperes law where e H =J + D t j i Ej ijk Hk,j = J i + is the charge density, J i is the current density, Di = j Hj i + Pi is the electric displacement vector, Hi is the magnetic eld, Bi = + Mi is the magnetic induction, Ei is the electric eld, Mi is the magnetization vector and Pi is the polarization vector. Taking the divergence of Amperes law produces the law of conservation of charge which requires that e +J = 0 t e 1i + ( gJ ) = 0. t g xi The gure 2.5-3 is constructed to suggest some of the interactions that can occur between various variables which dene the continuum. Pyroelectric eects occur when a change in temperature causes changes in the electrical properties of a material. Temperature changes can also change the mechanical properties of materials. Similarly, piezoelectric eects occur when a change in either stress or strain causes changes in the electrical properties of materials. Photoelectric eects are said to occur if changes in electric or mechanical properties eect the refractive index of a material. Such changes can be studied by modifying the constitutive equations to include the eects being considered. From gure 2.5-3 we see that there can exist a relationship between the displacement eld Di and electric eld Ei . When this relationship is linear we can write Di = ji Ej and Ej = jn Dn , where ji are 306 dielectric constants and jn are dielectric impermabilities. Similarly, when linear piezoelectric eects exist we can write linear relations between stress and electric elds such as ij = gkij Ek and Ei = eijk jk , where gkij and eijk are called piezoelectric constants. If there is a linear relation between strain and an electric elds, this is another type of piezoelectric eect whereby eij = dijk Ek and Ek = hijk ejk , where dijk and hijk are another set of piezoelectric constants. Similarly, entropy changes can cause pyroelectric eects. Piezooptical eects (photoelasticity) occurs when mechanical stresses change the optical properties of the material. Electrical and heat eects can also change the optical properties of materials. Piezoresistivity occurs when mechanical stresses change the electric resistivity of materials. Electric eld changes can cause variations in temperature, another pyroelectric eect. When temperature eects the entropy of a material this is known as a heat capacity eect. When stresses eect the entropy in a material this is called a piezocaloric eect. Some examples of the representation of these additional eects are as follows. The piezoelectric eects are represented by equations of the form ij = hmij Dm Di = dijk jk eij = gkij Dk Di = eijk ejk where hmij , dijk , gkij and eijk are piezoelectric constants. Knowledge of the material or electric interaction can be used to help modify the constitutive equations. For example, the constitutive equations can be modied to included temperature eects by expressing the constitutive equations in the form ij = cijkl ekl ij T and eij = sijkl kl + ij T where for isotropic materials the coecients ij and ij are constants. As another example, if the strain is modied by both temperature and an electric eld, then the constitutive equations would take on the form eij = sijkl kl + ij T + dmij Em . Note that these additional eects are additive under conditions of small changes. That is, we may use the principal of superposition to calculate these additive eects. If the electric eld and electric displacement are replaced by a magnetic eld and magnetic ux, then piezomagnetic relations can be found to exist between the variables involved. One should consult a handbook to determine the order of magnitude of the various piezoelectric and piezomagnetic eects. For a large majority of materials these eects are small and can be neglected when the eld strengths are weak. The Boltzmann Transport Equation The modeling of the transport of particle beams through matter, such as the motion of energetic protons or neutrons through bulk material, can be approached using ideas from the classical kinetic theory of gases. Kinetic theory is widely used to explain phenomena in such areas as: statistical mechanics, uids, plasma physics, biological response to high-energy radiation, high-energy ion transport and various types of radiation shielding. The problem is basically one of describing the behavior of a system of interacting particles and their distribution in space, time and energy. The average particle behavior can be described by the Boltzmann equation which is essentially a continuity equation in a six-dimensional phase space (x, y, z, Vx , Vy , Vz ). We 307 will be interested in examining how the particles in a volume element of phase space change with time. We introduce the following notation: (i) r the position vector of a typical particle of phase space and d = dxdydz the corresponding spatial volume element at this position. (ii) V the velocity vector associated with a typical particle of phase space and dv = dVx dVy dVz the corresponding velocity volume element. (iii) a unit vector in the direction of the velocity V = v . (iv) E = 1 mv 2 kinetic energy of particle. 2 (v) d is a solid angle about the direction and d dE d is a volume element of phase space involving the solid angle about the direction . (vi) n = n(r, E, , t) the number of particles in phase space per unit volume at position r per unit velocity at position V per unit energy in the solid angle d at time t and N = N (r, E, , t) = vn(r, E, , t) the number of particles per unit volume per unit energy in the solid angle d at time t. The quantity N (r, E, , t)d dE d represents the number of particles in a volume element around the position r with energy between E and E + dE having direction in the solid angle d at time t. (vii) (r, E, , t) = vN (r, E, , t) is the particle ux (number of particles/cm2 Mev sec). (viii) (E E, ) a scattering cross-section which represents the fraction of particles with energy E and direction that scatter into the energy range between E and E + dE having direction in the solid angle d per particle ux. (ix) s (E, r) fractional number of particles scattered out of volume element of phase space per unit volume per ux. (x) a (E, r) fractional number of particles absorbed in a unit volume of phase space per unit volume per ux. Consider a particle at time t having a position r in phase space as illustrated in the gure 2.5-4. This particle has a velocity V in a direction and has an energy E. In terms of d = dx dy dz, and E an element of volume of phase space can be denoted d dEd, where d = d(, ) = sin dd is a solid angle about the direction . The Boltzmann transport equation represents the rate of change of particle density in a volume element d dE d of phase space and is written d N (r, E, , t) d dE d = DC N (r, E, , t) dt (2.5.71) where DC is a collision operator representing gains and losses of particles to the volume element of phase space due to scattering and absorption processes. The gains to the volume element are due to any sources S(r, E, , t) per unit volume of phase space, with units of number of particles/sec per volume of phase space, together with any scattering of particles into the volume element of phase space. That is particles entering the volume element of phase space with energy E, which experience a collision, leave with some energy E E and thus will be lost from our volume element. Particles entering with energies E > E may, 308 Figure 2.5-4. Volume element and solid angle about position r. depending upon the cross-sections, exit with energy E E = E and thus will contribute a gain to the volume element. In terms of the ux the gains due to scattering into the volume element are denoted by d dE (E E, )(r, E , , t) d dE d and represents the particles at position r experiencing a scattering collision with a particle of energy E and direction which causes the particle to end up with energy between E and E + dE and direction in d. The summations are over all possible initial energies. In terms of the losses are due to those particles leaving the volume element because of scattering and are s (E, r)(r, E, , t)d dE d. The particles which are lost due to absorption processes are a (E, r)(r, E, , t) d dE d. The total change to the number of particles in an element of phase space per unit of time is obtained by summing all gains and losses. This total change is dN d dE d = dt d dE (E E, )(r, E , , t) d dE d (2.5.72) s (E, r)(r, E, , t)d dE d a (E, r)(r, E, , t) d dE d + S(r, E, , t)d dE d. The rate of change dN dt on the left-hand side of equation (2.5.72) expands to N N dx N dy N dz dN = + + + dt t x dt y dt z dt N dVx N dVy N dVz + + + Vx dt Vy dt Vz dt 309 which can be written as N dN F = + V r N + V N dt t m where dV dt (2.5.73) = F m represents any forces acting upon the particles. The Boltzmann equation can then be F N + V r N + V N = Gains Losses. t m expressed as (2.5.74) If the right-hand side of the equation (2.5.74) is zero, the equation is known as the Liouville equation. In the special case where the velocities are constant and do not change with time the above equation (2.5.74) can be written in terms of the ux and has the form 1 + r + s (E, r) + a (E, r) (r, E, , t) = DC v t where DC = d dE (E E, )(r, E , , t) + S(r, E, , t). (2.5.75) The above equation represents the Boltzmann transport equation in the case where all the particles are the same. In the case of atomic collisions of particles one must take into consideration the generation of secondary particles resulting from the collisions. Let there be a number of particles of type j in a volume element of phase space. For example j = p (protons) and j = n (neutrons). We consider steady state conditions and dene the quantities (i) j (r, E, ) as the ux of the particles of type j. (ii) jk (, , E, E ) the collision cross-section representing processes where particles of type k moving in direction with energy E produce a type j particle moving in the direction with energy E. (iii) j (E) = s (E, r) + a (E, r) the cross-section for type j particles. The steady state form of the equation (2.5.64) can then be written as j (r, E, )+j (E)j (r, E, ) = k jk (, , E, E )k (r, E , )d dE (2.5.76) where the summation is over all particles k = j. The Boltzmann transport equation can be represented in many dierent forms. These various forms are dependent upon the assumptions made during the derivation, the type of particles, and collision crosssections. In general the collision cross-sections are dependent upon three components. (1) Elastic collisions. Here the nucleus is not excited by the collision but energy is transferred by projectile recoil. (2) Inelastic collisions. Here some particles are raised to a higher energy state but the excitation energy is not sucient to produce any particle emissions due to the collision. (3) Non-elastic collisions. Here the nucleus is left in an excited state due to the collision processes and some of its nucleons (protons or neutrons) are ejected. The remaining nucleons interact to form a stable structure and usually produce a distribution of low energy particles which is isotropic in character. 310 Various assumptions can be made concerning the particle ux. The resulting form of Boltzmanns equation must be modied to reect these additional assumptions. As an example, we consider modications to Boltzmanns equation in order to describe the motion of a massive ion moving into a region lled with a homogeneous material. Here it is assumed that the mean-free path for nuclear collisions is large in comparison with the mean-free path for ion interaction with electrons. In addition, the following assumptions are made (i) All collision interactions are non-elastic. (ii) The secondary particles produced have the same direction as the original particle. This is called the straight-ahead approximation. (iii) Secondary particles never have kinetic energies greater than the original projectile that produced them. (iv) A charged particle will eventually transfer all of its kinetic energy and stop in the media. This stopping distance is called the range of the projectile. The stopping power Sj (E) = Rj (E) = E dE . 0 Sj (E ) 1 dE dx represents the energy dRj dE loss per unit length traveled in the media and determines the range by the relation Boltzmann equation for homogeneous materials takes on the form j (r, E, ) = k=j = 1 Sj (E) or Using the above assumptions Wilson, et.al. show that the steady state linearized 1 (Sj (E)j (r, E, )) + j (E)j (r, E, ) Aj E (2.5.77) dE d jk (, , E, E )k (r, E , ) where Aj is the atomic mass of the ion of type j and j (r, E, ) is the ux of ions of type j moving in the direction with energy E. Observe that in most cases the left-hand side of the Boltzmann equation represents the time rate of change of a distribution type function in a phase space while the right-hand side of the Boltzmann equation represents the time rate of change of this distribution function within a volume element of phase space due to scattering and absorption collision processes. Boltzmann Equation for gases Consider the Boltzmann equation in terms of a particle distribution function f (r, V , t) which can be written as F + V r + V t m f (r, V , t) = DC f (r, V , t) (2.5.78) for a single species of gas particles where there is only scattering and no absorption of the particles. An element of volume in phase space (x, y, z, Vx , Vy , Vz ) can be thought of as a volume element d = dxdydz for the spatial elements together with a volume element dv = dVx dVy dVz for the velocity elements. These elements are centered at position r and velocity V at time t. In phase space a constant velocity V1 can be 2 thought of as a sphere since V12 = Vx + Vy2 + Vz2 . The phase space volume element d dv changes with time since the position r and velocity V change with time. The position vector r changes because of velocity 1 John W. Wilson, Lawrence W. Townsend, Walter Schimmerling, Govind S. Khandelwal, Ferdous Kahn, John E. Nealy, Francis A. Cucinotta, Lisa C. Simonsen, Judy L. Shinn, and John W. Norbury, Transport Methods and Interactions for Space Radiations, NASA Reference Publication 1257, December 1991. 311 and the velocity vector changes because of the acceleration F m. Here f (r, V , t)d dv represents the expected number of particles in the phase space element d dv at time t. Assume there are no collisions, then each of the gas particles in a volume element of phase space centered at position r and velocity V1 move during a time interval dt to a phase space element centered at position r + V1 dt and V1 + F m dt. If there were no loss or gains of particles, then the number of particles must be conserved and so these gas particles must move smoothly from one element of phase space to another without any gains or losses of particles. Because of scattering collisions in d many of the gas particles move into or out of the velocity range V1 to V1 + dV1 . These collision scattering processes are denoted by the collision operator DC f (r, V , t) in the Boltzmann equation. Consider two identical gas particles which experience a binary collision. Imagine that particle 1 with velocity V1 collides with particle 2 having velocity V2 . Denote by (V1 V1 , V2 V2 ) dV1 dV2 the conditional probability that particle 1 is scattered from velocity V1 to between V1 and V1 + dV1 and the struck particle 2 is scattered from velocity V2 to between V2 and V2 + dV2 . We will be interested in collisions of the type (V1 , V2 ) (V1 , V2 ) for a xed value of V1 as this would represent the number of particles scattered into dV1 . Also of interest are collisions of the type (V1 , V2 ) (V1 , V2 ) for a xed value V1 as this represents particles scattered out of dV1 . Imagine a gas particle in d with velocity V1 subjected to a beam of particles with velocities V2 . The incident ux on the element d dV1 is |V1 V2 |f (r, V2 , t)dV2 and hence (V1 V1 , V2 V2 ) dV1 dV2 dt |V1 V2 |f (r, V2 , t) dV2 (2.5.79) represents the number of collisions, in the time interval dt, which scatter from V1 to between V1 and V1 + dV1 as well as scattering V2 to between V2 and V2 + dV2 . Multiply equation (2.5.79) by the density of particles in the element d dV1 and integrate over all possible initial velocities V1 ,V2 and nal velocities V2 not equal to V1 . This gives the number of particles in d which are scattered into dV1 dt as N sin = d dV1 dt dV2 dV2 dV1 (V1 V1 , V2 V2 )|V1 V2 |f (r, V1 , t)f (r, V2 , t). (2.5.80) In a similar manner the number of particles in d which are scattered out of dV1 dt is N sout = d dV1 dtf (r, V1 , t) Let W (V1 V1 , V2 V2 ) = |V1 V2 | (V1 V1 , V2 V2 ) (2.5.82) dV2 dV2 dV1 (V1 V1 , V2 V2 )|V2 V1 |f (r, V2 , t). (2.5.81) dene a symmetric scattering kernel and use the relation DC f (r, V , t) = N sin N sout to represent the Boltzmann equation for gas particles in the form F + V r + V t m dV 1 f (r, V1 , t) = (2.5.83) dV 2 dV2 W (V1 V1 , V2 V2 ) f (r, V1 , t)f (r, V2 , t) f (r, V1 , t)f (r, V2 , t) . Take the moment of the Boltzmann equation (2.5.83) with respect to an arbitrary function (V1 ). That is, multiply equation (2.5.83) by (V1 ) and then integrate over all elements of velocity space dV1 . Dene the following averages and terminology: 312 The particle density per unit volume + n = n(r, t) = dV f (r, V , t) = f (r, V , t)dVx dVy dVz (2.5.84) where = nm is the mass density. The mean velocity 1 V1 = V = n + V1 f (r, V1 , t)dV1x dV1y dV1z For any quantity Q = Q(V1 ) dene the barred quantity 1 Q = Q(r, t) = n(r, t) Further, assume that F m 1 Q(V )f (r, V , t) dV = n + Q(V )f (r, V , t)dVx dVy dVz . (2.5.85) is independent of V , then the moment of equation (2.5.83) produces the result Fi n + nV1i n =0 i t x m V1i i=1 i=1 3 3 (2.5.86) known as the Maxwell transfer equation. The rst term in equation (2.5.86) follows from the integrals f (r, V1 , t) (V1 )dV1 = t t f (r, V1 , t)(V1 ) dV1 = (n) t (2.5.87) where dierentiation and integration have been interchanged. The second term in equation (2.5.86) follows from the integral 3 V1 r f (V1 )dV1 = i=1 3 V1i xi f dV1 xi V1i f dV1 (2.5.88) = i=1 3 = i=1 nV1i . xi The third term in equation (2.5.86) is obtained from the following integral where integration by parts is employed F f dV1 = m V1 = i=1 3 i=1 + 3 Fi f m V1i dV1 Fi f m V1i dV1x dV1y dV1y (2.5.89) = = n V1i V1i Fi f dV1 m Fi m = Fi m V1i 313 since Fi does not depend upon V1 and f (r, V , t) equals zero for Vi equal to . The right-hand side of equation (2.5.86) represents the integral of (DC f ) over velocity space. This integral is zero because of the symmetries associated with the right-hand side of equation (2.5.83). Physically, the integral of (Dc f ) over velocity space must be zero since collisions with only scattering terms cannot increase or decrease the number of particles per cubic centimeter in any element of phase space. In equation (2.5.86) we write the velocities V1i in terms of the mean velocities (u, v, w) and random velocities (Ur , Vr , Wr ) with V11 = Ur + u, V12 = Vr + v, V13 = Wr + w (2.5.90) or V1 = Vr + V with V1 = Vr + V = V since Vr = 0 (i.e. the average random velocity is zero.) For future reference we write equation (2.5.86) in terms of these random velocities and the material derivative. Substitution of the velocities from equation (2.5.90) in equation (2.5.86) gives Fi (n) + n(Vr + v) + n(Wr + w) n n(Ur + u) + =0 t x y z m V1i i=1 or (n) + nu + nv + nw t x y z Fi + nUr + nVr + nWr n = 0. x y z m V1i i=1 Observe that nu = + 3 3 (2.5.91) (2.5.92) uf (r, V , t)dVx dVy dVz = nu (2.5.93) and similarly nv = nv, nw = nw. This enables the equation (2.5.92) to be written in the form n + nu + nv + nw t x y z n + + (nu) + (nv) + (nw) t x y z + Fi nUr + nVr + nWr n = 0. x y z m V1i i=1 3 (2.5.94) The middle bracketed sum in equation (2.5.94) is recognized as the continuity equation when multiplied by m and hence is zero. The moment equation (2.5.86) now has the form n D Fi + nUr + nVr + nWr n = 0. Dt x y z m V1i i=1 3 (2.5.95) Note that from the equations (2.5.86) or (2.5.95) one can derive the basic equations of uid ow from continuum mechanics developed earlier. We consider the following special cases of the Maxwell transfer equation. 314 (i) In the special case = m the equation (2.5.86) reduces to the continuity equation for uids. That is, equation (2.5.86) becomes (nm) + (nmV1 ) = 0 t which is the continuity equation + (V ) = 0 t where is the mass density and V is the mean velocity dened earlier. (ii) In the special case = mV1 is momentum, the equation (2.5.86) reduces to the momentum equation for uids. To show this, we write equation (2.5.86) in terms of the dyadic V1 V1 in the form nmV1 + (nmV1 V1 ) nF = 0 t or (Vr + V ) + ((Vr + V )(Vr + V )) nF = 0. t Let (2.5.99) (2.5.98) (2.5.97) (2.5.96) = Vr Vr denote a stress tensor which is due to the random motions of the gas particles and write equation (2.5.99) in the form The term V t V +V + V ( V ) + V ( (V )) nF = 0. t t (2.5.100) + (V ) = 0 because of the continuity equation and so equation (2.5.100) reduces V +VV t = nF + . to the momentum equation (2.5.101) For F = q E + q V B + mb, where q is charge, E and B are electric and magnetic elds, and b is a body force per unit mass, together with = 3 3 (pij + ij )ei ej i=1 j=1 (2.5.102) the equation (2.5.101) becomes the momentum equation DV = b p + + nq(E + V B). Dt (2.5.103) In the special case were E and B vanish, the equation (2.5.103) reduces to the previous momentum equation (2.5.25) . (iii) In the special case = m 2 V1 V1 = m 2 2 2 2 (V11 + V12 + V13 ) is the particle kinetic energy, the equation (2.5.86) simplies to the energy equation of uid mechanics. To show this we substitute into equation (2.5.95) and simplify. Note that m (Ur + u)2 + (Vr + v)2 + (Wr + w)2 2 m 2 2 = Ur + Vr2 + Wr + u2 + v 2 + w2 2 = (2.5.104) 315 2 2 2 since uUr = vVr = wWr = 0. Let V 2 = u2 + v 2 + w2 and Cr = Ur + Vr2 + Wr and write equation (2.5.104) in the form = Also note that nUr = nm Ur (Ur + u)2 + Ur (Vr + v)2 + Ur (Wr + w)2 2 2 nm Ur Cr 2 + uUr + vUr Vr + wUr Wr = 2 2 m 2 Cr + V 2 . 2 (2.5.105) (2.5.106) and that nm 2 Vr Cr + uVr Ur + vVr2 + wVr Wr 2 nm 2 2 nWr = Wr Cr + uWr Ur + vWr Vr + wWr 2 nVr = are similar results. We use V1i (2.5.107) (2.5.108) () = mV1i together with the previous results substituted into the equation (2.5.95), and nd that the Maxwell transport equation can be expressed in the form D Dt 2 V2 Cr + 2 2 = 2 [uUr + vUr Vr + wUr Wr ] x [uVr Ur + vVr2 + wVr Wr ] y 2 [uWr Ur + vWr Vr + wWr ] z 2 2 Ur Cr Vr Cr x 2 y 2 z (2.5.109) 2 Wr Cr 2 + nF V . Compare the equation (2.5.109) with the energy equation (2.5.48) De D + Dt Dt V2 2 = ( V ) q + b V 2 Cr 2 (2.5.110) where the internal heat energy has been set equal to zero. Let e = random motion of the gas particles, F = mb, and let q = x 2 Ur Cr 2 denote the internal energy due to y y 2 Vr Cr 2 z T k z 2 Wr Cr 2 (2.5.111) = x T k x T k y z represent the heat conduction terms due to the transport of particle energy 2 mCr 2 by way of the random particle motion. The remaining terms are related to the rate of change of work and surface stresses giving 2 [uUr + vUr Vr + wUr Wr ] = (uxx + vxy + wxz ) x x [uVr Ur + vVr2 + wVr Wr ] = (uyx + vyy + wyz ) y y 2 [uWr Ur + vWr Vr + wWr ] = (uzx + vzy + wzz ) . z z (2.5.112) 316 This gives the stress relations due to random particle motion 2 xx = Ur yx = Vr Ur yy = Vr2 yz = Vr Wr zx = Wr Ur zy = Wr Vr 2 zz = Wr . xy = Ur Vr xz = Ur Wr (2.5.113) The Boltzmann equation is a basic macroscopic model used for the study of individual particle motion where one takes into account the distribution of particles in both space, time and energy. The Boltzmann equation for gases assumes only binary collisions as three-body or multi-body collisions are assumed to rarely occur. Another assumption used in the development of the Boltzmann equation is that the actual time of collision is thought to be small in comparison with the time between collisions. The basic problem associated with the Boltzmann equation is to nd a velocity distribution, subject to either boundary and/or initial conditions, which describes a given gas ow. The continuum equations involve trying to obtain the macroscopic variables of density, mean velocity, stress, temperature and pressure which occur in the basic equations of continuum mechanics considered earlier. Note that the moments of the Boltzmann equation, derived for gases, also produced these same continuum equations and so they are valid for gases as well as liquids. In certain situations one can assume that the gases approximate a Maxwellian distribution f (r, V , t) n(r, t) m 2kT 3/2 exp m V V 2kT (2.5.114) thereby enabling the calculation of the pressure tensor and temperature from statistical considerations. In general, one can say that the Boltzmann integral-dierential equation and the Maxwell transfer equation are two important formulations in the kinetic theory of gases. The Maxwell transfer equation depends upon some gas-particle property which is assumed to be a function of the gas-particle velocity. The Boltzmann equation depends upon a gas-particle velocity distribution function f which depends upon position r, velocity V and time t. These formulations represent two distinct and important viewpoints considered in the kinetic theory of gases. 317 EXERCISE 2.5 1. Let p = p(x, y, z), [dyne/cm2 ] denote the pressure at a point (x, y, z) in a uid medium at rest (hydrostatics), and let V denote an element of uid volume situated at this point as illustrated in the gure 2.5-5. Figure 2.5-5. Pressure acting on a volume element. (a) Show that the force acting on the face ABCD is p(x, y, z)yz e1 . (b) Show that the force acting on the face EF GH is p(x + x, y, z)yz e1 = p(x, y, z) + 2 p (x)2 p x + 2 + yz e1 . x x 2! (c) In part (b) neglect terms with powers of x greater than or equal to 2 and show that the resultant force p in the x-direction is xyz e1 . x (d) What has been done in the x-direction can also be done in the y and z-directions. Show that the p p resultant forces in these directions are xyz e2 and xyz e3 . (e) Show that p = y z p p p e1 + e2 + e3 is the force per unit volume acting at the point (x, y, z) of the uid medium. x y z 2. Follow the example of exercise 1 above but use cylindrical coordinates and nd the force per unit volume at a point (r, , z). Hint: An element of volume in cylindrical coordinates is given by V = rrz. 3. Follow the example of exercise 1 above but use spherical coordinates and nd the force per unit volume at a point (, , ). Hint: An element of volume in spherical coordinates is V = 2 sin . 4. Show that if the density = (x, y, z, t) is a constant, then v r = 0. ,r 5. Assume that and are zero. Such a uid is called a nonviscous or perfect uid. (a) Show the Cartesian equations describing conservation of linear momentum are u u u u 1 p +u +v +w = bx t x y z x v v v v 1 p +u +v +w = by t x y z y w w w w 1 p +u +v +w = bz t x y z z where (u, v, w) are the physical components of the uid velocity. (b) Show that the continuity equation can be written + ( u) + ( v) + ( w) = 0 t x y z 318 6. Assume = = 0 so that the uid is ideal or nonviscous. Use the results given in problem 5 and make the following additional assumptions: The density is constant and so the uid is incompressible. The body forces are zero. Steady state ow exists. Only two dimensional ow in the x-yplane is considered such that u = u(x, y), v = v(x, y) and w = 0. (a) Employ the above assumptions and simplify the equations in problem 5 and verify the results u u 1 p u +v + =0 x y x v v 1 p u +v + =0 x y y u v + =0 x y (b) Make the additional assumption that the ow is irrotational and show that this assumption produces the results v u =0 x y and 12 1 u + v 2 + p = constant. 2 (c) Point out the Cauchy-Riemann equations and Bernoullis equation in the above set of equations. 7. Assume the body forces are derivable from a potential function such that bi = ,i . Show that for an ideal uid with constant density the equations of uid motion can be written in either of the forms 1 v r + v r v s = g rm p,m g rm ,m ,s t or vr 1 + vr,s v s = p,r ,r t (v ) v = 1 (v v) v ( v) are 2 used to express the Navier-Stokes-Duhem equations in alternate forms involving the vorticity = v. 8. The vector identities 2 v = ( v) ( v) and (a) Use Cartesian tensor notation and derive the above identities. (b) Show the second identity can be written v 2 in generalized coordinates as v j v m = g mj v k vk,j mnp ijk gpi vn vk,j . Hint: Show that = 2v k vk,j . ,j xj 9. Use problem 8 and show that the results in problem 7 can be written v r t vi t rnp or p v2 ++ m x 2 2 p v jk ++ ijk v = xi 2 p vn = g rm 10. In terms of physical components, show that in generalized orthogonal coordinates, for i = j, the rate v(i) v(j) 1 hi hj of deformation tensor Dij can be written D(ij) = + , no summations j i 2 hj x hi hi x hj and for i = j there results D(ii) = Problem 17 Exercise 2.1.) 1 v(i) v(i) hi 2 + hi xi hi xi 3 k=1 hi 1 v(k) k , hi hk x no summations. (Hint: See 319 Figure 2.5-6. Plane Couette ow 11. Find the physical components of the rate of deformation tensor Dij in Cartesian coordinates. (Hint: See problem 10.) 12. Find the physical components of the rate of deformation tensor in cylindrical coordinates. (Hint: See problem 10.) 13. (Plane Couette ow) Assume a viscous uid with constant density is between two plates as illustrated in the gure 2.5-6. (a) Dene = as the kinematic viscosity and show the equations of uid motion can be written v i 1 + v i,s v s = g im p,m + g jm v i,mj + g ij bj , t i = 1, 2, 3 (b) Let v = (u, v, w) denote the physical components of the uid ow and make the following assumptions u = u(y), v = w = 0 Steady state ow exists The top plate, with area A, is a distance p and are constants above the bottom plate. The bottom plate is xed and a constant force F is applied to the top plate to keep it moving with a velocity u0 = u( ). The body force components are zero. Find the velocity u = u(y) (c) Show the tangential stress exerted by the moving uid is u0 F = 21 = xy = yx = . This A example illustrates that the stress is proportional to u0 and inversely proportional to . 14. In the continuity equation make the change of variables t= t , = 0 , v= v , v0 x= x , L y= y , L z= z L and write the continuity equation in terms of the barred variables and the Strouhal parameter. 15. (Plane Poiseuille ow) Consider two at plates parallel to one another as illustrated in the gure 2.5-7. One plate is at y = 0 and the other plate is at y = 2 . Let v = (u, v, w) denote the physical components of the uid velocity and make the following assumptions concerning the ow The body forces are zero. The p p p = p0 is a constant and = = 0. The velocity in the x-direction is a function of y only derivative x y z 320 Figure 2.5-7. Plane Poiseuille ow with u = u(y) and v = w = 0 with boundary values u(0) = u(2 ) = 0. The density is constant and = / is the kinematic viscosity. d2 u p0 + = 0, u(0) = u(2 ) = 0 dy 2 (b) Find the velocity u = u(y) and nd the maximum velocity in the x-direction. (c) Let M denote the (a) Show the equation of uid motion is mass ow rate across the plane x = x0 = constant, , where 0 y 2 , and 0 z 1. 2 Show that M = p0 3 . Note that as increases, M decreases. 3 (cu) , where c is the t 3 specic heat [cal/gm C], is the volume density [gm/cm ], H is the rate of heat generation [cal/sec cm3 ], u 16. The heat equation (or diusion equation) can be expressed div ( k grad u)+ H = is the temperature [C], k is the thermal conductivity [cal/sec cm C]. Assume constant thermal conductivity, volume density and specic heat and express the boundary value problem 2u u = c , 0 < x < L x2 t u(L, t) = u1 , u(x, 0) = f (x) k u(0, t) = 0, in a form where all the variables are dimensionless. Assume u1 is constant. 17. Simplify the Navier-Stokes-Duhem equations using the assumption that there is incompressible ow. 18. (Rayleigh impulsive ow) The gure 2.5-8 illustrates uid motion in the plane where y > 0 above a plate located along the axis where y = 0. The plate along y = 0 has zero velocity for all negative time and at time t = 0 the plate is given an instantaneous velocity u0 in the positive x-direction. Assume the physical components of the velocity are v = (u, v, w) which satisfy u = u(y, t), v = w = 0. Assume that the density of the uid is constant, the gradient of the pressure is zero, and the body forces are zero. (a) Show that the velocity in the x-direction is governed by the dierential equation 2u u = 2, t y with = . Assume u satises the initial condition u(0, t) = u0 H(t) where H is the Heaviside step function. Also assume there exist a condition at innity limy u(y, t). This latter condition requires a bounded velocity at innity. (b) Use any method to show the velocity is u(y, t) = u0 u0 erf y 2 t = u0 erfc y 2 t 321 Figure 2.5-8. Rayleigh impulsive ow where erf and erfc are the error function and complimentary error function respectively. Pick a point on the line y = y0 = 2 and plot the velocity as a function of time. How does the viscosity eect the velocity of the uid along the line y = y0 ? 19. Simplify the Navier-Stokes-Duhem equations using the assumption that there is incompressible and irrotational ow. 20. Let = + 2 and show the constitutive equations (2.5.21) for uid motion can be written in the 3 form 2 ij = pij + vi,j + vj,i ij vk,k + ij vk,k . 3 21. (a) Write out the Navier-Stokes-Duhem equation for two dimensional ow in the x-y direction under the assumptions that + 2 = 0 3 (This condition is referred to as Stokes ow.) The uid is incompressible There is a gravitational force b = g h Hint: Express your answer as two scalar equations involving the variables v1 , v2 , h, g, , p, t, plus the continuity equation. (b) In part (a) eliminate the pressure and body force terms by cross dierentiation and subtraction. (i.e. take the derivative of one equation with respect to x and take the derivative of the other equation with respect to y v1 1 v2 and then eliminate any common terms.) (c) Assume that = e3 where = and 2 x y derive the vorticity-transport equation d = 2 dt where d = + v1 + v2 . dt t x y Hint: The continuity equation makes certain terms zero. (d) Dene a stream function = (x, y) satisfying v1 = and v2 = and show the continuity equation is identically satised. y x 1 Show also that = 2 2 and that 4 = 1 2 2 2 + . t y x x y If is very large, show that 4 0. 322 22. In generalized orthogonal coordinates, show that the physical components of the rate of deformation stress can be written, for i = j (ij) = and for i = j = k (ii) = p + 2 + h1 h2 h3 1 v(i) 1 hi 1 hi + v(j) j + v(k) k i hi x hi hj x hi hk x {h2 h3 v(1)} + 2 {h1 h3 v(2)} + 3 {h1 h2 v(3)} , x1 x x hi hj xj v(i) hi + hj hi xi v(j) hj , no summation, no summation 23. Find the physical components for the rate of deformation stress in Cartesian coordinates. Hint: See problem 22. 24. Find the physical components for the rate of deformations stress in cylindrical coordinates. Hint: See problem 22. 1 25. Verify the Navier-Stokes equations for an incompressible uid can be written vi = p,i + vi,mm + bi where = is called the kinematic viscosity. 26. Verify the Navier-Stokes equations for a compressible uid with zero bulk viscosity can be written 1 vi = p,i + vm,mi + vi,mm + bi with = the kinematic viscosity. 3 27. The constitutive equation for a certain non-Newtonian Stokesian uid is ij = pij +Dij +Dik Dkj . Assume that and are constants (a) Verify that ij,j = p,i + Dij,j + (Dik Dkj,j + Dik,j Dkj ) (b) Write out the Cauchy equations of motion in Cartesian coordinates. (See page 236). 28. Let the constitutive equations relating stress and strain for a solid material take into account thermal 1+ ij kk ij + T ij stresses due to a temperature T . The constitutive equations have the form eij = E E where is a coecient of linear expansion for the material and T is the absolute temperature. Solve for the stress in terms of strains. 29. Derive equation (2.5.53) and then show that when the bulk coecient of viscosity is zero, the NavierStokes equations, in Cartesian coordinates, can be written in the conservation form ( u) ( u2 + p xx ) ( uv xy ) ( uw xz ) + + + = bx t x y z ( v) ( uv xy ) ( v 2 + p yy ) ( vw yz ) + + + = by t x y z ( w) ( uw xz ) ( vw yz ) ( w2 + p zz ) + + + = bz t x y z 2 where v1 = u,v2 = v,v3 = w and ij = (vi,j + vj,i ij vk,k ). Hint: Alternatively, consider 2.5.29 and use 3 the continuity equation. 323 2 30. Show that for a perfect gas, where = 3 and = is a function of position, the vector form of equation (2.5.25) is 4 Dv = b p + ( v) + (v ) v 2 + () ( v) ( v) ( (v)) Dt 3 31. Derive the energy equation D p Q Dh = + q + . Hint: Use the continuity equation. Dt Dt t 32. Show that in Cartesian coordinates the Navier-Stokes equations of motion for a compressible uid can be written Du p u =bx + 2 + V Dt x x x Dv v p =by + + V 2 Dt y y y Dv w p =bz + + V 2 Dt z z z u v w u ( + )+ ( + ) y y x z x z w v w w + )+ + ) + ( ( z z y x y x w w u v + )+ + ) + ( ( x x z y z y + where (Vx , Vy , Vz ) = (u, v, w). 33. Show that in cylindrical coordinates the Navier-Stokes equations of motion for a compressible uid can be written V V Vr p 1 1 Vr + + V + + ) 2 ( r r r r r r r Vz 2 Vr 1 V Vr Vr + + )+ ( ) ( z z r r r r r DV Vr V Vr V 1 p 1 1 V 1 Vz + + + ) + V + + ) = b 2 ( ( Dt r r r r r z r z V V 2 1 Vr V V 1 Vr + + )+ ( + ) ( r r r r r r r r DVz Vz p 1 Vr Vz = bz + + V + + ) 2 r( Dt z z z r r z r 1 V 1 Vz + + ) ( r r z = br 34. Show that the dissipation function can be written as = 2 Dij Dij + 2 . 35. Verify the identities: (a) D et (et / ) = + (et V ) Dt t (b) D D De (et / ) = + V 2 /2 . Dt Dt Dt DVr V2 Dt r 36. Show that the conservation law for heat ow is given by T + (T v T ) = SQ t where is the thermal conductivity of the material, T is the temperature, Jadvection = T v, Jconduction = T and SQ is a source term. Note that in a solid material there is no ow and so v = 0 and 324 the above equation reduces to the heat equation. Assign units of measurements to each term in the above equation and make sure the equation is dimensionally homogeneous. 37. Show that in spherical coordinates the Navier-Stokes equations of motion for a compressible uid can be written 2 2 V p 1 1 V DV V + V ) = b + + V + (V /) + ) 2 ( Dt 1 1 V + + (V /)) ( sin sin V 2 V 4V 2 V 2V cot cot V + (4 + cot (V /) + ) sin 2 DV V V V cot 1 p 1 2 V ( + ) = b + ( + V ) + V Dt sin 1 V 1 V 1 (V / sin ) + )+ (V /) + ) + ( ( sin sin 1 V 1 V V cot 1 V + (V /) + 2 cot + 3 sin 1 V 1 p V V V V cot DV + + = b + + (V /) Dt sin sin 2 1 V 1 + V + V cot + V + sin sin sin 1 1 V + (V / sin ) + sin sin 1 V 1 V + (V /) + 2 cot (V / sin ) + + 3 sin sin ( 38. Verify all the equations (2.5.28). 39. Use the conservation of energy equation (2.5.47) together with the momentum equation (2.5.25) to derive the equation (2.5.48). 40. Verify the equation (2.5.55). 41. Consider nonviscous ow and write the 3 linear momentum equations and the continuity equation and make the following assumptions: (i) The density is constant. (ii) Body forces are zero. (iii) Steady state ow only. (iv) Consider only two dimensional ow with non-zero velocity components u = u(x, y) and v = v(x, y). Show that there results the system of equations u u u 1 P +v + = 0, x y x u v v 1 P +v + = 0, x y y u v + = 0. x y Recognize that the last equation in the above set as one of the Cauchy-Riemann equations that f (z) = u iv be an analytic function of a complex variable. Further assume that the uid ow is irrotational so that 12 v u P = 0. Show that this implies that u + v 2 + = Constant. If in addition u and v are derivable x y 2 from a potential function (x, y), such that u = and v = , then show that is a harmonic function. x y By constructing the conjugate harmonic function (x, y) the complex potential F (z) = (x, y) + i(x, y) is such that F (z) = u(x, y) iv(x, y) and F (z) gives the velocity. The family of curves (x, y) =constant are called equipotential curves and the family of curves (x, y) = constant are called streamlines. Show that these families are an orthogonal family of curves. 325 2.6 ELECTRIC AND MAGNETIC FIELDS Introduction In electromagnetic theory the mks system of units and the Gaussian system of units are the ones most often encountered. In this section the equations will be given in the mks system of units. If you want the equations in the Gaussian system of units make the replacements given in the column 3 of Table 1. Table 1. MKS AND GAUSSIAN UNITS MKS symbol E (Electric eld) B (Magnetic eld) D (Displacement eld) H (Auxiliary Magnetic eld) J (Current density) A (Vector potential) V (Electric potential) (Dielectric constant) (Magnetic permeability) Electrostatics A basic problem in electrostatic theory is to determine the force F on a charge Q placed a distance r from another charge q. The solution to this problem is Coulombs law 1 qQ er F= 4 0 r2 where q, Q are measured in coulombs, 0 MKS units volt/m weber/m2 coulomb/m2 ampere/m ampere/m2 weber/m volt Replacement symbol E B c D 4 cH 4 GAUSSIAN units statvolt/cm gauss statcoulomb/cm2 oersted statampere/cm2 gauss-cm statvolt J A c V 4 4 c2 (2.6.1) = 8.85 1012 coulomb2 /N m2 is called the permittivity in a vacuum, r is in meters, [F ] has units of Newtons and er is a unit vector pointing from q to Q if q, Q have the same sign or pointing from Q to q if q, Q are of opposite sign. The quantity E = F /Q is called the electric eld produced by the charges. In the special case Q = 1, we have E = F and so Q = 1 is called a test charge. This tells us that the electric eld at a point P can be viewed as the force per unit charge exerted on a test charge Q placed at the point P. The test charge Q is always positive and so is repulsed if q is positive and attracted if q is negative. The electric eld associated with many charges is obtained by the principal of superposition. For example, let q1 , q2 , . . . , qn denote n-charges having respectively the distances r1 , r2 , . . . , rn from a test charge Q placed at a point P. The force exerted on Q is F =F1 + F2 + + Fn F= or E = E(P ) = 1 4 0 q1 Q q2 Q qn Q 2 er1 + r 2 er2 + + r 2 ern r1 n 2 n (2.6.2) F 1 = Q 4 0 i=1 qi 2 er ri i 326 where E = E(P ) is the electric eld associated with the system of charges. The equation (2.6.2) can be generalized to other situations by dening other types of charge distributions. We introduce a line charge density , (coulomb/m), a surface charge density , (coulomb/m2 ), a volume charge density , (coulomb/m3 ), then we can calculate the electric eld associated with these other types of charge distributions. For example, if there is a charge distribution = (s) along a curve C, where s is an arc length parameter, then we would have E(P ) = 1 4 0 C er ds r2 (2.6.3) as the electric eld at a point P due to this charge distribution. The integral in equation (2.6.3) being a line integral along the curve C and where ds is an element of arc length. Here equation (2.6.3) represents a continuous summation of the charges along the curve C. For a continuous charge distribution over a surface S, the electric eld at a point P is E(P ) = 1 4 0 S er d r2 (2.6.4) where d represents an element of surface area on S. Similarly, if represents a continuous charge distribution throughout a volume V , then the electric eld is represented E(P ) = 1 4 0 er d r2 (2.6.5) V where d is an element of volume. In the equations (2.6.3), (2.6.4), (2.6.5) we let (x, y, z) denote the position of the test charge and let (x , y , z ) denote a point on the line, on the surface or within the volume, then r = (x x ) e1 + (y y ) e2 + (z z ) e3 (2.6.6) r represents the distance from the point P to an element of charge ds, d or d with r = |r| and er = . r If the electric eld is conservative, then E = 0, and so it is derivable from a potential function V by taking the negative of the gradient of V and E = V. (2.6.7) For these conditions note that V dr = E dr is an exact dierential so that the potential function can be represented by the line integral P V = V(P ) = E dr (2.6.8) where is some reference point (usually innity, where V() = 0). For a conservative electric eld the line integral will be independent of the path connecting any two points a and b so that b a b b V(b) V(a) = E dr E dr = a E dr = a V dr. (2.6.9) Let = in equation (2.6.8), then the potential function associated with a point charge moving in the radial direction er is r V(r) = E dr = q 4 0 r 1 q dr = r2 4 0 1r q |= . r 4 0 r 327 By superposition, the potential at a point P for a continuous volume distribution of charges is given by 1 1 d and for a surface distribution of charges V(P ) = d and for a line V(P ) = 4 0 4 0 Vr Sr 1 distribution of charges V(P ) = ds; and for a discrete distribution of point charges 4 0 C r N qi 1 V(P ) = . When the potential functions are dened from a common reference point, then the 4 0 i=1 ri principal of superposition applies. The potential function V is related to the work done W in moving a charge within the electric eld. The work done in moving a test charge Q from point a to point b is an integral of the force times distance moved. The electric force on a test charge Q is F = QE and so the force F = QE is in opposition to this force as you move the test charge. The work done is b b b W= a F dr = a QE dr = Q a V dr = Q[V(b) V(a)]. (2.6.10) The work done is independent of the path joining the two points and depends only on the end points and the change in the potential. If one moves Q from innity to point b, then the above becomes W = QV (b). An electric eld E = E(P ) is a vector eld which can be represented graphically by constructing vectors at various selected points in the space. Such a plot is called a vector eld plot. A eld line associated with a vector eld is a curve such that the tangent vector to a point on the curve has the same direction as the vector eld at that point. Field lines are used as an aid for visualization of an electric eld and vector elds in general. The tangent to a eld line at a point has the same direction as the vector eld E at that point. For example, in two dimensions let r = x e1 + y e2 denote the position vector to a point on a eld line. The tangent vector to this point has the direction dr = dx e1 + dy e2 . If E = E(x, y) = N (x, y) e1 + M (x, y) e2 is the vector eld constructed at the same point, then E and dr must be colinear. Thus, for each point (x, y) on a eld line we require that dr = K E for some constant K. Equating like components we nd that the eld lines must satisfy the dierential relation. dx dy = =K N (x, y) M (x, y) or M (x, y) dx + N (x, y) dy =0. (2.6.11) In two dimensions, the family of equipotential curves V(x, y) = C1 =constant, are orthogonal to the family of eld lines and are described by solutions of the dierential equation N (x, y) dx M (x, y) dy = 0 obtained from equation (2.6.11) by taking the negative reciprocal of the slope. The eld lines are perpendicular to the equipotential curves because at each point on the curve V = C1 we have V being perpendicular to the curve V = C1 and so it is colinear with E at this same point. Field lines associated with electric elds are called electric lines of force. The density of the eld lines drawn per unit cross sectional area are proportional to the magnitude of the vector eld through that area. 328 Figure 2.6-1. Electric forces due to a positive charge at (a, 0) and negative charge at (a, 0). EXAMPLE 2.6-1. Find the eld lines and equipotential curves associated with a positive charge q located at the point (a, 0) and a negative charge q located at the point (a, 0). Solution: With reference to the gure 2.6-1, the total electric force E on a test charge Q = 1 place at a general point (x, y) is, by superposition, the sum of the forces from each of the isolated charges and is E = E1 + E2 . The electric force vectors due to each individual charge are E1 = kq(x + a) e1 + kqy e2 2 with r1 = (x + a)2 + y 2 3 r1 kq(x a) e1 kqy e2 2 E2 = with r2 = (x a)2 + y 2 3 r2 (2.6.12) where k = 1 is a constant. This gives 4 0 E = E1 + E2 = kq(x + a) kq(x a) kqy kqy e1 + 3 3 3 r3 r1 r2 r1 2 e2 . This determines the dierential equation of the eld lines dx kq(x+a) 3 r1 kq(xa) 3 r2 = kqy 3 r1 dy . kqy r3 2 (2.6.13) To solve this dierential equation we make the substitutions cos 1 = x+a r1 and cos 2 = xa r2 (2.6.14) 329 Figure 2.6-2. Lines of electric force between two opposite sign charges. as suggested by the geometry from gure 2.6-1. From the equations (2.6.12) and (2.6.14) we obtain the relations sin 1 d1 = r1 dx (x + a) dr1 2 r1 r2 dx (x a)dr2 2 r2 2r1 dr1 =2(x + a) dx + 2ydy sin 2 d2 = 2r2 dr2 =2(x a) dx + 2y dy which implies that sin 1 d1 = (x + a)y dy y 2 dx +3 3 r1 r1 2 (x a)y dy y dx sin 2 d2 = +3 3 r2 r2 (2.6.15) Now compare the results from equation (2.6.15) with the dierential equation (2.6.13) and determine that y is an integrating factor of equation (2.6.13) . This shows that the dierential equation (2.6.13) can be written in the much simpler form of the exact dierential equation sin 1 d1 + sin 2 d2 = 0 in terms of the variables 1 and 2 . The equation (2.6.16) is easily integrated to obtain cos 1 cos 2 = C where C is a constant of integration. In terms of x, y the solution can be written x+a (x + a)2 + y 2 These eld lines are illustrated in the gure 2.6-2. xa = C. (x a)2 + y 2 (2.6.18) (2.6.17) (2.6.16) 330 The dierential equation for the equipotential curves is obtained by taking the negative reciprocal of the slope of the eld lines. This gives dy = dx This result can be written in the form (x + a)dx + ydy (x a)dx + ydy + =0 3 3 r1 r2 kq(xa) 3 r2 kqy 3 r1 kq(x+a) 3 r1 kqy 3 r2 . which simplies to the easily integrable form dr1 dr2 2 + r2 = 0 r1 2 in terms of the new variables r1 and r2 . An integration produces the equipotential curves 1 1 =C2 r1 r2 1 =C2 . (x a)2 + y 2 or 1 (x + a)2 + y 2 The potential function for this problem can be interpreted as a superposition of the potential functions kq kq V1 = and V2 = associated with the isolated point charges at the points (a, 0) and (a, 0). r1 r2 Observe that the electric lines of force move from positive charges to negative charges and they do not cross one another. Where eld lines are close together the eld is strong and where the lines are far apart the eld is weak. If the eld lines are almost parallel and equidistant from one another the eld is said to be uniform. The arrows on the eld lines show the direction of the electric eld E. If one moves along a eld line in the direction of the arrows the electric potential is decreasing and they cross the equipotential curves at right angles. Also, when the electric eld is conservative we will have E = 0. In three dimensions the situation is analogous to what has been done in two dimensions. If the electric eld is E = E(x, y, z) = P (x, y, z) e1 + Q(x, y, z) e2 + R(x, y, z) e3 and r = x e1 + y e2 + z e3 is the position vector to a variable point (x, y, z) on a eld line, then at this point dr and E must be colinear so that dr = K E for some constant K. Equating like coecients gives the system of equations dx dy dz = = = K. P (x, y, z) Q(x, y, z) R(x, y, z) (2.6.19) From this system of equations one must try to obtain two independent integrals, call them u1 (x, y, z) = c1 and u2 (x, y, z) = c2 . These integrals represent one-parameter families of surfaces. When any two of these surfaces intersect, the result is a curve which represents a eld line associated with the vector eld E. These type of eld lines in three dimensions are more dicult to illustrate. The electric ux E of an electric eld E over a surface S is dened as the summation of the normal component of E over the surface and is represented E = S E n d with units of N m2 C (2.6.20) 331 where n is a unit normal to the surface. The ux E can be thought of as being proportional to the number of electric eld lines passing through an element of surface area. If the surface is a closed surface we have by the divergence theorem of Gauss E = V E d = S E n d where V is the volume enclosed by S. Gauss Law Let d denote an element of surface area on a surface S. A cone is formed if all points on the boundary of d are connected by straight lines to the origin. The cone need not be a right circular cone. The situation is illustrated in the gure 2.6-3. Figure 2.6-3. Solid angle subtended by element of area. We let r denote a position vector from the origin to a point on the boundary of d and let n denote a unit outward normal to the surface at this point. We then have n r = r cos where r = |r| and is the angle between the vectors n and r. Construct a sphere, centered at the origin, having radius r. This sphere d intersects the cone in an element of area d. The solid angle subtended by d is dened as d = 2 . Note r that this is equivalent to constructing a unit sphere at the origin which intersect the cone in an element of area d. Solid angles are measured in steradians. The total solid angle about a point equals the area of the sphere divided by its radius squared or 4 steradians. The element of area d is the projection of d on the nr nr d d so that d = 3 d = 2 . Observe that sometimes the constructed sphere and d = d cos = r r r dot product n r is negative, the sign depending upon which of the normals to the surface is constructed. (i.e. the inner or outer normal.) The Gauss law for electrostatics in a vacuum states that the ux through any surface enclosing many charges is the total charge enclosed by the surface divided by E n d = S Qe 0 0. The Gauss law is written (2.6.21) 0 for charges inside S for charges outside S 332 where Qe represents the total charge enclosed by the surface S with n the unit outward normal to the surface. The proof of Gausss theorem follows. Consider a single charge q within the closed surface S. The electric 1q er and so the ux eld at a point on the surface S due to the charge q within S is represented E = 4 0 r2 integral is q er n d q q E n d = E = d = = (2.6.22) 4 0 r2 4 0 r2 0 S S S since er n cos d d = = 2 = d and 2 2 r r r d = 4. By superposition of the charges, we obtain a similar S n result for each of the charges within the surface. Adding these results gives Qe = i=1 qi . For a continuous distribution of charge inside the volume we can write Qe = V d , where is the charge distribution per unit volume. Note that charges outside of the closed surface do not contribute to the total ux across the surface. This is because the eld lines go in one side of the surface and go out the other side. In this case S E n d = 0 for charges outside the surface. Also the position of the charge or charges within the The equation (2.6.21) is the Gauss law in integral form. We can put this law in dierential form as volume does not eect the Gauss law. follows. Using the Gauss divergence theorem we can write for an arbitrary volume that E n d = S V E d = V 0 d = Qe 0 = 1 0 V d which for an arbitrary volume implies E = 0 . (2.6.23) The equations (2.6.23) and (2.6.7) can be combined so that the Gauss law can also be written in the form which is called Poissons equation. 2 V = 0 EXAMPLE 2.6-2 Find the electric eld associated with an innite plane sheet of positive charge. Solution: Assume there exists a uniform surface charge and draw a circle at some point on the plane surface. Now move the circle perpendicular to the surface to form a small cylinder which extends equal distances above and below the plane surface. We calculate the electric ux over this small cylinder in the limit as the height of the cylinder goes to zero. The charge inside the cylinder is A where A is the area of the circle. We nd that the Gauss law requires that E n d = S Qe 0 = A 0 (2.6.24) where n is the outward normal to the cylinder as we move over the surface S. By the symmetry of the situation the electric force vector is uniform and must point away from both sides to the plane surface in the direction of the normals to both sides of the surface. Denote the plane surface normals by en and en and assume that E = en on one side of the surface and E = en on the other side of the surface for some constant . Substituting this result into the equation (2.6.24) produces E n d = 2A S (2.6.25) 333 since only the ends of the cylinder contribute to the above surface integral. On the sides of the cylinder we will have n en = 0 and so the surface integral over the sides of the cylinder is zero. By equating the and consequently we can write results from equations (2.6.24) and (2.6.25) we obtain the result that = 20 en where en represents one of the normals to the surface. E= 20 Note an electric eld will always undergo a jump discontinuity when crossing a surface charge . As in en and Edown = en so that the dierence is the above example we have Eup = 20 2 Eup Edown = 0 en or E i ni (1) + E i ni (2) + 0 = 0. (2.6.26) It is this dierence which causes the jump discontinuity. EXAMPLE 2.6-3. Calculate the electric eld associated with a uniformly charged sphere of radius a. Solution: We proceed as in the previous example. Let denote the uniform charge distribution over the surface of the sphere and let er denote the unit normal to the sphere. The total charge then is written as q= Sa d = 4a2 . If we construct a sphere of radius r > a around the charged sphere, then we have E er d = Sr by the Gauss theorem Qe 0 = q 0 . (2.6.27) Again, we can assume symmetry for E and assume that it points radially outward in the direction of the surface normal er and has the form E = er for some constant . Substituting this value for E into the equation (2.6.27) we nd that E er d = Sr Sr d = 4r2 = q 0 . (2.6.28) 1q er where er is the outward normal to the sphere. This shows that the electric eld 4 0 r2 outside the sphere is the same as if all the charge were situated at the origin. This gives E = For S a piecewise closed surface enclosing a volume V and F i = F i (x1 , x2 , x3 ) i = 1, 2, 3, a continuous vector eld with continuous derivatives the Gauss divergence theorem enables us to replace a ux integral of F i over S by a volume integral of the divergence of F i over the volume V such that F i ni d = S V F i d ,i or S F n d = V div F d. (2.6.29) If V contains a simple closed surface where F i is discontinuous we must modify the above Gauss divergence theorem. EXAMPLE 2.6-4. We examine the modication of the Gauss divergence theorem for spheres in order to illustrate the concepts. Let V have surface area S which encloses a surface . Consider the gure 2.6-4 where the volume V enclosed by S and containing has been cut in half. 334 Figure 2.6-4. Sphere S containing sphere . Applying the Gauss divergence theorem to the top half of gure 2.6-4 gives F i nT d + i ST Sb1 F i nbT d + i T F i nT d = i VT i F ,i d (2.6.30) where the ni are the unit outward normals to the respective surfaces ST , Sb1 and T . Applying the Gauss divergence theorem to the bottom half of the sphere in gure 2.6-4 gives F i nB d + i SB Sb2 F i nbB d + i B F i nB d = i VB F i d ,i (2.6.31) Observe that the unit normals to the surfaces Sb1 and Sb2 are equal and opposite in sign so that adding the equations (2.6.30) and (2.6.31) we obtain F i ni d + S F i ni d = VT +VB (1) i F ,i d (2.6.32) 335 where S = ST + SB is the total surface area of the outside sphere and = T + B is the total surface area of the inside sphere, and ni (1) is the inward normal to the sphere when the top and bottom volumes are combined. Applying the Gauss divergence theorem to just the isolated small sphere we nd F i ni d = V (2) F i d ,i (2.6.33) where ni (2) is the outward normal to . By adding the equations (2.6.33) and (2.6.32) we nd that F i ni d + S F i ni (1) + F i ni (2) d = V F i d ,i (2.6.34) where V = VT + VB + V . The equation (2.6.34) can also be written as F i ni d = S V i F ,i d F i ni (1) + F i ni (2) d. (2.6.35) In the case that V contains a surface the total electric charge inside S is Qe = V d + d (2.6.36) where is the surface charge density on and is the volume charge density throughout V. The Gauss theorem requires that E i ni d = S Qe 0 = 1 0 V d + 1 0 d. (2.6.37) In the case of a jump discontinuity across the surface we use the results of equation (2.6.34) and write E i ni d = S V E i,i d E i ni (1) + E i ni (2) d. (2.6.38) Subtracting the equation (2.6.37) from the equation (2.6.38) gives Ei ,i V 0 d E i ni (1) + E i ni (2) + 0 d = 0. (2.6.39) For arbitrary surfaces S and , this equation implies the dierential form of the Gauss law Ei = ,i 0 . (2.6.40) Further, on the surface , where there is a surface charge distribution we have E i ni (1) + E i ni (2) + 0 =0 (2.6.41) which shows the electric eld undergoes a discontinuity when you cross a surface charge . 336 Electrostatic Fields in Materials When charges are introduced into materials it spreads itself throughout the material. Materials in which the spreading occurs quickly are called conductors, while materials in which the spreading takes a long time are called nonconductors or dielectrics. Another electrical property of materials is the ability to hold local charges which do not come into contact with other charges. This property is called induction. For example, consider a single atom within the material. It has a positively charged nucleus and negatively charged electron cloud surrounding it. When this atom experiences an electric eld E the negative cloud moves opposite to E while the positively charged nucleus moves in the direction of E. If E is large enough it can ionize the atom by pulling the electrons away from the nucleus. For moderately sized electric elds the atom achieves an equilibrium position where the positive and negative charges are oset. In this situation the atom is said to be polarized and have a dipole moment p. Denition: When a pair of charges +q and q are separated by a distance 2d the electric dipole moment is dened by p = 2dq, where p has dimensions of [C m]. In the special case where d has the same direction as E and the material is symmetric we say that p is proportional to E and write p = E, where is called the atomic polarizability. If in a material subject to an electric eld their results many such dipoles throughout the material then the dielectric is said to be polarized. The vector quantity P is introduced to represent this eect. The vector P is called the polarization vector having units of [C/m2 ], and represents an average dipole moment per unit volume of material. The vectors Pi and Ei are related through the displacement vector Di such that Pi = Di For an anisotropic material (crystal) Di = where j i j i Ej 0 Ei . (2.6.42) and Pi = j Ej i (2.6.43) is called the dielectric tensor and j is called the electric susceptibility tensor. Consequently, i Pi = j Ej = i j i Ej 0 Ei =( j i j 0 i )Ej so that j = i j i j 0 i . (2.6.44) A dielectric material is called homogeneous if the electric force and displacement vector are the same for any two points within the medium. This requires that the electric force and displacement vectors be constant parallel vector elds. It is left as an exercise to show that the condition for homogeneity is that direction. This requires that space and the quantity ke = j i i i = j where j is the Kronecker delta. The term 0 j i,k = 0. A dielectric material is called isotropic if the electric force vector and displacement vector have the same = 0 Ke is called the dielectric constant of the medium. The constant 0 = 8.85(10) 12 coul /N m is the permittivity of free 2 2 0 ). is called the relative dielectric constant (relative to j i = j 0 e i For free space ke = 1. Similarly for an isotropic material we have where e is called the electric susceptibility. For a linear medium the vectors P , D and E are related by Di = 0 Ei + Pi = 0 Ei + 0 e Ei = 0 (1 + e )Ei = 0 Ke Ei = Ei (2.6.45) 337 where Ke = 1 + e is the relative dielectric constant. The equation (2.6.45) are constitutive equations for dielectric materials. The eect of polarization is to produce regions of bound charges b within the material and bound surface charges b together with free charges f which are not a result of the polarization. Within dielectrics we have P = b for bound volume charges and P en = b for bound surface charges, where en is a unit normal to the bounding surface of the volume. In these circumstances the expression for the potential function is written V= and the Gauss law becomes 0 1 4 0 V 1 b d + r 4 0 S b d r (2.6.46) E = = b + f = P + f or ( 0 E + P ) = f . (2.6.47) Since D = 0E + P the Gauss law can also be written in the form D = f or i D,i = f . (2.6.48) When no confusion arises we replace f by . In integral form the Gauss law for dielectrics is written D n d = Qf e S (2.6.49) where Qf e is the total free charge density within the enclosing surface. Magnetostatics A stationary charge generates an electric eld E while a moving charge generates a magnetic eld B. Magnetic eld lines associated with a steady current moving in a wire form closed loops as illustrated in the gure 2.6-5. Figure 2.6-5. Magnetic eld lines. The direction of the magnetic force is determined by the right hand rule where the thumb of the right hand points in the direction of the current ow and the ngers of the right hand curl around in the direction of the magnetic eld B. The force on a test charge Q moving with velocity V in a magnetic eld is Fm = Q(V B). The total electromagnetic force acting on Q is the electric force plus the magnetic force and is F = Q E + (V B) (2.6.51) (2.6.50) 338 which is known as the Lorentz force law. The magnetic force due to a line charge density moving along a curve C is the line integral Fmag = C ds(V B) = C I Bds. (2.6.52) Similarly, for a moving surface charge density moving on a surface Fmag = S d(V B) = S K B d (2.6.53) and for a moving volume charge density Fmag = V d (V B) = V J B d (2.6.54) where the quantities I = V , K = V and J = V are respectively the current, the current per unit length, and current per unit area. A conductor is any material where the charge is free to move. The ow of charge is governed by Ohms law. Ohms law states that the current density vector Ji is a linear function of the electric intensity or Ji = im Em , where im is the conductivity tensor of the material. For homogeneous, isotropic conductors im = im so that Ji = Ei where is the conductivity and 1/ is called the resistivity. Surround a charge density with an arbitrary simple closed surface S having volume V and calculate the ux of the current density across the surface. We nd by the divergence theorem J n d = S V J d. (2.6.55) If charge is to be conserved, the current ow out of the volume through the surface must equal the loss due to the time rate of change of charge within the surface which implies J n d = S V J d = d dt t d = V V d t (2.6.56) or J + V d = 0. (2.6.57) This implies that for an arbitrary volume we must have J = . t (2.6.58) Note that equation (2.6.58) has the same form as the continuity equation (2.3.73) for mass conservation and so it is also called a continuity equation for charge conservation. For magnetostatics there exists steady line currents or stationary current so t = 0. This requires that J = 0. 339 Figure 2.6-6. Magnetic eld around wire. Biot-Savart Law The Biot-Savart law for magnetostatics describes the magnetic eld at a point P due to a steady line current moving along a curve C and is B(P ) = 0 4 I er ds r2 (2.6.59) C with units [N/amp m] and where the integration is in the direction of the current ow. In the Biot-Savart law we have the constant 0 = 4 107 N/amp2 which is called the permeability of free space, I = I et is the current owing in the direction of the unit tangent vector et to the curve C, er is a unit vector directed from a point on the curve C toward the point P and r is the distance from a point on the curve to the general point P. Note that for a steady current to exist along the curve the magnitude of I must be the same everywhere along the curve. Hence, this term can be brought out in front of the integral. For surface currents K and volume currents J the Biot-Savart law is written B(P ) = and EXAMPLE 2.6-5. Calculate the magnetic eld B a distance h perpendicular to a wire carrying a constant current I. Solution: The magnetic eld circles around the wire. For the geometry of the gure 2.6-6, the magnetic eld points out of the page. We can write I er = I et er = I sin e where e is a unit vector tangent to the circle of radius h which encircles the wire and cuts the wire perpendicularly. 0 4 K er d r2 J er d. r2 S 0 B(P ) = 4 V 340 For this problem the Biot-Savart law is B(P ) = 0 I 4 e ds. r2 In terms of we nd from the geometry of gure 2.6-6 tan = Therefore, B(P ) = 0 2 1 s h with ds = h sec2 d and cos = h . r I sin h sec2 e d. h2 / cos2 But, = /2 + so that sin = cos and consequently B(P ) = 0 I e 4h 2 cos d = 1 e 0 I (sin 2 sin 1 ). 4h e 0 I . 2h For a long straight wire 1 /2 and 2 /2 to give the magnetic eld B(P ) = For volume currents the Biot-Savart law is B(P ) = and consequently (see exercises) B = 0. Recall the divergence of an electric eld is E = 0 0 4 V J er d r2 (2.6.60) (2.6.61) is known as the Gausss law for electric elds and so in analogy the divergence B = 0 is sometimes referred to as Gausss law for magnetic elds. If B = 0, then there exists a vector eld A such that B = A. The vector eld A is called the vector potential of B. Note that B = ( A) = 0. Also the vector potential A is not unique since B is also derivable from the vector potential A + where is an arbitrary continuous and dierentiable scalar. Amperes Law Amperes law is associated with the work done in moving around a simple closed path. For example, consider the previous example 2.6-5. In this example the integral of B around a circular path of radius h which is centered at some point on the wire can be associated with the work done in moving around this path. The summation of force times distance is B dr = C C B e ds = 0 I 2h ds = 0 I C (2.6.62) where now dr = e ds is a tangent vector to the circle encircling the wire and C ds = 2h is the distance around this circle. The equation (2.6.62) holds not only for circles, but for any simple closed curve around the wire. Using the Stokes theorem we have B dr = C S ( B) en d = 0 I = S 0 J en d (2.6.63) 341 where S J en d is the total ux (current) passing through the surface which is created by encircling some curve about the wire. Equating like terms in equation (2.6.63) gives the dierential form of Amperes law B = 0 J. Magnetostatics in Materials Similar to what happens when charges are introduced into materials we have magnetic elds whenever there are moving charges within materials. For example, when electrons move around an atom tiny current loops are formed. These current loops create what are called magnetic dipole moments m throughout the material. When a magnetic eld B is applied to a material medium there is a net alignment of the magnetic dipoles. The quantity M , called the magnetization vector is introduced. Here M is associated with a dielectric medium and has the units [amp/m] and represents an average magnetic dipole moment per unit volume and is analogous to the polarization vector P used in electrostatics. The magnetization vector M acts a lot like the previous polarization vector in that it produces bound volume currents Jb and surface currents Kb where M = Jb is a volume current density throughout some volume and M en = Kb is a surface current on the boundary of this volume. From electrostatics note that the time derivative of free current and becomes B = 0 Jt = 0 (Jb + Jf + where J = Jb + Jf . The term E 0 t 0 E 0 t E 0 t (2.6.64) has the same units as current density. The E 0 t total current in a magnetized material is then Jt = Jb + Jf + where Jb is the bound current, Jf is the is the induced current. Amperes law, equation (2.6.64), in magnetized materials then E ) = 0 J + 0 t E t (2.6.65) 0 is referred to as a displacement current or as a Maxwell correction to the eld equation. This term implies that a changing electric eld induces a magnetic eld. An auxiliary magnet eld H dened by Hi = 1 Bi Mi 0 (2.6.66) is introduced which relates the magnetic force vector B and magnetization vector M . This is another constitutive equation which describes material properties. For an anisotropic material (crystal) Bi = j Hj i and M i = j Hj i (2.6.67) where j is called the magnetic permeability tensor and j is called the magnetic permeability tensor. Both i i of these quantities are dimensionless. For an isotropic material j j = i i where = 0 km . 0 (2.6.68) is the relative permeability Here 0 = 4 107 N/amp2 is the permeability of free space and km = coecient. Similarly, for an isotropic material we have j i = j m i where m is called the magnetic sus- ceptibility coecient and is dimensionless. The magnetic susceptibility coecient has positive values for 342 materials called paramagnets and negative values for materials called diamagnets. For a linear medium the quantities B, M and H are related by Bi = 0 (Hi + Mi ) = 0 Hi + 0 m Hi = 0 (1 + m )Hi = 0 km Hi = Hi where = 0 km = 0 (1 + m ) is called the permeability of the material. Note: The auxiliary magnetic vector H for magnetostatics in materials plays a role similar to the displacement vector D for electrostatics in materials. Be careful in using electromagnetic equations from dierent texts as many authors interchange the roles of B and H. Some authors call H the magnetic eld. However, the quantity B should be the fundamental quantity.1 Electrodynamics In the nonstatic case of electrodynamics there is an additional quantity Jp = current which satises Jp = and the current density has three parts J = Jb + Jf + Jp = M + Jf + consisting of bound, free and polarization currents. Faradays law states that a changing magnetic eld creates an electric eld. In particular, the electromagnetic force induced in a closed loop circuit C is proportional to the rate of change of ux of the magnetic eld associated with any surface S connected with C. Faradays law states E dr = C P t (2.6.69) called the polarization (2.6.70) b P = P = t t t P t (2.6.71) t B en d. S Using the Stokes theorem, we nd ( E) en d = S S B en d. t The above equation must hold for an arbitrary surface and loop. Equating like terms we obtain the dierential form of Faradays law E = B . t (2.6.72) This is the rst electromagnetic eld equation of Maxwell. Amperes law, equation (2.6.65), written in terms of the total current from equation (2.6.71) , becomes B = 0 ( M + Jf + which can also be written as ( 1 P ) + 0 t 0 E t (2.6.73) 1 B M ) = Jf + (P + 0 t 0 E) D.J. Griths, Introduction to Electrodynamics, Prentice Hall, 1981. P.232. 343 or H = Jf + This is Maxwells second electromagnetic eld equation. To the equations (2.6.74) and (2.6.73) we add the Gausss law for magnetization, equation (2.6.61) and Gausss law for electrostatics, equation (2.6.48). These four equations produce the Maxwells equations of electrodynamics and are now summarized. The general form of Maxwells equations involve the quantities Ei , Electric force vector, [Ei ] = Newton/coulomb Bi , Magnetic force vector, [Bi ] = Weber/m2 Hi , Auxilary magnetic force vector, [Hi ] = ampere/m Di , Displacement vector, [Di ] = coulomb/m2 Ji , Free current density, [Ji ] = ampere/m2 Pi , Polarization vector, [Pi ] = coulomb/m2 Mi , Magnetization vector, [Mi ] = ampere/m for i = 1, 2, 3. There are also the quantities , representing the free charge density, with units [ ] = coulomb/m3 0, D . t (2.6.74) Permittivity of free space, [ 0 ] = farads/m or coulomb2 /Newton m2 . 0 , Permeability of free space, [0 ] = henrys/m or kg m/coulomb2 In addition, there arises the material parameters: i , magnetic permeability tensor, which is dimensionless j i j, dielectric tensor, which is dimensionless i , electric susceptibility tensor, which is dimensionless j i , magnetic susceptibility tensor, which is dimensionless j These parameters are used to express variations in the electric eld Ei and magnetic eld Bi when acting in a material medium. In particular, Pi , Di , Mi and Hi are dened from the equations Di = j Ej = i Pi =j Ej , i 0 Ei + Pi i j = i 0 j + j i Bi =j Hj = 0 Hi + 0 Mi , i and i i = 0 (j + i ) j j M i = j Hj i for i = 1, 2, 3. The above quantities obey the following laws: Faradays Law This law states the line integral of the electromagnetic force around a loop is proportional to the rate of ux of magnetic induction through the loop. This gives rise to the rst electromagnetic eld equation: E = B t or ijk Ek,j = B i . t (2.6.75) 344 Amperes Law This law states the line integral of the magnetic force vector around a closed loop is proportional to the sum of the current through the loop and the rate of ux of the displacement vector through the loop. This produces the second electromagnetic eld equation: H = Jf + D t or ijk i Hk,j = Jf + Di . t (2.6.76) Gausss Law for Electricity This law states that the ux of the electric force vector through a closed surface is proportional to the total charge enclosed by the surface. This results in the third electromagnetic eld equation: D = f or i D,i = f or 1i gD = f . g xi (2.6.77) Gausss Law for Magnetism This law states the magnetic ux through any closed volume is zero. This produces the fourth electromagnetic eld equation: B =0 or i B,i = 0 or 1i gB = 0. g xi (2.6.78) When no confusion arises it is convenient to drop the subscript f from the above Maxwell equations. Special expanded forms of the above Maxwell equations are given on the pages 176 to 179. Electromagnetic Stress and Energy Let V denote the volume of some simple closed surface S. Let us calculate the rate at which electromagnetic energy is lost from this volume. This represents the energy ow per unit volume. Begin with the rst two Maxwells equations in Cartesian form ijk Ek,j ijk Hk,j Bi t Di . =Ji + t = (2.6.79) (2.6.80) Now multiply equation (2.6.79) by Hi and equation (2.6.80) by Ei . This gives two terms with dimensions of energy per unit volume per unit of time which we write Bi Hi t Di Ei . ijk Hk,j Ei =Ji Ei + t ijk Ek,j Hi = (2.6.81) (2.6.82) Subtracting equation (2.6.82) from equation (2.6.81) we nd ijk (Ek,j Hi Hk,j Ei ) = Ji Ei ijk [(Ek Hi ),j Ek Hi,j is the same as Di Ei t Di Ei + Hi,j Ek ] = Ji Ei t Bi Hi t Bi Hi t Observe that jki (Ek Hi ),j ijk (Ej Hk ),i so that the above simplies to Di Bi Ei Hi . t t (2.6.83) ijk (Ej Hk ),i + Ji Ei = 345 Now integrate equation (2.6.83) over a volume and apply Gausss divergence theorem to obtain ijk Ej Hk ni S d + V Ji Ei d = V ( Di Bi Ei + Hi ) d. t t (2.6.84) The rst term in equation (2.6.84) represents the outward ow of energy across the surface enclosing the volume. The second term in equation (2.6.84) represents the loss by Joule heating and the right-hand side is the rate of decrease of stored electric and magnetic energy. The equation (2.6.84) is known as Poyntings theorem and can be written in the vector form (E H) n d = S V (E B D H E J) d. t t (2.6.85) For later use we dene the quantity Si = ijk Ej Hk or S = E H [Watts/m2 ] (2.6.86) as Poyntings energy ux vector and note that Si is perpendicular to both Ei and Hi and represents units of energy density per unit time which crosses a unit surface area within the electromagnetic eld. Electromagnetic Stress Tensor Instead of calculating energy ow per unit volume, let us calculate force per unit volume. Consider a region containing charges and currents but is free from dielectrics and magnetic materials. To obtain terms with units of force per unit volume we take the cross product of equation (2.6.79) with Di and the cross product of equation (2.6.80) with Bi and subtract to obtain irs ijk (Ek,j Ds + Hk,j Bs ) = ris Ji Bs + ris Di Bs Bs + Di t t which simplies using the e identity to (rj sk rk sj )(Ek,j Ds + Hk,j Bs ) = which further simplies to Es,r Ds + Er,s Ds Hs,r Bs + Hr,s Bs = ris Ji Bs ris Ji Bs + ris (Di Bs ) t + ( t ris Di Bs ). (2.6.87) Observe that the rst two terms in the equation (2.6.87) can be written Er,s Ds Es,r Ds =Er,s Ds 0 Es,r Es 0( =(Er Ds ),s Er Ds,s 1 Es Es ),r 2 1 =(Er Ds ),s Er (Ej Dj sr ),s 2 1 =(Er Ds Ej Dj rs ),s Er 2 which can be expressed in the form E Er,s Ds Es,r Ds = Trs,s Er 346 where 1 E Trs = Er Ds Ej Dj rs 2 is called the electric stress tensor. In matrix form the stress tensor is written E1 D2 E1 D3 E1 D1 1 Ej Dj 2 E . Trs = E2 D1 E2 D2 1 Ej Dj E2 D3 2 E3 D1 E3 D2 E3 D3 1 Ej Dj 2 (2.6.88) (2.6.89) By performing similar calculations we can transform the third and fourth terms in the equation (2.6.87) and obtain M Hr,s Bs Hs,r Bs = Trs,s (2.6.90) where 1 M Trs = Hr BS Hj Bj rs 2 is the magnetic stress tensor. In matrix form the magnetic stress tensor is written M Trs (2.6.91) B1 H1 1 Bj Hj 2 = B2 H1 B3 H1 B1 H2 B2 H2 1 Bj Hj 2 B3 H2 B1 H3 . B2 H3 B3 H3 1 Bj Hj 2 (2.6.92) The total electromagnetic stress tensor is E M Trs = Trs + Trs . (2.6.93) Then the equation (2.6.87) can be written in the form Trs,s Er = or Er + For free space Di = 0 Ei ris Ji BS ris Ji Bs + ( t ( t ris Di Bs ) = Trs,s ris Di Bs ). (2.6.94) and Bi = 0 Hi so that the last term of equation (2.6.94) can be written in terms 0 0 of the Poynting vector as Sr =( t t ris Di Bs ). (2.6.95) Now integrate the equation (2.6.94) over the volume to obtain the total electromagnetic force Er d + V V ris Ji Bs d = V Trs,s d 0 0 V Sr d. t Applying the divergence theorem of Gauss gives Er d + V V ris Ji Bs d = S Trs ns d 0 0 V Sr d. t (2.6.96) The left side of the equation (2.6.96) represents the forces acting on charges and currents contained within the volume element. If the electric and magnetic elds do not vary with time, then the last term on the right is zero. In this case the forces can be expressed as an integral of the electromagnetic stress tensor. 347 EXERCISE 2.6 1. Find the eld lines and equipotential curves associated with a positive charge q located at (a, 0) and a positive charge q located at (a, 0). The eld lines are illustrated in the gure 2.6-7. Figure 2.6-7. Lines of electric force between two charges of the same sign. 2. Calculate the lines of force and equipotential curves associated with the electric eld E = E(x, y) = 2y e1 + 2x e2 . Sketch the lines of force and equipotential curves. Put arrows on the lines of force to show direction of the eld lines. 3. A right circular cone is dened by x = u sin 0 cos , y = u sin 0 sin , z = u cos 0 A r2 with 0 2 and u 0. Show the solid angle subtended by this cone is = 4. = 2(1 cos 0 ). A charge +q is located at the point (0, a) and a charge q is located at the point (0, a). Show that 2aq 1 e2 . the electric force E at the position (x, 0), where x > a is E = 4 0 (a2 + x2 )3/2 Let the circle x2 + y 2 = a2 carry a line charge . Show the electric eld at the point (0, 0, z) is 1 az(2) e3 E= . 4 0 (a2 + z 2 )3/2 5. 6. Use superposition to nd the electric eld associated with two innite parallel plane sheets each carrying an equal but opposite sign surface charge density . Find the eld between the planes and outside of each plane. Hint: Fields are of magnitude 2 0 and perpendicular to plates. 7. For a volume current J the Biot-Savart law gives B = 0 4 r r Hint: Let er = and consider (J 3 ). Then use numbers 13 and 10 of the appendix C. Also note that r r J = 0 because J does not depend upon position. V J er d. Show that B = 0. r2 348 8. A homogeneous dielectric is dened by Di and Ei having parallel vector elds. Show that for a j i,k homogeneous dielectric 9. 10. = 0. is a constant. Show that for a homogeneous, isotropic dielectric medium that Show that for a homogeneous, isotropic linear dielectric in Cartesian coordinates Pi,i = e f . 1 + e 11. Verify the Maxwells equations in Gaussian units for a charge free isotropic homogeneous dielectric. 1 E = D = 0 B =H = 0 1 B H = c t c t 1 D 4 E 4 H = + J= + E c t c c t c E = 12. charge. Verify the Maxwells equations in Gaussian units for an isotropic homogeneous dielectric with a 1 B c t 1 D 4 H = J + c c t E = D =4 B =0 13. For a volume charge in an element of volume d located at a point (, , ) Coulombs law is E(x, y, z) = 2 2 2 2 1 4 0 V er d r2 (a) Show that r = (x ) + (y ) + (z ) . 1 (b) Show that er = ((x ) e1 + (y ) e2 + (z ) e3 ) . r (c) Show that 1 er (x ) e1 + (y ) e2 + (z ) e3 1 E(x, y, z) = ddd = ddd 2 + (y )2 + (z )2 ]3/2 4 0 4 0 r2 V [(x ) V (, , ) 1 ddd (d) Show that the potential function for E is V = 2 + (y )2 + (z )2 ]1/2 4 0 V [(x ) (e) Show that E = V. (f) Show that 2 V = Hint: Note that the integrand is zero everywhere except at the point where (, , ) = (x, y, z). Consider the integral split into two regions. One region being a small sphere about the point (x, y, z) in the limit as the radius of this sphere approaches zero. Observe the identity er er = (, , ) enables one to employ the Gauss divergence theorem to obtain a (x,y,z) r2 r2 er surface integral. Use a mean value theorem to show ndS = 4 since n = er . 4 0 r2 4 0 S 14. Show that for a point charge in space = q(x x0 )(y y0 )(z z0 ), where is the Dirac delta function, the equation (2.6.5) can be reduced to the equation (2.6.1). 15. (a) Show the electric eld E = 1 r2 er is irrotational. Here er = r r is a unit vector in the direction of r. (b) Find the potential function V such that E = V which satises V(r0 ) = 0 for r0 > 0. 349 16. (a) If E is a conservative electric eld such that E = V, then show that E is irrotational and satises E = curl E = 0. (b) If E = curl E = 0, show that E is conservative. (i.e. Show E = V.) Hint: The work done on a test charge Q = 1 along the straight line segments from (x0 , y0 , z0 ) to (x, y0 , z0 ) and then from (x, y0 , z0 ) to (x, y, z0 ) and nally from (x, y, z0 ) to (x, y, z) can be written x y z V = V(x, y, z) = x0 E1 (x, y0 , z0 ) dx y0 E2 (x, y, z0 ) dy z0 E3 (x, y, z) dz. Now note that V = E2 (x, y, z0 ) y and from E = 0 we nd for 17. V x and z z0 E3 (x, y, z) dz y E3 E2 V = , which implies = E2 (x, y, z). Similar results are obtained y z y V . Hence show V = E. z (a) Show that if B = 0, then there exists some vector eld A such that B = A. The vector eld A is called the vector potential of B. 1 Hint: Let A(x, y, z) = 1 0 sB(sx, sy, sz) r ds where r = x e1 + y e2 + z e3 dBi 2 s ds by parts. ds (b) Show that ( A) = 0. and integrate 0 18. Use Faradays law and Amperes law to show g im (E j ),m g jm E i,mj = 0 ,j Ji + t 0 E i t 19. Assume that J = E where is the conductivity. Show that for = 0 Maxwells equations produce E + 0 t B + 0 0 t 0 2E =2 E t2 2B =2 B. 0 t2 0 and Here both E and B satisfy the same equation which is known as the telegraphers equation. 20. Show that Maxwells equations (2.6.75) through (2.6.78) for the electric eld under electrostatic E =0 D =f Now E is irrotational so that E = V. Show that 2 V = f . conditions reduce to 350 21. Show that Maxwells equations (2.6.75) through (2.6.78) for the magnetic eld under magnetostatic conditions reduce to H = J and B = 0. The divergence of B being zero implies B can be derived from a vector potential function A such that B = A. Here A is not unique, see problem 24. If we select A such that A = 0 then show for a homogeneous, isotropic material, free of any permanent magnets, that 2 A = J. 22. Show that under nonsteady state conditions of electrodynamics the Faraday law from Maxwells equations (2.6.75) through (2.6.78) does not allow one to set E = V. Why is this? Observe that B = 0 so we can write B = A for some vector potential A. Using this vector potential show that A = 0. This shows that the quantity inside the parenthesis is Faradays law can be written E + t conservative and so we can write E + A = V for some scalar potential V. The representation t E = V A t A t is a more general representation of the electric potential. Observe that for steady state conditions so that this potential representation reduces to the previous one for electrostatics. 23. Using the potential formulation E = V A = t 0 =0 A derived in problem 22, show that in a vacuum t (a) Gauss law can be written 2 V + (b) Amperes law can be written A = 0 J 0 0 V t 0 0 2A t2 (c) Show the result in part (b) can also be expressed in the form 2 A 0 24. 0 A t A + 0 0 V t = 0 J The Maxwell equations in a vacuum have the form E = B t 0 H = D + V t D = 0 B = 0 where D = 0 = 1/c2 where c is the speed of light. A V. Introduce the vector potential A and scalar potential V dened by B = A and E = t Note that the vector potential is not unique. For example, given as a scalar potential we can write 0 E, B = 0 H with and 0 constants satisfying B = A = (A + ), since the curl of a gradient is zero. Therefore, it is customary to impose some kind of additional requirement on the potentials. These additional conditions are such that E and B are 1 V not changed. One such condition is that A and V satisfy A + 2 = 0. This relation is known as the c t Lorentz relation or Lorentz gauge. Find the Maxwells equations in a vacuum in terms of A and V and show that 2 1 2 c2 t2 V= 0 and 2 1 2 c2 t2 A = 0 V . 351 25. In a vacuum show that E and B satisfy 2 E = 26. (a) Show that the wave equations in problem 25 have solutions in the form of waves traveling in the x- direction given by E = E(x, t) = E0 ei(kxt) and B = B(x, t) = B0 ei(kxt) 1 2E c2 t2 2 B = 1 2B c2 t2 E =0 B = 0 where E0 and B0 are constants. Note that wave functions of the form u = Aei(kxt) are called plane harmonic waves. Sometimes they are called monochromatic waves. Here i2 = 1 is an imaginary unit. Eulers identity shows that the real and imaginary parts of these type wave functions have the form A cos(kx t) and A sin(kx t). These represent plane waves. The constant A is the amplitude of the wave , is the angular frequency, and k/2 is called the wave number. The motion is a simple harmonic motion both in time and space. That is, at a xed point x the motion is simple harmonic in time and at a xed time t, the motion is harmonic in space. By examining each term in the sine and cosine terms we nd that x has dimensions of length, k has dimension of reciprocal length, t has dimensions of time and has dimensions of reciprocal time or angular velocity. The quantity c = /k is the wave velocity. The value = 2/k has dimension of length and is called the wavelength and 1/ is called the wave number. The wave number represents the number of waves per unit of distance along the x-axis. The period of the wave is T = /c = 2/ and the frequency is f = 1/T. The frequency represents the number of waves which pass a xed point in a unit of time. (b) Show that = 2f (c) Show that c = f (d) Is the wave motion u = sin(kx t) + sin(kx + t) a traveling wave? Explain. 1 2 (e) Show that in general the wave equation 2 = 2 2 have solutions in the form of waves traveling in c t either the +x or x direction given by = (x, t) = f (x + ct) + g(x ct) where f and g are arbitrary twice dierentiable functions. (f) Assume a plane electromagnetic wave is moving in the +x direction. Show that the electric eld is in the xyplane and the magnetic eld is in the xzplane. Hint: Assume solutions Ex = g1 (x ct), Ey = g2 (x ct), Ez = g3 (x ct), Bx = g4 (x ct), By = g5 (x ct), Bz = g6 (x ct) where gi ,i = 1, ..., 6 are arbitrary functions. Then show that Ex does not satisfy E = 0 which implies g1 must be independent of x and so not a wave function. Do the same for the components of B. Since both E = B = 0 then Ex = Bx = 0. Such waves are called transverse waves because the electric and magnetic elds are perpendicular to the direction of propagation. Faradays law implies that the E and B waves must be in phase and be mutually perpendicular to each other. 352 BIBLIOGRAPHY Abramowitz, M. and Stegun, I.A., Handbook of Mathematical Functions, 10th ed, New York:Dover, 1972. Akivis, M.A., Goldberg, V.V., An Introduction to Linear Algebra and Tensors, New York:Dover, 1972. Aris, Rutherford, Vectors, Tensors, and the Basic Equations of Fluid Mechanics, Englewood Clis, N.J.:Prentice-Hall, 1962. Atkin, R.J., Fox, N., An Introduction to the Theory of Elasticity, London:Longman Group Limited, 1980. Bishop, R.L., Goldberg, S.I.,Tensor Analysis on Manifolds, New York:Dover, 1968. Borisenko, A.I., Tarapov, I.E., Vector and Tensor Analysis with Applications, New York:Dover, 1968. Chorlton, F., Vector and Tensor Methods, Chichester,England:Ellis Horwood Ltd, 1976. Dodson, C.T.J., Poston, T., Tensor Geometry, London:Pittman Publishing Co., 1979. Eisenhart, L.P., Riemannian Geometry, Princeton, N.J.:Univ. Princeton Press, 1960. Eringen, A.C., Mechanics of Continua, Huntington, N.Y.:Robert E. Krieger, 1980. D.J. Griths, Introduction to Electrodynamics, Prentice Hall, 1981. Flgge, W., Tensor Analysis and Continuum Mechanics, New York:Springer-Verlag, 1972. u Fung, Y.C., A First Course in Continuum Mechanics, Englewood Clis,N.J.:Prentice-Hall, 1969. Goodbody, A.M., Cartesian Tensors, Chichester, England:Ellis Horwood Ltd, 1982. Hay, G.E., Vector and Tensor Analysis, New York:Dover, 1953. Hughes, W.F., Gaylord, E.W., Basic Equations of Engineering Science, New York:McGraw-Hill, 1964. Jereys, H., Cartesian Tensors, Cambridge, England:Cambridge Univ. Press, 1974. Lass, H., Vector and Tensor Analysis, New York:McGraw-Hill, 1950. Levi-Civita, T., The Absolute Dierential Calculus, London:Blackie and Son Limited, 1954. Lovelock, D., Rund, H. ,Tensors, Dierential Forms, and Variational Principles, New York:Dover, 1989. Malvern, L.E., Introduction to the Mechanics of a Continuous Media, Englewood Clis, N.J.:Prentice-Hall, 1969. McConnell, A.J., Application of Tensor Analysis, New York:Dover, 1947. Newell, H.E., Vector Analysis, New York:McGraw Hill, 1955. Schouten, J.A., Tensor Analysis for Physicists,New York:Dover, 1989. Scipio, L.A., Principles of Continua with Applications, New York:John Wiley and Sons, 1967. Sokolniko, I.S., Tensor Analysis, New York:John Wiley and Sons, 1958. Spiegel, M.R., Vector Analysis, New York:Schaum Outline Series, 1959. Synge, J.L., Schild, A., Tensor Calculus, Toronto:Univ. Toronto Press, 1956. Bibliography 353 APPENDIX A UNITS OF MEASUREMENT The following units, abbreviations and prexes are from the Syst`me International dUnit`s e e Prexes. Prex tera giga mega kilo hecto deka deci centi milli micro nano pico Abreviations Multiplication factor 1012 109 106 103 102 10 101 102 103 106 109 1012 Symbol T G M K h da d c m n p (designated SI in all Languages.) Basic Units. Basic units of measurement Name Length meter Mass kilogram Time second Electric current ampere Temperature degree Kelvin Luminous intensity candela Unit Supplementary units Name radian steradian Symbol m kg s A K cd Unit Plane angle Solid angle Symbol rad sr 354 Name Area Volume Frequency Density Velocity Angular velocity Acceleration Angular acceleration Force Pressure Kinematic viscosity Dynamic viscosity Work, energy, quantity of heat Power Electric charge Voltage, Potential dierence Electromotive force Electric force eld Electric resistance Electric capacitance Magnetic ux Inductance Magnetic ux density Magnetic eld strength Magnetomotive force DERIVED UNITS Units square meter cubic meter hertz kilogram per cubic meter meter per second radian per second meter per second squared radian per second squared newton newton per square meter square meter per second newton second per square meter joule watt coulomb volt volt volt per meter ohm farad weber henry tesla ampere per meter ampere Symbol m2 m3 1 Hz (s ) kg/m3 m/s rad/s m/s2 rad/s2 N (kg m/s2 ) N/m2 m2 /s N s/m2 J (N m) W (J/s) C (A s) V (W/A) V (W/A) V/m (V/A) F (A s/V) Wb (V s) H (V s/A) T (Wb/m2 ) A/m A Physical constants. 4 arctan 1 = = 3.14159 26535 89793 23846 2643 . . . n lim 1+ 1 n n = e = 2.71828 18284 59045 23536 0287 . . . Eulers constant = 0.57721 56649 01532 86060 6512 . . . = lim n 1+ 1 11 + + + log n 23 n speed of light in vacuum = 2.997925(10)8 m s1 electron charge = 1.60210(10)19 C Avogadros constant = 6.02252(10)23 mol1 Planks constant = 6.6256(10)34 J s Universal gas constant = 8.3143 J K 1 mol1 = 8314.3 J Kg 1 K 1 Boltzmann constant = 1.38054(10)23 J K 1 StefanBoltzmann constant = 5.6697(10)8 W m2 K 4 Gravitational constant = 6.67(10)11 N m2 kg 2 355 APPENDIX B CHRISTOFFEL SYMBOLS OF SECOND KIND 1. Cylindrical coordinates (r, , z) = (x1 , x2 , x3 ) x = r cos y = r sin z=z r0 0 2 <z < h1 = 1 h2 = r h3 = 1 The coordinate curves are formed by the intersection of the coordinate surfaces x2 + y 2 = r2 , y/x = tan z = Constant 1 22 2 12 Cylinders Planes Planes. 2 21 1 r = r = = 2. Spherical coordinates (, , ) = (x1 , x2 , x3 ) x = sin cos y = sin sin z = cos 0 0 0 2 h1 = 1 h2 = h3 = sin The coordinate curves are formed by the intersection of the coordinate surfaces x2 + y 2 + z 2 = 2 x + y = tan z y = x tan 1 22 1 33 2 33 2 2 2 Spheres Cones Planes. 2 21 3 31 3 23 1 1 = = cot = = sin2 = sin cos 2 12 3 13 3 32 = = = = 356 3. Parabolic cylindrical coordinates (, , z) = (x1 , x2 , x3 ) x = 1 y = ( 2 2 ) 2 z=z < < <z < 0 h1 = h2 = h3 = 1 2 + 2 2 + 2 The coordinate curves are formed by the intersection of the coordinate surfaces x2 = 2 2 (y x2 = 2 2 (y + 2 ) 2 Parabolic cylinders Parabolic cylinders Planes. 1 22 1 21 2 12 + 2 =2 + 2 =2 + 2 = 2 2 ) 2 z = Constant + 2 =2 + 2 =2 + 2 2 1 11 2 22 2 11 = 1 12 2 21 = = 4. Parabolic coordinates (, , ) = (x1 , x2 , x3 ) x = cos y = sin 1 z = ( 2 2 ) 2 0 0 0 < < 2 h1 = h2 = h3 = 2 + 2 2 + 2 The coordinate curves are formed by the intersection of the coordinate surfaces x2 + y 2 = 2 2 (z x2 + y 2 = 2 2 (z + y = x tan 1 11 2 22 1 22 2 11 2 33 2 + 2 2 + 2 2 + 2 2 + 2 2 2 + 2 2 ) 2 Paraboloids Paraboloids Planes. 1 33 1 21 2 12 3 23 3 31 2 2 + 2 2 + 2 2 + 2 1 1 2 ) 2 = = = = = = = = = = 1 21 2 21 3 32 3 13 = = = = 357 5. Elliptic cylindrical coordinates (, , z) = (x1 , x2 , x3 ) x = cosh cos y = sinh sin z=z 0 0 2 <z < h1 = h2 = h3 = 1 sinh2 + sin2 sinh2 + sin2 The coordinate curves are formed by the intersection of the coordinate surfaces x2 y2 + =1 cosh2 sinh2 y2 x2 =1 cos2 sin2 z = Constant 1 11 1 22 1 21 sinh cosh sinh2 + sin2 sinh cosh = sinh2 + sin2 sin cos = sinh2 + sin2 = Elliptic cylinders Hyperbolic cylinders Planes. 2 22 2 11 2 21 sin cos sinh2 + sin2 sin cos = sinh2 + sin2 sinh cosh = sinh2 + sin2 = 1 12 = 2 12 = 6. Elliptic coordinates (, , ) = (x1 , x2 , x3 ) (1 2 )( 2 1) cos (1 2 )( 2 1) sin 1< 1 1 0 < 2 h1 = h2 = h3 = 2 2 2 1 2 2 1 2 (1 2 )( 2 1) x= y= z = The coordinate curves are formed by the intersection of the coordinate surfaces y2 z2 x2 +2 + 2 =1 2 1 1 z2 x2 y2 =1 2 1 2 1 2 y = x tan 1 11 2 22 1 22 1 33 2 11 +2 1 + 2 2 2 = 2 1 2 = = = = 1 + 2 (1 2 ) ( 2 2 ) 1 + 2 1 2 2 2 Prolate ellipsoid Two-sheeted hyperboloid Planes 2 33 1 12 2 21 3 31 3 32 1 + 2 1 2 2 2 = 2 2 =2 2 = 1 + 2 = 1 2 = 1 2 (1 + 2 ) ( 2 2 ) 358 7. Bipolar coordinates (u, v, z) = (x1 , x2 , x3 ) x= a sinh v , 0 u < 2 cosh v cos u a sin u , < v < y= cosh v cos u z=z <z < h2 = h2 1 2 h2 = 2 a2 (cosh v cos u)2 h2 = 1 3 The coordinate curves are formed by the intersection of the coordinate surfaces (x a coth v)2 + y 2 = a2 sinh2 v a2 x2 + (y a cot u)2 = sin2 u z = Constant 2 11 1 12 2 21 = Cylinders Cylinders Planes. sinh v cos u + cosh v sinh v = cos u cosh v sin u = cos u cosh v 1 11 2 22 1 22 = sin u cos u cosh v sinh v = cos u cosh v sin u = cos u + cosh v 8. Conical coordinates (u, v, w) = (x1 , x2 , x3 ) x= uvw , b 2 > v 2 > a2 > w 2 , ab u (v 2 a2 )(w2 a2 ) y= a a2 b 2 v (v 2 b2 )(w2 b2 ) z= b b 2 a2 u0 h2 = 1 1 u2 (v 2 w2 ) a2 )(b2 v 2 ) u2 (v 2 w2 ) h2 = 3 2 a2 )(w2 b2 ) (w h2 = 2 (v 2 The coordinate curves are formed by the intersection of the coordinate surfaces x2 + y 2 + z 2 = u2 x y z +2 +2 = 0, v2 v a2 v b2 x2 y2 z2 +2 +2 = 0, 2 2 w w a w b2 2 22 3 33 1 22 1 33 2 33 v v v +2 b2 v 2 a2 + v 2 v w2 w w w = 2 v w2 a2 + w2 b2 + w2 = = = = u v 2 w2 (b2 v 2 ) (a2 + v 2 ) u v 2 w2 (a2 + w2 ) (b2 + w2 ) (v 2 v b2 v 2 a2 + v 2 w2 ) (a2 + w2 ) (b2 + w2 ) 3 22 2 21 2 23 3 31 3 32 2 2 2 Spheres Cones Cones. w a2 + w2 b2 + w2 (b2 v 2 ) (a2 + v 2 ) (v 2 w2 ) 1 = u w = 2 v w2 1 = u v =2 v w2 = 359 9. Prolate spheroidal coordinates (u, v, ) = (x1 , x2 , x3 ) x = a sinh u sin v cos , y = a sinh u sin v sin , z = a cosh u cos v, u0 0v h2 = h2 1 2 h2 = a2 (sinh2 u + sin2 v) 2 h2 = a2 sinh2 u sin2 v 3 0 < 2 The coordinate curves are formed by the intersection of the coordinate surfaces x2 y2 z2 + + = 1, 2 2 (a sinh u) a sinh u) a cosh u)2 y2 z2 x2 = 1, (a cos v)2 (a sin v)2 (a cos v)2 y = x tan , 1 11 2 22 1 22 1 33 2 11 cosh u sinh u sin2 v + sinh2 u cos v sin v = sin2 v + sinh2 u cosh u sinh u = 2 sin v + sinh2 u sin2 v cosh u sinh u = sin2 v + sinh2 u cos v sin v = 2 sin v + sinh2 u = 2 33 1 12 2 21 3 31 3 32 = = = = = Prolate ellipsoids Two-sheeted hyperpoloid Planes. cos v sin vsinh2 u sin2 v + sinh2 u cos v sin v sin2 v + sinh2 u cosh u sinh u sin2 v + sinh2 u cosh u sinh u cos v sin v 10. Oblate spheroidal coordinates (, , ) = (x1 , x2 , x3 ) x = a cosh cos cos , y = a cosh cos sin , z = a sinh sin , 0 2 2 0 2 h2 = h2 1 2 h2 = a2 (sinh2 + sin2 ) 2 h2 = a2 cosh2 cos2 3 The coordinate curves are formed by the intersection of the coordinate surfaces x2 y2 z2 + + = 1, (a cosh )2 (a cosh )2 (a sinh )2 y2 z2 x2 + = 1, (a cos )2 (a cos )2 (a sin )2 y = x tan , 1 11 2 22 1 22 1 33 2 11 cosh sinh sin2 + sinh2 cos sin = sin2 + sinh2 cosh sinh = 2 sin + sinh2 cos2 cosh sinh = sin2 + sinh2 cos sin = 2 sin + sinh2 = 2 33 1 12 2 21 3 31 3 32 = Oblate ellipsoids One-sheet hyperboloids Planes. cos sin cosh2 sin2 + sinh2 cos sin = sin2 + sinh2 cosh sinh = sin2 + sinh2 sinh = cosh sin = cos 360 11. Toroidal coordinates (u, v, ) = (x1 , x2 , x3 ) x= a sinh v cos , cosh v cos u a sinh v sin y= , cosh v cos u a sin u z= , cosh v cos u 0 u < 2 < v < 0 < 2 h2 = h2 1 2 h2 = 2 h2 = 3 a2 (cosh v cos u)2 a2 sinh2 v (cosh v cos u)2 The coordinate curves are formed by the intersection of the coordinate surfaces a cos u 2 a2 , = sin u sin2 u 2 cosh v a2 , x2 + y 2 a + z2 = sinh v sinh2 v y = x tan , x2 + y 2 + z 1 11 2 22 1 22 1 33 2 11 sin u cos u cosh v sinh v = cos u cosh v sin u = cos u + cosh v = sin usinh v 2 cos u + cosh v sinh v = cos u + cosh v = 2 33 1 12 2 21 3 31 3 32 Spheres Tores planes = = = = = sinh v (cos u cosh v 1) cos u cosh v sinh v cos u cosh v sin u cos u cosh v sin u cos u cosh v cos u cosh v 1 cos u sinh v cosh v sinh v 361 12. Confocal ellipsoidal coordinates (u, v, w) = (x1 , x2 , x3 ) x2 = (a2 u)(a2 v)(a2 w) , (a2 b2 )(a2 c2 ) (b2 u)(b2 v)(b2 w) , y2 = (b2 a2 )(b2 c2 ) (c2 u)(c2 v)(c2 w) , z2 = (c2 a2 )(c2 b2 ) h2 = 1 u < c2 < b 2 < a2 c2 < v < b 2 < a 2 c2 < b 2 < v < a2 (u v)(u w) 4(a2 u)(b2 u)(c2 u) (v u)(v w) h2 = 2 4(a2 v)(b2 v)(c2 v) (w u)(w v) h2 = 3 4(a2 w)(b2 w)(c2 w) 1 11 2 22 3 33 1 22 1 33 2 11 2 33 3 11 3 22 = = = = = = 1 1 1 1 1 + + + + 2 u) 2 u) u) 2 (b 2 (c 2 (u v) 2 (u w) 1 1 1 1 1 + + + + = 2 v) 2 v) 2 v) 2 (a 2 (b 2 (c 2 (u + v) 2 (v w) 1 1 1 1 1 + + + + = 2 w) 2 w) 2 w) 2 (a 2 (b 2 (c 2 (u + w) 2 (v + w) = 2 (a2 a2 u b2 u c2 u (v w) v) (b2 v) (c2 v) (u v) (u w) 1 12 1 13 2 21 2 23 3 31 3 32 = = = = = = 1 2 (u v) 1 2 (u w) 1 2 (u + v) 1 2 (v w) 1 2 (u + w) 1 2 (v + w) 2 (a2 a2 u b2 u c2 u (v + w) 2 (u v) (a2 w) (b2 w) (c2 w) (u w) a2 v b2 v c2 v (u w) 2 (a2 u) (b2 u) (c2 u) (u + v) (v w) a2 v b2 v c2 v (u + w) 2 (u + v) (a2 w) (b2 w) (c2 w) (v w) (u v) a2 w b2 w c2 w 2 (a2 u) (b2 u) (c2 u) (u + w) (v + w) (u + v) a2 w b2 w c2 w 2 (a2 v) (b2 v) (c2 v) (u + w) (v + w) 362 APPENDIX C VECTOR IDENTITIES The following identities assume that A, B, C, D are dierentiable vector functions of position while f, f1 , f2 are dierentiable scalar functions of position. 1. 2. 3. 4. 5. A (B C) = B (C A) = C (A B) A (B C) = B(A C) C(A B) (A B) (C D) = (A C)(B D) (A D)(B C) A (B C) + B (C A) + C (A B) = 0 (A B) (C D) = B(A C D) A(B C D) = C(A B C) D(A B C) 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. (A B) (B C) (C A) = (A B C)2 (f1 + f2 ) = f1 + f2 (A + B) = A + B (A + B) = A + B (f A) = (f ) A + f A (f1 f2 ) = f1 f2 + f2 f1 (f A) =)f ) A + f ( A) (A B) = B ( A) A ( B) (A )A = |A|2 2 A ( A) (A B) = (B )A + (A )B + B ( A) + A ( B) (A B) = (B )A B( A) (A )B + A( B) (f ) = 2 f (f ) = 0 ( A) = 0 ( A) = ( A) 2 A INDEX A Absolute dierentiation 120 Absolute scalar eld 43 Absolute tensor 45,46,47,48 Acceleration 121, 190, 192 Action integral 198 Addition of systems 6, 51 Addition of tensors 6, 51 Adherence boundary condition 294 Aelotropic material 245 Ane transformation 86, 107 Airy stress function 264 Almansi strain tensor 229 Alternating tensor 6,7 Amperes law 176,301,337,341 Angle between vectors 80, 82 Angular momentum 218, 287 Angular velocity 86,87,201,203 Arc length 60, 67, 133 Associated tensors 79 Auxiliary Magnetic eld 338 Axis of symmetry 247 B Basic equations elasticity 236, 253, 270 Basic equations for a continuum 236 Basic equations of uids 281, 287 Basis vectors 1,2,37,48 Beltrami 262 Bernoullis Theorem 292 Biharmonic equation 186, 265 Bilinear form 97 Binormal vector 130 Biot-Savart law 336 Bipolar coordinates 73 Boltzmann equation 302,306 Boundary conditions 257, 294 Bulk modulus 251 Bulk coecient of viscosity 285 C Cartesian coordinates 19,20,42, 67, 83 Cartesian tensors 84, 87, 226 Cauchy stress law 216 Cauchy-Riemann equations 293,321 Charge density 323 Christoel symbols 108,110,111 Circulation 293 Codazzi equations 139 Coecient of viscosity 285 Cofactors 25, 26, 32 Compatibility equations 259, 260, 262 Completely skew symmetric system 31 Compound pendulum 195,209 Compressible material 231 Conic sections 151 Conical coordinates 74 Conjugate dyad 49 Conjugate metric tensor 36, 77 363 Conservation of angular momentum 218, 295 Conservation of energy 295 Conservation of linear momentum 217, 295 Conservation of mass 233, 295 Conservative system 191, 298 Conservative electric eld 323 Constitutive equations 242, 251,281, 287 Continuity equation 106,234, 287, 335 Contraction 6, 52 Contravariant components 36, 44 Contravariant tensor 45 Coordinate curves 37, 67 Coordinate surfaces 37, 67 Coordinate transformations 37 Coulomb law 322 Covariant components 36, 47 Covariant dierentiation 113,114,117 Covariant tensor 46 Cross product 11 Curl 21, 173 Curvature 130, 131, 134, 149 Curvature tensor 134, 145 Curvilinear coordinates 66, 81 Cylindrical coordinates 18, 42, 69 364 D Deformation 222 Derivative of tensor 108 Derivatives and indicial notation 18, 31 Determinant 10, 25, 32, 33 Dielectric tensor 333 Dierential geometry 129 Diusion equation 303 Dilatation 232 Direction cosines 85 Displacement vector 333 Dissipation function 297 Distribution function 302 Divergence 21, 172 Divergence theorem 24 Dot product 5 Double dot product 50, 62 Dual tensor 100 Dummy index 4, 5 Dyads 48,62,63 Dynamics 187 E e Permutation symbol 6, 7, 12 e- identity 12 Eigenvalues 179,189 Eigenvectors 179,186 Einstein tensor 156 Elastic constants 248 Elastic stiness 242 Elasticity 211,213 Electrostatic eld 322,333 Electric ux 327 Electric units 322 Electrodynamics 339 Electromagnetic energy 341 Electromagnetic stress 341,342 Elliptic coordinates 72 Elliptical cylindrical coordinates 71 Enthalpy 298 Entropy 300 Epsilon permutation symbol 83 Equation of state 300 INDEX Equilibrium equations 273,300 Elastic constants 243,248 Equipotential curves 325 Euler number 294 Euler-Lagrange equations 192 Eulerian angles 201, 209 Eulerian form 287 Eulerian system 227 Eulers equations of motion 204 F Faradays law 176,301, 340 Field lines 324, 327 Field electric 322 First fundamental form 133,143 Fourier law 297, 299 Free indices 3 Frenet-Serret formulas 131, 188 Froude number 294 Fluids 281 G Gas law 300 Gauss divergence theorem 24, 330 Gauss equations 138 Gausss law for electricity 176,301,328 Gausss law for magnetism 176,301,341 Gaussian curvature 137,139, 149 Geodesics 140, 146 Geodesic curvature 135, 140 General tensor 48 Generalized e identity 84, 104 Generalized Hookes law 242 Generalized Kronecker delta 13, 31 Generalized stress strain 242 Geometry in Riemannian Space 80 Gradient 20, 171 Gradient basis 37 Greens theorem 24 Group properties 41, 54 Generalized velocity 121 Generalized acceleration 121 INDEX H Hamiltonian 208 Heat equation 316 Hexagonal material 247 Higher order tensors 47, 93 Hookes law 212, 242, 252 Hydrodynamic equations 283 I Ideal uid 283 Idemfactor 50 Incompressible material 231 Index notation 1, 2, 14 Indicial notation 1, 2, 14,24 Inner product 52 Inertia 30 Integral theorems 24 Intrinsic derivative 120 Invariant 43 Inviscid uid 283 Isotropic material 248 Isotropic tensor 104 N J Naviers equations 254, 257 Jacobian 17, 30, 40, 101, 127 Jump discontinuity 330 K Kronecker delta 3, 8, 13, 31, 76 Kinetic energy 201 Kinematic viscosity 302 L Lagranges equation of motion 191, 196 Lagrangian 209 Laplacian 174 Linear form 96 Linear momentum 209, 287 Linear transformation 86 Linear viscous uids 284 Lorentz transformation 57 Lames constants 251 Oblate Spheroidal coordinates 75 Oblique coordinates 60 Oblique cylindrical coordinates 102 Order 2 Orthogonal coordinates 78, 86 Orthotropic material 246 Outer product 6, 51 Osculating plane 188 Navier-Stokes equations 288, 290 Newtonian uids 286 Nonviscous uid 283 Normal curvature 135, 136 Normal plane 188 Normal stress 214 Normal vector 130, 132 Notation for physical components 92 O M Magnitude of vector 80 Magnetostatics 334,338 Magnetic eld 334 Magnetization vector 337 Magnetic permeability 337 Material derivative 234, 288 Material symmetry 244, 246 Maxwell equations 176, 339 Maxwell transfer equation 308 365 Maximum, minimum curvature 130, 140 Mean curvature 137, 148 Metric tensor 36, 65 Meusniers Theorem 150 Mixed tensor 49 Mohrs circle 185 Moment of inertia 30, 184, 200 Momentum 217, 218 Multilinear forms 96, 98 Multiplication of tensors 6, 51 366 P Parallel vector eld 122 Pappovich-Neuber solution 263 Parabolic coordinates 70 Parabolic cylindrical coordinates 69 Particle motion 190 Pendulum system 197, 210 Perfect gas 283, 299 Permutations 6 Phase space 302 Physical components 88, 91,93 Piezoelectric 300 Pitch,roll, Yaw 209 Plane Couette ow 315 Plane Poiseuille ow 316 Plane strain 263 Plane stress 264 Poissons equation 329 Poissons ratio 212 Polar element 273 Polarization vector 333 Polyads 48 Potential energy 191 Potential function 323 Poyntings vector 341 Pressure 283 Principal axes 183 Projection 35 Prolated Spheroidal coordinates 74 Pully system 194, 207 Q Quotient law 53 R Radius of curvature 130, 136 Range convention 2, 3 Rate of deformation 281, 286 Rate of strain 281 INDEX Rayleigh implusive ow 317 Reciprocal basis 35, 38 Relative scalar 127 Relative tensor 50, 121 Relative motion 202 Relativity 151 Relative motion 155 Reynolds number 294 Riccis theorem 119 Riemann Christoel tensor 116, 129,139, 147 Riemann space 80 Rectifying plane 188 Rigid body rotation 199 Rotation of axes 85, 87, 107 Rules for indices 2 S Scalar 40, 43 Scalar invariant 43, 62, 105 Scalar potential 191 Scaled variables 293 Second fundamental form 135, 145 Second order tensor 47 Shearing stresses 214 Simple pulley system 193 Simple pendulum 194 Skew symmetric system 3, 31 Skewed coordinates 60, 102 Solid angle 328 Space curves 130 Special tensors 65 Spherical coordinates 18, 43, 56, 69, 103,194 Stokes ow 318 Stokes hypothesis 285 Stokes theorem 24 Straight line 60 Strain 218, 225, 228 Strain deviator 279 INDEX 367 U Unit binormal 131, 192 Unit normal 131, 191 Unit tangent 131, 191 Unit vector 81, 105 V Vector identities 15, 20, 315 Vector transformation 45, 47 Vector operators 20, 175 Vector potential 188 Velocity 95, 121, 190, 193 Velocity strain tensor 281 Viscosity 285 Viscosity table 285 Viscous uid 283 Viscous forces 288 Viscous stress tensor 285 Vorticity 107, 292 W Wave equation 255, 269 Weighted tensor 48, 127 Weingartens equation 138, 153 Work 191, 279 Work done 324 Y Youngs modulus 212 Stress 214 Stress deviator 279 Strong conservative form 298 Strouhal number 294 St Venant 258 Subscripts 2 Subtraction of tensors 51, 62 Summation convention 4, 9 Superscripts 2 Surface 62, 131 Surface area 59 Surface curvature 149 Surface metric 125, 133 Susceptibility tensor 333 Sutherland formula 285 Symmetric system 3, 31, 51, 101 Symmetry 243 System 2, 31 T Tangential basis 37 Tangent vector 130 Tensor and vector forms 40, 150 Tensor derivative 141 Tensor general 48 Tensor notation 92, 160 Tensor operations 6, 51, 175 Test charge 322 Thermodynamics 299 Third fundamental form 146 Third order systems 31 Toroidal coordinates 75, 103 Torus 124 Transformation equations 17, 37, 86 Transitive property 45,46 Translation of coordinates 84 Transport equation 302 Transposition 6 Triad 50 Trilinear form 98 Triple scalar product 15 ... 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