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i i “CM˙Final” — 2015/3/13 — 11:09 — page 1 — #20 i i 1 Introduction to Tensors In elementary physics, we often come across two classes of quantities, namely scalars and vectors. Mass, density and temperature are examples of scalar quantities, while velocity and acceleration are examples of vector quantities. A scalar quantity’s value does not de- pend on the choice of the coordinate system. Similarly, although the components of a vector depend on a particular choice of coordinate system, the vector itself is invariant, and has an existence independent of the choice of coordinate system. In this chapter, we generalize the concept of a scalar and vector, to that of a tensor. In this general framework, scalars are considered as zeroth-order tensors and vectors as first-order tensors. Tensor quantities of order two and greater, similar to scalars and vectors, have an existence independent of the coordinate system. Their components, however, just as in the case of vectors, de- pend on the choice of coordinate system. We will see that the governing field equations of continuum mechanics can be written as tensorial equations. The advantage of writing the field equations in such ‘coordinate-free’ notation is that it is immediately obvious that these equations are valid no matter what the choice of coordinate system is. A particular coordinate system is invoked only while solving a particular problem, whence the appro- priate form of the differential operators and the components of the tensors with respect to the chosen coordinate system are used. It must be borne in mind, however, that although using tensorial notation shows the ‘coordinate-free’ nature of the governing equations in a given frame of reference, it does not address the issue of how the equations transform under a change of frame of reference. This aspect will be discussed in greater detail later in this book. We now present a review of tensors. Throughout the text, scalars are denoted by light- face letters, vectors are denoted by boldface lower-case letters, while second and higher- order tensors are denoted by boldface capital letters. As a notational issue, summation over repeated indices is assumed, with the indices ranging from 1 to 3. Thus, for example, uivi represents u1v1 + u2v2 + u3v3, and Tijnj represents Ti1n1 + Ti2n2 + Ti3n3. The quantity on the right-hand side of a ‘:=’ symbol defines the quantity on its left-hand side. A function on ‘V V to V’ means that the function is defined in terms of two elements that belong to V, and× the result is also in V. 1.1 Vector Spaces In what follows, we consider only real vector spaces. We denote the set of real numbers by .A vector space (or linear space) is a set, say V, equipped with an addition function on V< V to V (denoted by +), and a scalar multiplication function on V to V, which satisfy× the following conditions: < × Downloaded from https://www.cambridge.org/core. The University of North Carolina Chapel Hill Libraries, on 20 Aug 2019 at 17:43:31, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316134054.002 i i i i i i “CM˙Final” — 2015/3/13 — 11:09 — page 2 — #21 i i 2 Continuum Mechanics 1. Commutativity: For all u, v V, 2 u + v = v + u. 2. Associativity: For all u, v, w V, 2 (u + v) + w = u + (v + w). 3. Existence of a zero element: There exists 0 V such that 2 u + 0 = u. 4. Existence of negative elements: For each u V, there exists a negative element de- noted u in V such that 2 − u u = 0. − 5. Distributivity with respect to addition of vectors: For all a , and u, v V, 2 < 2 a(u + v) = au + av. 6. Distributivity with respect to scalar addition: For all a, b , and for all u V, 2 < 2 (a + b)u = au + bu. 7. Associativity: For all a, b , and for all u V, 2 < 2 a(bu) = (ab)u. 8. Identity in scalar multiplication: For all u V, 2 1u = u. Since a vector space is a group with respect to addition (see Section 1.12) with 0 and u playing the roles of the neutral and reverse elements, respectively, all the results derived− for groups are applicable for vector spaces. In particular, the zero element of V, and the negative element u corresponding to a given u V are unique. Note also that au + bv V for all a, b ,− and all u, v V. 2 2 Perhaps, the2 < most famous2 example of a vector space is the n-dimensional coordinate space n, which is defined by < n := (u , u ,..., un) : u . < f 1 2 i 2 <g For n, addition and scalar multiplication are defined by the relations < u + v := (u1 + v1, u2 + v2,..., un + vn), au := (au1, au2,..., aun). Downloaded from https://www.cambridge.org/core. The University of North Carolina Chapel Hill Libraries, on 20 Aug 2019 at 17:43:31, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316134054.002 i i i i i i “CM˙Final” — 2015/3/13 — 11:09 — page 3 — #22 i i Introduction to Tensors 3 If p is a positive real number, then another example of a vector space is the space Z 1 Lp[0, 1] := f : f p dx < ¥ , 0 j j with addition and scalar multiplication defined by ( f + g)(x) := f (x) + g(x), x [0, 1], 2 (a f )(x) := a[ f (x)], x [0, 1]. 2 To show that Lp[0, 1] is a vector space, we need to show that f + g Lp[0, 1] if f , g Lp[0, 1]. This follows from the inequality 2 2 f + g p [2 max( f , g )]p 2p f p + g p . j j ≤ j j j j ≤ j j j j A subset u , u ,..., u of V is said to be linearly dependent if and only if there exist f 1 2 mg scalars a1, a2,..., am, not all zero, such that a u + a u + + a u = 0. 1 1 2 2 ··· m m Thus, a subset u , u ,..., u of V is linearly independent if and only if the equation f 1 2 mg a u + a u + + a u = 0, 1 1 2 2 ··· m m implies that a = a = = a = 0. A subset, say u , u ,..., u , 0 , which includes the 1 2 ··· m f 1 2 m g zero element is always linearly dependent even when the subset u1, u2,..., um is linearly independent, since the coefficient of the zero element can be takenf to be nonzero.g A subset e , e ,..., e of V is said to be a basis for V if f 1 2 ng 1. e , e ,..., e is linearly independent, and f 1 2 ng 2. Any element of V can be expressed as a linear combination of e1, e2,..., en , i.e., if u V, then f g 2 u = u e + u e + + u e , 1 1 2 2 ··· n n where the scalars u1, u2,..., un are known as the components of u with respect to the basis e , e ,..., e . f 1 2 ng If the bases have a finite number of elements, we have the following theorem: Theorem 1.1.1. All bases for a given vector space contain the same number of elements. Proof. Suppose that e , e ,..., e and e , e ,..., e are bases for a vector f 1 2 ng f 1∗ 2∗ m∗ g space. Every ei∗, since it is an element of the vector space, can be expressed in terms of the basis e , e ,..., e as b e , b . Thus, f 1 2 ng ij j ij 2 < aiei∗ = bijaiej = 0 Downloaded from https://www.cambridge.org/core. The University of North Carolina Chapel Hill Libraries, on 20 Aug 2019 at 17:43:31, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316134054.002 i i i i i i “CM˙Final” — 2015/3/13 — 11:09 — page 4 — #23 i i 4 Continuum Mechanics implies, by the linear independence of e , e ,..., e , that f 1 2 ng bijai = 0 i = 1, 2, . , m, j = 1, 2, . , n. Let m > n. Then, the number of unknowns ai is more than the number of equa- tions, so that it is possible to find a nontrivial solution. Thus, there exist a1, a2, . ., am, not all zero such that aiei∗ = 0, i.e., e , e ,..., e is linearly dependent, contradicting the fact that it is a basis. f 1∗ 2∗ m∗ g Hence, m n. Next, suppose that m < n. Now reverse the roles of ei∗ and e in the≤ above argument to conclude that m n. Hence, m = n. f g f ig ≥ In view of the above result, a vector space V is said to be n-dimensional if it contains a basis with n elements. If no such finite integer n exists, then the vector space is said to be infinite-dimensional. For example, n is finite-dimensional with < e1 = (1, 0, 0, . , 0), e2 = (0, 1, 0, . , 0), ... en = (0, 0, 0, . , 1), as the ‘canonical’ or natural basis. On the other hand, Lp[0, 1] is an infinite-dimensional vector space. Note that even in the finite-dimensional case, the basis need not be unique.
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