Continuum Mechanics: Concise Theory and Problems

Chapter 1

VECTOR AND TENSOR THEORY

The theory of scalar-, vector- and tensor-valued functions defined on subsets of a three-dimensional Euclidean space is a major part of the mathematical framework upon which continuum mechanics is built. This chapter is intended to provide a concise survey of basic results needed in the rest of the book and its contents will be found to be closely integrated into the subsequent text. It is not advisable, however, for the reader to postpone his study of continuum mechanics until the whole of this material has been mastered. Rather he should use Sections 1 to 3, 9 and 10 to refresh, and perhaps reorientate, his knowledge of vector algebra and analysis, and then turn back to the topics discussed in the remaining sections as the need arises.

1VECTOR ALGEBRA

The scalar and vector products, with which the reader is already assumed to be familiar, can be defined by a system of axioms and we follow this approach here as a means of reviewing briefly the essential facts of vector algebra and, at the same time, providing a natural starting point for the development of tensor algebra. Contact with the geometrical viewpoint customarily adopted in elementary treatments of vector theory is made in Section 9.

Let be a three-dimensional vector space over the field of real numbers. We say that is a Euclidean vector space if, to each pair of vectors a, b in , there corresponds a scalar (in ), denoted by a . b and called the scalar product of a and b, and a vector (in ), written a Λ b and referred to as the vector product of a and b, with the following properties:

The norm (or magnitude), |a|, of a vector a is defined by

and a vector with unit norm is termed a unit vector. Two vectors a and b are said to be orthogonal if a . b = 0.

Problem 1 Prove that a Λ b = 0 if and only if a and b are linearly dependent.

Solution. (a) If a and b are linearly dependent, either a = 0 or there is a scalar α such that b = αa. In the first case it follows from axiom (5) that a Λ b = 0, and in the second case the same conclusion is reached via axioms (4) and (5).

(b) If a Λ b = 0, axioms (7) and (3) and equation (8) give a. b = ±|a| |b|. Suppose that the plus sign applies. Then, using axioms (2) and (1),

and, in consequence of (3), |b|a = |a|b. When the minus sign holds, |b|a = – |a|b by a similar argument. In both cases either a = 0 or b is a scalar multiple of a. Thus a and b are linearly dependent.

The scalar triple product of three vectors a, b, c, denoted by [a, b, c], is defined by

Henceforth the abbreviated term triple product is used.

Problem 2 Establish the following properties of the triple product :

Solution. (i) In view of axioms (4), (2) and (1) the sign of a triple product is reversed when the second and third members of the product are exchanged. From axiom (6), with appeal to (2) and (5),

whence it is clear that sign reversal also results from interchanging the first and second members of a triple product. Repeated application of these properties leads to the identities (10).

(ii) Replace c by c Λ d in axiom (2).

(iii) (a) First observe that axiom (2), in conjunction with the properties (10), implies that a triple product having the zero vector as one of its members vanishes. If a, b, c are linearly dependent there exist scalars α, β, γ, not all zero, such that αa + βb + γc = 0. Hence the triple products

are all zero and, with the use of equations (10) and (11), they reduce in turn to α[a, b, c], β[a, b, c], γ[a, b, c]. Since at least one of α, β, γ is non-zero, [a, b, c] = 0.

(b) The converse result is proved by contradiction. Suppose that [a, b, c] = 0 and that a, b, c are linearly independent vectors. Problem 1 tells us that b Λ c ≠ 0, and we see from equations (9) and (10) that a, b and c are each orthogonal to b Λ c. Since a, b, c form a basis of it follows that every vector is orthogonal to b Λ c. This conclusion is plainly false, however, as b Λ c is not orthogonal to itself, so a, b, c must be linearly dependent.

A fundamental property of a finite-dimensional vector space equipped with a scalar product is the existence of an orthonormal basis.¹ Thus, in the Euclidean vector space defined above there is a set of three vectors, e = {e1, e2, e3}, such that

where δij, the Kronecker delta, takes the value 1 when i = j and 0 when i ≠ j. Corresponding to an arbitrary vector a there is an ordered triplet of scalars, (a1, a2, a3), such that

a1, a2, a3 are called the components of a relative to the basis e. On the right of (13) use has been made of the summation convention according to which any expression in which a suffix appears twice is understood to be summed over the range, 1, 2, 3 of that suffix. Only the letters p, q, r, s and π, ρ, σ are used in this book for repeated subscripts and summation over indices appearing more than twice is indicated explicitly. In view of the orthonormality condition (12),

whence

Problem 3 Show that

the plus or minus signs holding together.

Solution. On setting a = e2 Λ e3 in equation (14) we obtain

use being made of equations (9) and (10). Similarly,

With the aid of (12) we deduce from (A) and (B) that

while axiom (7), in combination with (12), yields the same results with [e1, e2, e3]² replaced by 1. Hence [e1, e2, e3] = ±1 and equations (15) follow directly from (A) and (B).

The relations (15) are contained in the compact expression

where εijk, the alternator, takes the value 1 when i, j, k is a cyclic permutation of 1, 2, 3, – 1 when i, j, k is a non-cyclic permutation of 1, 2, 3, and is otherwise zero.

Let a, b be an arbitrary pair of vectors having components ai, bi relative to e. Then, with the use of axioms (1) and (2) and equation (12),

This formula for the scalar product of two vectors in terms of their components relative to an orthonormal basis confirms that there is precisely one scalar product on satisfying axioms (1) to (3). When b = a, equations (17) and (8) give

The component form of the vector product a Λ b relative to the basis e is found with the aid of axioms (4) and (5) and equation (16) to be

It is seen from (19) that there are two vector products on obeying the axioms (4) to (7), one being the negative of the other. If c is a third arbitrary vector with components ci relative to e, the component form of the triple product [a, b, c], obtained by combining equation (9) with (17) and (19), is

Two ordered bases of are said to be similar if their triple products have the same sign.² Similarity is an equivalence relation which partitions the collection of all ordered bases of into two classes, one containing the members with positive triple products and the other those with negative triple products. The pairing of with a rule³ identifying the equivalence class to which each of its ordered bases belongs is called an orientation of . There are two such orientations, denoted by and , and an ordered basis of is said to be positive in if its triple product is positive and positive in if its triple product is negative. The conferment of an orientation on removes the ambiguity of sign in equations (15), (16), (19) and (20), the upper or lower sign being chosen according as e is positive in or . Evidently an orientated Euclidean vector space has a unique vector product meeting the axioms (4) to (7).

2TENSOR ALGEBRA

A tensor⁴ A is a linear transformation of the Euclidean vector space into itself. Specifically, A assigns to an arbitrary vector a a vector, denoted by Aa, in such a way that

The set of all tensors on is denoted by .

Two tensors are equal if and only if their actions on an arbitrary vector are identical, and the rules for the addition, scalar multiplication and multiplication (or composition) of tensors are

The zero tensor O assigns to a the zero vector and the identity tensor I assigns to a the vector a itself:

We leave it to the reader to establish, on the basis of equations (21) to (23), the following properties:

Associated with an arbitrary tensor A there is a unique tensor AT, called the transpose of A, such that⁵

It follows from this definition that (AT)T = A and also, with the use of equations (22) and axioms (1) and (2), that

A tensor A such that AT = A is said to be symmetric, and if AT = –A, A is called a skew-symmetric tensor. The identity

demonstrates that an arbitrary tensor can be expressed as the sum of symmetric and skew-symmetric parts. This decomposition is unique.

Problem 4 Let {f, g, h} and {l, m, n} be arbitrarily chosen bases of and let A be an arbitrary tensor. Show that

Solution. Let fi, gi, hi be the components of f, g, h relative to the orthonormal basis e. Then, using equations (13), (21), (10) and (11), we find that

and it follows that the left-hand side of (A) can be put into the form

The expression in curly brackets is unchanged by a cyclic permutation of the suffixes p, q, r; its sign is reversed by a non-cyclic permutation of these subscripts; and if any two of p, q, r have equal values the expression is zero. Recalling the definition of the alternator and equation (20), we can therefore rewrite (D) as

If e is positive in , the upper sign applies in (E) and [e1, e2, e3] = 1 ; if e is positive in , the lower sign must be chosen and [e1, e2, e3] = – 1. Thus, regardless of orientation,

and since the choice of basis on the left of (F) is arbitrary, we arrive immediately at the result (A).

The steps which have been used to establish (A) also yield derivations of equations (B) and (C) and the reader should check his understanding of the solution by writing out the details.

It is a consequence of Problem 4 that, corresponding to an arbitrary tensor A, there are scalars, IA, IIA, IIIA such that

These scalars are called the principal invariants of A, but IA is more commonly referred to as the trace of A, written tr A, and IIIA as the determinant of A, denoted by det A. Thus

From equations (22)1, 2, with (11) and (10), we see that

the trace therefore being a linear function from to . Immediate consequences of the definition (30) are

It also follows from (30) that a tensor preserves the orientation of if and only if its determinant is positive.

Problem 5 Given a tensor A, show that there exists a non-zero vector n such that An = 0 if and only if det A = 0.

Solution. (a) Suppose that det A = IIIA = 0 and let {f, g, h} be an arbitrary basis of . Then, from (30) and property (iii) of Problem 2 (p. 12), the vectors Af, Ag, Ah are linearly dependent This means that there are scalars α, β, γ, not all zero, such that

use being made of (21). Thus An = 0 where n = αf + βg + γh ≠ 0.

(b) Conversely, if there is a non-zero n such that An = 0, on choosing vectors l, m which form with n a linearly independent set and then replacing {a, b, c} by {l, m, n) in equation (30), we conclude, with further appeal to property (iii) of Problem 2, that IIIA = det A = 0.

If det A ≠ 0, A is said to be invertible since there then exists⁶ a unique